Problem 1.1
Given:
[Difficulty: 3]
Common Substances
Tar
Sand
“Silly Putty”
Jello
Modeling clay
Toothpaste
Wax
Shaving cream
Some of these substances exhibit characteristics of solids and fluids under different conditions.
Find:
Explain and give examples.
Solution:
Tar, Wax, and Jello behave as solids at room temperature or below at ordinary pressures. At high
pressures or over long periods, they exhibit fluid characteristics. At higher temperatures, all three
liquefy and become viscous fluids.
Modeling clay and silly putty show fluid behavior when sheared slowly. However, they fracture
under suddenly applied stress, which is a characteristic of solids.
Toothpaste behaves as a solid when at rest in the tube. When the tube is squeezed hard, toothpaste
“flows” out the spout, showing fluid behavior. Shaving cream behaves similarly.
Sand acts solid when in repose (a sand “pile”). However, it “flows” from a spout or down a steep
incline.
Problem 1.2
Given:
Five basic conservation laws stated in Section 1-4.
Write:
A word statement of each, as they apply to a system.
Solution:
Assume that laws are to be written for a system.
[Difficulty: 2]
a.
Conservation of mass — The mass of a system is constant by definition.
b.
Newton's second law of motion — The net force acting on a system is directly proportional to the product of the
system mass times its acceleration.
c.
First law of thermodynamics — The change in stored energy of a system equals the net energy added to the
system as heat and work.
d.
Second law of thermodynamics — The entropy of any isolated system cannot decrease during any process
between equilibrium states.
e.
Principle of angular momentum — The net torque acting on a system is equal to the rate of change of angular
momentum of the system.
Problem 1.3
[Difficulty: 3]
Open-Ended Problem Statement: The barrel of a bicycle tire pump becomes quite warm during use.
Explain the mechanisms responsible for the temperature increase.
Discussion: Two phenomena are responsible for the temperature increase: (1) friction between the pump piston
and barrel and (2) temperature rise of the air as it is compressed in the pump barrel.
Friction between the pump piston and barrel converts mechanical energy (force on the piston moving through a
distance) into thermal energy as a result of friction. Lubricating the piston helps to provide a good seal with the
pump barrel and reduces friction (and therefore force) between the piston and barrel.
Temperature of the trapped air rises as it is compressed. The compression is not adiabatic because it occurs during a
finite time interval. Heat is transferred from the warm compressed air in the pump barrel to the cooler surroundings.
This raises the temperature of the barrel, making its outside surface warm (or even hot!) to the touch.
Problem 1.4
[Difficulty: 3]
Open-Ended Problem Statement: Consider the physics of “skipping” a stone across the water surface
of a lake. Compare these mechanisms with a stone as it bounces after being thrown along a roadway.
Discussion: Observation and experience suggest two behaviors when a stone is thrown along a water surface:
1.
If the angle between the path of the stone and the water surface is steep the stone may penetrate the water
surface. Some momentum of the stone will be converted to momentum of the water in the resulting splash.
After penetrating the water surface, the high drag* of the water will slow the stone quickly. Then, because the
stone is heavier than water it will sink.
2.
If the angle between the path of the stone and the water surface is shallow the stone may not penetrate the water
surface. The splash will be smaller than if the stone penetrated the water surface. This will transfer less
momentum to the water, causing less reduction in speed of the stone. The only drag force on the stone will be
from friction on the water surface. The drag will be momentary, causing the stone to lose only a portion of its
kinetic energy. Instead of sinking, the stone may skip off the surface and become airborne again.
When the stone is thrown with speed and angle just right, it may skip several times across the water surface. With
each skip the stone loses some forward speed. After several skips the stone loses enough forward speed to penetrate
the surface and sink into the water.
Observation suggests that the shape of the stone significantly affects skipping. Essentially spherical stones may be
made to skip with considerable effort and skill from the thrower. Flatter, more disc-shaped stones are more likely to
skip, provided they are thrown with the flat surface(s) essentially parallel to the water surface; spin may be used to
stabilize the stone in flight.
By contrast, no stone can ever penetrate the pavement of a roadway. Each collision between stone and roadway will
be inelastic; friction between the road surface and stone will affect the motion of the stone only slightly. Regardless
of the initial angle between the path of the stone and the surface of the roadway, the stone may bounce several times,
then finally it will roll to a stop.
The shape of the stone is unlikely to affect trajectory of bouncing from a roadway significantly.
Problem 1.5
Given:
Dimensions of a room
Find:
Mass of air
[Difficulty: 1]
Solution:
p
Basic equation:
ρ=
Given or available data
p = 14.7psi
Rair⋅ T
T = ( 59 + 460)R
V = 10⋅ ft × 10⋅ ft × 8⋅ ft
Then
ρ =
p
Rair⋅ T
M = ρ⋅ V
ρ = 0.076⋅
Rair = 53.33 ⋅
V = 800⋅ ft
lbm
ft
3
M = 61.2⋅ lbm
ft⋅ lbf
lbm⋅ R
3
ρ = 0.00238⋅
slug
ft
M = 1.90⋅ slug
3
ρ = 1.23
kg
3
m
M = 27.8kg
Problem 1.6
[Difficulty: 1]
Given:
Data on oxygen tank.
Find:
Mass of oxygen.
Solution:
Compute tank volume, and then use oxygen density (Table A.6) to find the mass.
The given or available
data is:
D = 16⋅ ft
p = 1000⋅ psi
RO2 = 48.29 ⋅
ft⋅ lbf
lbm⋅ R
Tc = 279 ⋅ R
so the reduced temperature and pressure are:
TR =
Z = 0.948
T
Tc
= 1.925
Hence
π⋅ D
V =
6
M = V⋅ ρ =
p⋅ V
RO2⋅ T
M = 11910 ⋅ lbm
(data from NIST WebBook)
pR =
p = ρ⋅ RO2⋅ T
3
V=
p c = 725.2 ⋅ psi
p
pc
= 1.379
Since this number is close to 1, we can assume ideal gas behavior.
Therefore, the governing equation is the ideal gas equation
where V is the tank volume
T = 537 ⋅ R
(Table A.6)
For oxygen the critical temperature and pressure are:
Using a compressiblity factor chart:
T = ( 77 + 460 ) ⋅ R
π
6
× ( 16⋅ ft)
M = 1000⋅
ρ=
and
lbf
2
in
3
V = 2144.7⋅ ft
3
× 2144.7⋅ ft ×
1
M
V
3
⋅
lbm⋅ R
48.29 ft⋅ lbf
×
1
⋅
1
537 R
×
⎛ 12⋅ in ⎞
⎜ ft
⎝
⎠
2
Problem 1.7
Given:
[Difficulty: 3]
Small particle accelerating from rest in a fluid. Net weight is W, resisting force FD = kV, where V
is speed.
Find:
Time required to reach 95 percent of terminal speed, Vt.
Solution:
Consider the particle to be a system. Apply Newton's second law.
Basic equation: ∑Fy = may
FD = kV
Assumptions:
1.
W is net weight
2.
Resisting force acts opposite to V
Particle
y
Then
W
∑F
y
= W − kV = ma y = m
dV W dV
=
g dt
dt
or
dV
k
= g(1 − V)
W
dt
Separating variables,
dV
= g dt
k
1− W
V
V
t
dV
W
k
(
)
V
=
−
ln
1
−
=
W
∫0 gdt = gt
k
1 − Wk V
Integrating, noting that velocity is zero initially,
∫
or
kgt
kgt
−
−
⎞
W ⎛⎜
k
W
1 − V = e ; V = ⎜1 − e W ⎟⎟
k ⎝
W
⎠
0
kgt
But V→Vt as t→∞, so Vt =
When
V
Vt
= 0.95 , then e
−
kgt
W
W
k
. Therefore
= 0.05 and
kgt
W
−
V
= 1− e W
Vt
= 3. Thus t = 3W/gk
Problem 1.8
Given:
[Difficulty: 2]
Small particle accelerating from rest in a fluid. Net weight is W,
resisting force is FD = kV, where V is speed.
Find:
FD = kV
Distance required to reach 95 percent of terminal speed, Vt.
Solution:
Particle
Consider the particle to be a system. Apply Newton's second law.
Basic equation: ∑Fy = may
W
Assumptions:
1.
W is net weight.
2.
Resisting force acts opposite to V.
Then,
∑F
y
= W − kV = ma y = m dV
dt =
At terminal speed, ay = 0 and V = Vt =
Separating variables
y
W
k
W
g
V dV
dy
or
. Then 1 −
V
Vg
1 − Wk V =
V dV
g dy
= g1 V dV
dy
V dV
= g dy
1 − V1t V
Integrating, noting that velocity is zero initially
gy = ∫
0.95Vt
0
⎡
⎛ V
V dV
= ⎢ −VVt − Vt 2 ln ⎜1 −
1
⎝ Vt
1 − V ⎢⎣
Vt
gy = −0.95Vt 2 − Vt 2 ln (1 − 0.95) − Vt 2 ln (1)
gy = −Vt 2 [ 0.95 + ln 0.05] = 2.05 Vt 2
∴y =
2.05 2
W2
Vt = 2.05 2
g
gt
0.95Vt
⎞⎤
⎟⎥
⎠ ⎥⎦ 0
Problem 1.9
[Difficulty: 2]
Given:
Mass of nitrogen, and design constraints on tank dimensions.
Find:
External dimensions.
Solution:
Use given geometric data and nitrogen mass, with data from Table A.6.
The given or available data is: M = 5 ⋅ kg
T = ( 20 + 273 ) ⋅ K
p = ( 200 + 1 ) ⋅ atm
p = 20.4⋅ MPa
T = 293 ⋅ K
RN2 = 296.8 ⋅
p = ρ⋅ RN2⋅ T
The governing equation is the ideal gas equation
ρ=
and
J
kg⋅ K
(Table A.6)
M
V
2
where V is the tank volume
V=
π⋅ D
4
⋅L
L = 2⋅ D
where
Combining these equations:
Hence
M = V⋅ ρ =
p⋅ V
RN2⋅ T
=
p
RN2⋅ T
2
⋅
π⋅ D
4
⋅L =
3
2
p
RN2⋅ T
⋅
π⋅ D
4
⋅ 2⋅ D =
p ⋅ π⋅ D
2 ⋅ RN2⋅ T
1
1
Solving for D
⎛ 2 ⋅ RN2⋅ T⋅ M ⎞
D= ⎜
p⋅ π
⎝
⎠
D = 0.239 ⋅ m
3
2
⎛2
N⋅ m
1
m ⎞
D = ⎜ × 296.8 ⋅
× 293 ⋅ K × 5 ⋅ kg ×
⋅
kg⋅ K
6 N
⎜π
20.4 × 10
⎝
⎠
L = 2⋅ D
These are internal dimensions; the external ones are 2 x 0.5 cm larger:
L = 0.477 ⋅ m
L = 0.249 ⋅ m
D = 0.487 ⋅ m
3
Problem 1.10
[Difficulty: 4]
NOTE: Drag formula is in error: It should be:
FD = 3 ⋅ π⋅ V⋅ d
Given:
Data on sphere and formula for drag.
Find:
Diameter of gasoline droplets that take 1 second to fall 10 in.
Solution:
Use given data and data in Appendices; integrate equation of
motion by separating variables.
FD = 3πVd
a = dV/dt
Mg
The data provided, or available in the Appendices, are:
− 7 lbf ⋅ s
μ = 4.48 × 10
⋅
ft
2
ρw = 1.94⋅
slug
ft
3
SG gas = 0.72
ρgas = SGgas⋅ ρw
dV
g−
Integrating twice and using limits
slug
ft
M⋅
Newton's 2nd law for the sphere (mass M) is (ignoring buoyancy effects)
so
ρgas = 1.40⋅
3 ⋅ π⋅ μ⋅ d
V( t) =
M
dV
dt
3
= M ⋅ g − 3 ⋅ π⋅ μ⋅ V⋅ d
= dt
⋅V
− 3⋅ π⋅ μ ⋅ d ⎞
⎛
⋅t
⎜
M⋅ g
M
⋅⎝1 − e
⎠
3⋅ π⋅ μ ⋅ d
Replacing M with an expression involving diameter d
π⋅ d
M = ρgas ⋅
6
⎡
⎛ − 3⋅ π⋅ μ ⋅ d ⋅ t
⎢
⎜
M⋅ g
M
M
x( t) =
⋅⎝e
⋅ ⎢t +
−
3⋅ π⋅ μ ⋅ d ⎣
3⋅ π⋅ μ ⋅ d
⎞⎤
⎥
1⎠⎥
⎦
⎡⎢
⎛ − 18⋅ μ ⋅ t
2 ⎜
ρgas ⋅ d ⋅ g ⎢
ρgas ⋅ d ⎜ ρgas⋅ d2
x( t) =
⋅⎝e
⋅ ⎢t +
−
18⋅ μ
18⋅ μ
⎣
⎞⎤⎥
⎥
1⎠⎥
⎦
2
3
This equation must be solved for d so that x ( 1 ⋅ s) = 10⋅ in. The answer can be obtained from manual iteration, or by using
Excel's Goal Seek.
−3
⋅ in
1
10
0.75
7.5
x (in)
x (in)
d = 4.30 × 10
0.5
0.25
5
2.5
0
0.025
0.05
t (s)
0.075
0.1
0
0.25
0.5
0.75
1
t (s)
Note That the particle quickly reaches terminal speed, so that a simpler approximate solution would be to solve Mg = 3πµVd for d,
with V = 0.25 m/s (allowing for the fact that M is a function of d)!
Problem 1.11
[Difficulty: 3]
Given:
Data on sphere and formula for drag.
Find:
Maximum speed, time to reach 95% of this speed, and plot speed as a function of time.
Solution:
Use given data and data in Appendices, and integrate equation of motion by separating variables.
The data provided, or available in the Appendices, are:
kg
ρair = 1.17⋅
3
m
μ = 1.8 × 10
− 5 N⋅ s
⋅
2
m
Then the density of the sphere is
kg
ρw = 999 ⋅
3
m
SG Sty = 0.016
ρSty = SGSty⋅ ρw
ρSty = 16
d = 0.3⋅ mm
kg
3
m
3
π⋅ d
kg
( 0.0003⋅ m)
M = ρSty⋅
= 16⋅
× π×
6
3
6
m
The sphere mass is
3
M = 2.26 × 10
− 10
kg
M ⋅ g = 3 ⋅ π⋅ V⋅ d
Newton's 2nd law for the steady state motion becomes (ignoring buoyancy effects)
so
M⋅ g
− 10
1
Vmax =
=
× 2.26 × 10
3 ⋅ π⋅ μ⋅ d
3⋅π
⋅ kg × 9.81⋅
m
2
s
2
×
m
1.8 × 10
−5
Newton's 2nd law for the general motion is (ignoring buoyancy effects)
dV
so
g−
3 ⋅ π⋅ μ⋅ d
M
Integrating and using limits
×
⋅ N⋅ s
1
0.0003⋅ m
m
Vmax = 0.0435
s
M⋅
dV
dt
= M ⋅ g − 3 ⋅ π⋅ μ⋅ V⋅ d
= dt
FD = 3πVd
⋅V
V( t) =
− 3⋅ π⋅ μ⋅ d ⎞
⎛
⋅t
⎜
M⋅ g
M
⋅⎝1 − e
⎠
3 ⋅ π⋅ μ⋅ d
a = dV/dt
Mg
Using the given data
0.05
V (m/s)
0.04
0.03
0.02
0.01
0
0.01
0.02
t (s)
The time to reach 95% of maximum speed is obtained from
so
t=−
M
3 ⋅ π⋅ μ⋅ d
⎛
⋅ ln⎜ 1 −
0.95⋅ Vmax⋅ 3 ⋅ π⋅ μ⋅ d ⎞
⎝
The plot can also be done in Excel.
M⋅ g
⎠
− 3⋅ π⋅ μ⋅ d ⎞
⎛
⋅t
⎜
M⋅ g
M
⋅⎝1 − e
⎠ = 0.95⋅ Vmax
3 ⋅ π⋅ μ⋅ d
Substituting values
t = 0.0133 s
Problem 1.12
[Difficulty: 3]
Given:
Data on sphere and terminal speed.
Find:
Drag constant k, and time to reach 99% of terminal speed.
Solution:
Use given data; integrate equation of motion by separating variables.
kVt
− 13
M = 1 × 10
The data provided are:
mg
ft
Vt = 0.2⋅
s
⋅ slug
M⋅
Newton's 2nd law for the general motion is (ignoring buoyancy effects)
dV
dt
M ⋅ g = k ⋅ Vt
Newton's 2nd law for the steady state motion becomes (ignoring buoyancy effects)
− 13
k = 1 × 10
⋅ slug × 32.2⋅
ft
2
2
s
×
0.2⋅ ft
s
×
lbf ⋅ s
= M⋅ g − k⋅ V
⋅
M
k
− 13
⋅ slug ×
M⋅ g
− 11
×
⋅ lbf ⋅ s
Vt
lbf ⋅ s
slug⋅ ft
g−
= dt
k
M
⋅V
⋅ V⎞
⎠
V = 0.198 ⋅
2
ft
1.61 × 10
t = 0.0286 s
⎝
k
V = 0.99⋅ Vt
We must evaluate this when
t = −1 × 10
⋅ ln⎜⎛ 1 −
M⋅ g
ft
dV
t=−
k =
so
− 11 lbf ⋅ s
k = 1.61 × 10
slug⋅ ft
To find the time to reach 99% of Vt, we need V(t). From 1, separating variables
Integrating and using limits
(1)
⎛
⋅ ln⎜ 1 − 1.61 × 10
⎜
⎝
ft
s
− 11 lbf ⋅ s
⋅
ft
×
2
1
− 13
1 × 10
×
⋅ slug
s
32.2⋅ ft
×
0.198 ⋅ ft
s
×
slug⋅ ft ⎞
2
lbf ⋅ s
⎠
Problem 1.13
[Difficulty: 5]
Given:
Data on sphere and terminal speed from Problem 1.12.
Find:
Distance traveled to reach 99% of terminal speed; plot of distance versus time.
Solution:
Use given data; integrate equation of motion by separating variables.
− 13
M = 1 × 10
The data provided are:
ft
Vt = 0.2⋅
s
⋅ slug
mg
M⋅
Newton's 2nd law for the general motion is (ignoring buoyancy effects)
dV
dt
= M⋅ g − k⋅ V
Newton's 2nd law for the steady state motion becomes (ignoring buoyancy effects)
k = 1 × 10
− 13
⋅ slug × 32.2⋅
ft
2
×
s
2
s
0.2⋅ ft
×
lbf ⋅ s
V⋅ dV
g−
k
M
(
− 13
y = 1 ⋅ 10
)
2
⋅ slug ⋅
2
M⋅ g
Vt
ft
dV
dt
= M⋅
dy dV
dV
⋅
= M ⋅ V⋅
= M⋅ g − k⋅ V
dt dy
dy
M ⋅g
2
⋅ ln⎜⎛ 1 −
⎝
k
M⋅ g
⋅ V⎞ −
⎠
M
k
⋅V
V = 0.198 ⋅
ft
s
2
2⎛
2
⎛
ft
1
s
0.198 ⋅ ft slug⋅ ft ⎞
⎞ ⋅ ⎜ lbf ⋅ s ⎞ ⋅ ln⎜ 1 − 1.61⋅ 10− 11⋅ lbf ⋅ s ⋅
⋅
⋅
⋅
⋅⎛
...
⎜
− 11
− 13
32.2⋅ ft
s
2
⎜
slug⋅ ft ⎠
ft
⎝
1 ⋅ 10
⋅ slug
lbf ⋅ s ⎠
⎝ 1.61⋅ 10 ⋅ lbf ⋅ s ⎠
⎝
ft
⋅ slug ×
− 11
1.61⋅ 10
−3
⋅
M⋅
V = 0.99⋅ Vt
− 13
y = 4.49 × 10
k =
2
32.2⋅ ft
s
+ 1 ⋅ 10
so
= dy
k
We must evaluate this when
M ⋅ g = k ⋅ Vt
⋅V
2
y=−
Integrating and using limits
(1)
− 11 lbf ⋅ s
k = 1.61 × 10
slug⋅ ft
To find the distance to reach 99% of Vt, we need V(y). From 1:
Separating variables
kVt
×
⋅ lbf ⋅ s
2
0.198 ⋅ ft
lbf ⋅ s
×
s
slug⋅ ft
⋅ ft
Alternatively we could use the approach of Problem 1.12 and first find the time to reach terminal speed, and use this time in y(t) to
find the above value of y:
dV
From 1, separating variables
Integrating and using limits
g−
k
M
t=−
M
k
= dt
⋅V
⋅ ln⎜⎛ 1 −
⎝
k
M⋅ g
⋅ V⎞
⎠
(2)
V = 0.99⋅ Vt
We must evaluate this when
t = 1 × 10
− 13
2
ft
⋅ slug ×
1.61 × 10
− 11
lbf ⋅ s
⋅
slug⋅ ft
⋅ lbf ⋅ s
V = 0.198 ⋅
⎛
⋅ ln⎜ 1 − 1.61 × 10
ft
s
− 11 lbf ⋅ s
⋅
⎜
⎝
ft
×
2
1
− 13
1 × 10
×
⋅ slug
s
32.2⋅ ft
×
0.198 ⋅ ft
s
×
slug⋅ ft ⎞
2
lbf ⋅ s
⎠
t = 0.0286 s
V=
From 2, after rearranging
− 13
⋅ slug ×
32.2⋅ ft
2
s
−3
y = 4.49 × 10
dt
=
M⋅ g
k
⎛
−
⎜
⋅⎝1 − e
⎡
⎛ −
M⋅ g ⎢
M ⎜
y=
⋅⎝e
⋅ ⎢t +
k ⎣
k
Integrating and using limits
y = 1 × 10
dy
− 11
1.61 × 10
M
⋅ t⎞
M
⎠
⋅t
⎞⎤
⎥
− 1⎠⎥
⎦
2
ft
×
k
k
⋅
⋅ lbf ⋅ s
lbf ⋅ s
slug⋅ ft
⋅ ⎡0.0291⋅ s ...
⎢
⎛
⎜ −
⎢
2
ft
lbf ⋅ s ⎜
⎢+ 10− 13⋅ slug⋅
⋅
⋅⎝e
⎢
− 11
slug⋅ ft
1.61 × 10
⋅ lbf ⋅ s
⎣
1.61× 10
− 11
− 13
1× 10
⋅ ft
5
y (0.001 ft)
3.75
2.5
1.25
0
5
10
15
t (ms)
This plot can also be presented in Excel.
20
25
⎤
⎞ ⎥
⋅ .0291
⎥
⎥
− 1⎠
⎥
⎦
Problem 1.14
[Difficulty: 4]
2
Given:
M  70 kg
Data on sky diver:
k  0.25
N s
2
m
Find:
Maximum speed; speed after 100 m; plot speed as function of time and distance.
Solution:
Use given data; integrate equation of motion by separating variables.
Treat the sky diver as a system; apply Newton's 2nd law:
M
Newton's 2nd law for the sky diver (mass M) is (ignoring buoyancy effects):
2
(a) For terminal speed Vt, acceleration is zero, so M  g  k  V  0
dV
dt
(1)
M g
Vt 
so
2
 M g  k V
FD = kV2
k
1
2
2

m
m
N s 
Vt   70 kg  9.81 

2
2 kg  m

s
0.25 N s


(b) For V at y = 100 m we need to find V(y). From (1) M 






Separating variables and integrating:
dV
dt
V
 M
m
Vt  52.4
s
dV dy
dV
2

 M  V
 M g  k V
dy dt
dt
y
M g
0
so
2

k V 
2 k
ln 1 

y
M

g
M


Mg

dV   g dy

2
0
k V
V
1
a = dV/dt
2
or
2
V 
M g
k



1  e
2 k  y 
M

1
Hence



V( y )  Vt  1  e
2 k  y 
M
2

1
For y = 100 m:
2

N s
1
kg m 

 2 0.25
 100 m

2
70 kg s 2 N
m 
m
V( 100  m)  52.4   1  e

s
2
V( 100  m)  37.4
m
s
V(m/s)
60
40
20
0
100
200
300
400
500
y(m)
(c) For V(t) we need to integrate (1) with respect to t:
M





Separating variables and integrating:
dV
dt
V
2
 M g  k V
t

dV   1 dt

M g
2
0
V
k
V
0

1
M 
t 
 ln
2 k g 

so

 2
e
V( t)  Vt

 2
e
Rearranging
k g
k g
M
V

t
M

V V 
k
  1  M  ln t
 2 k g  Vt  V 
M g
V
k

M g
 1

t
or
k
V( t)  Vt tanh Vt  t
M


 1
V(m/s)
60
40
V ( t)
20
0
5
10
t
t(s)
The two graphs can also be plotted in Excel.
15
20
Problem 1.15
[Difficulty: 5]
2
Given:
M = 70⋅ kg
Data on sky diver:
k vert = 0.25⋅
N⋅ s
2
2
k horiz = 0.05⋅
m
Find:
Plot of trajectory.
Solution:
Use given data; integrate equation of motion by separating variables.
N⋅ s
m
U0 = 70⋅
s
2
m
Treat the sky diver as a system; apply Newton's 2nd law in horizontal and vertical directions:
Vertical: Newton's 2nd law for the sky diver (mass M) is (ignoring buoyancy effects):
M⋅
dV
2
= M ⋅ g − k vert⋅ V (1)
dt
For V(t) we need to integrate (1) with respect to t:
⌠
⎮
⎮
⎮
⎮
⌡
Separating variables and integrating:
V
t
⌠
dV = ⎮ 1 dt
⌡
M⋅ g
2
0
−V
k vert
V
0
⎛
⎜
1
M
t= ⋅
⋅ ln⎜
2 k vert⋅ g ⎜
⎜
⎝
⎛
⎜ 2⋅
⎜
M⋅ g ⎝ e
oV( t) =
⋅
k vert ⎛
r
⎜ 2⋅
⎜e
⎝
so
Rearranging
M⋅ g
k vert
M⋅ g
k vert
kvert ⋅ g
kvert ⋅ g
M
dt
⌠
y ( t) = ⎮
⌡
t
0
⌠
⎮
V( t) dt = ⎮
⎮
⌡
t
M⋅ g
k vert
M⋅ g
k vert
⎛
⎛ kvert⋅ g ⎞ ⎞
⋅t
M
⎝
⎠⎠
⋅ ln⎜ cosh⎜
⎝
⋅t
=V
⎞
⎟
⎟
⎠
⎞
− 1⎠
⋅t
⎞
so
V( t) =
M⋅ g
k vert
+ 1⎠
or
⌠
y = ⎮ V dt
⌡
⎛ kvert⋅ g ⎞
⎛ k vert⋅ g ⎞ ⎞
M⋅ g ⎛
⋅ t dt =
⋅t
⋅ ln⎜ cosh⎜
k vert ⎝
M
M
⎝
⎠
⎝
⎠⎠
⋅ tanh⎜
0
y ( t) =
−V
M
dy
For y(t) we need to integrate again:
+V
⎛ k vert⋅ g ⎞
⋅ tanh⎜
⎝
M
⋅t
⎠
30
600
20
400
y(m)
y(m)
After the first few seconds we reach steady state:
y( t)
y( t)
10
200
0
1
2
3
4
5
0
20
40
t
t
t(s)
t(s)
M⋅
Horizontal: Newton's 2nd law for the sky diver (mass M) is:
dU
dt
2
= −k horiz⋅ U
For U(t) we need to integrate (2) with respect to t:
U
Separating variables and integrating:
⌠
1
⎮
dU =
⎮
2
U
⎮
⌡U
t
⌠ k
horiz
⎮
−
dt
⎮
M
⌡
so
0
−
k horiz
M
⋅t = −
1
U
+
1
U0
0
Rearranging
U0
U( t) =
1+
For x(t) we need to integrate again:
dx
dt
=U
k horiz⋅ U0
M
⋅t
⌠
x = ⎮ U dt
⌡
or
t
⌠
t
U0
⌠
⎛ k horiz⋅ U0
⎞
M
⎮
dt =
x ( t) = ⎮ U( t) dt = ⎮
⋅t + 1
⋅ ln⎜
⌡
k horiz ⎝
M
k
⋅U
⎠
0
⎮ 1 + horiz 0 ⋅ t
⎮
M
⌡
0
x ( t) =
M
k horiz
⎞
⎛ khoriz⋅ U0
⋅t + 1
⎝ M
⎠
⋅ ln⎜
60
(2)
2
x(km)
1.5
1
0.5
0
20
40
60
t(s)
Plotting the trajectory:
y(km)
0
1
2
−1
−2
−3
x(km)
These plots can also be done in Excel.
3
Problem 1.16
[Difficulty: 3]
Given:
Long bow at range, R = 100 m. Maximum height of arrow is h = 10 m. Neglect air resistance.
Find:
Estimate of (a) speed, and (b) angle, of arrow leaving the bow.
Plot:
(a) release speed, and (b) angle, as a function of h
Solution:
Let V0 = u 0 i + v 0 j = V0 (cos θ 0 i + sin θ 0 j)
y
ΣFy = m dv
= − mg , so v = v0 – gt, and tf = 2tv=0 = 2v0/g
dt
v2
dv
= − mg, v dv = −g dy, 0 − 0 = − gh
dy
2
Also,
mv
Thus
h = v 20 2g
ΣFx = m
v 20 = 2gh
From Eq. 2:
u0 =
x
θ0
R
(1)
2u v
du
= 0, so u = u 0 = const, and R = u 0 t f = 0 0
g
dt
From Eq. 1:
h
V0
(2)
(3)
gR
gR
=
2v 0 2 2gh
∴ u 20 =
gR 2
8h
1
Then
⎛
gR 2 ⎞ 2
gR 2
⎟
+ 2 gh and V0 = ⎜⎜ 2 gh +
V =u +v =
8h
8h ⎟⎠
⎝
2
0
2
0
2
0
(4)
1
⎛
m
1 ⎞2
9.81 m
m
⎟⎟ = 37.7
V0 = ⎜⎜ 2 × 9.81 2 × 10 m +
× 100 2 m 2 ×
2
s
10
m
8
s
s
⎝
⎠
From Eq. 3:
v 0 = 2gh = V0 sin θ , θ = sin −1
2gh
V0
(5)
1
⎡
⎤
2
s ⎥
m
⎛
⎞
θ = sin ⎢⎜ 2 × 9.81 2 ×10 m ⎟ ×
= 21.8°
⎢⎝
s
⎠ 37.7 m ⎥
⎣
⎦
−1
Plots of V0 = V0(h) (Eq. 4) and θ0 = θ 0(h) (Eq. 5) are presented below:
V 0 (m/s)
Initial Speed vs Maximum Height
80
70
60
50
40
30
20
10
0
0
5
10
15
20
25
30
25
30
h (m)
Initial Angle vs Maximum Height
60
50
o
θ ( )
40
30
20
10
0
0
5
10
15
h (m)
20
Problem 1.17
[Difficulty: 2]
Given: Basic dimensions M, L, t and T.
Find:
Dimensional representation of quantities below, and typical units in SI and English systems.
Solution:
(a) Power
Power 
Energy
Time
Force  Distance

Time
F L

t
Force  Mass  Acceleration
From Newton's 2nd law
F
so
M L
t
Power 
Hence
(b) Pressure
(c) Modulus of elasticity
(d) Angular velocity
(e) Energy
Pressure 
Pressure 
Force
Area
Force
Area
F L
t
F

L
F

AngularVelocity 
2
L
2
M  L L


2
M L
t t
M L

2
t L
2
M L

2
t L
Radians

Time
2
t
2
kg m
3
L t
3
L t
(h) Shear stress
(i) Strain
(j) Angular momentum
Strain 
Area
LengthChange
Length
F
L
2


kg
slug
2
1
1
1
t
s
s
Momentum  Mass  Velocity  M 

ft s
2
M  L L
M L

2
t
MomentOfForce  Force  Length  F L 
Force
2
m s
ft s
Energy  Force  Distance  F L 
ShearStress 
slug
m s
M L
2
t L
2
L
t


2
2
kg m
2
M  L L
t
(g) Momentum
3
2
2
2
slug ft
2

t
2
s
2
kg m
2
slug ft
2
s
M L
kg m
slug ft
t
s
s
kg
slug
L t
2
2
2
m s
L
2
2
s
M
2
2
s
M L
2
s
kg
2
M

slug ft
s
M

t
(f) Moment of a force
2
2
ft s
Dimensionless
L
AngularMomentum  Momentum  Distance 
M L
t
L 
M L
t
2
2
kg m
slugs ft
s
s
2
Problem 1.18
[Difficulty: 2]
Given: Basic dimensions F, L, t and T.
Find:
Dimensional representation of quantities below, and typical units in SI and English systems.
Solution:
(a) Power
(b) Pressure
(c) Modulus of elasticity
(d) Angular velocity
Power =
Energy
Pressure =
Pressure =
Time
Force
Area
Force
Area
Force × Distance
=
Time
L
L
s
N
lbf
Radians
Time
=
Force
Area
=
ft
N
lbf
ft
1
1
1
t
s
s
N⋅ m
lbf ⋅ ft
or
M=
L
t
L
t
F = M⋅
=
F⋅ t ⋅ L
L⋅ t
L
t
2
= F⋅ t
F
L
2
m
Force = Mass × Acceleration so
Momentum = M ⋅
2
m
2
From Newton's 2nd law
SpecificHeat =
s
2
Momentum = Mass × Velocity = M ⋅
(h) Specific heat
t
F
=
(f) Momentum
ShearStress =
lbf ⋅ ft
2
Energy = Force × Distance = F⋅ L
(g) Shear stress
N⋅ m
2
(e) Energy
Hence
F⋅ L
F
=
AngularVelocity =
=
2
2
N⋅ s
lbf ⋅ s
N
lbf
2
m
Energy
Mass × Temperature
=
F⋅ L
M⋅ T
=
F⋅ L
⎛ F⋅ t2 ⎞
⎜
⋅T
⎝ L ⎠
=
L
2
2
t ⋅T
2
ft
L
2
2
m
ft
2
s ⋅R
s ⋅K
F⋅ t
2
LengthChange
(i) Thermal expansion coefficient ThermalExpansionCoefficient =
(j) Angular momentum
Length
Temperature
=
1
1
1
T
K
R
N⋅ m⋅ s
lbf ⋅ ft⋅ s
AngularMomentum = Momentum × Distance = F⋅ t⋅ L
2
Problem 1.19
[Difficulty: 1]
Given:
Viscosity, power, and specific energy data in certain units
Find:
Convert to different units
Solution:
Using data from tables (e.g. Table G.2)
2
(a)
⎛ 1 ⋅ ft ⎞
2
2 ⎜
2
12
m
m
ft
1⋅
= 10.76 ⋅
= 1⋅
×⎜
s
s
s
⎝ 0.0254⋅ m ⎠
(b)
100 ⋅ W = 100 ⋅ W ×
(c)
1⋅
kJ
kg
= 1⋅
kJ
kg
×
1 ⋅ hp
746 ⋅ W
1000⋅ J
1 ⋅ kJ
×
= 0.134 ⋅ hp
1 ⋅ Btu
1055⋅ J
×
0.454 ⋅ kg
1 ⋅ lbm
= 0.43⋅
Btu
lbm
Problem 1.20
[Difficulty: 1]
Given:
Pressure, volume and density data in certain units
Find:
Convert to different units
Solution:
Using data from tables (e.g. Table G.2)
6895⋅ Pa
1⋅ psi = 1⋅ psi ×
(b)
1⋅ liter = 1⋅ liter ×
(c)
⎛ 1 ⋅ ft ⎞
lbf ⋅ s
lbf ⋅ s
4.448⋅ N ⎜ 12
N⋅s
1⋅
×
= 47.9⋅
= 1⋅
×⎜
⋅ ⎠
2
2
1⋅ lbf
2
⎝ 0.0254m
ft
ft
m
1⋅ psi
×
1⋅ kPa
(a)
1⋅ quart
0.946⋅ liter
1000⋅ Pa
×
= 6.89⋅ kPa
1⋅ gal
4⋅ quart
= 0.264⋅ gal
2
Problem 1.21
[Difficulty: 1]
Given:
Specific heat, speed, and volume data in certain units
Find:
Convert to different units
Solution:
Using data from tables (e.g. Table G.2)
(a)
4.18⋅
(b)
30⋅
kJ
kg⋅ K
m
s
= 4.18⋅
= 30⋅
m
s
×
kJ
kg⋅ K
3.281 ⋅ ft
1⋅ m
3
(c)
5⋅ L = 5⋅ L ×
×
1⋅ m
1000⋅ L
⋅
1 ⋅ Btu
1.055 ⋅ kJ
1 ⋅ mi
5280⋅ ft
⋅
×
1 ⋅ kg
2.2046⋅ lbm
3600⋅ s
hr
= 67.1⋅
3
×
×
1⋅ K
1.8⋅ R
mi
hr
⎛ 100 ⋅ cm × 1⋅ in ⎞ = 305 ⋅ in3
⎜ 1⋅ m
2.54⋅ cm ⎠
⎝
= 0.998 ⋅
Btu
lbm⋅ R
Problem 1.22
[Difficulty: 1]
Given:
Quantities in English Engineering (or customary) units.
Find:
Quantities in SI units.
Solution:
Use Table G.2 and other sources (e.g., Machinery's Handbook, Mark's Standard Handbook)
(a)
3.7⋅ acre⋅ ft = 3.7⋅ acre ×
(b)
150 ⋅
2
3
(c)
(d)
in
s
3
= 150 ⋅
in
gal
×
3 ⋅ gpm = 3 ⋅
3⋅
mph
s
min
= 3⋅
1 ⋅ acre
hr⋅ s
231 ⋅ in
×
0.3048⋅ m
1 ⋅ ft
3
3
= 4.56 × 10 ⋅ m
3
⎛ 0.0254⋅ m ⎞ = 0.00246 ⋅ m
⎜ 1 ⋅ in
s
⎝
⎠
3
mile
×
3
×
s
4047⋅ m
1 ⋅ gal
1609⋅ m
1 ⋅ mile
3
3
×
⎛ 0.0254⋅ m ⎞ ⋅ 1⋅ min = 0.000189⋅ m
⎜ 1 ⋅ in
s
⎝
⎠ 60⋅ s
×
1 ⋅ hr
3600⋅ s
= 1.34⋅
m
2
s
Problem 1.23
[Difficulty: 1]
Given:
Quantities in English Engineering (or customary) units.
Find:
Quantities in SI units.
Solution:
Use Table G.2 and other sources (e.g., Google)
(a)
100 ⋅
ft
3
m
= 100 ⋅
ft
3
min
3
(b)
5 ⋅ gal = 5 ⋅ gal ×
(c)
65⋅ mph = 65⋅
231 ⋅ in
1 ⋅ gal
mile
hr
×
3
×
⎛ 0.0254⋅ m ⎞ = 0.0189⋅ m3
⎜ 1⋅ in
⎝
⎠
1852⋅ m
1 ⋅ mile
×
3
(d)
3
3
⎛ 0.0254⋅ m × 12⋅ in ⎞ × 1 ⋅ min = 0.0472⋅ m
⎜ 1 ⋅ in
s
1 ⋅ ft ⎠
60⋅ s
⎝
×
5.4⋅ acres = 5.4⋅ acre ×
4047⋅ m
1 ⋅ acre
1 ⋅ hr
3600⋅ s
m
= 29.1⋅
4
s
2
= 2.19 × 10 ⋅ m
Problem 1.24
Given:
Quantities in SI (or other) units.
Find:
Quantities in BG units.
Solution:
Use Table G.2.
(a)
50⋅ m = 50⋅ m ×
(b)
250⋅ cc = 250⋅ cm ×
(c)
100⋅ kW = 100⋅ kW ×
(d)
5⋅
2
2
2
⎛ 1⋅ in × 1⋅ ft ⎞ = 538⋅ ft 2
⎜ 0.0254m
⋅
12⋅ in ⎠
⎝
3
kg
2
m
= 5⋅
kg
2
m
×
3
⎛ 1⋅ m × 1⋅ in × 1⋅ ft ⎞ = 8.83 × 10− 3⋅ ft 3
⎜ 100⋅ cm 0.0254m
12⋅ in ⎠
⋅
⎝
1000⋅ W
1⋅ kW
×
1⋅ hp
746⋅ W
= 134⋅ hp
2
⋅
slug
12⋅ in ⎞
1⋅ slug
⎛ 0.0254m
= 0.0318⋅
⎜ 1⋅ in × 1⋅ ft ×
2
14.95⋅ kg
⎝
⎠
ft
[Difficulty: 1]
Problem 1.25
Given:
Quantities in SI (or other) units.
Find:
Quantities in BG units.
Solution:
Use Table G.2.
(a)
180⋅ cc = 180⋅ cm ×
(b)
300⋅ kW = 300⋅ kW ×
(c)
50⋅
3
N⋅s
2
m
(d)
2
= 50⋅
N⋅s
2
×
m
2
40⋅ m ⋅ hr = 40⋅ m ×
3
⎛ 1⋅ m × 1⋅ in × 1⋅ ft ⎞ = 6.36 × 10− 3⋅ ft 3
⎜ 100⋅ cm 0.0254m
12⋅ in ⎠
⋅
⎝
1000⋅ W
1⋅ kW
1⋅ lbf
4.448⋅ N
×
×
1⋅ hp
746⋅ W
= 402⋅ hp
2
⋅
12⋅ in ⎞
lbf ⋅ s
⎛ 0.0254m
= 1.044⋅
⎜ 1⋅ in × 1⋅ ft
2
⎝
⎠
ft
2
⎛ 1⋅ in × 1⋅ ft ⎞ ⋅ hr = 431⋅ ft 2⋅ hr
⎜ 0.0254m
⋅
12⋅ in ⎠
⎝
[Difficulty: 1]
Problem 1.26
[Difficulty: 2]
Given:
Geometry of tank, and weight of propane.
Find:
Volume of propane, and tank volume; explain the discrepancy.
Solution:
Use Table G.2 and other sources (e.g., Google) as needed.
The author's tank is approximately 12 in in diameter, and the cylindrical part is about 8 in. The weight of propane specified is 17 lb.
The tank diameter is
D = 12⋅ in
The tank cylindrical height is
L = 8⋅ in
The mass of propane is
mprop = 17⋅ lbm
The specific gravity of propane is
SG prop = 0.495
The density of water is
ρ = 998⋅
kg
3
m
The volume of propane is given by
mprop
mprop
Vprop =
=
ρprop
SGprop⋅ ρ
3
1
m
Vprop = 17⋅ lbm ×
×
×
0.495
998 ⋅ kg
0.454 ⋅ kg
1 ⋅ lbm
×
⎛ 1⋅ in ⎞
⎜ 0.0254⋅ m
⎝
⎠
3
3
Vprop = 953 ⋅ in
The volume of the tank is given by a cylinder diameter D length L, πD2L/4 and a sphere (two halves) given by πD3/6
2
Vtank =
Vtank =
The ratio of propane to tank volumes is
Vprop
Vtank
π⋅ D
4
3
⋅L +
π⋅ ( 12⋅ in)
4
π⋅ D
6
2
⋅ 8 ⋅ in + π⋅
( 12⋅ in)
6
3
3
Vtank = 1810⋅ in
= 53⋅ %
This seems low, and can be explained by a) tanks are not filled completely, b) the geometry of the tank gave an overestimate of
the volume (the ends are not really hemispheres, and we have not allowed for tank wall thickness).
Problem 1.27
[Difficulty: 1]
Given:
Acreage of land, and water needs.
Find:
Water flow rate (L/min) to water crops.
Solution:
Use Table G.2 and other sources (e.g., Machinery's Handbook, Mark's Standard Handbook) as needed.
The volume flow rate needed is Q 
Performing unit conversions
Q
4 cm
week
 10 hectare
4 cm  10 hectare
Q  397
week
L
min

0.04 m  10 hectare
week
4

2
1  10  m
1 hectare

1000 L
3
m

1 week
7 day

1 day
24 hr

1 hr
60 min
Problem 1.28
Given:
Data in given units
Find:
Convert to different units
[Difficulty: 1]
Solution:
3
(a)
1⋅
3
in
= 1⋅
min
3
(b)
(c)
(d)
1⋅
1⋅
m
s
3
3
⋅
mm
1000⋅ mm⎞
1⋅ min
⎛ 0.0254m
×
×
= 273⋅
⎜
min ⎝ 1⋅ in
s
1⋅ m ⎠
60⋅ s
in
×
3
= 1⋅
liter
min
m
s
= 1⋅
1⋅ gal
×
3
×
4 × 0.000946⋅ m
liter
min
×
3
1 ⋅ gal
4 × 0.946 ⋅ liter
3
×
60⋅ s
1⋅ min
= 15850⋅ gpm
60⋅ s
1 ⋅ min
= 0.264 ⋅ gpm
3
0.0254⋅ m ⎞
m
60⋅ min
1 ⋅ SCFM = 1 ⋅
×
= 1.70⋅
× ⎛⎜
min
1
hr
1 ⋅ hr
⋅ ft
⎜
12
⎝
⎠
ft
Problem 1.29
Given:
Density of mercury.
Find:
Specific gravity, volume and weight.
Solution:
Use basic definitions
SG =
v=
ρ
ρw = 1.94⋅
From Appendix A
ρw
1
ρ
[Difficulty: 1]
slug
ft
so
v =
1
⋅
ft
3
so
3
SG =
26.3
SG = 13.6
1.94
3
−5m
3
×
26.3 slug
⋅ ⎞
1⋅ lbm
1⋅ slug
⎛ 0.3048m
×
×
⎜ 1⋅ ft
0.4536kg
⋅
32.2
⋅
lbm
⎝
⎠
v = 7.37 × 10
γ = ρ⋅ g
Hence on earth
γE = 26.3⋅
slug
ft
On the moon
γM = 26.3⋅
3
× 32.2⋅
slug
ft
3
ft
s
× 5.47⋅
×
2
ft
2
s
1⋅ lbf ⋅ s
2
1⋅ slug ⋅ ft
γE = 847
lbf
ft
2
×
1 ⋅ lbf ⋅ s
1 ⋅ slug⋅ ft
Note that mass-based quantities are independent of gravity
γM = 144
3
lbf
ft
3
kg
Problem 1.30
Given:
Definition of kgf.
Find:
Conversion from psig to kgf/cm2.
Solution:
Use Table G.2.
Define kgf
kgf = 1 ⋅ kg × 9.81⋅
m
2
kgf = 9.81N
s
Then
32⋅
lbf
2
in
×
4.448⋅ N
1⋅ lbf
×
1⋅ kgf
9.81⋅ N
×
2
⎛ 12⋅ in × 1⋅ ft × 1⋅ m ⎞ = 2.25 kgf
⎜ 1⋅ ft
100⋅ cm ⎠
0.3048m
⋅
2
⎝
cm
[Difficulty: 1]
Problem 1.31
[Difficulty: 3]
Given:
Information on canal geometry.
Find:
Flow speed using the Manning equation, correctly and incorrectly!
Solution:
Use Table G.2 and other sources (e.g., Google) as
needed.
The Manning equation is
The given data is
V=
2
1
3
2
Rh ⋅ S0
which assumes Rh in meters and V in m/s.
n
1
S0 =
10
Rh = 7.5⋅ m
1
2
7.5 ⋅ ⎛⎜
3
V=
Hence
Using the equation incorrectly:
⎞
⎝ 10 ⎠
1
2
V = 86.5⋅
0.014
Rh = 7.5⋅ m ×
1 ⋅ in
0.0254⋅ m
24.6 ⋅ ⎛⎜
3
V =
×
m
(Note that we don't cancel units; we just write m/s
next to the answer! Note also this is a very high
speed due to the extreme slope S0.)
s
1 ⋅ ft
Rh = 24.6⋅ ft
12⋅ in
1
2
Hence
n = 0.014
⎞
⎝ 10 ⎠
1
2
V = 191 ⋅
0.014
ft
V = 191 ⋅
This incorrect use does not provide the correct answer
(Note that we again don't cancel units; we just
write ft/s next to the answer!)
s
ft
s
×
12⋅ in
1 ⋅ ft
×
0.0254⋅ m
1 ⋅ in
This demonstrates that for this "engineering" equation we must be careful in its use!
To generate a Manning equation valid for Rh in ft and V in ft/s, we need to do the following:
2
1
2
3
Rh ( m) ⋅ S0
m⎞
1 ⋅ in
1 ⋅ in
1 ⋅ ft ⎞
ft ⎞
1 ⋅ ft
⎛
⎛
=
× ⎛⎜
V⎜
= V⎜
×
×
×
n
⎝s⎠
⎝ s ⎠ 0.0254⋅ m 12⋅ in
⎝ 0.0254⋅ m 12⋅ in ⎠
V = 58.2
m
s
which is wrong!
V⎛⎜
ft ⎞
⎝s⎠
=
2
1
3
2
Rh ( ft) ⋅ S0
n
−
×
⎛ 1⋅ in × 1 ⋅ ft ⎞
⎜ 0.0254⋅ m 12⋅ in
⎝
⎠
2
2
1
3
3
2
×
1
3
⎛ 1 ⋅ in × 1⋅ ft ⎞ = Rh ( ft) ⋅ S0 × ⎛ 1⋅ in × 1 ⋅ ft ⎞
⎜ 0.0254⋅ m 12⋅ in
⎜ 0.0254⋅ m 12⋅ in
n
⎝
⎠
⎝
⎠
1
3
⎛ 1 ⋅ 1 ⎞ = 1.49
⎜ .0254 12
⎝
⎠
In using this equation, we ignore the units and just evaluate the conversion factor
2
Hence
3
ft ⎞ 1.49⋅ Rh ( ft) ⋅ S0
⎛
V⎜
=
n
⎝s⎠
1
2
Handbooks sometimes provide this form of the Manning equation for direct use with BG units. In our case
we are asked to instead define a new value for n:
1
2
n BG =
n
1.49
n BG = 0.0094
where
24.6 ⋅ ⎛⎜
Using this equation with Rh = 24.6 ft: V =
Converting to m/s
⎞
⎝ 10 ⎠
1
2
0.0094
V = 284 ⋅
ft
s
2
1
2
3
3
ft ⎞ Rh ( ft) ⋅ S0
⎛
V⎜
=
n BG
⎝s⎠
×
12⋅ in
1 ⋅ ft
×
0.0254⋅ m
1 ⋅ in
V = 284
ft
V = 86.6
m
s
s
which is the correct
answer!
[Difficulty: 2]
Problem 1.32
Given:
Equation for COPideal and temperature data.
Find:
COPideal, EER, and compare to a typical Energy Star compliant EER value.
Solution:
Use the COP equation. Then use conversions from Table G.2 or other sources (e.g., www.energystar.gov) to
find the EER.
The given data is
The COPIdeal is
TL  ( 20  273)  K
TL  293 K
TH  ( 40  273)  K
TH  313 K
293
COPIdeal 
 14.65
313  293
The EER is a similar measure to COP except the cooling rate (numerator) is in BTU/hr and the electrical input (denominator) is in W:
BTU
EER Ideal  COPIdeal 
hr
W
2545
EER Ideal  14.65 
BTU
hr
746  W
 50.0
BTU
hr W
This compares to Energy Star compliant values of about 15 BTU/hr/W! We have some way to go! We can define the isentropic
efficiency as
ηisen 
EER Actual
EER Ideal
Hence the isentropic efficiency of a very good AC is about 30%.
Problem 1.33
[Difficulty: 2]
Given:
Equation for maximum flow rate.
Find:
Whether it is dimensionally correct. If not, find units of 2.38 coefficient. Write a SI version of the equation
Solution:
Rearrange equation to check units of 0.04 term. Then use conversions from Table G.2 or other sources (e.g., Google)
mmax⋅ T0
2.38 =
"Solving" the equation for the constant 2.38:
At ⋅ p 0
Substituting the units of the terms on the right, the units of the constant are
1
slug
s
×R
2
1
1
1
×
ft
2
×
1
psi
=
slug
s
×R
2
1
×
ft
2
2
×
in
lbf
×
lbf ⋅ s
2
slug ⋅ ft
2
=
2
R ⋅ in ⋅ s
ft
3
1
2
c = 2.38⋅
Hence the constant is actually
2
R ⋅ in ⋅ s
ft
3
For BG units we could start with the equation and convert each term (e.g., At), and combine the result into a new constant, or simply
convert c directly:
1
1
2
c = 2.38⋅
2
2
R ⋅ in ⋅ s
ft
3
1
= 2.38⋅
2
R ⋅ in ⋅ s
ft
3
2
×
2
⎛ K ⎞ × ⎛ 1⋅ ft ⎞ × 1⋅ ft
⎜ 1.8⋅ R
⎜ 12⋅ in
0.3048m
⋅
⎝
⎠
⎝
⎠
1
2
c = 0.04⋅
K ⋅s
m
so
mmax = 0.04⋅
At ⋅ p 0
T0
with At in m2, p 0 in Pa, and T0 in K.
Problem 1.34
[Difficulty: 1]
Given:
Equation for mean free path of a molecule.
Find:
Dimensions of C for a diemsionally consistent equation.
Solution:
Use the mean free path equation. Then "solve" for C and use dimensions.
The mean free path equation is
"Solving" for C, and using dimensions
m
λ  C
C
ρ d
2
λ ρ d
2
m
L
C
M
3
L
M
2
L
0
The constant C is dimensionless.
Problem 1.35
[Difficulty: 1]
Given:
Equation for drag on a body.
Find:
Dimensions of CD.
Solution:
Use the drag equation. Then "solve" for CD and use dimensions.
The drag equation is
1
2
FD = ⋅ ρ⋅ V ⋅ A ⋅ CD
2
"Solving" for CD, and using dimensions
CD =
2⋅ FD
2
ρ⋅ V ⋅ A
F
CD =
M
3
2
×
L
But, From Newton's 2nd law
⎛ L ⎞ × L2
⎜t
⎝ ⎠
Force = Mass ⋅ Acceleration
F = M⋅
or
L
t
Hence
F
CD =
M
3
L
The drag coefficient is dimensionless.
×
⎛ L⎞
2
=
2
⎜t ×L
⎝ ⎠
M⋅ L
t
2
3
×
L
M
×
t
2
2
L
×
1
2
L
=0
2
Problem 1.36
Given:
Data on a container and added water.
Find:
Weight and volume of water added.
Solution:
Use Appendix A.
For the empty container
Wc  3.5 lbf
For the filled container
M total  2.5 slug
The weight of water is then
Ww  M total  g  Wc
[Difficulty: 1]
2
The temperature is
ft
1 lbf  s
Ww  2.5 slug  32.2

 3.5 lbf
2
1 slug  ft
s
Ww  77.0 lbf
90°F  32.2°C
ρ  1.93
and from Table A.7
slug
ft
Hence
Vw 
Mw
Ww
Vw 
g ρ
or
ρ
2
3
3
1 s
1 ft
1 slug  ft
Vw  77.0 lbf 




32.2 ft
1.93 slug
2
1 lbf  s
Vw  1.24ft
3
Problem 1.37
[Difficulty: 1]
Given:
Equation for vibrations.
Find:
Dimensions of c, k and f for a dimensionally consistent equation. Also, suitable units in SI and BG systems.
Solution:
Use the vibration equation to find the diemsions of each quantity
2
m
The first term of the equation is
d x
2
dt
L
M
The dimensions of this are
t
2
Each of the other terms must also have these dimensions.
c
Hence
dx
dt
M L

t
t
f 
c:
L
t

t
k L
so
2
M L
and
2
M L
t
and
2
c
M
k
M
t
t
2
M L
t
Suitable units for c, k, and f are
2
M L
k x 
c
so
2
kg
slug
s
s
k:
kg
slug
2
kg m
f:
2
s
2
s
slug ft
2
s
s
Note that c is a damping (viscous) friction term, k is a spring constant, and f is a forcing function. These are more typically expressed
using F (force) rather than M (mass). From Newton's 2nd law:
F  M
L
t
2
c
Using this in the dimensions and units for c, k, and f we find
c:
N s
lbf  s
m
ft
M
or
k:
F t
2
L t

F t
L
N
lbf
m
ft
F t
2
L
k
F t
2
L t
2

F
f F
L
f:
N
lbf
Problem 1.38
Given:
Specific speed in customary units
Find:
Units; Specific speed in SI units
[Difficulty: 1]
Solution:
1
The units are
rpm⋅ gpm
3
2
or
ft
3
ft
4
3
4
s
2
Using data from tables (e.g. Table G.2)
1
NScu = 2000⋅
rpm⋅ gpm
2
3
4
ft
3
1
1
⎛ 1 ⋅ ft ⎞
2
3
⎜ 12
rpm⋅ gpm
2 ⋅ π⋅ rad 1 ⋅ min ⎛ 4 × 0.000946⋅ m 1 ⋅ min ⎞
NScu = 2000 ×
×
×
×⎜
⋅
×⎜
1 ⋅ rev
60⋅ s
1 ⋅ gal
3
60⋅ s ⎠
⎝
⎝ 0.0254⋅ m ⎠
2
ft
4
1
⎛ m3 ⎞
⋅⎜
s ⎝ s ⎠
NScu = 4.06⋅
3
rad
m
4
2
4
Problem 1.39
Given:
"Engineering" equation for a pump
Find:
SI version
[Difficulty: 1]
Solution:
The dimensions of "1.5" are ft.
The dimensions of "4.5 x 10-5" are ft/gpm2.
Using data from tables (e.g. Table G.2), the SI versions of these coefficients can be obtained
0.0254m
⋅
1.5⋅ ft = 1.5⋅ ft ×
1
12
−5
4.5 × 10
−5
ft
⋅
= 0.457⋅ m
⋅ ft
2
= 4.5⋅ 10
ft
⋅
gpm
2
gpm
×
0.0254m
⋅
1
12
4.5⋅ 10
−5
⋅
ft
2
gpm
= 3450⋅
m
⎛ m3 ⎞
⎜
⎝ s ⎠
2
The equation is
⎛ ⎛ m3 ⎞ ⎞
H( m) = 0.457 − 3450⋅ ⎜ Q⎜
⎝ ⎝ s ⎠⎠
2
⋅ ft
×
60⋅ s ⎞
1quart
⎛ 1⋅ gal
⋅
⎜ 4⋅ quart ⋅
3 1min
0.000946m
⋅
⎝
⎠
2
Problem 1.40
Given:
[Difficulty: 2]
Air at standard conditions – p = 29.9 in Hg, T = 59°F
Uncertainty in p is ± 0.1 in Hg, in T is ± 0.5°F
Note that 29.9 in Hg corresponds to 14.7 psia
Find:
Air density using ideal gas equation of state; Estimate of uncertainty in calculated value.
Solution:
ρ=
in 2
lbf
lb ⋅o R
1
p
144
×
×
= 14.7 2 ×
ft 2
in 53.3 ft ⋅ lbf 519o R
RT
The uncertainty in density is given by
1
⎡⎛ p ∂ρ ⎞ 2 ⎛ T ∂ρ ⎞ 2 ⎤ 2
u ρ = ⎢⎜⎜
u p ⎟⎟ + ⎜⎜
uT ⎟⎟ ⎥
⎢⎣⎝ ρ ∂p ⎠ ⎝ ρ ∂T ⎠ ⎥⎦
p ∂ρ
RT
1
= RT
=
= 1;
ρ ∂p
RT RT
T ∂ρ T
p
p
= ⋅−
=−
= −1;
2
ρ ∂T ρ RT
ρRT
± 0.1
= ± 0.334%
29.9
± 0.5
= ± 0.0963%
uT =
460 + 59
up =
Then
[
]
1
2 2
u ρ = u 2p + (− uT )
[
u ρ = ± 0.348% = ± 2.66 × 10 − 4
]
1
2 2
= ± 0.334% 2 + (− 0.0963% )
lbm
ft 3
Problem 1.41
Given:
[Difficulty: 2]
Air in hot air balloon
p = 759 ± 1 mm Hg T = 60 ± 1°C
Find:
Solution:
(a) Air density using ideal gas equation of state
(b) Estimate of uncertainty in calculated value
We will apply uncertainty concepts.
Governing Equations:
ρ=
p
(Ideal gas equation of state)
R⋅ T
1
2
⎡⎛ x ∂R
⎤2
⎞
1
u R = ± ⎢⎜⎜
u x1 ⎟⎟ + L⎥
⎢⎣⎝ R ∂x1
⎥⎦
⎠
We can express density as:
3
ρ = 101⋅ 10 ×
N
2
×
m
kg⋅ K
287⋅ N ⋅ m
×
(Propagation of Uncertainties)
1
333⋅ K
kg
= 1.06
3
kg
ρ = 1.06
m
3
m
1
So the uncertainty in the density is:
⎡⎛ p ∂ρ ⎞ 2 ⎛ T ∂ρ ⎞ 2 ⎤ 2
u ρ = ± ⎢⎜⎜
u p ⎟⎟ + ⎜⎜
uT ⎟⎟ ⎥
⎢⎣⎝ ρ ∂p ⎠ ⎝ ρ ∂T ⎠ ⎥⎦
Solving each term separately:
1
p ∂ρ
= RT
=1
ρ ∂p
RT
T ∂ρ T ⎛ − p
= ⎜
ρ ∂T ρ ⎝ RT 2
Therefore:
[
p
⎞
= −1
⎟=−
RT
⎠
u ρ = ± (u p ) + (− uT )
2
up =
]
1
2 2
[
uT =
1
759
1
333
= ± (0.1318% ) + (− 0.3003% )
kg ⎞
⎛
u ρ = ±0.328% ⎜ ± 3.47 × 10 −3 3 ⎟
m ⎠
⎝
2
= 0.1318%
⋅
= 0.3003%
⋅
]
1
2 2
Problem 1.42
[Difficulty: 2]
m = 1.62 ± 0.01oz (20 to 1)
D = 1.68 ± 0.01in. (20 to 1)
Given:
Standard American golf ball:
Find:
Density and specific gravity; Estimate uncertainties in calculated values.
Solution:
Density is mass per unit volume, so
ρ=
m
m
m
3
6 m
=4 3 =
=
3
4π (D 2) π D 3
V 3 πR
ρ=
6
π
×1.62 oz ×
1
0.4536 kg
in.3
×
×
= 1130 kg/m 3
3
3
3
3
16 oz
(1.68) in.
(0.0254) m
SG =
and
ρ
ρH O
= 1130
2
kg
m3
×
= 1.13
m 3 1000 kg
1
2
2
⎡⎛ m ∂ρ
⎞ ⎛ D ∂ρ
⎞ ⎤2
u ρ = ⎢⎜⎜
u m ⎟⎟ + ⎜⎜
u D ⎟⎟ ⎥
⎣⎢⎝ ρ ∂m ⎠ ⎝ ρ ∂D ⎠ ⎥⎦
The uncertainty in density is given by
m ∂ρ m 1 ∀
=
= = 1;
ρ ∂m ρ ∀ ∀
um =
D ∂ρ D ⎛
6m ⎞
6 m
= −3;
= ⋅ ⎜ − 3 4 ⎟ = −3
ρ ∂D ρ ⎝ πD ⎠
π ρD 4
± 0.01
= ± 0.617%
1.62
uD =
± 0.1
= ± 0.595%
1.68
Thus
[
]
1
2 2
u ρ = ± u + (− 3u D )
2
m
[
]
1
2 2
= ± 0.617% + (− 3 × 0.595% )
2
u ρ = ±1.89% = ± 21.4
u SG = u ρ = ±1.89% = ± 0.0214
Finally,
ρ = 1130 ± 21.4 kg/m 3
SG = 1.13 ± 0.0214
(20 to 1)
(20 to 1)
kg
m3
Problem 1.43
Given:
[Difficulty: 2]
Pet food can
H = 102 ± 1 mm (20 to 1)
D = 73 ± 1 mm (20 to 1)
m = 397 ± 1 g
(20 to 1)
Find:
Magnitude and estimated uncertainty of pet food density.
Solution:
Density is
ρ=
m
m
4 m
or ρ = ρ ( m, D, H )
=
=
2
∀ πR H π D 2 H
1
2
From uncertainty analysis:
⎡⎛ m ∂ρ
⎞ ⎛ D ∂ρ
⎞ ⎛ H ∂ρ
⎞ ⎤
u ρ = ± ⎢⎜⎜
u m ⎟⎟ + ⎜⎜
u D ⎟⎟ + ⎜⎜
u H ⎟⎟ ⎥
⎢⎣⎝ ρ ∂m ⎠ ⎝ ρ ∂D ⎠ ⎝ ρ ∂H
⎠ ⎥⎦
Evaluating:
m ∂ρ m 4 1
±1
1 4m
um =
=
=
= 1;
= ±0.252%
ρ ∂m ρ π D 2 H ρ πD 2 H
397
D ∂ρ D
4m
1 4m
±1
.
= ( −2)
= ( −2)
= −2; u D =
= ±137%
3
2
ρ ∂D ρ
ρ
73
πD H
πD H
4m
1 4m
H ∂ρ H
±1
= ( −1)
= ( −1)
= −1; u H =
= ±0.980%
2 2
2
ρ ∂H ρ
ρ πD H
102
πD H
2
Substituting:
[
2
2
2
2
u ρ = ±2.92%
∀=
π
4
D2 H =
π
4
× (73) 2 mm 2 × 102 mm ×
m3
9
10 mm
397 g
kg
m
×
= 930 kg m 3
ρ= =
−4 3
∀ 4.27 × 10 m
1000 g
Thus:
]
1
2 2
u ρ = ± (1 × 0.252 ) + (− 2 × 1.37 ) + (− 1 × 0.980)
ρ = 930 ± 27.2 kg m 3 (20 to 1)
3
= 4.27 × 10 −4 m 3
Problem 1.44
Given:
[Difficulty: 2]
Mass flow rate of water determine by collecting discharge over a timed interval is 0.2 kg/s.
Scales can be read to nearest 0.05 kg.
Stopwatch can be read to nearest 0.2 s.
Find:
Estimate precision of flow rate calculation for time intervals of (a) 10 s, and (b) 1 min.
Solution:
Apply methodology of uncertainty analysis, Appendix F:
m =
∆m
∆t
1
Computing equations:
2
2
⎡⎛ ∆m ∂m
⎞ ⎛ ∆t ∂m
⎞ ⎤2
u ∆t ⎟ ⎥
u ∆m ⎟ + ⎜
u m = ± ⎢⎜
⎠ ⎝ m ∂∆t
⎠ ⎦⎥
⎣⎢⎝ m ∂∆m
∆m ∂m
1
= ∆t
= 1 and
m ∂∆m
∆t
Thus
∆t ∂m ∆t 2
∆m
=
⋅ − 2 = −1
m ∂∆t ∆m ∆t
The uncertainties are expected to be ± half the least counts of the measuring instruments.
Tabulating results:
Water
Time
Interval, ∆t
Uncertainty
Error in ∆t
in ∆t
(s)
(s)
Collected,
Error in ∆m
∆m
(kg)
(%)
Uncertainty
Uncertainty
in ∆m

in m
(%)
(%)
(kg)
10
± 0.10
± 1.0
2.0
± 0.025
± 1.25
± 1.60
60
± 0.10
± 0.167
12.0
± 0.025
± 0.208
± 0.267
A time interval of about 15 seconds should be chosen to reduce the uncertainty in results to ± 1 percent.
Problem 1.45
[Difficulty: 3]
Given:
Nominal mass flow rate of water determined by collecting discharge (in a beaker) over a timed
 = 100 g s ; Scales have capacity of 1 kg, with least count of 1 g; Timer has least
interval is m
count of 0.1 s; Beakers with volume of 100, 500, 1000 mL are available – tare mass of 1000 mL
beaker is 500 g.
Find:
Estimate (a) time intervals, and (b) uncertainties, in measuring mass flow rate from using each of
the three beakers.
Solution:
To estimate time intervals assume beaker is filled to maximum volume in case of 100 and 500 mL
beakers and to maximum allowable mass of water (500 g) in case of 1000 mL beaker.
 =
m
Then
Tabulating results
∆m
∆t
∆t =
and
∆m ρ∆∀
=


m
m
∆∀ = 100 mL 500 mL 1000 mL
∆t =
1s
5s
5 s
Apply the methodology of uncertainty analysis, Appendix E. Computing equation:
1
2
2
⎡⎛ ∆m ∂m
⎞ ⎛ ∆t ∂m
⎞ ⎤2
u ∆t ⎟ ⎥
u ∆m ⎟ + ⎜
u m = ± ⎢⎜
⎠ ⎝ m ∂∆t
⎠ ⎥⎦
⎣⎢⎝ m ∂∆m
The uncertainties are ± half the least counts of the measuring instruments: δ∆m = ±0.5 g
1
∆m ∂m
= ∆t = 1 and
m ∂∆m
∆t
∆t ∂m ∆t 2 ∆m
= −1
⋅−
=
m ∂∆t ∆m ∆t 2
δ∆t = 0.05 s
[
2
]
1
2 2
∴ u m = ± u ∆m + (− u ∆t )
Tabulating results:
Beaker
Volume ∆∀
(mL)
100
500
1000
Water
Collected
∆m(g)
100
500
500
Error in ∆m
(g)
Uncertainty
in ∆m (%)
± 0.50
± 0.50
± 0.50
± 0.50
± 0.10
± 0.10
Time
Interval ∆t
(s)
1.0
5.0
5.0
Error in ∆t
(s)
Uncertainty
in ∆t (%)
Uncertainty
 (%)
in m
± 0.05
± 0.05
± 0.05
± 5.0
± 1.0
± 1.0
± 5.03
± 1.0
± 1.0
Since the scales have a capacity of 1 kg and the tare mass of the 1000 mL beaker is 500 g, there is no advantage in
 could be reduced to ± 0.50 percent by using the large beaker if a scale
using the larger beaker. The uncertainty in m
with greater capacity the same least count were available
Problem 1.46
Given:
[Difficulty: 2]
Standard British golf ball:
m = 45.9 ± 0.3 g (20 to 1)
D = 411
. ± 0.3 mm (20 to 1)
Find:
Density and specific gravity; Estimate of uncertainties in calculated values.
Solution:
Density is mass per unit volume, so
ρ=
m
=
∀
ρ=
6
π
m
4
πR 3
3
=
m
3
6 m
=
3
π D3
4π ( D 2)
× 0.0459 kg ×
1
m 3 = 1260 kg m 3
(0.0411) 3
and
ρ
SG =
ρH 2 O
= 1260
kg
m3
×
m3
= 126
.
1000 kg
The uncertainty in density is given by
⎡⎛ m
uρ = ± ⎢⎜⎜
⎢⎣⎝ ρ
m ∂ρ m
=
ρ ∂m ρ
1
2
∂ρ ⎞ ⎛ D ∂ρ ⎞
um ⎟ + ⎜
uD ⎟
∂m ⎟⎠ ⎜⎝ ρ ∂D ⎟⎠
2
⎤2
⎥
⎥⎦
1 ∀
= = 1;
∀ ∀
um = ±
6 m⎞
D ∂D D ⎛
⎛ 6m ⎞
= −3⎜ 4 ⎟ = −3;
= ⎜− 3
4 ⎟
ρ ∂m ρ ⎝ π D ⎠
⎝ πD ⎠
Thus
[
u ρ = ± u m + (− 3u D )
2
]
1
2 2
uD = ±
[
0.3
= ±0.730%
41.1
= ± 0.654 2 + (− 3 × 0.730 )
u ρ = ± 2.29% = ± 28.9 kg m 3
u SG = u ρ = ± 2.29% = ± 0.0289
Summarizing
0.3
= ±0.654%
45.9
ρ = 1260 ± 28.9 kg m 3 (20 to 1)
. ± 0.0289 (20 to 1)
SG = 126
]
1
2 2
Problem 1.47
[Difficulty: 3]
Given:
Soda can with estimated dimensions D = 66.0 ± 0.5 mm, H = 110 ± 0.5 mm. Soda has SG = 1.055
Find:
Volume of soda in the can (based on measured mass of full and empty can); Estimate average
depth to which the can is filled and the uncertainty in the estimate.
Solution:
Measurements on a can of coke give
m f = 386.5 ± 0.50 g, m e = 17.5 ± 0.50 g ∴ m = m f − m e = 369 ± u m g
⎡⎛ m ∂m
f
um
u m = ⎢⎜
⎢⎜⎝ m ∂m f f
⎣
1
2
⎞ ⎛ me ∂m
⎞ ⎤
⎟ +⎜
⎥
⎟
u
m
e ⎟
⎟ ⎜ m ∂m
e
⎠ ⎥⎦
⎠ ⎝
0.5 g
0.50
=±
= ±0.00129, u me = ±
= 0.0286
386.5 g
17.5
u mf
2
2
1
2
2
⎡⎛ 386.5
⎞ ⎛ 17.5
⎞ ⎤2
u m = ± ⎢⎜
×1× 0.00129 ⎟ + ⎜
×1× 0.0286 ⎟ ⎥ = ±0.0019
⎠ ⎝ 369
⎠ ⎦⎥
⎣⎢⎝ 369
Density is mass per unit volume and SG = ρ/ρΗ2Ο so
∀=
m
ρ
=
1
m
m3
kg
= 369 g ×
×
×
= 350 × 10 −6 m 3
ρH 2 O SG
.
1000 kg 1055
1000 g
The reference value ρH2O is assumed to be precise. Since SG is specified to three places beyond the decimal point,
assume uSG = ± 0.001. Then
1
2
⎡⎛ m ∂v ⎞ 2 ⎛ SG ∂v
⎞ ⎤2
u SG ⎟ ⎥
um ⎟ + ⎜
u v = ⎢⎜
⎠ ⎥⎦
⎢⎣⎝ v ∂m ⎠ ⎝ v ∂SG
[
∀=
πD
4
2
2
L or
L=
]
1
2 2
u v = ± (1× 0.0019 ) + (− 1× 0.001)
= ±0.0021 = ±0.21%
4∀ 4 350 ×10 −6 m 3 10 3 mm
= ×
×
= 102 mm
πD 2 π
0.066 2 m 2
m
1
⎡⎛ ∀ ∂L ⎞ 2 ⎛ D ∂L ⎞ 2 ⎤ 2
u L = ⎢⎜
uD ⎟ ⎥
u∀ ⎟ + ⎜
⎢⎣⎝ L ∂∀ ⎠ ⎝ L ∂D ⎠ ⎥⎦
∀ ∂L
4 πD 2
=
=1
L ∂∀ πD 2 4
4∀
4∀
0.5
D ∂L D
= ⋅ −2 3 = −2 2 = −2; u D = ±
= ±0.0076
πD
πD L
66
L ∂D L
[
2
]
1
2 2
u L = ± (1× 0.0021) + (− 2 × 0.0076 )
= ±0.0153 = ±1.53%
Notes:
1.
2.
Printing on the can states the content as 355 ml. This suggests that the implied accuracy of the SG value may be
over stated.
Results suggest that over seven percent of the can height is void of soda.
Problem 1.48
Given:
Data on water
Find:
Viscosity; Uncertainty in viscosity
[Difficulty: 3]
Solution:
The data is:
 5 Ns
A  2.414  10

2
B  247.8 K
C  140  K
T  303  K
m
0.5 K
uT 
The uncertainty in temperature is
u T  0.171 %
293 K
B
Also
μ( T)  A 10
( T C)
 3 Ns
μ ( 293 K )  1.005  10
Evaluating

2
m
A  B ln( 10)
d
μ ( T)  
dT
For the uncertainty
B
10
Hence
u μ( T) 
T
d
μ( T)  u T
μ( T) dT


C T
ln( 10)  B T u T

CT

2
 ( C  T)
2
Evaluating
u μ( T)  1.11 %
Problem 1.49
[Difficulty: 4]
Given:
Dimensions of soda can: D = 66 mm, H = 110 mm
Find:
Measurement precision needed to allow volume to be estimated with an
uncertainty of ± 0.5 percent or less.
Solution:
H
Use the methods of Appendix F:
Computing equations:
D
∀=
πD H
2
4
1
⎡⎛ H ∂∀ ⎞2 ⎛ D ∂∀ ⎞2 ⎤ 2
u ∀ = ± ⎢⎜
uH ⎟ + ⎜
uD ⎟ ⎥
⎣⎢⎝ ∀ ∂H ⎠ ⎝ ∀ ∂D ⎠ ⎥⎦
Since ∀ =
π D2 H
4
, then
∂∀
∂H
= π D4 and
2
∂∀
∂D
δx
δx
= π DH
2 . Letting u D = ± D and u H = ± H , and substituting,
1
1
⎡⎛ 4H π D 2 δ x ⎞2 ⎛ 4D π DH δ x ⎞ 2 ⎤ 2
⎡⎛ δ x ⎞ 2 ⎛ 2δ x ⎞ 2 ⎤ 2
u ∀ = ± ⎢⎜
⎟ +⎜
⎟ ⎥ = ± ⎢⎜ ⎟ + ⎜
⎟ ⎥
2
2
⎢⎣⎝ π D H 4 H ⎠ ⎝ π D H 2 D ⎠ ⎥⎦
⎢⎣⎝ H ⎠ ⎝ D ⎠ ⎦⎥
⎛ δ x ⎞ ⎛ 2δ x ⎞
2
=⎜ ⎟ +⎜
⎟ = (δ x)
⎝H⎠ ⎝ D ⎠
2
Solving,
u∀
2
δx=±
2
u∀
1
1 2
2 2 2
⎣⎡( H ) + ( D ) ⎦⎤
=±
⎡
⎣⎢
(
⎡⎛ 1 ⎞ 2 ⎛ 2 ⎞ 2 ⎤
⎢⎜ ⎟ + ⎜ ⎟ ⎥
⎣⎢⎝ H ⎠ ⎝ D ⎠ ⎦⎥
0.005
1
110 mm
) +(
2
2
66 mm
)
2
1
⎤2
⎦⎥
= ±0.158 mm
Check:
uH = ±
uD = ±
δx
H
δx
D
=±
0.158 mm
= ±1.44 × 10−3
110 mm
=±
0.158 mm
= ±2.39 × 10−3
66 mm
u ∀ = ±[(u H ) 2 + (2u D ) 2 ] 2 = ±[(0.00144) 2 + (0.00478) 2 ] 2 = ±0.00499
1
1
If δx represents half the least count, a minimum resolution of about 2 δx ≈ 0.32 mm is needed.
Problem 1.50
[Difficulty: 3]
Given:
Lateral acceleration, a = 0.70 g, measured on 150-ft diameter skid pad; Uncertainties in Path
deviation ±2 ft; vehicle speed ±0.5 mph
Find:
Estimate uncertainty in lateral acceleration; ow could experimental procedure be improved?
Solution:
Lateral acceleration is given by a = V2/R.
From Appendix F, u a = ±[(2 u v ) 2 + ( u R ) 2 ]1/ 2
From the given data,
V 2 = aR; V = aR = 0.70 × 32.2
Then
uv = ±
and
uR = ±
δV
V
δR
R
= ±0.5
ft
ft
× 75 ft = 41.1
2
s
s
mi
s
ft
hr
×
× 5280
×
= ±0.0178
mi 3600 s
hr 41.1 ft
= ±2 ft ×
1
= ±0.0267
75 ft
so
u a = ± (2 × 0.0178) 2 + (0.0267) 2
1/ 2
= ±0.0445
u a = ±4.45 percent
Experimental procedure could be improved by using a larger circle, assuming the absolute errors in measurement are
constant.
For
D = 400 ft; R = 200 ft
V 2 = aR; V = aR = 0.70 × 32.2
ft
ft
× 200 ft = 67.1 = 45.8 mph
2
s
s
0.5
2
= ±0.0109; u R = ±
= ± 0.0100
45.8
200
2
u a = ± (2 × 0.0109) + 0.0100 2 = ± 0.0240 = ± 2.4%
uV = ±
[
]
Given data:
H=
δL =
δθ =
57.7
0.5
0.2
ft
ft
deg
For this building height, we are to vary θ (and therefore L ) to minimize the uncertainty u H.
Plotting u H vs θ
5
10
15
20
25
30
35
40
45
50
55
60
65
70
75
80
85
uH
4.02%
2.05%
1.42%
1.13%
1.00%
0.95%
0.96%
1.02%
1.11%
1.25%
1.44%
1.70%
2.07%
2.62%
3.52%
5.32%
10.69%
Uncertainty in Height (H = 57.7 ft) vs θ
12%
10%
8%
uH
θ (deg)
6%
4%
2%
0%
0
10
20
30
40
50
60
70
80
90
θ (o)
Optimizing using Solver
θ (deg)
31.4
uH
0.947%
To find the optimum θ as a function of building height H we need a more complex Solver
θ (deg)
50
75
100
125
175
200
250
300
400
500
600
700
800
900
1000
29.9
34.3
37.1
39.0
41.3
42.0
43.0
43.5
44.1
44.4
44.6
44.7
44.8
44.8
44.9
uH
0.992%
0.877%
0.818%
0.784%
0.747%
0.737%
0.724%
0.717%
0.709%
0.705%
0.703%
0.702%
0.701%
0.700%
0.700%
Use Solver to vary ALL θ's to minimize the total u H!
Total u H's:
11.3%
Optimum Angle vs Building Height
50
40
θ (deg)
H (ft)
30
20
10
0
0
100
200
300
400
500
H (ft)
600
700
800
900
1000
Problem 1.52
[Difficulty: 4]
Given:
American golf ball, m = 1.62 ± 0.01 oz, D = 1.68 in.
Find:
Precision to which D must be measured to estimate density within uncertainty of ± 1percent.
Solution:
Apply uncertainty concepts
Definition: Density,
ρ≡
m
∀
∀ = 34 π R 3 = π D6
3
1
2
⎡⎛ x ∂R
⎤2
⎞
u R = ± ⎢⎜ 1
u x1 ⎟ + L⎥
⎢⎣⎝ R ∂x1
⎥⎦
⎠
Computing equation:
From the definition,
ρ = π Dm = π6Dm = ρ (m, D)
3/6
Thus
m ∂ρ
ρ ∂m
= 1 and
D ∂ρ
ρ ∂D
3
= 3 , so
u ρ = ±[(1 u m ) 2 + (3 u D ) 2 ] 2
1
u 2ρ = u m 2 + 9 u 2D
Solving,
u D = ± 13 [u ρ 2 − u m2 ] 2
1
From the data given,
u ρ = ±0.0100
um =
±0.01 oz
= ±0.00617
1.62 oz
1
1
u D = ± [(0.0100) 2 − (0.00617) 2 ] 2 = ±0.00262 or ± 0.262%
3
Since u D = ± δDD , then
δ D = ± D u D = ±1.68 in.x 0.00262 = ± 0.00441 in.
The ball diameter must be measured to a precision of ± 0.00441 in.( ± 0.112 mm) or better to estimate density
within ± 1percent. A micrometer or caliper could be used.
Problem 1.53
[Difficulty: 5]
δV = 0.001⋅
in
δD = 0.0005⋅ in
Given:
Syringe pump to deliver 100 mL/min
Find:
(a) Plot uncertainty in flow rate as a function of bore.
(b) Find combination of piston speed and bore resulting in minimum uncertainty in flow rate.
Solution:
min
We will apply uncertainty concepts.
Q=
Governing Equations:
π
4
2
⋅D ⋅V
(Flow rate in syringe pump)
1
2
⎡⎛ x ∂R
⎤2
⎞
1
u R = ± ⎢⎜⎜
u x1 ⎟⎟ + L⎥
⎢⎣⎝ R ∂x1
⎥⎦
⎠
Now solving for the piston speed in terms of the bore:
V( D) =
(Propagation of Uncertainties)
4⋅ Q
2
π⋅ D
So the uncertainty in the flow rate is:
[
u Q = ± (2u D ) + (uV )
2
]
1
2 2
where
1
2
⎡⎛ D ∂Q ⎞ ⎛ V ∂Q ⎞ ⎤
⎡⎛ D 2Q ⎞ ⎛ V Q
u Q = ± ⎢⎜⎜
u D ⎟⎟ + ⎜⎜
uV ⎟⎟ ⎥ = ± ⎢⎜⎜
u D ⎟⎟ + ⎜⎜
uV
Q
D
Q
V
⎢⎣⎝ Q ∂D ⎠ ⎝ Q ∂V
⎥
⎢
⎠ ⎦
⎠ ⎝
⎣⎝
2
uD =
δD
D
uv =
δV
2
2
The uncertainty is minimized when
V
∂u Q
∂D
⎞
⎟⎟
⎠
2
⎤
⎥
⎥⎦
1
2
=0
1
Substituting expressions in terms of bore we get:
⎡ 32 δD⋅ Q ⎞ 2⎤
⎥
Dopt = ⎢ ⋅ ⎛⎜
⎢ π2 ⎝ δV ⎠ ⎥
⎣
⎦
Substituting all known values yields
Dopt = 1.76⋅ in
Plugging this into the expression for the piston speed yields
in
Vopt = 2.50⋅
min
6
and the uncertainty is
u opt = 0.0694⋅ %
Graphs of the piston speed and the uncertainty in the flowrate as a function of the bore are shown on the following page.
10
0.3
6
0.2
4
0.1
2
Piston Speed
Uncertainty
0
0
1
2
3
Bore (in)
4
0
5
Uncertainty in Flowrate (%)
Piston Speed (in/min)
8
Problem 2.1
Given:
Velocity fields
Find:
Whether flows are 1, 2 or 3D, steady or unsteady.
[Difficulty: 1]
Solution:
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
→ →
V = V ( x , y)
→ →
V = V ( x , y)
→ →
V = V ( x)
→ →
V = V ( x)
→ →
V = V ( x)
→ →
V = V ( x , y)
→ →
V = V ( x , y)
→ →
V = V ( x , y , z)
2D
2D
1D
1D
1D
2D
2D
3D
→ →
V = V ( t)
→ →
V ≠ V ( t)
→ →
V ≠ V ( t)
→ →
V ≠ V ( t)
→ →
V = V ( t)
→ →
V ≠ V ( t)
→ →
V = V ( t)
→ →
V ≠ V ( t)
Unsteady
Steady
Steady
Steady
Unsteady
Steady
Unsteady
Steady
Problem 2.2
Given:
Velocity fields
Find:
Whether flows are 1, 2 or 3D, steady or unsteady.
[Difficulty: 1]
Solution:
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
→ →
V = V ( y)
→ →
V = V ( x)
→ →
V = V ( x , y)
→ →
V = V ( x , y)
→ →
V = V ( x)
→ →
V = V ( x , y , z)
→ →
V = V ( x , y)
→ →
V = V ( x , y , z)
1D
1D
2D
2D
1D
3D
2D
3D
→ →
V = V ( t)
→ →
V ≠ V ( t)
→ →
V = V ( t)
→ →
V = V ( t)
→ →
V = V ( t)
→ →
V ≠ V ( t)
→ →
V = V ( t)
→ →
V ≠ V ( t)
Unsteady
Steady
Unsteady
Unsteady
Unsteady
Steady
Unsteady
Steady
Problem 2.3
Given:
Viscous liquid sheared between parallel disks.
Upper disk rotates, lower fixed.
Velocity field is:

rω z
V = eˆθ
h
Find:
a.
Dimensions of velocity field.
b.
Satisfy physical boundary conditions.


To find dimensions, compare to V = V ( x, y , z ) form.
Solution:


The given field is V = V (r , z ) . Two space coordinates are included, so the field is 2-D.
Flow must satisfy the no-slip condition:

1.
At lower disk, V = 0 since stationary.

z = 0, so V = eˆθ
rω 0
= 0 , so satisfied.
h

2.
At upper disk, V = eˆθ rω since it rotates as a solid body.

z = h, so V = eˆθ
rω h
= eˆθ rω , so satisfied.
h
[Difficulty: 2]
Problem 2.4
Given:
Velocity field
Find:
Equation for streamlines
[Difficulty: 1]
Streamline Plots
Solution:
v
u
So, separating variables
dy

dy
y
dx


B x y
2
2
A x  y

C=1
C=2
C=3
C=4
B y
4
A x
B dx

A x
y (m)
For streamlines
5
3
2
Integrating
The solution is
ln( y ) 
y
B
A
1
 ln( x )  c    ln( x )  c
2
1
C
x
0
1
2
3
x (m)
The plot can be easily done in Excel.
4
5
Problem 2.5
[Difficulty: 2]
Given:
Velocity field
Find:
Equation for streamlines; Plot several in the first quadrant, including one that passes through point (0,0)
Solution:
Governing equation: For streamlines
v
u

dy

dy
dx
Assumption: 2D flow
Hence
v
u
So, separating variables
dy
dx


A y
A x

y
Streamline Plots
x
5
4
Integrating
ln( y )  ln( x )  c
3
The solution is
ln( x  y )  c
or
y
C
y (m)
x
y
C=1
C=2
C=3
C=4
dx
2
1
x
0
1
The plot can be easily done in Excel.
The streamline passing through (0,0) is given by the vertical axis, then the horizontal axis.
The value of A is irrelevant to streamline shapes but IS relevant for computing the velocity at each point.
2
3
x (m)
4
5
Problem 2.6
[Difficulty: 1]
Given:
Velocity field
Find:
Whether field is 1D, 2D or 3D; Velocity components at (2,1/2); Equation for streamlines; Plot
Solution:
The velocity field is a function of x and y. It is therefore 2D.
u = a⋅ x ⋅ y = 2 ⋅
At point (2,1/2), the velocity components are
1
1
2
v = b ⋅ y = −6 ⋅
v
For streamlines
=
u
dy
So, separating variables
y
=
dy
=
dx
× 2⋅ m ×
m⋅ s
b⋅ y
×
m⋅ s
2
a⋅ x ⋅ y
1
2
⎛ 1 ⋅ m⎞
⎜2
⎝
⎠
⋅m
2
u = 2⋅
m
s
3 m
v=− ⋅
2 s
b⋅ y
=
a⋅ x
b dx
⋅
a x
b
b
ln( y ) =
Integrating
a
⋅ ln( x) + c
y = C⋅ x
a
−3
y = C⋅ x
The solution is
The streamline passing through point (2,1/2) is given by
1
2
−3
= C⋅ 2
C =
1 3
⋅2
2
C= 4
y=
4
3
x
20
Streamline for C
Streamline for 2C
Streamline for 3C
Streamline for 4C
16
12
8
4
1
This can be plotted in Excel.
1.3
1.7
2
t=0
x
0.05
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.10
1.20
1.30
1.40
1.50
1.60
1.70
1.80
1.90
2.00
c=1
y
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
c=2
y
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
c=3
y
3.00
3.00
3.00
3.00
3.00
3.00
3.00
3.00
3.00
3.00
3.00
3.00
3.00
3.00
3.00
3.00
3.00
3.00
3.00
3.00
3.00
t =1 s
(### means too large to view)
c=1 c=2 c=3
x
y
y
y
0.05 20.00 40.00 60.00
0.10 10.00 20.00 30.00
0.20
5.00 10.00 15.00
0.30
3.33
6.67 10.00
0.40
2.50
5.00
7.50
0.50
2.00
4.00
6.00
0.60
1.67
3.33
5.00
0.70
1.43
2.86
4.29
0.80
1.25
2.50
3.75
0.90
1.11
2.22
3.33
1.00
1.00
2.00
3.00
1.10
0.91
1.82
2.73
1.20
0.83
1.67
2.50
1.30
0.77
1.54
2.31
1.40
0.71
1.43
2.14
1.50
0.67
1.33
2.00
1.60
0.63
1.25
1.88
1.70
0.59
1.18
1.76
1.80
0.56
1.11
1.67
1.90
0.53
1.05
1.58
2.00
0.50
1.00
1.50
t = 20 s
x
0.05
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.10
1.20
1.30
1.40
1.50
1.60
1.70
1.80
1.90
2.00
c=1
y
######
######
######
######
######
######
######
######
86.74
8.23
1.00
0.15
0.03
0.01
0.00
0.00
0.00
0.00
0.00
0.00
0.00
c=2
y
######
######
######
######
######
######
######
######
173.47
16.45
2.00
0.30
0.05
0.01
0.00
0.00
0.00
0.00
0.00
0.00
0.00
c=3
y
######
######
######
######
######
######
######
######
260.21
24.68
3.00
0.45
0.08
0.02
0.00
0.00
0.00
0.00
0.00
0.00
0.00
Streamline Plot (t = 0)
3.5
c=1
3.0
c=2
c=3
2.5
y
2.0
1.5
1.0
0.5
0.0
0.0
0.5
1.0
1.5
2.0
x
Streamline Plot (t = 1 s)
70
c=1
60
c=2
50
c=3
y
40
30
20
10
0
0.0
0.5
1.0
1.5
2.0
x
Streamline Plot (t = 20 s)
20
18
c=1
16
c=2
14
c=3
y
12
10
8
6
4
2
0
0.0
0.2
0.4
0.6
x
0.8
1.0
1.2
a=
b=
C=
x
0.05
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.10
1.20
1.30
1.40
1.50
1.60
1.70
1.80
1.90
2.00
1
1
0
y
0.16
0.22
0.32
0.39
0.45
0.50
0.55
0.59
0.63
0.67
0.71
0.74
0.77
0.81
0.84
0.87
0.89
0.92
0.95
0.97
1.00
2
y
0.15
0.20
0.27
0.31
0.33
0.35
0.37
0.38
0.39
0.40
0.41
0.41
0.42
0.42
0.43
0.43
0.44
0.44
0.44
0.44
0.45
4
y
0.14
0.19
0.24
0.26
0.28
0.29
0.30
0.30
0.31
0.31
0.32
0.32
0.32
0.32
0.33
0.33
0.33
0.33
0.33
0.33
0.33
6
y
0.14
0.18
0.21
0.23
0.24
0.25
0.26
0.26
0.26
0.27
0.27
0.27
0.27
0.27
0.27
0.27
0.27
0.28
0.28
0.28
0.28
Streamline Plot
1.2
c=0
1.0
c=2
c=4
0.8
c=6
y 0.6
0.4
0.2
0.0
0.0
0.5
1.0
x
1.5
2.0
A = 10
B = 20
C=
1
y
0.50
0.48
0.45
0.43
0.42
0.40
0.38
0.37
0.36
0.34
0.33
0.32
0.31
0.30
0.29
0.29
0.28
0.27
0.26
0.26
0.25
2
y
1.00
0.95
0.91
0.87
0.83
0.80
0.77
0.74
0.71
0.69
0.67
0.65
0.63
0.61
0.59
0.57
0.56
0.54
0.53
0.51
0.50
4
y
2.00
1.90
1.82
1.74
1.67
1.60
1.54
1.48
1.43
1.38
1.33
1.29
1.25
1.21
1.18
1.14
1.11
1.08
1.05
1.03
1.00
6
y
3.00
2.86
2.73
2.61
2.50
2.40
2.31
2.22
2.14
2.07
2.00
1.94
1.88
1.82
1.76
1.71
1.67
1.62
1.58
1.54
1.50
Streamline Plot
3.5
c=1
c=2
3.0
c=4
2.5
c = 6 ((x,y) = (1.2)
2.0
y
x
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.10
1.20
1.30
1.40
1.50
1.60
1.70
1.80
1.90
2.00
1.5
1.0
0.5
0.0
0.0
0.5
1.0
x
1.5
2.0
Problem 2.10
[Difficulty: 2]
Given:
Velocity field
Find:
Equation for streamline through (1,3)
Solution:
For streamlines
v
u
So, separating variables
y
A
dy
y


dy
dx
x

2
A

y
x
x
dx
x
Integrating
ln( y )  ln( x )  c
The solution is
y  C x
which is the equation of a straight line.
For the streamline through point (1,3)
3  C 1
C3
and
y  3 x
For a particle
up 
or
x  dx  A dt
x
dx
dt

A
x
2  A t  c
t
x
2
2 A

c
2 A
Hence the time for a particle to go from x = 1 to x = 2 m is
2
∆t  t( x  2 )  t( x  1 )
∆t 
( 2  m)  c
2 A
2
2

( 1  m)  c
2 A

2
4 m  1 m
2
2  2
m
s
∆t  0.75 s
Problem 2.11
[Difficulty: 3]
Given:
Flow field
Find:
Plot of velocity magnitude along axes, and y = x; Equation for streamlines
Solution:
u
On the x axis, y = 0, so
Plotting
M y
2 π
0
v
M x
2 π
200
v (m/s)
150
100
50
0
0.2
0.4
0.6
0.8
1
x (km)
The velocity is perpendicular to the axis and increases linearly with distance x.
This can also be plotted in Excel.
u
On the y axis, x = 0, so
M y
v
2 π
M x
2 π
0
Plotting
0
0.2
0.4
0.6
u (m/s)
 50
 100
 150
 200
y (km)
The velocity is perpendicular to the axis and increases linearly with distance y.
This can also be plotted in Excel.
0.8
1
u
On the y = x
axis
The flow is perpendicular to line y = x:
M y

2 π
M x
2 π
Slope of line y =
x:
2
r
x y
2 π
u
v
 1
2
2
2
then along y =
r x  x 
x
M
M 2 x
M r
2
2
2
2
u v 
 x x 

2 π
2 π
2 π
Then the magnitude of the velocity along y = x is V 
Plotting
M x
1
Slope of trajectory of
motion:
If we define the radial position:
v
2 x
200
V(m/s)
150
100
50
0
0.2
0.4
0.6
0.8
1
r (km)
This can also be plotted in
Excel.
For
streamlines
So, separating
variables
Integrati
ng
M x
v

u
dy
dx
2 π


M y

x
y
2 π
y  dy  x  dx
y
2
2
2

2
x
2
2
c
The solution
x y C
is
The streamlines form a set of concentric circles.
which is the equation of a
circle.
This flow models a rigid body vortex flow. See Example 5.6 for streamline plots. Streamlines are circular, and the velocity
approaches zero as we approach the center. In Problem 2.10, we see that the streamlines are also circular. In a real tornado, at
large distances from the center, the velocities behave as in Problem 2.10; close to the center, they behave as in this problem.
[Difficulty: 3]
Problem 2.12
Given:
Flow field
Find:
Plot of velocity magnitude along axes, and y = x; Equation of streamlines
Solution:
K y
u
On the x axis, y = 0, so
2
2  π x  y
2

Plotting
0
K x
v
2
2  π x  y

2

K
2  π x
160
v( m/s)
80
1
 0.5
0
0.5
1
 80
 160
x (km)
The velocity is perpendicular to the axis, is very high close to the origin, and falls off to zero.
This can also be plotted in Excel.
u
On the y axis, x = 0, so
K y
2
2
2  π  x  y 
Plotting

K
2  π y
v
K x
2
2
2  π  x  y 
0
160
v( m/s)
80
1
 0.5
0
 80
 160
y (km)
0.5
1
The velocity is perpendicular to the axis, is very high close to the origin, and falls off to zero.
This can also be plotted in Excel.
K x
u
On the y = x axis

2
2  π x  x
The flow is perpendicular to line y = x:
2


K
u
v
2
x y
Then the magnitude of the velocity along y = x is
V
2
2
K
2
4 π
Plotting
2
2

K
4  π x
 1
r
then along y = x
u v 

2
1
Slope of trajectory of motion:
r

2
2  π x  x
Slope of line y = x:
If we define the radial position:
K x
v
4  π x
1

x
2

1
x
2

K
2  π 2  x

x x 
2 x
K
2  π r
160
v( m/s)
80
1
 0.5
0
0.5
1
 80
 160
x (km)
This can also be plotted in Excel.
K x
For streamlines
v

u
dy
dx

2
2 π x  y

2

K y

2
2  π x  y
So, separating variables
Integrating
2

x
y
y  dy  x  dx
y
2
2
The solution is

2

2
x
2
2
c
x y C
which is the equation of a
circle.
Streamlines form a set of concentric circles.
This flow models a vortex flow. See Example 5.6 for streamline plots. Streamlines are circular, and the velocity approaches infinity
as we approach the center. In Problem 2.11, we see that the streamlines are also circular. In a real tornado, at large distances from
the center, the velocities behave as in this problem; close to the center, they behave as in Problem 2.11.
Problem 2.13
[Difficulty: 3]
Given:
Flow field
Find:
Plot of velocity magnitude along axes, and y = x; Equations of streamlines
Solution:
q x
u
On the x axis, y = 0, so
2
2  π x  y
2


Plotting
q
2  π x
v
q y
2
2  π x  y

2
0
100
u (m/s)
50
1
 0.5
0
0.5
1
 50
 100
x (km)
The velocity is very high close to the origin, and falls off to zero. It is also along the axis. This can be plotted in Excel.
q x
u
On the y axis, x = 0, so
2
2  π x  y
Plotting
2

0
v
q y
2
2  π x  y

2

q
2  π y
100
v (m/s)
60
20
1
 0.5
 20 0
 60
 100
y (km)
The velocity is again very high close to the origin, and falls off to zero. It is also along the axis.
This can also be plotted in Excel.
0.5
1
u
On the y = x axis

q x
2
2  π x  x
The flow is parallel to line y = x:
2


q
v
4  π x

v
u
2
x y
2
q
2
u v 
Plotting

q
4  π x
4 π

1
x
2
1
r
then along y = x
2
V
Then the magnitude of the velocity along y = x is

2
1
Slope of trajectory of motion:
r
2
2  π x  x
Slope of line y = x:
If we define the radial position:
q x

1
x
2

2
2
x x 
q
2  π 2  x

2 x
q
2  π r
100
V(m/s)
60
20
1
 0.5
 20 0
0.5
 60
 100
r (km)
This can also be plotted in Excel.
q y

For streamlines
v
u

dy
dx
2
2  π x  y



q x
2
2  π x  y
So, separating variables
dy
y

2



y
x
2
dx
x
Integrating
ln( y )  ln( x )  c
The solution is
y  C x
This flow field corresponds to a sink (discussed in Chapter 6).
which is the equation of a straight line.
1
Problem 2.14
[Difficulty: 2]
Given:
Velocity field
Find:
Proof that the parametric equations for particle motion are x p = c1 ⋅ e
A⋅ t
and y p = c2 ⋅ e
− A⋅ t
; pathline that was at
(2,2) at t = 0; compare to streamline through same point, and explain why they are similar or not.
Solution:
Governing equations:
up =
For pathlines
dx
vp =
dt
dy
v
For streamlines
dt
u
=
dy
dx
Assumption: 2D flow
Hence for pathlines
So, separating variables
dx
Eliminating t
vp =
dy
y
ln( x ) = A⋅ t + C1
x=e
The pathlines are
= A⋅ x
dt
= A⋅ dt
x
Integrating
dx
up =
A⋅ t+ C1
x = c1 ⋅ e
t=
=e
C1 A ⋅ t
⋅e
= c1 ⋅ e
A⋅ t
v
u
So, separating variables
dy
y
=
dy
dx
=−
A⋅ y
A⋅ x
=
=e
C2 − A ⋅ t
⋅e
A A
x ⋅y
= c2 ⋅ e
= const or
x⋅ y = 4
y
x
x
ln( y ) = −ln( x ) + c
The solution is
ln( x ⋅ y ) = c
or
x ⋅ y = const
or
x⋅ y = 4
for given data
The streamline passing through (2,2) and the pathline that started at (2,2) coincide because the flow is steady!
( A A) = const
ln x ⋅ y
dx
Integrating
− A⋅ t
− A⋅ t
⎛ 1 1⎞
⎜ A A
ln⎝ x ⋅ y ⎠ = const or
x
=−
− A⋅ t+ C2
y = c2 ⋅ e
so
For streamlines
= −A⋅ y
= −A⋅ dt
y=e
A⋅ t
⋅ ln⎛⎜
dt
ln( y ) = −A⋅ t + C2
⎞ = − 1 ⋅ ln⎛ y ⎞
A
A ⎜ c2
⎝ c1 ⎠
⎝ ⎠
1
dy
for given data
Problem 2.15
[Difficulty: 2]
Given:
Velocity field
Find:
Proof that the parametric equations for particle motion are x p = c1 ⋅ e
A⋅ t
and y p = c2 ⋅ e
2⋅ A ⋅ t
; pathline that was at
(2,2) at t = 0; compare to streamline through same point, and explain why they are similar or not.
Solution:
Governing equations:
up =
For pathlines
dx
vp =
dt
dy
dt
v
For
streamlines
u
=
dy
dx
Assumption: 2D flow
Hence for pathlines
So, separating variables
up =
dx
Eliminating t
For streamlines
x = c1 ⋅ e
y = c2 ⋅ e
v
=
u
So, separating variables
dy
y
A⋅ t+ C1
dy
y
=
dy
dx
ln⎛
or
y = C⋅ x
= 2 ⋅ A⋅ dt
C1 A ⋅ t
⋅e
= c1 ⋅ e
A⋅ t
y=e
2⋅ A⋅ t+ C2
y = c2 ⋅ e
2⋅ A ⋅ t
=
= 2 ⋅ A⋅ y
dt
ln( y ) = 2 ⋅ A⋅ t + C2
A⋅ t
x
= c2 ⋅ ⎛⎜ ⎞
⎝ c1 ⎠
2 ⋅ A⋅ y
2 ⋅ dx
The solution is
=e
dy
vp =
ln( x ) = A⋅ t + C1
x=e
The pathlines are
= A⋅ x
dt
= A⋅ dt
x
Integrating
dx
A⋅ x
=
2
so
y = c⋅ x
=e
C2 2⋅ A ⋅ t
⋅e
or
y=
2⋅ y
x
ln( y ) = 2 ⋅ ln( x ) + c
Integrating
y ⎞
⎜ 2 =c
⎝x ⎠
2
or
y=
1
2
⋅x
2⋅ A ⋅ t
2⋅ A ⋅ t
2
x
= c2 ⋅ e
2
for given data
The streamline passing through (2,2) and the pathline that started at (2,2) coincide because the flow is steady!
1
2
⋅x
2
for given data
t=0
x
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.10
1.20
1.30
1.40
1.50
1.60
1.70
1.80
1.90
2.00
t =1 s
C=1
y
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
C=2
y
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
C=3
y
3.00
3.00
3.00
3.00
3.00
3.00
3.00
3.00
3.00
3.00
3.00
3.00
3.00
3.00
3.00
3.00
3.00
3.00
3.00
3.00
3.00
x
0.000
0.025
0.050
0.075
0.100
0.125
0.150
0.175
0.200
0.225
0.250
0.275
0.300
0.325
0.350
0.375
0.400
0.425
0.450
0.475
0.500
t = 20 s
C=1
y
1.00
1.00
0.99
0.99
0.98
0.97
0.95
0.94
0.92
0.89
0.87
0.84
0.80
0.76
0.71
0.66
0.60
0.53
0.44
0.31
0.00
C=2
y
1.41
1.41
1.41
1.41
1.40
1.39
1.38
1.37
1.36
1.34
1.32
1.30
1.28
1.26
1.23
1.20
1.17
1.13
1.09
1.05
1.00
C=3
y
1.73
1.73
1.73
1.73
1.72
1.71
1.71
1.70
1.69
1.67
1.66
1.64
1.62
1.61
1.58
1.56
1.54
1.51
1.48
1.45
1.41
x
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.10
1.20
1.30
1.40
1.50
1.60
1.70
1.80
1.90
2.00
C=1
y
1.00
1.00
1.00
0.99
0.98
0.97
0.96
0.95
0.93
0.92
0.89
0.87
0.84
0.81
0.78
0.74
0.70
0.65
0.59
0.53
0.45
C=2
y
1.41
1.41
1.41
1.41
1.40
1.40
1.39
1.38
1.37
1.36
1.34
1.33
1.31
1.29
1.27
1.24
1.22
1.19
1.16
1.13
1.10
C=3
y
1.73
1.73
1.73
1.73
1.72
1.72
1.71
1.70
1.69
1.68
1.67
1.66
1.65
1.63
1.61
1.60
1.58
1.56
1.53
1.51
1.48
Streamline Plot (t = 0)
3.5
c=1
c=2
c=3
3.0
2.5
y
2.0
1.5
1.0
0.5
0.0
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
x
Streamline Plot (t = 1s)
2.0
c=1
c=2
c=3
1.8
1.6
1.4
y
1.2
1.0
0.8
0.6
0.4
0.2
0.0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
x
Streamline Plot (t = 20s)
2.0
c=1
c=2
c=3
1.8
1.6
1.4
y
1.2
1.0
0.8
0.6
0.4
0.2
0.0
0.0
0.5
1.0
1.5
x
2.0
2.5
Problem 2.17
[Difficulty: 4]
Given:
Pathlines of particles
Find:
Conditions that make them satisfy Problem 2.10 flow field; Also Problem 2.11 flow field; Plot pathlines
Solution:
The given pathlines are
x p  a  sin ( ω t)
The velocity field of Problem 2.12 is
u
Ky

2
2 π x  y
y p  a cos( ω t)
K x
v

2
2
2  π x  y

2
If the pathlines are correct we should be able to substitute xp and y p into the velocity field to find the velocity as a function of time:
Ky
u

2
2 π x  y
v
Kx
2 π  x
2
y 
2

2

K  a  cos ( ω t)

2
2
2
2 π a  sin ( ω t)  a  cos ( ω t)

2

K  ( a  sin ( ω t) )
2 π  a  sin ( ω t)
2
2
 a  cos ( ω t) 
2
2


K  cos ( ω t)
2 π a
K  sin ( ω t)
(1)
(2)
2 π a
We should also be able to find the velocity field as a function of time from the pathline equations (Eq. 2.9):
dxp
dt
u
dxp
u
dxp
dt
dt
 a ω cos( ω t)
Comparing Eqs. 1, 2 and 3
u  a ω cos( ω t)  
Hence we see that
a ω 
K
2  π a
K cos( ω t)
2  π a
or
v
(2.9)
v
dyp
dt
 a ω sin( ω t)
v  a ω sin( ω t)  
ω
K
2  π a
(3)
K sin( ω t)
2  π a
for the pathlines to be correct.
2
The pathlines are
a = 300 m
a = 400 m
a = 500 m
400
To plot this in Excel, compute x p and y p
for t ranging from 0 to 60 s, with ω given
by the above formula. Plot y p versus xp.
Note that outer particles travel much
slower!
200
 400
 200
0
200
This is the free vortex flow discussed in
Example 5.6
400
 200
 400
u
The velocity field of Problem 2.11 is
M y
v
2 π
M x
2 π
If the pathlines are correct we should be able to substitute xp and y p into the velocity field to find the velocity as a function of time:
u
v
Recall that
u
M y
2 π
M x
2 π
dxp
dt


M  ( a cos( ω t) )
2 π
M  ( a sin( ω t) )
2 π
u  a ω cos( ω t)  
Hence we see that
ω
2 π

M  a cos( ω t)
(4)
2 π
M  a sin( ω t)
(5)
2 π
 a ω cos( ω t)
Comparing Eqs. 1, 4 and 5
M

v
M  a cos( ω t)
2 π
for the pathlines to be correct.
dyp
dt
 a ω sin( ω t)
v  a ω sin( ω t)  
(3)
M  a sin( ω t)
2 π
The pathlines
To plot this in Excel, compute x p and y p
for t ranging from 0 to 75 s, with ω given
by the above formula. Plot y p versus xp.
Note that outer particles travel faster!
400
200
 400
 200
0
200
400
This is the forced vortex flow discussed in
Example 5.6
 200
 400
 600
a = 300 m
a = 400 m
a = 500 m
Note that this is rigid
body rotation!
Problem 2.18
[Difficulty: 2]
Given:
Time-varying velocity field
Find:
Streamlines at t = 0 s; Streamline through (3,3); velocity vector; will streamlines change with time
Solution:
v
For streamlines
u

dy
At t = 0 (actually all times!)
dx
dy
So, separating variables
y
dy
dx


y

dx
a y  ( 2  cos( ω t) )
a x  ( 2  cos( ω t) )
y
x
x
x
Integrating
ln( y )  ln( x )  c
The solution is
y
C
C 
3
For the streamline through point (3,3)

which is the equation of a hyperbola.
x
3
C1
y
and
1
x
The streamlines will not change with time since dy/dx does not change with time.
At t = 0
5
u  a x  ( 2  cos( ω t) )  5 
1
m
u  45
3
v  a y  ( 2  cos( ω t) )  5 
y
4
2
s
v  45
1
 3 m  3
s
1
s
 3 m  3
m
s
The velocity vector is tangent to the curve;
0
1
2
3
x
4
5
Tangent of curve at (3,3) is
dx
Direction of velocity at (3,3) is
This curve can be plotted in Excel.
dy
v
u

 1
y
x
 1
Problem 2.19
[Difficulty: 3]
Given:
Velocity field
Find:
Plot of pathline traced out by particle that passes through point (1,1) at t = 0; compare to streamlines through same
point at the instants t = 0, 1 and 2s
Solution:
Governing equations:
up =
For pathlines
dx
dt
vp =
dy
v
For streamlines
dt
u
=
dy
dx
Assumption: 2D flow
Hence for pathlines
So, separating variables
Integrating
up =
dx
dt
= A⋅ ( 1 + B⋅ t)
A = 1⋅
⎛
t
⎝
2
2⎞
⎠
s
B = 1⋅
dy
dx = A⋅ ( 1 + B⋅ t) ⋅ dt
x = A⋅ ⎜ t + B⋅
m
y
1
1
2
y=e
The pathlines are
Using given data
2
⎛
t ⎞
x = A⋅ ⎜ t + B⋅
+1
2⎠
⎝
For streamlines
v
u
So, separating variables
Integrating
=
dy
dx
1+ B⋅ t
=
2
s
1
y = c2 ⋅ e
1
y=e
1
2
=e
C2
⋅e
2
1
2
⋅ C⋅ t
= c2 ⋅ e
dy
y
=
C
A
⋅ t⋅ dx
C
A
2
2
⋅ C⋅ t
2
2
⋅ C⋅ t
2
which we can integrate for any given t (t is treated as a constant)
⋅ t⋅ x + c
C
A
⋅ t⋅ x + const
y=
⎛ C ⋅ t⋅ x + const⎞
⎜
⎝A
⎠
( 1+ B⋅ t)
2
⋅ C⋅ t
A⋅ ( 1 + B⋅ t)
( 1 + B⋅ t) ⋅ ln( y ) =
y
1
2
⋅ C⋅ t + C2
2
1
The solution is
C = 1⋅
C⋅ y ⋅ t
=
( 1 + B⋅ t) ⋅
= C⋅ t ⋅ y
⋅ C⋅ t + C2
1
2
⎛
t ⎞
x = A⋅ ⎜ t + B⋅
+ C1
2⎠
⎝
dt
= C⋅ t⋅ dt
ln( y ) =
+ C1
dy
vp =
s
1
y=1
For particles at (1,1) at t = 0, 1, and 2s, using A, B, and C data:
y=x
2
1
y = (2⋅ x − 1)
Streamline and Pathline Plots
5
Streamline (t=0)
Streamline (t=1)
Streamline (t=2)
Pathline
4
y (m)
3
2
1
0
1
2
3
x (m)
4
5
3
Problem 2.20
[Difficulty: 3]
Given:
Velocity field
Find:
Plot of pathline traced out by particle that passes through point (1,1) at t = 0; compare to streamlines through
same point at the instants t = 0, 1 and 2s
Solution:
up =
dx
dt
vp =
= B⋅ x ⋅ ( 1 + A⋅ t)
A = 0.5⋅
For pathlines
Governing equations:
dy
v
For streamlines
dt
u
=
dy
dx
Assumption: 2D flow
up =
Hence for pathlines
dx
So, separating variables
dx
dt
⎛
x=e ⎝
2⎞
t
2⎠
⎝
2
⎠
+ C1
B⋅ ⎜t+ A⋅
t
C1
= e ⋅e ⎝
x = c1 ⋅ e ⎝
⎛
B⋅ ⎜t+ A⋅
For streamlines
x=e ⎝
v
u
So, separating variables
Integrating
=
2⎞
⎛
⎛
Using given data
1
vp =
s
dy
dx
=
( 1 + A⋅ t) ⋅
y
t
B⋅ ⎜t+ A ⋅
The pathlines are
B = 1⋅
dy
⎛
ln( x ) = B⋅ ⎜ t + A⋅
B⋅ ⎜t+ A⋅
s
= B⋅ ( 1 + A⋅ t) ⋅ dt
x
Integrating
1
2⎠
dt
= C⋅ y
C = 1⋅
1
s
= C⋅ dt
ln( y ) = C⋅ t + C2
+ C1
2⎞
dy
⎛
B⋅ ⎜t+ A⋅
= c1 ⋅ e ⎝
2⎞
t
2⎠
y=e
C⋅ t+ C2
=e
C2 C⋅ t
⋅e
= c2 ⋅ e
C⋅ t
2⎞
t
2⎠
y = c2 ⋅ e
C⋅ t
2⎞
t
2⎠
y=e
C⋅ t
C⋅ y
B⋅ x ⋅ ( 1 + A⋅ t)
dy
y
=
C dx
⋅
B x
( 1 + A⋅ t) ⋅ ln( y ) =
C
B
⋅ ln( x ) + c
which we can integrate for any given t (t is treated as a constant)
C
The solution is
y
1+ A ⋅ t
= const ⋅ x
B
y = const ⋅ x
or
C
y=x
For particles at (1,1) at t = 0, 1, and 2s
C
B
y=x
C
( 1+ A )B
y=x
Streamline and Pathline Plots
5
Streamline (t=0)
Streamline (t=1)
Streamline (t=2)
Pathline
4
y (m)
3
2
1
0
1
2
3
x (m)
4
5
( 1+ 2⋅ A )B
Problem 2.21
[Difficulty: 3]
Given:
Eulerian Velocity field
Find:
Lagrangian position function that was at point (1,1) at t = 0; expression for pathline; plot pathline and compare to
streamlines through same point at the instants t = 0, 1 and 2s
Solution:
Governing equations:
up =
For pathlines (Lagrangian description)
dx
dt
vp =
dy
vp =
dy
v
For streamlines
dt
u
Assumption: 2D flow
Hence for pathlines
up =
dx
dt
=A
So, separating variables
dx = A⋅ dt
Integrating
x = A⋅ t + x 0
A = 2
Using given data
The pathlines are given by combining the equations t =
For streamlines
y ( x ) = y 0 − B⋅
v
u
So, separating variables
=
dy
dx
dy = −
=
B⋅ t
A
s
= −B⋅ t
dt
m
B = 2
2
s
dy = −B⋅ t⋅ dt
The Lagrangian description is
Hence
m
( x − x0) 2
2
2⋅ A
t
2
x0 = 1 m
y = −B⋅
x ( t) = A⋅ t + x 0
y ( t) = −B⋅
x ( t) = 2 ⋅ t + 1
y ( t) = 1 − t
x − x0
A
or, using given data
y = −B⋅
2
t
+ y0
t
y0 = 1 m
2
+ y0
2
2
2
2
+ y 0 = −B⋅
y(x) = 1 −
( x − 1)
(x − x0)2
2
2⋅ A
+ y0
2
4
−B⋅ t
A
⋅ dx
which we can integrate for any given t (t is treated as a constant)
=
dy
dx
y=−
The solution is
B⋅ t
y = −
A
⋅x + c
B⋅ t
A
and for the one through (1,1)
⋅ ( x − 1) + 1
1=−
B⋅ t
A
⋅1 + c
c=1+
y = 1 − t⋅ ( x − 1)
x = 1 , 1.1 .. 20
Streamline Plots
20
−40
5
10
15
y (m)
− 28
− 52
− 76
Streamline (t=0)
Streamline (t=1)
Streamline (t=2)
Pathline
− 100
x (m)
20
25
B⋅ t
A
Problem 2.22
[Difficulty: 3]
Given:
Velocity field
Find:
Plot of pathline of particle for t = 0 to 1.5 s that was at point (1,1) at t = 0; compare to streamlines through same
point at the instants t = 0, 1 and 1.5 s
Solution:
Governing equations:
up =
For pathlines
dx
dt
vp =
dy
v
For streamlines
dt
u
=
dy
dx
Assumption: 2D flow
Hence for pathlines
So, separating variables
up =
dx
dt
= ax
ln⎛⎜
⎞ = a⋅ t
x0
⎝ ⎠
x
x ( t) = x 0⋅ e
Using given data
x ( t) = e
v
u
So, separating variables
dy
y
Hence
s
=
=
ln⎛⎜
dy
dx
x0 = 1 m
a⋅ t
vp =
dy
= b ⋅ y ⋅ ( 1 + c⋅ t )
dt
2⋅ t
=
ln⎛⎜
b = 2
1
2
c = 0.4
s
⎞ = b ⋅ ⎛ t + 1 ⋅ c⋅ t2⎞
⎜ 2
⎝
⎠
⎝ y0 ⎠
y
y ( t) = e
dy
y
= b ⋅ ( 1 + c⋅ t) ⋅ dt
y0 = 1 m
⎛ 1 2⎞
b⋅ ⎜t+ ⋅ c⋅ t
⎝ 2 ⎠
y ( t) = e
2
2⋅ t+ 0.4⋅ t
b ⋅ y ⋅ ( 1 + c⋅ t )
a⋅ x
b ⋅ ( 1 + c⋅ t )
a⋅ x
⋅ dx
which we can integrate for any given t (t is treated as a constant)
⎞ = b ⋅ ( 1 + c⋅ t) ⋅ ln⎛ x ⎞
⎜x
⎝ y0 ⎠ a
⎝ 0⎠
y
b
The solution is
1
dy = b ⋅ y ⋅ ( 1 + c⋅ t) ⋅ dt
Hence
For streamlines
a = 2
= a⋅ dt
x
Integrating
dx
x
y = y 0 ⋅ ⎛⎜ ⎞
⎝ x0 ⎠
a
⋅ ( 1+ c⋅ t)
1
s
b
t = 0
x
y = y 0 ⋅ ⎛⎜ ⎞
⎝ x0 ⎠
a
= x
x
t = 1 y = y 0 ⋅ ⎛⎜ ⎞
⎝ x0 ⎠
b
⋅ ( 1+ c⋅ t)
a
= x
1.4
t = 1.5
x
y = y 0 ⋅ ⎛⎜ ⎞
⎝ x0 ⎠
Streamline and Pathline Plots
10
Streamline (t=0)
Streamline (t=1)
Streamline (t=1.5)
Pathline
8
6
y (m)
For
b
⋅ ( 1+ c⋅ t)
4
2
0
2
4
6
x (m)
8
10
⋅ ( 1+ c⋅ t)
a
= x
1.6
Problem 2.23
[Difficulty: 3]
Given:
Velocity field
Find:
Plot of pathline of particle for t = 0 to 1.5 s that was at point (1,1) at t = 0; compare to streamlines through same
point at the instants t = 0, 1 and 1.5 s
Solution:
Governing equations:
For pathlines
up =
dx
a =
1 1
vp =
dt
dy
v
For streamlines
dt
u
Assumption: 2D flow
Hence for pathlines
So, separating variables
up =
dx
= a⋅ x
dt
vp =
5 s
= a⋅ dt
x
Integrating
dx
ln⎛⎜
dy
dt
= b⋅ y⋅ t
⎞ = a⋅ t
x0
⎝ ⎠
x
ln⎛⎜
x ( t) = x 0⋅ e
For streamlines
x ( t) = e
5
v
=
u
So, separating variables
dy
y
Hence
=
=
ln⎛⎜
dy
dx
b⋅ t
a⋅ x
a⋅ t
y
y ( t) = y 0⋅ e
1
25
1
2
s
= b ⋅ t⋅ dt
y0 = 1 m
2
⋅ b⋅ t
2
2
t
y ( t) = e
50
b⋅ y⋅ t
a⋅ x
⋅ dx
which we can integrate for any given t (t is treated as a constant)
⎞ = b ⋅ t⋅ ln⎛ x ⎞
⎜x
⎝ y0 ⎠ a
⎝ 0⎠
y
b
The solution is
y
⎞ = b ⋅ 1 ⋅ t2
2
⎝ y0 ⎠
x0 = 1 m
t
Using given data
dy
dy = b ⋅ y ⋅ t⋅ dt
1
Hence
b =
x
y = y 0 ⋅ ⎛⎜ ⎞
⎝ x0 ⎠
a
⋅t
b
a
= 0.2
x0 = 1
y0 = 1
=
dy
dx
b
x
y = y 0 ⋅ ⎛⎜ ⎞
⎝ x0 ⎠
t = 0
= 1
b
x
y = y 0 ⋅ ⎛⎜ ⎞
x0
⎝ ⎠
t = 5
x
y = y 0 ⋅ ⎛⎜ ⎞
x0
⎝ ⎠
b
= x
b
t = 10
⋅t
a
a
⋅t = 1
⋅t
a
= x
2
b
a
⋅t = 2
Streamline and Pathline Plots
10
8
6
y (m)
For
⋅t
a
4
2
Streamline (t=0)
Streamline (t=1)
Streamline (t=1.5)
Pathline
0
2
4
6
x (m)
8
10
Pathline
t
0.00
0.25
0.50
0.75
1.00
1.25
1.50
1.75
2.00
2.25
2.50
2.75
3.00
3.25
3.50
3.75
4.00
4.25
4.50
4.75
5.00
x
1.00
1.00
1.01
1.03
1.05
1.08
1.12
1.17
1.22
1.29
1.37
1.46
1.57
1.70
1.85
2.02
2.23
2.47
2.75
3.09
3.49
Streamlines
t=0
x
y
1.00
1.00
1.00
0.78
1.00
0.61
1.00
0.47
1.00
0.37
1.00
0.29
1.00
0.22
1.00
0.17
1.00
0.14
1.00
0.11
1.00
0.08
1.00
0.06
1.00
0.05
1.00
0.04
1.00
0.03
1.00
0.02
1.00
0.02
1.00
0.01
1.00
0.01
1.00
0.01
1.00
0.01
y
1.00
0.78
0.61
0.47
0.37
0.29
0.22
0.17
0.14
0.11
0.08
0.06
0.05
0.04
0.03
0.02
0.02
0.01
0.01
0.01
0.01
t=1s
x
1.00
1.00
1.01
1.03
1.05
1.08
1.12
1.17
1.22
1.29
1.37
1.46
1.57
1.70
1.85
2.02
2.23
2.47
2.75
3.09
3.49
t=2s
x
1.00
1.00
1.01
1.03
1.05
1.08
1.12
1.17
1.22
1.29
1.37
1.46
1.57
1.70
1.85
2.02
2.23
2.47
2.75
3.09
3.49
y
1.00
0.97
0.88
0.75
0.61
0.46
0.32
0.22
0.14
0.08
0.04
0.02
0.01
0.01
0.00
0.00
0.00
0.00
0.00
0.00
0.00
y
1.00
0.98
0.94
0.87
0.78
0.68
0.57
0.47
0.37
0.28
0.21
0.15
0.11
0.07
0.05
0.03
0.02
0.01
0.01
0.00
0.00
Pathline and Streamline Plots
1.0
Pathline
Streamline (t = 0)
Streamline (t = 1 s)
Streamline (t = 2 s)
0.8
y
0.6
0.4
0.2
0.0
0.0
0.5
1.0
1.5
2.0
x
2.5
3.0
3.5
Problem 2.25
[Difficulty: 3]
Given:
Flow field
Find:
Pathline for particle starting at (3,1); Streamlines through same point at t = 1, 2, and 3 s
Solution:
dx
For particle paths
dt
Separating variables and integrating
dx
x
dy
= u = a⋅ x ⋅ t
an
d
= a⋅ t⋅ dt
or
ln( x ) =
or
y = b ⋅ t + c2
dy = b ⋅ dt
dt
=v=b
1
2
⋅ a⋅ t + c1
2
Using initial condition (x,y) = (3,1) and the given values for a and b
c1 = ln( 3 ⋅ m)
x = 3⋅ e
The pathline is then
For streamlines (at any time t)
v
u
dx
and
y = 4⋅ t + 1
b
=
a⋅ x ⋅ t
So, separating variables
dy =
b dx
⋅
a⋅ t x
Integrating
y=
b
a⋅ t
c2 = 1 ⋅ m
2
0.05⋅ t
dy
=
an
d
⋅ ln( x ) + c
We are interested in instantaneous streamlines at various times that always pass through point (3,1). Using a and b values:
c=y−
y= 1+
The streamline equation is
b
a ⋅t
40
t
⋅ ln( x) = 1 −
⋅ ln⎛⎜
4
0.1⋅ t
⋅ ln( 3)
x⎞
⎝ 3⎠
30
Pathline
Streamline (t=1)
Streamline (t=2)
Streamline (t=3)
20
y
10
0
1
2
3
4
5
− 10
− 20
These curves can be plotted in
Excel.
x
Problem 2.26
[Difficulty: 4]
Given:
Velocity field
Find:
Plot streamlines that are at origin at various times and pathlines that left origin at these times
Solution:
v
For streamlines
u
=
dy
dx
v 0 ⋅ sin⎡⎢ω⋅ ⎜⎛ t −
⎣ ⎝
=
u0
v 0 ⋅ sin⎡⎢ω⋅ ⎛⎜ t −
So, separating variables (t=const)
x
⎣ ⎝
dy =
u0
u0
v 0 ⋅ cos⎡⎢ω⋅ ⎛⎜ t −
⎞⎤
u0 ⎥
⎠⎦ + c
ω
v 0 ⋅ ⎡⎢cos⎡⎢ω⋅ ⎜⎛ t −
Using condition y = 0 when x = 0
For particle paths, first find x(t)
y=
dx
dt
⎞⎤
⎥
⎠⎦ ⋅ dx
x
⎣ ⎝
y=
Integrating
⎞⎤
u0 ⎥
⎠⎦
x
⎞⎤ − cos( ω⋅ t)⎤
⎥
u0 ⎥
⎠⎦
⎦
x
⎣ ⎣ ⎝
= u = u0
Separating variables and integrating
dx = u 0 ⋅ dt
Using initial condition x = 0 at t = τ
c1 = −u 0 ⋅ τ
x = u 0 ⋅ t + c1
o
r
x = u 0 ⋅ ( t − τ)
x ⎞⎤
= v = v 0 ⋅ sin⎡⎢ω⋅ ⎜⎛ t −
⎥
dt
⎣ ⎝ u 0 ⎠⎦
dy
For y(t) we have
and
dy
dt
This gives streamlines y(x) at each time t
ω
so
dy
⎡ ⎡
= v = v 0 ⋅ sin⎢ω⋅ ⎢t −
dt
⎣ ⎣
u 0 ⋅ ( t − τ) ⎤⎤
u0
⎥⎥
⎦⎦
= v = v 0 ⋅ sin( ω⋅ τ)
Separating variables and integrating
dy = v 0 ⋅ sin( ω⋅ τ) ⋅ dt
y = v 0 ⋅ sin( ω⋅ τ) ⋅ t + c2
Using initial condition y = 0 at t = τ
c2 = −v 0 ⋅ sin( ω⋅ τ) ⋅ τ
y = v 0 ⋅ sin( ω⋅ τ) ⋅ ( t − τ)
The pathline is then
x ( t , τ) = u 0 ⋅ ( t − τ)
y ( t , τ) = v 0 ⋅ sin( ω⋅ τ) ⋅ ( t − τ)
These terms give the path of a particle (x(t),y(t)) that started at t = τ.
0.5
0.25
0
1
2
− 0.25
− 0.5
Streamline t = 0s
Streamline t = 0.05s
Streamline t = 0.1s
Streamline t = 0.15s
Pathline starting t = 0s
Pathline starting t = 0.05s
Pathline starting t = 0.1s
Pathline starting t = 0.15s
The streamlines are sinusoids; the pathlines are straight (once a water particle is fired it travels in a straight line).
These curves can be plotted in Excel.
3
Problem 2.27
Given:
Velocity field
Find:
Plot streakline for first second of flow
[Difficulty: 5]
Solution:
Following the discussion leading up to Eq. 2.10, we first find equations for the pathlines in form
(
x p( t) = x t , x 0 , y 0 , t0
)
and
(
y p( t) = y t , x 0 , y 0 , t0
)
where x 0, y 0 is the position of the particle at t = t0, and re-interprete the results as streaklines
( )
(
x st t0 = x t , x 0 , y 0 , t0
)
and
( )
(
y st t0 = y t , x 0 , y 0 , t0
)
which gives the streakline at t, where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t)
For particle paths, first find x(t)
dx
dt
Separating variables and integrating
= u = u0
dx = u 0 ⋅ dt
(
x = x0 + u0 ⋅ t − t0
o
r
x ⎞⎤
so
= v = v 0 ⋅ sin⎡⎢ω⋅ ⎜⎛ t −
u0 ⎥
dt
⎣ ⎝
⎠⎦
x
⎞⎤
⎡
⎛
0
dy
= v = v 0 ⋅ sin⎢ω⋅ ⎜ t0 −
⎥
u0
dt
⎣ ⎝
⎠⎦
x 0 ⎞⎤
⎡ ⎛
dy = v 0 ⋅ sin⎢ω⋅ ⎜ t0 −
⎥ ⋅ dt
u0
⎣ ⎝
⎠⎦
dy
For y(t) we have
and
Separating variables and integrating
( )
(
The streakline is then
x st t0 = x 0 + u 0 t − t0
With
x0 = y0 = 0
( )
(
x st t0 = u 0 ⋅ t − t0
)
dy
)
(
⎡ ⎡
x 0 + u 0⋅ t − t0
⎣ ⎣
u0
= v = v 0 ⋅ sin⎢ω⋅ ⎢t −
dt
(
( )
( ) (
y st t0 = v 0 ⋅ sin⎡ω⋅ t0 ⎤ ⋅ t − t0
⎣
⎦
Streakline for First Second
y (m)
1
2
4
6
−1
−2
x (m)
This curve can be plotted in Excel. For t = 1, t0 ranges from 0 to t.
)
(
2
0
⎥⎥
⎦⎦
x 0 ⎞⎤
⎡ ⎛
y = y 0 + v 0 ⋅ sin⎢ω⋅ ⎜ t0 −
⎥ ⋅ t − t0
u0
⎣ ⎝
⎠⎦
x 0 ⎞⎤
⎡ ⎛
y st t0 = y 0 + v 0 ⋅ sin⎢ω⋅ ⎜ t0 −
⎥ ⋅ t − t0
u0
⎣ ⎝
⎠⎦
( )
)
)⎤⎤
8
10
)
)
Problem 2.28
[Difficulty: 4]
Given:
Velocity field
Find:
Plot of streakline for t = 0 to 3 s at point (1,1); compare to streamlines through same point at the instants t = 0, 1
and 2 s
Solution:
Governing equations:
For pathlines
up =
dx
vp =
dt
dy
v
For streamlines
dt
u
=
dy
dx
Following the discussion leading up to Eq. 2.10, we first find equations for the pathlines in form
(
x p( t) = x t , x 0 , y 0 , t0
( )
(
)
x st t0 = x t , x 0 , y 0 , t0
)
(
)
and
y p( t) = y t , x 0 , y 0 , t0
and
y st t0 = y t , x 0 , y 0 , t0
( )
(
)
which gives the streakline at t, where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t)
Assumption: 2D flow
For pathlines
So, separating variables
up =
dx
x
Integrating
dx
dt
= B⋅ x ⋅ ( 1 + A⋅ t)
A = 0.5
1
s
B = 1
1
s
dy
= B⋅ ( 1 + A⋅ t) ⋅ dt
y
2
2
t − t0 ⎞
⎜⎛
x ⎞
⎛
ln⎜
= B⋅ ⎜ t − t0 + A⋅
2
x0
⎝
⎠
⎝ ⎠
ln⎛⎜
⎝
2
2
t − t0 ⎞
⎜⎛
B⋅ ⎜t− t0+ A⋅
2 ⎠
x = x0⋅ e ⎝
The pathlines are
vp =
dy
dt
= C⋅ y
C = 1
= C⋅ dt
y
y0
⎞ = C⋅ t − t
( 0)
⎠
y = y0⋅ e
2
2
⎛⎜
t − t0 ⎞
B⋅ ⎜t− t0+ A ⋅
2 ⎠
x p( t) = x 0⋅ e ⎝
( )
C⋅ t− t0
y p( t) = y 0⋅ e
( )
C⋅ t− t0
where x 0, y 0 is the position of the particle at t = t0. Re-interpreting the results as streaklines:
The streaklines are then
2
2
t − t0 ⎞
⎜⎛
B⋅ ⎜t− t0+ A⋅
2 ⎠
x st( t0 ) = x 0 ⋅ e ⎝
( )
y st t0 = y 0 ⋅ e
( )
C⋅ t− t0
where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t)
1
s
v
For streamlines
u
So, separating variables
=
dy
dx
=
( 1 + A⋅ t) ⋅
C⋅ y
B⋅ x ⋅ ( 1 + A⋅ t)
dy
y
=
C dx
⋅
B x
C
( 1 + A⋅ t) ⋅ ln( y ) =
Integrating
B
which we can integrate for any given t (t is treated as a constant)
⋅ ln( x ) + const
C
The solution is
y
1+ A ⋅ t
= const ⋅ x
B
2
For particles at (1,1) at t = 0, 1, and 2s
y=x
y=x
1
3
y=x
2
Streamline and Pathline Plots
10
Streamline (t=0)
Streamline (t=1)
Streamline (t=2)
Streakline
8
y (m)
6
4
2
0
2
4
6
x (m)
8
10
Problem 2.29
[Difficulty: 4]
Given:
Velocity field
Find:
Plot of streakline for t = 0 to 3 s at point (1,1); compare to streamlines through same point at the instants t = 0, 1
and 2 s
Solution:
Governing equations:
For pathlines
up =
dx
vp =
dt
dy
v
For streamlines
dt
u
=
dy
dx
Following the discussion leading up to Eq. 2.10, we first find equations for the pathlines in form
(
x p( t) = x t , x 0 , y 0 , t0
( )
(
)
x st t0 = x t , x 0 , y 0 , t0
)
(
)
and
y p( t) = y t , x 0 , y 0 , t0
and
y st t0 = y t , x 0 , y 0 , t0
( )
(
)
which gives the streakline at t, where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t)
Assumption: 2D flow
For pathlines
So, separating variables
up =
dx
x
Integrating
dx
dt
= a⋅ x ⋅ ( 1 + b ⋅ t )
a = 1
= a⋅ ( 1 + b ⋅ t) ⋅ dt
2
2
⎛⎜
t − t0 ⎞
x ⎞
⎛
ln⎜
= a⋅ ⎜ t − t 0 + b ⋅
2
⎝
⎠
⎝ x0 ⎠
2
2
⎛⎜
t − t0 ⎞
a⋅ ⎜t− t0+ b⋅
2 ⎠
x = x0⋅ e ⎝
1
s
b =
1
1
5
s
vp =
dy
y
ln⎛⎜
⎝
dy
dt
= c⋅ y
c = 1
= c⋅ dt
y
y0
⎞ = c⋅ t − t
( 0)
⎠
y = y0⋅ e
( )
c⋅ t− t0
1
s
2
2
⎛⎜
t − t0 ⎞
a⋅ ⎜t− t0+ b⋅
2 ⎠
x p( t) = x 0⋅ e ⎝
The pathlines are
y p( t) = y 0⋅ e
( )
c⋅ t− t0
where x 0, y 0 is the position of the particle at t = t0. Re-interpreting the results as streaklines:
The streaklines are then
2
2
⎛⎜
t − t0 ⎞
a⋅ ⎜t− t0+ b⋅
2 ⎠
x st( t0 ) = x 0 ⋅ e ⎝
( )
y st t0 = y 0 ⋅ e
( )
c⋅ t− t0
where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t)
v
For streamlines
u
So, separating variables
=
dy
dx
=
( 1 + b ⋅ t) ⋅
c⋅ y
a⋅ x ⋅ ( 1 + b ⋅ t )
dy
y
=
c dx
⋅
a x
( 1 + b ⋅ t) ⋅ ln( y ) =
Integrating
c
a
which we can integrate for any given t (t is treated as a constant)
⋅ ln( x ) + const
c
The solution is
y
1+ b⋅ t
= const ⋅ x
a
2
y=x
For particles at (1,1) at t = 0, 1, and 2s
y=x
3
1
y=x
2
Streamline and Pathline Plots
5
Streamline (t=0)
Streamline (t=1)
Streamline (t=2)
Streakline
4
y (m)
3
2
1
0
1
2
3
x (m)
4
5
Problem 2.30
[Difficulty: 4]
Given:
Velocity field
Find:
Plot of pathline for t = 0 to 3 s for particle that started at point (1,2) at t = 0; compare to streakline through same
point at the instant t = 3
Solution:
Governing equations:
up =
For pathlines
dx
vp =
dt
dy
dt
Following the discussion leading up to Eq. 2.10, we first find equations for the pathlines in form
(
x p( t) = x t , x 0 , y 0 , t0
( )
(
)
x st t0 = x t , x 0 , y 0 , t0
)
(
)
and
y p( t) = y t , x 0 , y 0 , t0
and
y st t0 = y t , x 0 , y 0 , t0
( )
(
)
which gives the streakline at t, where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t)
Assumption: 2D flow
For pathlines
So, separating variables
up =
dx
dt
= a⋅ x ⋅ t
ln⎛⎜
⎞ = a ⋅ ⎛ t2 − t 2⎞
0 ⎠
⎝
⎝ x0 ⎠ 2
x = x0⋅ e
⋅ ⎛t − t0
2 ⎝
2
a
x p( t) = x 0⋅ e
1
4
1
2
s
b =
1
m
3
s
vp =
dy
dt
=b
dy = b ⋅ dt
x
a
The pathlines are
a =
= a⋅ t⋅ dt
x
Integrating
dx
2
2⎞
⎠
⋅ ⎛t − t0
⎝
2
(
)
(
)
y − y0 = b⋅ t − t0
y = y0 + b⋅ t − t0
2⎞
⎠
(
y p( t) = y 0 + b ⋅ t − t0
)
where x 0, y 0 is the position of the particle at t = t0. Re-interpreting the results as streaklines:
a
The pathlines are then
( )
x st t0 = x 0 ⋅ e
2
⋅ ⎛t − t0
⎝
2
2⎞
⎠
( )
(
y st t0 = y 0 + b ⋅ t − t0
where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t)
)
Streakline and Pathline Plots
2
Streakline
Pathline
y (m)
1.5
1
0.5
0
1
2
x (m)
3
4
Problem 2.31
[Difficulty: 4]
Given:
2D velocity field
Find:
Streamlines passing through (6,6); Coordinates of particle starting at (1,4); that pathlines, streamlines and
streaklines coincide
Solution:
v
For streamlines
u
=
a⋅ y
Integrating
3
dy
dx
b
=
a⋅ y
⌠
⌠
⎮
2
⎮ a ⋅ y dy = ⎮ b dx
⌡
⌡
or
2
3
= b⋅ x + c
For the streamline through point (6,6)
c = 60 and
For particle that passed through (1,4) at t = 0
u=
dx
v=
dy
dt
dt
= a⋅ y
3
y = 6 ⋅ x + 180
⌠
⌠
⎮
2
⎮ 1 dx = x − x 0 = ⎮ a ⋅ y dt
⌡
⌡
2
⌠
⌠
⎮ 1 dy = ⎮ b dt
⌡
⌡
=b
t
⎛
⌠
2
x − x 0 = ⎮ a ⋅ y 0 + b ⋅ t dt
⌡
Then
(
)
x0 = 1
y0 = 4
2
x = 1 + 16⋅ t + 8 ⋅ t +
y = y0 + b⋅ t = y0 + 2⋅ t
x = x 0 + a⋅ ⎜ y 0 ⋅ t + b ⋅ y 0 ⋅ t +
2
2
4 3
⋅t
3
t
⌠
2
x − x 0 = ⎮ a ⋅ y 0 + b ⋅ t dt
⎮
⌡t
(
)
y = 6⋅ m
⌠
⌠
⎮ 1 dy = ⎮ b dt
⌡
⌡
⎡
(
y = y0 + b⋅ t − t0
x = x 0 + a⋅ ⎢y 0 ⋅ t − t0 + b ⋅ y 0 ⋅ ⎛ t − t0
⎝
⎣
2
(
)
2
2⎞
⎠
+
2
)
⋅ ⎛ t − t0
3 ⎝
b
3
3
(
)
3
4
x = −3 +
Evaluating at t = 3
x = 31.7⋅ m
⋅ t −1 =
(
3
1
3
⋅ 4 ⋅ t − 13
This is a steady flow, so pathlines, streamlines and streaklines always coincide
)
y = 2⋅ ( t − 1)
y = 4⋅ m
⎤
⎥
⎠⎦
3⎞
0
Hence, with x 0 = -3, y 0 = 0 at t0 = 1
3
x = 26.3⋅ m
At t = 1 s
y = 4 + 2⋅ t
For particle that passed through (-3,0) at t = 1
2 3⎞
b ⋅t
⎝
0
Hence, with
We need y(t)
⎠
Problem 2.32
Solution
Pathlines:
t
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
1.80
2.00
2.20
2.40
2.60
2.80
3.00
3.20
3.40
3.60
3.80
4.00
[Difficulty: 3]
The particle starting at t = 3 s follows the particle starting at t = 2 s;
The particle starting at t = 4 s doesn't move!
Starting at t = 0
x
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
1.80
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
2.00
Starting at t = 1 s
y
0.00
0.40
0.80
1.20
1.60
2.00
2.40
2.80
3.20
3.60
4.00
3.80
3.60
3.40
3.20
3.00
2.80
2.60
2.40
2.20
2.00
Starting at t = 2 s
x
y
x
y
0.00
0.20
0.40
0.60
0.80
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
0.00
0.40
0.80
1.20
1.60
2.00
1.80
1.60
1.40
1.20
1.00
0.80
0.60
0.40
0.20
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
-0.20
-0.40
-0.60
-0.80
-1.00
-1.20
-1.40
-1.60
-1.80
-2.00
Streakline at t = 4 s
x
2.00
1.80
1.60
1.40
1.20
1.00
0.80
0.60
0.40
0.20
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
Pathline and Streakline Plots
4
3
2
1
y
0
-0.5
0.0
0.5
1.0
1.5
Pathline starting at t = 0
-1
Pathline starting at t = 1 s
Pathline starting at t = 2 s
-2
Streakline at t = 4 s
-3
x
2.0
2.5
y
2.00
1.60
1.20
0.80
0.40
0.00
-0.40
-0.80
-1.20
-1.60
-2.00
-1.80
-1.60
-1.40
-1.20
-1.00
-0.80
-0.60
-0.40
-0.20
0.00
Problem 2.33
[Difficulty: 3]
Given:
Velocity field
Find:
Equation for streamline through point (1.1); coordinates of particle at t = 5 s and t = 10 s that was at (1,1) at t = 0;
compare pathline, streamline, streakline
Solution:
Governing equations:
v
For streamlines
u
=
dy
For pathlines
dx
up =
dx
vp =
dt
dy
dt
Assumption: 2D flow
Given data
For streamlines
a =
1 1
v
dy
So, separating variables
a
b
Integrating
The solution is then
5 s
=
u
b = 1
⋅ dy =
Hence
t0 = 0
dx
x
(
)
b
x
y = y 0 + ⋅ ln⎛⎜ ⎞ = 5 ⋅ ln( x ) + 1
a
x0
up =
dx
ln⎛⎜
⎝
The pathlines are
y0 = 1
x
⋅ y − y 0 = ln⎛⎜ ⎞
b
⎝ x0 ⎠
a
x
Integrating
x0 = 1
a⋅ x
⎝
Hence for pathlines
s
b
=
dx
m
dx
dt
⎠
= a⋅ x
= a⋅ dt
x
x0
dy
dt
=b
dy = b ⋅ dt
⎞ = a⋅ t − t
( 0)
⎠
x = x0⋅ e
vp =
( )
a⋅ t− t0
(
)
(
)
y − y0 = b⋅ t − t0
y = y0 + b⋅ t − t0
or
b
x
y = y 0 + ⋅ ln⎛⎜ ⎞
a
x0
⎝
⎠
For a particle that was at x 0 = 1 m, y 0 = 1 m at t0 = 0 s, at time t = 1 s we find the position is
x = x0⋅ e
( )
a⋅ t− t0
1
= e
5
(
For a particle that was at x 0 = 1 m, y 0 = 1
x = x0⋅ e
( )
a⋅ t− t0
= e
( )
a⋅ t− t0
= e
2
m
m at t0 = 0 s, at time t = 5 s we find the position is
(
)
y = y 0 + b ⋅ t − t0 = 6
m
For a particle that was at x 0 = 1 m, y 0 = 1
x = x0⋅ e
)
y = y 0 + b ⋅ t − t0 = 2
m
m
at t0 = 0 s, at time t = 10 s we find the position is
(
)
y = y 0 + b ⋅ t − t0 = 11 m
m
For this steady flow streamlines, streaklines and pathlines coincide
Streamline and Position Plots
15
Streamline
Position at t = 1 s
Position at t = 5 s
Position at t = 10 s
12
y (m)
9
6
3
0
2
4
6
x (m)
8
10
Problem 2.34
[Difficulty: 3]
Given:
Velocity field
Find:
Equation for streamline through point (2.5); coordinates of particle at t = 2 s that was at (0,4) at t = 0; coordinates of
particle at t = 3 s that was at (1,4.25) at t = 1 s; compare pathline, streamline, streakline
Solution:
Governing equations:
v
For streamlines
u
=
dy
dx
up =
For pathlines
dx
dt
vp =
dy
dt
Assumption: 2D flow
Given data
For streamlines
a = 2
v
u
So, separating variables
a
b
Integrating
=
m
b = 1
s
dy
dx
1
s
x0 = 2
y0 = 5
x = 1
x = x
b⋅ x
=
a
⋅ dy = x ⋅ dx
1
2
2
⋅ y − y0 = ⋅ ⎛ x − x0 ⎞
⎝
⎠
2
b
a
(
)
2
The solution is then
x
2
2
y = y0 +
⋅ ⎛ x − x0 ⎞ =
+4
⎝
⎠
2⋅ a
4
Hence for pathlines
up =
b
dx
dt
=a
Hence
dx = a⋅ dt
Integrating
x − x 0 = a⋅ t − t 0
vp =
dy
dt
= b⋅ x
dy = b ⋅ x ⋅ dt
(
)
(
)
dy = b ⋅ ⎡x 0 + a⋅ t − t0 ⎤ ⋅ dt
⎣
⎦
a
2
2
y − y 0 = b ⋅ ⎡⎢x 0 ⋅ t − t0 + ⋅ ⎛ ⎛ t − t0 ⎞ ⎞ − a⋅ t0 ⋅ t − t0 ⎤⎥
⎝
⎝
⎠⎠
2
⎣
⎦
(
The pathlines are
(
x = x 0 + a⋅ t − t 0
)
)
(
)
a
2
2
y = y 0 + b ⋅ ⎡⎢x 0 ⋅ t − t0 + ⋅ ⎛ ⎛ t − t0 ⎞ ⎞ − a⋅ t0 ⋅ t − t0 ⎤⎥
⎝
⎝
⎠⎠
2
⎣
⎦
(
)
(
)
For a particle that was at x 0 = 0 m, y 0 = 4 m at t0 = 0s, at time t = 2 s we find the position is
(
)
x = x 0 + a⋅ t − t 0 = 4 m
a
2
2
y = y 0 + b ⋅ ⎡⎢x 0 ⋅ t − t0 + ⋅ ⎛ ⎛ t − t0 ⎞ ⎞ − a⋅ t0 ⋅ t − t0 ⎥⎤ = 8m
⎝
⎝
⎠⎠
2
⎣
⎦
(
)
(
)
For a particle that was at x 0 = 1 m, y 0 = 4.25 m at t0 = 1 s, at time t = 3 s we find the position is
(
)
x = x 0 + a⋅ t − t 0 = 5 m
a
2
2
y = y 0 + b ⋅ ⎡⎢x 0 ⋅ t − t0 + ⋅ ⎛ ⎛ t − t0 ⎞ ⎞ − a⋅ t0 ⋅ t − t0 ⎥⎤ = 10.25
m
⎝
⎝
⎠⎠
2
⎣
⎦
(
)
(
)
For this steady flow streamlines, streaklines and pathlines coincide; the particles refered to are the same particle!
Streamline and Position Plots
15
Streamline
Position at t = 1 s
Position at t = 5 s
Position at t = 10 s
12
y (m)
9
6
3
0
1.2
2.4
3.6
x (m)
4.8
6
Problem 2.35
[Difficulty: 4]
Given:
Velocity field
Find:
Coordinates of particle at t = 2 s that was at (1,2) at t = 0; coordinates of particle at t = 3 s that was at (1,2) at t = 2 s;
plot pathline and streakline through point (1,2) and compare with streamlines through same point at t = 0, 1 and 2 s
Solution
:
Governing equations:
For pathlines
up =
dx
dy
vp =
dt
For
streamlines
dt
v
u
=
dy
dx
Following the discussion leading up to Eq. 2.10, we first find equations for the pathlines in form
(
x p( t) = x t , x 0 , y 0 , t0
( )
)
(
x st t0 = x t , x 0 , y 0 , t0
)
(
)
and
y p( t) = y t , x 0 , y 0 , t0
and
y st t0 = y t , x 0 , y 0 , t0
( )
(
)
which gives the streakline at t, where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t)
Assumption: 2D flow
Given data
Hence for pathlines
a = 0.2
up =
dx
dt
1
s
b = 0.4
m
2
s
= a⋅ y
vp =
dy
dt
= b⋅ t
Hence
dx = a⋅ y ⋅ dt
dy = b ⋅ t⋅ dt
For x
b 2
2
dx = ⎡⎢a⋅ y 0 + a⋅ ⋅ ⎛ t − t0 ⎞⎤⎥ ⋅ dt
⎝
⎠⎦
2
⎣
Integrating
⎡⎢ 3 t 3
⎥⎤
t
0
2
x − x 0 = a⋅ y 0 ⋅ ( t − t 0 ) + a⋅ ⋅ ⎢ −
− t0 ⋅ ( t − t0 )⎥
3
2 ⎣3
⎦
The pathlines are
⎡⎢ 3 t 3
⎥⎤
t
0
2
x ( t ) = x 0 + a⋅ y 0 ⋅ ( t − t 0 ) + a⋅ ⋅ ⎢ −
− t0 ⋅ ( t − t0 )⎥
3
2 ⎣3
⎦
b 2
2
y − y0 = ⋅ ⎛ t − t0 ⎞
⎠
2 ⎝
b
b
These give the position (x,y) at any time t of a particle that was at (x 0,y 0) at time t0
Note that streaklines are obtained using the logic of the Governing equations, above
b 2
2
y ( t) = y0 + ⋅ ⎛ t − t0 ⎞
⎠
2 ⎝
The streaklines are
⎡⎢ 3 t 3
⎥⎤
t
0
2
x ( t 0 ) = x 0 + a⋅ y 0 ⋅ ( t − t 0 ) + a⋅ ⋅ ⎢ −
− t0 ⋅ ( t − t0 )⎥
3
2 ⎣3
⎦
b 2
2
y t0 = y 0 + ⋅ ⎛ t − t0 ⎞
⎝
⎠
2
( )
b
These gives the streakline at t, where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t)
For a particle that was at x 0 = 1 m, y 0 = 2 m at t0 = 0s, at time t = 2 s we find the position is (from pathline equations)
⎡⎢ 3 t 3
⎥⎤
t
0
2
x = x 0 + a⋅ y 0 ⋅ ( t − t 0 ) + a⋅ ⋅ ⎢ −
− t0 ⋅ ( t − t0 )⎥ = 1.91m
3
2 ⎣3
⎦
b 2
2
y = y 0 + ⋅ ⎛ t − t0 ⎞ = 2.8 m
⎝
⎠
2
b
For a particle that was at x 0 = 1 m, y 0 = 2 m at t0 = 2 s, at time t = 3 s we find the position is
⎡⎢ 3 t 3
⎥⎤
t
0
2
x = x 0 + a⋅ y 0 ⋅ ( t − t 0 ) + a⋅ ⋅ ⎢ −
− t0 ⋅ ( t − t0 )⎥ = 1.49m
3
2 ⎣3
⎦
b 2
2
y = y 0 + ⋅ ⎛ t − t0 ⎞ = 3.0
⎠
2 ⎝
b
For streamlines
v
u
So, separating variables
=
dy
y ⋅ dy =
2
Integrating
=
dx
b
a
y − y0
⋅ t⋅ dx
2
=
2
The streamlines are then
y =
b⋅ t
a⋅ y
2
y0 +
where we treat t as a constant
b⋅ t
a
(
⋅ x − x0
2⋅ b⋅ t
a
(
)
and we have
)
⋅ x − x0 =
x0 = 1 m
4 ⋅ t⋅ ( x − 1) + 4
y0 = 2
m
m
Streamline Plots
Pathline Plots
5
15
Pathline (t0=0)
Pathline (t0=2)
Streakline
12
3
y (m)
y (m)
4
9
2
6
1
3
0
0.6
1.2
x (m)
Streamline (t=0)
Streamline (t=1)
Streamline (t=2)
Streamline (t=3)
1.8
2.4
3
0
2
4
6
x (m)
8
10
Problem 2.36
[Difficulty: 4]
Given:
Velocity field
Find:
Coordinates of particle at t = 2 s that was at (2,1) at t = 0; coordinates of particle at t = 3 s that was at (2,1) at t = 2 s;
plot pathline and streakline through point (2,1) and compare with streamlines through same point at t = 0, 1 and 2 s
Solution:
Governing equations:
For pathlines
up =
dx
vp =
dt
dy
v
For
streamlines
dt
u
=
dy
dx
Following the discussion leading up to Eq. 2.10, we first find equations for the pathlines in form
(
x p( t) = x t , x 0 , y 0 , t0
( )
)
(
x st t0 = x t , x 0 , y 0 , t0
)
(
)
and
y p( t) = y t , x 0 , y 0 , t0
and
y st t0 = y t , x 0 , y 0 , t0
( )
(
)
which gives the streakline at t, where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t)
Assumption: 2D flow
Given data
m
a = 0.4
2
b = 2
s
Hence for pathlines
up =
dx
dt
m
2
s
= a⋅ t
vp =
dy
dt
=b
Hence
dx = a⋅ t⋅ dt
dy = b ⋅ dt
Integrating
a 2
2
x − x0 = ⋅ ⎛ t − t0 ⎞
⎝
⎠
2
y − y0 = b⋅ t − t0
The pathlines are
a 2
2
x ( t) = x0 + ⋅ ⎛ t − t0 ⎞
⎠
2 ⎝
y ( t) = y0 + b⋅ t − t0
(
)
(
)
(
)
These give the position (x,y) at any time t of a particle that was at (x 0,y 0) at time t0
Note that streaklines are obtained using the logic of the Governing equations, above
The streaklines are
a 2
2
x t0 = x 0 + ⋅ ⎛ t − t0 ⎞
⎠
2 ⎝
( )
( )
y t0 = y 0 + b ⋅ t − t0
These gives the streakline at t, where x 0, y 0 is the point at which dye is released (t0 is varied from 0 to t)
For a particle that was at x 0 = 2 m, y 0 = 1 m at t0 = 0s, at time t = 2 s we find the position is (from pathline equations)
a 2
2
x = x 0 + ⋅ ⎛ t − t0 ⎞ = 2.8 m
⎠
2 ⎝
(
)
y = y 0 + b ⋅ t − t0 = 5 m
For a particle that was at x 0 = 2 m, y 0 = 1 m at t0 = 2 s, at time t = 3 s we find the position is
a 2
2
x = x 0 + ⋅ ⎛ t − t0 ⎞ = 3 m
⎝
⎠
2
v
For streamlines
u
=
dy
dx
b
=
(
b
a⋅ t
So, separating variables
dy =
Integrating
b
y − y0 =
⋅ x − x0
a⋅ t
The streamlines are then
b
5⋅ ( x − 2)
y = y0 +
⋅ x − x0 =
+1
a⋅ t
t
a⋅ t
)
y = y 0 + b ⋅ t − t0 = 3 m
⋅ dx
where we treat t as a constant
(
(
)
and we have
x0 = 2 m
m
)
Pathline Plots
Streamline Plots
8
8
Pathline (t0=0)
Pathline (t0=2)
Streakline
Streamline (t=0)
Streamline (t=1)
Streamline (t=2)
6
y (m)
6
y (m)
y0 = 1
4
2
4
2
0
1
2
3
x (m)
4
5
0
1
2
3
x (m)
4
5
Problem 2.37
[Difficulty: 2]
Given:
Sutherland equation
Find:
Corresponding equation for kinematic viscosity
1
Solution:
Governing equation:
μ
b T
2
1
S
p  ρ R T
Sutherland equation
Ideal gas equation
T
Assumptions: Sutherland equation is valid; air is an ideal gas
The given data is
6
b  1.458  10
kg

m s K
The kinematic viscosity is
where
ν
b' 
μ
ρ

μ R T

p
S  110.4  K
1
R  286.9 
1
3
3
2
2
2
R T b  T
R b T
b' T




p
p
S
S
S
1
1
1
T
T
T
b'  4.129  10
p
2
9
m
1.5
K
N m
kg K
 1.458  10
6

s
2
kg
1
m s K
m

3
 4.129  10
ν
b' T
2
1
S
with
T
b'  4.129  10
9

2
m
3
101.3  10  N
2
s K
3
Hence
p  101.3  kPa
2
R b
b'  286.9 
J
kg K
9

2
m
3
s K
2
S  110.4 K
2
Check with Appendix A, Table A.10. At T  0 °C we find
2
5 m
T  273.1 K
ν  1.33  10

s
3
2
9
m
4.129  10
3
s K
ν 
1
 ( 273.1  K)
2
2
2
5 m
ν  1.33  10
110.4

Check!
s
273.1
At T  100 °C we find
2
5 m
T  373.1 K
ν  2.29  10

s
3
2
9
m
4.129  10
3
s K
ν 
1
 ( 373.1  K)
2
2
2
5 m
ν  2.30  10
110.4

Check!
s
373.1
Viscosity as a Function of Temperature
5
2.5 10
Kinematic Viscosity (m2/s)
Calculated
Table A.10
2 10
5
5
1.5 10
0
20
40
60
Temperature (C)
80
100
Problem 2.38
Given:
Sutherland equation with SI units
Find:
Corresponding equation in BG units
1
Solution:
Governing equation:
[Difficulty: 2]
μ=
b⋅ T
2
1+
S
Sutherland equation
T
Assumption: Sutherland equation is valid
The given data is
−6
b = 1.458 × 10
kg
⋅
S = 110.4 ⋅ K
1
m⋅ s⋅ K
2
1
Converting constants
−6
b = 1.458 × 10
kg
⋅
1
m⋅ s⋅ K
Alternatively
b = 2.27 × 10
−8
Also
S = 110.4 ⋅ K ×
lbm
0.454 ⋅ kg
×
slug
32.2⋅ lbm
μ=
b⋅ T
2
1+
S
ft
×
⎛ 5⋅ K ⎞
⎜ 9⋅ R
⎝
⎠
2
−8
b = 2.27 × 10
1
⋅
×
slug
1
ft⋅ s⋅ R
lbf ⋅ s
−8
b = 2.27 × 10
slug⋅ ft
2
⋅
1
2
S = 198.7 ⋅ R
5⋅ K
lbf ⋅ s
ft
2
2
lbf ⋅ s
ft ⋅ R
9⋅ R
with T in Rankine, µ in
T
0.3048⋅ m
2
1
and
×
2
slug
ft⋅ s⋅ R
×
2
Check with Appendix A, Table A.9. At T = 68 °F we find
T = 527.7 ⋅ R
μ = 3.79 × 10
− 7 lbf ⋅ s
⋅
ft
1
−8
2.27 × 10
lbf ⋅ s
1
2
ft ⋅ R
μ =
1+
× ( 527.7 ⋅ R)
2
2
2
μ = 3.79 × 10
198.7
− 7 lbf ⋅ s
⋅
ft
Check!
2
527.7
At T = 200 °F we find
T = 659.7 ⋅ R
μ = 4.48 × 10
− 7 lbf ⋅ s
⋅
ft
2
1
−8
2.27 × 10
lbf ⋅ s
1
2
μ =
ft ⋅ R
1+
× ( 659.7 ⋅ R)
2
2
198.7
659.7
μ = 4.48 × 10
− 7 lbf ⋅ s
⋅
ft
2
Check!
Data:
Using procedure of Appendix A.3:
T (oC)
0
100
200
300
400
µ(x10 )
1.86E-05
2.31E-05
2.72E-05
3.11E-05
3.46E-05
5
T (K)
273
373
473
573
673
T (K)
273
373
473
573
673
3/2
T /µ
2.43E+08
3.12E+08
3.78E+08
4.41E+08
5.05E+08
The equation to solve for coefficients
S and b is
3 2
T
µ
S
⎛1 ⎞
= ⎜ ⎟T +
b
b
⎝ ⎠
From the built-in Excel
Linear Regression functions:
Hence:
b = 1.531E-06
S = 101.9
Slope = 6.534E+05
Intercept = 6.660E+07
. .
1/2
kg/m s K
K
2
R = 0.9996
Plot of Basic Data and Trend Line
6.E+08
Data Plot
5.E+08
Least Squares Fit
4.E+08
T3/2/µ 3.E+08
2.E+08
1.E+08
0.E+00
0
100
200
300
400
T
500
600
700
800
Problem 2.40
[Difficulty: 2]
Given:
Velocity distribution between flat plates
Find:
Shear stress on upper plate; Sketch stress distribution
Solution:
Basic equation
du
τyx = μ⋅
dy
τyx = −
du
dy
Hence
y=
dy
⎡
u max⋅ ⎢1 −
⎣
2
⎛ 2 ⋅ y ⎞ ⎥⎤ = u ⋅ ⎛ − 4 ⎞ ⋅ 2⋅ y = − 8 ⋅ umax⋅ y
⎜ h
max ⎜ 2
2
⎝ ⎠⎦
h
⎝ h ⎠
8 ⋅ μ⋅ u max⋅ y
h
At the upper surface
d
=
h
2
and
2
τyx = −8 × 1.14 × 10
− 3 N⋅ s
⋅
2
h = 0.1⋅ mm
× 0.1⋅
m
m
s
×
0.1
2
u max = 0.1⋅
⋅ mm ×
1⋅ m
1000⋅ mm
×
m
s
− 3 N⋅ s
μ = 1.14 × 10
⋅
(Table A.8)
2
m
2
⎛ 1 × 1000⋅ mm ⎞
⎜ 0.1⋅ mm
1⋅ m ⎠
⎝
N
τyx = −4.56⋅
2
m
The upper plate is a minus y surface. Since τyx < 0, the shear stress on the upper plate must act in the plus x direction.
⎛ 8 ⋅ μ⋅ umax ⎞
⋅y
⎜ h2
⎝
⎠
τyx( y ) = −⎜
The shear stress varies linearly with y
0.05
0.04
0.03
y (mm)
0.02
0.01
−5
−4
−3
−2
−1
0
1
− 0.01
− 0.02
− 0.03
− 0.04
− 0.05
Shear Stress (Pa)
2
3
4
5
Problem 2.41
[Difficulty: 2]
Given:
Velocity distribution between parallel plates
Find:
Force on lower plate
Solution:
Basic equations
du
dy
so
du
τyx = μ⋅
dy
F = τyx⋅ A
=
⎡
d
dy
τyx = −
u max⋅ ⎢1 −
⎣
8 ⋅ μ⋅ u max⋅ y
h
At the lower surface
y=−
h
F=−
and
2
2
m
2
A = 1⋅ m
− 3 N⋅ s
μ = 1.14 × 10
− 3 N⋅ s
⋅
2
m
F = 2.28⋅ N
⋅
2
m
F = −8 × 1 ⋅ m × 1.14 × 10
(to the right)
2
2
h = 0.1⋅ mm
s
8 ⋅ A⋅ μ⋅ u max⋅ y
h
and
u max = 0.05⋅
Hence
2
⎛ 2 ⋅ y ⎞ ⎥⎤ = u ⋅ ⎛ − 4 ⎞ ⋅ 2⋅ y = − 8 ⋅ umax⋅ y
⎜ h
max ⎜ 2
2
⎝ ⎠⎦
h
⎝ h ⎠
× 0.05⋅
(Table
A.8)
m
s
×
−0.1
2
⋅ mm ×
1⋅ m
1000⋅ mm
×
2
⎛ 1 ⋅ 1 × 1000⋅ mm ⎞
⎜ 0.1 mm
1⋅ m ⎠
⎝
Problem 2.42
[Difficulty: 2]
Open-Ended Problem Statement: Explain how an ice skate interacts with the ice surface.
What mechanism acts to reduce sliding friction between skate and ice?
Discussion: The normal freezing and melting temperature of ice is 0°C (32°F) at atmospheric
pressure. The melting temperature of ice decreases as pressure is increased. Therefore ice can be caused to
melt at a temperature below the normal melting temperature when the ice is subjected to increased pressure.
A skater is supported by relatively narrow blades with a short contact against the ice. The blade of a typical
skate is less than 3 mm wide. The length of blade in contact with the ice may be just ten or so millimeters.
With a 3 mm by 10 mm contact patch, a 75 kg skater is supported by a pressure between skate blade and
ice on the order of tens of megaPascals (hundreds of atmospheres). Such a pressure is enough to cause ice
to melt rapidly.
When pressure is applied to the ice surface by the skater, a thin surface layer of ice melts to become liquid
water and the skate glides on this thin liquid film. Viscous friction is quite small, so the effective friction
coefficient is much smaller than for sliding friction.
The magnitude of the viscous drag force acting on each skate blade depends on the speed of the skater, the
area of contact, and the thickness of the water layer on top of the ice.
The phenomenon of static friction giving way to viscous friction is similar to the hydroplaning of a
pneumatic tire caused by a layer of water on the road surface.
Problem 2.43
[Difficulty: 2]
Given: Velocity profile
Find:
Plot of velocity profile; shear stress on surface
Solution:
2
⎛
y ⎞
⎜
⋅ h⋅ y −
⋅ sin( θ)
u=
μ ⎝
2 ⎠
2
ρ⋅ g
The velocity profile is
u
Hence we can plot
u max
⎡y
= 2⋅ ⎢
⎣h
−
1
2
⋅ ⎛⎜
y⎞
u max =
so the maximum velocity is at y = h
ρ⋅ g h
⋅ ⋅ sin( θ)
μ 2
2⎤
⎥
⎝h⎠ ⎦
1
y/h
0.75
0.5
0.25
0
0.25
0.5
0.75
1
u/umax
This graph can be plotted in Excel
The given data is
− 3 lbf ⋅ s
h = 0.1⋅ in
μ = 2.15 × 10
⋅
ft
τyx = μ⋅
dy
At the surface y = 0
τyx = ρ⋅ g ⋅ h ⋅ sin( θ)
Hence
τyx = 0.85 × 1.94⋅
θ = 45⋅ deg
⎛
y ⎞
τyx = μ⋅
= μ⋅
⋅ ⎜ h⋅ y −
⋅ sin( θ) = ρ⋅ g ⋅ ( h − y ) ⋅ sin( θ)
dy
dy μ ⎝
2 ⎠
du
Basic equation
2
du
slug
ft
3
× 32.2⋅
ft
2
s
d ρ⋅ g
× 0.1⋅ in ×
1 ⋅ ft
12⋅ in
2
2
× sin( 45⋅ deg) ×
lbf ⋅ s
slug⋅ ft
The surface is a positive y surface. Since τyx > 0, the shear stress on the surface must act in the plus x direction.
lbf
τyx = 0.313 ⋅
2
ft
Problem 2.44
Given:
Ice skater and skate geometry
Find:
Deceleration of skater
[Difficulty: 2]
τ yx = µ
y
Solution:
Governing equation:
du
τyx = μ⋅
dy
ΣFx = M ⋅ ax
du
dy
V = 20 ft/s
h
x
L
Assumptions: Laminar flow
The given data is
W = 100 ⋅ lbf
V = 20⋅
− 5 lbf ⋅ s
μ = 3.68 × 10
⋅
ft
Then
ft
L = 11.5⋅ in
s
w = 0.125 ⋅ in
Table A.7 @32oF
2
du
V
ft
1
12⋅ in
− 5 lbf ⋅ s
τyx = μ⋅
= μ⋅ = 3.68 × 10 ⋅
× 20⋅ ×
×
dy
h
2
s
0.0000575 ⋅ in
ft
ft
lbf
τyx = 154 ⋅
2
ft
Equation of motion
ΣFx = M ⋅ ax
ax = −
τyx⋅ A⋅ g
W
ax = −154
lbf
ft
ax = −0.495 ⋅
−W
τyx⋅ A =
⋅a
g x
or
2
ft
2
s
=−
τyx⋅ L⋅ w⋅ g
W
× 11.5⋅ in × 0.125 ⋅ in × 32.2⋅
ft
2
s
×
1
100 ⋅ lbf
×
ft
2
( 12⋅ in)
2
h = 0.0000575 ⋅ in
Problem 2.45
Given:
Block pulled up incline on oil layer
Find:
Force required to pull the block
[Difficulty: 2]
Solution:
Governing equations:
U
du
τyx = μ⋅
dy
y
x
x
f
N
W
d
ΣFx = M ⋅ ax
θ
Assumptions: Laminar flow
The given data is
W = 10⋅ lbf
U = 2⋅
μ = 3.7 × 10
− 2 N⋅ s
⋅
ft
w = 10⋅ in
s
d = 0.001 ⋅ in
θ = 25⋅ deg
Fig. A.2 @100 oF (38oC)
2
m
Equation of motion
ΣFx = M ⋅ ax = 0
The friction force is
du
U 2
f = τyx⋅ A = μ⋅ ⋅ A = μ⋅ ⋅ w
dy
d
Hence
F = f + W⋅ sin( θ) = μ⋅
− 2 N⋅ s
F = 3.7 × 10
F = 17.1⋅ lbf
F − f − W⋅ sin( θ) = 0
s
o
U
d
2
⋅ w + W⋅ sin( θ)
2
lbf ⋅ s m
ft
1
ft
2
⋅
× 0.0209⋅
⋅
× 2⋅ ×
× ( 10⋅ in) ×
+ 10⋅ lbf ⋅ sin( 25⋅ deg)
2
2 N⋅ s
s
0.001 ⋅ in
12⋅ in
m
ft
Problem 2.46
Given:
Block moving on incline on oil layer
Find:
Speed of block when free, pulled, and pushed
[Difficulty: 2]
Solution:
y
U
Governing equations:
x
x
du
τyx = μ⋅
dy
f
N
W
ΣFx = M ⋅ ax
d
θ
Assumptions: Laminar flow
The given data is
M = 10⋅ kg
W = M⋅ g
W = 98.066 N
d = 0.025 ⋅ mm
θ = 30⋅ deg
F = 75⋅ N
− 1 N⋅s
μ = 10
⋅
w = 250 ⋅ mm
Fig. A.2 SAE 10-39 @30oC
2
m
Equation of motion
ΣFx = M ⋅ ax = 0
The friction force is
du
U 2
f = τyx⋅ A = μ ⋅ ⋅ A = μ ⋅ ⋅ w
dy
d
Hence for uphill motion
F = f + W ⋅ sin ( θ) = μ ⋅
For no force:
U =
d ⋅ W⋅ sin( θ)
2
F − f − W ⋅ sin ( θ) = 0
so
U
d
U =
d ⋅ ( F − W⋅ sin( θ) )
2
μ⋅ w
U=
d ⋅ ( F − W⋅ sin( θ) )
(For downpush change
sign of W)
2
μ⋅ w
U = 0.196
m
U = 0.104
m
μ⋅ w
Pushing up:
2
⋅ w + W ⋅ sin ( θ)
s
s
Pushing down:
U =
d ⋅ ( F + W ⋅ sin ( θ) )
2
μ⋅w
U = 0.496
m
s
Problem 2.47
[Difficulty: 2]
Given:
Data on tape mechanism
Find:
Maximum gap region that can be pulled without breaking tape
Solution:
Basic equation
du
τyx  μ
dy
F  τyx A
and
FT  2  F  2  τyx A
Here F is the force on each side of the tape; the total force is then
The velocity gradient is linear as shown
du
dy

V 0
c

V
c
A  w L
Combining these results
V
FT  2  μ  w L
c
L
Solving for L
c
y
The area of contact is
t
F,V
x
c
FT c
2  μ V w
L
The given data is
Hence
FT  25 lbf
c  0.012  in
L  25 lbf  0.012  in 
1  ft
12 in

μ  0.02
1
2

1

ft s
0.02 slug
slug
ft s

V  3
ft
s
w  1  in
1 s
1 1
12 in slug ft
 


3 ft 1 in
2
1  ft
s  lbf
L  2.5 ft
Problem 2.48
Given:
Flow data on apparatus
Find:
The terminal velocity of mass m
[Difficulty: 2]
Solution:
Given data:
Dpiston = 73⋅ mm
Dtube = 75⋅ mm
Mass = 2 ⋅ kg
Reference data:
kg
ρwater = 1000⋅
3
m
(maximum density of water)
L = 100 ⋅ mm
μ = 0.13⋅
From Fig. A.2:, the dynamic viscosity of SAE 10W-30 oil at 25oC is:
SG Al = 2.64
N⋅ s
2
m
The terminal velocity of the mass m is equivalent to the terminal velocity of the piston. At that terminal speed, the acceleration of
the piston is zero. Therefore, all forces acting on the piston must be balanced. This means that the force driving the motion
(i.e. the weight of mass m and the piston) balances the viscous forces acting on the surface of the piston. Thus, at r = Rpiston:
2 ⎞⎤
⎡⎢
⎛⎜ π⋅ D
piston ⋅ L ⎥
⎢Mass + SGAl⋅ ρwater⋅ ⎜
⎥ ⋅ g = τrz⋅ A =
4
⎣
⎝
⎠⎦
⎛ μ⋅ d V ⎞ ⋅ π⋅ D
⎜
piston⋅ L)
z (
⎝ dr ⎠
The velocity profile within the oil film is linear ...
d
Vz =
dr
Therefore
V
⎛ Dtube − Dpiston ⎞
⎜
⎝
⎠
2
Thus, the terminal velocity of the piston, V, is:
g ⋅ ⎛ SG Al⋅ ρwater⋅ π⋅ Dpiston ⋅ L + 4 ⋅ Mass⎞ ⋅ Dtube − Dpiston
⎝
⎠
2
V =
or
V = 10.2
8 ⋅ μ⋅ π⋅ Dpiston⋅ L
m
s
(
)
Problem 2.49
[Difficulty: 3]
Given:
Flow data on apparatus
Find:
Sketch of piston speed vs time; the time needed for the piston to reach 99% of its new terminal speed.
Solution:
Given data:
Dpiston  73 mm
Dtube  75 mm
L  100  mm
Reference data:
kg
ρwater  1000
3
m
(maximum density of water)
(From Problem 2.48)
μ  0.13
From Fig. A.2, the dynamic viscosity of SAE 10W-30 oil at 25oC is:
m
V0  10.2
s
SG Al  2.64
N s
2
m
The free body diagram of the piston after the cord is cut is:
Piston weight:
2
 π D
piston
Wpiston  SGAl ρwater g  
L
4


Viscous force:
Fviscous( V)  τrz A
or
Fviscous( V)  μ 
1

   π Dpiston L
   Dtube  Dpiston
2

dV
mpiston
 Wpiston  Fviscous( V)
dt
Applying Newton's second law:
Therefore
dV
dt
If
 g  a V
V  g  a V
V
where
then
The differential equation becomes
a 
dX
dt
dX
dt
The solution to this differential equation is:
8 μ

SGAl ρwater Dpiston Dtube  Dpiston
 a

dV
dt
 a X
X( t)  X0  e
 a t
where
X( 0 )  g  a V0
or
g  a V( t)  g  a V0  e


 a t
Therefore
g (  a t) g
V( t)   V0    e

a
a

Plotting piston speed vs. time (which can be done in Excel)
Piston speed vs. time
12
10
8
V ( t) 6
4
2
0
1
2
t
The terminal speed of the piston, Vt, is evaluated as t approaches infinity
g
Vt 
a
or
m
Vt  3.63
s
The time needed for the piston to slow down to within 1% of its terminal velocity is:
 V  g 
 0 a
t   ln

a
g
 1.01 Vt 
a

1
or
t  1.93 s
3
Problem 2.50
[Difficulty: 3]
Given:
Block on oil layer pulled by hanging weight
Find:
Expression for viscous force at speed V; differential equation for motion; block speed as function of time; oil viscosity
Mg
Solution:
Governing equations:
x
y
Ft
du
τyx = μ⋅
dy
ΣFx = M ⋅ ax
Ft
Fv
mg
N
Assumptions: Laminar flow; linear velocity profile in oil layer
M = 5 ⋅ kg
Equation of motion (block)
ΣFx = M ⋅ ax
so
dV
Ft − Fv = M ⋅
dt
( 1)
Equation of motion (block)
ΣFy = m⋅ ay
so
dV
m⋅ g − Ft = m⋅
dt
( 2)
Adding Eqs. (1) and (2)
dV
m⋅ g − Fv = ( M + m) ⋅
dt
The friction force is
du
V
Fv = τyx⋅ A = μ⋅ ⋅ A = μ⋅ ⋅ A
dy
h
Hence
m⋅ g −
To solve separate variables
W = m⋅ g = 9.81⋅ N
μ⋅ A
h
M+m
dt =
m⋅ g −
t=−
Hence taking antilogarithms
1−
⋅ V = ( M + m) ⋅
μ⋅ A
h
( M + m) ⋅ h
μ⋅ A
μ⋅ A
m⋅ g ⋅ h
A = 25⋅ cm
2
The given data is
h = 0.05⋅ mm
dV
dt
⋅ dV
⋅V
⋅ ⎛⎜ ln⎜⎛ m⋅ g −
⎝ ⎝
−
⋅V = e
μ⋅ A
μ⋅ A
( M+ m) ⋅ h
h
⋅t
⋅ V⎞ − ln( m⋅ g ) ⎞ = −
⎠
⎠
( M + m) ⋅ h
μ⋅ A
⋅ ln⎛⎜ 1 −
⎝
μ⋅ A
m⋅ g ⋅ h
⋅ V⎞
⎠
⎡
−
m⋅ g ⋅ h ⎢
V=
⋅ ⎣1 − e
Finally
μ⋅ A
( M + m) ⋅ h
μ⋅ A
⎤
⎥
⎦
⋅t
The maximum velocity is V =
m⋅ g ⋅ h
μ⋅ A
In Excel:
The data is
M=
m=
5.00
1.00
kg
kg
To find the viscosity for which the speed is 1 m/s after 1 s
use Goal Seek with the velocity targeted to be 1 m/s by varying
g=
0=
9.81
the viscosity in the set of cell below:
1.30
m/s2
N.s/m2
A=
h=
25
0.5
cm 2
mm
t (s)
1.00
V (m/s)
1.000
Speed V of Block vs Time t
t (s)
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.10
1.20
1.30
1.40
1.50
1.60
1.70
1.80
1.90
2.00
2.10
2.20
2.30
2.40
2.50
2.60
2.70
2.80
2.90
3.00
V (m/s)
0.000
0.155
0.294
0.419
0.531
0.632
0.722
0.803
0.876
0.941
1.00
1.05
1.10
1.14
1.18
1.21
1.25
1.27
1.30
1.32
1.34
1.36
1.37
1.39
1.40
1.41
1.42
1.43
1.44
1.45
1.46
1.6
1.4
1.2
1.0
V (m/s) 0.8
0.6
0.4
0.2
0.0
0.0
0.5
1.0
1.5
t (s)
2.0
2.5
3.0
Problem 2.51
[Difficulty: 4]
Ff = τ⋅ A
x, V, a
M⋅ g
Given:
Data on the block and incline
Find:
Initial acceleration; formula for speed of block; plot; find speed after 0.1 s. Find oil viscosity if speed is 0.3 m/s after
0.1 s
Solution:
Given data
M = 5 ⋅ kg
From Fig. A.2
μ = 0.4⋅
A = ( 0.1⋅ m)
2
d = 0.2⋅ mm
θ = 30⋅ deg
N⋅ s
2
m
Applying Newton's 2nd law to initial instant (no friction)
so
M ⋅ a = M ⋅ g ⋅ sin( θ) − Ff = M ⋅ g ⋅ sin( θ)
ainit = g ⋅ sin( θ) = 9.81⋅
m
2
× sin( 30⋅ deg)
s
M ⋅ a = M ⋅ g ⋅ sin( θ) − Ff
Applying Newton's 2nd law at any instant
so
M⋅ a = M⋅
dV
g ⋅ sin( θ) −
−
Integrating and using limits
or
V = 5 ⋅ kg × 9.81⋅
m
2
s
V( 0.1⋅ s) = 0.404 ⋅
m
s
M⋅ d
μ⋅ A
μ⋅ A
M⋅ d
⋅ ln⎛⎜ 1 −
⎝
m
2
s
du
V
Ff = τ⋅ A = μ⋅ ⋅ A = μ⋅ ⋅ A
dy
d
and
μ⋅ A
= M ⋅ g ⋅ sin( θ) −
dV
Separating variables
At t = 0.1 s
dt
ainit = 4.9
d
⋅V
= dt
⋅V
μ⋅ A
M ⋅ g ⋅ d ⋅ sin( θ)
⋅ V⎞ = t
⎠
− μ⋅ A ⎞
⎛
⋅t
⎜
M ⋅ g ⋅ d ⋅ sin( θ)
M⋅ d
V( t) =
⋅⎝1 − e
⎠
μ⋅ A
2
× 0.0002⋅ m⋅ sin( 30⋅ deg) ×
m
0.4⋅ N⋅ s⋅ ( 0.1⋅ m)
⎛ 0.4⋅ 0.01 ⋅ 0.1⎞⎤
⎡
−⎜
⎢
N⋅ s
5⋅ 0.0002
⎠⎥
×
× ⎣1 − e ⎝
⎦
2
2
kg⋅ m
The plot looks like
V (m/s)
1.5
1
0.5
0
0.2
0.4
0.6
0.8
t (s)
To find the viscosity for which V(0.1 s) = 0.3 m/s, we must solve
− μ⋅ A
⎤
⎡
⋅ ( t= 0.1⋅ s )
⎢
⎥
M ⋅ g ⋅ d ⋅ sin( θ)
M⋅ d
V( t = 0.1⋅ s) =
⋅ ⎣1 − e
⎦
μ⋅ A
The viscosity µ is implicit in this equation, so solution must be found by manual iteration, or by any of a number of classic
root-finding numerical methods, or by using Excel's Goal Seek
Using Excel:
μ = 1.08⋅
N⋅ s
2
m
1
Problem 2.52
[Difficulty: 3]
Given:
Block sliding on oil layer
Find:
Direction of friction on bottom of block and on plate; expression for speed U versus time; time required to lose 95%
of initial speed
Solution:
U
Governing equations:
du
τyx = μ⋅
dy
ΣFx = M ⋅ ax
Fv
y
h
Assumptions: Laminar flow; linear velocity profile in oil layer
x
The bottom of the block is a -y surface, so τyx acts to the left; The plate
is a +y surface, so τyx acts to the right
Equation of motion
ΣFx = M ⋅ ax
The friction force is
du
U 2
Fv = τyx⋅ A = μ⋅ ⋅ A = μ⋅ ⋅ a
dy
h
Hence
−
2
1
U
dU
⋅ U = M⋅
h
⋅ dU = −
μ⋅ a
dt
2
⋅ dt
M⋅ h
2
⎞ = − μ⋅ a ⋅ t
M⋅ h
⎝ U0 ⎠
ln⎛⎜
dU
Fv = M ⋅
dt
U
U
To solve separate variables
μ⋅ a
so
2
−
Hence taking antilogarithms
U = U0 ⋅ e
μ⋅ a
M⋅ h
⋅t
t
t=−
Solving for t
M⋅ h
μ⋅ a
Hence for
U
U0
= 0.05
t = 3.0⋅
2
⋅ ln⎛⎜
⎞
⎝ U0 ⎠
M⋅ h
μ⋅ a
U
2
Problem 2.53
Given:
Varnish-coated wire drawn through die
Find:
Force required to pull wire
[Difficulty: 2]
r
F
x
D
d
Solution:
Governing equations:
du
τyx = μ⋅
dy
ΣFx = M ⋅ ax
L
Assumptions: Laminar flow; linear velocity profile in varnish layer
The given data is
D = 1 ⋅ mm
d = 0.9⋅ mm
L = 50⋅ mm
Equation of motion
ΣFx = M ⋅ ax
F − Fv = 0
The friction force is
du
V
Fv = τyx⋅ A = μ⋅ ⋅ A = μ⋅
⋅ π⋅ d ⋅ L
dr
⎛D − d⎞
so
V = 50⋅
−2
m
μ = 20 × 10
s
poise
for steady speed
⎜
⎝ 2 ⎠
Hence
F − Fv = 0
so
F =
2 ⋅ π⋅ μ⋅ V⋅ d ⋅ L
D− d
−2
F = 2 ⋅ π × 20 × 10
F = 2.83 N
poise ×
0.1⋅ kg
m⋅ s⋅ poise
× 50⋅
m
s
× 0.9⋅ mm × 50⋅ mm ×
1
( 1 − 0.9) ⋅ mm
×
m
1000⋅ mm
Problem 2.54
[Difficulty: 3]
Given:
Data on annular tube
Find:
Whether no-slip is satisfied; location of zeroshear stress; viscous forces
Solution:
The velocity profile is
Check the no-slip condition. When
2
2


Ro  Ri
r
2
2

u z( r) 

 ln 
 Ri  r 
4 μ L 
 Ri  
 Ri 
ln


 Ro 

∆p
1
2
2

Ro  Ri
 Ro  
2
2

u z R o  

 ln
 Ri  Ro 
4 μ L 
 Ri  
 Ri 
ln

 Ro 


1
r  Ro
∆p
1 ∆p  2
2
2
2
u z Ro 

 R  Ro   Ro  Ri   0


4 μ L  i
 
When
r  Ri
2
2

Ro  Ri
 Ri  
2
2

u z R i  

 ln
 Ri  Ri 
0
Ri 
4 μ L 
 Ri 
 
ln

Ro

 

1
∆p
The no-slip condition is satisfied.
The given data is
The viscosity of the honey is
Ri  5  mm
Ro  25 mm
μ  5
N s
2
m
∆p  125  kPa
L  2 m
The plot looks like
Radial Position (mm)
25
20
15
10
5
0
0.25
0.5
0.75
Velocity (m/s)
For each, shear stress is given by
du
τrx  μ
dr
τrx  μ
duz( r)
dr
2
2



Ro  Ri
1 ∆p  2
r
2


 ln  
 μ
 Ri  r 
dr  4  μ L 
 Ri  
 Ri 
ln





 Ro 

d

Ro  Ri
1 ∆p 
τrx  
 2  r 
4 L 
 Ri 
ln
r

 Ro 

2
Hence
2
For zero stress
2  r 
Ro  Ri
2


2
 Ri 
ln
r
 Ro 
2
0
r 
or
2
2

Ro  Ri 
2

Fo  ∆p π Ro 

 Ri  
2  ln


 Ro  
2
2
 Ri 
2  ln
 Ro 

Ro  Ri 
1 ∆p 
Fo  τrx A  
 2  Ro 
 2  π R o  L
4 L 
 Ri  
ln
 Ro


 Ro  
2
On the outer surface
Ri  Ro
r  13.7 mm

2
2  1  m 


2 ( 25 mm)  ( 5  mm)   
1 m 

3 N
 1000 mm 
Fo  125  10 
 π   25 mm 


2
1000 mm 
5
m
2  ln 


 25 
Fo  172 N

Ro  Ri 
1 ∆p 
Fi  τrx A  
 2  Ri 
 2  π R i  L
4 L 
 Ri  
ln
 Ri


 Ro  
2
On the inner surface
2
2
2

Ro  Ri 
2

Fi  ∆p π Ri 

 Ri  
2  ln


 Ro  
Hence
2

2
2  1  m 



(
25

mm
)

(
5

mm
)

 
2
1 m 

3 N
 1000 mm 
Fi  125  10 
 π   5  mm 


2
1000 mm 
5
m
2  ln 


 25 
Fi  63.4 N
Note that
Fo  Fi  236 N
and
∆p π  Ro  Ri

2
2
  236 N
The net pressure force just balances the net
viscous force!
Problem 2.55
[Difficulty: 3]
Given:
Data on flow through a tube with a filament
Find:
Whether no-slip is satisfied; location of zero stress;stress on tube and filament
Solution:
V( r) 
The velocity profile is
Check the no-slip condition.
When
r
r
d
2

∆p
2
ln
V
D

2
1

∆p
16 μ L

∆p
d
 ln
2
2

2
2
D d
 d d 
2
2
2 r  
 d 


16 μ L 

1
 d  D  D  d

16 μ L 

1
2
 D
2
V( d ) 
2
D d
2
 d  4 r 
D
V( D) 
When

16 μ L 

1
∆p
2
2
 d D 
2
D d
ln
2
d
 D
 ln
D 
 d 

  0
2
2
d
ln 
 D

 ln
d 
 d 

0
The no-slip condition is satisfied.
The given data is
d  1  μm
The viscosity of SAE 10-30 oil at 100 oC is (Fig. A.2)
D  20 mm
∆p  5  kPa
 2 N s
μ  1  10

2
m
L  10 m
The plot looks like
Radial Position (mm)
10
8
6
4
2
0
0.25
0.5
0.75
1
Velocity (m/s)
du
τrx  μ
dr
For each, shear stress is given by
dV( r)
d
τrx  μ
 μ
dr
dr
2
2
 1 ∆p  2
D d
2  r  
2

 ln
 d  4 r 
 16 μ L 

d
 Di  
ln 


 D

1 ∆p 
D d 
τrx( r) 

 8  r 

16 L 
d
ln   r

 D 
2
2
8  r 
For the zero-stress point
D d
2
2
d
ln   r
 D
2
0
or
r 
2
d D
d
8  ln 
 D
r  2.25 mm
Radial Position (mm)
10
7.5
5
2.5
3
2
1
0
1
2
3
4
Stress (Pa)
Using the stress formula
D
τrx   2.374 Pa
2
 
d
τrx   2.524  kPa
2
 
Problem 2.56
Given:
Flow between two plates
Find:
Force to move upper plate; Interface velocity
[Difficulty: 2]
Solution:
The shear stress is the same throughout (the velocity gradients are linear, and the stresses in the fluid at the interface must be
equal and opposite).
Hence
du1
du2
τ  μ1 
 μ2 
dy
dy
Solving for the interface
velocity V i
Then the force
required is
Vi 

Vi
V  Vi
μ1 
 μ2 
h1
h2
or
V
μ1 h 2
1

μ2 h 1
1


where V i is the interface velocity
m
s
0.1 0.3
1

0.15 0.5
m
Vi  0.714
s
Vi
N s
m
1
1000 mm
2
F  τ A  μ1   A  0.1
 0.714  

 1 m
h1
2
s
0.5 mm
1 m
m
F  143 N
Problem 2.57
[Difficulty: 2]
Given:
Flow of three fluids between two plates
Find:
Upper plate velocity; Interface velocities; plot velocity distribution
Solution:
The shear stress is the same throughout (the velocity gradients are linear, and the stresses in the fluids at the interfaces must be
equal and opposite).
F  100  N
Given data
h 1  0.5 mm
2
A  1 m
μ1  0.15
h 2  0.25 mm
N s
2
μ2  0.5
m
τ 
The (constant) stress is
τ  μ
For each fluid
∆V
or
∆y
τ h 1
V12 
μ1
τ h 2
V23 
 V12
μ2
τ h 3
V 
 V23
μ3
Hence
Hence
Hence
F
2
μ3  0.2
m
N s
2
m
τ  100 Pa
A
∆V 
N s
h 3  0.2 mm
τ ∆y
μ
where ΔV is the overall change in velocity over distance Δy
m
V12  0.333
s
where V 12 is the velocity at the 1 - 2 interface
m
V23  0.383
s
where V 23 is the velocity at the 2 - 3 interface
V  0.483
m
where V is the velocity at the upper plate
s
1
Position (mm)
0.75
0.5
0.25
0
0.1
0.2
0.3
Velocity (m/s)
0.4
0.5
Problem 2.58
[Difficulty: 2]
Problem 2.59
[Difficulty: 2]
Problem 2.60
[Difficulty: 2]
Problem 2.61
Given:
Data on the viscometer
Find:
Time for viscometer to lose 99% of speed
[Difficulty: 3]
Solution:
The given data is
R = 50⋅ mm
H = 80⋅ mm
2
a = 0.20⋅ mm
I = 0.0273⋅ kg⋅ m
μ = 0.1⋅
N⋅ s
2
m
I⋅ α = Torque = −τ⋅ A⋅ R
The equation of motion for the slowing viscometer is
where α is the angular acceleration and τ is the viscous stress, and A is the surface area of the viscometer
The stress is given by
τ = μ⋅
du
dy
= μ⋅
V− 0
μ⋅ V
=
a
a
=
μ⋅ R⋅ ω
a
where V and ω are the instantaneous linear and angular velocities.
Hence
Separating variables
μ⋅ R⋅ ω
I⋅ α = I⋅
dω
dω
μ⋅ R ⋅ A
ω
dt
=−
a
2
μ⋅ R ⋅ A
⋅ A⋅ R =
a
⋅ω
2
=−
a⋅ I
⋅ dt
2
−
Integrating and using IC ω = ω0
μ⋅ R ⋅ A
ω( t) = ω0 ⋅ e
a⋅ I
⋅t
2
−
A = 2 ⋅ π⋅ R ⋅ H
a⋅ I
0.01⋅ ω0 = ω0 ⋅ e
The time to slow down by 99% is obtained from solving
Note that
μ⋅ R ⋅ A
t=−
so
a⋅ I
2
⋅ ln( 0.01)
μ⋅ R ⋅ A
a⋅ I
t=−
so
⋅t
⋅ ln( 0.01)
3
2 ⋅ π⋅ μ⋅ R ⋅ H
2
t = −
0.0002⋅ m⋅ 0.0273⋅ kg⋅ m
2⋅ π
2
⋅
m
0.1⋅ N⋅ s
⋅
1
( 0.05⋅ m)
⋅
1
2
⋅
N⋅ s
3 0.08⋅ m kg⋅ m
⋅ ln( 0.01)
t = 4.00 s
Problem 2.62
Difficulty: [2]
Problem 2.63
[Difficulty: 4]
Problem 2.64
[Difficulty: 3]
Given: Shock-free coupling assembly
Find:
Required viscosity
Solution:
du
τrθ = μ⋅
dr
Basic equation
Shear force
F = τ⋅ A
V2 = ω2(R + δ)
δ
τrθ = μ⋅
V1 = ω1R
P = T⋅ ω2 = F⋅ R⋅ ω2 = τ⋅ A2 ⋅ R⋅ ω2 =
P=
Hence
(
P = T⋅ ω
Power
⎡⎣ω1⋅ R − ω2 ⋅ ( R + δ)⎤⎦
du
∆V
τrθ = μ⋅
= μ⋅
= μ⋅
dr
∆r
δ
Assumptions: Newtonian fluid, linear velocity
profile
Then
Torque T = F⋅ R
)
(
)
μ⋅ ω1 − ω2 ⋅ R
δ
(ω1 − ω2)⋅ R
Because δ << R
δ
⋅ 2 ⋅ π⋅ R⋅ L⋅ R⋅ ω2
3
2 ⋅ π⋅ μ⋅ ω2 ⋅ ω1 − ω2 ⋅ R ⋅ L
δ
P⋅ δ
μ=
(
)
3
2 ⋅ π⋅ ω2 ⋅ ω1 − ω2 ⋅ R ⋅ L
μ =
10⋅ W × 2.5 × 10
μ = 0.202 ⋅
2⋅ π
N⋅ s
2
m
−4
⋅m
×
1
⋅
min
9000 rev
μ = 2.02⋅ poise
×
1
⋅
min
1000 rev
×
1
( .01⋅ m)
3
×
1
0.02⋅ m
×
N⋅ m
s⋅ W
2
×
⎛ rev ⎞ × ⎛ 60⋅ s ⎞
⎜ 2 ⋅ π⋅ rad
⎜
⎝
⎠ ⎝ min ⎠
which corresponds to SAE 30 oil at 30oC.
2
Problem 2.65
[Difficulty: 4] Part 1/2
Problem 2.65
[Difficulty: 4] Part 2/2
Problem 2.66
[Difficulty: 4]
Problem 2.67
[Difficulty: 4]
The data is
N (rpm) µ (N·s/m )
10
0.121
20
0.139
30
0.153
40
0.159
50
0.172
60
0.172
70
0.183
80
0.185
2
The computed data is
ω (rad/s) ω/θ (1/s) η (N·s/m x10 )
1.047
120
121
2.094
240
139
3.142
360
153
4.189
480
159
5.236
600
172
6.283
720
172
7.330
840
183
8.378
960
185
2
3
From the Trendline analysis
k = 0.0449
n - 1 = 0.2068
n = 1.21
The fluid is dilatant
The apparent viscosities at 90 and 100 rpm can now be computed
N (rpm) ω (rad/s)
90
9.42
100
10.47
ω/θ (1/s)
1080
1200
η (N·s/m x10 )
191
195
2
3
Viscosity vs Shear Rate
2
3
η (N.s/m x10 )
1000
Data
Power Trendline
100
η = 44.94(ω/θ)
2
R = 0.9925
0.2068
10
100
1000
Shear Rate ω/θ (1/s)
Problem 2.69
[Difficulty: 4]
Given:
Data on insulation material
Find:
Type of material; replacement material
Solution:
The velocity gradient is
du/dy = U/ δ
Data and
computations
where δ =
τ (Pa)
50
100
150
163
171
170
202
246
349
444
U (m/s)
0.000
0.000
0.000
0.005
0.01
0.03
0.05
0.1
0.2
0.3
0.001 m
du/dy (s-1)
0
0
0
5
10
25
50
100
200
300
Hence we have a Bingham plastic, with
τy =
154
µp =
0.963
Pa
2
N·s/m
At τ = 450 Pa, based on the linear fit
du/dy =
307
s
For a fluid with
τy =
250
Pa
-1
we can use the Bingham plastic formula to solve for µ p given τ , τ y and du/dy from above
µp =
N·s/m
0.652
2
Shear Stress vs Shear Strain
500
450
τ (Pa)
400
350
300
250
Linear data fit:
τ = 0.9632(du/dy ) + 154.34
2
R = 0.9977
200
150
100
50
0
0
50
100
150
200
du/dy (1/s)
250
300
350
Problem 2.70
[Difficulty: 3]
Given: Viscometer data
Find:
Value of k and n in Eq. 2.17
Solution:
τ (Pa)
du/dy (s-1)
0.0457
0.119
0.241
0.375
0.634
1.06
1.46
1.78
5
10
25
50
100
200
300
400
Shear Stress vs Shear Strain
10
Data
Power Trendline
τ (Pa)
The data is
1
1
10
100
τ = 0.0162(du/dy)0.7934
2
R = 0.9902
0.1
0.01
du/dy (1/s)
k = 0.0162
n = 0.7934
Hence we have
The apparent viscosity from
Blood is pseudoplastic (shear thinning)
η =
-1
du/dy (s ) η (N·s/m )
2
5
10
25
50
100
200
300
400
0.0116
0.0101
0.0083
0.0072
0.0063
0.0054
0.0050
0.0047
k (du/dy )n -1
2
o
µ water = 0.001 N·s/m at 20 C
Hence, blood is "thicker" than water!
1000
Problem 2.71
[Difficulty: 5]
Problem 2.72
[Difficulty: 5]
Problem 2.73
[Difficulty: 4]
Given:
Conical bearing geometry
Find:
Expression for shear stress; Viscous torque on shaft
Solution:
Basic equation
ds
τ = μ⋅
du
dT = r⋅ τ⋅ dA
dy
dz
z
Infinitesimal shear torque
AA
r
Assumptions: Newtonian fluid, linear velocity profile (in narrow clearance gap), no slip condition
Section AA
tan( θ) =
r
r = z⋅ tan( θ)
so
z
U = ωr
a
Then
τ = μ⋅
du
dy
= μ⋅
∆u
∆y
= μ⋅
( ω⋅ r − 0 )
(a − 0)
=
μ⋅ ω⋅ z⋅ tan( θ)
a
As we move up the device, shear stress increases linearly (because rate of shear strain does)
But from the sketch
dz = ds ⋅ cos( θ)
The viscous torque on the element of area is
dA = 2 ⋅ π⋅ r⋅ ds = 2 ⋅ π⋅ r⋅
dT = r⋅ τ⋅ dA = r⋅
T=
Solving for µ
μ=
Using given data
cos( θ)
μ⋅ ω⋅ z⋅ tan( θ)
a
3
Integrating and using limits z = H and z = 0
dz
⋅ 2 ⋅ π⋅ r⋅
3
dz
dT =
cos( θ)
2 ⋅ π⋅ μ⋅ ω⋅ z ⋅ tan( θ)
a⋅ cos( θ)
4
π⋅ μ⋅ ω⋅ tan( θ) ⋅ H
2 ⋅ a⋅ cos( θ)
2 ⋅ a⋅ cos( θ) ⋅ T
3
4
π⋅ ω⋅ tan( θ) ⋅ H
H = 25⋅ mm
θ = 30⋅ deg
μ =
a = 0.2⋅ mm
2 ⋅ a⋅ cos( θ) ⋅ T
3
4
π⋅ ω⋅ tan( θ) ⋅ H
From Fig. A.2, at 20oC, CASTOR OIL has this viscosity!
ω = 75⋅
rev
T = 0.325 ⋅ N⋅ m
s
μ = 1.012 ⋅
N⋅ s
2
m
3
⋅ dz
Problem 2.74
[Difficulty: 5]
Problem 2.75
[Difficulty: 5]
Problem 2.76
Given:
[Difficulty: 5]
Geometry of rotating bearing
Find:
Expression for shear stress; Maximum shear stress; Expression for total torque; Total torque
Solution:
τ = μ⋅
Basic equation
du
dT = r⋅ τ⋅ dA
dy
Assumptions: Newtonian fluid, narrow clearance gap, laminar motion
From the figure
h = a + R⋅ ( 1 − cos( θ) )
dA = 2 ⋅ π⋅ r⋅ dr = 2 ⋅ π R⋅ sin( θ) ⋅ R⋅ cos( θ) ⋅ dθ
To find the maximum τ set
du
dy
=
dy
=
u−0
u = ω⋅ r = ω⋅ R⋅ sin( θ)
τ = μ⋅
Then
du
r = R⋅ sin( θ)
h
u
=
h
μ⋅ ω⋅ R⋅ sin( θ)
a + R⋅ ( 1 − cos( θ) )
d ⎡ μ⋅ ω⋅ R⋅ sin( θ) ⎤
⎢
⎥=0
dθ ⎣ a + R⋅ ( 1 − cos( θ) ) ⎦
R⋅ cos( θ) − R + a⋅ cos( θ) = 0
R⋅ μ⋅ ω⋅ ( R⋅ cos( θ) − R + a⋅ cos( θ) )
so
( R + a − R⋅ cos( θ) )
θ = acos⎛⎜
2
⎞ = acos⎛ 75 ⎞
⎜ 75 + 0.5
⎝ R + a⎠
⎝
⎠
R
=0
θ = 6.6⋅ deg
kg
τ = 12.5⋅ poise × 0.1⋅
τ = 79.2⋅
m⋅ s
poise
2
× 2 ⋅ π⋅
70 rad
1
N⋅ s
⋅
× 0.075 ⋅ m × sin( 6.6⋅ deg) ×
×
60 s
[ 0.0005 + 0.075 ⋅ ( 1 − cos( 6.6⋅ deg) ) ] ⋅ m m⋅ kg
N
2
m
The torque is
⌠
T = ⎮ r⋅ τ⋅ A dθ =
⌡
θ
⌠ max
4
2
μ⋅ ω⋅ R ⋅ sin( θ) ⋅ cos( θ)
⎮
dθ
⎮
a + R⋅ ( 1 − cos( θ) )
⌡
wher
e
⎛ R0 ⎞
θmax = asin⎜
⎝R⎠
0
This integral is best evaluated numerically using Excel, Mathcad, or a good calculator
T = 1.02 × 10
−3
⋅ N⋅ m
θmax = 15.5⋅ deg
Problem 2.77
[Difficulty: 2]
Problem 2.78
Given:
Data on size of various needles
Find:
Which needles, if any, will float
[Difficulty: 2]
Solution:
For a steel needle of length L, diameter D, density ρs, to float in water with surface tension σ and contact angle θ, the vertical
force due to surface tension must equal or exceed the weight
2
2 ⋅ L⋅ σ⋅ cos( θ) ≥ W = m⋅ g =
4
⋅ ρs⋅ L⋅ g
π⋅ SG ⋅ ρ⋅ g
θ = 0 ⋅ deg
m
8 ⋅ σ⋅ cos( θ)
π⋅ ρs⋅ g
and for water
ρ = 1000⋅
kg
3
m
SG = 7.83
From Table A.1, for steel
8 ⋅ σ⋅ cos( θ)
⋅
D≤
or
−3 N
σ = 72.8 × 10
From Table A.4
Hence
π⋅ D
=
8
π⋅ 7.83
× 72.8 × 10
−3 N
⋅
m
3
×
m
999 ⋅ kg
2
×
s
9.81⋅ m
Hence D < 1.55 mm. Only the 1 mm needles float (needle length is irrelevant)
×
kg⋅ m
2
N⋅ s
−3
= 1.55 × 10
⋅ m = 1.55⋅ mm
Problem 2.79
Given:
Caplillary rise data
Find:
Values of A and b
[Difficulty: 3]
Solution:
D (in.)
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
h (in.)
0.232
0.183
0.090
0.059
0.052
0.033
0.017
0.010
0.006
0.004
0.003
A = 0.403
b = 4.530
2
The fit is a good one (R = 0.9919)
Capillary Rise vs. Tube Diameter
0.3
h (in.)
h = 0.403e-4.5296D
R2 = 0.9919
0.2
0.1
0.0
0.0
0.2
0.4
0.6
D (in.)
0.8
1.0
1.2
Problem 2.80
[Difficulty: 2]
Open-Ended Problem Statement: Slowly fill a glass with water to the maximum possible
level before it overflows. Observe the water level closely. Explain how it can be higher than the rim of the
glass.
Discussion: Surface tension can cause the maximum water level in a glass to be higher than the rim of
the glass. The same phenomenon causes an isolated drop of water to “bead up” on a smooth surface.
Surface tension between the water/air interface and the glass acts as an invisible membrane that allows
trapped water to rise above the level of the rim of the glass. The mechanism can be envisioned as forces
that act in the surface of the liquid above the rim of the glass. Thus the water appears to defy gravity by
attaining a level higher than the rim of the glass.
To experimentally demonstrate that this phenomenon is the result of surface tension, set the liquid level
nearly as far above the glass rim as you can get it, using plain water. Add a drop of liquid detergent (the
detergent contains additives that reduce the surface tension of water). Watch as the excess water runs over
the side of the glass.
Problem 2.81
[Difficulty: 5]
Open-Ended Problem Statement: Plan an experiment to measure the surface tension of a
liquid similar to water. If necessary, review the NCFMF video Surface Tension for ideas. Which method
would be most suitable for use in an undergraduate laboratory? What experimental precision could be
expected?
Discussion: Two basic kinds of experiment are possible for an undergraduate laboratory:
1.
Using a clear small-diameter tube, compare the capillary rise of the unknown liquid with that of a
known liquid (compare with water, because it is similar to the unknown liquid).
This method would be simple to set up and should give fairly accurate results. A vertical
traversing optical microscope could be used to increase the precision of measuring the liquid
height in each tube.
A drawback to this method is that the specific gravity and co ntact angle of the two liquids must be
the same to allow the capillary rises to be compared.
The capillary rise would be largest and therefore easiest to measure accurately in a tube with the
smallest practical diameter. Tubes of several diameters could be used if desired.
2.
Dip an object into a pool of test liquid and measure the vertical force required to pull the object
from the liquid surface.
The object might be made rectangular (e.g., a sheet of plastic material) or circular (e.g., a metal
ring). The net force needed to pull the same object from each liquid should be proportional to the
surface tension of each liquid.
This method would be simple to set up. However, the force magnitudes to be measured would be
quite small.
A drawback to this method is that the contact angles of the two liquids must be the same.
The first method is probably best for undergraduate laboratory use. A quantitative estimate of experimental
measurement uncertainty is impossible without knowing details of the test setup. It might be reasonable to
expect results accurate to within ± 10% of the true surface tension.
*Net force is the total vertical force minus the weight of the object. A buoyancy correction would be
necessary if part of the object were submerged in the test liquid.
Problem 2.82
[Difficulty: 2]
Problem 2.83
[Difficulty: 2]
Given:
Boundary layer velocity profile in terms of constants a, b and c
Find:
Constants a, b and c
Solution:
u = a + b ⋅ ⎛⎜
Basic equation
y⎞
⎝δ⎠
+ c⋅ ⎛⎜
y⎞
2
⎝δ⎠
Assumptions: No slip, at outer edge u = U and τ = 0
At y = 0
0=a
a=0
At y = δ
U= a+ b+ c
b+c=U
(1)
At y = δ
τ = μ⋅
b + 2⋅ c = 0
(2)
0=
du
dy
d
dy
a + b ⋅ ⎛⎜
y⎞
⎝δ⎠
From 1 and 2
c = −U
Hence
u = 2 ⋅ U⋅ ⎛⎜
Dimensionless Height
=0
+ c⋅ ⎛⎜
y⎞
⎝δ⎠
2
=
b
δ
+ 2 ⋅ c⋅
y
2
=
δ
b
δ
+ 2⋅
c
δ
b = 2⋅ U
y⎞
⎝δ⎠
− U⋅ ⎛⎜
y⎞
⎝δ⎠
2
u
U
= 2 ⋅ ⎛⎜
y⎞
⎝δ⎠
−
⎛y⎞
⎜δ
⎝ ⎠
2
1
0.75
0.5
0.25
0
0.25
0.5
Dimensionless Velocity
0.75
1
Problem 2.84
[Difficulty: 2]
Given:
Boundary layer velocity profile in terms of constants a, b and c
Find:
Constants a, b and c
Solution:
Basic equation
u = a + b ⋅ ⎛⎜
+ c⋅ ⎛⎜
y⎞
⎝δ⎠
y⎞
3
⎝δ⎠
Assumptions: No slip, at outer edge u = U and τ = 0
At y = 0
0=a
a=0
At y = δ
U= a+ b+ c
b+c=U
(1)
At y = δ
τ = μ⋅
b + 3⋅ c = 0
(2)
0=
From 1 and 2
Dimensionless Height
dy
d
dy
c=−
u=
Hence
du
=0
a + b ⋅ ⎛⎜
y⎞
⎝δ⎠
U
2
⋅ ⎛⎜
y⎞
⎝δ⎠
−
U
2
y⎞
⎝δ⎠
b=
2
3⋅ U
+ c⋅ ⎛⎜
3
2
⋅ ⎛⎜
3
=
b
δ
+ 3 ⋅ c⋅
y
2
3
=
b
⋅ ⎛⎜
y⎞
δ
δ
+ 3⋅
c
δ
⋅U
y⎞
⎝δ⎠
3
u
U
=
3
2
⎝δ⎠
−
1
2
⋅ ⎛⎜
y⎞
3
⎝δ⎠
1
0.75
0.5
0.25
0
0.25
0.5
Dimensionless Velocity
0.75
1
Problem 2.85
Given:
Local temperature
Find:
Minimum speed for compressibility effects
[Difficulty: 1]
Solution:
Basic equation
V = M⋅ c
c=
Hence
and
k⋅ R⋅ T
M = 0.3
for compressibility effects
For air at STP, k = 1.40 and R = 286.9J/kg.K (53.33 ft.lbf/lbmoR).
V = M ⋅ c = M ⋅ k ⋅ R⋅ T
1
V = 0.3 × ⎡1.4 × 53.33 ⋅
⎢
⎣
ft⋅ lbf
lbm⋅ R
×
32.2⋅ lbm⋅ ft
2
lbf ⋅ s
2
× ( 60 + 460 ) ⋅ R⎤ ⋅
⎥
⎦
60⋅ mph
88⋅
ft
s
V = 229 ⋅ mph
Problem 2.86
[Difficulty: 3]
Given:
Geometry of and flow rate through tapered nozzle
Find:
At which point becomes turbulent
Solution:
Basic equation
Re =
For pipe flow (Section 2-6)
ρ⋅ V⋅ D
μ
= 2300
for transition to turbulence
2
π⋅ D
Q=
Also flow rate Q is given by
4
⋅V
We can combine these equations and eliminate V to obtain an expression for Re in terms of D and Q
Re =
ρ⋅ V⋅ D
μ
=
ρ⋅ D 4 ⋅ Q
4 ⋅ Q⋅ ρ
⋅
=
μ
π⋅ μ⋅ D
2
π⋅ D
Re =
4 ⋅ Q⋅ ρ
π⋅ μ⋅ D
For a given flow rate Q, as the diameter is reduced the Reynolds number increases (due to the velocity increasing with A -1 or D -2).
Hence for turbulence (Re = 2300), solving for D
The nozzle is tapered:
Carbon tetrachloride:
Din = 50⋅ mm
μCT = 10
D=
4 ⋅ Q⋅ ρ
2300⋅ π⋅ μ
Dout =
− 3 N⋅ s
⋅
Din
Dout = 22.4⋅ mm
5
(Fig A.2)
For water
2
ρ = 1000⋅
3
m
m
SG = 1.595
kg
ρCT = SG⋅ ρ
(Table A.2)
ρCT = 1595
kg
3
m
For the given flow rate
Q = 2⋅
L
4 ⋅ Q⋅ ρCT
min
π⋅ μCT⋅ Din
For the diameter at which we reach turbulence
But
L = 250 ⋅ mm
D =
= 1354
4 ⋅ Q⋅ ρCT
2300⋅ π⋅ μCT
LAMINAR
4 ⋅ Q⋅ ρCT
π⋅ μCT⋅ Dout
D − Din
Dout − Din
Lturb = 186 ⋅ mm
TURBULENT
D = 29.4⋅ mm
and linear ratios leads to the distance from D in at which D = 29.4⋅ mm
Lturb = L⋅
= 3027
Lturb
L
=
D − Din
Dout − Din
Problem 2.87
[Difficulty: 2]
Given:
Data on water tube
Find:
Reynolds number of flow; Temperature at which flow becomes turbulent
Solution:
Basic equation
At 20oC, from Fig. A.3 ν  9  10
For the heated pipe
Hence
Re 
For pipe flow (Section 2-6)
Re 
ν
V D
ν
V D
2300
2
7 m

and so
s
 2300

1
2300
ρ V D
μ
Re  0.25

m
s
V D
ν
 0.005  m 
9  10
for transition to turbulence
 0.25
m
s
 0.005  m
From Fig. A.3, the temperature of water at this viscosity is approximately
ν  5.435  10
T  52 C
1
2
7m
s
7

s
2
m
Re  1389
Problem 2.88
[Difficulty: 3]
Given:
Data on supersonic aircraft
Find:
Mach number; Point at which boundary layer becomes turbulent
Solution:
Basic equation
V = M⋅ c
Hence
M=
V
c
c=
and
k⋅ R⋅ T
For air at STP, k = 1.40 and R = 286.9J/kg.K (53.33 ft.lbf/lbmoR).
V
=
k ⋅ R⋅ T
At 27 km the temperature is approximately (from Table A.3)
T = 223.5 ⋅ K
1
2
2
⎛
⎞ ⋅ ⎜ 1 × 1 ⋅ kg⋅ K × 1⋅ N⋅ s × 1 ⋅ 1 ⎞ M = 2.5
M = ⎜⎛ 2700 × 10 ⋅ ×
hr 3600⋅ s ⎠ ⎝ 1.4 286.9 N⋅ m
kg⋅ m
223.5 K ⎠
⎝
3 m
For boundary layer transition, from Section 2-6
Then
Retrans =
ρ⋅ V⋅ x trans
1 ⋅ hr
Retrans = 500000
μ ⋅ Retrans
x trans =
so
μ
ρ⋅ V
We need to find the viscosity and density at this altitude and pressure. The viscosity depends on temperature only, but at 223.5
K = - 50oC, it is off scale of Fig. A.3. Instead we need to use formulas as in Appendix A
μ=
b ⋅T
2
1+
S
where
3
−6
b = 1.458 × 10
1
m⋅ s ⋅ K
3
S = 110.4 ⋅ K
2
− 5 N⋅s
− 5 kg
μ = 1.459 × 10
m⋅ s
⋅
2
m
− 5 kg
x trans = 1.459 × 10
kg
m
kg
⋅
T
μ = 1.459 × 10
Hence
ρ = 0.0297
m
1
For µ
kg
ρ = 0.02422× 1.225⋅
At this altitude the density is (Table A.3)
⋅
m⋅ s
× 500000×
3
m
1
1 hr
3600⋅ s
⋅
×
×
⋅
×
0.0297 kg
2700
3 m
1⋅ hr
10
1
x trans = 0.327m
Problem 2.89
Given:
Type of oil, flow rate, and tube geometry
Find:
Whether flow is laminar or turbulent
[Difficulty: 2]
Solution:
ν=
Data on SAE 30 oil SG or density is limited in the Appendix. We can Google it or use the following
At 100 oC, from Figs. A.2 and A.3
− 3 N⋅ s
μ = 9 × 10
⋅
ν = 1 × 10
2
− 3 N⋅ s
⋅
2
1 × 10
m
Hence
The specific weight is
SG =
1
×
ρ
−5
⋅
s
2
×
⋅
s
kg⋅ m
ρ = 900
2
s ⋅N
m
γ = ρ⋅ g
γ = 900 ⋅
kg
3
2
Q=
π⋅ D
4
⋅V
−6
Q = 100 ⋅ mL ×
Then
Hence
V =
Re =
4
π
V=
so
10
m
2
2
N⋅ s
×
3 N
γ = 8.829 × 10 ⋅
kg⋅ m
s
4⋅ Q
2
⋅m
1 1
× ⋅
9 s
3
−5 m
⋅
s
×
Q = 1.111 × 10
2
⎛ 1 ⋅ 1 × 1000⋅ mm ⎞
⎜ 12 mm
1⋅ m ⎠
⎝
V = 0.0981
ρ⋅ V⋅ D
μ
Re = 900 ⋅
kg
3
m
Flow is laminar
3
m
π⋅ D
3
1 ⋅ mL
× 1.11 × 10
3
SG = 0.9
× 9.81⋅
m
For pipe flow (Section 2-6)
kg
m
kg
ρwater = 1000⋅
3
m
ρwater
ρ=
so
ρ
2
−5 m
m
ρ = 9 × 10
μ
× 0.0981⋅
m
s
× 0.012 ⋅ m ×
1
9 × 10
2
⋅
m
− 3 N⋅ s
2
×
N⋅ s
kg⋅ m
Re = 118
m
s
3
−5m
s
μ
ν
Problem 2.90
Given:
Data on seaplane
Find:
Transition point of boundary layer
[Difficulty: 2]
Solution:
For boundary layer transition, from Section 2-6
Retrans  500000
Then
Retrans 
At 45oF = 7.2 oC (Fig A.3)
ρ V x trans
μ
2
5 m
ν  0.8  10

s
V x trans

ν
10.8

 5 ft

ft
V
 5 ft
s
ν  8.64  10
m

2
s
s
2
 500000 
s
ν Retrans
2
2
1
x trans  8.64  10
x trans 
so
1
100  mph

60 mph
88
x trans  0.295  ft
ft
s
As the seaplane touches down:
At 45oF = 7.2 oC (Fig A.3)
2
5 m
ν  1.5  10

s
10.8

 4 ft

2
 4 ft
s
ν  1.62  10
2
1
x trans  1.62  10
ft
m
2
s
s
2
s

 500000 
1
100  mph

60 mph
88
ft
s
x trans  0.552  ft
Problem 2.91
[Difficulty: 3]
Given:
Data on airliner
Find:
Sketch of speed versus altitude (M = const)
Solution:
Data on temperature versus height can be obtained from Table A.3
At 5.5 km the temperature is approximately
252
c
The speed of sound is obtained from
where
k = 1.4
R = 286.9
J/kg·K
c = 318
m/s
V = 700
km/hr
V = 194
m/s
K
k  R T
(Table A.6)
We also have
or
Hence M = V/c or
M = 0.611
V = M · c = 0.611·c
To compute V for constant M , we use
V = 677
At a height of 8 km:
km/hr
NOTE: Realistically, the aiplane will fly to a maximum height of about 10 km!
T (K)
4
262
5
259
5
256
6
249
7
243
8
236
9
230
10
223
11
217
12
217
13
217
14
217
15
217
16
217
17
217
18
217
19
217
20
217
22
219
24
221
26
223
28
225
30
227
40
250
50
271
60
256
70
220
80
181
90
181
c (m/s) V (km/hr)
325
322
320
316
312
308
304
299
295
295
295
295
295
295
295
295
295
295
296
298
299
300
302
317
330
321
297
269
269
Speed vs. Altitude
713
709
750
704
695
686
677
668
658
700
649
649
649
649
649
649
649
649
Speed V (km/hr)
z (km)
650
649
649
651
654
600
657
660
663
697
725
705
653
592
592
550
0
20
40
60
Altitude z (km)
80
100
Problem 2.92
[Difficulty: 4]
Open-Ended Problem Statement: How does an airplane wing develop lift?
Discussion: The sketch shows the cross-section of a typical airplane wing. The airfoil section is
rounded at the front, curved across the top, reaches maximum thickness about a third of the way back, and
then tapers slowly to a fine trailing edge. The bottom of the airfoil section is relatively flat. (The discussion
below also applies to a symmetric airfoil at an angle of incidence that produces lift.)
NACA 2412 Wing Section
It is both a popular expectation and an experimental fact that air flows more rapidly over the curved top
surface of the airfoil section than along the relatively flat bottom. In the NCFMF video Flow Visualization,
timelines placed in front of the airfoil indicate that fluid flows more rapidly along the top of the section
than along the bottom.
In the absence of viscous effects (this is a valid assumption outside the boundary layers on the airfoil)
pressure falls when flow speed increases. Thus the pressures on the top surface of the airfoil where flow
speed is higher are lower than the pressures on the bottom surface where flow speed does not increase.
(Actual pressure profiles measured for a lifting section are shown in the NCFMF video Boundary Layer
Control.) The unbalanced pressures on the top and bottom surfaces of the airfoil section create a net force
that tends to develop lift on the profile.
Problem 3.1
[Difficulty: 2]
Given:
Data on nitrogen tank
Find:
Pressure of nitrogen; minimum required wall thickness
Assumption:
Ideal gas behavior
Solution:
Ideal gas equation of state:
p ⋅V = M⋅R⋅T
where, from Table A.6, for nitrogen
R = 55.16⋅
Then the pressure of nitrogen is
p =
ft⋅ lbf
lbm⋅ R
⎛ 6 ⎞
3⎟
⎝ π⋅ D ⎠
M⋅ R⋅ T
= M⋅ R⋅ T⋅ ⎜
V
p = 140⋅ lbm × 55.16⋅
p = 3520⋅
ft⋅ lbf
lbm⋅ R
⎡
⎤ × ⎛ ft ⎞
⎜
⎟
3⎥ ⎝ 12⋅ in ⎠
⎣ π × ( 2.5⋅ ft) ⎦
× ( 77 + 460) ⋅ R × ⎢
6
lbf
2
in
To determine wall thickness, consider a free body diagram for one hemisphere:
π⋅ D
ΣF = 0 = p ⋅
2
− σc ⋅ π ⋅ D ⋅ t
pπD2/4
p⋅ D
σcπDt
4
where σc is the circumferential stress in the container
Then
t=
p⋅ π⋅ D
2
4 ⋅ π ⋅ D ⋅ σc
t = 3520 ⋅
lbf
2
in
t = 0.0733⋅ ft
×
=
4 ⋅ σc
2.5 ⋅ ft
4
2
×
in
3
30 × 10 ⋅ lbf
t = 0.880⋅ in
2
Problem 3.2
[Difficulty: 2]
Given: Pure water on a standard day
Find:
Boiling temperature at (a) 1000 m and (b) 2000 m, and compare with sea level value.
Solution:
We can determine the atmospheric pressure at the given altitudes from table A.3, Appendix A
The data are
Elevation
(m)
0
1000
2000
p/p o
p (kPa)
1.000
0.887
0.785
101.3
89.9
79.5
We can also consult steam tables for the variation of saturation temperature with pressure:
p (kPa)
70
80
90
101.3
T sat (°C)
90.0
93.5
96.7
100.0
We can interpolate the data from the steam tables to correlate saturation temperature with altitude:
Elevation
(m)
0
1000
2000
p/p o
p (kPa) T sat (°C)
1.000
0.887
0.785
101.3
89.9
79.5
The data are plotted here. They
show that the saturation temperature
drops approximately 3.4°C/1000 m.
100.0
96.7
93.3
Saturation
Temperature (°C)
Variation of Saturation Temperature with
Pressure
Sea Level
100
1000 m
98
96
2000 m
94
92
90
88
70
75
80
85
90
95
Absolute Pressure (kPa)
100
105
Problem 3.3
[Difficulty: 2]
Given:
Data on flight of airplane
Find:
Pressure change in mm Hg for ears to "pop"; descent distance from 8000 m to cause ears to "pop."
Solution:
Assume the air density is approximately constant constant from 3000 m to 2900 m.
From table A.3
ρSL  1.225
kg
ρair  0.7423  ρSL
3
m
ρair  0.909
kg
3
m
We also have from the manometer equation, Eq. 3.7
∆p  ρair  g  ∆z
Combining
∆hHg 
ρair
ρHg
∆hHg 
 ∆z 
0.909
13.55  999
∆p  ρHg  g  ∆hHg
and also
ρair
SGHg  ρH2O
SGHg  13.55 from Table A.2
 ∆z
 100  m
∆hHg  6.72 mm
For the ear popping descending from 8000 m, again assume the air density is approximately constant constant, this time at 8000 m.
From table A.3
ρair  0.4292  ρSL
ρair  0.526
kg
3
m
We also have from the manometer equation
ρair8000  g  ∆z8000  ρair3000  g  ∆z3000
where the numerical subscripts refer to conditions at 3000m and 8000m.
Hence
∆z8000 
ρair3000  g
ρair8000  g
 ∆z3000 
ρair3000
ρair8000
 ∆z3000
∆z8000 
0.909
0.526
 100  m
∆z8000  173 m
Problem 3.4
[Difficulty: 3]
Given: Boiling points of water at different elevations
Find: Change in elevation
Solution:
From the steam tables, we have the following data for the boiling point (saturation temperature) of water
Tsat (oF)
p (psia)
10.39
8.39
195
185
The sea level pressure, from Table A.3, is
pSL =
14.696
psia
Hence
Altitude vs Atmospheric Pressure
o
p/pSL
195
185
0.707
0.571
From Table A.3
p/pSL
0.7372
0.6920
0.6492
0.6085
0.5700
15000
12500
Altitude (ft)
Tsat ( F)
Altitude (m)
2500
3000
3500
4000
4500
Altitude (ft)
8203
9843
11484
13124
14765
Data
10000
Linear Trendline
7500
z = -39217(p/pSL) + 37029
5000
R2 = 0.999
2500
0.55
0.60
0.65
0.70
p/pSL
Then, any one of a number of Excel functions can be used to interpolate
(Here we use Excel 's Trendline analysis)
p/pSL
0.707
0.571
Altitude (ft)
9303
14640
Current altitude is approximately
The change in altitude is then 5337 ft
Alternatively, we can interpolate for each altitude by using a linear regression between adjacent data points
p/pSL
p/pSL
For
0.7372
0.6920
Altitude (m)
2500
3000
Altitude (ft)
8203
9843
0.6085
0.5700
Altitude (m)
4000
4500
Altitude (ft)
13124
14765
Then
0.7070
2834
9299
0.5730
4461
14637
The change in altitude is then 5338 ft
9303 ft
0.75
Problem 3.5
Given:
Data on system
Find:
Force on bottom of cube; tension in tether
[Difficulty: 2]
Solution:
dp
Basic equation
dy
= − ρ⋅ g
or, for constant ρ
∆p = ρ⋅ g⋅ h
where h is measured downwards
The absolute pressure at the interface is
pinterface = patm + SGoil⋅ ρ⋅ g⋅ hoil
Then the pressure on the lower surface is
pL = pinterface + ρ⋅ g⋅ hL = patm + ρ⋅ g⋅ SGoil⋅ hoil + hL
For the cube
(
V = 125⋅ mL
1
3
V = 1.25 × 10
Then the size of the cube is
d = V
d = 0.05 m
Hence
hL = hU + d
hL = 0.35 m
The force on the lower surface is
F L = pL ⋅ A
where
(
−4
)
3
⋅m
and the depth in water to the upper surface is hU = 0.3⋅ m
where hL is the depth in water to the lower surface
A = d
2
2
A = 0.0025 m
)
FL = ⎡patm + ρ⋅ g⋅ SGoil⋅ hoil + hL ⎤ ⋅ A
⎣
⎦
⎡
kg
m
N⋅ s ⎥⎤
3 N
2
FL = ⎢101 × 10 ⋅
+ 1000⋅
× 9.81⋅ × ( 0.8 × 0.5⋅ m + 0.35⋅ m) ×
× 0.0025⋅ m
2
3
2
⎢
kg⋅ m⎥
m
m
s
⎣
⎦
2
FL = 270.894 N
For the tension in the tether, an FBD gives
Note: Extra decimals needed for computing T later!
ΣFy = 0
FL − FU − W − T = 0
(
)
where FU = ⎡patm + ρ⋅ g⋅ SGoil⋅ hoil + hU ⎤ ⋅ A
⎣
⎦
or
T = FL − FU − W
(
)
Note that we could instead compute
∆F = FL − FU = ρ⋅ g⋅ SGoil⋅ hL − hU ⋅ A
Using FU
⎡
kg
m
N⋅ s ⎥⎤
3 N
2
FU = ⎢101 × 10 ⋅
+ 1000⋅
× 9.81⋅ × ( 0.8 × 0.5⋅ m + 0.3⋅ m) ×
× 0.0025⋅ m
2
3
2
⎢
kg⋅ m⎥
m
m
s
⎣
⎦
T = ∆F − W
2
FU = 269.668 N
For the oak block (Table A.1)
and
Note: Extra decimals needed for computing T later!
SGoak = 0.77
W = 0.77 × 1000⋅
W = SGoak⋅ ρ⋅ g⋅ V
so
kg
3
m
T = FL − FU − W
× 9.81⋅
m
2
× 1.25 × 10
s
T = 0.282 N
−4
3
⋅m ×
2
N⋅ s
kg⋅ m
W = 0.944 N
Problem 3.6
Given:
Data on system before and after applied force
Find:
Applied force
[Difficulty: 2]
Solution:
Basic equation
dp
dy
For initial state
p1  patm  ρ g h
For the initial FBD
ΣFy  0
For final state
p2  patm  ρ g H
For the final FBD
ΣFy  0
F1  p1 A  ρ g h A
and
F2  p2 A  ρ g H A
F2  W  F  0
4
 
p y0  patm
with
(Gage; F1 is hydrostatic upwards force)
W  F 1  ρ  g h A
and
π D

p  patm  ρ g y  y0
F1  W  0
F  ρH2O SG g
From Fig. A.1

or, for constant ρ
  ρ g
(Gage; F2 is hydrostatic upwards force)
F  F2  W  ρ g H A  ρ g h  A  ρ g A ( H  h )
2
 ( H  h)
SG  13.54
F  1000
kg
3
m
F  45.6 N
 13.54  9.81
m
2
s

π
4
2
 ( 0.05 m)  ( 0.2  0.025)  m 
2
N s
kg m
Problem 3.7
[Difficulty: 1]
Given: Pressure and temperature data from balloon
Find: Plot density change as a function of elevation
Assumption: Ideal gas behavior
Solution:
Density Distribution
0.078
p (psia)
14.71
14.62
14.53
14.45
14.36
14.27
14.18
14.10
14.01
13.92
13.84
T (oF)
53.6
52.0
50.9
50.4
50.2
50.0
50.5
51.4
52.9
54.0
53.8
 (lbm/ft3)
0.07736
0.07715
0.07685
0.07647
0.07604
0.07560
0.07506
0.07447
0.07380
0.07319
0.07276
Density (lbm/ft3)
Using the ideal gas equation,  = p/RT
0.077
0.076
0.075
0.074
0.073
0.072
0
1
2
3
4
5
6
Elevation Point
7
8
9
10
Problem 3.8
[Difficulty: 2]
Given:
Properties of a cube floating at an interface
Find:
The pressures difference between the upper and lower surfaces; average cube density
Solution:
The pressure difference is obtained from two applications of Eq. 3.7
pU  p0  ρSAE10 g ( H  0.1 d)
pL  p0  ρSAE10 g H  ρH2O g 0.9 d
where pU and pL are the upper and lower pressures, p0 is the oil free surface pressure, H is the depth of the interface, and d
is the cube size
Hence the pressure difference is

∆p  pL  pU  ρH2O  g 0.9 d  ρSAE10  g  0.1 d
From Table A.2
∆p  ρH2O  g d  0.9  SGSAE10  0.1
SGSAE10  0.92
kg
∆p  999
3
m
 9.81
2
m
2
 0.1 m  ( 0.9  0.92  0.1) 
s
Ns
kg  m
∆p  972 Pa
For the cube density, set up a free body force balance for the cube
ΣF  0  ∆p  A  W
Hence
2
W  ∆p A  ∆p d
ρcube 
m
d
3
ρcube  972
2
W

3

d g
N
2
m

∆p  d

3
d g
1
0.1 m
∆p
d g
2

s
9.81 m

kg  m
2
Ns
ρcube  991
kg
3
m

Problem 3.9
Given:
Data on tire at 3500 m and at sea level
Find:
Absolute pressure at 3500 m; pressure at sea level
[Difficulty: 2]
Solution:
At an elevation of 3500 m, from Table A.3:
pSL  101 kPa
patm  0.6492  pSL
patm  65.6 kPa
and we have
pg  0.25 MPa
pg  250 kPa
p  pg  patm
At sea level
patm  101  kPa
p  316 kPa
Meanwhile, the tire has warmed up, from the ambient temperature at 3500 m, to 25oC.
At an elevation of 3500 m, from Table A.3
Tcold  265.4  K
and
Thot  ( 25  273)  K
Thot  298 K
Hence, assuming ideal gas behavior, pV = mRT, and that the tire is approximately a rigid container, the absolute pressure of the
hot tire is
phot 
Thot
Tcold
p
phot  354 kPa
Then the gage pressure is
pg  phot  patm
pg  253 kPa
Problem 3.10
Given:
Data on air bubble
Find:
Bubble diameter as it reaches surface
[Difficulty: 2]
Solution:
dp
Basic equation
dy
 ρsea g
and the ideal gas equation
M
p  ρ R  T 
V
 R T
We assume the temperature is constant, and the density of sea water is constant
For constant sea water density
p  patm  SGsea ρ g h
Then the pressure at the initial depth is
p1  patm  SGsea ρ g h1
The pressure as it reaches the surface is
p2  patm
M R T
V
For the bubble
p
Hence
p1 V 1  p2 V 2
Then the size of the bubble at the surface is
From Table A.2
but M and T are constant
 p1 
D 2  D 1  
 p2 
SGsea  1.025
1
3
M R T  const  p V
P1
V2  V1
p2
or
or
1
3
3
3 p1
D2  D1 
p2
 patm  ρsea g h1
 ρsea g h1 
  D1  1 

patm
patm




1
3
 D 1 
(This is at 68oF)

slug


ft
D2  0.3 in  1  1.025  1.94
D2  0.477 in
where p is the pressure at any depth h
3
 32.2 
ft
2
s
2
2
 1 ft   lbf s 

14.7 lbf  12 in 
slugft
 

2
 100 ft 
in

1
3
Problem 3.11
[Difficulty: 2]
Given:
Properties of a cube suspended by a wire in a fluid
Find:
The fluid specific gravity; the gage pressures on the upper and lower surfaces
Solution:
From a free body analysis of the cube:
(
)
2
ΣF = 0 = T + pL − pU ⋅ d − M⋅ g
where M and d are the cube mass and size and pL and pU are the pressures on the lower and upper surfaces
For each pressure we can use Eq. 3.7
p = p 0 + ρ ⋅ g⋅ h
Hence
pL − pU = ⎡p0 + ρ⋅ g⋅ ( H + d)⎤ − p0 + ρ⋅ g⋅ H = ρ⋅ g⋅ d = SG⋅ ρH2O⋅ d
⎣
⎦
(
)
where H is the depth of the upper surface
2 ⋅ slug× 32.2⋅
M⋅ g − T
SG =
Hence the force balance gives
ρH2O ⋅ g ⋅ d
2
×
s
SG =
3
2
ft
1.94 ⋅
slug
ft
3
× 32.2⋅
lbf ⋅ s
− 50.7 ⋅ lbf
slug⋅ ft
SG = 1.75
2
ft
2
×
s
lbf⋅ s
slug⋅ ft
× ( 0.5 ⋅ ft)
3
From Table A.1, the fluid is Meriam blue.
The individual pressures are computed from Eq 3.7
p = p 0 + ρ ⋅ g⋅ h
pg = 1.754 × 1.94⋅
For the upper surface
pg = ρ⋅ g⋅ h = SG⋅ ρH2O⋅ h
or
slug
ft
pg = 1.754 × 1.94⋅
For the lower surface
3
slug
ft
3
× 32.2⋅
ft
2
s
× 32.2⋅
ft
2
s
×
2
3
2
⋅ ft ×
⎛ 1⋅ ft ⎞
⎟
slug⋅ ft ⎝ 12⋅ in ⎠
lbf ⋅ s
2
×⎜
2
⎛ 2 + 1 ⎞ ⋅ ft × lbf ⋅ s × ⎛ 1⋅ ft ⎞
⎟
⎜
⎟
slug⋅ ft ⎝ 12⋅ in ⎠
⎝ 3 2⎠
×⎜
pg = 0.507⋅ psi
2
pg = 0.888⋅ psi
Note that the SG calculation can also be performed using a buoyancy approach (discussed later in the chapter):
Consider a free body diagram of the cube:
ΣF = 0 = T + FB − M ⋅ g
where M is the cube mass and FB is the buoyancy force
FB = SG ⋅ ρH2O ⋅ L ⋅ g
Hence
3
T + SG ⋅ ρH2O ⋅ L ⋅ g − M ⋅ g = 0
3
or
SG =
M⋅ g − T
ρH2O ⋅ g ⋅ L
3
as before
SG = 1.75
Problem 3.12
[Difficulty: 4]
Given:
Model behavior of seawater by assuming constant bulk modulus
Find:
(a) Expression for density as a function of depth h.
(b) Show that result may be written as
ρ = ρo + bh
(c) Evaluate the constant b
(d) Use results of (b) to obtain equation for p(h)
(e) Determine depth at which error in predicted pressure is 0.01%
Solution:
5
SGo = 1.025
From Table A.2, App. A:
dp
Governing Equations:
Ev = 2.42⋅ GPa = 3.51 × 10 ⋅ psi
= ρ⋅ g
dh
Ev =
(Hydrostatic Pressure - h is positive downwards)
dp
(Definition of Bulk Modulus)
dρ
ρ
ρ
Then
dρ
dp = ρ⋅ g⋅ dh = Ev⋅
ρ
or
dρ
ρ
2
=
g
Ev
dh
h
⌠
⌠ g
1
⎮
dρ = ⎮
dh
⎮
2
⎮ Ev
⎮ ρ
⌡0
⌡ρ
Now if we integrate:
o
After integrating:
Now for
ρ o⋅ g⋅ h
Ev
ρ − ρo
ρ⋅ ρo
=
g⋅ h
Ev
Therefore: ρ =
E v⋅ ρ o
E v − g⋅ h ⋅ ρ o
ρ
and
ρo
1
=
1−
<<1, the binomial expansion may be used to approximate the density:
ρ o⋅ g⋅ h
Ev
ρ
ρo
2
In other words, ρ = ρo + b⋅ h where b =
Since
ρo ⋅ g
= 1+
ρ o⋅ g⋅ h
Ev
(Binomial expansion may
be found in a host of
sources, e.g. CRC
Handbook of
Mathematics)
Ev
dp = ρ⋅ g⋅ dh then an approximate expression for the pressure as a function of depth is:
h
⌠
papprox − patm = ⎮
⌡0
( ρo + b ⋅ h )⋅ g dh → papprox − patm =
(
g⋅ h ⋅ 2⋅ ρo + b ⋅ h
2
)
Solving for papprox we get:
papprox = patm +
(
g⋅ h ⋅ 2⋅ ρ o + b⋅ h
2
)
= patm + ρo⋅ g⋅ h +
b⋅ g ⋅ h
2
2
⎛
2
b⋅ h ⎞
⎝
2 ⎠
= patm + ⎜ ρo⋅ h +
⎟⋅g
Now if we subsitiute in the expression for b and simplify, we get:
2
⎛
ρo ⋅ g h2 ⎞⎟
⎛ ρ o⋅ g⋅ h ⎞
⎜
⋅ g = patm + ρo⋅ g⋅ h⋅ ⎜ 1 +
papprox = patm + ρo⋅ h +
⋅
⎟
⎜
Ev 2 ⎟
2⋅ E v
⎝
⎠
⎝
⎠
⎛ ρ o⋅ g⋅ h ⎞
papprox = patm + ρo⋅ g⋅ h⋅ ⎜ 1 +
⎟
2Ev
⎝
⎠
The exact soution for p(h) is obtained by utilizing the exact solution for ρ(h). Thus:
ρ
⌠ E
v
⎛ρ⎞
pexact − patm = ⎮
dρ = Ev⋅ ln ⎜ ⎟
⎮
ρ
⎝ ρo ⎠
⌡ρ
Subsitiuting for
ρ
ρo
we get:
o
If we let x =
ρ o⋅ g⋅ h
Ev
For the error to be 0.01%:
∆pexact − ∆papprox
∆pexact
= 1−
This equation requires an iterative solution, e.g. Excel's Goal Seek. The result is:
h=
x⋅ E v
ρ o⋅ g
⎛ ρ o⋅ g⋅ h ⎞
pexact = patm + Ev⋅ ln ⎜ 1 −
⎟
Ev
⎝
⎠
⎛ x⎞
ρo⋅ g⋅ h⋅ ⎜ 1 + ⎟
⎝ 2⎠
Ev⋅ ln ⎡⎣( 1 − x)
− 1⎤
= 1−
⎦
ln ⎡⎣( 1 − x)
− 1⎤
= 0.0001
⎦
x = 0.01728 Solving x for h:
3
2
2
ft
s
5 lbf
⎛ 12⋅ in ⎞ × slug⋅ ft
×
×
×⎜
⎟
2 1.025 × 1.94⋅ slug 32.2⋅ ft ⎝ ft ⎠
2
h = 0.01728 × 3.51 × 10 ⋅
in
⎛ x⎞
x⋅ ⎜ 1 + ⎟
⎝ 2⎠
−1
4
h = 1.364 × 10 ⋅ ft
lbf ⋅ s
This depth is over 2.5 miles, so the
incompressible fluid approximation is a
reasonable one at all but the lowest depths
of the ocean.
Problem 3.13
[Difficulty: 3]
Given:
Model behavior of seawater by assuming constant bulk modulus
Find:
The percent deviations in (a) density and (b) pressure at depth h = 6.5
mi, as compared to values assuming constant density.
Plot results over the range of 0 mi - 7 mi.
Solution:
5
SGo  1.025
From Table A.2, App. A:
dp
Governing Equations:
dh
Ev  2.42 GPa  3.51  10  psi
 ρ g
(Hydrostatic Pressure - h is positive downwards)
dp
Ev 
(Definition of Bulk Modulus)
dρ
ρ
ρ
dρ
dp  ρ g dh  Ev
ρ
Then
dρ
or
ρ
2

g
Ev
h  6.5 mi
Now if we integrate:
dh
h

 g
1

dρ  
dh

2
 Ev
 ρ
0
ρ
o
ρ  ρo
After integrating:
ρ ρo

g h
Therefore: ρ 
Ev

∆ρ
ρo

ρ  ρo
ρo
1  1 

ρ
ρo
1
1 
1
ρ o g h
1 

1
Ev
E v ρ o
and
Ev  ρo g h
ρ o g h 

Ev
 
ρ o g h
To determine an expression for the percent deviation in pressure, we find
ρ
For variable density and constant bulk modulus:
For constant density:
1
ρo g h
Ev

pconstρ  patm   ρo g dh  ρo g h
0
ρ o g h
Ev
∆ρ
ρo
ρo g h
Ev

1
p  patm for variable ρ, and then for constant ρ.
 E
v
ρ
p  patm  
dρ  Ev ln  

ρ
 ρo 
ρ
o
h
ρo
1

ρo g h
Ev
1
Ev
ρ
ρ o g h
Ev
∆p
pconstρ

p  pconstρ
If we let x 
pconstρ
Ev
 ρ   ρ  g h
 o
 ρo 

Ev ln 

ρ o g h
x  3.51  10
ρ o g
Ev
ρ o g h
 1

 ρo g h  

 ln  1 
 1
pconstρ
ρ o g h 
Ev

 
∆p
 ρ 1

 ρo 
 ln 
Ev
2
3
2
1
1 ft
1 s
5 lbf
 12 in   slug ft  mi





2 5280 ft
2 1.025 1.94 slug 32.2 ft  ft 
x  149.5 mi
lbf  s
in
Substituting into the expressions for the deviations we get:
∆ρ
h
x
h
h
devρ 



ρo
h
xh
149.5 mi  h
1
x
∆p
x 
devp 
  ln  1 
pconstρ
h 
h
 1
 1 
x 
149.5 mi
h


 ln  1 
1
   1

149.5 mi  
h
devρ  4.55 %
For h = 6.5 mi we get:
The plot below shows the deviations in density and pressure as a function of depth from 0 mi to 7 mi:
Errors in Density and Pressure Assuming Constant Density
8
Errors in Density and Pressure (%)
Density
Pressure
6
4
2
0
2
4
Depth (mi)
6
devp  2.24 %
Problem 3.14
Given:
[Difficulty: 3]
Cylindrical cup lowered slowly beneath pool surface
Air
Find:
H
Expression for y in terms of h and H.
Plot y/H vs. h/H.
D
y
Air
H–y
Solution:
y
Governing Equations:
dp
dh
 ρ g
(Hydrostatic Pressure - h is positive downwards)
p V  M R T
Assumptions:
(Ideal Gas Equation)
(1) Constant temperature compression of air inside cup
(2) Static liquid
(3) Incompressible liquid
First we apply the ideal gas equation (at constant temperature) for the pressure of the air in the cup:
Therefore:
π 2
π 2
p V  pa  D  H  p  D  ( H  y)
4
4
and upon simplification:
p V  constant
pa H  p ( H  y)
Now we look at the hydrostatic pressure equation for the pressure exerted by the water. Since ρ is constant, we integrate:
p  pa  ρ g ( h  y) at the water-air interface in the cup.
Since the cup is submerged to a depth of h, these pressures must be equal:
pa H  pa  ρ g ( h  y)  ( H  y)  pa H  pa y  ρ g ( h  y)  ( H  y)


Explanding out the right hand side of this expression:
2
0  pa y  ρ g ( h  y)  ( H  y)  ρ g h H  ρ g h y  ρ g H y  ρ g y  pa y
2
2
 pa
y 
ρ g y  pa  ρ g ( h  H)  y  ρ g h H  0


 ρ g

 ( h  H)  y  h H  0

2
We now use the quadratic equation:
 pa

 pa


 ( h  H)  
 ( h  H)  4 h H
 ρ g

 ρ g

y
2
we only use the minus sign because y
can never be larger than H.
Now if we divide both sides by H, we get an expression for y/H:
2
 pa

 pa

h
h
h


 1  

 1  4
y
H
 ρ  g H H 
 ρ  g H H 

H
2
The exact shape of this curve will depend upon the height of the cup. The plot below was generated assuming:
pa  101.3 kPa
H  1 m
Height Ratio, y/H
0.8
0.6
0.4
0.2
0
20
40
60
Depth Ratio, h/H
80
100
Problem 3.15
Given:
Geometry of straw
Find:
Pressure just below the thumb
Assumptions:
[Difficulty: 1]
(1) Coke is incompressible
(2) Pressure variation within the air column is negligible
(3) Coke has density of water
Solution:
Basic equation
dp
dy
  ρ g
or, for constant ρ
∆p  ρ g h
where h is measured downwards
This equation only applies in the 15 cm of coke in the straw - in the other 30 cm of air the pressure is essentially constant.
The gage pressure at the coke surface is
Hence, with
hcoke  15 cm
pcoke  ρ g hcoke
because h is measured downwards
pcoke  1000
kg
3
m
pcoke  1.471 kPa
pcoke  99.9 kPa
 9.81
m
2
s
gage
2
 15 cm 
2
m
N s
kPa m


100 cm kg m 1000 N
Problem 3.16
[Difficulty: 2]
Given:
Data on water tank and inspection cover
Find:
If the support bracket is strong enough; at what water depth would it fail
pbaseA
Assumptions:
Water is incompressible and static
Cover
Solution:
Basic equation
patmA
dp
dy
  ρ g
or, for constant ρ
∆p  ρ g h
where h is measured downwards
The absolute pressure at the base is
pbase  patm  ρ g h
h  16 ft
The gage pressure at the base is
pbase  ρ g h
This is the pressure to use as we have patm on the outside of the cover.
The force on the inspection cover is
F  pbase A
where
where
2
A  1 in  1 in
A  1 in
F  ρ  g h A
F  1.94
slug
ft
3
 32.2
ft
2
2
2
2
 ft   lbf  s

 12 in  slug ft
 16 ft  1 in  
s
F  6.94 lbf
The bracket is strong enough (it can take 9 lbf).
To find the maximum depth we start with F  9.00 lbf
h
F
ρ  g A
h  9 lbf 
h  20.7 ft
1

ft
3
1.94 slug

1
2

s
32.2 ft

1
2
in
2
 12 in   slug ft

 ft  lbf  s2

Problem 3.17
Given:
[Difficulty: 4]
Container of mercury with vertical tubes of known diameter, brass
cylinder of known dimensions introduced into larger tube, where it floats.
d1  39.5 mm
d2  12.7 mm D  37.5 mm
H  76.2 mm
SGHg  13.55 SGb  8.55
Find:
(a) Pressureon the bottom of the cylinder
(b) New equlibrium level, h, of the mercury
Solution:
We will analyze a free body diagram of the cylinder, and apply the hydrostatics equation.
ΣFz  0
Governing equations:
dp
dz
(Vertical Equilibrium)
  ρ g
(Hydrostatic Pressure - z is positive upwards)
ρ  SG ρwater
Assumptions:
(Definition of Specific Gravity)
(1) Static liquid
(2) Incompressible liquid
If we take a free body diagram of the cylinder:
ΣFz  p
π
4
π 2
2
 D  ρ b g   D  H  0
4
p  8.55  1000
kg
3
m
 9.81
m
2
thus: p  ρb g H  SGb ρwater g H
m
 76.2 mm 
p  6.39 kPa (gage)
3
10  mm
s
pA
This pressure must be generated by a column of mercury h+x in height. Thus:
p  ρHg g ( h  x)  SGHg ρwater g ( h  x)  SGb ρwater g H
Thus:
hx 
SGb
SGHg
H
The value of x can be found by realizing that the volume of mercury in the system remains constant. Therefore:
π
π 2
2
2
 D  x    d1  D   h   d2  h


4
4
4
π
2
These expressions now allow us to solve for h:
h 
8.55
13.55

( 37.5 mm)
2
z
mg
2
2
 d2  
 d1 
x     1      h
 D 
D 
Now if we solve for x:
h
SGb

D
2
2
2
SGHg
d1  d2
H
Substituting in values:
2
( 39.5 mm)  ( 12.7 mm)
2
 76.2 mm
h  39.3 mm
Problem 3.18
[Difficulty: 2]
Given:
Data on partitioned tank
Find:
Gage pressure of trapped air; pressure to make water and mercury levels equal
Solution:
The pressure difference is obtained from repeated application of Eq. 3.7, or in other words, from Eq. 3.8. Starting
from the right air chamber
pgage  SGHg  ρH2O  g  ( 3  m  2.9  m)  ρH2O  g  1  m


pgage  ρH2O  g  SGHg  0.1  m  1.0  m
2
kg
m
Ns
pgage  999
 9.81  ( 13.55  0.1  m  1.0  m) 
3
2
kg  m
m
s
pgage  3.48 kPa
If the left air pressure is now increased until the water and mercury levels are now equal, Eq. 3.8 leads to
pgage  SGHg  ρH2O  g  1.0  m  ρH2O  g  1.0  m


pgage  ρH2O  g  SGHg  1  m  1.0  m
2
kg
m
Ns
pgage  999
 9.81  ( 13.55  1  m  1.0  m) 
3
2
kg  m
m
s
pgage  123 kPa
Problem 3.19
[Difficulty: 2]
Given:
Data on partitioned tank
Find:
Pressure of trapped air required to bring water and mercury levels equal if right air opening is sealed
Solution:
First we need to determine how far each free surface moves.
In the tank of Problem 3.18, the ratio of cross section areas of the partitions is 0.75/3.75 or 1:5. Suppose the water surface (and
therefore the mercury on the left) must move down distance x to bring the water and mercury levels equal. Then by mercury volume
conservation, the mercury free surface (on the right) moves up (0.75/3.75)x = x/5. These two changes in level must cancel the original
discrepancy in free surface levels, of (1m + 2.9m) - 3 m = 0.9 m. Hence x + x/5 = 0.9 m, or x = 0.75 m. The mercury level thus moves
up x/5 = 0.15 m.
Assuming the air (an ideal gas, pV=RT) in the right behaves isothermally, the new pressure there will be
pright 
Aright  Lrightold
Lrightold
 patm 
 patm 
p
Vrightnew
Aright  Lrightnew
Lrightnew atm
Vrightold
where V, A and L represent volume, cross-section area, and vertical length
Hence
pright 
3
3  0.15
 101  kPa
pright  106 kPa
When the water and mercury levels are equal application of Eq. 3.8 gives:
pleft  pright  SGHg  ρH2O  g  1.0  m  ρH2O  g  1.0  m

pleft  pright  ρH2O  g  SGHg  1.0  m  1.0  m
pleft  106  kPa  999
kg
m
pgage  pleft  patm
3
 9.81
m
2

2
 ( 13.55 1.0  m  1.0  m ) 
s
pgage  229  kPa  101  kPa
N s
kg  m
pleft  229 kPa
pgage  128 kPa
Problem 3.20
Given:
[Difficulty: 2]
Two-fluid manometer as shown
l  10.2 mm SGct  1.595 (From Table A.1, App. A)
Find:
Pressure difference
Solution:
We will apply the hydrostatics equation.
Governing equations:
dp
dh
 ρ g
(Hydrostatic Pressure - h is positive downwards)
ρ  SG ρwater
Assumptions:
(Definition of Specific Gravity)
(1) Static liquid
(2) Incompressible liquid
z
Starting at point 1 and progressing to point 2 we have:
d
p1  ρwater g ( d  l)  ρct g l  ρwater g d  p2
Simplifying and solving for p2  p1 we have:


∆p  p2  p1  ρct g l  ρwater g l  SGct  1  ρwater g l
Substituting the known data:
∆p  ( 1.591  1)  1000
kg
3
m
 9.81
m
2
s
 10.2 mm 
m
3
10  mm
∆p  59.1 Pa
Problem 3.21
Given:
[Difficulty: 2]
U-tube manometer, partiall filled with water, then a given volume of
Meriam red oil is added to the left side
3
D  6.35 mm
Voil  3.25 cm
SGoil  0.827
(From Table A.1, App. A)
Find:
Equilibrium height, H, when both legs are open to atmosphere.
Solution:
We will apply the basic pressure-height relation.
Governing Equations:
dp
dh
 ρ g
(Hydrostatic Pressure - h is positive downwards)
ρ  SG ρwater
Assumptions:
(Definition of Specific Gravity)
(1) Static liquid
(2) Incompressible liquid
Integration of the pressure equation gives:
Thus: pB  pA  ρoil g L and

p2  p1  ρ  g h2  h1

A
pD  pC  ρwater g ( L  H)
B
Since
pA  pC  patm and pB  pD since they are at the same height:
ρoil g L  ρwater g ( L  H) or
Solving for H:

H  L 1  SGoil
SGoil L  L  H

The value of L comes from the volume of the oil:
Solving for L:
L
4 Voil
π D
We can now calculate H:
2
Voil 
3
L 
4  3.25 cm
π  ( 6.35 mm)
2
H  102.62 mm ( 1  0.827)
π
4
2
D L
 10 mm 

 cm 

3
L  102.62 mm
H  17.75 mm
C
L
L–H
D
Problem 3.22
Given:
[Difficulty: 2]
Two fluid manometer contains water and kerosene. With both tubes
open to atmosphere, the difference in free surface elevations is known
Ho = 20⋅ mm SGk = 0.82 (From Table A.1, App. A)
Find:
The elevation difference, H, between the free surfaces of the fluids
when a gage pressure of 98.0 Pa is applied to the right tube.
Solution:
We will apply the hydrostatics equation.
Governing Equations:
dp
dh
= ρ⋅ g
(Hydrostatic Pressure - h is positive downwards)
ρ = SG⋅ ρwater
Assumptions:
(Definition of Specific Gravity)
(1) Static liquid
(2) Incompressible liquid
When the gage pressure Δp is applied to the right tube, the water in the
right tube is displaced downward by a distance, l. The kerosene in the
left tube is displaced upward by the same distance, l.
Δp
l
Under the applied gage pressure Δp, the elevation difference, H, is:
H0
H = H o + 2⋅ l
l
H1
Since points A and B are at the same elevation in the same fluid, their
pressures are the same. Initially:
(
pA = ρ k⋅ g⋅ H o + H 1
)
A
pB = ρwater⋅ g⋅ H1
B
Setting these pressures equal:
(
)
ρk⋅ g⋅ Ho + H1 = ρwater⋅ g⋅ H1
Solving for H1
H1 =
ρ k⋅ H o
ρwater − ρk
=
SGk⋅ Ho
1 − SGk
H1 =
0.82 × 20⋅ mm
1 − 0.82
Now under the applied gage pressure:
(
)
pA = ρk⋅ g⋅ Ho + H1 + ρwater⋅ g⋅ l
H
h
(
)
pB = ∆p + ρwater⋅ g⋅ H1 − l
H1 = 91.11⋅ mm
A
B
Setting these pressures equal:
(
)
(
)
∆p
SGk⋅ Ho + H1 + l =
+ H1 − l
ρwater⋅ g
l=
∆p
⎤
+ H1 − SGk⋅ (Ho + H1)⎥
⎢
2 ρwater⋅ g
⎣
⎦
1⎡
Substituting in known values we get:
l =
1
2
⎡
N
⎢
⎣
m
× ⎢98.0⋅
2
3
×
1 m
999 kg
×
1
2
⋅
s
9.81 m
×
kg⋅ m
2
N⋅ s
+ [ 91.11⋅ mm − 0.82 × ( 20⋅ mm + 91.11⋅ mm) ] ×
⎤
⎥
3
⎥
10 ⋅ mm⎦
m
l = 5.000⋅ mm
Now we solve for H:
H = 20⋅ mm + 2 × 5.000⋅ mm
H = 30.0⋅ mm
Problem 3.23
Given:
Data on manometer
Find:
Deflection due to pressure difference
[Difficulty: 2]
Solution:
Basic equation
dp
dy
= − ρ⋅ g
or, for constant ρ
∆p = ρ⋅ g⋅ ∆h
where h is measured downwards
Starting at p1
pA = p1 + SGA⋅ ρ⋅ g⋅ ( h + l )
Next, from A to B
pB = pA − SGB⋅ ρ⋅ g⋅ h
Finally, from A to the location of p2
p2 = pB − SGA⋅ ρ⋅ g⋅ l
Combining the three equations
p2 = pA − SGB⋅ ρ⋅ g⋅ h − SGA⋅ ρ⋅ g⋅ l = ⎡p1 + SGA⋅ ρ⋅ g⋅ ( h + l) − SGB⋅ ρ⋅ g⋅ h⎤ − SGA⋅ ρ⋅ g⋅ l
⎣
⎦
(
where l is the (unknown) distance from the level of the right
interface
)
(
)
p2 − p1 = SGA − SGB ⋅ ρ⋅ g⋅ h
h=
p1 − p2
(SGB − SGA)⋅ ρ⋅ g
h = 18⋅
lbf
ft
2
h = 0.139⋅ ft
×
1
( 2.95 − 0.88)
×
1
⋅
ft
3
1.94 slug
h = 1.67⋅ in
×
1
2
⋅
s
32.2 ft
×
slug⋅ ft
2
s ⋅ lbf
Problem 3.24
Given:
Data on manometer
Find:
Gage pressure at point a
Assumption:
[Difficulty: 2]
e
c
Water, liquids A and B are static and incompressible
d
Solution:
Basic equation
dp
dy
or, for constant ρ
  ρ g
∆p  ρ g ∆h
where ∆h is height difference
Starting at point a
p1  pa  ρH2O g h1
where
h1  0.125 m  0.25 m
Next, in liquid A
p2  p1  SGA ρH2O g h2
where
h2  0.25 m
Finally, in liquid B
patm  p2  SGB ρH2O g h3
where
h3  0.9 m  0.4 m
h1  0.375 m
h3  0.5 m
Combining the three equations


patm  p1  SGA ρH2O g h2  SGB ρH2O g h3  pa  ρH2O g h1  SGA ρH2O g h2  SGB ρH2O g h3

pa  patm  ρH2O g h1  SGA h2  SGB h3
or in gage pressures

pa  ρH2O g h1  SGA h2  SGB h3
pa  1000
kg
3
 9.81
m
3
pa  4.41  10 Pa
m
2


2
 [ 0.375  ( 1.20  0.25)  ( 0.75  0.5) ]  m 
s
pa  4.41 kPa
(gage)
N s
kg m
Problem 3.25
[Difficulty: 2]
SGoil  0.827 from Table A.1
Given:
Two fluid manometer, Meriam red oil is the second fluid
Find:
The amplification factor which will be seen in this demonstrator
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dh
 ρ g
(Hydrostatic Pressure - h is positive downwards)
ρ  SG ρwater
Assumptions:
(Definition of Specific Gravity)
(1) Static liquid
(2) Incompressible liquid
h
Integrating the hydrostatic pressure equation we get:
p  p o  ρ  g h
For the left leg of the manometer:
b
hA
hB
pa  patm  ρwater g hA


pb  pa  ρwater g l  patm  ρwater g hA  l
l
a
pa  patm  ρwater g hB
For the right leg:


pb  pa  ρoil g l  patm  ρwater g hB  SGoil l
Combining the right hand sides of these two equations:
Upon simplification:
hA  l  hB  SGoil l
AF 
l
∆h

1
1  SGoil




patm  ρwater g hA  l  patm  ρwater g hB  SGoil l

∆h  hA  hB  l 1  SGoil
For Meriam red
AF 

1
1  0.827
so the amplification factor would be:
 5.78
AF  5.78
Problem 3.26
[Difficulty: 2]
Given:
Water flow in an inclined pipe as shown. The pressure difference is
measured with a two-fluid manometer
L  5 ft
h  6 in SGHg  13.55 (From Table A.1, App. A)
Find:
Pressure difference between A and B
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dh
 ρ g
(Hydrostatic Pressure - h is positive downwards)
ρ  SG ρwater
Assumptions:
(Definition of Specific Gravity)
(1) Static liquid
(2) Incompressible liquid
(3) Gravity is constant
Integrating the hydrostatic pressure equation we get:
∆p  ρ g ∆h
Progressing through the manometer from A to B:
pA  ρwater g L sin( 30 deg)  ρwater g a  ρwater g h  ρHg g h  ρwater g a  pB
Simplifying terms and solving for the pressure difference:


∆p  pA  pB  ρwater g h  SGHg  1  L sin( 30 deg)


Substituting in values:
∆p  1.94
slug
ft
3
 32.2
ft
2
s


 6 in 
ft
12 in


 ( 13.55  1)  5 ft  sin( 30 deg) 
2
 ft 

slugft

 12 in 
lbf s

2
∆p  1.638 psi
Problem 3.27
Given:
Data on fluid levels in a tank
Find:
Air pressure; new equilibrium level if opening appears
[Difficulty: 2]
Solution:
Using Eq. 3.8, starting from the open side and working in gage pressure
pair = ρH2O × g × ⎡SGHg × ( 0.3 − 0.1) ⋅ m − 0.1 ⋅ m − SGBenzene× 0.1 ⋅ m⎤
⎣
⎦
Using data from Table A.2
pair = 999⋅
kg
m
3
× 9.81⋅
m
2
2
× ( 13.55 × 0.2 ⋅ m − 0.1 ⋅ m − 0.879 × 0.1 ⋅ m ) ×
s
N ⋅s
kg ⋅ m
pair = 24.7⋅ kPa
To compute the new level of mercury in the manometer, assume the change in level from 0.3 m is an increase of x. Then, because the
volume of mercury is constant, the tank mercury level will fall by distance (0.025/0.25)2x. Hence, the gage pressure at the bottom of the tan
can be computed from the left and the right, providing a formula for x
⎡
⎛ 0.025 ⎞ ⎤⎥ ⋅ m ...
SGHg × ρH2O × g × ( 0.3⋅ m + x) = SGHg × ρH2O × g × ⎢0.1⋅ m − x⋅ ⎜
⎟
⎣
⎝ 0.25 ⎠ ⎦
+ ρH2O × g × 0.1 ⋅ m + SGBenzene × ρH2O × g × 0.1 ⋅ m
2
Hence
The new manometer height is
x =
[ 0.1⋅ m + 0.879 × 0.1⋅ m + 13.55 × ( 0.1 − 0.3) ⋅ m]
2
⎡
0.025 ⎞ ⎤
⎢1 + ⎛⎜
⎟ ⎥ × 13.55
⎣ ⎝ 0.25 ⎠ ⎦
h = 0.3⋅ m + x
x = −0.184 m
(The negative sign indicates the
manometer level actually fell)
h = 0.116 m
Problem 3.28
Given:
[Difficulty: 2]
Reservoir manometer with vertical tubes of knowm diameter. Gage liquid is Meriam red oil
D = 18⋅ mm d = 6⋅ mm
SGoil = 0.827 (From Table A.1, App. A)
Find:
The manometer deflection, L when a gage pressure equal to 25 mm of
water is applied to the reservoir.
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dh
= ρ⋅ g
(Hydrostatic Pressure - h is positive downwards)
ρ = SG⋅ ρwater
Assumptions:
(Definition of Specific Gravity)
(1) Static liquid
(2) Incompressible liquid
Integrating the hydrostatic pressure equation we get:
∆p = ρ⋅ g⋅ ∆h
Beginning at the free surface of the reservoir, and accounting for the changes in pressure with elevation:
patm + ∆p + ρoil⋅ g⋅ ( x + L) = patm
∆p
x+L =
Upon simplification:
The gage pressure is defined as:
ρoil⋅ g
Combining these two expressions:
x+L =
ρwater⋅ g⋅ h
ρoil⋅ g
x and L are related through the manometer dimensions:
L=
Therefore:
∆h
2⎤
⎡
d
SGoil⋅ ⎢1 + ⎛⎜ ⎞⎟ ⎥
D
⎣
(Note:
s =
L
∆h
which yields
=
π
4
∆h
SGoil
2
⋅D ⋅x =
π 2
⋅d ⋅L
4
2
⎛d⎞ L
⎟
⎝D⎠
x=⎜
Substituting values into the expression:
L =
25⋅ mm
⎡
6⋅ mm ⎞ ⎤
⎟⎥
⎝ 18⋅ mm ⎠ ⎦
0.827⋅ ⎢1 + ⎛⎜
⎝ ⎠⎦
s = 1.088
∆p = ρwater⋅ g⋅ ∆h where ∆h = 25⋅ mm
⎣
for this manometer.)
2
L = 27.2⋅ mm
Problem 3.29
[Difficulty: 2]
Given:
A U-tube manometer is connected to the open tank filled with water as
shown (manometer fluid is Meriam blue)
D1  2.5 m D2  0.7 m d  0.2 m SGoil  1.75 (From Table A.1, App. A)
Find:
The manometer deflection, l
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dh
 ρ g
(Hydrostatic Pressure - h is positive downwards)
ρ  SG ρwater
Assumptions:
(Definition of Specific Gravity)
(1) Static liquid
(2) Incompressible liquid
Integrating the hydrostatic pressure equation we get:
∆p  ρ g ∆h
When the tank is filled with water, the oil in the left leg of the manometer is displaced
downward by l/2. The oil in the right leg is displaced upward by the same distance, l/2.
D1
Beginning at the free surface of the tank, and accounting for the changes in pressure with
elevation:
d 
D2
l

patm  ρwater g  D1  D2  d    ρoil g l  patm
2


Upon simplification:


ρwater g  D1  D2  d 
l
  ρoil g l
2
D1  D2  d 
l 
l
 SGoil l
2
2.5 m  0.7 m  0.2 m
1.75 
1
2
l 
D1  D2  d
1
SGoil 
2
l  1.600 m

Problem 3.30
Given:
[Difficulty: 2]
Reservoir manometer with dimensions shown. The manometer fluid
specific gravity is given.
D 
5
8
 in
d 
3
16
 in SGoil  0.827
Find:
The required distance between vertical marks on the scale
corresponding to Δp of 1 in water.
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dz
  ρ g
(Hydrostatic Pressure - z is positive upwards)
ρ  SG ρwater
Assumptions:
(Definition of Specific Gravity)
(1) Static liquid
(2) Incompressible liquid
Integrating the hydrostatic pressure equation we get:
∆p  ρ g ∆z
h
Beginning at the free surface of the tank, and accounting for the changes in pressure with
elevation:
patm  ∆p  ρoil g ( x  h)  patm
Upon simplification:
∆p  ρoil g ( x  h)
Therefore:
ρwater g l  ρoil g ( x  h)
The applied pressure is defined as:
x and h are related through the manometer dimensions:
Solving for h:
h
l
2
 d
SGoil 1    
D

 
x
xh 
π
4
2
D x 
∆p  ρwater g l
where
l
SGoil
π 2
d h
4
Substituting values into the expression:
2
d h

D
x
h 
1 in

2
 0.1875 in  

 0.625 in  
0.827 1  

h  1.109 in
l  1 in
Problem 3.31
Given:
[Difficulty: 2]
A U-tube manometer is connected to the open tank filled with water as
shown (manometer fluid is mercury). The tank is sealed and pressurized.
D1  2.5 m D2  0.7 m d  0.2 m po  0.5 atm
Find:
The manometer deflection, l
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dh
 ρ g
(Hydrostatic Pressure - h is positive downwards)
ρ  SG ρwater
Assumptions:
SGHg  13.55 (From Table A.1, App. A)
(Definition of Specific Gravity)
(1) Static liquid
(2) Incompressible liquid
Integrating the hydrostatic pressure equation we get:
∆p  ρ g ∆h
When the tank is filled with water and pressurized, the mercury in the left leg of the
manometer is displaced downward by l/2. The mercury in the right leg is displaced
upward by the same distance, l/2.
D1
Beginning at the free surface of the tank, and accounting for the changes in pressure with
elevation:
d 
D2

l

patm  po  ρwater g  D1  D2  d    ρHg g l  patm
2

Upon simplification:

po  ρwater g  D1  D2  d 

po
l
  ρHg g l
2
l
ρwater g
 D1  D2  d
1
SGHg 
2
Substituting values into the expression:
5
3
2

 0.5 atm  1.013  10  N  1  m  1  s   2.5 m  0.7 m  0.2 m
2

1000 kg 9.8 m 
m  atm


l 
13.55 
1
2
l  0.549 m
Problem 3.32
[Difficulty: 3]
Given:
Inclined manometer as shown.
D  96 mm d  8 mm
Angle θ is such that the liquid deflection L is five times that of a regular
U-tube manometer.
Find:
Angle θ and manometer sensitivity.
Solution:
We will apply the hydrostatics equations to this system.
Governing Equation:
dp
dz
Assumptions:
  ρ g
(Hydrostatic Pressure - z is positive upwards)
(1) Static liquid
(2) Incompressible liquid
Integrating the hydrostatic pressure equation we get:
∆p  ρ g ∆z
Applying this equation from point 1 to point 2:
p1  ρ g ( x  L sin ( θ) )  p2
Upon simplification:
x
p1  p2  ρ g ( x  L sin ( θ) )
Since the volume of the fluid must remain constant:
π
4
2
D x 
π 2
d L
4
2
 d  L

D
x
 d

Therefore: p1  p2  ρ g L    sin ( θ)
D
2
 

Now for a U-tube manometer:
p1  p2  ρ  g h
For equal applied pressures:
L 
Hence:
 d  2

  sin ( θ)  h
 D 

p1incl  p2incl
p1U  p2U
Since L/h = 5:
 d  2

  sin ( θ)
 D 

ρ g L 

sin ( θ) 
ρ  g h
h
L
2
 d   1   8 mm 



5  96 mm 
D
2

θ  11.13 deg
The sensitivity of the manometer:
s
L
∆he

L
SG h
s
5
SG
Problem 3.33
Given:
Data on inclined manometer
Find:
Angle θ for given data; find sensitivity
[Difficulty: 3]
Solution:
Basic equation
dp
dy
  ρ g
or, for constant ρ
where Δh is height difference
∆p  ρ g ∆h
Under applied pressure
∆p  SGMer ρ g ( L sin( θ)  x)
From Table A.1
SGMer  0.827
and Δp = 1 in. of water, or
∆p  ρ g h
∆p  1000
kg
3
 9.81
h  0.025 m
2
m
2
 0.025 m 
s
The volume of liquid must remain constant, so x Ares  L Atube
Solving for θ
h  25 mm
where
m
Combining Eqs 1 and 2
(1)
x  L
Atube
Ares
N s
∆p  245 Pa
kg m
d

 D
2
 L 
(2)
2

d 
∆p  SGMer ρ g L sin ( θ)  L   
D

sin ( θ) 
 
∆p
SGMer ρ g L
sin ( θ)  245
N
2

m
d

D
2

1
0.827

1
The sensitivity is the ratio of manometer deflection to a vertical water manometer
L
h

0.15 m
0.025 m
m
1000 kg
θ  11 deg
s
3

s6

1
2

s
9.81 m

1

1
0.15 m

kg m
2
s N
2
 8   0.186

 76 

Problem 3.34
Given:
[Difficulty: 4]
Barometer with water on top of the mercury column, Temperature is
known:
h2  6.5 in
h1  28.35 in
SGHg  13.55
T  70 °F
(From Table A.2, App. A)
pv  0.363 psi (From Table A.7, App. A)
Find:
(a) Barometric pressure in psia
(b) Effect of increase in ambient temperature on length of mercury
column for the same barometric pressure:
Tf  85 °F
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dh
  ρ g
(Hydrostatic Pressure - h is positive downwards)
ρ  SG ρwater
Assumptions:
(Definition of Specific Gravity)
(1) Static liquid
(2) Incompressible liquid
Water vapor
Water
Integrating the hydrostatic pressure equation we get:
h2
∆p  ρ g ∆h
Mercury
Start at the free surface of the mercury and progress through the barometer to the vapor
pressure of the water:
h1
patm  ρHg g h1  ρwater g h2  pv

patm  pv  ρwater g SGHg h1  h2
patm  0.363
lbf
2
in
 1.93 
slug
ft
3
 32.2
ft
2
s

2

lbf  s
slug ft
 ft 

 12 in 
 ( 13.55  28.35 in  6.5 in)  
3
patm  14.41
At the higher temperature, the vapor pressure of water increases to 0.60 psi. Therefore, if the atmospheric pressure
were to remain constant, the length of the mercury column would have to decrease - the increased water vapor would
push the mercury out of the tube!
lbf
2
in
Problem 3.35
Given:
[Difficulty: 3]
U-tube manometer with tubes of different diameter and two liquids, as shown.
d1 = 10⋅ mm d2 = 15⋅ mm SGoil = 0.85
Find:
Solution:
∆p = 250⋅
(a) the deflection, h, corresponding to
(b) the sensitivity of the manometer
N
2
m
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dz
= − ρ⋅ g
(Hydrostatic Pressure - z is positive upwards)
ρ = SG⋅ ρwater
Assumptions:
(Definition of Specific Gravity)
(1) Static liquid
(2) Incompressible liquid
patm
patm + Δp
patm
patm
Integrating the hydrostatic pressure equation we get:
(
)
(
)
p − po = −ρ⋅ g⋅ z − zo = ρ⋅ g⋅ zo − z
h
From the left diagram:
l2
pA − patm = ρwater⋅ g⋅ l1 = ρoil⋅ g⋅ l2
l3
lw
l1
( 1)
l4
A
B
From the right diagram:
pB − patm + ∆p = ρwater⋅ g⋅ l3
(
)
( 2)
pB − patm = ρwater⋅ g⋅ l4 + ρoil⋅ g⋅ l2
( 3)
Combining these three equations:
From the diagram we can see
(
)
(
)
∆p = ρwater⋅ g⋅ l4 − l3 + ρoil⋅ g⋅ l2 = ρwater⋅ g⋅ l4 + l1 − l3
lw = l1 − l3
(
∆p = ρwater⋅ g⋅ h + lw
)
and
h = l4
Therefore:
( 4)
2
We can relate lw to h since the volume of water in the manometer is constant:
π
4
2
π
2
⋅ d1 ⋅ lw = ⋅ d2 ⋅ h
4
⎛ d2 ⎞
lw = ⎜ ⎟ ⋅ h
⎝ d1 ⎠
2⎤
⎡
⎢ ⎛ d2 ⎞ ⎥
∆p = ρwater⋅ g⋅ h⋅ 1 + ⎜ ⎟
⎢
⎥
⎣ ⎝ d1 ⎠ ⎦
Substituting this into (4) yields:
Substituting values into the equation:
N
h = 250⋅
2
×
m
The sensitivity for the manometer is defined as:
Therefore:
s=
1
⎛ d2 ⎞
1+⎜ ⎟
⎝ d1 ⎠
2
s=
s =
1
3
⋅
m
999 kg
h
×
Solving for h:
2
1
s
9.81 m
×
where
∆he
1
⎛ 15⋅ mm ⎞
1+⎜
⎟
⎝ 10⋅ mm ⎠
∆p
h=
1
⎛ 15⋅ mm ⎞
1+⎜
⎟
⎝ 10⋅ mm ⎠
2
2⎤
⎡
⎢ ⎛ d2 ⎞ ⎥
ρwater⋅ g⋅ 1 + ⎜ ⎟
⎢
⎥
⎣ ⎝ d1 ⎠ ⎦
×
kg⋅ m
2
N⋅ s
3
×
10 ⋅ mm
h = 7.85⋅ mm
m
∆p = ρwater⋅ g⋅ ∆he
s = 0.308
2
The design is a poor one. The sensitivity could be improved by interchanging
d2 and d1 , i.e., having d2 smaller than d1
A plot of the manometer sensitivity is shown below:
Sensitivity
1
0.5
0
1
2
Diameter Ratio, d2/d1
3
4
5
Problem 3.36
Given:
[Difficulty: 3]
Water column standin in glass tube
∆h = 50⋅ mm D = 2.5⋅ mm σ = 72.8 × 10
−3N
m
(From Table A.4, App. A)
Find:
(a) Column height if surface tension were zero.
(b) Column height in 1 mm diameter tube
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dh
= ρ⋅ g
(Hydrostatic Pressure - h is positive downwards)
ΣFz = 0
Assumptions:
(Static Equilibrium)
(1) Static, incompressible liquid
(2) Neglect volume under meniscus
(3) Applied pressure remains constant
(4) Column height is sum of capillary rise and pressure
difference
Δhc
Δh
Δhp
∆h = ∆hc + ∆hp
Assumption #4 can be written as:
Choose a free-body diagram of the capillary rise portion of the column for analysis:
ΣFz = π⋅ D⋅ σ⋅ cos ( θ) −
4⋅ σ
Therefore: ∆hc =
⋅ cos ( θ)
ρ⋅ g⋅ D
π 2
⋅ D ⋅ ρ⋅ g⋅ ∆hc = 0
4
θ
Substituting values:
⎛ 103⋅ mm ⎞
⎟
×⎜
∆hc = 4 × 72.8 × 10 ⋅ ×
⋅
×
⋅ ×
⋅
×
2 ⎝
m 999 kg 9.81 m 2.5 mm
m ⎠
N⋅ s
−3 N
1
3
m
1
2
s
1
kg⋅ m
1
2
Δhc
∆hc = 11.89⋅ mm
Therefore: ∆hp = ∆h − ∆hc
∆hp = 50⋅ mm − 11.89⋅ mm
π Dδ
∆hp = 38.1⋅ mm
Mg = ρgV
(result for σ = 0)
For the 1 mm diameter tube:
⎛ 103⋅ mm ⎞
⎟
∆hc = 4 × 72.8 × 10 ⋅ ×
⋅
×
⋅ × ⋅
×
×⎜
m 999 kg 9.81 m 1 mm N⋅ s2 ⎝ m ⎠
−3 N
∆h = 29.7⋅ mm + 38.1⋅ mm
1
3
m
1
2
s
1
1
kg⋅ m
2
∆hc = 29.71⋅ mm
∆h = 67.8⋅ mm
Problem 3.37
Given:
[Difficulty: 4]
Sealed tank is partially filled with water. Water drains slowly from the
tank until the system attains equilibrium. U-tube manometer is connected
to the tank as shown. (Meriam blue in manometer)
L  3 m
D1  2.5 m
D2  0.7 m
d  0.2 m
Find:
The manometer deflection, l, under equilibrium conditions
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dh
Assumptions:
 ρ g
SGoil  1.75
(From Table A.2, App. A)
(Hydrostatic Pressure - h is positive downwards)
p  V  M R T
(Ideal gas equation of state)
ρ  SG ρwater
(Definition of Specific Gravity)
(1) Static liquid
(2) Incompressible liquid
(3) Air in tank behaves ideally
patm
p0
Integrating the hydrostatic pressure equation we get:
L
∆p  ρ g ∆h
D1
d
H
D2
To determine the surface pressure po
under equilibrium conditions
po V o

M R Ta
Thus,
M R To

po 

L  D1
Simplifying: po 
p
( L  H) a
 L  D 1  p
( L  H)
a  ρwater g H  pa
Va
Vo
 pa 
 L  D 1  A  p
( L  H)  A
a
po  ρwater g H  pa
Now under equilibrium conditions:
Upon rearranging:
Now we apply the quadratic formula to solve for H:
l

we assume that the air expands at constant temperature:
pa Va

2


Combining these expressions:
ρwater g H  pa  ρwater g L  H  D1 pa  0
a  ρwater g
a  999
kg
3
 9.81
m

b   pa  ρwater g L

m
3 Pa
a  9.8  10 
2
s
kg
m
5 N


b   1.013  10 
 999
 9.81  3 m
2
3
2
m
m
s


5 N
2
c  D 1 pa
m
5
b  1.307  10 Pa
5
c  2.5 m  1.013  10 
c  2.532  10  Pa m
m
Hupper 
b 

2
b  4 a c
2 a
 1.307  105 Pa2  4  9.8  103 Pa  2.532  105 Pa m
5
 1.307  10  Pa 
m
Hupper 
3 Pa
2  9.8  10 
m
Hupper  10.985 m
Hlower 
b 

b  4 a c
2 a
 1.307  105 Pa2  4  9.8  103 Pa  2.532  105 Pa m
5
 1.307  10  Pa 
2
m
Hlower 
3 Pa
2  9.8  10 
m
Hlower  2.352 m
H  2.352 m
Since H can not be greater than 3 m (otherwise the tank would overflow!), we must select the lower value for H:
Solving for the pressure inside the tank:
po 
( 3 m  2.5 m)
5
 1.013  10  Pa
( 3 m  2.352 m)
l

po  ρwater g  H  D2  d    ρoil g l  pa
2

Applying the hydrostatic pressure equation to the manometer:
Solving for the manometer deflection:
 pa  po
l
 ρwater g

 H  D2  d 
1
 SGoil  1
2


1 m
1 s
kg m
1
5 N
4 N 
l   1.013  10 
 7.816  10   


 
 2.352 m  0.7 m  0.2 m 
2
2
2

999 kg 9.81 m

1
m
m 
N s

 1.75 
2
3
4
po  7.816  10 Pa
2
l  0.407 m
Problem 3.38
[Difficulty :2]
Fluid 1
Fluid 2
Given:
Two fluids inside and outside a tube
Find:
(a) An expression for height Δh
(b) Height difference when D =0.040 in for water/mercury
Assumptions:
ρ1gΔhπD2/4
(1) Static, incompressible fluids
(2) Neglect meniscus curvature for column height and
volume calculations
Solution:
A free-body vertical force analysis for the section of fluid 1 height Δh in the tube below
the "free surface" of fluid 2 leads to
∑
2
F = 0 = ∆p⋅
π⋅ D
2
− ρ1⋅ g⋅ ∆h⋅
4
π⋅ D
4
+ π⋅ D⋅ σ⋅ cos ( θ)
where Δp is the pressure difference generated by fluid 2 over height Δh,
2
π⋅ D
Hence
∆p⋅
Solving for Δh
∆h = −
4
2
− ρ1⋅ g⋅ ∆h⋅
4⋅ σ⋅ cos ( θ)
(
σπDcosθ
π⋅ D
4
∆p = ρ2⋅ g⋅ ∆h
2
2
π⋅ D
π⋅ D
= ρ2⋅ g⋅ ∆h⋅
− ρ1⋅ g⋅ ∆h⋅
= −π⋅ D⋅ σ⋅ cos ( θ)
4
4
)
g⋅ D ⋅ ρ 2 − ρ 1
For fluids 1 and 2 being water and mercury (for mercury σ = 375 mN/m and θ = 140o, from Table A.4), solving for Δh when
D = 0.040 in
∆h = −4 × 0.375⋅
∆h = 0.360⋅ in
N
m
×
lbf
4.448⋅ N
×
0.0254m
in
2
× cos ( 140⋅ deg) ×
s
32.2⋅ ft
×
1
0.040⋅ in
×
3
3
1
slugft
⋅
⎛ 12⋅ in ⎞ ×
×
⎟
2
1.94⋅ slug ⎝ ft ⎠
( 13.6 − 1)
lbf⋅ s
ft
×⎜
Problem 3.39
[Difficulty: 2]
h1
Oil
Air
h4
h2
Hg
Given:
Data on manometer before and after an "accident"
Find:
Change in mercury level
Assumptions:
h3
x
(1) Liquids are incompressible and static
(2) Pressure change across air in bubble is negligible
(3) Any curvature of air bubble surface can be neglected in volume calculations
Solution:
Basic equation
dp
dy
  ρ g
or, for constant ρ
For the initial state, working from right to left
where ∆h is height difference
∆p  ρ g ∆h

patm  patm  SGHg ρ g h3  SGoil ρ g h1  h2

SGHg ρ g h3  SGoil ρ g h1  h2


(1)
Note that the air pocket has no effect!
For the final state, working from right to left


patm  patm  SGHg ρ g h3  x  SGoil ρ g h4


SGHg ρ g h3  x  SGoil ρ g h4
(2)
The two unknowns here are the mercury levels before and after (i.e., h3 and x)

Combining Eqs. 1 and 2
SGHg ρ g x  SGoil ρ g h1  h2  h4
From Table A.1
SGHg  13.55
The term
h1  h2  h4
h1  h2  h4 
Then from Eq. 3
x 
1.4
13.55

x
SGoil
SGHg

 h1  h2  h4

(3)
is the difference between the total height of oil before and after the
accident
∆V
 π  d2 


 4 
 1.019 in

2
 1   0.2 in3  1.019 in

π  0.5 in 
4

x  0.1053 in
Problem 3.40
[Difficulty: 2]
Water
Given:
Water in a tube or between parallel plates
Find:
Height Δh for each system
Solution:
a) Tube: A free-body vertical force analysis for the section of water height Δh above the "free surface" in the tube, as
shown in the figure, leads to
∑
2
F = 0 = π⋅ D⋅ σ⋅ cos ( θ) − ρ⋅ g⋅ ∆h⋅
π⋅ D
4
Assumption: Neglect meniscus curvature for column height and volume calculations
Solving for Δh
∆h =
4⋅ σ⋅ cos ( θ)
ρ ⋅ g⋅ D
b) Parallel Plates: A free-body vertical force analysis for the section of water height Δh above the "free surface" between
plates arbitrary width w (similar to the figure above), leads to
∑ F = 0 = 2⋅ w⋅ σ⋅ cos(θ) − ρ⋅ g⋅ ∆h⋅ w⋅ a
Solving for Δh
∆h =
2⋅ σ⋅ cos ( θ)
ρ ⋅ g⋅ a
For water σ = 72.8 mN/m and θ = 0o (Table A.4), so
4 × 0.0728⋅
a) Tube
∆h =
999⋅
kg
3
× 9.81⋅
m
m
2
∆h =
999⋅
kg
3
m
m
×
kg⋅ m
2
−3
∆h = 5.94 × 10
m
∆h = 5.94⋅ mm
m
∆h = 2.97⋅ mm
N⋅ s
× 0.005⋅ m
s
2 × 0.0728⋅
b) Parallel Plates
N
× 9.81⋅
m
2
s
N
m
× 0.005⋅ m
×
kg⋅ m
2
N⋅ s
−3
∆h = 2.97 × 10
σ=
ρ=
0.005
1.94
lbf/ft
3
slug/ft
Using the formula above
Δh (in)
0.0400
0.0200
0.0133
0.0100
0.0080
0.0067
0.0057
0.0050
0.0044
0.0040
0.0036
0.0033
0.0031
0.0029
0.0027
0.0025
0.0024
0.0022
0.0020
Capillary Height Between Vertical Plates
0.045
Height Δh (in)
a (in)
0.004
0.008
0.012
0.016
0.020
0.024
0.028
0.032
0.036
0.040
0.044
0.048
0.052
0.056
0.060
0.064
0.068
0.072
0.080
0.040
0.035
0.030
0.025
0.020
0.015
0.010
0.005
0.000
0.00
0.01
0.02
0.03
0.04
Gap a (in)
0.05
0.06
0.07
0.08
p SL =
R =
=
101
286.9
999
kPa
J/kg.K
kg/m3
The temperature can be computed from the data in the figure.
The pressures are then computed from the appropriate equation.
From Table A.3
Atmospheric Pressure vs Elevation
1.00000
0
10
20
30
40
50
60
70
80
90
0.10000
Pressure Ratio p /p SL
0.01000
0.00100
Computed
0.00010
Table A.3
0.00001
0.00000
Elevation (km)
Agreement between calculated and tabulated data is very good (as it should be, considering the table data are also computed!)
100
z (km)
T (oC)
T (K)
0.0
2.0
4.0
6.0
8.0
11.0
12.0
14.0
16.0
18.0
20.1
22.0
24.0
26.0
28.0
30.0
32.2
34.0
36.0
38.0
40.0
42.0
44.0
46.0
47.3
50.0
52.4
54.0
56.0
58.0
60.0
61.6
64.0
66.0
68.0
70.0
72.0
74.0
76.0
78.0
80.0
82.0
84.0
86.0
88.0
90.0
15.0
2.0
-11.0
-24.0
-37.0
-56.5
-56.5
-56.5
-56.5
-56.5
-56.5
-54.6
-52.6
-50.6
-48.7
-46.7
-44.5
-39.5
-33.9
-28.4
-22.8
-17.2
-11.7
-6.1
-2.5
-2.5
-2.5
-5.6
-9.5
-13.5
-17.4
-20.5
-29.9
-37.7
-45.5
-53.4
-61.2
-69.0
-76.8
-84.7
-92.5
-92.5
-92.5
-92.5
-92.5
-92.5
288.0
275.00
262.0
249.0
236.0
216.5
216.5
216.5
216.5
216.5
216.5
218.4
220.4
222.4
224.3
226.3
228.5
233.5
239.1
244.6
250.2
255.8
261.3
266.9
270.5
270.5
270.5
267.4
263.5
259.5
255.6
252.5
243.1
235.3
227.5
219.6
211.8
204.0
196.2
188.3
180.5
180.5
180.5
180.5
180.5
180.5
m =
0.0065
(K/m)
T = const
m =
-0.000991736
(K/m)
m =
-0.002781457
(K/m)
T = const
m =
0.001956522
(K/m)
m =
0.003913043
(K/m)
T = const
p /p SL
z (km)
p /p SL
1.000
0.784
0.608
0.465
0.351
0.223
0.190
0.139
0.101
0.0738
0.0530
0.0393
0.0288
0.0211
0.0155
0.0115
0.00824
0.00632
0.00473
0.00356
0.00270
0.00206
0.00158
0.00122
0.00104
0.000736
0.000544
0.000444
0.000343
0.000264
0.000202
0.000163
0.000117
0.0000880
0.0000655
0.0000482
0.0000351
0.0000253
0.0000180
0.0000126
0.00000861
0.00000590
0.00000404
0.00000276
0.00000189
0.00000130
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
6.0
7.0
8.0
9.0
10.0
11.0
12.0
13.0
14.0
15.0
16.0
17.0
18.0
19.0
20.0
22.0
24.0
26.0
28.0
30.0
40.0
50.0
60.0
70.0
80.0
90.0
1.000
0.942
0.887
0.835
0.785
0.737
0.692
0.649
0.609
0.570
0.533
0.466
0.406
0.352
0.304
0.262
0.224
0.192
0.164
0.140
0.120
0.102
0.0873
0.0747
0.0638
0.0546
0.0400
0.0293
0.0216
0.0160
0.0118
0.00283
0.000787
0.000222
0.0000545
0.0000102
0.00000162
Problem 3.43
[Difficulty: 3]
Given:
Data on isothermal atmosphere
Find:
Elevation changes for 3% pressure change and 5% density change; plot of pressure and density versus elevation
Solution:
Assumptions:
Static, isothermal fluid,; g = constant; ideal gas behavior
dp
Basic equations
dz
dp
Then
dz
For an ideal gas with T constant
p2
p1

g
ρ1 Rair T
Rair  287
Evaluating
C
For a 3% reduction in pressure
For a 5% reduction in density
To plot
p2
p1
and
ρ2
ρ1
we rearrange Eq. 1
p2
p1
ρ2
ρ1
ρ2
ρ1
 p2 

 p1 
 ln 
ρ2 Rair T
From Table A.6
and
Rair T
Rair T0
p  ρ R  T
and
p g
  ρ g  
∆z  
Integrating
  ρ g

dp
p

where
T  T0
so
∆z  
ρ2
ρ1
g
Rair T
 dz
Rair T0
g
 ρ2 
 ρ2 
  C ln 
 ρ1 
 ρ1 
 ln 
(1)
N m
kg K
Rair T0
 287
g
N m
kg K
 ( 30  273)  K 
1
2

s
9.81 m

kg m
2
C  8865 m
N s
 0.97
so from Eq. 1
∆z  8865 m ln ( 0.97)
∆z  270 m
 0.95
so from Eq. 1
∆z  8865 m ln ( 0.95)
∆z  455 m

p2
p1

e
∆z
C
5000
Elevation (m)
4000
3000
2000
1000
0.5
0.6
0.7
0.8
Pressure or Density Ratio
This plot can be plotted in Excel
0.9
1
Problem 3.44
[Difficulty: 3]
Given:
Atmospheric conditions at ground level (z = 0) in Denver, Colorado are p0 = 83.2 kPa, T0 = 25°C.
Pike's peak is at elevation z = 2690 m.
Find:
p/p0 vs z for both cases.
Solution:
dp
Governing Equations:
Assumptions:
  ρ g
dz
p  ρ R  T
(1) Static fluid
(2) Ideal gas behavior
(a) For an incompressible atmosphere:
dp
dz
At
z
  ρ g

p  p 0    ρ  g dz
0
becomes
p
(b) For an adiabatic atmosphere:
ρ
dp
dz

m


s
p  83.2 kPa   1  9.81
z  2690 m
k
g z 

p  p0  ρ0 g z  p0  1 
R  T0 


or
2
 2690 m 
 p 
ρ  ρ 0  
 p0 
 const
 p 
dp  ρ0    g dz
 p0 
becomes
287 N  m
or
1
p
p
But
k1

1
k
k

dp 
 p  p0

1
k1

k
 p
p



1
298 K

2
N  s 
p  57.5 kPa
kg m 

1
k
1
k
  ρ g
kg K
(1)
1
k
dp  
ρ 0 g
p0
1
k
 dz
k1 
 k1
ρ 0 g
k  k
k 
 p
 p0

 g z

1
k1 
hence
p0
k
0
Solving for the pressure ratio
 k  1 ρ0 
p
 1 
  g z
p0
k p0

At
z  2690 m
k
k1
or


1.4  1


1.4
p  83.2 kPa   1 
 9.81
p
 k  1  g z 
 1 
p0
k R  T0 

m
2
s
 2690 m 
kg K
287 N m

k
k1
(2)

1
298 K

2
N s 
kg m 

1.4
1.41
p  60.2 kPa
Elevation above Denver (m)
Equations 1 and 2 can be plotted:
510
3
410
3
310
3
210
3
110
3
Temperature Variation with Elevation
Incompressible
Adiabatic
0
0.4
0.6
0.8
Pressure Ratio (-)
1
Problem 3.45
Given:
[Difficulty: 3]
Martian atmosphere behaves as an idel gas, constant temperature
m
Mm  32.0 T  200 K g  3.92
2
s
ρo  0.015
kg
3
m
Find:
Density at z = 20 km
Plot the ratio of density to sea level density versus altitude, compare to
that of earth.
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
  ρ g
dz
(Hydrostatic Pressure - z is positive upwards)
p  ρ R  T
Ru
R
Mm
Assumptions:
(Ideal Gas Equation of State)
(Definition of Gas Constant)
(1) Static fluid
(2) Constant gravitational acceleration
(3) Ideal gas behavior
dp  R T dρ
Taking the differential of the equation of state (constant temperature):
R  T
Substituting into the hydrostatic pressure equation:
ρ
Integrating this expression:
dρ
z
 1

g

dρ   
dz
 R T
 ρ
0
ρ
N m
kg mol K



1
32.0
  3.92
ρ  0.015
kg
m
3
e
For the Martian atmosphere, let
m
s
2

 ρ    g z

R T
 ρo 
kg mol
R  259.822
kg
3
 20 10  m 
dρ
Therefore:
ln 
o
Evaluating: R  8314.3
dz
  ρ g
ρ
ρ
or
ρo

g
R T

e
 dz
g z
RT
( 1)
N m
kg K
1
kg K 1 1 N  s 


 
259.822 N  m 200 K kg m 
2

 3 kg
3
ρ  3.32  10
m
x
g
R T
x  3.92
m
2
s

1

kg K
259.822 N  m

1

1
200 K
2

N s
kg m
x  0.07544
1
km
ρ
Therefore:
ρo
 x z
e
These data are plotted along with the data for Earth's atmosphere from Table A.3.
Density Ratios of Earth and Mars versus Elevation
20
Elevation z (km)
15
10
5
Earth
Mars
0
0
0.2
0.4
0.6
Density Ratio (-)
0.8
1
Problem 3.46
Given:
[Difficulty: 3]
Door located in plane vertical wall of water tank as shown
a = 1.5⋅ m b = 1⋅ m
c = 1⋅ m
ps
Atmospheric pressure acts on outer surface of door.
Find:
c
Resultant force and line of action:
(a) for
(b) for
y
ps = patm
y’
a
psg = 0.3⋅ atm
Plot F/Fo and y'/yc over range of ps/patm (Fo is force
determined in (a), yc is y-ccordinate of door centroid).
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dy
b
= ρ⋅ g
(Hydrostatic Pressure - y is positive downwards)
⌠
⎮
p dA
⎮
⌡
⌠
⎮
y'⋅ FR =
y⋅ p dA
⎮
⌡
FR =
Assumptions:
(Hydrostatic Force on door)
(First moment of force)
(1) Static fluid
(2) Incompressible fluid
We will obtain a general expression for the force and line of action, and then simplify for parts (a) and (b).
Since
dp = ρ⋅ g⋅ dh
Now because
patm
it follows that p = ps + ρ⋅ g⋅ y
acts on the outside of the door,
psg is the surface gage pressure:
c+ a
c+ a
⌠
⌠
⌠
ρ⋅ g 2
⎮
⎡
⎤
FR =
p dA = ⎮
p⋅ b dy = ⎮
psg + ρ⋅ g⋅ y ⋅ b dy = b⋅ ⎢psg⋅ a +
⋅ a + 2⋅ a⋅ c ⎥
⎮
⌡
2
⌡c
⎣
⎦
⌡
c
(
⌠
⎮
y'⋅ FR =
y⋅ p dA
⎮
⌡
Therefore:
Evaluating the integral:
y' =
(
)
)
c+ a
1 ⌠
1 ⌠
⎮
y' =
y⋅ p dA =
⋅⎮
y⋅ psg + ρ⋅ g⋅ y ⋅ b dy
FR ⎮
FR ⌡
⌡
c
(
b ⎡ psg ⎡
ρ⋅ g ⎡
2
2
3
3⎤
⎢ ⎣( c + a) − c ⎤⎦ +
⋅ ⎣( c + a) − c ⎤⎦⎥
FR ⎣ 2
3
⎦
)
p = psg + ρ⋅ g⋅ y
( 1)
(
psg = 0
For part (a) we know
Fo =
)
⎤
b ⎡ psg 2
ρ⋅ g ⎡ 3
⋅⎢
a + 2⋅ a⋅ c +
⋅ ⎣a + 3⋅ a⋅ c⋅ ( a + c)⎤⎦⎥
FR ⎣ 2
3
⎦
Simplifying: y' =
1
2
kg
× 999⋅
3
so substituting into (1) we get:
× 9.81⋅
m
m
1
3
× 999⋅
kg
3
Fo =
2
2
s
m
× 9.81⋅
2
m
ρ ⋅ g⋅ b
2
(2
y' =
1
× 1⋅ m⋅
2
N⋅ s
Fo = 25.7⋅ kN
kg⋅ m
ρ ⋅ g⋅ b ⎡ 3
⋅ ⎣a + 3⋅ a⋅ c⋅ ( a + c)⎤⎦
3⋅ F o
⋅
25.7 × 10
s
)
⋅ a + 2⋅ a⋅ c
× 1⋅ m × ⎡⎣( 1.5⋅ m) + 2 × 1.5⋅ m × 1⋅ m⎤⎦ ×
Substituting into (2) for the line of action we get:
y' =
( 2)
1
× ⎡⎣( 1.5⋅ m) + 3 × 1.5⋅ m × 1⋅ m × ( 1.5⋅ m + 1⋅ m)⎤⎦ ×
3
3 N
2
N⋅ s
kg⋅ m
y' = 1.9 m
psg = 0.3⋅ atm . Substituting into (1) we get:
For part (b) we know
⎡
1.013 × 10 ⋅ N
⎢
⎣
m ⋅ atm
5
FR = 1⋅ m × ⎢0.3⋅ atm ×
2
× 1.5⋅ m +
1
2
× 999⋅
kg
3
× 9.81⋅
m
m
2
× ⎡⎣( 1.5⋅ m) + 2 × 1.5⋅ m × 1⋅ m⎤⎦ ×
2
s
2
N⋅ s ⎤⎥
kg⋅ m⎥
⎦
FR = 71.3⋅ kN
kg
m
⎤
999⋅
× 9.81⋅
⎢
5
3
2⎥
2
m
s
3
3 N⋅ s ⎥
⎢ 0.3⋅ atm × 1.013 × 10 ⋅ N × ⎡( 1.5) 2 + 2⋅ 1.5⋅ 1⎤ ⋅ m2 +
1⋅ m ×
× ⎡⎣( 1.5) + 3⋅ 1.5⋅ 1⋅ ( 1.5 + 1)⎤⎦ ⋅ m ×
⎣
⎦
⎢ 2
2
kg⋅ m⎥
3
m ⋅ atm
⎣
⎦
y' =
Substituting ⎡into (2) for the line of action we get:
3
71.3 × 10 ⋅ N
y' = 1.789 m
The value of F/Fo is obtained from Eq. (1) and our result from part (a):
F
Fo
=
⎡
⎣
b⋅ ⎢psg⋅ a +
ρ ⋅ g⋅ b
2
For the gate
yc = c +
a
2
ρ⋅ g
2
(2
(2
)⎤
⋅ a + 2⋅ a⋅ c ⎥
)
⋅ a + 2⋅ a⋅ c
⎦ = 1+
2⋅ psg
ρ⋅ g⋅ ( a + 2⋅ c)
Therefore, the value of y'/yc is obtained from Eqs. (1) and (2):
⎡ psg 2
ρ⋅ g ⎡ 3
⎤⎦⎥⎤
(
)
⎢
a
+
2
⋅
a
⋅
c
+
⋅
a
+
3
⋅
a
⋅
c
⋅
(
a
+
c
)
⎣
⎡ psg 2
y'
2⋅ b
3
⎦
(a + 2⋅ a⋅ c) + ρ⋅ g ⋅ ⎡⎣a3 + 3⋅ a⋅ c⋅ (a + c)⎤⎦⎤⎥ = 2⋅ b ⋅ ⎣ 2
=
⋅⎢
yc
FR⋅ ( 2⋅ c + a) ⎣ 2
3
ρ
⋅
g
(
2
⋅
c
+
a
)
2
⎦
⎡ ⎡
⎤⎤
⋅ (a + 2⋅ a⋅ c)⎥⎥
⎢b⋅ ⎢psg⋅ a +
2
⎣ ⎣
⎦⎦
Simplifying this expression we get:
y'
yc
=
2
( 2⋅ c + a)
(
⋅
)
psg 2
ρ⋅ g ⎡ 3
a + 2⋅ a⋅ c +
⋅ ⎣a + 3⋅ a⋅ c⋅ ( a + c)⎤⎦
2
3
psg⋅ a +
ρ⋅ g
2
(2
)
⋅ a + 2⋅ a⋅ c
Based on these expressions we see that the force on the gate varies linearly with the increase in surface pressure, and that the line of
action of the resultant is always below the centroid of the gate. As the pressure increases, however, the line of action moves closer to
the centroid.
Plots of both ratios are shown below:
Force Ratio vs. Surface Pressure
40
Force Ratio F/Fo
30
20
10
0
0
1
2
3
4
5
4
5
Surface Pressure (atm)
Line of Action Ratio vs. Surface Pressure
1.05
Line of Action Ratio y'/yc
1.04
1.03
1.02
1.01
1
0
1
2
Surface Pressure (atm)
3
Problem 3.47
Given:
[Difficulty: 2]
Door of constant width, located in plane vertical wall of water tank is
hinged along upper edge.
b  1 m
D  1 m
ps
L  1.5 m
D
x
y
(a) Force F, if ps  patm
Find:
Hinge
h
Atmospheric pressure acts on outer surface of door; force F is applied
at lower edge to keep door closed.
L
(b) Force F, if p  0.5 atm
sg
Plot F/Fo over tange of ps/patm (Fo is force determined in (a)).
pdA
F
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
 ρ g
dh
FR 
(Hydrostatic Pressure - h is positive downwards)


p dA


(Hydrostatic Force on door)
ΣMz  0
Assumptions:
(Rotational Equilibrium)
(1) Static fluid
(2) Constant density
(3) Door is in equilibrium
Taking moments about the hinge:
 F L 


y p dA  0


dA  b dy
L
1 
Solving for the force: F    b y p dy
L 0
Since
dp  ρ g dh
and hence
it follows that
p  p s  ρ  g ( D  y )
L
( 1)
We will obtain a general expression for F
and then simplify for parts (a) and (b).
p  p s  ρ  g h
Now because
patm
1 
From Equation (1): F    b y psg  ρ g ( D  y) dy


L 
0
where
h  Dy
acts on the outside of the door,
L
psg
is the surface gage pressure.
b 
2
F     psg  ρ g D  y  ρ g y  dy


L 
0


After integrating: F 
2
3
b 
L
L 
  psg  ρ g D 
 ρ  g 
L 
2
3


(a) For ps  patm it follows that psg  0
Fo  999
kg
3
 9.81
m
m
2


F  b psg
or
L
2
 D  L

 2 3
F o  ρ  g b L  
Therefore:
 D  L 

 2 3 
 ρ  g L  
( 2)
( 3)
2
 1 m  1.5 m   N s

3  kg m
 2
 1 m  1.5 m  
s
Fo  14.7 kN
(b) For psg  0.5 atm we substitute variables:

101 kPa


atm
F  1 m  0.5 atm 

1.5 m
2
kg
3
 9.81
m
F
From Equations (2) and (3) we have:
 999
Fo



m
2
2
 1 m  1.5 m   N s 

3  kg m
 2

 1.5 m  
s
F  52.6 kN
 D  L 

psg
2
 2 3   1 
 D L
 D L
ρ  g b L    
2 ρ g   
 2 3
 2 3
b psg
L
 ρ  g L  
Here is a plot of the force ratio as a function of the surface pressure:
Force Ratio, F/Fo
30
20
10
0
0
1
2
3
Surface Pressure Ratio, ps/patm
4
5
Problem 3.48
[Difficulty: 5]
Discussion: The design requirements are specified except that a typical floor height is about 12 ft, making the total required lift
about 36 ft. A spreadsheet was used to calculate the system properties for various pressures. Results are presented on the next page,
followed by a sample calculation. Total cost dropped quickly as system pressure was increased. A shallow minimum was reached in
the 100-110 psig range. The lowest-cost solution was obtained at a system pressure of about 100 psig. At this pressure, the reservoir
of 140 gal required a 3.30 ft diameter pressure sphere with a 0.250 in wall thickness. The welding cost was $155 and the material cost
$433, for a total cost of $588. Accumulator wall thickness was constrained at 0.250 in for pressures below 100 psi; it increased for
higher pressures (this caused the discontinuity in slope of the curve at 100 psig). The mass of steel became constant above 110 psig.
No allowance was made for the extra volume needed to pressurize the accumulator. Fail-safe design is essential for an elevator to be
used by the public. The control circuitry should be redundant. Failures must be easy to spot. For this reason, hydraulic actuation is
good: leaks will be readily apparent. The final design must be reviewed, approved, and stamped by a professional engineer since the
design involves public safety. The terminology used in the solution is defined in the following table:
Symbol
Definition
Units
p
System pressure
psig
Ap
Area of lift piston
in2
Voil
Volume of oil
gal
Ds
Diameter of spherical accumulator
ft
t
Wall thickness of accumulator
in
Aw
Area of weld
in2
Cw
Cost of weld
$
Ms
Mass of steel accumulator
lbm
Cs
Cost of steel
$
Ct
Total Cost
$
A sample calculation and the results of the system simulation in Excel are presented below.
p
πD S2
4
πD S tσ
Results of system simulation:
Problem 3.49
Given:
Geometry of chamber system
Find:
Pressure at various locations
Assumptions:
[Difficulty: 2]
(1) Water and Meriam Blue are static and incompressible
(2) Pressure gradients across air cavities are negligible
Solution:
Basic equation
dp
or, for constant ρ
  ρ g
dy
where ∆h is height difference
∆p  ρ g ∆h
For point A
pA  patm  ρH2O g h1 or in gage pressure
pA  ρH2O g h1
Here we have
h1  8 in
h1  0.667 ft
pA  1.94
slug
ft
3
 32.2
ft
2
2
 0.667 ft 
s
lbf s
 ft 


slugft

 12 in 
For the first air cavity
pair1  pA  SGMB ρH2O g h2 where
From Table A.1
SGMB  1.75
lbf
pair1  0.289
2
 1.75  1.94
slug
in
ft
3
 32.2
ft
2
2
pA  0.289 psi
h2  4 in
h2  0.333 ft
2
 0.333 ft 
 ft 

slug ft  12 in 
where
h3  6 in
s
lbf  s
(gage)

2
pair1  0.036 psi
(gage)
Note that p = constant throughout the air pocket
For point B
pB  pair1  SGHg ρH2O g h3
pB  0.036
lbf
2
 1.75  1.94
in
For point C
slug
ft
3
 32.2
ft
2
2
 0.5 ft 
s
pC  pair2  SGHg ρH2O g h4
pC  0.416
lbf
2
 1.75  1.94
in
slug
ft
3
 32.2
ft
2
lbf
2
in
 1.75  1.94
ft
3
 32.2
pB  0.416 psi
h4  10 in
 0.833 ft 
 ft 

slug ft  12 in 
2
s
slug
2

where
For the second air cavity pair2  pC  SGHg ρH2O h5
pair2  1.048
 ft 

slug ft  12 in 
lbf  s
h3  0.5 ft
ft
2
s
lbf  s
h5  6 in
 0.5 ft 
 ft 

slug ft  12 in 
2
lbf  s
h4  0.833 ft
2

where

(gage)
pC  1.048 psi
(gage)
h5  0.5 ft
2
pair2  0.668 psi
(gage)
Problem 3.50
Given:
Geometry of gate
Find:
Force FA for equilibrium
[Difficulty: 3]
h
H = 25 ft
FA
A
y
R = 10 ft
y
B
x
z
Solution:
⌠
⎮
F R = ⎮ p dA
⌡
Basic equation
or, use computing equations
dp
= ρ⋅ g
dh
ΣMz = 0
FR = pc⋅ A
Ixx
y' = yc +
A ⋅ yc
where y would be measured
from the free surface
Assumptions: static fluid; ρ = constant; patm on other side; door is in equilibrium
Instead of using either of these approaches, we note the following, using y as in the sketch
ΣMz = 0
FA =
FA ⋅ R =
1 ⌠
⎮
⋅
y⋅ ρ ⋅ g⋅ h dA
R ⎮
⌡
⌠
⎮
y⋅ p d A
⎮
⌡
with
with
dA = r⋅ dr⋅ dθ
p = ρ ⋅ g⋅ h
and
(Gage pressure, since p =
patm on other side)
y = r⋅ sin ( θ)
h = H−y
π
Hence
⌠
π R
3
4
1 ⌠ ⌠
ρ⋅ g ⎮ ⎛ H ⋅ R
R
2⎞
FA = ⋅ ⎮ ⎮ ρ⋅ g⋅ r⋅ sin ( θ) ⋅ ( H − r⋅ sin ( θ) ) ⋅ r dr dθ =
⋅⎮ ⎜
⋅ sin ( θ) −
⋅ sin ( θ) ⎟ dθ
R ⌡0 ⌡0
R
3
4
⎠
⌡0 ⎝
FR =
Using given data
3
4
⎛ 2⋅ H ⋅ R 2 π ⋅ R 3 ⎞
ρ ⋅ g ⎛ 2⋅ H ⋅ R
π⋅ R ⎞
⎟ = ρ ⋅ g⋅ ⎜
⎟
⋅⎜
−
−
R ⎝ 3
8 ⎠
8 ⎠
⎝ 3
FR = 1.94⋅
slug
ft
3
× 32.2⋅
2
2 π
3⎤ lbf ⋅ s
⎡2
× ⎢ × 25⋅ ft × ( 10⋅ ft) − × ( 10⋅ ft) ⎥ ×
2 ⎣3
8
⎦ slug⋅ ft
s
ft
4
FR = 7.96 × 10 ⋅ lbf
Problem 3.51
Given:
Geometry of access port
Find:
Resultant force and location
[Difficulty: 2]
y’
y
a = 1.25 ft
dy
w
FR
SG = 2.5
b = 1 ft
Solution:
Basic equation
FR =
⌠
⎮
p dA
⎮
⌡
dp
dy
⌠
⌠
⎮
⎮
ΣMs = y'⋅ FR =
y dF R =
y⋅ p dA
⎮
⎮
⌡
⌡
= ρ⋅ g
Ixx
y' = yc +
A ⋅ yc
FR = pc⋅ A
or, use computing equations
We will show both methods
Assumptions:
Static fluid; ρ = constant; patm on other side
FR =
⌠
⌠
⎮
⎮
p dA =
SG⋅ ρ⋅ g⋅ y dA
⎮
⎮
⌡
⌡
a
w
dA = w⋅ dy and
but
b
y
=
w =
a
b
a
⋅y
a
Hence
2
⌠
⌠
b
b 2
SG⋅ ρ⋅ g⋅ b⋅ a
FR = ⎮ SG⋅ ρ⋅ g⋅ y⋅ ⋅ y dy = ⎮ SG⋅ ρ⋅ g⋅ ⋅ y dy =
⎮
⎮
a
a
3
⌡0
⌡0
Alternatively
FR = pc⋅ A
Hence
FR =
For y'
3
⌠
⌠
b 3
SG⋅ ρ⋅ g⋅ b⋅ a
⎮
y'⋅ FR =
y⋅ p dA = ⎮ SG⋅ ρ⋅ g⋅ ⋅ y dy =
⎮
⎮
a
4
⌡
⌡0
Alternatively
Ixx
y' = yc +
A ⋅ yc
2
pc = SG⋅ ρ⋅ g⋅ yc = SG⋅ ρ⋅ g⋅ ⋅ a
3
and
A =
with
1
2
⋅ a⋅ b
2
SG⋅ ρ⋅ g⋅ b⋅ a
3
a
Using given data, and SG = 2.5 (Table A.1)
and
3
y' =
3
b⋅ a
Ixx =
36
and
2.5
FR =
y' =
3
3
4
⋅a
⋅ 1.94⋅
slug
ft
3
(Google it!)
× 32.2⋅
ft
2
y' =
SG⋅ ρ⋅ g⋅ b⋅ a
4⋅ F R
2
3
2
y' = 0.938⋅ ft
3
4
⋅a
3
⋅a +
× 1⋅ ft × ( 1.25⋅ ft) ×
s
=
b⋅ a
2 3
3
⋅
⋅
= ⋅a
36 a⋅ b 2⋅ a
4
2
lbf ⋅ s
slug⋅ ft
FR = 81.3⋅ lbf
Problem 3.52
Given:
Geometry of plane gate
Find:
Minimum weight to keep it closed
[Difficulty: 3]
L=3m
h
y
L/2
dF
W
w=2m
Solution:
FR =
Basic equation
⌠
⎮
p dA
⎮
⌡
dp
dh
= ρ⋅ g
ΣMO = 0
Ixx
y' = yc +
A ⋅ yc
F R = pc ⋅ A
or, use computing equations
Assumptions: static fluid; ρ = constant; patm on other side; door is in equilibrium
Instead of using either of these approaches, we note the following, using y as in the sketch
⌠
L
⎮
W⋅ ⋅ cos ( θ) =
y dF
⎮
2
⌡
ΣMO = 0
We also have
dF = p⋅ dA
Hence
W=
with
p = ρ⋅ g⋅ h = ρ⋅ g⋅ y⋅ sin ( θ)
(Gage pressure, since p = patm on other side)
⌠
⌠
2
⎮
⎮
y⋅ p dA =
⋅
y⋅ ρ⋅ g⋅ y⋅ sin ( θ) ⋅ w dy
⎮
L⋅ cos ( θ) ⌡
L⋅ cos ( θ) ⎮
⌡
2
⋅
L
⌠
2⋅ ρ⋅ g⋅ w⋅ tan( θ) ⌠ 2
2
2
⎮
W=
⋅
y⋅ p dA =
⋅ ⎮ y dy = ⋅ ρ⋅ g⋅ w⋅ L ⋅ tan( θ)
⎮
⌡
L⋅ cos ( θ) ⌡
L
3
0
2
Using given data
W =
2
3
⋅ 1000⋅
kg
3
m
× 9.81⋅
m
2
s
2
× 2⋅ m × ( 3⋅ m) × tan( 30⋅ deg) ×
2
N⋅ s
kg⋅ m
W = 68⋅ kN
Problem 3.53
[Difficulty: 4]
Given:
Semicylindrical trough, partly filled with water to depth d.
Find:
(a) General expressions for
FR and y' on end of trough, if open to the atmosphere.
(b) Plots of results vs. d/R between 0 and 1.
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dy
= ρ⋅ g
FR =
⌠
⎮
p dA
⎮
⌡
y'⋅ FR =
Assumptions:
(Hydrostatic Force on door)
⌠
⎮
y⋅ p dA
⎮
⌡
(First moment of force)
R–d
(1) Static fluid
(2) Incompressible fluid
Integrating the pressure equation:
Therefore:
(Hydrostatic Pressure - y is positive downwards)
p = ρ ⋅ g⋅ h
where h = y − ( R − d)
⎡ y − ⎛1 −
⎜
⎣R ⎝
p = ρ⋅ g⋅ [ y − ( R − d) ] = ρ⋅ g⋅ R⋅ ⎢
Expressing this in terms of θ and α in the figure:
For the walls at the end of the trough:
d ⎞⎤
⎟⎥
R ⎠⎦
θ
y
h
d
dy
α
p = ρ⋅ g⋅ R⋅ ( cos ( θ) − cos ( α) )
dA = w⋅ dy = 2⋅ R⋅ sin ( θ) ⋅ dy Now since y = R⋅ cos ( θ) it follows that dy = −R⋅ sin ( θ) ⋅ dθ
Substituting this into the hydrostatic force equation:
R
0
⌠
⌠
FR = ⎮
p⋅ w dy = ⎮ ρ⋅ g⋅ R⋅ ( cos ( θ) − cos ( α) ) ⋅ 2⋅ R⋅ sin ( θ) ⋅ ( −R⋅ sin ( θ) ) dθ
⌡R−d
⌡α
Upon simplification:
3⌠
2
2
3 ⎡ ( sin ( α) )
⎛ α sin ( α) ⋅ cos ( α) ⎞⎥⎤
FR = 2⋅ ρ⋅ g⋅ R ⎮ ⎡⎣sin ( θ) ⋅ cos ( θ) − ( sin ( θ) ) ⋅ cos ( α)⎤⎦ dθ = 2⋅ ρ⋅ g⋅ R ⋅ ⎢
− cos ( α) ⋅ ⎜ −
⎟
⌡0
2
⎣ 3
⎝2
⎠⎦
α
3
3 ⎡ ( sin ( α) )
FR = 2⋅ ρ⋅ g⋅ R ⋅ ⎢
⎣
Non-dimensionalizing the force:
⎡ ( sin ( α) ) 3
FR
ρ⋅ g⋅ R
3
3
= 2⋅ ⎢
⎣
3
⎛ α − sin ( α) ⋅ cos ( α) ⎞⎥⎤
⎟
2
⎝2
⎠⎦
3
− cos ( α) ⋅ ⎜
⎛ α − sin ( α) ⋅ cos ( α) ⎞⎥⎤
⎟
2
⎝2
⎠⎦
− cos ( α) ⋅ ⎜
To find the line of action of the force:
R
0
⌠
⌠
y'⋅ FR = ⎮
y⋅ p⋅ w dy = ⎮ R⋅ cos ( θ) ⋅ ρ⋅ g⋅ R⋅ ( cos ( θ) − cos ( α) ) ⋅ 2⋅ R⋅ sin ( θ) ⋅ ( −R⋅ sin ( θ) ) dθ
⌡R−d
⌡α
Upon simplification:
3
( sin ( α) ) ⎤
sin ( 4⋅ α) ⎞
4 ⎡1 ⎛
2
2
2
⎥
y'⋅ FR = 2⋅ ρ⋅ g⋅ R ⋅ ⎮ ⎡⎣( sin ( θ) ) ⋅ ( cos ( θ) ) − cos ( α) ⋅ ( sin ( θ) ) ⋅ cos ( θ)⎤⎦ dθ = 2⋅ ρ⋅ g⋅ R ⋅ ⎢ ⋅ ⎜ α −
− cos ( α) ⋅
⎟
⌡0
4
3
⎣8 ⎝
⎠
⎦
4⌠
α
3
( sin ( α) ) ⎤
sin ( 4⋅ α) ⎞
4 ⎡1 ⎛
⎥
y'⋅ FR = 2⋅ ρ⋅ g⋅ R ⋅ ⎢ ⋅ ⎜ α −
⎟ − cos ( α) ⋅
4
3
⎣8 ⎝
⎠
⎦
and therefore
y' =
Simplifying the expression:
y'
R
=
y'⋅ FR
FR
or
y'
R
=
y'⋅ FR
R ⋅ FR
sin ( 4⋅ α) ⎞
( sin ( α) )
1 ⎛
⋅ ⎜α −
⎟ − cos ( α) ⋅
4
8 ⎝
3
⎠
( sin ( α) )
3
3
⎛ α − sin ( α) ⋅ cos ( α) ⎞
⎟
2
⎝2
⎠
− cos ( α) ⋅ ⎜
Plots of the non-dimensionalized force and the line of
action of the force are shown in the plots below:
Non-dimensional Force
0.8
0.6
0.4
0.2
0
0
0.5
1
d/R
1
0.8
y'/R
0.6
0.4
0.2
0
0
0.5
d/R
3
1
Problem 3.54
Given:
Gate geometry
Find:
Depth H at which gate tips
[Difficulty: 3]
Solution:
This is a problem with atmospheric pressure on both sides of the plate, so we can first determine the location of the
center of pressure with respect to the free surface, using Eq.3.11c (assuming depth H)
3
Ixx
y' = yc +
A ⋅ yc
w⋅ L
Ixx =
and
with
12
yc = H −
L
2
where L = 1 m is the plate height and w is the plate width
Hence
⎛
⎝
3
L⎞
y' = ⎜ H −
w⋅ L
⎟+
2⎠
⎛
12⋅ w⋅ L⋅ ⎜ H −
⎝
L⎞
⎟
⎛
⎝
= ⎜H −
2⎠
2
L⎞
L
⎟+
2⎠
⎛
12⋅ ⎜ H −
⎝
L⎞
⎟
2⎠
But for equilibrium, the center of force must always be at or below the level of the hinge so that the stop can hold the gate in
place. Hence we must have
y' > H − 0.45⋅ m
L
Combining the two equations ⎛⎜ H − ⎞⎟ +
2⎠
⎝
Solving for H
H ≤
L
2
+
L
2
⎛
⎝
12⋅ ⎜ H −
L
L⎞
≥ H − 0.45⋅ m
⎟
2⎠
2
⎛L
⎞
12⋅ ⎜ − 0.45⋅ m⎟
⎝2
⎠
H ≤
1⋅ m
2
+
( 1⋅ m)
2
⎛ 1⋅ m − 0.45⋅ m⎞
12 × ⎜
⎟
⎝ 2
⎠
H ≤ 2.17⋅ m
Problem 3.55
Given:
Geometry of cup
Find:
Force on each half of cup
Assumptions:
[Difficulty: 1]
(1) Tea is static and incompressible
(2) Atmospheric pressure on outside of cup
Solution:
⌠
⎮
p dA
⎮
⌡
Basic equation
FR =
or, use computing equation
FR = pc⋅ A
dp
dh
= ρ⋅ g
The force on the half-cup is the same as that on a rectangle of size
FR =
⌠
⌠
⎮
⎮
p dA =
ρ⋅ g⋅ y dA
⎮
⎮
⌡
⌡
h
Hence
⌠
ρ⋅ g⋅ w⋅ h
FR = ⎮ ρ⋅ g⋅ y⋅ w dy =
⌡0
2
Alternatively
FR = pc⋅ A
Using given data
FR =
1
2
× 999⋅
3
m
× 9.81⋅
w = 6.5⋅ cm
and
dA = w⋅ dy
but
2
h
ρ⋅ g⋅ w⋅ h
FR = pc ⋅ A = ρ⋅ g⋅ yc⋅ A = ρ⋅ g⋅ ⋅ h ⋅ w =
2
2
and
kg
h = 8⋅ cm
m
2
s
2
3
2
⎛ m ⎞ × N⋅ s
⎟
⎝ 100⋅ cm ⎠ kg⋅ m
× 6.5⋅ cm × ( 8⋅ cm) × ⎜
Hence a teacup is being forced apart by about 2 N: not much of a force, so a paper cup works!
2
FR = 2.04⋅ N
Problem 3.56
Given:
Geometry of lock system
Find:
Force on gate; reactions at hinge
[Difficulty: 3]
Ry
Rx
Solution:
Basic equation
or, use computing equation
FR =
⌠
⎮
p dA
⎮
⌡
dp
= ρ⋅ g
dh
FR
FR = pc⋅ A
Assumptions: static fluid; ρ = constant; patm on other side
The force on each gate is the same as that on a rectangle of size
h = D = 10⋅ m
w =
and
W
2⋅ cos ( 15⋅ deg)
⌠
⌠
⎮
⎮
FR = ⎮ p dA = ⎮ ρ⋅ g⋅ y dA
⌡
⌡
h
but
Hence
⌠
ρ ⋅ g⋅ w ⋅ h
F R = ⎮ ρ ⋅ g⋅ y⋅ w dy =
⌡0
2
Alternatively
FR = pc⋅ A
Using given data
FR =
1
2
⋅ 1000⋅
3
× 9.81⋅
m
dA = w⋅ dy
2
h
ρ ⋅ g⋅ w ⋅ h
FR = pc⋅ A = ρ⋅ g⋅ yc⋅ A = ρ⋅ g⋅ ⋅ h⋅ w =
2
2
and
kg
Fn
m
2
s
×
34⋅ m
2⋅ cos ( 15⋅ deg)
2
2
2
× ( 10⋅ m) ×
N⋅ s
kg⋅ m
FR = 8.63⋅ MN
For the force components Rx and Ry we do the following
FR
w
ΣMhinge = 0 = FR⋅ − Fn⋅ w⋅ sin ( 15⋅ deg)
2
Fn =
ΣFx = 0 = FR⋅ cos ( 15⋅ deg) − Rx = 0
Rx = FR⋅ cos ( 15⋅ deg)
Rx = 8.34⋅ MN
ΣFy = 0 = −Ry − FR⋅ sin ( 15⋅ deg) + Fn = 0
Ry = Fn − FR⋅ sin ( 15⋅ deg)
Ry = 14.4⋅ MN
R = ( 8.34⋅ MN , 14.4⋅ MN)
R = 16.7⋅ MN
2⋅ sin ( 15⋅ deg)
Fn = 16.7⋅ MN
Problem 3.57
[Difficulty: 2]
Given:
Liquid concrete poured between vertical forms as shown
t = 0.25⋅ m H = 3⋅ m W = 5⋅ m SGc = 2.5 (From Table A.1, App. A)
Find:
(a) Resultant force on form
(b) Line of application
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dy
= ρ⋅ g
(Hydrostatic Pressure - y is positive downwards)
FR = pc⋅ A
(Hydrostatic Force)
Ixx
y' = yc +
A ⋅ yc
(Location of line of action)
Ixy
x' = xc +
A ⋅ yc
Assumptions:
For a rectangular plate:
Liquid Concrete
(1) Static fluid
(2) Incompressible fluid
(3) Atmospheric pressure acts at free surface
and on the outside of the form.
W⋅ H
Ixx =
t = 0.25 m
3
12
H=3m
y’
x’
Ixy = 0
xc = 2.5⋅ m yc = 1.5⋅ m
Integrating the hydrostatic pressure equation:
The density of concrete is:
ρ = 2.5 × 1000⋅
kg
p = ρ ⋅ g⋅ y
ρ = 2.5 × 10
3
m
W=5m
FR
3 kg
3
m
Therefore, the force is: FR = ρ⋅ g⋅ yc⋅ H⋅ W
Substituting in values gives us:
m
3 kg
× 9.81⋅ × 1.5⋅ m × 3⋅ m × 5⋅ m
3
2
FR = 2.5 × 10 ⋅
m
FR = 552⋅ kN
s
To find the line of action of the resultant force:
3
2
W⋅ H
H
y' = yc +
= yc +
12⋅ W⋅ H⋅ yc
12⋅ yc
Since Ixy = 0
it follows that x' = xc
y' = 1.5⋅ m +
( 3⋅ m)
2
12⋅ 1.5⋅ m
y' = 2.00 m
x' = 2.50⋅ m
Problem 3.58
Given:
[Difficulty: 4]
Window, in shape of isosceles triangle and hinged at the top is located in
the vertical wall of a form that contains concrete.
a  0.4 m b  0.3 m c  0.25 m SGc  2.5 (From Table A.1, App. A)
Find:
The minimum force applied at D needed to keep the window closed.
Plot the results over the range of concrete depth between 0 and a.
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dh
 ρ g
FR 
Assumptions:


p dA


(Hydrostatic Force on door)


y' FR   y p dA

(First moment of force)
ΣM  0
(Rotational equilibrium)
b
(1) Static fluid
(2) Incompressible fluid
(3) Atmospheric pressure acts at free surface and on the
outside of the window.
Integrating the pressure equation yields: p  ρ g ( h  d)
p0
Summing moments around the hinge:
h
d
a
w
for h > d
for h < d
d  ac
where
FD 
(Hydrostatic Pressure - h is positive downwards)
 FD  a 
d  0.15 m
dA
D


h p dA  0


a
a
1 
1 
ρ g 

  h p dA    h ρ  g ( h  d )  w dh 
  h ( h  d)  w dh
a 
a d
a d
From the law of similar triangles:
w
b

ah
a
h
Therefore: w 
b
a
a
( a  h)
dF = pdA
FD
a
a
ρ g 
b
ρ  g b   3
2
  h  ( a  d)  h  a d h dh
FD 

 h ( h  d)  ( a  h) dh 
2 
a  a
d
a
d
Into the expression for the force at D:
Evaluating this integral we get:
FD 






4
4
3
3
2
2
ρ  g b  a  d
( a  d)  a  d
a d a  d 

 


2 
4
3
2

a
4
2 1 
 d  1 
FD  ρ g b a    1        1 
 4  a  3 
The density of the concrete is:
and after collecting terms:
3
2
d    d   1 d   d  




1





1

 
  
a    a   2 a   a  
ρ  2.5  1000
kg
3
ρ  2.5  10
m
3 kg
3
m
( 1)
d
a

0.15
0.4
 0.375
Substituting in values for the force at D:
2
m
1
0.375 
3 kg
2 1
4
3
2  N s
 9.81  0.3 m ( 0.4 m)    1  ( 0.375)    ( 1  0.375)  1  ( 0.375)  
 1  ( 0.375)  
3
2
3
2
 4
 kg m
FD  2.5  10 
m
s
To plot the results for different values of c/a, we use Eq. (1) and remember that
Therefore, it follows that
d
a
 1
c
d  ac
FD  32.9 N
In addition, we can maximize the force by the maximum force
a
(when c = a or d = 0):
2
2
 1  1   ρ  g b a

12
 4 3
Fmax  ρ g b a   
and so
 1   d  4 1  d    d  3 1 d   d  2
 12   1        1    1        1    
Fmax
 4   a   3  a    a   2 a   a  
FD
1.0
Force Ratio (FD/Fmax)
0.8
0.6
0.4
0.2
0.0
0.0
0.5
Concrete Depth Ratio (c/a)
1.0
Problem 3.59
[Difficulty: 2]
Given:
Door as shown; Data from Example 3.6.
Find:
Force to keep door shut using the two seperate pressures method.
Solution:
We will apply the computing equations to this system.
F R  pc  A
Governing Equations:
p0
3
Ixx
y'  yc 
yc A
bL
Ixx 
12
p0
h1 ’
h2 ’
F1
F2
F 1  p0 A
F1  100
lbf
ft
2
 3 ft  2 ft
F2  pc A  ρ g hc L b  γ hc L b
F1  600 lbf
F2  100
lbf
ft
For the rectangular door
1
3
Ixx 
 b L
12
2
Ixx
1 L
h'2  hc 
 hc 

b L  hc
12 hc
x'  1 ft
z'  1.5 ft
 1.5 ft  3 ft  2 ft
3
F2  900 lbf
2
h'2  1.5 m 
1 ( 3 m)

12 1.5 m
h'2  2 m
The free body diagram of the door is then
h1’
h2’
F1


ΣMAx  0  L Ft  F1 L  h'1  F2 L  h'2
L
 h'1 
 h'2 
Ft  F1  1 
  F2  1  
L 
L 


F2
Az

Ft
Ay


Ft  600 lbf   1 
1.5 
 2
  900 lbf   1  
3 
 3

Ft  600 lbf
Problem 3.60
Given:
[Difficulty: 2]
γ = 62.4⋅
Plug is used to seal a conduit.
lbf
ft
3
Find:
Magnitude, direction and location of the force of water on the plug.
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dh
Assumptions:
=γ
(Hydrostatic Pressure - y is positive downwards)
F R = pc ⋅ A
(Hydrostatic Force)
Ixx
y' = yc +
A ⋅ yc
(Location of line of action)
(1) Static fluid
(2) Incompressible fluid
(3) Atmospheric pressure acts on the outside of the plug.
Integrating the hydrostatic pressure equation:
π 2
FR = pc⋅ A = γ⋅ hc⋅ ⋅ D
4
p = γ⋅ h
FR = 62.4⋅
lbf
ft
π
For a circular area:
⋅D
4
3
× 12⋅ ft ×
π
4
× ( 6⋅ ft)
2
2
π 4
64
D
Ixx =
⋅ D Therefore: y' = yc +
= yc +
64
π 2
16⋅ yc
⋅ D ⋅ yc
4
y' = 12⋅ ft +
4
FR = 2.12 × 10 ⋅ lbf
( 6⋅ ft)
2
16 × 12⋅ ft
y' = 12.19⋅ ft
The force of water is to the right and
perpendicular to the plug.
Problem 3.61
Given:
Description of car tire
Find:
Explanation of lift effect
[Difficulty: 1]
Solution:
The explanation is as follows: It is true that the pressure in the entire tire is the same everywhere. However, the tire at the top of the hub
will be essentially circular in cross-section, but at the bottom, where the tire meets the ground, the cross section will be approximately a
flattened circle, or elliptical. Hence we can explain that the lower cross section has greater upward force than the upper cross section has
downward force (providing enough lift to keep the car up) two ways. First, the horizontal projected area of the lower ellipse is larger than
that of the upper circular cross section, so that net pressure times area is upwards. Second, any time you have an elliptical cross section
that's at high pressure, that pressure will always try to force the ellipse to be circular (thing of a round inflated balloon - if you squeeze it it
will resist!). This analysis ignores the stiffness of the tire rubber, which also provides a little lift.
Problem 3.62
Given:
[Difficulty: 2]
Circular access port of known diameter in side of water standpipe of
known diameter. Port is held in place by eight bolts evenly spaced
around the circumference of the port.
Center of the port is located at a know distance below the free surface of
the water.
d  0.6 m D  7 m L  12 m
Find:
(a) Total force on the port
(b) Appropriate bolt diameter
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
 ρ g
dh
(Hydrostatic Pressure - y is positive downwards)
FR  pc A
F
σ
A
Assumptions:
(Hydrostatic Force)
h
(Normal Stress in bolt)
L
(1) Static fluid
(2) Incompressible fluid
(3) Force is distributed evenly over all bolts
(4) Appropriate working stress in bolts is 100 MPa
(5) Atmospheric pressure acts at free surface of water and on
outside of port.
D
p  ρ  g h
Integrating the hydrostatic pressure equation:
The resultant force on the port is:
d
π 2
FR  pc A  ρ g L  d
4
FR  999
kg
3
 9.81
m
m
2
 12 m 
π
s
4
2
 ( 0.6 m) 
2
N s
kg m
FR  33.3 kN
To find the bolt diameter we consider:
2
Therefore: 2 π db 
FR
σ
σ
FR
A
where A is the area of all of the bolts:
Solving for the bolt diameter we get:
 1
3
db  
 33.3  10  N 
2 π

 FR 

 2 π σ 
A  8
π
4
2
 db  2 π  db
2
1
2
db  
1
100  10

2
m 
6 N


1
2
3

10  mm
m
db  7.28 mm
Problem 3.63
Given:
Geometry of rectangular gate
Find:
Depth for gate to open
[Difficulty: 3]
Solution:
Basic equation
Computing equations
L
dp
dh
 ρ g
FR  pc A
y’
D
ΣMz  0
Ixx
y'  yc 
A  yc
Ixx 
b D
3
F1
12
F2
Assumptions: Static fluid; ρ = constant; patm on other side; no friction in hinge
For incompressible fluid
p  ρ  g h
where p is gage pressure and h is measured downwards
The force on the vertical gate (gate 1) is the same as that on a rectangle of size h = D and width w
Hence
D
ρ  g w  D
F1  pc A  ρ g yc A  ρ g  D w 
2
2
The location of this force is
3
Ixx
D w D
1
2
2
y'  yc 




 D
12
w D D
A  yc
2
3
2
The force on the horizontal gate (gate 2) is due to constant pressure, and is at the centroid
F 2  p ( y  D )  A  ρ  g D  w  L
Summing moments about the hinge
L
2 
L

ΣMhinge  0  F1 ( D  y')  F2  F1  D   D  F2
2
3 
2

F 1
D
3

ρ  g w  D
2
2

D
L
L
 F2  ρ g D w L
3
2
2
3
ρ  g w  D
ρ  g D  w  L

6
2
D 
3 L 
D  8.66 ft
3  5ft
2
Problem 3.64
Given:
[Difficulty: 3]
Gate AOC, hinged along O, has known width;
Weight of gate may be neglected. Gate is sealed at C.
b = 6⋅ ft
Find:
Force in bar AB
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dh
Assumptions:
= ρ⋅ g
(Hydrostatic Pressure - h is positive downwards)
FR = pc⋅ A
(Hydrostatic Force)
Ixx
y' = yc +
A ⋅ yc
(Location of line of action)
ΣMz = 0
(Rotational equilibrium)
(1) Static fluid
(2) Incompressible fluid
(3) Atmospheric pressure acts at free surface of water and on
outside of gate
(4) No resisting moment in hinge at O
(5) No vertical resisting force at C
FAB
L1
h1 ’
L1
p = ρ ⋅ g⋅ h
Integrating the hydrostatic pressure equation:
F1
L2
The free body diagram of the gate is shown here:
F1 is the resultant of the distributed force on AO
F2 is the resultant of the distributed force on OC
x2’
FAB is the force of the bar
Cx is the sealing force at C
First find the force on AO:
F1 = 1.94⋅
slug
ft
3
× 32.2⋅
ft
2
s
F1 = pc ⋅ A1 = ρ⋅ g⋅ hc1⋅ b ⋅ L1
2
× 6⋅ ft × 6⋅ ft × 12⋅ ft ×
lbf⋅ s
slugft
⋅
F1 = 27.0⋅ kip
F2
b⋅ L 1
Ixx
3
L1
2
h'1 = hc1 +
= hc1 +
= hc1 +
A⋅ hc1
12⋅ b⋅ L1⋅ hc1
12⋅ hc1
Next find the force on OC:
F2 = 1.94⋅
slug
ft
3
× 32.2⋅
ft
2
h'1 = 6⋅ ft +
( 12⋅ ft)
2
12 × 6⋅ ft
h'1 = 8⋅ ft
2
× 12⋅ ft × 6⋅ ft × 6⋅ ft ×
s
lbf ⋅ s
slug⋅ ft
Since the pressure is uniform over OC, the force acts at the centroid of OC, i.e.,
(
)
(
F2 = 27.0⋅ kip
)
FAB⋅ L1 + L3 − F1⋅ L1 − h'1 + F2⋅ x'2 = 0
Summing moments about the hinge gives:
FAB
x'2 = 3⋅ ft
L1
h1 ’
L1
Solving for the force in the bar:
Substituting in values:
FAB =
FAB =
1
12⋅ ft + 3⋅ ft
FAB = 1800⋅ lbf
(
)
F1⋅ L1 − h'1 − F2⋅ x'2
F1
L2
L1 + L3
3
3
⋅ ⎡⎣27.0 × 10 ⋅ lbf × ( 12⋅ ft − 8⋅ ft) − 27.0 × 10 ⋅ lbf × 3⋅ ft⎤⎦
Thus bar AB is in compression
x2’
F2
Problem 3.65
Given:
[Difficulty: 3]
Gate shown with fixed width, bass of gate is negligible.
Gate is in equilibrium.
b = 3⋅ m
Find:
Water depth, d
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dh
Assumptions:
= ρ⋅ g
(Hydrostatic Pressure - h is positive downwards)
FR = pc⋅ A
(Hydrostatic Force)
Ixx
y' = yc +
A ⋅ yc
(Location of line of action)
ΣMz = 0
(Rotational equilibrium)
(1) Static fluid
(2) Incompressible fluid
(3) Atmospheric pressure acts at free surface of water and on
outside of gate
M
h
p = ρ ⋅ g⋅ h
Integrating the hydrostatic pressure equation:
FR = pc⋅ A = ρ⋅ g⋅ hc⋅ A
hc =
d
A = b⋅
2
d
y
d
l
sin ( θ)
L
2
ρ ⋅ g⋅ b⋅ d
Therefore, FR =
2⋅ sin ( θ)
To find the line of application of this force:
θ
3
Ixx
y' = yc +
A ⋅ yc
Since
3
b⋅ l
and
Ixx =
12
A = b⋅ l it follows that
2
b⋅ l
l
y' = yc +
= yc +
12⋅ b⋅ l⋅ yc
12⋅ yc
where l is the length of the gate in contact with the water (as seen in diagram)
l=
l and d are related through:
d
l
d
Therefore, yc =
and
=
sin ( θ)
2
2⋅ sin ( θ)
y' =
d
2⋅ sin ( θ)
d
+
2
( sin ( θ) )
2
⋅
2⋅ sin ( θ)
12⋅ d
The free body diagram of the gate is shown here:
=
2⋅ d
3⋅ sin ( θ)
T
y
d
FR
y’
Summing moments about the hinge gives:
T⋅ L − ( l − y') ⋅ FR = 0
Solving for l:
l=
d
sin ( θ)
=
M⋅ g⋅ L
FR
where T = M⋅ g
+ y'
⎛ 2⋅ M⋅ g⋅ L ⋅ sin ( θ) + 2⋅ d ⎞ ⋅ sin ( θ) or
d=⎜
2
3⋅ sin ( θ) ⎟
⎝ ρ ⋅ g⋅ b⋅ d
⎠
Solving for d:
⎡ 6⋅ M⋅ L ⋅ ( sin ( θ) ) 2⎤
⎥
⎣ ρ⋅ b
⎦
d=⎢
θ
So upon further substitution we get:
d
3
=
2⋅ M⋅ L⋅ ( sin ( θ) )
ρ ⋅ b⋅ d
Ahoriz
A vertical
2
2
1
3
Substituting in values:
⎡
d = ⎢6 × 2500⋅ kg × 5⋅ m ×
⎣
d = 2.66 m
1
3
⋅
m
999 kg
×
1
3m
2⎤
× ( sin ( 60⋅ deg) ) ⎥
⎦
1
3
Problem 3.66
Given:
Geometry of gate
Find:
Force at A to hold gate closed
[Difficulty: 3]
y
h
Solution:
Basic equation
Computing equations
D
y’
dp
dh
 ρ g
ΣMz  0
FR  pc A
FR
Ixx
y'  yc 
A  yc
Ixx 
w L
FA
3
12
Assumptions: Static fluid; ρ = constant; patm on other side; no friction in hinge
For incompressible fluid
p  ρ  g h
where p is gage pressure and h is measured downwards
The hydrostatic force on the gate is that on a rectangle of size L and width w.
Hence
L


FR  pc A  ρ g hc A  ρ g  D   sin ( 30 deg)  L w
2


FR  1000
kg
3
 9.81
m
m
2
s


  1.5 
3
2
2


sin ( 30 deg)  m  3 m  3 m 
N s
kg m
FR  199 kN
Ixx
where y' and y are measured along the plane of the gate to the free surface
The location of this force is given by y'  yc 
A  yc
c
yc 
D
sin ( 30 deg)

L
2
yc 
1.5 m
sin ( 30 deg)

3 m
2
yc  4.5 m
3
2
2
Ixx
w L
1 1
L
( 3 m)
y'  yc 
 yc 


 yc 
 4.5 m 
A  yc
12 w L yc
12 yc
12 4.5 m
Taking moments about the hinge
y'  4.67 m
D

  F L
ΣMH  0  FR  y' 
 A
sin
(
30

deg
)


D
 y' 



sin ( 30 deg) 

FA  FR 
L
1.5
 4.67 



sin ( 30 deg) 

FA  199 kN
3
FA  111 kN
Problem 3.67
Given:
Block hinged and floating
Find:
SG of the wood
[Difficulty: 3]
Solution:
dp
Basic equation
dh
 ρ g
ΣMz  0
Ixx
y'  yc 
A  yc
FR  pc A
Computing equations
Assumptions: Static fluid; ρ = constant; patm on other side; no friction in hinge
p  ρ  g h
For incompressible fluid
where p is gage pressure and h is measured downwards
The force on the vertical section is the same as that on a rectangle of height d and width L
Hence
d
ρ  g L  d
F1  pc A  ρ g yc A  ρ g  d L 
2
2
2
Mg
y’
y
The location of this force is
F1
3
Ixx
d L d
1
2
2
y'  yc 
 

  d
A  yc
2
12
L d d
3
x
F2
The force on the horizontal section is due to constant pressure, and is at the centroid
F2  p ( y  d)  A  ρ g d L L
Summing moments about the hinge
L
L
ΣMhinge  0  F1 ( d  y')  F2  M g
2
2
Hence
F 1  d 


SG ρ g L
2
4

ρ g L d
2
2

d
3
2 L
 ρ  g d L 
2
2 
L
3 L
 d  F2  SG ρ L  g
3 
2
2
3
SG 
1  d
d
  
3  L
L
3
SG 
1  0.5 
0.5

 
3  1 
1
SG  0.542
Problem 3.68
[Difficulty: 4]
Given:
Various dam cross-sections
Find:
Which requires the least concrete; plot cross-section area A as a function of α
Solution:
For each case, the dam width b has to be large enough so that the weight of the dam exerts enough moment to balance the
moment due to fluid hydrostatic force(s). By doing a moment balance this value of b can be found
a) Rectangular dam
Straightforward application of the computing equations of Section 3-5 yields
D
1
2
FH = p c ⋅ A = ρ⋅ g ⋅ ⋅ w⋅ D = ⋅ ρ⋅ g ⋅ D ⋅ w
2
2
D
3
Ixx
D
2
w⋅ D
y' = y c +
=
+
= ⋅D
A⋅ y c
2
3
D
12⋅ w⋅ D⋅
2
so
Also
y = D − y' =
FH
y
mg
O
D
b
3
m = ρcement⋅ g ⋅ b ⋅ D⋅ w = SG ⋅ ρ⋅ g ⋅ b ⋅ D⋅ w
Taking moments about O
∑ M0. = 0 = −FH⋅y + 2 ⋅m⋅g
so
⎛ 1 ⋅ ρ⋅ g⋅ D2⋅ w⎞ ⋅ D = b ⋅ ( SG⋅ ρ⋅ g ⋅ b⋅ D⋅ w)
⎜2
⎝
⎠ 3 2
Solving for b
b=
The minimum rectangular cross-section area is
A = b⋅ D =
For concrete, from Table A.1, SG = 2.4, so
A=
b
D
3 ⋅ SG
2
D
3 ⋅ SG
2
D
3 ⋅ SG
2
=
D
3 × 2.4
2
A = 0.373 ⋅ D
b) Triangular dams
FV
D
Instead of analysing right-triangles, a general analysis is made, at the end of
which right triangles are analysed as special cases by setting α = 0 or 1.
x
FH
Straightforward application of the computing equations of Section 3-5 yields
y
m 1g
m 2g
O
D
1
2
FH = p c⋅ A = ρ⋅ g ⋅ ⋅ w⋅ D = ⋅ ρ⋅ g ⋅ D ⋅ w
2
2
αb
b
3
Ixx
w⋅ D
D
2
y' = y c +
=
+
= ⋅D
A⋅ y c
D
2
3
12⋅ w⋅ D⋅
2
so
Also
y = D − y' =
D
3
FV = ρ⋅ V⋅ g = ρ⋅ g ⋅
α⋅ b ⋅ D
2
⋅w =
1
2
⋅ ρ⋅ g ⋅ α⋅ b ⋅ D⋅ w
x = ( b − α⋅ b ) +
2
3
⋅ α⋅ b = b ⋅ ⎛⎜ 1 −
α⎞
⎝
3⎠
For the two triangular masses
1
m1 = ⋅ SG ⋅ ρ⋅ g ⋅ α⋅ b ⋅ D⋅ w
2
x 1 = ( b − α⋅ b ) +
1
m2 = ⋅ SG ⋅ ρ⋅ g ⋅ ( 1 − α) ⋅ b ⋅ D⋅ w
2
x2 =
2
3
1
3
⋅ α⋅ b = b ⋅ ⎛⎜ 1 −
⎝
2⋅ α ⎞
⋅ b ( 1 − α)
Taking moments about O
∑ M0. = 0 = −FH⋅y + FV⋅x + m1⋅g⋅x1 + m2⋅g⋅x2
so
Solving for b
1
1
D
α
2
−⎛⎜ ⋅ ρ⋅ g ⋅ D ⋅ w⎞ ⋅ + ⎛⎜ ⋅ ρ⋅ g ⋅ α⋅ b ⋅ D⋅ w⎞ ⋅ b ⋅ ⎛⎜ 1 − ⎞ ...
3⎠
⎝2
⎠ 3 ⎝2
⎠ ⎝
2⋅ α ⎞ ⎡ 1
2
1
⎞
⎛
⎛
+ ⎜ ⋅ SG ⋅ ρ⋅ g ⋅ α⋅ b ⋅ D⋅ w ⋅ b ⋅ ⎜ 1 −
+ ⎢ ⋅ SG ⋅ ρ⋅ g ⋅ ( 1 − α) ⋅ b ⋅ D⋅ w⎥⎤ ⋅ ⋅ b ( 1 − α)
3
2
2
⎝
⎠ ⎝
⎠ ⎣
⎦ 3
b=
D
(3⋅α − α2) + SG⋅(2 − α)
For a right triangle with the hypotenuse in contact with the water, α = 1 ,
b=
The cross-section area is
=0
D
3 − 1 + SG
=
D
b = 0.477 ⋅ D
3 − 1 + 2.4
A=
b⋅ D
2
and
2
= 0.238 ⋅ D
For a right triangle with the vertical in contact with the water, α = 0, and
2
A = 0.238 ⋅ D
3
⎠
b=
A=
The cross-section area is
A=
For a general triangle
D
2 ⋅ SG
b⋅ D
2
b⋅ D
2
D
=
b = 0.456 ⋅ D
2 ⋅ 2.4
2
2
= 0.228 ⋅ D
A = 0.228 ⋅ D
2
2
D
=
(3⋅α − α2) + SG⋅(2 − α)
2⋅
D
A=
2⋅
(3⋅α − α2) + 2.4⋅(2 − α)
2
D
A=
The final result is
2
2 ⋅ 4.8 + 0.6⋅ α − α
The dimensionless area, A /D 2, is plotted
A /D 2
0.2282
0.2270
0.2263
0.2261
0.2263
0.2270
0.2282
0.2299
0.2321
0.2349
0.2384
Dam Cross Section vs Coefficient
Dimensionless Area A /D 2
Alpha
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
0.240
0.238
0.236
0.234
0.232
0.230
0.228
0.226
Solver can be used to
find the minimum area
Alpha
0.300
0.224
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Coefficient
A /D 2
0.2261
From the Excel workbook, the minimum area occurs at α = 0.3
2
Amin =
D
2
A = 0.226 ⋅ D
2
2 ⋅ 4.8 + 0.6 × 0.3 − 0.3
The final results are that a triangular cross-section with α = 0.3 uses the least concrete; the next best is a right triangle with the
vertical in contact with the water; next is the right triangle with the hypotenuse in contact with the water; and the cross-section
requiring the most concrete is the rectangular cross-section.
1.0
Problem 3.69
Given:
Geometry of dam
Find:
Vertical force on dam
Assumption:
[Difficulty: 2]
Water is static and incompressible
Solution:
Basic equation:
For incompressible fluid
dp
dh
= ρ⋅ g
p = patm + ρ⋅ g⋅ h
where h is measured downwards from the free surface
The force on each horizontal section (depth d = 0.5 m and width w = 3 m) is
(
)
F = p⋅ A = patm + ρ⋅ g⋅ h ⋅ d⋅ w
Hence the total force is
(
) (
) (
) (
)
FT = ⎡patm + patm + ρ⋅ g⋅ h + patm + ρ⋅ g⋅ 2⋅ h + patm + ρ⋅ 3⋅ g⋅ h + patm + ρ⋅ g⋅ 4⋅ h ⎤ ⋅ d⋅ w
⎣
⎦
where we have used h as the height of the steps
(
FT = d⋅ w⋅ 5⋅ patm + 10⋅ ρ⋅ g⋅ h
⎛
)
2
kg
m
N⋅ s ⎟⎞
3 N
+ 10 × 999⋅
× 9.81⋅ × 0.5⋅ m ×
2
3
2
kg⋅ m ⎟
FT = 0.5⋅ m × 3⋅ m × ⎜ 5 × 101 × 10 ⋅
⎜
⎝
FT = 831⋅ kN
m
m
s
⎠
Problem 3.70
Given:
Geometry of dam
Find:
Vertical force on dam
Assumptions:
[Difficulty: 2]
(1) water is static and incompressible
(2) since we are asked for the force of the water, all pressures will be written as gage
Solution:
dp
Basic equation:
dh
 ρ g
p  ρ  g h
For incompressible fluid
where p is gage pressure and h is measured downwards from the free surface
The force on each horizontal section (depth d and width w) is
F  p  A  ρ  g h d w
(Note that d and w will change in terms of x and y for each section of the dam!)
Hence the total force is (allowing for the fact that some faces experience an upwards (negative) force)
F T  p A  Σ ρ  g h d w  ρ  g d Σ h w
Starting with the top and working downwards
FT  1.94
slug
ft
3
 32.2
ft
2
s
4
FT  2.70  10  lbf
2
 3 ft  [ ( 3 ft  12 ft)  ( 3 ft  6 ft)  ( 9 ft  6 ft)  ( 12 ft  12 ft) ] 
lbf  s
slug ft
The negative sign indicates a net upwards force (it's actually a buoyancy effect on the three middle sections)
Problem 3.71
Given:
[Difficulty: 3]
Parabolic gate, hinged at O has a constant width.
2
b  1.5 m a  1.0 m
D  1.2 m H  1.4 m
Find:
(a) Magnitude and moment of the vertical force on the gate due to water
(b) Horizontal force applied at A required to maintain equilibrium
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dh
 ρ g
(Hydrostatic Pressure - h is positive downwards)
ΣMz  0
Fv 
Assumptions:
(Rotational equilibrium)


p dA y


(Vertical Hydrostatic Force)
x' Fv 


x dF v


(Moment of vertical force)
y' FH 


y dF H


(Moment of Horizontal Hydrostatic Force)
y
(1) Static fluid
(2) Incompressible fluid
(3) Atmospheric pressure acts at free surface of water
and on outside of gate
Integrating the hydrostatic pressure equation:
FA
x’
FV
p  ρ  g h
FH
(a) The magnitude and moment of the vertical component of hydrostatic force:
Ox
y’
x
Oy


3
2


Fv 
p dA y 
ρ g h b dx where h  D  y x  a y dx  3 a y  dy




D
Substituting back into the relation for the force:
D


2
Fv   ρ g ( D  y)  b 3 a y dy  3 ρ g b a 
0
0
4
 D4 D4 
D
  ρ g b a
Evaluating the integral: Fv  3 ρ g b a 

4 
4
 3
H
D y2  y3 dy
y
Substituting values we calculate the force:
Fv  999
kg
 9.81
3
m
m
 1.5 m  1.0
2
s
1
2

( 1.2 m)
4
2

4
m
FA
x’
N s
FH
Fv  7.62 kN




x' Fv   x dFv   x p dAy


D
D

3
2
2 
x' Fv   a y  ρ g ( D  y)  b 3 a y dy  3 ρ g a  b 
0
0
 D7
2
x' Fv  3 ρ g a  b 
 6
kg
x'Fv  999
3
 9.81
m
7
D 
2
m
2
s
D y5  y6 dy
D
3
7
x
Using the derivation for the force:
Evaluating the integral:
7
   ρ g a2 b D7  ρ g a2 b
7 
42
14

y’
Oy
Ox
To find the associated moment:
H
FV
kg m
Now substituting values into this equation:
2
 1.0   1.5 m  ( 1.20 m)  N s
2
14
kg m
m 
x'Fv  3.76 kN m

(positive indicates
counterclockwise)
(b) Horizontal force at A to maintain equilibrium: we take moments at O:
x' FV  y' FH  H FA  0
Solving for the force at A:
FA 
1
H

 x' Fv  y' FH

To get the moment of the horizontal hydrostatic force:
y' FH 
D
D





2



y dF H 
y p dA x 
y ρ g h b dy  ρ g b  y ( D  y) dy  ρ g b  D y  y dy








0
0

D D 
D
  ρ g b
Evaluating the integral: y' FH  ρ g b 

3
6
2


3
y'FH  999
kg
3
 9.81
m
Therefore:
FA 
m
2
 1.5 m 
s
1

1
1.4 m
( 1.20 m)
6
3
3

3
Now substituting values into this equation:
2

N s
kg m
 ( 3.76 kN m  4.23 kN m)
y'FH  4.23 kN m (counterclockwise)
FA  5.71 kN
Problem 3.72
Given:
[Difficulty: 3]
Parabolic gate, hinged at O has a constant width.
1
b  2 m c  0.25 m
D  2 m H  3 m
Find:
(a) Magnitude and line of action of the vertical force on the gate due to water
(b) Horizontal force applied at A required to maintain equilibrium
(c) Vertical force applied at A required to maintain equilibrium
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dh
 ρ g
(Hydrostatic Pressure - h is positive downwards)
ΣMz  0
Fv 
Assumptions:
(Rotational equilibrium)


p dA y


(Vertical Hydrostatic Force)


x' Fv   x dFv

(Location of line of action)
F H  pc  A
(Horizontal Hydrostatic Force)
Ixx
h'  hc 
A  hc
(Location of line of action)
(1) Static fluid
(2) Incompressible fluid
(3) Atmospheric pressure acts at free surface of water and on
outside of gate
Integrating the hydrostatic pressure equation:


Fv  

x’
h’
FH
p  ρ g h
Ox
(a) The magnitude and line of action of the vertical component of hydrostatic force:
D
y
D
D
D

( 1)

B
x
Oy
 c
 c
 c
 c




2
p dA y  
ρ g h  b dx  
ρ g ( D  y) b dx  
ρ g D  c x b dx  ρ g b  
0
0
0
0
3
3
 3
 2

2
2
D
1 D 
2 ρ g b D
Evaluating the integral: Fv  ρ g b  
 


1
 1 3 1
3
 2

2
2
c 
c
c
FV
 D  c x2 dx
2
Fv 
Substituting values:
3
kg
 999
3
 9.81
m
2
1
2
2
 1  m  N s

kg m
 0.25 
 2 m  ( 2 m)  
s
x' Fv 
To find the line of action of this force:
m
3
2

1 
1 



x dFv Therefore, x' 

x dF v 

x p dA y


Fv 
Fv 


D
Using the derivation for the force:
x'  999
kg
3
m
 9.81
m
2
D
 c
 c
1 
ρ  g b 
2
3
x' 

x ρ g D  c x  b dx 

D x  c x dx
Fv 0
Fv 0



2
2
ρ  g b D
ρ  g b  D D c  D  
      

Fv  2 c 4  c  
F v 4 c
Evaluating the integral: x' 
 2 m 
1

73.9  10
s
Fv  73.9 kN
1
3 N

1
4
Now substituting values into this equation:
2
1
2
 ( 2 m) 

m 
0.25
N s
x'  1.061 m
kg m
To find the required force at A for equilibrium, we need to find the horizontal force of the water on the gate and its
line of action as well. Once this force is known we take moments about the hinge (point O).
2
D
D
FH  pc A  ρ g hc b D  ρ g  b D  ρ g b
2
2
FH  999
kg
3
 9.81
m
m
2
 2 m 
s
hc 
since
( 2 m)
2
2
D
2
Therefore the horizontal force is:
2

N s
FH  39.2 kN
kg m
To calculate the line of action of this force:
3
Ixx
D b D
D D
2
1 2
h'  hc 



 

 D
2
2
3
12 b D D
6
A  hc
h' 
2
 2 m
3
h'  1.333 m
y
Now we have information to solve parts (b) and (c):
(b) Horizontal force applied at A for equilibrium: take moments about O:
FA H  Fv x'  FH ( D  h')  0
Solving for FA
FA 
Fv x'  FH ( D  h')
D
h’
FH
H
FV
H
1 1
  [ 73.9 kN  1.061 m  39.2 kN  ( 2 m  1.333 m) ]
3 m
x
Oy
Ox
FA 
FA
x’
FA  34.9 kN
(c) Vertical force applied at A for equilibrium: take moments about O:
FA L  Fv x'  FH ( D  h')  0
Solving for FA
FA 
y
Fv x'  FH ( D  h')
L
D
L is the value of x at y = H. Therefore: L 
FA 
H
c
L 
3 m 
1
0.25
L
x’
h’
FH
 m L  3.464 m
1
1
  [ 73.9 kN  1.061 m  39.2 kN  ( 2 m  1.333 m) ]
3.464 m
Ox
FA  30.2 kN
Oy
FA
FV
x
Problem 3.73
[Difficulty: 2]
Given:
Liquid concrete is poured into the form shown
R = 2⋅ ft w = 15⋅ ft SGc = 2.5 (Table A.1, App. A)
Find:
Magnitude and line of action of the vertical force on the form
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dh
= ρ⋅ g
Fv =
⌠
⎮
p dA y
⎮
⌡
x'⋅ Fv =
Assumptions:
(Vertical Hydrostatic Force)
⌠
⎮
x dF v
⎮
⌡
(Moment of vertical force)
(1) Static fluid
(2) Incompressible fluid
(3) Atmospheric pressure acts at free surface of concrete
and on outside of gate
p = ρ ⋅ g⋅ h
Integrating the hydrostatic pressure equation:
Fv =
(Hydrostatic Pressure - h is positive downwards)
⌠
⌠
⎮
⎮
p dA y =
ρ⋅ g⋅ h⋅ sin ( θ) dA
⎮
⎮
⌡
⌡
where
dA = w⋅ R⋅ dθ
and
π
⌠2
h = R − y = R − R⋅ sin ( θ)
π
⌠2
⎮
2⎮
2
Therefore, Fv = ⎮ ρ⋅ g⋅ ( R − R⋅ sin ( θ) ) ⋅ w⋅ R⋅ sin ( θ) dθ = ρ⋅ g⋅ w⋅ R ⋅ ⎮ ⎡⎣sin ( θ) − ( sin ( θ) ) ⎤⎦ dθ
⌡0
⌡0
Evaluating the integral:
The density of concrete is:
2
⎡
⎣
⎛ π − 0⎞ + ( 0 − 0)⎤ = ρ⋅ g⋅ w⋅ R2⋅ ⎛ 1 − π ⎞
⎟
⎥
⎜
⎟
4⎠
⎝4
⎠
⎦
⎝
Fv = ρ⋅ g⋅ w⋅ R ⋅ ⎢−( 0 − 1) − ⎜
ρ = 2.5 × 1.94⋅
slug
ft
Substituting values we calculate the force:
ρ = 4.85⋅
3
Fv = 4.85⋅
ft
slug
ft
To find the line of action:
x'⋅ Fv =
slug
3
⌠
⌠
⎮
⎮
x dF v =
x⋅ p dAy
⎮
⎮
⌡
⌡
× 32.2⋅
ft
2
s
3
2
⎛
⎝
× 15⋅ ft × ( 2⋅ ft) × ⎜ 1 −
π⎞
2
lbf⋅ s
⎟×
4 ⎠ slugft
⋅
Using the derivation for the force:
Fv = 2011⋅ lbf
π
⌠2
x'⋅ Fv =
⌠
3⎮
2
⎮
R⋅ cos ( θ) ⋅ ρ⋅ g⋅ ( R − R⋅ sin ( θ) ) ⋅ w⋅ R⋅ sin ( θ) dθ = ρ⋅ g⋅ w⋅ R ⋅ ⎮ ⎡⎣sin ( θ) ⋅ cos ( θ) − ( sin ( θ) ) ⋅ cos ( θ)⎤⎦ dθ
⎮
⌡
⌡
0
Evaluating the integral:
x' =
x'⋅ Fv
Fv
ρ ⋅ g⋅ w ⋅
=
R
Therefore the line of action of the force is:
3
6
⎛
ρ ⋅ g⋅ w ⋅ R ⋅ ⎜ 1 − ⎟
4⎠
⎝
2
3
R
1⎞
− ⎟ = ρ ⋅ g⋅ w ⋅
2
3
6
⎝
⎠
3 ⎛1
x'⋅ Fv = ρ⋅ g⋅ w⋅ R ⋅ ⎜
π⎞
=
R
π⎞
⎛
6⋅ ⎜ 1 − ⎟
4⎠
⎝
Substituting values:
x' =
2⋅ ft
⎛
6⋅ ⎜ 1 −
⎝
π⎞
⎟
4⎠
x' = 1.553⋅ ft
Problem 3.74
[Difficulty: 2]
Given:
Open tank as shown. Width of curved surface b = 10⋅ ft
Find:
(a) Magnitude of the vertical force component on the curved surface
(b) Line of action of the vertical component of the force
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dh
=γ
Fv = −
⌠
⎮
p dA y
⎮
⌡
x’
(Vertical Hydrostatic Force)
⌠
⎮
x dF v
⎮
⌡
x'⋅ Fv =
Assumptions:
(Hydrostatic Pressure - h is positive downwards)
FRy
y
(1) Static fluid
(2) Incompressible fluid
(3) Atmospheric pressure acts at free surface of water
and on outside of wall
p = γ⋅ h
Integrating the hydrostatic pressure equation:
L
(Moment of vertical force)
x
(
dAy = b ⋅ dx
We also define the incremental area on the curved surface as:
2
2
h = L− R −x
We can define along the surface
)
1
2
Substituting these into the force equation we get:
R
Fv = −
⌠
⎮
⎮
⌡
⌠
1⎤
⎡⎢
⎮
⎥
R
2
⎮
⌠
2
2 ⎥
⎢
L−
p dAy = −⎮ γ⋅ ⎣L − R − x ⎦ ⋅ b dx = −γ⋅ b ⋅ ⎮
⌡0
⌡0
(
(
)
lbf
π ⎞⎤
⎡
⎛
Fv = −⎢62.4⋅
× 10⋅ ft × 4⋅ ft × ⎜ 10⋅ ft − 4⋅ ft × ⎟⎥
3
4⎠
⎝
ft
⎣
⎦
To find the line of action of the force:
Therefore:
x' =
x'⋅ Fv
Fv
Evaluating the integral:
Substituting known values:
x'⋅ Fv =
⌠
⎮
x dF v
⎮
⌡
⎛
γ ⋅ b ⋅ R⋅ ⎜ L − R⋅ ⎟
4⎠
⎝
x' =
)
⎛
⎝
3
R
=
2
Fv = −17.12 × 10 ⋅ lbf
(
dFv = −γ⋅ b ⋅ L −
where
(
⌠
⋅ ⎮ x⋅ γ⋅ b ⋅ L −
π ⎞ ⌡0
1
2
R − x dx = − γ ⋅ b ⋅ R ⋅ ⎜ L − R ⋅
R
2
− x ) dx =
⎟
4⎠
(negative indicates downward)
2
2
)
R − x ⋅ dx
R
1
2
π⎞
⌠
⋅⎮
π ⎞ ⌡0
⎛
R⋅ ⎜ L − R⋅ ⎟
4⎠
⎝
( L⋅ x − x⋅
2
2
R −x
) dx
2
4⋅ R
4⋅ R
2 1 3⎞
⎛1
⎛L R⎞
⎛L R⎞
⋅ ⎜ ⋅ L⋅ R − ⋅ R ⎟ =
⋅⎜ − ⎟ =
⋅⎜ − ⎟
R ⋅ ( 4⋅ L − π ⋅ R ) ⎝ 2
3
⎠ R ⋅ ( 4 ⋅ L − π ⋅ R ) ⎝ 2 3 ⎠ 4⋅ L − π ⋅ R ⎝ 2 3 ⎠
4
x' =
4⋅ 4⋅ ft
⎛ 10⋅ ft − 4⋅ ft ⎞
⋅⎜
⎟
4⋅ 10⋅ ft − π⋅ 4⋅ ft ⎝ 2
3 ⎠
x' = 2.14⋅ ft
Problem 3.75
[Difficulty: 2]
Given:
Gate formed in the shape of a circular arc has width w. Liquid is water;
depth h = R
Find:
(a) Magnitude of the net vertical force component due to fluids acting on the gate
(b) Line of action of the vertical component of the force
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dy
= ρ⋅ g
Fv = −
⌠
⎮
p dA y
⎮
⌡
x'⋅ Fv =
Assumptions:
(Hydrostatic Pressure - y is positive downwards)
(Vertical Hydrostatic Force)
⌠
⎮
x dF v
⎮
⌡
(Moment of vertical force)
(1) Static fluid
(2) Incompressible fluid
(3) Atmospheric pressure acts at free surface of water
and on outside of gate
Integrating the hydrostatic pressure equation:
p = ρ ⋅ g⋅ y
Instead of y, we use θ as our variable of integration:
Therefore, dy = R⋅ cos ( θ) ⋅ dθ
In addition,
y = R⋅ sin ( θ)
dAy = w⋅ R⋅ sin ( θ) ⋅ dθ
π
π
⌠2
⌠2
⎮
π
2 ⎮
2
2
Therefore, Fv = −⎮ ρ⋅ g⋅ R⋅ sin ( θ) ⋅ w⋅ R⋅ sin ( θ) dθ = −ρ⋅ g⋅ R ⋅ w⋅ ⎮ ( sin ( θ) ) dθ = −ρ⋅ g⋅ R ⋅ w⋅
⌡0
⌡0
4
2
Fv = −
π ⋅ ρ ⋅ g⋅ R ⋅ w
4
(negative indicates downward)
To find the line of action of the vertical component of the force:
2
2
dFv = −ρ⋅ g⋅ R ⋅ w⋅ ( sin ( θ) ) ⋅ dθ
x' =
x'⋅ Fv
Fv
=−
π
⌠2
4
2
π⋅ ρ⋅ g⋅ R ⋅ w
x'⋅ Fv =
⌠
⎮
x dFv where x = R⋅ cos ( θ) and the elemental force is
⎮
⌡
Substituting into the above integral yields:
π
⌠2
⎮
4⋅ R ⎮
4⋅ R 1
2
2
2
⋅ ⎮ −( R⋅ cos ( θ) ) ⋅ ⎡⎣ρ⋅ g⋅ R ⋅ w⋅ ( sin ( θ) ) ⎤⎦ dθ =
⋅ ⎮ ( sin ( θ) ) ⋅ cos ( θ) dθ =
⋅
⌡0
π ⌡0
π 3
x' =
4⋅ R
3⋅ π
Problem 3.76
[Difficulty: 3]
Given:
Dam with cross-section shown. Width of dam
b  160 ft
Find:
(a) Magnitude and line of action of the vertical force component on the dam
(b) If it is possible for the water to overturn dam
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dh
 ρ g
Fv 
(Hydrostatic Pressure - h is positive downwards from
free surface)


p dA y


(Vertical Hydrostatic Force)
FH  pc A
(Horizontal Hydrostatic Force)


x dF v


Ixx
h'  hc 
hc A
x' Fv 
(Moment of vertical force)
(Line of action of vertical force)
ΣMz  0
Assumptions:
y
(Rotational Equilibrium)
A
x’
(1) Static fluid
(2) Incompressible fluid
(3) Atmospheric pressure acts at free surface of water
and on outside of dam
FV
B
Integrating the hydrostatic pressure equation:
y’
p  ρ  g h
x
x
Into the vertical force equation:
h’
FH
x
B
B




Fv   p dAy   ρ g h b dx  ρ g b  ( H  y) dx
x
x

A
A
From the definition of the dam contour:
x y  A y  B Therefore: y 
B
xA
and
xA 
10 ft
9 ft
2
 1 ft
xA  2.11 ft
xB

F v  ρ  g b  

x
Into the force equation:
A
Fv  1.94
slug
ft
3


ft
 32.2
2
s
x' 
Fv
1 


Fv 

xB
x' Fv 


x dFv where




dFv  ρ g b  H 
xB


 xB  A  
H xB  xA  B ln 
 xA
xA  A
1



5
x' 




 xB  A 

Fv  2.71  10  lbf
B 
  dx
x  A
Therefore:
 H x  B x  dx


x  A


 xB  A 
H  2
2
 xB  xA   B xB  xA  B A ln 


2 
xA  A

Evaluating the integral:
2
 7.0  1   lbf  s

 2.11  1  slug ft
2
B 

x ρ  g b   H 
 dx 
x

A

xA
Substituting known values:
 160 ft  9 ft  ( 7.0 ft  2.11 ft)  10 ft  ln 
To find the line of action of the force:
x' Fv
 H  B  dx  ρ g b H x  x  B ln xB  A 


  B A


x  A


 xA  A 
Substituting known values we get:
H xB  xA  B ln 
x
9 ft
x' 
2
2

2
2
2

 A  A
 71 

 2.11  1 
2
 7  2.11  ft  10 ft  ( 7  2.11)  ft  10 ft  1 ft  ln 
x'  4.96 ft
 71 

 2.11  1 
2
9 ft  ( 7  2.11)  ft  10 ft  ln 
To determine whether or not the water can overturn the dam, we need the horizontal force and its line of action:
H
ρ  g b H
FH  pc A  ρ g  H b 
2
2
FH 
Substituting values:
For the line of action:
Therefore: h' 
H
2

1
2
2
 1.94
ft
Ixx
h'  hc 
hc A
b H
3
12

slug
3
 32.2
ft
2
2
 160 ft  ( 9 ft) 
s
where
hc 
2 1
H H
2



 H
H b H
2
6
3
H
2
3
5
FH  4.05  10  lbf
slug ft
A  H b
2
h' 
2
lbf  s
Ixx 
 9 ft
b H
3
12
h'  6.00 ft
Taking moments of the hydrostatic forces about the origin:
Mw  FH ( H  h')  Fv x'
5
5
Mw  4.05  10  lbf  ( 9  6)  ft  2.71  10  lbf  4.96 ft
5
Mw  1.292  10  lbf  ft
The negative sign indicates that this is a clockwise moment about the origin. Since the weight of the dam will also contribute a clockwise
moment about the origin, these two moments should not cause the dam to tip to the left.
Therefore, the water can not overturn the dam.
Problem 3.77
[Difficulty: 3]
w  35 m
Given:
Tainter gate as shown
Find:
Force of the water acting on the gate
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dh
 ρ g
(Hydrostatic Pressure - h is positive downwards from
free surface)
dF  p  dA
Assumptions:
(Hydrostatic Force)
(1) Static fluid
(2) Incompressible fluid
(3) Atmospheric pressure acts at free surface of water
and on outside of gate
Integrating the hydrostatic pressure equation:
p  ρ g h  ρ g R sin( θ)
Resolving the hydrostatic force into horizontal and vertical components:
dFH  dF cos ( θ)  p dA cos ( θ)  ρ g R sin ( θ)  w R dθ cos ( θ)
since
θ1

2
FH   ρ g R  w sin ( θ)  cos ( θ) dθ
0
Integrating this expression:
30 deg

F H  ρ  g R  w  
0
2
FH 
1
8
 999
kg
3
2
sin ( θ)  cos ( θ) dθ  ρ g R  w
 9.81
m
m
2
( sin ( 30 deg) )
2
10 m 
where θ1  asin 
  30 deg
 20 m 
2
2

30 deg
s
2
8
Substituting known values:
N s
7
FH  1.715  10  N
kg m
2
π
3


12
8


( sin ( θ) ) dθ  ρ g R  w 
2
2
dFv  dF sin ( θ)  p dA sin ( θ)  ρ g R  w ( sin ( θ) )  dθ
Substituting known values:
2
π
3
kg
m
N s
2

  999 3  9.81 2  ( 20 m)  35 m 
12
8
kg
m


m
s
Fv  
ρ  g R  w
2
2
 ( 20 m)  35 m 
Similarly, we can calculate the vertical component of the hydrostatic force:
2 
F v  ρ  g R  w  
0
dA  w R dθ
6
Fv  6.21  10  N
Now since the gate surface in contact with the water is a circular arc, all elements dF of the force, and hence the line of action of the resulta
must pass through the pivot. Thus:
Magnitude of the resultant force:
FR 
2
FH  Fv
2
FR 
1.715  107 N2  6.21  106 N2
7
FR  1.824  10 N
The line of action of the force:
 Fv 

 FH 
α  atan 
 6.21  106 N 

 1.715  107 N 


α  atan 
α  19.9 deg
The force passes through the pivot at an
angle α to the horizontal.
Problem 3.78
Given:
Gate geometry
Find:
Force on stop B
[Difficulty: 4]
x
y’
Solution:
Basic equations
4R/3π
R/2
D
FV
dp
dh
= ρ⋅ g
W1
A
R
FB
ΣMA = 0
WGate
FH
y
W2
x
Weights for computing FV
F1
Assumptions: static fluid; ρ = constant; patm on other side
p = ρ⋅ g⋅ h
For incompressible fluid
where p is gage pressure and h is measured downwards
We need to compute force (including location) due to water on curved surface and underneath. For curved surface we could integrate
pressure, but here we use the concepts that FV (see sketch) is equivalent to the weight of fluid above, and FH is equivalent to the force on
a vertical flat plate. Note that the sketch only shows forces that will be used to compute the moment at A
For FV
FV = W1 − W2
with
kg
m
N⋅ s
W1 = ρ⋅ g⋅ w⋅ D⋅ R = 1000⋅
× 9.81⋅ × 3⋅ m × 4.5⋅ m × 3⋅ m ×
3
2
kg⋅ m
m
s
2
W2 = ρ⋅ g⋅ w⋅
π⋅ R
2
4
= 1000⋅
397
189
×
3⋅ m
2
m
2
× 3⋅ m ×
π
s
4
2
× ( 3⋅ m) ×
2
N⋅ s
kg⋅ m
FV = 189⋅ kN
R
4⋅ R
FV⋅ x = W1⋅ − W2⋅
2
3⋅ π
x =
For FH
3
× 9.81⋅
m
FV = W1 − W2
with x given by
kg
−
208
189
Computing equations
×
or
4
3⋅ π
× 3⋅ m
FH = pc⋅ A
x=
W1 R W2 4⋅ R
⋅ −
⋅
Fv 2
F v 3⋅ π
x = 1.75 m
Ixx
y' = yc +
A ⋅ yc
W1 = 397⋅ kN
W2 = 208⋅ kN
Hence
R⎞
⎛
FH = pc⋅ A = ρ⋅ g⋅ ⎜ D − ⎟ ⋅ w⋅ R
2⎠
⎝
kg
FH = 1000⋅
× 9.81⋅
3
m
⎛
⎝
m
× ⎜ 4.5⋅ m −
2
s
2
3⋅ m ⎞
N⋅ s
2 ⎠
kg⋅ m
⎟ × 3⋅ m × 3⋅ m ×
FH = 265⋅ kN
The location of this force is
3
2
Ixx
R ⎞ w⋅ R
1
R
R
⎛
y' = yc +
= ⎜D − ⎟ +
×
= D− +
A ⋅ yc ⎝
2⎠
12
R⎞
2
R⎞
⎛
⎛
w ⋅ R⋅ ⎜ D − ⎟
12⋅ ⎜ D − ⎟
2
2⎠
⎝
⎠
⎝
y' = 4.5⋅ m −
3⋅ m
2
( 3⋅ m)
+
⎛
⎝
2
12 × ⎜ 4.5⋅ m −
y' = 3.25 m
3⋅ m ⎞
⎟
2 ⎠
The force F1 on the bottom of the gate is F1 = p⋅ A = ρ⋅ g⋅ D⋅ w⋅ R
F1 = 1000⋅
kg
3
× 9.81⋅
m
2
m
× 4.5⋅ m × 3⋅ m × 3⋅ m ×
2
s
N⋅ s
F1 = 397⋅ kN
kg⋅ m
For the concrete gate (SG = 2.4 from Table A.2)
WGate = SG⋅ ρ⋅ g⋅ w⋅
FB =
4
3⋅ π
2
= 2.4⋅ 1000⋅
4
⋅ WGate +
4
3⋅ π
x
R
× 499⋅ kN +
FB = 278⋅ kN
kg
3
× 9.81⋅
m
F B ⋅ R + F 1⋅
Hence, taking moments about A
FB =
π⋅ R
R
− WGate⋅
2
⋅ FV +
1.75
3
4⋅ R
3⋅ π
[ y' − ( D − R) ]
R
× 189⋅ kN +
m
2
× 3⋅ m ×
s
π
4
2
2
× ( 3⋅ m) ×
N⋅ s
kg⋅ m
− FV⋅ x − FH⋅ [ y' − ( D − R) ] = 0
⋅ FH −
1
2
⋅ F1
[ 3.25 − ( 4.5 − 3) ]
3
× 265⋅ kN −
1
2
× 397⋅ kN
WGate = 499⋅ kN
Problem 3.79
Given:
Sphere with different fluids on each side
Find:
Resultant force and direction
[Difficulty: 4]
Solution:
The horizontal and vertical forces due to each fluid are treated separately. For each, the horizontal force is equivalent to that
on a vertical flat plate; the vertical force is equivalent to the weight of fluid "above".
For horizontal forces, the computing equation of Section 3-5 is FH  pc A where A is the area of the equivalent vertical
plate.
For vertical forces, the computing equation of Section 3-5 is FV  ρ g V where V is the volume of fluid above the curved
surface.
The data is
ρ  999
For water
kg
3
m
For the fluids
SG1  1.6
SG2  0.8
For the weir
D  3 m
L  6 m
(a) Horizontal Forces
For fluid 1 (on the left)
D
1
2

FH1  pc A   ρ1 g   D L   SG1 ρ g D  L
2
2


FH1 
For fluid 2 (on the right)
2
 1.6 999
kg
3
 9.81
m
m
2
2
 ( 3 m)  6 m
s
2
N s
kg m
FH1  423 kN
D D
1
2

FH2  pc A   ρ2 g    L   SG2 ρ g D  L
4 2
8

FH2 
The resultant horizontal force is
1
1
8
 0.8 999
kg
3
m
 9.81
m
2
s
2
 ( 3 m)  6 m
2
N s
kg m
FH  FH1  FH2
(b) Vertical forces
For the left geometry, a "thought experiment" is needed to obtain surfaces with fluid "above"
FH2  52.9 kN
FH  370 kN
Hence
π D
4
FV1  SG1 ρ g
2
L
2
kg
FV1  1.6  999
3
 9.81
m
m
2

s
π ( 3 m)
2
2
 6 m 
8
N s
kg m
FV1  333 kN
(Note: Use of buoyancy leads to the same result!)
For the right side, using a similar logic
π D
4
FV2  SG2 ρ g
2
L
4
FV2  0.8  999
kg
3
 9.81
m
The resultant vertical force is
FV  FV1  FV2
m
2
s

π ( 3 m)
16
2
2
 6 m 
N s
kg m
FV2  83.1 kN
FV  416 kN
Finally the resultant force and direction can be computed
F 
2
FH  FV
 FV 

 FH 
α  atan 
2
F  557 kN
α  48.3 deg
Problem 3.80
[Difficulty: 3]
Given:
Cylindrical weir as shown; liquid is water
Find:
Magnitude and direction of the resultant force of the water on the weir
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
Assumptions:
dp
= ρ⋅ g
dh
⎯→
⎯
→
dFR = −p ⋅ dA
(Hydrostatic Pressure - h is positive downwards from
free surface)
(Hydrostatic Force)
(1) Static fluid
(2) Incompressible fluid
(3) Atmospheric pressure acts on free surfaces and on the
first quadrant of the cylinder
Using the coordinate system shown in the diagram at the right:
h1
3⋅ π
⌠ 2
⎮
FRx = ⎮
⌡0
y
x
D1
⎯
→→
→→
⌠
⌠
⌠
⎮
⎮
⎮
FRx = FR⋅ i = −
p dA ⋅ i = −
p ⋅ cos ( θ + 90⋅ deg) dA = ⎮ p ⋅ sin( θ) dA
⎮
⎮
⌡
⌡
⌡
⎯
→→
→→
⌠
⌠
⎮
⎮
FRy = FR⋅ j = −
p dA⋅ j = −
p⋅ cos ( θ) dA
⎮
⎮
⌡
⌡
θ
h2
Now since dA = L⋅ R⋅ dθ it follows that
3⋅ π
⌠ 2
p⋅ L⋅ R⋅ sin ( θ) dθ
and
⎮
FRy = −⎮
⌡0
Next, we integrate the hydrostatic pressure equation:
p⋅ L⋅ R⋅ cos ( θ) dθ
p = ρ ⋅ g⋅ h
Now over the range
Over the range
0≤θ≤π
π≤θ≤
3⋅ π
2
h1 = R ( 1 − cos ( θ) )
h2 = −R⋅ cos ( θ)
Therefore we can express the pressure in terms of θ and substitute into the force equations:
3⋅ π
3⋅ π
⌠ 2
⌠ 2
π
⎮
⎮
⌠
⎮
ρ⋅ g⋅ R⋅ ( 1 − cos ( θ) ) ⋅ L⋅ R⋅ sin ( θ) dθ − ⎮
FRx = ⎮
p⋅ L⋅ R⋅ sin ( θ) dθ =
ρ⋅ g⋅ R⋅ cos ( θ) ⋅ L⋅ R⋅ sin ( θ) dθ
⌡0
⌡0
⌡π
π
3⋅ π
⌠ 2
2 ⌠
2 ⎮
FRx = ρ⋅ g⋅ R ⋅ L⋅ ⎮ ( 1 − cos ( θ) ) ⋅ sin ( θ) dθ − ρ⋅ g⋅ R ⋅ L⋅ ⎮
⌡0
⌡π
cos ( θ) ⋅ sin ( θ) dθ
D2
3⋅ π
⎤
⎡
⎢ π
⎥
⌠ 2
⎮
3
1⎞
2 ⎛
2 ⎢⌠
⎥
2
FRx = ρ⋅ g⋅ R ⋅ L⋅ ⎮ ( 1 − cos ( θ) ) ⋅ sin ( θ) dθ − ⎮
cos ( θ) ⋅ sin ( θ) dθ = ρ⋅ g⋅ R ⋅ L⋅ ⎜ 2 − ⎟ = ⋅ ρ⋅ g⋅ R ⋅ L
⎢⌡
⎥
⌡
⎝ 2⎠ 2
π
⎣ 0
⎦
Substituting known values:
FRx =
3
2
× 999⋅
kg
3
× 9.81⋅
m
m
2
2
2
× ( 1.5⋅ m) × 6⋅ m ×
s
N⋅ s
FRx = 198.5⋅ kN
kg⋅ m
Similarly we can calculate the vertical force component:
3⋅ π
3⋅ π
⎡
⎤
⎢ π
⎥
⌠ 2
⌠ 2
⎮
⎮
⎢⌠
⎥
FRy = −⎮
p⋅ L⋅ R⋅ cos ( θ) dθ = − ⎮ ρ⋅ g⋅ R⋅ ( 1 − cos ( θ) ) ⋅ L⋅ R⋅ cos ( θ) dθ − ⎮
ρ⋅ g⋅ R⋅ cos ( θ) ⋅ L⋅ R⋅ cos ( θ) dθ
⎢⌡
⎥
⌡0
⌡
π
⎣ 0
⎦
3⋅ π
⎡
⎤
⎢ π
⎥
⌠ 2
⎮
3⋅ π
2 ⎢⌠
2 ⎥
2 ⎛ π 3⋅ π π ⎞
2
FRy = −ρ⋅ g⋅ R ⋅ L⋅ ⎮ ( 1 − cos ( θ) ) ⋅ cos ( θ) dθ − ⎮
( cos ( θ) ) dθ = ρ⋅ g⋅ R ⋅ L⋅ ⎜ +
− ⎟ =
⋅ ρ ⋅ g⋅ R ⋅ L
⎢⌡
⎥
⌡
2
4
2
4
⎝
⎠
π
⎣ 0
⎦
Substituting known values:
FRy =
3⋅ π
4
× 999⋅
kg
3
m
× 9.81⋅
m
2
2
× ( 1.5⋅ m) × 6⋅ m ×
s
2
N⋅ s
kg⋅ m
FRy = 312⋅ kN
Now since the weir surface in contact with the water is a circular arc, all elements dF of the force, and hence the line of action of the
resultant force, must pass through the pivot. Thus:
2
Magnitude of the resultant force:
FR =
( 198.5⋅ kN) + ( 312⋅ kN)
The line of action of the force:
α = atan ⎜
⎛ 312⋅ kN ⎞
⎟
⎝ 198.5⋅ kN ⎠
2
FR = 370⋅ kN
α = 57.5⋅ deg
Problem 3.81
[Difficulty: 3]
Given:
Cylindrical log floating against dam
Find:
(a) Mass per unit length of the log (b) Contact force per unit length between log and dam
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dh
= ρ⋅ g
(Hydrostatic Pressure - h is positive downwards from
free surface)
⎯
⎯
→
→
dF = p ⋅ dA
Assumptions:
(Hydrostatic Force)
(1) Static fluid
(2) Incompressible fluid
(3) Atmospheric pressure acts on free surfaces and on the
first quadrant of the log
Integrating the hydrostatic pressure equation:
dFH
R = D/2
dFV
dF
h
p = ρ⋅ g⋅ h = ρ⋅ g⋅ R⋅ ( 1 − cos ( θ) )
θ
Resolving the incremental force into horizontal and vertical components:
2
dF = p ⋅ dA = p ⋅ w⋅ R⋅ dθ = ρ⋅ g⋅ R⋅ ( 1 − cos ( θ) ) ⋅ w⋅ R⋅ dθ = ρ⋅ g⋅ R ⋅ w⋅ ( 1 − cos ( θ) )
2
dFH = dF⋅ sin( θ) = ρ⋅ g⋅ R ⋅ w⋅ ( 1 − cos ( θ) ) ⋅ dθ⋅ sin( θ)
2
dFv = dF⋅ cos ( θ) = ρ⋅ g⋅ R ⋅ w⋅ ( 1 − cos ( θ) ) ⋅ dθ⋅ cos ( θ)
Integrating the expression for the horizontal force will provide us with the contact force per unit length:
3⋅ π
⌠ 2
⎮
FH = ⎮
⌡0
3⋅ π
⌠ 2
2
2 ⎮
ρ⋅ g⋅ R ⋅ w⋅ ( 1 − cos ( θ) ) ⋅ sin( θ) dθ = ρ⋅ g⋅ R ⋅ w⋅ ⎮
⌡0
2
⎛ 1 + 1 ⎞ = ρ⋅ g⋅ R ⋅ w
⎟
2
⎝ 2
⎠
2
( sin( θ) − sin( θ) ⋅ cos ( θ) ) dθ = ρ⋅ g⋅ R ⋅ w⋅ ⎜ −
Therefore:
FH
w
=
ρ⋅ g⋅ R
2
2
Integrating the expression for the vertical force will provide us with the mass per unit length of the log:
3⋅ π
3⋅ π
⌠ 2
⌠ 2
⎮
3⋅ π ⎞
2
2 ⎮
2 ⎛
Fv = ⎮
ρ⋅ g⋅ R ⋅ w⋅ ( 1 − cos ( θ) ) ⋅ cos ( θ) dθ = ρ⋅ g⋅ R ⋅ w⋅ ⎮
( 1 − cos ( θ) ) ⋅ cos ( θ) dθ = ρ⋅ g⋅ R ⋅ w⋅ ⎜ −1 −
⎟
⌡0
⌡0
4 ⎠
⎝
Therefore:
Fv
w
2
⎛
⎝
= − ρ ⋅ g⋅ R ⋅ ⎜ 1 +
3⋅ π ⎞
⎟
4 ⎠
From a free-body diagram for the log:
ΣFy = 0
−
m
w
⋅g −
Solving for the mass of the log:
Fv
w
m
=0
m
w
w
2
⎛
⎝
=−
= ρ⋅ R ⋅ ⎜ 1 +
Fv
w⋅ g
3⋅ π ⎞
⎟
4 ⎠
Problem 3.82
Given:
[Difficulty: 3]
Curved surface, in shape of quarter cylinder, with given radius R and width w; water stands to depth H.
R = 0.750⋅ m w = 3.55⋅ m
H = 0.650⋅ m
Find:
Magnitude and line of action of (a) vertical force and (b) horizontal force on the curved
surface
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dh
= ρ⋅ g
(Hydrostatic Pressure - h is positive downwards from
free surface)
⌠
⎮
p dA y
⎮
⌡
Fv =
(Vertical Hydrostatic Force)
FH = pc⋅ A
(Horizontal Hydrostatic Force)
⌠
⎮
x dF v
⎮
⌡
Ixx
h' = hc +
hc⋅ A
x'⋅ Fv =
Assumptions:
(Moment of vertical force)
(Line of action of horizontal force)
dF
(1) Static fluid
(2) Incompressible fluid
(3) Atmospheric pressure acts on free surface of the
water and on the left side of the curved surface
Integrating the hydrostatic pressure equation:
From the geometry: h = H − R⋅ sin ( θ)
⎛H⎞
θ1 = asin ⎜ ⎟
⎝R⎠
R
θ
h
H
p = ρ ⋅ g⋅ h
y = R⋅ sin ( θ)
⎛ 0.650 ⎞
θ1 = asin ⎜
⎟
⎝ 0.750 ⎠
x = R⋅ cos ( θ)
dA = w⋅ R⋅ dθ
x’
dF
FV
θ1 = 1.048⋅ rad
h’
R
Therefore the vertical component of the hydrostatic force is:
θ
1
⌠
⌠
⌠
⎮
⎮
⎮
Fv = ⎮ p dAy = ⎮ ρ⋅ g⋅ h⋅ sin ( θ) dA =
ρ⋅ g⋅ ( H − R⋅ sin ( θ) ) ⋅ sin ( θ) ⋅ w⋅ R dθ
⌡0
⌡
⌡
θ
(
)
1
⌠
⎡
⎛ θ1 sin 2⋅ θ1 ⎞⎤
2
Fv = ρ⋅ g⋅ w⋅ R⋅ ⎮ ⎡⎣H⋅ sin ( θ) − R⋅ ( sin ( θ) ) ⎤⎦ dθ = ρ⋅ g⋅ w⋅ R⋅ ⎢H⋅ 1 − cos θ1 − R⋅ ⎜
−
⎟⎥
⌡0
4
⎣
⎝2
⎠⎦
(
( ))
θ
FH
y’
H
kg
Fv = 999⋅
3
× 9.81⋅
m
m
2
2
⎡
⎣
⎛ 1.048 − sin ( 2 × 1.048⋅ rad) ⎞⎤ × N⋅ s
⎟⎥
4
⎝ 2
⎠⎦ kg⋅ m
× 3.55⋅ m × 0.750⋅ m × ⎢0.650⋅ m × ( 1 − cos ( 1.048⋅ rad) ) − 0.750⋅ m × ⎜
s
Fv = 2.47⋅ kN
To calculate the line of action of this force:
θ
1
⌠
2⌠ ⎡
2
⎮
⎮
x'⋅ Fv =
R⋅ cos ( θ) ⋅ ρ⋅ g⋅ h⋅ sin ( θ) dA = ρ⋅ g⋅ w⋅ R ⋅
⎣H⋅ sin ( θ) ⋅ cos ( θ) − R⋅ ( sin ( θ) ) ⋅ cos ( θ)⎤⎦ dθ
⎮
⌡
⌡
0
2 H
2 R
3
Evaluating the integral: x'⋅ Fv = ρ⋅ g⋅ w⋅ R ⋅ ⎡⎢ ⋅ sin θ1 − ⋅ sin θ1 ⎥⎤
2
3
⎣
⎦
( ( ))
x' =
x'⋅ Fv
Fv
x' = 999⋅
2
=
kg
3
( ( ))
ρ ⋅ g⋅ w ⋅ R ⎡ H
2 R
3⎤
⋅ ⎢ ⋅ sin θ1 − ⋅ sin θ1 ⎥
Fv
2
3
⎣
⎦
× 9.81⋅
m
m
2
( ( ))
( ( ))
2
× 3.55⋅ m × ( 0.750⋅ m) ×
s
Therefore we may find the line of action:
Substituting in known values:
( )
0.650
sin θ1 =
0.750
⎡ 0.650⋅ m ⎛ 0.650 ⎞ 2 0.750⋅ m ⎛ 0.650 ⎞ 3⎤ N⋅ s2
×⎜
×⎜
⎟ −
⎟ ⎥×
3 N ⎣
2
3
⎝ 0.750 ⎠
⎝ 0.750 ⎠ ⎦ kg⋅ m
2.47 × 10
1
⋅
1
×⎢
x' = 0.645 m
2
For the horizontal force:
FH =
1
2
× 999⋅
kg
3
m
H
ρ ⋅ g⋅ H ⋅ w
FH = pc⋅ A = ρ⋅ g⋅ hc⋅ H⋅ w = ρ⋅ g⋅ ⋅ H⋅ w =
2
2
× 9.81⋅
m
2
2
2
× ( 0.650⋅ m) × 3.55⋅ m ×
s
For the line of action of the horizontal force:
N⋅ s
FH = 7.35⋅ kN
kg⋅ m
Ixx
h' = hc +
hc⋅ A
3
Ixx
H w⋅ H 2 1
H H
2
h' = hc +
=
+
⋅ ⋅
=
+
= ⋅H
12 H w⋅ H
6
hc⋅ A
2
2
3
where
h' =
2
3
Ixx =
w⋅ H
12
× 0.650⋅ m
3
A = w⋅ H
Therefore:
h' = 0.433 m
Problem 3.83
Given:
Canoe floating in a pond
Find:
What happens when an anchor with too short of a line is thrown from canoe
[Difficulty: 2]
Solution:
Governing equation:
FB   w gVdisp  W
Before the anchor is thrown from the canoe the buoyant force on the canoe balances out the weight of the canoe and anchor:
FB1  Wcanoe  Wanchor   w gVcanoe1
The anchor weight can be expressed as
Wanchor   a gVa
so the initial volume displaced by the canoe can be written as
Vcanoe1 
Wcanoe  a

Va
w g w
After throwing the anchor out of the canoe there will be buoyant forces acting on the canoe and the anchor. Combined, these buoyant
forces balance the canoe weight and anchor weight:
FB2  Wcanoe  Wanchor   w gVcanoe2   w gVa
Vcanoe 2 
Wcanoe Wa

 Va
w g w g
Vcanoe 2 
Wcanoe  a
Va  Va

w g w
Using the anchor weight,
Hence the volume displaced by the canoe after throwing the anchor in is less than when the anchor was in the canoe, meaning that the
canoe is floating higher.
Problem 3.84
Given:
[Difficulty: 3]
Curved surface, in shape of quarter cylinder, with given radius R and width w; liquid concrete stands to depth H.
R = 1⋅ ft
w = 4⋅ ft
Fvmax = 350⋅ lbf
SG = 2.50 From Table A.1, App A
Find:
(a) Maximum depth of concrete to avoid cracking
(b) Line of action on the form.
(c) Plot the vertical force and line of action over H ranging from 0 to R.
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dh
= ρ⋅ g
Fv =
(Hydrostatic Pressure - h is positive downwards from
free surface)
⌠
⎮
p dA y
⎮
⌡
(Vertical Hydrostatic Force)
x
⌠
⎮
x'⋅ Fv =
x dF v
⎮
⌡
Assumptions:
d
(Moment of vertical force)
(1) Static fluid
(2) Incompressible fluid
(3) Atmospheric pressure acts on free surface of the concrete
θ1
θ
x’
h
FV
p = ρ ⋅ g⋅ h
Integrating the hydrostatic pressure equation:
From the geometry: y = R⋅ sin ( θ)
y
x = R⋅ cos ( θ)
h = y−d
d = R−H
dA = w⋅ R⋅ dθ
Therefore the vertical component of the hydrostatic force is:
π
Fv =
⌠
⌠
⎮
⎮
p dA y =
⎮
⎮
⌡
⌡
⌠2
⎮
ρ⋅ g⋅ h⋅ sin ( θ) dA = ⎮ ρ⋅ g⋅ ( R⋅ sin ( θ) − d) ⋅ sin ( θ) ⋅ w⋅ R dθ
⌡θ
where
1
π
⌠2
(
)
⎛d⎞
θ1 = asin ⎜ ⎟
⎝ R⎠
⎮
⎡ ⎛ π θ1 sin 2⋅ θ1 ⎞
⎤
2
Fv = ρ⋅ g⋅ w⋅ R⋅ ⎮ ⎡⎣R⋅ ( sin ( θ) ) − d⋅ ( sin ( θ) )⎤⎦ dθ = ρ⋅ g⋅ w⋅ R⋅ ⎢R⋅ ⎜ −
+
⎟ − d⋅ cos θ1 ⎥
4
⌡θ
⎣ ⎝4 2
⎠
⎦
( )
In terms of H:
1
( )
sin θ1 =
R−H
R
( )
cos θ1 =
2
R − ( R − H)
R
2
=
2⋅ R ⋅ H − H
R
2
(
)
( ) ( )
sin 2⋅ θ1 = 2⋅ sin θ1 ⋅ cos θ1 =
2⋅ ( R − H ) ⋅ 2⋅ R ⋅ H − H
R
2
2
⎡⎢ ⎡⎢
⎛ H⎞
⎤
⎤
asin ⎜ 1 − ⎟
2⎥
2⎥
R
π
(
R
−
H
)
⋅
2
⋅
R
⋅
H
−
H
2
⋅
R
⋅
H
−
H
⎝
⎠
⎥ − ( R − H) ⋅
⎥
Fv = ρ⋅ g⋅ w⋅ R⋅ ⎢R⋅ ⎢ −
+
2
2
R
⎢ ⎢4
⎥
⎥
2R
⎣ ⎣
⎦
⎦
This equation can be solved iterative
for H:
H = 0.773⋅ ft
To calculate the line of action of this force:
π
x'⋅ Fv =
⌠
⎮
⎮
⌡
⌠2
2 ⎮ ⎡
2
x⋅ ρ⋅ g⋅ h⋅ sin ( θ) dA = ρ⋅ g⋅ R ⋅ w⋅ ⎮ ⎣R⋅ ( sin ( θ) ) ⋅ cos ( θ) − d⋅ sin ( θ) ⋅ cos ( θ)⎤⎦ dθ
⌡θ
1
Evaluating the integral:
R
d
2
3
2⎤
x'⋅ Fv = ρ⋅ g⋅ R ⋅ w⋅ ⎡⎢ ⋅ ⎡1 − sin θ1 ⎤ − ⋅ cos θ1 ⎥
⎣
⎦
2
⎣3
⎦
Therefore we may find the line of action:
x' =
( ( ))
( )
sin θ1 =
Substituting in known values:
⎛
slug ⎞
⎝
ft ⎠
x' = ⎜ 2.5 × 1.94⋅
3
⎟ × 32.2⋅
ft
2
x'⋅ Fv
Fv
1 − 0.773
1
2
× ( 1⋅ ft) × 4⋅ ft ×
s
2
=
ρ ⋅ g⋅ R ⋅ w ⎡ R ⎡
d
3
2⎤
⋅ ⎢ ⋅ 1 − sin θ1 ⎤ − ⋅ cos θ1 ⎥
⎣
⎦
Fv
2
⎣3
⎦
⋅
1
350 lbf
( ( ))
( )
= 0.227
1
( ( ))
cos θ1 =
( ( ))
2
1 − 0.227 = 0.9739
2
⎡ 1⋅ ft × ⎡1 − ( 0.227) 3⎤ − 0.227⋅ ft × ( 0.9739) 2⎤ × lbf ⋅ s
⎣
⎦
⎥
2
⎣ 3
⎦ slug⋅ ft
×⎢
x' = 0.396⋅ ft
We may use the equations we developed above to plot the vertical force and line of action as a function of the height of the concrete in the
Vertical Force vs. Depth Ratio
Line of Action vs. Depth Ratio
500.0
0.4
Line of Action (ft)
Vertical Force (lbf)
400.0
300.0
200.0
0.2
100.0
0.0
0.0
0.5
Depth Ratio H/R
1.0
0.0
0.0
0.5
Depth Ratio H/R
1.0
Problem 3.85
Given:
[Difficulty: 3]
Model cross section of canoe as a parabola. Assume constant width W over entire length L
2
y  a x
a  1.2 ft
1
W  2 ft
L  18 ft
Find:
Expression relating the total mass of canoe and contents to distance d. Determine maximum
allowable total mass without swamping the canoe.
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dh
 ρ g
Fv 
Assumptions:
(Hydrostatic Pressure - h is positive downwards from
free surface)


p dA y


(Vertical Hydrostatic Force)
(1) Static fluid
(2) Incompressible fluid
(3) Atmospheric pressure acts on free surface of the water and inner
surface of the canoe.
At any value of d the weight of the canoe and its contents is balanced by the net vertical force of the water on the canoe.
Integrating the hydrostatic pressure equation:
Fv 
p  ρ  g h




p dA y 
ρ g h L dx where h  ( H  d)  y




y  a x
To determine the upper limit of integreation we remember that
y  Hd


F v  2 
0
Therefore, x 
Hd
a
Hd
a
At the surface
and so the vertical force is:


2

ρ g ( H  d)  a x   L dx  2 ρ g L 
0
Upon simplification: Fv  2 ρ g L
2
( H  d)
a
3
2
Hd
a
3
3


2
2
(
H

d
)
a
(
H

d
)
2




( H  d)  a x  dx  2 ρ g L
 
 

3  a  
a

3
 1  4 ρ g L  ( H  d) 2  M g
 1   
 3
3 a
or
M
4 ρ  L
 ( H  d)
3 a
3
2
where M is the
mass of the canoe.
3
ft
2 32.174 lb
 ( 2.4 ft) 
1.2
slug
4
slug
The limit for no swamping is d=0, and so: M   1.94
 18 ft 
3
3
ft
This leaves us no margin, so if we set d=0.2 ft we get
M 
4
3
 1.94
slug
ft
3
 18 ft 
ft
1.2
3
M  5.08  10  lb
3
2 32.174 lb
 ( 2.2 ft) 
slug
3
M  4.46  10  lb
Clearly the answer is highly dependent upon the allowed risk of swamping!
Problem 3.86
[Difficulty: 4]
Given:
Cylinder of mass M, length L, and radius R is hinged along its length and immersed in an incompressilble liquid to depth
Find:
General expression for the cylinder specific gravity as a function of α=H/R needed to hold
the cylinder in equilibrium for α ranging from 0 to 1.
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dh
= ρ⋅ g
Fv =
(Hydrostatic Pressure - h is positive downwards from free surface)
⌠
⎮
p dA y
⎮
⌡
(Vertical Hydrostatic Force)
ΣM = 0
Assumptions:
H = αR
(Rotational Equilibrium)
h
(1) Static fluid
(2) Incompressible fluid
(3) Atmospheric pressure acts on free surface of the liquid.
dFV
θ
dF
dFH
The moments caused by the hydrostatic force and the weight of the cylinder about the hinge need to balance each other.
Integrating the hydrostatic pressure equation:
p = ρ ⋅ g⋅ h
dFv = dF⋅ cos ( θ) = p⋅ dA⋅ cos ( θ) = ρ⋅ g⋅ h⋅ w⋅ R⋅ dθ⋅ cos ( θ)
Now the depth to which the cylinder is submerged is
H = h + R⋅ ( 1 − cos ( θ) )
h = H − R⋅ ( 1 − cos ( θ) ) and into the vertical force equation:
Therefore
2 ⎡H
dFv = ρ⋅ g⋅ [ H − R⋅ ( 1 − cos ( θ) ) ] ⋅ w⋅ R⋅ cos ( θ) ⋅ dθ = ρ⋅ g⋅ w⋅ R ⋅ ⎢
⎣R
⎤
⎦
− ( 1 − cos ( θ) )⎥ ⋅ cos ( θ) ⋅ dθ
1 + cos ( 2⋅ θ)⎤
2
2
2⎡
dFv = ρ⋅ g⋅ w⋅ R ⋅ ⎡⎣( α − 1) ⋅ cos ( θ) + ( cos ( θ) ) ⎤⎦ ⋅ dθ = ρ⋅ g⋅ w⋅ R ⋅ ⎢( α − 1) ⋅ cos ( θ) +
⎥ ⋅ dθ
2
⎣
⎦
Now as long as α is not greater than 1, the net horizontal hydrostatic force will be zero due to symmetry, and the vertical force is:
θmax
⌠
Fv = ⎮
⌡− θ
max
θmax
⌠
1 dF v = ⎮
⌡0
2 dF v
where
(
)
cos θmax =
R−H
R
= 1−α
or
θmax = acos ( 1 − α)
2⌠
⎮
Fv = 2ρ⋅ g⋅ w⋅ R ⋅
θmax
⎮
⌡0
⎡( α − 1) ⋅ cos ( θ) + 1 + 1 ⋅ cos ( 2⋅ θ)⎤ dθ
⎢
⎥
2 2
⎣
⎦
Now upon integration of this expression we have:
2
Fv = ρ⋅ g⋅ w⋅ R ⋅ ⎡⎣acos ( 1 − α) − ( 1 − α) ⋅ α⋅ ( 2 − α)⎤⎦
The line of action of the vertical force due to the liquid is through the centroid of the displaced liquid, i.e., through the center of the cylinde
2
The weight of the cylinder is given by: W = M⋅ g = ρc⋅ V⋅ g = SG⋅ ρ⋅ π⋅ R ⋅ w⋅ g
where ρ is the density of the fluid and
SG =
ρc
ρ
The line of action of the weight is also throught the center of the cylinder. Taking moment about the hinge we get:
ΣMo = W⋅ R − Fv⋅ R = 0
2
or in other words
W = Fv
and therefore:
2
SG⋅ ρ⋅ π⋅ R ⋅ w⋅ g = ρ⋅ g⋅ w⋅ R ⋅ ⎡⎣acos ( 1 − α) − ( 1 − α) ⋅ α⋅ ( 2 − α)⎤⎦
SG =
1
π
⋅ ⎡⎣acos ( 1 − α) − ( 1 − α) ⋅ α⋅ ( 2 − α)⎤⎦
Specific Gravity, SG
0.6
0.4
0.2
0
0
0.5
alpha (H/R)
1
Problem 3.87
Given:
[Difficulty: 4]
Canoe, modeled as a right semicircular cylindrical shell, floats in water of depth d. The shell has outer radius R and leng
R = 1.2⋅ ft
L = 17⋅ ft
d = 1⋅ ft
Find:
(a) General expression for the maximum total mass that can be floated, as a function of depth,
(b) evaluate for the given conditions
(c) plot for range of water depth between 0 and R.
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
dp
dy
= ρ⋅ g
Fv =
Assumptions:
(Hydrostatic Pressure - y is positive downwards from
free surface)
⌠
⎮
p dA y
⎮
⌡
(Vertical Hydrostatic Force)
(1) Static fluid
(2) Incompressible fluid
(3) Atmospheric pressure acts on free surface of the liquid.
θmax
y is a function of θ for a given depth d:
y = d − ( R − R⋅ cos ( θ) ) = d − R + R⋅ cos ( θ)
θ
y
The maximum value of θ:
⎡ ( R − d) ⎤
⎥
⎣ R ⎦
d
dF
θmax = acos ⎢
A free-body diagram of the canoe gives: ΣFy = 0 = M⋅ g − Fv
where
is the vertical force of the water on the canoe.
Fv
θ
θ
max
max
⌠
⌠
⌠
⌠
⎮
⎮
Fv =
p dA y =
p⋅ cos ( θ) dA = ⎮
ρ⋅ g⋅ y⋅ L⋅ R⋅ cos ( θ) dθ = 2⋅ ρ⋅ g⋅ L⋅ R⋅ ⎮
( d − R + R⋅ cos ( θ) ) ⋅ cos ( θ) dθ
⎮
⎮
⌡0
⌡− θ
⌡
⌡
max
θmax
⌠
F v = 2⋅ ρ ⋅ g⋅ L ⋅ R ⋅ ⎮
⌡0
Since
For
M=
Fv
g
⎛ θmax sin (2⋅ θmax) ⎞⎤
⎡⎣( d − R) ⋅ cos ( θ) + R⋅ ( cos ( θ) ) 2⎤⎦ dθ = 2⋅ ρ⋅ g⋅ L⋅ R⋅ ⎡⎢( d − R) ⋅ sin θ
+
R
⋅
⎜
+
⎟⎥
(
)
max
4
⎣
⎝ 2
⎠⎦
it follows that
R = 1.2⋅ ft L = 17⋅ ft
and d = 1⋅ ft
⎡
(
)
⎛ θmax
M = 2⋅ ρ⋅ L⋅ R⋅ ⎢( d − R) ⋅ sin θmax + R⋅ ⎜
⎣
we can determine the mass:
⎝ 2
+
(
) ⎟⎥
sin 2⋅ θmax ⎞⎤
4
⎠⎦
⎡ ( 1.2 − 1)⎤
⎥
⎣ 1.2 ⎦
θmax = acos ⎢
θmax = 1.403⋅ rad
M = 2 × 1.94⋅
slug
ft
3
⎡
⎣
⎛ 1.403⋅ rad + sin ( 2 × 1.403⋅ rad) ⎞⎤ × 32.2⋅ lbm
⎟⎥
2
4
slug
⎝
⎠⎦
× 17⋅ ft × 1.2⋅ ft × ⎢( 1⋅ ft − 1.2⋅ ft) × sin ( 1.403⋅ rad) + 1.2⋅ ft × ⎜
M = 1895⋅ lbm
When we enter the values of d/R into the expressions for θmax and M, we get the following graph:
Mass versus Submersion Depth Ratio
Mass, M (kg)
1000
500
0
0
0.5
Submersion Depth Ratio (d/R)
1
Problem 3.88
Given:
Geometry of glass observation room
Find:
Resultant force and direction
Assumptions:
[Difficulty: 4]
Water in aquarium is static and incompressible
Solution:
The x, y and z components of force due to the fluid are treated separately. For the x, y components, the horizontal force is equivalent to that
on a vertical flat plate; for the z component, (vertical force) the force is equivalent to the weight of fluid above.
For horizontal forces, the computing equation of Section 3-5 is FH  pc A where A is the area of the equivalent vertical plate.
For the vertical force, the computing equation of Section 3-5 is FV  ρ g V where V is the volume of fluid above the curved surface.
The data are
ρ  1.94
For water
slug
ft
For the fluid (Table A.2)
SG  1.025
For the aquarium
R  5 ft
3
H  35 ft
(a) Horizontal Forces
Consider the x component
yc  H 
The center of pressure of the glass is
Hence


FHx  pc A  SG ρ g yc 
FHx  1.025  1.94
slug
ft
3
π R
4 R
3 π
yc  32.88 ft
2
4
 32.2
ft
2
 32.88 ft 
s
π ( 5 ft)
4
2
2

lbf  s
slug ft
4
FHx  4.13  10  lbf
The y component is of the same magnitude as the x component
4
FHy  FHx
FHy  4.13  10  lbf
The resultant horizontal force (at 45o to the x and y axes) is
FH 
2
FHx  FHy
2
4
FH  5.85  10  lbf
(b) Vertical forces
The vertical force is equal to the weight of fluid above (a volume defined by a rectangular column minus a segment of a sphere)
3
π R
2
The volume is
V 
H 
Then
FV  SG ρ g V
4
4 π R
3
8
V  621.8 ft
3
FV  1.025  1.94
slug
ft
3
 32.2
ft
2
s
3
 621.8 ft 
2
lbf  s
slug ft
4
FV  3.98  10  lbf
Finally the resultant force and direction can be computed
F 
2
FH  FV
 FV 

 FH 
α  atan 
2
4
F  7.07  10  lbf
α  34.3 deg
Note that α is the angle the resultant force makes with the horizontal
Problem *3.89
[Difficulty: 2]
Given:
Hydrometer as shown, submerged in nitric acid. When submerged in
water, h = 0 and the immersed volume is 15 cubic cm.
SG  1.5 d  6 mm
Find:
The distance h when immersed in nitric acid.
Solution:
We will apply the hydrostatics equations to this system.
Fbuoy  ρ g Vd
Governing Equations:
Assumptions:
(1) Static fluid
(2) Incompressible fluid
ΣFz  0 M g  Fbuoy  0
Taking a free body diagram of the hydrometer:
Solving for the mass of the hydrometer:
When immersed in water:
M  ρw  V w
M
Fbuoy
g
 ρ V d
When immersed in nitric acid:
Since the mass of the hydrometer is the same in both cases:
π 2
When the hydrometer is in the nitric acid: Vn  Vw   d  h
4
π 2
Therefore: ρw Vw  SG ρw  Vw   d  h
4




Vw  SG  Vw 
π 2 
 d  h
4

 SG  1   4
h  Vw 

 SG  π d2
(Buoyant force is equal to weight of displaced fluid)
ρ w  V w  ρ n V n
ρn  SG ρw
Solving for the height h:
Vw ( 1  SG)  SG
3
M  ρ n V n
π 2
d h
4
4
 1.5  1  
 10 mm 



2
 1.5  π  ( 6 mm)  cm 
h  15 cm  
3
h  177 mm
Problem *3.90
[Difficulty: 3]
Given:
Data on sphere and weight
Find:
SG of sphere; equilibrium position when freely floating
T
Solution:
Basic equation
FB
F B  ρ  g V
where
Hence
T  M g
M g  ρ g
V
2
γ 
Weight
Volume
ΣFz  0  T  FB  W
M  10 kg
 SG ρ g V  0
SG  10 kg 
The specific weight is
ΣFz  0
and
m
3
1000 kg

SG 
1

0.025 m
SG ρ g V
V
FB  ρ g

3
M
ρ V

V
W
1
2
1
SG  0.9
2
 SG ρ g
W  SG ρ g V
2
γ  0.9  1000
kg
3
 9.81
m
2
m
2

s
N s
kg m
γ  8829
W  FB
with
FB  ρ g Vsubmerged
Vsubmerged 
From references (trying Googling "partial sphere volume")
 3 V 

 4 π 
R
where h is submerged depth and R is the sphere radius
π h
2
3
1
3
 3  0.025 m3

 4 π

2
Hence
π h
W  SG ρ g V  FB  ρ g
 ( 3  R  h)
3
2
h  ( 3 0.181 m  h) 
3
3 0.9 .025 m
π
 ( 3  R  h)
1
3
R  
2
h  ( 3  R  h) 
R  0.181 m
3 SG V
π
2
h  ( 0.544  h)  0.0215
This is a cubic equation for h. We can keep guessing h values, manually iterate, or use Excel's Goal Seek to find
3
m
For the equilibriul position when floating, we repeat the force balance with T = 0
FB  W  0
N
h  0.292 m
Problem *3.91
[Difficulty: 2]
Given:
Specific gravity of a person is to be determined from measurements of weight in air and the met weight when
totally immersed in water.
Find:
Expression for the specific gravity of a person from the measurements.
Solution:
We will apply the hydrostatics equations to this system.
Fbuoy  ρ g Vd
Governing Equation:
Assumptions:
(Buoyant force is equal to weight of displaced fluid)
(1) Static fluid
(2) Incompressible fluid
Fnet
Taking a free body diagram of the body:
Fnet
ΣFy  0
Fnet  M g  Fbuoy  0
Fbuoy
is the weight measurement for the immersed body.
Fnet  M g  Fbuoy  M g  ρw g Vd
Therefore the weight measured in water is:
However in air:
Fair  M g
Fnet  Fair  ρw g Vd
and
Vd 
Fair  Fnet
Mg
ρw g
Now in order to find the specific gravity of the person, we need his/her density:
Fair  M g  ρ g Vd  ρ g
Fair  Fnet
ρw  g
Now if we call the density of water at 4 deg C

ρ
Simplifying this expression we get: Fair 
F  Fnet
ρw air
ρw4C
then:

 ρ 
ρ

SG
 w4C  F  F
Fair 

air
net  SG  Fair  Fnet
ρ
 w 
w


 ρw4C 
Solving this expression for the specific gravity of the person SG, we get:
SG  SGw
F
Fair
air  Fnet
Problem *3.92
[Difficulty: 2]
Given:
Iceberg floating in seawater
Find:
Quantify the statement, "Only the tip of an iceberg shows (in seawater)."
Solution:
We will apply the hydrostatics equations to this system.
Fbuoy  ρ g Vd
Governing Equations:
Assumptions:
(Buoyant force is equal to weight of displaced fluid)
(1) Static fluid
(2) Incompressible fluid
Taking a free body diagram of the iceberg:
M g  Fbuoy  ρsw g Vd
Combining these expressions:
ΣFz  0 M g  Fbuoy  0
M  ρice Vtot
But the mass of the iceberg is also:
ρice Vtot g  ρsw g Vd
The volume of the iceberg above the water is:
Therefore we may define a volume fraction:
Mg
ρice
SGice
Fbuoy
Vd  Vtot
 Vtot
ρsw
SGsw
 SGice 
Vshow  Vtot  Vd  Vtot  1 

 SGsw 
VF 
Vshow
Vtot
 1
Substituting in data from Tables A.1 and A.2 we get: VF  1 
SGice
SGsw
0.917
1.025
VF  0.1054
Only 10% of the iceberg is above water
Problem *3.93
[Difficulty: 2]
Given:
Geometry of steel cylinder
Find:
Volume of water displaced; number of 1 kg wts to make it sink
Solution:
The data is
For water
ρ  999
kg
3
m
For steel (Table A.1)
SG  7.83
For the cylinder
D  100 mm
The volume of the cylinder is
Vsteel  δ 
The weight of the cylinder is
W  SG ρ g Vsteel
H  1 m
 π D 2

 π D  H 
 4
Vsteel  3.22  10

kg
W  7.83  999
3
 9.81
m
δ  1 mm
m
2
 3.22  10
4
3
m 
s
4
3
m
2
N s
W  24.7 N
kg m
At equilibium, the weight of fluid displaced is equal to the weight of the cylinder
Wdisplaced  ρ g Vdisplaced  W
Vdisplaced 
W
ρ g
3
 24.7 N 
m
999 kg
2

s
9.81 m

kg m
Vdisplaced  2.52 L
2
N s
To determine how many 1 kg wts will make it sink, we first need to find the extra volume that will need to be dsiplaced
Distance cylinder sank
x1 
Vdisplaced
x1  0.321 m
 π D 2 


 4 
x2  H  x1
Hence, the cylinder must be made to sink an additional distance
We deed to add n weights so that
1 kg n g  ρ g
π D
4
x2  0.679 m
2
 x2
2
n
ρ  π  D  x2
4  1 kg
Hence we need n  6 weights to sink the cylinder
 999
kg
3
m

π
4
2
 ( 0.1 m)  0.679 m 
1
1 kg
2

N s
kg m
n  5.33
Problem *3.94
[Difficulty: 2]
Given:
Experiment performed by Archimedes to identify the material conent of King
Hiero's crown. The crown was weighed in air and in water.
Find:
Expression for the specific gravity of the crown as a function of the weights in water and air.
Solution:
We will apply the hydrostatics equations to this system.
F b  ρ  g V d
(Buoyant force is equal to weight of displaced fluid)
(1) Static fluid
(2) Incompressible fluid
Ww
Governing Equations:
Assumptions:
ΣFz  0
Taking a free body diagram of the body:
Ww  M g  Fb  0
Ww is the weight of the crown in water.
Mg
Ww  M g  Fbuoy  M g  ρw g Vd
However in air:
Therefore the weight measured in water is:
so the volume is:
Vd 
Wa  Ww
ρw  g
Wa  M g
Fb
Ww  Wa  ρw g Vd
M ρw g
Wa
M
Now the density of the crown is: ρc 


ρ
Vd
Wa  Ww
Wa  Ww w
Therefore, the specific gravity of the crown is:
SG 
ρc
ρw

Wa
Wa  Ww
SG 
Wa
Wa  Ww
Note: by definition specific gravity is the density of an object divided by the density of water at 4 degrees Celsius, so the measured
temperature of the water in the experiment and the data from tables A.7 or A.8 may be used to correct for the variation in density of the
water with temperature.
Problem *3.95
[Difficulty: 2]
Open-Ended Problem Statement: Gas bubbles are released from the regulator of a submerged
Scuba diver. What happens to the bubbles as they rise through the seawater?
Discussion: Air bubbles released by a submerged diver should be close to ambient pressure at the
depth where the diver is swimming. The bubbles are small compared to the depth of submersion, so each
bubble is exposed to essentially constant pressure. Therefore the released bubbles are nearly spherical in
shape.
The air bubbles are buoyant in water, so they begin to rise toward the surface. The bubbles are quite light,
so they reach terminal speed quickly. At low speeds the spherical shape should be maintained. At higher
speeds the bubble shape may be distorted.
As the bubbles rise through the water toward the surface, the hydrostatic pressure decreases. Therefore the
bubbles expand as they rise. As the bubbles grow larger, one would expect the tendency for distorted
bubble shape to be exaggerated.
Problem *3.96
[Difficulty: 2]
Given:
Balloons with hot air, helium and hydrogen. Claim lift per cubic foot of 0.018, 0.066, and 0.071 pounds force per cubic f
for respective gases, with the air heated to 150 deg. F over ambient.
Find:
(a) evaluate the claims of lift per unit volume
(b) determine change in lift when air is heated to 250 deg. F over ambient.
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
Assumptions:
L  ρa g V  ρg g V
(Net lift force is equal to difference in weights of air and gas)
p  ρ R  T
(Ideal gas equation of state)
(1) Static fluid
(2) Incompressible fluid
(3) Ideal gas behavior
The lift per unit volume may be written as: LV 
 ρg 
 g ρa  ρg  ρa g  1 

V
 ρa 
L


we take into account that the pressure inside and outside the balloon are equal:
lbf
At standard conditions the specific weight of air is: γa  0.0765
3
ft
Rg  386.1
For helium:
For hydrogen:
ft lbf
lbm R
Rg  766.5
Tg  Ta
ft lbf
Tg  Ta
lbm R
and therefore:
now if we take the ideal gas equation and
Ra Ta 
Ra Ta 


L
 ρa g  1 
  γa  1 

V
 R g T g 
 R g T g 
the gas constant is:
Ra  53.33
ft lbf
lbm R
lbf 
53.33 
LVHe  0.0765
 1 

3 
386.1 
ft
and therefore:
and
Ta  519 R
lbf
LVHe  0.0659
3
ft
lbf 
53.33 
LVH2  0.0765
 1 

3 
766.5 
ft
lbf
LVH2  0.0712
3
ft
For hot air at 150 degrees above ambient:
Rg  Ra
Tg  Ta  150 R and therefore:
lbf 
519 
lbf
LVair150  0.0765
 1 
 LVair150  0.0172 3
3 
519  150 
ft
ft
The agreement with the claims stated above is good.
For hot air at 250 degrees above ambient:
Rg  Ra
LVair250
LVair150
Tg  Ta  250 R and therefore:
 1.450
lbf 
519 
LVair250  0.0765
 1 

3 
519  250 
ft
lbf
LVair250  0.0249
3
ft
Air at ΔT of 250 deg. F gives 45% more lift than air at ΔT of 150 deg.F!
Problem *3.97
[Difficulty: 2]
V
FB
y
FD
W = Mg
Given:
Data on hydrogen bubbles
Find:
Buoyancy force on bubble; terminal speed in water
Solution:
Basic equation
F B  ρ  g V  ρ  g
FB  1.94
slug
ft
3
π 3
d
6
 32.2
FB  FD  W  0
Hence
V 
2

π
6
2
1 ft 
lbf  s
 
12 in 
slug ft
 11
FB  1.89  10
 5 lbf  s

2
from Table A.7 at 68oF
ft
V  1.89  10
 11
 lbf 
 3 ft

s
1
3 π

1

2.10  10
V  0.825
ft
2
 5 lbf  s
for terminal speed
 lbf
where we have ignored W, the weight of the bubble (at
STP most gases are about 1/1000 the density of water)
μ  2.10  10
with
3 π  μ  d
ΣFy  0  FB  FD  W
3


  0.001 in 
F D  3 π  μ  V  d  F B
FB
V  1.15  10
ft
s
For terminal speed
ΣFy  M ay
and

1
0.001 in
in
min
As noted by Professor Kline in the film "Flow Visualization", bubbles rise slowly!

12 in
1 ft
Problem *3.98
[Difficulty: 3]
Given:
Data on hot air balloon
Find:
Maximum mass of balloon for neutral buoyancy; mass for initial acceleration of 2.5 ft/s2.
Assumptions:
Fbuoyancy
Whot air
Air is treated as static and incompressible, and an ideal gas
Solution:
y
Basic equation
Hence
FB  ρatm g V
ΣFy  M ay
and
Wload
ΣFy  0  FB  Whotair  Wload  ρatm g V  ρhotair g V  M g


M  V ρatm  ρhotair 
3
M  320000 ft  14.7
lbf
2
in
V patm
R
 1


Tatm

1
for neutral buoyancy

Thotair 

2
1
1
 12 in   lbm R  





53.33

ft

lbf
(
160

460
)

R
ft
(
48

460
)

R








M  4517 lbm

Initial acceleration
ΣFy  FB  Whotair  Wload  ρatm  ρhotair  g V  Mnew g  Maccel a  Mnew  2 ρhotair V  a
Solving for Mnew
ρatm  ρhotair g V  Mnew g  Mnew  2 ρhotair V a
Mnew  V
ρatm  ρhotair g  2 ρhotair a  V patm  g 
1
1 
2 a 






ag
  Tatm Thotair  Thotair
ag
2
2
lbf  12 in 
lbm R
s
1
1
1
3



 ft
Mnew  320000 ft  14.7


 32.2 

 
  2 2.5
 2
2  ft  53.33 ft lbf ( 2.5  32.2)  ft 
( 160  460)
 ( 48  460) ( 160  460)
in
s R
Mnew  1239 lbm
To make the balloon move up or down during flight, the air needs to be heated to a higher temperature, or let cool (or let in ambient air).
Problem *3.99
[Difficulty: 4]
Given:
Spherical balloon filled with helium lifted a payload of mass M=230 kg.
At altitude, helium and air were in thermal equilibrium. Balloon diameter is
120 m and specific gravity of the skin material is 1.28.
Find:
The altitude to which the balloon rose.
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
Assumptions:
Fbuoy  ρ g Vd
(Buoyant force is equal to mass of displaced fluid)
p  ρ R  T
(Ideal gas equation of state)
(1) Static, incompressible fluid
(2) Static equilibrium at 49 km altitude
(3) Ideal gas behavior
D
t
Taking a free body diagram of the balloon and payload: ΣFz  Fbuoy  MHe g  Ms g  M g  0
z
M
Substituting for the buoyant force and knowing that mass is density times volume:
ρair g Vb  ρHe g Vb  ρs g Vs  M g  0
The volume of the balloon:
ρair  ρHe 
p
T

6
π D
p 6
 
T π
3
6
π D
3
π
Vb 
6
2

  π ρs t D  M 
2

1
( 120 m)
3
1
kg
3
m
2
V s  π D  t
Substituting these into the force equation:
From the ideal gas equation of state and remembering that pressure and temperature of the air
and helium are equal:
Substituting known values and consulting Appendix A for gas constants:
  1
1 
R  R 
He 
 air
 π  1280

3
 D The volume of the skin:
  π ρs t D  M

ρair Vb  ρHe Vb  ρs Vs  M  0
 0.013 10
 m  ( 120 m)  230 kg 
3

2
1
2
1
287

1

N m Pa m
 4 kPa

 3.616  10 
kg K
N
K
2080
To determine the altitude, we need to check this ratio against data from Table A.3. We find that
the ratio of pressure to temperature matches the result above at:
h  48.3 km
Problem *3.100
[Difficulty: 3]
Given:
A pressurized balloon is to be designed to lift a payload of mass M to an altitude of 40 km, where p = 3.0 mbar
and T = -25 deg C. The balloon skin has a specific gravity of 1.28 and thickness 0.015 mm. The gage pressure of
the helium is 0.45 mbar. The allowable tensile stress in the balloon is 62 MN/m2
Find:
(a) The maximum balloon diameter
(b) The maximum payload mass
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
Assumptions:
t
D
Fbuoy = ρ⋅ g ⋅ Vd
(Buoyant force is equal to mass
of displaced fluid)
p = ρ⋅ R⋅ T
(Ideal gas equation of state)
M
(1) Static, incompressible fluid
(2) Static equilibrium at 40 km altitude
(3) Ideal gas behavior
πD tσ
The diameter of the balloon is limited by the allowable tensile stress in the skin:
ΣF =
π
4
2
⋅ D ⋅ ∆p − π⋅ D⋅ t⋅ σ = 0
−3
Dmax = 4 × 0.015 × 10
4⋅ t⋅ σ
Dmax =
∆p
Solving this expression for the diameter:
6 N
⋅ m × 62 × 10 ⋅
2
m
2
1
×
−3
0.45⋅ 10
πD 2∆p/4
×
⋅ bar
bar ⋅ m
Fbuoyant
Dmax = 82.7m
5
10 ⋅ N
z
To find the maximum allowable payload we perform a force balance on the system:
ΣFz = Fbuoy − M He⋅ g − M b ⋅ g − M ⋅ g = 0
Solving for M:
(
The air density:
ρa ⋅ g ⋅ Vb − ρHe⋅ g ⋅ Vb − ρs ⋅ g ⋅ Vs − M ⋅ g = 0
Mg
)
M = ρa − ρHe ⋅ Vb − ρs ⋅ Vs
The volume of the skin is:
2
Vs = π⋅ D ⋅ t
pa
ρa =
Ra⋅ T
Repeating for helium:
The payload mass is:
M =
6
⋅ bar ×
287⋅ N ⋅ m
×
π
6
1
( 273 − 25) ⋅ K
(
)
3
2
⋅ ρa − ρHe ⋅ D − π⋅ ρs⋅ D ⋅ t
5
×
10 ⋅ N
2
bar ⋅ m
− 3 kg
ρa = 4.215 × 10
3
m
− 4 kg
ρHe = 6.688 × 10
3
m
− 3 kg
× ( 4.215 − 0.6688) × 10
M = 638 kg
kg⋅ K
3
Vb = ⋅ D
6
M=
Therefore, the mass is:
−3
π
π
The volume of the balloon is:
ρa = 3.0 × 10
p
ρHe =
R⋅ T
M bg
⋅
3
m
3
3 kg
× ( 82.7⋅ m) − π × 1.28 × 10 ⋅
3
m
2
−3
× ( 82.7⋅ m) × 0.015 × 10
⋅m
Problem *3.101
[Difficulty: 3]
Given: Geometry of block and rod
F hinge,y
(L + c)/2
Find:
Angle for equilibrium
Fhinge,x
L/2
c
θ
Assumptions: Water is static and incompressible
a
F BR
Solution:
F BB
ΣM Hinge = 0
Basic
equations
FB = ρ⋅ g ⋅ V
WR
(Buoyancy)
L
WB
The free body diagram is as shown. FBB and F BR are the buoyancy of the
block and rod, respectively; c is the (unknown) exposed length of the rod
Taking moments about the hinge
( L + c)
(WB − FBB)⋅ L⋅ cos( θ) − FBR⋅
with
WB = M B⋅ g
FBB = ρ⋅ g ⋅ VB
(
MB =
(
2⋅ L
1
2
2
⋅ L −c
× 1000⋅
M B = 29.1 kg
2
) + ρ⋅VB − 12 ⋅MR
kg
3
m
2
ρ⋅ A⋅ L − c
We can solve for M B
ρ⋅ A
FBR = ρ⋅ g ⋅ ( L − c) ⋅ A
(MB − ρ⋅ VB)⋅ L − ρ⋅ A⋅ ( L − c)⋅
Combining equations
MB =
2
L
⋅ cos( θ) + WR⋅ ⋅ cos( θ) = 0
2
2
× 20⋅ cm ×
2
( L + c)
2
WR = M R⋅ g
L
+ MR⋅ = 0
2
) = 2⋅⎛⎜ MB − ρ⋅VB + 12 ⋅MR⎞ ⋅L
⎝
and since
⎠
c=
2
⎛ m ⎞ × 1 ⋅ ⎡⎢( 5⋅ m) 2 −
⎜
5⋅ m ⎣
⎝ 100 ⋅ cm ⎠
a
sin( θ)
MB =
ρ⋅ A
2⋅ L
⎡
⋅ ⎢L −
⎣
2
2
⎛ a ⎞ ⎤⎥ + ρ⋅ V − 1 ⋅ M
⎜ sin( θ)
B 2 R
⎝
⎠⎦
2
⎛ 0.25⋅ m ⎞ ⎥⎤ + 1000⋅ kg × 0.025 ⋅ m3 − 1 × 1.25⋅ kg
⎜
2
3
⎝ sin( 12⋅ deg) ⎠ ⎦
m
Problem *3.102
[Difficulty: 3]
Given:
Glass hydrometer used to measure SG of liquids. Stem has diameter D=5 mm, distance between marks on stem is
d=2 mm per 0.1 SG. Hydrometer floats in kerosene (Assume zero contact angle between glass and kerosene).
Find:
Magnitude of error introduced by surface tension.
Solution:
We will apply the hydrostatics equations to this system.
Fbuoy = ρ⋅ g ⋅ Vd
Governing Equations:
Assumptions:
(Buoyant force is equal to weight of displaced fluid)
D = 5 mm
(1) Static fluid
(2) Incompressible fluid
(3) Zero contact angle between ethyl alcohol and glass
The surface tension will cause the hydrometer to sink ∆h lower into the liquid. Thus for
this change:
ΣFz = ∆Fbuoy − Fσ = 0
∆Fbuoy = ρ⋅ g ⋅ ∆V = ρ⋅ g ⋅
The change in buoyant force is:
ρ⋅ g ⋅
π
4
Solving for ∆h:
2
⋅ D ⋅ ∆h = π⋅ D⋅ σ
∆h =
4
y
Fσ
Kerosene
2
⋅ D ⋅ ∆h
∆F B
Fσ = π⋅ D⋅ σ⋅ cos( θ) = π⋅ D⋅ σ
The force due to surface tension is:
Thus,
π
d=
2 mm/0.1 SG
ρ⋅ g ⋅ D⋅ ∆h
Upon simplification:
4
4⋅ σ
=σ
From Table A.2, SG = 1.43 and from Table A.4, σ = 26.8 mN/m
ρ⋅ g ⋅ D
−3 N
Therefore, ∆h = 4 × 26.8 × 10
⋅
m
3
×
m
1430⋅ kg
2
×
s
9.81⋅ m
So the change in specific gravity will be: ∆SG = 1.53 × 10
1
×
−3
−3
5 × 10
⋅m ×
×
⋅m
kg⋅ m
2
−3
m
s ⋅N
0.1
∆SG = 0.0765
−3
2 × 10
∆h = 1.53 × 10
⋅m
From the diagram, surface tension acts to cause the hydrometer to float lower in the liquid. Therefore, surface tension results in an
indicated specific gravity smaller than the actual specific gravity.
Problem *3.103
[Difficulty:4]
Given:
Sphere partially immersed in a liquid of specific gravity SG.
Find:
(a) Formula for buoyancy force as a function of the submersion depth d
(b) Plot of results over range of liquid depth
Solution:
We will apply the hydrostatics equations to this system.
Fbuoy = ρ⋅ g ⋅ Vd
Governing Equations:
Assumptions:
(Buoyant force is equal to weight of displaced fluid)
(1) Static fluid
(2) Incompressible fluid
(3) Atmospheric pressure acts everywhere
d
We need an expression for the displaced volume of fluid at an arbitrary
depth d. From the diagram we see that:
(
(
))
d = R 1 − cos θmax
at an arbitrary depth h:
h = d − R⋅ ( 1 − cos( θ) )
R
dθ
r = R⋅ sin( θ)
Rsin θ
So if we want to find the volume of the submerged portion of the sphere we calculate:
θmax
θ
θ
⌠ max 2
⌠ max 2
3
2
3⌠
( sin( θ) ) dθ
R ⋅ ( sin( θ) ) ⋅ R⋅ sin( θ) dθ = π⋅ R ⋅ ⎮
Vd = ⎮
π r dh = π⋅ ⎮
⌡
⌡
⌡
Evaluating the integral we get:
0
0
0
h
θmax
⎡⎢ cos θ
( ( max))3
Vd = π⋅ R ⋅ ⎢
− cos( θmax) +
3
⎣
3
⎤
3
⎡
d
d⎞
2⎤
d
3 1⎛
we
get:
Now
since:
⎢
cos
θ
=
1
−
V
π
⋅
R
1
−
=
⋅
− ⎛⎜ 1 − ⎞ + ⎥
(
)
⎜
⎥
max
d
3⎦
R
R⎠
R ⎠ 3⎦
⎣3 ⎝
⎝
2⎥
⎡
d
2⎤
d
3 1
Fbuoy = ρw⋅ SG⋅ g ⋅ π⋅ R ⋅ ⎢ ⋅ ⎛⎜ 1 − ⎞ − ⎜⎛ 1 − ⎞ + ⎥
R⎠
R ⎠ 3⎦
⎣3 ⎝
⎝
3
Thus the buoyant force is:
If we non-dimensionalize by the force on a fully submerged sphere:
Fd =
Fbuoy
4
3
ρw⋅ SG⋅ g ⋅ ⋅ π⋅ R
3
=
3 ⎡1
3 ⎡1
d
2⎤
d
Fd = ⎢ ⋅ ⎛⎜ 1 − ⎞ − ⎜⎛ 1 − ⎞ + ⎥
4 ⎣3 ⎝
R⎠
R ⎠ 3⎦
⎝
3
⎤
d
⎢ ⋅ ⎛⎜ 1 − ⎞ − ⎜⎛ 1 − d ⎞ + 2⎥
4 ⎣3 ⎝
R⎠
R ⎠ 3⎦
⎝
3
Force Ratio Fd
1.0
0.5
0.0
0.0
0.5
1.0
Submergence Ratio d/R
1.5
2.0
Problem 3.104
[Difficulty: 2]
Given: Geometry of rod
Find:
(L + c)/2
How much of rod is submerged; force to lift rod out of water
L/2
c
Solution:
Basic equations
θ
ΣM Hinge = 0
FB = ρ⋅ g ⋅ V
(Buoyancy)
FBR
WR
The free body diagram is as shown. FBR is the buoyancy of the rod; c is
the (unknown) exposed length of the rod
L
Taking moments about the hinge
−FBR⋅
( L + c)
2
L
⋅ cos( θ) + WR⋅ ⋅ cos( θ) = 0
2
with
FBR = ρ⋅ g ⋅ ( L − c) ⋅ A
Hence
−ρ⋅ A⋅ ( L − c) ⋅
We can solve for c
(
2
2
2
L⋅ M R
L −
ρ⋅ A
3
2
c =
L
+ M R⋅ = 0
2
) = MR⋅ L
2
ρ⋅ A⋅ L − c
c=
( L + c)
WR = M R⋅ g
( 5 ⋅ m) − 5 ⋅ m ×
m
1000⋅ kg
×
1
⋅
20
2
1
cm
2
⎛ 100 ⋅ cm ⎞ × 1.25⋅ kg
⎜ 1⋅ m
⎝
⎠
×
c = 4.68 m
Then the submerged length is
L − c = 0.323 m
To lift the rod out of the water requires a force equal to half the rod weight (the reaction also takes half the weight)
F=
1
2
⋅ MR⋅ g =
1
2
× 1.25⋅ kg × 9.81⋅
m
2
s
2
×
N⋅ s
kg⋅ m
F = 6.1 N
a
Problem *3.105
[Difficulty: 2]
FB
y
x
H = 60 cm
W
θ
h = 5 cm
Given:
Data on river
Find:
Largest diameter of log that will be transported
Solution:
Basic equation
FB = ρ⋅ g ⋅ Vsub
ΣFy = 0
where
FB = ρ⋅ g ⋅ Vsub = ρ⋅ g ⋅ Asub ⋅ L
and
ΣFy = 0 = FB − W
W = SG⋅ ρ⋅ g ⋅ V = SG⋅ ρ⋅ g ⋅ A ⋅ L
2
R
Asub =
⋅ ( θ − sin ( θ) )
2
From references (e.g. CRC Mathematics Handbook)
Hence
ρ⋅ g ⋅
R
2
2
where R is the radius and θ is
the included angle
2
⋅ ( θ − sin ( θ) ) ⋅ L = SG⋅ ρ⋅ g ⋅ π⋅ R ⋅ L
θ − sin( θ) = 2 ⋅ SG ⋅ π = 2 × 0.8 × π
This equation can be solved by manually iterating, or by using a good calculator, or by using Excel's Goal Seek
θ = 239 ⋅ deg
From geometry the submerged amount of a log is H − h
Hence
H − h = R + R⋅ cos⎛⎜ π −
Solving for R
R=
⎝
R + R⋅ cos⎛⎜ π −
⎝
θ⎞
2⎠
θ⎞
2⎠
H− h
θ⎞
1 + cos⎛⎜ 180deg −
2⎠
⎝
D = 2⋅ R
and also
R =
( 0.6 − 0.05) ⋅ m
239 ⎞
⋅ deg⎥⎤
1 + cos⎡⎢⎜⎛ 180 −
2 ⎠
⎣⎝
⎦
D = 0.737 m
R = 0.369 m
Problem *3.106
[Difficulty: 4]
Given:
Data on sphere and tank bottom
Find:
Expression for SG of sphere at which it will float to surface;
minimum SG to remain in position
y
FU
FB
x
Assumptions: (1) Water is static and incompressible
(2) Sphere is much larger than the hole at the bottom of the tank
Solution:
FL
FB = ρ⋅ g ⋅ V
Basic equations
and
FL = p atm⋅ π⋅ a
where
ΣFy = FL − FU + FB − W
2
FU = ⎡p atm + ρ⋅ g ⋅ ( H − 2 ⋅ R)⎤ ⋅ π⋅ a
⎣
⎦
2
4
3
2
Vnet = ⋅ π⋅ R − π⋅ a ⋅ 2 ⋅ R
3
FB = ρ⋅ g ⋅ Vnet
W = SG ⋅ ρ⋅ g ⋅ V
W
V=
with
4
3
⋅ π⋅ R
3
Now if the sum of the vertical forces is positive, the sphere will float away, while if the sum is zero or negative the sphere will stay
at the bottom of the tank (its weight and the hydrostatic force are greater than the buoyant force).
Hence
4
4
2
2
3
2
3
ΣFy = p atm⋅ π⋅ a − ⎡p atm + ρ⋅ g ⋅ ( H − 2 ⋅ R)⎤ ⋅ π⋅ a + ρ⋅ g ⋅ ⎜⎛ ⋅ π⋅ R − 2 ⋅ π⋅ R⋅ a ⎞ − SG⋅ ρ⋅ g ⋅ ⋅ π⋅ R
⎣
⎦
3
3
⎝
4 3
2
ΣFy = π⋅ ρ⋅ g ⋅ ⎢⎡( 1 − SG ) ⋅ ⋅ R − H⋅ a ⎥⎤
3
⎣
⎦
This expression simplifies to
ΣFy = π × 1.94⋅
slug
ft
ΣFy = −0.012 ⋅ lbf
⎠
3
× 32.2⋅
ft
2
s
×
2
3
2⎤
⎡4
⎢ × ( 1 − 0.95) × ⎛⎜ 1⋅ in × ft ⎞ − 2.5⋅ ft × ⎛⎜ 0.075 ⋅ in × ft ⎞ ⎥ × lbf ⋅ s
12⋅ in ⎠
12⋅ in ⎠ ⎦ slug⋅ ft
⎣3
⎝
⎝
Therefore, the sphere stays at the bottom of the tank.
Problem *3.107
[Difficulty: 4]
Given:
Cylindrical timber, D = 1 ft and L = 15 ft, is weighted on the lower end so that is floats vertically with 10 ft
submerged in sea water. When displaced vertically from equilibrium, the timber oscillates in a vertical direction
upon release.
Find:
Estimate the frequency of the oscillation. Neglect viscous forces or water motion.
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
Assumptions:
Fbuoy  ρ g  Vd
(1) Static fluid
(2) Incompressible fluid
(3) Atmospheric pressure acts everywhere
(4) Viscous effects and water motion are negligible.
ΣFy  Fbuoy  M  g  0
At equilibrium:
(Buoyant force is equal to weight of displaced fluid)
D = 1 ft
M  ρ Vd  ρ A  d
2
ΣFy  Fbuoy  M  g  M 
Once the timber is displaced:
d y
dt
2
2
ρ g  A ( d  y )  M  g  M 
d y
2
2
ρ g  A  d  ρ g  A  y  ρ A  d  g  M 
dt
d y
dt
2
Thus we have the equation:
M
2
2
d y
dt
 ρ g  A  y  0
2
or:
d y
2
dt

ρ g  A
ρ A d
ω
g
d
To express this as a frequency:
ω 
32.2 ft
s
f 
ω
2 π
2

2
y  0
d y
2

dt
2
This ODE describes simple harmonic motion with the natural frequency ω described by:
Solving for ω:
d =10 ft
(Equilibrium
Depth)
L
ω 
g
d
y  0
g
d
1
ω  1.7944
10 ft
1.7944
f 
2 π
rad
1
s
f  0.286 Hz
s
Problem *3.108
[Difficulty: 3]
Given:
Data on boat
Find:
Effective density of water/air bubble mix if boat sinks
Floating
H = 8 ft
Solution:
Basic equations
Sinking
h = 7 ft
FB = ρ⋅ g ⋅ V
ΣFy = 0
and
θ = 60 o
We can apply the sum of forces for the "floating" free body
ΣFy = 0 = FB − W
FB = SGsea⋅ ρ⋅ g ⋅ Vsubfloat
where
2
1
2⋅ h ⎞
L⋅ h
Vsubfloat = ⋅ h ⋅ ⎛⎜
⋅L =
2 ⎝ tan⋅ θ ⎠
tan( θ)
Hence
W=
SGsea⋅ ρ⋅ g ⋅ L⋅ h
SGsea = 1.024
(Table A.2)
2
(1)
tan( θ)
We can apply the sum of forces for the "sinking" free body
2
ΣFy = 0 = FB − W
1
2⋅ H ⎞
L⋅ H
Vsubsink = ⋅ H⋅ ⎛⎜
⋅L =
2 ⎝ tan⋅ θ ⎠
tan( θ)
FB = SGmix⋅ ρ⋅ g ⋅ Vsub
where
2
Hence
W=
Comparing Eqs. 1 and 2
SGmix⋅ ρ⋅ g ⋅ L⋅ H
(2)
tan( θ)
W=
SGsea⋅ ρ⋅ g ⋅ L⋅ h
2
tan( θ)
h
SGmix = SGsea ⋅ ⎛⎜ ⎞
H
⎝ ⎠
The density is
ρmix = SGmix⋅ ρ
2
=
SGmix⋅ ρ⋅ g ⋅ L⋅ H
tan( θ)
2
SGmix = 1.024 ×
⎛7⎞
⎜8
⎝ ⎠
ρmix = 0.784 × 1.94⋅
2
SGmix = 0.784
slug
ft
3
ρmix = 1.52⋅
slug
ft
3
Problem *3.109
[Difficulty: 2]
FB
F
20 cm
D = 10 cm
8 cm
2 cm
W
y
x
Given:
Data on inverted bowl and dense fluid
Find:
Force to hold in place
Assumption: Fluid is static and incompressible
Solution:
Basic equations
FB = ρ⋅ g ⋅ V
Hence
F = FB − W
For the buoyancy force
FB = SG fluid ⋅ ρH2O⋅ g ⋅ Vsub
For the weight
W = SG bowl⋅ ρH2O⋅ g ⋅ Vbowl
Hence
(
and
ΣFy = 0
ΣFy = 0 = FB − F − W
with
Vsub = Vbowl + Vair
)
F = SGfluid⋅ ρH2O⋅ g ⋅ Vbowl + Vair − SG bowl⋅ ρH2O⋅ g ⋅ Vbowl
(
)
F = ρH2O⋅ g ⋅ ⎡SG fluid⋅ Vbowl + Vair − SGbowl⋅ Vbowl⎤
⎣
⎦
3
2
3 ⎞⎤
2
⎡
⎛
⎡
m
π⋅ ( 0.1⋅ m) ⎤
⎥ − 5.7 × ⎜ 0.9⋅ L × m ⎥ × N ⋅ s
F = 999⋅
× 9.81⋅
+ ( 0.08 − 0.02) ⋅ m⋅
× ⎢15.6 × ⎢0.9⋅ L ×
3
2 ⎣
1000⋅ L
4
1000⋅ L ⎠⎦ kg⋅ m
⎣
⎦
⎝
m
s
kg
F = 159.4 N
m
Problem *3.110
[Difficulty: 4]
Open-Ended Problem Statement: In the “Cartesian diver” child's toy, a miniature “diver” is
immersed in a column of liquid. When a diaphragm at the top of the column is pushed down, the diver
sinks to the bottom. When the diaphragm is released, the diver again rises. Explain how the toy might
work.
Discussion: A possible scenario is for the toy to have a flexible bladder that contains air. Pushing
down on the diaphragm at the top of the liquid column would increase the pressure at any point in the
liquid. The air in the bladder would be compressed slightly as a result. The volume of the bladder, and
therefore its buoyancy, would decrease, causing the diver to sink to the bottom of the liquid column.
Releasing the diaphragm would reduce the pressure in the water column. This would allow the bladder to
expand again, increasing its volume and therefore the buoyancy of the diver. The increased buoyancy
would permit the diver to rise to the top of the liquid column and float in a stable, partially submerged
position, on the surface of the liquid.
Problem *3.111
[Difficulty: 4]
Open-Ended Problem Statement: Consider a conical funnel held upside down and submerged
slowly in a container of water. Discuss the force needed to submerge the funnel if the spout is open to the
atmosphere. Compare with the force needed to submerge the funnel when the spout opening is blocked by a
rubber stopper.
Discussion: Let the weight of the funnel in air be Wa. Assume the funnel is held with its spout vertical
and the conical section down. Then Wa will also be vertical.
Two possible cases are with the funnel spout open to atmosphere or with the funnel spout sealed.
With the funnel spout open to atmosphere, the pressures inside and outside the funnel are equal, so no net
pressure force acts on the funnel. The force needed to support the funnel will remain constant until it first
contacts the water. Then a buoyancy force will act vertically upward on every element of volume located
beneath the water surface.
The first contact of the funnel with the water will be at the widest part of the conical section. The buoyancy
force will be caused by the volume formed by the funnel thickness and diameter as it begins to enter the
water. The buoyancy force will reduce the force needed to support the funnel. The buoyancy force will
increase as the depth of submergence of the funnel increases until the funnel is fully submerged. At that
point the buoyancy force will be constant and equal to the weight of water displaced by the volume of the
material from which the funnel is made.
If the funnel material is less dense than water, it would tend to float partially submerged in the water. The
force needed to support the funnel would decrease to zero and then become negative (i.e., down) to fully
submerge the funnel.
If the funnel material were denser than water it would not tend to float even when fully submerged. The
force needed to support the funnel would decrease to a minimum when the funnel became fully submerged,
and then would remain constant at deeper submersion depths.
With the funnel spout sealed, air will be trapped inside the funnel. As the funnel is submerged gradually
below the water surface, it will displace a volume equal to the volume of the funnel material plus the
volume of trapped air. Thus its buoyancy force will be much larger than when the spout is open to
atmosphere. Neglecting any change in air volume (pressures caused by submersion should be small
compared to atmospheric pressure) the buoyancy force would be from the entire volume encompassed by
the outside of the funnel. Finally, when fully submerged, the volume of the rubber stopper (although small)
will also contribute to the total buoyancy force acting on the funnel.
Problem *3.112
[Difficulty: 2]
Given:
Steel balls resting in floating plastic shell in a bucket of water
Find:
What happens to water level when balls are dropped in water
Solution:
Basic equation FB  ρ Vdisp g  W
for a floating body weight W
When the balls are in the plastic shell, the shell and balls displace a volume of water equal to their own weight - a large volume
because the balls are dense. When the balls are removed from the shell and dropped in the water, the shell now displaces only a
small volume of water, and the balls sink, displacing only their own volume. Hence the difference in displaced water before and
after moving the balls is the difference between the volume of water that is equal to the weight of the balls, and the volume of the
balls themselves. The amount of water displaced is significantly reduced, so the water level in the bucket drops.
Volume displaced before moving balls: V1 
Wplastic  Wballs
ρ g
Wplastic
Volume displaced after moving balls:
V2 
Change in volume displaced
∆V  V2  V1  Vballs 
ρ g

 Vballs
∆V  Vballs  1  SG balls
Wballs
ρ g
 Vballs 
SGballs  ρ g  Vballs
ρ g

Hence initially a large volume is displaced; finally a small volume is displaced (ΔV < 0 because SGballs > 1)
Problem *3.113
[Difficulty: 4]
Open-Ended Problem Statement: A proposed ocean salvage scheme involves pumping air
into “bags” placed within and around a wrecked vessel on the sea bottom. Comment on the practicality of
this plan, supporting your conclusions with analyses.
Discussion: This plan has several problems that render it impractical. First, pressures at the sea bottom
are very high. For example, Titanic was found in about 12,000 ft of seawater. The corresponding pressure
is nearly 6,000 psi. Compressing air to this pressure is possible, but would require a multi-stage compressor
and very high power.
Second, it would be necessary to manage the buoyancy force after the bag and object are broken loose from
the sea bed and begin to rise toward the surface. Ambient pressure would decrease as the bag and artifact
rise toward the surface. The air would tend to expand as the pressure decreases, thereby tending to increase
the volume of the bag. The buoyancy force acting on the bag is directly proportional to the bag volume, so
it would increase as the assembly rises. The bag and artifact thus would tend to accelerate as they approach
the sea surface. The assembly could broach the water surface with the possibility of damaging the artifact
or the assembly.
If the bag were of constant volume, the pressure inside the bag would remain essentially constant at the
pressure of the sea floor, e.g., 6,000 psi for Titanic. As the ambient pressure decreases, the pressure
differential from inside the bag to the surroundings would increase. Eventually the difference would equal
sea floor pressure. This probably would cause the bag to rupture.
If the bag permitted some expansion, a control scheme would be needed to vent air from the bag during the
trip to the surface to maintain a constant buoyancy force just slightly larger than the weight of the artifact in
water. Then the trip to the surface could be completed at low speed without danger of broaching the surface
or damaging the artifact.
Problem *3.114
Given:
[Difficulty: 3]
Cylindrical container rotating as in Example 3.10
R = 0.25⋅ m h o = 0.3⋅ m
f = 2 ⋅ Hz
Find:
(a) height of free surface at the entrance
(b) if solution depends on ρ
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
Assumptions:
(Hydrostatic equation)
(1) Incompressible fluid
(2) Atmospheric pressure acts everywhere
In order to obtain the solution we need an expression for the shape of the free surface in terms of ω, r, and h o. The required
expression was derived in Example 3.10. The equation is:
z = ho −
( ω⋅ R)
2
2⋅ g
⎡1
⋅⎢
⎣2
−
2
⎛ r ⎞ ⎥⎤
⎜R
⎝ ⎠⎦
The angular velocity ω is related to the frequency of rotation through:
h1 = ho −
Now since h1 is the z value which corresponds to r = 0:
Substituting known values:
h 1 = 0.3⋅ m −
1
4
× ⎛⎜ 12.57 ⋅
⎝
ω = 2 ⋅ π⋅ f
rad
s
( ω⋅ R)
rad
s
= 12.57 ⋅
2
4⋅ g
2
× 0.25⋅ m⎞ ×
⎠
ω = 2⋅ π × 2⋅
2
s
9.81⋅ m
The solution is independent of ρ because the equation of the free surface is independent of ρ as well.
h 1 = 0.05 m
rad
s
Problem *3.115
[Difficulty: 2]
Given:
U-tube accelerometer
Find:
Acceleration in terms of h and L
Solution:
We will apply the hydrostatics equations to this system.
Governing Equations:
(Hydrostatic equation in x-direction)
(Hydrostatic equation in y-direction)
Assumptions:
(1) Incompressible fluid
(2) Neglect sloshing
(3) Ignore corners
(4) Both ends of U-tube are open to atmosphere
In the coordinate system we are using, we can see that:
Thus,
∂p
= − ρa
∂x
∂p
= − ρg
∂y
ax = a
ay = 0 g x = 0 g y = −g
dp =
Now if we evaluate Δp from left to right in the U-tube:
∂p
∂p
∆x +
∆y
∂x
∂y
We may also write this expression as:
∆p =
Simplifying this expression:
∆p = ρ⋅ a⋅ L − ρ⋅ g ⋅ h = 0
∂p
∂p
dy
dx +
∂x
∂y
∆p = ( −ρ⋅ g ) ⋅ ( −b ) + ( −ρ⋅ a) ⋅ ( −L) + ( −ρ⋅ g ) ⋅ ( b + h ) = 0
Solving for h:
h=
a⋅ L
g
Problem *3.116
[Difficulty: 2]
Given:
Rectangular container with constant acceleration
Find:
Slope of free surface
Solution:
Basic equation
In components

 p  ρ g x  ρ ax
x

 p  ρ g y  ρ ay
y

 p  ρ g z  ρ az
z
We have
ay  az  0
g x  g  sin( θ)
g y  g  cos( θ)
Hence

 p  ρ g  sin( θ)  ρ ax
x

 p  ρ g  cos( θ)  0
y
(1)
From Eq. 3 we can simplify from
p  p ( x y z)
Hence a change in pressure is given by
dp 
At the free surface p = const., so

x
p  dx 
dp  0 

x
y
p  dy


p  dx 
y
p  dy
dy
or
dx

x

y

Hence at the free surface, using Eqs 1 and 2
dy
dx

x

y
p

ρ g  sin( θ)  ρ ax
dy
dx
ρ g  cos( θ)
p
9.81 ( 0.5) 
m
2
 3
s

9.81 ( 0.866 ) 
dy
dx
 0.224
m
2
s
m
2
s
At the free surface, the slope is
(3)
p  p ( x y )
to


 p 0
z
(2)
gz  0

g  sin( θ)  ax
g  cos( θ)
p
at the free surface
p
Problem *3.117
Given:
Spinning U-tube sealed at one end
Find:
Maximum angular speed for no cavitation
[Difficulty: 2]
Assumptions: (1) water is incompressible
(2) constant angular velocity
Solution:
Basic equation
2
⎛∂ ⎞
V
2
= −ρ⋅ ω ⋅ r
−⎜ p = ρ⋅ ar = −ρ⋅
r
⎝∂r ⎠
In components
∂
Between D and C, r = constant, so
∂z
∂
Between B and A, r = constant, so
∂z
∂
∂z
p = −ρ⋅ g
p = −ρ⋅ g
and so
p D − p C = −ρ⋅ g ⋅ H
(1)
p = −ρ⋅ g
and so
p A − p B = −ρ⋅ g ⋅ H
(2)
and so
⌠ C
⌠
2
⎮ 1 dp = ⎮ ρ⋅ ω ⋅ r dr
⌡
⌡p
0
p
∂
Between B and C, z = constant, so
∂r
2
p = ρ⋅ ω ⋅ r
B
2
Integrating
2 L
p C − p B = ρ⋅ ω ⋅
2
Since p D = p atm, then from Eq 1
p C = p atm + ρ⋅ g ⋅ H
From Eq. 3
From Eq. 2
2 L
L
(3)
2
2 L
2
p B = p C − ρ⋅ ω ⋅
2
so
p B = p atm + ρ⋅ g ⋅ H − ρ⋅ ω ⋅
2
p A = p B − ρ⋅ g ⋅ H
so
2 L
p A = p atm − ρ⋅ ω ⋅
2
2
Thus the minimum pressure occurs at point A (not B). Substituting known data to find the pressure at A:
p A = 14.7⋅
lbf
2
in
− 1.94⋅
slug
ft
3
× ⎛⎜ 1600⋅
⎝
rev
min
×
2 ⋅ π⋅ rad
rev
At 68oF from steam tables, the vapor pressure of water is
×
min ⎞
60⋅ s ⎠
2
×
1
2
× ⎛⎜ 3 ⋅ in ×
⎝
p v = 0.339 ⋅ psi
2
2
2
⎞ × lbf ⋅ s × ⎛ ft ⎞ = 2.881 ⋅ lbf
⎜
12⋅ in ⎠
slug⋅ ft ⎝ 12⋅ in ⎠
2
in
ft
which is less than the pressure at A.
Therefore, cavitation does not occur.:
Problem *3.118
[Difficulty: 2]
Given:
Spinning U-tube sealed at one end
Find:
Pressure at A; water loss due to leak
Assumption:
Water is incompressible; centripetal acceleration is constant
Solution:
Basic equation
From the analysis of Example Problem 3.10, solving the basic equation, the pressure p at any point (r,z) in a continuous
rotating fluid is given by
2
p = p0 +
ρ⋅ ω
⋅ ⎛ r − r0
2 ⎝
2
2⎞
⎠ − ρ⋅ g⋅ ( z − z0)
(1)
In this case
p = pA
The speed of rotation is
ω = 300 ⋅ rpm
ω = 31.4⋅
p D = 0 ⋅ kPa
(gage)
The pressure at D is
2
Hence
pA =
ρ⋅ ω
2
p0 = pD
where p 0 is a reference pressure at point (r0,z0)
z = zA = zD = z0 = H
( 2) − ρ⋅g⋅(0) = − ρ⋅ω2 ⋅L
2
2
⋅ −L
p A = −0.42⋅ psi
=−
r= 0
r0 = rD = L
rad
1
2
s
× 1.94⋅
slug
ft
3
× ⎛⎜ 31.4⋅
⎝
rad ⎞
s
2
⎠
2
× ( 3 ⋅ in) ×
4
2
⎛ 1 ⋅ ft ⎞ × lbf ⋅ s
⎜
slug⋅ ft
⎝ 12⋅ in ⎠
(gage)
When the leak appears,the water level at A will fall, forcing water out at point D. Once again, from the analysis
of Example Problem 3.10, we can use Eq 1
In this case
p = pA = 0
Hence
0=
2
ρ⋅ ω
2
p0 = pD = 0
The amount of water lost is
z0 = zD = H
r= 0
r0 = rD = L
( 2) − ρ⋅g⋅(zA − H)
⋅ −L
2
zA = H −
z = zA
ω ⋅L
2⋅ g
2
= 12in −
1
2
× ⎛⎜ 31.4⋅
⎝
rad ⎞
s
⎠
∆h = H − zA = 12⋅ in − 0.52⋅ in
2
2
× ( 3 ⋅ in) ×
2
s
32.2⋅ ft
×
1 ⋅ ft
12⋅ in
∆h = 11.48 ⋅ in
zA = 0.52⋅ in
Problem *3.119
[Difficulty: 2]
ω

R

Given:
Centrifugal manometer consists of pair of parallel disks that rotate to develop a
radial pressure difference. There is no flow between the disks.
Find:
(a) an expression for the pressure difference, ∆p, as a function of ω, R, and ρ.
(b) find ω if ∆p = 8 µm H2O and R = 50 mm
Solution:
We will apply the hydrostatics equations to this system.
G
G
− ∇ p + ρg = ρa
Governing Equations:
∂p
+ ρ g r = ρa r
∂r
−
Assumptions:
(1) Incompressible fluid
(2) Standard air between disks
(3) Rigid body motion
(4) Radial direction is horizontal
For rigid body motion:
ar = −
2
V
r
(Hydrostatic equation)
=−
( r⋅ ω)
2
r
2
= −r⋅ ω
(Hydrostatic equation in radial direction)
In addition, since r is horizontal:
gr = 0
∂p
= ρ rω 2
∂r
Thus, the hydrostatic equation becomes:
We can solve this expression by separating variables and integrating:
2⌠
R
∆p = ρ⋅ ω ⋅ ⎮
⌡
2
Evaluating the integral on the right hand side:
r dr
0
ω=
Solving for the rotational frequency:
2 ⋅ ∆p
ρ⋅ R
Therefore:
Substituting in values:
ω=
ω =
2⋅
2×
The pressure differential can be expressed as:
2
∆p =
ρ⋅ ω ⋅ R
2
2
∆p = ρ⋅ g ⋅ ∆h
ρw g ⋅ ∆h
⋅
ρair
2
R
999
1.225
× 9.81⋅
m
2
s
−6
× 8 × 10
⋅m ×
1
(50 × 10− 3⋅m)
2
ω = 7.16⋅
rad
s
Problem *3.120
[Difficulty: 2]
ω = 1000 s-1
ρ
r1
r2
Given:
Test tube with water
Find:
(a) Radial acceleration
(b) Radial pressure gradient
(c) Rotational speed needed to generate 250 MPa pressure at the bottom of the tube
Solution:
We will apply the hydrostatics equations to this system.
G
G
− ∇ p + ρg = ρa
Governing Equations:
−
∂p
+ ρ g r = ρa r
∂r
(1) Incompressible fluid
(2) Rigid body motion
(3) Radial direction is horizontal
For rigid body motion:
ar = −
2
In addition, since r is horizontal:
r
=−
( r⋅ ω)
2
r
gr = 0
r2 = 130 mm
(Hydrostatic equation)
Assumptions:
V
r1 = 50 mm
(Hydrostatic equation in radial direction)
2
2
= −r⋅ ω
ar = −r⋅ ω
∂p
= ρ rω 2
∂r
Thus, the hydrostatic equation becomes:
We can solve this expression by separating variables and integrating:
r2
2⌠
∆p = ρ⋅ ω ⋅ ⎮ r dr
⌡r
1
2
∆p =
Evaluating the integral on the right hand side:
Substituting in values:
ω =
6 N
2 × 250 × 10 ⋅
2
m
ω = 938 ⋅ Hz
ρ⋅ ω
⋅ ⎛ r − r1
2 ⎝2
2
2⎞
⎠
Solving for ω:
3
×
m
999 ⋅ kg
×
ω=
1
(130 × 10− 3⋅m) − (50 × 10− 3⋅m)
2
2
×
2 ⋅ ∆p
ρ⋅ ⎛ r2 − r1
⎝
2
kg⋅ m
2
N⋅ s
×
2⎞
⎠
rev
2 ⋅ π⋅ rad
Problem *3.121
[Difficulty: 3]
Given:
Rectangular container of base dimensions 0.4 m x 0.2 m and a height of 0.4 m is filled with water to a depth of d =
0.2 m. Mass of empty container is M c = 10 kg. The container slides down an incline of θ = 30 deg with respect to
the horizontal. The coefficient of sliding friction is 0.30.
Find:
The angle of the water surface relative to the horizontal.
y
x
Solution:
α
We will apply the hydrostatics equations to this system.
Governing Equations:
Assumptions:
θ
G
G
− ∇p + ρg = ρa (Hydrostatic equation)
G
G
(Newton's Second Law)
F = Ma
(1) Incompressible fluid
(2) Rigid body motion
Writing the component relations:
−
∂p
= ρa x
∂x
We write the total differential of pressure as:
∂p
= − ρa x
∂x
dp =
∂p
∂p
dy
dx +
∂x
∂y
dy
a x and
∂p ∂x
dy
α = atan⎛⎜ − ⎞
=−
=−
⎝ dx ⎠
dx
∂p ∂y
g + ay
M = M c + M w = M c + ρw⋅ Vw
−
∂p
− ρg = ρa y
∂y
∂p
= − ρ (g + a y )
∂y
Now along the free surface of the water dp = 0. Thus:
To determine the acceleration components we analyze a free-body diagram:
M = 10⋅ kg + 999 ⋅
kg
3
× 0.4⋅ m × 0.2⋅ m × 0.2⋅ m
M = 25.98 kg
m
ΣFy' = 0 = N − M ⋅ g ⋅ cos( θ)
N = M ⋅ g ⋅ cos( θ)
N = 25.98 ⋅ kg × 9.81⋅
m
2
2
N⋅ s
× cos( 30⋅ deg) ×
kg⋅ m
s
ΣFx' = M ⋅ ax' = M ⋅ g ⋅ sin( θ) − Ff = M ⋅ g ⋅ sin( θ) − μ⋅ N
ax' = 9.81⋅
m
2
× sin( 30⋅ deg) − 0.30 × 220.7 ⋅ N ×
s
ax' = g ⋅ sin( θ) − μ⋅
1
25.98 ⋅ kg
×
kg⋅ m
2
N⋅ s
y
N
x
M
ax' = 2.357
m
2
× cos( 30⋅ deg)
ax = 2.041
s
Thus,
m
2
s
α = atan⎛⎜
2.041
⎞
⎝ 9.81 − 1.178 ⎠
α = 13.30 ⋅ deg
ay = −2.357 ⋅
2
s
m
2
s
F f = µN
y’
m
Now that we have the acceleration in the x'-y' system, we transform it to the x-y system:
ax = 2.357 ⋅
N = 220.7 N
x’
ax = ax'⋅ cos( θ)
× sin( 30⋅ deg)
θ
N
Mg
ay = −ax'⋅ sin( θ)
ay = −1.178
m
2
s
Problem *3.122
[Difficulty: 3]
Given:
Rectangular container of base dimensions 0.4 m x 0.2 m and a height of 0.4 m is filled with water to a depth of d =
0.2 m. Mass of empty container is M c = 10 kg. The container slides down an incline of θ = 30 deg with respect to
the horizontal without friction.
Find:
(a) The angle of the water surface relative to the horizontal.
(b) The slope of the free surface for the same acceleration up the plane.
Solution:
We will apply the hydrostatics equations to this system.
G
G
− ∇ p + ρg = ρa
G
G
F = Ma
Governing Equations:
Assumptions:
y
x
α
(Hydrostatic equation)
(Newton's Second Law)
θ
(1) Incompressible fluid
(2) Rigid body motion
−
Writing the component relations:
∂p
= ρa x
∂x
We write the total differential of pressure as:
ax
∂p ∂x
dy
=−
=−
dx
∂p ∂y
g + ay
M = M c + M w = M c + ρw⋅ Vw
dp =
α = atan⎛⎜ −
and
∂p
= − ρa x
∂x
−
∂p
∂p
dy
dx +
∂x
∂y
dy ⎞
Now along the free surface of the water dp = 0. Thus:
To determine the acceleration components we analyze a free-body diagram:
⎝ dx ⎠
M = 10⋅ kg + 999 ⋅
∂p
= − ρ (g + a y )
∂y
∂p
− ρg = ρa y
∂y
kg
3
× 0.4⋅ m × 0.2⋅ m × 0.2⋅ m
M = 25.98 kg
m
ΣFx' = M ⋅ ax' = M ⋅ g ⋅ sin( θ) ax' = g ⋅ sin( θ)
ax = ax'⋅ cos( θ) = g ⋅ sin( θ) ⋅ cos( θ)
ay = −ax'⋅ sin( θ) = g ⋅ ( sin( θ) )
Thus,
dy
dx
=−
g ⋅ sin( θ) ⋅ cos( θ)
2
g ⎡⎣1 − ( sin( θ) ) ⎤⎦
=−
sin( θ) ⋅ cos( θ)
( cos( θ) )
2
slope =
g ⋅ sin( θ) ⋅ cos( θ)
2⎤
g ⎡⎣1 + ( sin( θ) ) ⎦
=−
sin( θ) ⋅ cos( θ)
1 + ( sin( θ) )
2
x
2
Ff = µN
y’
=−
sin( θ)
cos( θ)
For the acceleration up the incline: ax = −g ⋅ sin( θ) ⋅ cos( θ)
Thus,
y
= −tan( θ)
α = 30⋅ deg
x’
ay = g ⋅ ( sin( θ) )
slope =
θ
N
Mg
2
sin( 30⋅ deg) ⋅ cos( 30⋅ deg)
1 + ( sin( 30⋅ deg) )
2
slope = 0.346
Problem *3.123
[Difficulty: 3]
Given:
Cubical box with constant acceleration
Find:
Slope of free surface; pressure along bottom of box
Solution:
Basic equation
In components
∂
− p + ρ⋅ g x = ρ⋅ ax
∂x
∂
− p + ρ⋅ g y = ρ⋅ ay
∂y
∂
− p + ρ⋅ g z = ρ⋅ az
∂z
We have
ax = ax
ay = 0
az = 0
gz = 0
Hence
∂
∂
(3)
∂x
gx = 0
∂
p = −SG⋅ ρ⋅ ax (1)
∂y
p = −SG⋅ ρ⋅ g
From Eq. 3 we can simplify from
p = p ( x , y , z)
Hence a change in pressure is given by
dp =
dy
Hence at the free surface
dx
L = 80⋅ cm at the midpoint x =
L
y=
2
On the bottom y = 0 so
∂x
p ⋅ dy
p ⋅ dx +
x=0
y=
5
8
⋅L
∂
∂
∂y
p ⋅ dy
dy
or
=−
dx
∂x
∂
p
=−
p
ax
=−
g
0.25⋅ g
g
= −0.25
x
4
+C
and through volume conservation the fluid rise in the rear
balances the fluid fall in the front, so at the midpoint the
free surface has not moved from the rest position
(box is half filled)
2
so
p =0
(4)
1 L
=− ⋅ +C
2
4 2
L
dp = −SG ⋅ ρ⋅ ax ⋅ dx − SG ⋅ ρ⋅ g ⋅ dy
Combining Eqs 1, 2, and 4
p = p atm when
L
∂
∂y
∂z
p = p( x , y)
to
∂
(2)
∂y
y=−
The equation of the free surface is then
We have
∂x
p ⋅ dx +
dp = 0 =
At the free surface p = const., so
For size
∂
g y = −g
or
5
p atm = −SG⋅ ρ⋅ g ⋅ ⋅ L + c
8
C=
5
8
⋅L
y=
5
8
⋅L −
x
4
p = −SG ⋅ ρ⋅ ax ⋅ x − SG ⋅ ρ⋅ g ⋅ y + c
5
c = p atm + SG ⋅ ρ⋅ g ⋅ ⋅ L
8
5
5
x
p ( x , y ) = p atm + SG⋅ ρ⋅ ⎜⎛ ⋅ g ⋅ L − ax ⋅ x − g ⋅ y⎞ = p atm + SG⋅ ρ⋅ g ⋅ ⎛⎜ ⋅ L −
−y
4
⎝8
⎠
⎝8
2
kg
5
x
N⋅ s
m
p ( x , 0 ) = p atm + SG⋅ ρ⋅ g ⋅ ⎛⎜ ⋅ L − ⎞ = 101 + 0.8 × 1000⋅
×
× 9.81⋅ ×
3
4⎠
kg⋅ m
2
⎝8
m
s
p ( x , 0 ) = 105 − 1.96⋅ x
kPa
x⎞
⎛5
×
⎜ × 0.8⋅ m −
3
4⎠
⎝8
10 ⋅ Pa
(p in kPa, x in m)
Problem *3.124
[Difficulty: 3]
Given:
Gas centrifuge, with maximum peripheral speed Vmax = 950 ft/s contains
uranium hexafluoride gas (M = 352 lb/lbmol) at 620 deg F.
Find:
(a) Ratio of maximum pressure to pressure at the centrifuge axis
(b) Evaluate pressure ratio at 620 deg F.
Solution:
We will apply the hydrostatics equations to this system.
G
G
− ∇ p + ρg = ρa
Governing Equations:
−
Assumptions:
Vmax = ωr 2
r2
(Hydrostatic equation)
∂p
+ ρg r = ρa r (Hydrostatic equation radial component)
∂r
(1) Incompressible fluid
(2) Rigid body motion
(3) Ideal gas behavior, constant temperature
2
2
V
( r⋅ ω)
2
For rigid body motion: ar = −
=−
= −r⋅ ω
r
r
Thus:
∂p
p
= − ρa r =
rω 2
∂r
Rg T
Separating variables and integrating:
p
r
2
2
2 r
⎛ p2 ⎞
ω
2
ln⎜
=
⋅
⎝ p1 ⎠ Rg⋅ T 2
2
2
⌠ 2 1
ω ⌠
⎮
⋅ ⎮ r dr
dp =
Rg ⋅ T ⌡0
⎮ p
⌡p
where we define:
Vmax = ω⋅ r2
therefore:
⎛ p2 ⎞ Vmax
ln⎜
=
p1
⎝ ⎠ 2 ⋅ Rg ⋅ T
1
⎛ V 2⎞
⎜ max
⎜ 2⋅ R ⋅ T
p2
g ⎠
p rat =
= e⎝
p
Solving for the pressure ratio:
1
The gas constant:
Rg =
1545 ft⋅ lbf
⋅
352 lbm⋅ R
Substituting in all known values:
Rg = 4.39⋅
ft⋅ lbf
lbm⋅ R
2
2 ⎤
⎡⎛
ft
1
lbm ⋅ R
1
lbf ⋅ s
⎢⎜950⋅ ⎞ × ×
×
×
⎥
s ⎠ 2 4.39⋅ ft⋅ lbf ( 620+ 460) ⋅ R 32.2⋅ lbm⋅ ft⎦
p rat = e⎣⎝
p rat = 19.2
Problem *3.125
Given:
[Difficulty: 3]
Pail is swung in a vertical circle. Water moves as a rigid body.
Point of interest is the top of the trajectory.
Find:
d
V = 5 m/s
d
(a) Tension in the string
(b) Pressure on pail bottom from the water.
Solution:
R= 1 m
T
We will apply the hydrostatics equations to this system.
G
G
− ∇ p + ρg = ρa
Governing Equations:
−
Assumptions:
(Hydrostatic equation radial component)
(1) Incompressible fluid
(2) Rigid body motion
(3) Center of mass of bucket and water are located at a radius
of 1 m where V = rω = 5 m/s
(
)
(
(
)
−T − M b + M w ⋅ g = M b + M w ar
⎛V
⎞
T = Mb + Mw ⋅ ⎜
−g
⎝ r
⎠
π 2
and: M w = ρ⋅ V = ρ⋅ ⋅ d ⋅ h
4
Now we find T:
(Hydrostatic equation)
∂p
+ ρ g r = ρa r
∂r
Summing the forces in the radial direction:
Thus the tension is:
r
2
2
)
M w = 999 ⋅
2
kg
3
m
×
π
4
T = ( 1.529 + 25.11 ) ⋅ kg ×
ar = −
where
M b = 15⋅ N ×
where:
2
× ( 0.4⋅ m) × 0.2⋅ m
s
9.81⋅ m
×
V
r
kg⋅ m
2
N⋅ s
M w = 25.11 ⋅ kg
2
⎡⎛ m 2 1
m ⎤ N⋅ s
⎢⎜ 5⋅ ⎞ ×
− 9.81⋅ ⎥ ×
2⎥
⎢⎝ s ⎠
1⋅ m
kg⋅ m
s ⎦
⎣
If we apply this information to the radial hydrostatic equation we get:
−
M b = 1.529 ⋅ kg
∂p
V2
− ρg = − ρ
∂r
r
T = 405 ⋅ N
Thus:
⎞
⎛V 2
∂p
= ρ ⎜⎜
− g ⎟⎟
∂r
⎝ r
⎠
If we assume that the radial pressure gradient is constant throughout the water, then the pressure gradient is equal to:
p r = 999 ⋅
kg
3
m
×
2
2
⎡
⎤
⎢⎜⎛ 5⋅ m ⎞ × 1 − 9.81⋅ m ⎥ × N⋅ s
⎢⎝ s ⎠
2⎥
1⋅ m
kg⋅ m
s ⎦
⎣
and we may calculate the pressure at the bottom of the bucket:
p r = 15.17 ⋅
kPa
m
∆p = p r⋅ ∆r
∆p = 15.17 ⋅
kPa
m
× 0.2⋅ m
∆p = 3.03⋅ kPa
Problem *3.126
[Difficulty: 3]
D = 2.5 in.
z
H/2
r
R = 5 ft
Given:
rev
s
Find:
(a) Slope of free surface
(b) Spin rate necessary for spillage
(c) Likelihood of spilling versus slipping
Solution:
We will apply the hydrostatics equations to this system.
G
G
− ∇ p + ρg = ρa
Governing Equations:
Assumptions:
2
ar = −
r
SG = 1.05
Half-filled soft drink can at the outer edge of a merry-go-round
ω = 0.3⋅
V
H = 5 in.
=−
( r⋅ ω)
r
(Hydrostatic equation)
−
∂p
+ ρ g r = ρa r
∂r
−
∂p
+ ρg z = ρa z
∂z
(Hydrostatic equation radial component)
(Hydrostatic equation z component)
(1) Incompressible fluid
(2) Rigid body motion
(3) Merry-go-round is horizontal
2
2
= −r⋅ ω
∂p
∂p
dr +
dp =
dz
∂r
∂z
az = 0
gr = 0
g z = −g
Thus:
∂p
= ρ rω 2
∂r
∂p ∂r
ρrω 2 rω 2
dz
=−
=
=−
∂p ∂z
− ρg
dr
g
For the free surface the pressure is constant. Therefore:
slope = 5 ⋅ ft × ⎛⎜ 20⋅
So the slope at the free surface is
⎝
H
To spill, the slope must be
slope sp =
rev
min
H
×
min
60⋅ s
×
slope sp =
D
2 ⋅ π⋅ rad ⎞
rev
5
2.5
⎠
∂p
= − ρg So p = p(r,z)
∂z
2
2
×
s
32.2⋅ ft
slope = 0.681
slope sp = 2.000
D
Thus,
ωsp =
g dz
⋅
r dr
ωsp =
32.2⋅
ft
2
s
×
1
5 ⋅ ft
×2
rad
ωsp = 3.59⋅
s
This is nearly double the original speed (2.09 rad/s). Now the coefficient of static friction between the can and the surface of the
merry-go-round is probably less than 0.5.Thus the can would not likely spill or tip; it would slide off!
Problem *3.127
Discussion:
[Difficulty: 2]
Separate the problem into two parts: (1) the motion of the ball in water below the pool surface, and (2) the
motion of the ball in air above the pool surface.
Below the pool water surface the motion of each ball is controlled by buoyancy force and inertia. For small depths of submersion
ball speed upon reaching the surface will be small. As depth is increased, ball speed will increase until terminal speed in water is
approached. For large depths, the actual depth will be irrelevant because the ball will reach terminal speed before reaching the pool
water surface. All three balls are relatively light for their diameters, so terminal speed in water should be reached quickly. The depth
of submersion needed to reach terminal speed should be fairly small, perhaps 1 meter or less (The initial water depth required to
reach terminal speed may be calculated using the methods of Chapter 9).
Buoyancy is proportional to volume and inertia is proportional to mass. The ball with the largest volume per unit mass should
accelerate most quickly to terminal speed. This will probably be the beach ball, followed by the table-tennis ball and the water polo
ball.
The ball with the largest diameter has the smallest frontal area per unit volume; the terminal speed should be highest for this ball.
Therefore, the beach ball should have the highest terminal speed, followed by the water polo ball and the table-tennis ball.
Above the pool water surface the motion of each ball is controlled by aerodynamic drag force, gravity force, and inertia (see the
equation below). Without aerodynamic drag, the height above the pool water surface reached by each ball will depend only on its
initial speed (The maximum height reached by a ball in air with aerodynamic drag may be calculated using the methods of Chapter 9).
Aerodynamic drag reduces the height reached by the ball.
Aerodynamid drag is proportional to frontal area. The heaviest ball per unit frontal area (probably the water polo ball) should reach
the maximum height and the lightest ball per unit area (probably the beach ball) should reach the minimum height.
dV
1
2
ΣFy = −FD − m⋅ g = m⋅ ay = m⋅
= −CD⋅ A⋅ ⋅ ρ⋅ V − m⋅ g since
dt
2
V0
y
Thus, −
1
2
CD⋅ A⋅ ⋅ ρ⋅ V
2
m
−g=
dV
dt
= V⋅
dV
(1)
1
2
FD = CD⋅ A⋅ ⋅ ρ⋅ V
2
We solve this by separating variables:
dy
FD
V⋅ dV
W = mg
Solving for the maximum height:
CD⋅ A⋅ ρ V2
⋅
1+
m⋅ g
2
= −g ⋅ dy
Integrating this expression over the flight of the ball yields:
2
⎛⎜
ρ⋅ CD⋅ A Vo ⎞
−
⋅
⋅ ln⎜ 1 +
= −g ⋅ y max
m⋅ g
2 ⎠
ρ⋅ CD⋅ A ⎝
m⋅ g
2
⎛⎜
ρ⋅ CD⋅ A Vo ⎞
FDo ⎞
⎛
m
Simplifying: y max = −
(2)
⋅ ln⎜ 1 +
⋅ ln⎜ 1 +
y max = −
⋅
2 ⎠
m⋅ g
ρ⋅ CD⋅ A ⎝
ρ⋅ CD⋅ A ⎝
m⋅ g ⎠
m
If we neglect drag, equation (1) becomes:
−g ⋅ dy = V⋅ dV
Checking the limiting value predicted by Eq (2) as
lim y max
C D →0
Integrating and solving for the maximum height:
CD → 0
y max = −
Vo
2⋅ g
: we remember that for small x that ln(1+x) = -x. Thus:
⎛ m ρC D A Vo2 ⎞ Vo2
⎟⎟ =
= lim ⎜⎜
C D →0 ρC A mg
2
⎝ D
⎠ 2g
2
which is the result in Equation (3).
(3)
Problem *3.128
Given:
[Difficulty: 4]
A steel liner is to be formed in a spinning horizontal mold. To insure uniform thickness
the minimum angular velocity should be at least 300 rpm. For steel, SG = 7.8
θ
y
x
ri
Find:
(a) The resulting radial acceleration on the inside surface of the liner
(b) the maximum and minimum pressures on the surface of the mold
Solution:
We will apply the hydrostatics equations to this system.
G
G
− ∇ p + ρg = ρa
Governing Equations:
Assumptions:
2
ar = −
Hence:
Thus:
dp =
V
r
=−
ro
(gravity is
downward in
this diagram)
( r⋅ ω)
r
−
∂p
+ ρ g r = ρa r
∂r
−
1 ∂p
+ ρg θ = ρaθ
r ∂θ
−
∂p
+ ρg z = ρa z
∂z
(Hydrostatic equation)
(Hydrostatic equation radial component)
(Hydrostatic equation transeverse component)
(Hydrostatic equation z component)
(1) Incompressible fluid
(2) Rigid body motion
2
2
= −r⋅ ω
ar = 4 ⋅ in × ⎜⎛ 300 ×
⎝
aθ = 0
rev
min
×
2 ⋅ π⋅ rad
rev
az = 0
×
min ⎞
2
60⋅ s ⎠
∂p
= ρg r − ρa r = ρrω 2 − ρg cos θ
∂r
×
g r = −g ⋅ cos( θ)
g θ = g ⋅ sin( θ)
ft
ar = 329 ⋅
12⋅ in
2
ar = 10.23 ⋅ g
s
∂p
= ρrg θ − ρraθ = ρrg sin θ
∂θ
∂p
∂p
dr +
dθ = ρrω 2 − ρg cos θ dr + (ρrg sin θ )dθ
∂r
∂θ
(
ft
gz = 0
∂p
= ρg z − ρa z = 0
∂z
)
( )
We can integrate to find pressure as a function of r and θ.
r
(
p ri , θ = p atm
)
⌠
2
Therefore, we integrate: p − p atm = ⎮ ρ⋅ r⋅ ω − ρ⋅ g ⋅ cos( θ) dr + f ( θ)
⌡r
i
∂p ⎞
2
⎟ = ρrω − ρg cos θ
∂r ⎠θ
⎛ r2 − r 2⎞
i ⎠
⎝
p = p atm + ρ⋅ ω ⋅
− ρ⋅ g ⋅ cos( θ) ⋅ ( r − ri) + f ( θ)
2
2
df
∂p ⎞
= ρrg sin θ
⎟ = ρ (r − ri )g sin θ +
dθ
∂θ ⎠ r
Taking the derivative of pressure with respect to θ:
Thus, the integration function f(θ) is:
⎛ r2 − r 2⎞
i ⎠
⎝
p = p atm + ρ⋅ ω ⋅
− ρ⋅ g ⋅ ( r − ri) ⋅ cos( θ) − ρ⋅ g ⋅ ri⋅ cos( θ) + C
2
2
Therefore, the pressure is:
The integration constant is determined from the boundary condition:
( )
p ri , θ = p atm
⎛ r 2 − r 2⎞
i ⎠
⎝i
p atm = p atm + ρ⋅ ω ⋅
− ρ⋅ g ⋅ ( ri − ri) ⋅ cos( θ) − ρ⋅ g ⋅ ri⋅ cos( θ) + C
2
2
Therefore, the pressure is:
f ( θ) = −ρ⋅ g ⋅ ri⋅ cos( θ) + C
−ρ⋅ g ⋅ ri⋅ cos( θ) + C = 0
C = ρ⋅ g ⋅ ri⋅ cos( θ)
⎛ r2 − r 2⎞
i ⎠
⎝
p = p atm + ρ⋅ ω ⋅
− ρ⋅ g ⋅ ( r − ri) ⋅ cos( θ)
2
2
The maximum pressure should occur on the mold surface at θ = π:
p maxgage =
⎡⎛
⎢ 7.8⋅ 1.94⋅ slug ⎞ ×
3
⎢⎜
ft ⎠
⎣⎝
2
2
2
⎛ 31.42 ⋅ rad ⎞ × 1 ⋅ ⎛⎜ 6 − 4 ⎞ ⋅ ft2 −
⎜
2
s ⎠
2 ⎜
⎝
⎝ 12 ⎠
2
⎛ 7.8⋅ 1.94⋅ slug ⎞ × 32.2⋅ ft × ⎛ 6 − 4 ⎞ ⋅ ft ⋅ cos( π)⎥⎤ ⋅ lbf ⋅ s
⎜
⎜
⎥ slug⋅ ft
3
2 ⎝ 12 ⎠
ft ⎠
s
⎝
⎦
p maxgage = 1119⋅ psf
p maxgage = 7.77⋅ psi
The minimum pressure should occur on the mold surface at θ = 0:
2
2
2
⎡⎛
slug ⎞ ⎛
rad ⎞
1 ⎛⎜ 6 − 4 ⎞ 2
⎢
p mingage = 7.8⋅ 1.94⋅
× ⋅
× ⎜ 31.42 ⋅
⋅ ft −
3
2
⎢⎜
s ⎠
2 ⎜
⎝
ft
12
⎣⎝
⎠
⎝
⎠
2
⎛ 7.8⋅ 1.94⋅ slug ⎞ × 32.2⋅ ft × ⎛ 6 − 4 ⎞ ⋅ ft ⋅ cos( 0 )⎥⎤ ⋅ lbf ⋅ s
⎜
⎜
⎥ slug⋅ ft
3
2 ⎝ 12 ⎠
ft ⎠
s
⎝
⎦
p mingage = 956 ⋅ psf
(In both results we divided by 144 to convert from psf to psi.)
p mingage = 6.64⋅ psi
Problem *3.129
Discussion:
[Difficulty: 4]
A certain minimum angle of inclination would be needed to overcome static friction and start the container
into motion down the incline. Once the container is in motion, the retarding force would be provided by
sliding (dynamic) friction. the coefficient of dynamic friction usually is smaller than the static friction
coefficient. Thus the container would continue to accelerate as it moved down the incline. This acceleration
would procide a non-zero slope to the free surface of the liquid in the container.
In principle the slope could be measured and the coefficent of dynamic friction calculated. In practice several problems would arise.
To calculate dynamic friction coefficient one must assume the liquid moves as a solid body, i.e., that there is no sloshing. This
condition could only be achieved if there were nminimum initial disturbance and the sliding distance were long.
It would be difficult to measure the slope of the free surface of liquid in the moving container. Images made with a video camera or a
digital still camera might be processed to obtain the required slope information.
α
Ff = µN
y
x
θ
N
mg
ΣFy = 0 = N − M ⋅ g ⋅ cos( θ)
N = M ⋅ g ⋅ cos( θ)
ΣFx = M ⋅ ax = M ⋅ g ⋅ sin( θ) − Ff
Ff = μk ⋅ N = μk ⋅ M ⋅ g ⋅ cos⋅ ( θ)
Thus the acceleration is:
ax = g ⋅ sin( θ) − μk ⋅ g ⋅ cos( θ)
G
G
− ∇ p + ρg = ρa
−
∂p
+ ρg sin θ = ρa x = ρ ( g sin θ − µ k g cos θ )
∂x
∂p
= ρgµ k cos θ
∂x
−
∂p
− ρg cos θ = ρa x = 0
∂y
∂p
= − ρg cos θ
∂y
dp =
∂p
∂p
dy
dx +
∂x
∂y
ρgµ k cosθ
∂p ∂x
dy
=−
= µk
=−
∂p ∂y
− ρg cos θ
dx
For the free surface the pressure is constant. Therefore:
So the free surface angle is:
Now for a static liquid:
( )
α = atan μk
Now since it is necessary to make the container slip along the surface,
( )
( )
θ > atan μs > atan μk = α
Thus, α < θ, as shown in the sketch above.
Problem 4.1
Given:
Data on mass and spring
Find:
Maximum spring compression
[Difficulty: 2]
Solution:
The given data is
M = 5 ⋅ lb
h = 5 ⋅ ft
k = 25⋅
lbf
ft
Apply the First Law of Thermodynamics: for the system consisting of the mass and the spring (the spring has gravitional potential
energy and the spring elastic potential energy)
E1 = 0
Total mechanical energy at initial state
Note: The datum for zero potential is the top of the uncompressed spring
E2 = M ⋅ g ⋅ ( −x ) +
Total mechanical energy at instant of maximum compression x
But
E1 = E2
so
0 = M ⋅ g ⋅ ( −x ) +
Solving for x
x =
1
2
⋅ k⋅ x
x = 2 × 5 ⋅ lb × 32.2⋅
k
ft
2
s
x =
When just resting on the spring
M⋅ g
×
ft
25⋅ lbf
×
32.2⋅ lb⋅ ft
2
s ⋅ lbf
x = 0.200 ft
k
When dropped from height h:
E1 = M ⋅ g ⋅ h
Total mechanical energy at initial state
Total mechanical energy at instant of maximum compression x
E2 = M ⋅ g ⋅ ( −x ) +
Note: The datum for zero potential is the top of the uncompressed spring
E1 = E2
so
M ⋅ g ⋅ h = M ⋅ g ⋅ ( −x ) +
2
⋅ k⋅ x
2
2
2⋅ M⋅ g
But
1
1
2
⋅ k⋅ x
2
1
2
⋅ k⋅ x
2
x = 0.401 ⋅ ft
2
x −
Solving for x
x =
x = 5 ⋅ lb × 32.2⋅
ft
2
s
×
ft
25⋅ lbf
2⋅ M⋅ g
k
M⋅ g
k
×
+
⋅x −
2⋅ M⋅ g⋅ h
k
=0
2
⎛ M⋅ g ⎞ + 2⋅ M⋅ g⋅ h
⎜ k
k
⎝
⎠
32.2⋅ lb⋅ ft
2
s ⋅ lbf
2
+
⎛ 5 ⋅ lb × 32.2⋅ ft × ft × 32.2⋅ lb⋅ ft ⎞ + 2 × 5⋅ lb × 32.2⋅ ft × ft × 32.2⋅ lb⋅ ft × 5 ⋅ f
⎜
2
25⋅ lbf
2
25⋅ lbf
2
2
s
s ⋅ lbf ⎠
s ⋅ lbf
s
⎝
x = 1.63⋅ ft
Note that ignoring the loss of potential of the mass due to spring compression x gives
x =
2⋅ M⋅ g⋅ h
k
x = 1.41⋅ ft
Problem 4.2
Given:
An ice-cube tray with water at 15oC is frozen at –5oC.
Find:
Change in internal energy and entropy
Solution:
Apply the Tds and internal energy equations
Governing equations:
Assumption:
Tds = du + pdv
du = cv dT
Neglect volume change
Liquid properties similar to water
The given or available data is:
T1 = (15 + 273) K = 288 K
cv = 1
kcal
kg ⋅ K
T2 = (− 5 + 273) K = 268 K
ρ = 999
Then with the assumption:
Tds = du + pdv = du = c v dT
or
ds = cv
Integrating
⎛T
s 2 − s1 = cv ln⎜⎜ 2
⎝ T1
kg
m3
dT
T
∆S = 999
⎞
⎟⎟
⎠
or
⎛T ⎞
∆S = m(s 2 − s1 ) = ρVcv ln⎜⎜ 2 ⎟⎟
⎝ T1 ⎠
J
kcal
10 −6 m 3
kg
⎛ 268 ⎞
×
×
×1
× ln⎜
250
mL
⎟ × 4190
3
kcal
kg ⋅ K
mL
m
⎝ 288 ⎠
∆S = −0.0753
Also
[Difficulty: 2]
kJ
K
u 2 − u1 = cv (T2 − T1 )
∆U = 999
or
∆U = mcv (T2 − T1 ) = ρVcv ∆T
kcal
J
10 −6 m 3
kg
250
mL
×
×
×1
× (− 268 − 288)K × 4190
3
mL
kg ⋅ K
kcal
m
∆U = −20.9 kJ
Problem 4.3
[Difficulty: 2]
Given:
Data on ball and pipe
Find:
Speed and location at which contact is lost
θ
V
Fn
Solution:
The given data is
r = 1 ⋅ mm
M
R = 50⋅ mm
∑ Fn = Fn − m⋅g⋅cos(θ) = m⋅an
2
an = −
V
R+r
2
Contact is lost when
Fn = 0
or
2
V = g ⋅ ( R + r) ⋅ cos( θ)
−m⋅ g ⋅ cos ( θ) = −m⋅
V
R+r
(1)
2
For no resistance energy is conserved, so
E = m⋅ g ⋅ z + m⋅
V
2
2
= m⋅ g ⋅ ( R + r) ⋅ cos( θ) + m⋅
2
V = 2 ⋅ g ⋅ ( R + r) ⋅ ( 1 − cos( θ) )
Hence from energy considerations
2
Combining 1 and 2,
V = 2 ⋅ g ⋅ ( R + r) ⋅ ( 1 − cos( θ) ) = g ⋅ ( R + r) ⋅ cos( θ)
Hence
θ = acos⎛⎜
Then from 1
V =
2⎞
⎝3⎠
( R + r) ⋅ g ⋅ cos( θ)
θ = 48.2⋅ deg
V = 0.577
m
s
V
2
= E0 = m⋅ g ⋅ ( R + r)
(2)
or
2 ⋅ ( 1 − cos( θ) ) = cos( θ)
Problem 4.4
Given:
Data on Boeing 777-200 jet
Find:
Minimum runway length for takeoff
[Difficulty: 2]
Solution:
Basic equation
dV
dV
ΣFx = M ⋅
= M ⋅ V⋅
= Ft = constant
dt
dx
Separating variables
M ⋅ V⋅ dV = Ft⋅ dx
Integrating
x=
Note that the "weight" is already in mass units!
2
x =
For time calculation
Integrating
M⋅
t=
M⋅ V
2 ⋅ Ft
1
2
× 325 × 10 kg × ⎛⎜ 225
km
3
⎝
dV
= Ft
dt
dV =
hr
Ft
M
×
1 ⋅ km
1000⋅ m
×
2
2
1
N⋅ s
1
⎞ ×
⋅ ×
3600⋅ s ⎠
kg⋅ m
3 N
2 × 425 × 10
1 ⋅ hr
x = 747 m
⋅ dt
M⋅ V
Ft
3
t = 325 × 10 kg × 225
km
hr
×
1 ⋅ km
1000⋅ m
×
1 ⋅ hr
3600⋅ s
1
×
2 × 425 × 10
Aerodynamic and rolling resistances would significantly increase both these results
⋅
1
3 N
2
×
N⋅ s
kg⋅ m
t = 23.9 s
Problem 4.5
Given:
Car sliding to a stop
Find:
Initial speed; skid time
[Difficulty: 2]
Solution:
Governing equations:
ΣFx = M ⋅ ax
Ff = μ ⋅ W
Assumptions: Dry friction; neglect air resistance
Given data
L = 200 ⋅ ft
μ = 0.7
W
W d2
ΣFx = −Ff = −μ ⋅ W = M ⋅ ax =
⋅ ax =
⋅
x
g
g dt 2
or
d
2
dt
2
x = −μ ⋅ g
Integrating, and using I.C. V = V0 at t = 0
Hence
dx
dt
Integrating again
(1)
1
1
2
2
x = − ⋅ g ⋅ t + V0 ⋅ t + c2 = − ⋅ g ⋅ t + V0 ⋅ t
2
2
We have the final state, at which
From Eq. 1
= −μ⋅ g ⋅ t + c1 = −μ⋅ g ⋅ t + V0
dx
x f = L and
dt
=0
dx
= 0 = −μ⋅ g ⋅ tf + V0
dt
since x = 0 at t = 0
at
t = tf
so
tf =
(2)
V0
μ⋅ g
2
2
Substituting into Eq. 2
V0
V0
⎛ V0 ⎞
x = x f = L = − ⋅ g ⋅ t + V0 ⋅ t = − ⋅ g ⋅ tf + V0 ⋅ tf = − ⋅ g ⋅ ⎜
+ V0 ⋅
=
μ⋅ g
2 ⋅ μ⋅ g
2
2
2 ⎝ μ⋅ g ⎠
Solving
L=
Using the data
V0 = 64.7⋅ mph
1
V0
2 ⋅ μ⋅ g
2
1
1
2
or
The skid time is
V0 =
tf =
2 ⋅ μ⋅ g ⋅ L
V0
μ⋅ g
tf = 4.21 s
Problem 4.6
[Difficulty: 2]
Given:
Block sliding to a stop
Find:
Distance and time traveled; new coeeficient of friction
Solution:
Governing equations:
ΣFx = M ⋅ ax
Ff = μ⋅ W
Assumptions: Dry friction; neglect air resistance
m
V0 = 5 ⋅
s
μ = 0.6
Given data
M = 2 ⋅ kg
W
W d2
ΣFx = −Ff = −μ⋅ W = M ⋅ ax =
⋅ ax =
⋅
x
g
g dt 2
L = 2⋅ m
d
or
2
dt
2
x = −μ⋅ g
Integrating, and using I.C. V = V0 at t = 0
Hence
dx
dt
Integrating again
(1)
1
1
2
2
x = − ⋅ g ⋅ t + V0 ⋅ t + c2 = − ⋅ g ⋅ t + V0 ⋅ t
2
2
We have the final state, at which
From Eq. 1
= −μ⋅ g ⋅ t + c1 = −μ⋅ g ⋅ t + V0
dx
x f = L and
dt
dx
= 0 = −μ⋅ g ⋅ tf + V0
dt
=0
since x = 0 at t = 0
(2)
t = tf
at
tf =
so
V0
μ⋅ g
Using given data
2
Substituting into Eq. 2
Solving
tf = 0.850 s
2
V0
V0
⎛ V0 ⎞
x = x f = L = − ⋅ g ⋅ t + V0 ⋅ t = − ⋅ g ⋅ tf + V0 ⋅ tf = − ⋅ g ⋅ ⎜
+ V0 ⋅
=
μ⋅ g
2 ⋅ μ⋅ g
2
2
2 ⎝ μ⋅ g ⎠
1
x =
V0
1
2
1
2
2
2 ⋅ μ⋅ g
For rough surface, using Eq. 3 with x = L
(3)
μ =
Using give data
V0
2
2⋅ g⋅ L
μ = 0.637
tf =
V0
μ⋅ g
x = 2.12 m
tf = 0.800 s
Problem 4.7
Given:
Car entering a curve
Find:
Maximum speed
[Difficulty: 2]
Solution:
2
Governing equations:
ΣFr = M ⋅ ar
V
Ff = μ⋅ W
ar =
μwet = 0.3
r = 100 ⋅ ft
r
Assumptions: Dry friction; neglect air resistance
Given data
μdry = 0.7
2
V
ΣFr = −Ff = −μ⋅ W = −μ⋅ M ⋅ g = M ⋅ ax = M ⋅
r
or
V=
μ⋅ r⋅ g
Hence, using given data
V =
μdry⋅ r⋅ g
V = 32.4⋅ mph
V =
μwet⋅ r⋅ g
V = 21.2⋅ mph
Problem 4.8
Given:
Data on air compression process
Find:
Work done
[Difficulty: 2]
Solution:
Basic equation
δw = p ⋅ dv
Assumptions: 1) Adiabatic 2) Frictionless process δw = pdv
Given data
p 1 = 1 ⋅ atm
p 2 = 4 ⋅ atm
From Table A.6 R = 286.9 ⋅
J
kg⋅ K
T1 = 20 °C
T1 = 293 K
and
k = 1.4
Before integrating we need to relate p and v. An adiabatic frictionless (reversible) process is isentropic, which for an ideal gas gives
k
p⋅ v = C
δw = p ⋅ dv = C⋅ v
Integrating
w=
w=
C
k−1
k=
where
⋅ ⎛ v2
−k
1 −k
⎝
cp
cv
⋅ dv
1 −k ⎞
− v2
1
k 1 −k
k
1 −k ⎞
⎛
− p1⋅ v1 ⋅ v2
⎠ = ( k − 1) ⋅ ⎝ p 2⋅ v 2 v 2
⎠
R ⋅ T1 ⎛ T2
⎞
⋅ T2 − T1 =
⋅⎜
−1
( k − 1) T1
( k − 1)
R
(
)
⎝
(1)
⎠
k
But
k
p⋅ v = C
means
k
p1⋅ v1 = p2⋅ v2
k
k− 1
Rearranging
⎛ p2 ⎞
=⎜
T1
⎝ p1 ⎠
T2
or
⎛ R ⋅ T1 ⎞
⎛ R ⋅ T2 ⎞
p1⋅ ⎜
= p2⋅ ⎜
⎝ p1 ⎠
⎝ p2 ⎠
k
k− 1
⎤
⎡
⎢
⎥
k
R⋅ T1 ⎢⎛ p 2 ⎞
⎥
Combining with Eq. 1 w =
⋅ ⎢⎜
− 1⎥
k−1
⎣⎝ p 1 ⎠
⎦
1.4− 1
⎡⎢
⎤⎥
1.4
⎢ 4
⎥
1
J
w =
× 286.9 ⋅
× ( 20 + 273 ) K × ⎢⎛⎜ ⎞
− 1⎥
0.4
kg⋅ K
⎣⎝ 1 ⎠
⎦
w = 102
kJ
kg
k
Problem 4.9
[Difficulty: 2]
Given:
Data on cooling of a can of soda in a refrigerator
Find:
How long it takes to warm up in a room
Solution:
The First Law of Thermodynamics for the can (either warming or cooling) is
M ⋅ c⋅
dT
dt
(
)
= −k ⋅ T − Tamb
dT
or
dt
(
)
= −k ⋅ T − Tamb
where M is the can mass, c is the average specific heat of the can and its contents, T is the temperature, and Tamb is the ambient
temperature
Separating variables
Integrating
dT
T − Tamb
= −A⋅ dt
(
)
T( t) = Tamb + Tinit − Tamb ⋅ e
− At
where Tinit is the initial temperature. The available data from the coolling can now be used to obtain a value for constant A
Given data for cooling
Hence
Tinit = 80 °F
Tinit = 540 ⋅ R
Tamb = 35 °F
Tamb = 495 ⋅ R
T = 45 °F
T = 505 ⋅ R
when
τ = 2 ⋅ hr
k=
1
τ
⎛ Tinit − Tamb ⎞
⋅ ln⎜
⎝
T − Tamb
⎠
=
1
2 ⋅ hr
×
1 ⋅ hr
3600⋅ s
× ln⎛⎜
540 − 495 ⎞
⎝ 505 − 495 ⎠
−4 −1
k = 2.09 × 10
s
Then, for the warming up process
Tinit = 45 °F
Tinit = 505 ⋅ R
Tend = 60 °F
Tend = 520 ⋅ R
(
)
with
Tend = Tamb + Tinit − Tamb ⋅ e
Hence the time τ is
τ=
1
k
⎛ Tinit − Tamb ⎞
⋅ ln⎜
⎝ Tend − Tamb ⎠
=
Tamb = 72 °F
Tamb = 532 ⋅ R
− kτ
s
2.09⋅ 10
−4
⋅ ln⎛⎜
505 − 532 ⎞
⎝ 520 − 532 ⎠
3
τ = 3.88 × 10 s
τ = 1.08⋅ hr
Problem 4.10
Given:
Data on heating and cooling a copper block
Find:
Final system temperature
[Difficulty: 2]
Solution:
Basic equation
Q  W  ∆E
Assumptions: 1) Stationary system ΔE = ΔU 2) No work W = 0 3) Adiabatic Q = 0
Then for the system (water and copper)
∆U  0
or


M copper  ccopper  Tcopper  M w cw Tw  M copper  ccopper  M w cw  Tf
where Tf is the final temperature of the water (w) and copper (copp er)
The given data is
M copper  5  kg
ccopper  385 
J
kg K
Tcopper  ( 90  273 )  K
Also, for the water
ρ  999 
kg
3
so
Tf 
M w  ρ V
M copper  ccopper  Tcopper  M w cw Tw
Mcopper ccopper  Mw cw
Tf  291 K
J
kg K
Tw  ( 10  273 )  K
m
Solving Eq. 1 for Tf
cw  4186
Tf  18.1 °C
M w  4.00 kg
V  4 L
(1)
Problem 4.11
[Difficulty: 2]
Given:
Data on heat loss from persons, and people-filled auditorium
Find:
Internal energy change of air and of system; air temperature rise
Solution:
Basic equation
Q  W  ∆E
Assumptions: 1) Stationary system ΔE =ΔU 2) No work W = 0
Then for the air
For the air and people
∆U  Q  85
W
person
 6000 people  15 min 
60 s
∆U  459  MJ
min
∆U  Qsurroundings  0
The increase in air energy is equal and opposite to the loss in people energy
For the air
Hence
From Table A.6
∆U  Q
∆T 
but for air (an ideal gas)
Q
Rair  286.9 
∆T 
286.9
717.4
M  ρ V 
with
p V
Rair T
Rair Q T

M  cv
∆U  M  cv  ∆T
cv  p  V
J
kg K
and
cv  717.4 
 459  10  J  ( 20  273 ) K 
This is the temperature change in 15 min. The rate of change is then
kg K
2
1
6
J

m
3 N

101  10
∆T
15 min
 6.09
1
3.5  10
K
hr
5

1
3
m
∆T  1.521 K
Problem 4.12
[Difficulty: 3]
Given:
Data on velocity field and control volume geometry
Find:
Several surface integrals
Solution:

dA1 = wdzˆj − wdykˆ

dA1 = dzˆj − dykˆ

dA2 = − wdykˆ

dA2 = − dykˆ
(

V = aˆj + bykˆ
(a)
(b)
)
(

V = 10 ˆj + 5 ykˆ
(
)(
)
)

V ⋅ dA1 = 10 ˆj + 5 ykˆ ⋅ dzˆj − dykˆ = 10dz − 5 ydy
1
1
1

5
1
V ⋅ dA1 = 10dz − 5 ydy = 10 z 0 − y 2 = 7.5
A1
2 0
∫
∫
∫
0
0
(
)(
)
(c)

V ⋅ dA2 = 10 ˆj + 5 ykˆ ⋅ − dykˆ = −5 ydy
(d)
 
V V ⋅ dA2 = − 10 ˆj + 5 ykˆ 5 ydy
(e)
z
(
) (
)
) ∫(
)
1
1
 
1
25 3 ˆ
V V ⋅ dA2 = − 10 ˆj + 5 ykˆ 5 ydy = − 25 y 2 ˆj −
y k = −25 ˆj − 8.33kˆ
0
A2
3
0
∫ (
0


Control
volume
y
Problem 4.13
[Difficulty: 3]
Given:
Data on velocity field and control volume geometry
Find:
Volume flow rate and momentum flux
z
Solution:
3m
First we define the area and velocity vectors
y

dA = dydziˆ + dydyxkˆ

V = axiˆ + byˆj
4m
5m

V = xiˆ + yˆj
or
x
We will need the equation of the surface: z = 3 −
4
3
x or x = 4 − z
4
3
Then
a)
Volume flow rate
)(
)
3
3 5
3
 
2 ⎞
4 ⎞
⎛
⎛
Q = ∫ V ⋅ dA = ∫ xiˆ + yˆj ⋅ dydziˆ + dxdykˆ = ∫∫ xdydz = 5∫ ⎜ 4 − z ⎟dz = 5⎜ 4 z − z 2 ⎟
A
A
3 ⎠
3 ⎠0
⎝
0 0
0⎝
(
m3
m3
= 30
s
s
Q = (60 − 30 )
b) Momentum flux
(

)
(
)
ρ ∫ V V ⋅ dA = ρ ∫ (xiˆ + yˆj )(xiˆ + yˆj )⋅ dydziˆ + dxdykˆ = ρ ∫ (xiˆ + yˆj )( xdydz )
 
A
A
A
2
4 ⎞
y2
⎛
= ρ ∫ ∫ x dydziˆ + ρ ∫ ∫ xydydzˆj = 5∫ ⎜ 4 − z ⎟ dziˆ +
3 ⎠
2
0 0
0 0
0⎝
3 5
3 5
3
2
3
5 3
⎛
4 ⎞
∫ ⎜⎝ 4 − 3 z ⎟⎠dzˆj
00
3
32
16 ⎞
25 ⎛
2 ⎞
16
16 3 ⎞ ˆ 25
⎛
⎛
= 5∫ ⎜16 − z + z 2 ⎟dziˆ + ⎜ 4 z − z 2 ⎟ ˆj == 5⎜16 z − z 2 +
z ⎟ i + (12 − 6) ˆj
3
9
2
3
3
27
2
⎝
⎠0
⎝
⎠0
⎠
0⎝
= 5(48 − 48 + 16)iˆ + 75 ˆj
3
Momentum flux = 80iˆ + 75 ˆj N
Problem 4.14
[Difficulty: 3]
Given:
Data on velocity field and control volume geometry
Find:
Surface integrals
z
4m
3m
Solution:
5m
First we define the area and velocity vectors

dA = dydziˆ + dxdzˆj


V = axiˆ + byˆj + ckˆ or V = 2 xiˆ + 2 yˆj + kˆ
We will need the equation of the surface: y =
y
3
2
x or x = y
3
2
x
Then
∫V ⋅ dA = ∫ (− axiˆ + byˆj + ckˆ )⋅ (dydziˆ − dxdzˆj )

A
A
2 3
2 2
2
3
2
3
2
2
3
1
3
= ∫∫ − axdydz − ∫∫ bydxdz = −a ∫ dz ∫ ydy − b ∫ dz ∫ xdx = − 2a y 2 − 2b x 2
3
2
3 0
4
0 0
0 0
0
0
0
0
Q = (− 6a − 6b ) = −24
m
s
∫ (
A
3
2
3
x or x = y , and also dy = dx and a = b
2
3
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
− axi + byj + ck − axi + byj + ck ⋅ dydzi − dxdzj
) (
)(
)(
= ∫ (− axiˆ + byˆj + ckˆ )(− axdydz − bydxdz )
A
0
3
We will again need the equation of the surface: y =
  
V V ⋅ dA = ∫
2
)
A
3
3
3
⎞
⎛
⎞⎛
= ∫ ⎜ − axiˆ + axˆj + ckˆ ⎟⎜ − ax dxdz − a xdxdz ⎟
A
2
2
2
⎠
⎝
⎠⎝
3
⎞
⎛
= ∫ ⎜ − axiˆ + axˆj + ckˆ ⎟(− 3axdxdz )
A
2
⎠
⎝
2 2
2 2
2 2
9
= 3∫ ∫ a x dxdziˆ − ∫ ∫ a 2 x 2 dxdzˆj − 3∫ ∫ acxdxdzkˆ
200
0 0
0 0
2
2
3
⎛
⎞
⎟iˆ − (9)⎜ a 2 x
⎜
⎟
3
0⎠
⎝
m4
= 64iˆ − 96 ˆj − 60kˆ
s2
⎛ x3
= (6)⎜ a 2
⎜
3
⎝
2
2
⎛
⎞
⎟ ˆj − (6)⎜ ac x
⎜
⎟
2
0⎠
⎝
2
⎞
⎟ = 16a 2 iˆ − 24a 2 ˆj − 12ackˆ
⎟
0⎠
2
Problem 4.15
[Difficulty: 2]
Given:
Control Volume with linear velocity distribution
Find:
Volume flow rate and momentum flux
Solution:
Apply the expressions for volume and momentum flux
Governing equations:
Assumption:
∫ (

Q = V ⋅ dA
  
mf = ρ V V ⋅ dA
A
A
∫
)
(1) Incompressible flow
For a linear velocity profile
 V
V = yiˆ
h

dA = − w dyiˆ
and also
For the volume flow rate:
h
Q=
∫
(
)
Vw
V ˆ
i ⋅ − w dyiˆ = −
h
h
y =0
h
∫
Vw y 2
y dy = −
h 2
y =0
h
0
1
Q = − Vhw
2
The momentum flux is
h
mf =
h
V ˆ ⎛
V 2w y3
Vw
V 2w
⎞
i ⋅⎜− ρ
ydy ⎟ = − ρ 2 iˆ y 2 dy = − ρ 2 iˆ
3
h ⎝
h
h
h
⎠
y =0
y =0
∫
1
mf = − ρV 2 whiˆ
3
∫
h
0
Problem 4.16
Given:
Control Volume with linear velocity distribution
Find:
Kinetic energy flux
Solution:
Apply the expression for kinetic energy flux
V2  
ρV ⋅ dA
A 2
∫
Governing equation:
kef =
Assumption:
(1) Incompressible flow
For a linear velocity profile
[Difficulty: 2]
 V
V = yiˆ
h
V (y) =
V
y
h

dA = − w dyiˆ
and also
The kinetic energy flux is
h
2
V 3w
Vw
1 ⎛V ⎞ ⎛
⎞
ydy ⎟ = − ρ
kef =
⎜ y⎟ ⎜− ρ
h
2⎝h ⎠ ⎝
2h 3
⎠
y =0
∫
1
kef = − ρV 3 wh
8
h
∫
V 3w y 4
y dy = − ρ
2h 3 4
y =0
h
3
0
Problem 4.17
[Difficulty: 2]
Given:
Control Volume with parabolic velocity distribution
Find:
Volume flow rate and momentum flux
Solution:
Apply the expressions for volume and momentum flux
Governing equations:
Assumption:
∫ (

Q = V ⋅ dA
  
mf = ρ V V ⋅ dA
A
A
∫
)
(1) Incompressible flow
For a linear velocity profile
⎡ ⎛ r ⎞2 ⎤

V = uiˆ = umax ⎢1 − ⎜ ⎟ ⎥iˆ and also
⎢⎣ ⎝ R ⎠ ⎥⎦

dA = 2πrdriˆ
For the volume flow rate:
h
R
⎡ ⎛ r ⎞2 ⎤
⎡
⎡ r2
⎡ R2 R2 ⎤
r3 ⎤
r4 ⎤
ˆ
ˆ
2
u
Q = umax ⎢1 − ⎜ ⎟ ⎥i ⋅ 2πrdri = 2πumax
=
−
π
⎢ r − 2 ⎥ dy = 2πumax ⎢ −
⎥
⎢
⎥
max
2
4 ⎦⎥
R ⎦⎥
⎢
⎢⎣ ⎝ R ⎠ ⎥⎦
⎣⎢ 2 4 R ⎦⎥ 0
⎣⎢ 2
r =0
y =0 ⎣
R
(
∫
Q=
)
∫
1
πumax R 2
2
The momentum flux is
R
⎫
⎫⎪
⎡ ⎛ r ⎞ 2 ⎤ ⎧⎪
⎡ ⎛ r ⎞2 ⎤
⎡ ⎛ r ⎞ 2 ⎤ ⎧⎪
⎡
r3 ⎤ ⎪
ˆ
ˆ
ˆ
ˆ
mf =
umax ⎢1 − ⎜ ⎟ ⎥i ⎨umax ⎢1 − ⎜ ⎟ ⎥i ⋅ 2πrdri ⎬ = umax ⎢1 − ⎜ ⎟ ⎥i ⎨2πumax
⎢ r − 2 ⎥ dr ⎬
R ⎥⎦ ⎪
⎢
⎢⎣ ⎝ R ⎠ ⎥⎦ ⎪⎩
⎢⎣ ⎝ R ⎠ ⎥⎦
⎢⎣ ⎝ R ⎠ ⎥⎦ ⎪
⎪⎭
r =0
y =0 ⎣
⎭
⎩
R
(
∫
R
⎡ 2r 3 r 5 ⎤
⎢ r − 2 + 4 ⎥ dr
R
R ⎥⎦
⎢
y =0 ⎣
∫
=
2 ˆ
2πumax
i
=
2 ˆ⎡ r
2πumax
i⎢
h
r6 ⎤
−
+
⎥
2
6 R 4 ⎥⎦ 0
⎢⎣ 2 4 R
2
r4
2
R2 R2 ⎤
2 ˆ⎡ R
= 2πumax
−
+
i⎢
⎥
4
6 ⎦⎥
⎣⎢ 2
1 2
mf = πumax
R 2iˆ
3
)
∫
Problem 4.18
Given:
Control Volume with parabolic velocity distribution
Find:
Kinetic energy flux
Solution:
Apply the expressions for kinetic energy flux
V2  
ρV ⋅ dA
A 2
∫
Governing equation:
kef =
Assumption:
(1) Incompressible flow
For a linear velocity profile
[Difficulty: 2]
⎡ ⎛ r ⎞2 ⎤

V = uiˆ = umax ⎢1 − ⎜ ⎟ ⎥iˆ
⎣⎢ ⎝ R ⎠ ⎦⎥
⎡ ⎛ r ⎞2 ⎤
V = u = umax ⎢1 − ⎜ ⎟ ⎥
⎣⎢ ⎝ R ⎠ ⎦⎥
For the volume flow rate:
2
⎫⎪
⎡ ⎛ r ⎞ 2 ⎤ ⎫⎪ ⎧⎪
⎡ ⎛ r ⎞2 ⎤
1 ⎧⎪
kef =
⎨umax ⎢1 − ⎜ ⎟ ⎥ ⎬ ρ ⎨umax ⎢1 − ⎜ ⎟ ⎥iˆ ⋅ 2πrdriˆ ⎬
2⎪
⎢⎣ ⎝ R ⎠ ⎥⎦ ⎪⎭ ⎩⎪
⎢⎣ ⎝ R ⎠ ⎥⎦
⎪⎭
r =0 ⎩
R
(
∫
2
4
⎡ ⎛ r ⎞ 2 ⎤ ⎫⎪
1 2 ⎡
⎛ r ⎞ ⎛ r ⎞ ⎤ ⎧⎪
umax ⎢1 − 2⎜ ⎟ + ⎜ ⎟ ⎥ ρ ⎨2πumax ⎢1 − ⎜ ⎟ ⎥ rdr ⎬
2
⎝ R ⎠ ⎝ R ⎠ ⎥⎦ ⎩⎪
⎢⎣
⎢⎣ ⎝ R ⎠ ⎥⎦ ⎪⎭
r =0
R
=
∫
R
=
⎢⎣
⎛r⎞
⎝R⎠
⎡
r3
⎢⎣
R2
2
⎛r⎞
⎝R⎠
4
6
⎛r⎞ ⎤
⎝ R ⎠ ⎥⎦
∫
∫
3
πρumax
⎢r − 3
r =0
R
=
⎡
3
πρumax
⎢1 − 3⎜ ⎟ + 3⎜ ⎟ − ⎜ ⎟ ⎥ rdr
r =0
+3
r5
R4
−
r7 ⎤
⎥ dr
R 6 ⎥⎦
h
2
r6
r8 ⎤
3r 4
3 ⎡r
= πρumax
+
− 6⎥
⎢ −
2
4
2R
8R ⎦⎥ 0
⎣⎢ 2 4 R
1
3
kef = πρumax
R2
8
)

and also dA = 2πrdriˆ
Problem 4.19
Given:
Data on flow through nozzles
Find:
Exit velocity in each jet; velocity in pipe
[Difficulty: 1]
Solution:
Basic equation
→→
(
∑ V⋅A) = 0
CS
Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow
The given data is
Area of each nozzle
Q = 2.2⋅ gpm
A =
π
4
⋅d
2
d =
1
32
⋅ in
n = 50
−4
A = 7.67 × 10
2
in
Area of the pipe
π 2
Apipe = ⋅ D
4
Apipe = 0.442in
Total area of nozzles
Atotal = n ⋅ A
Atotal = 0.0383in
Then for the pipe flow
2
2
V =
The jet speeds are then
Q
Atotal
V = 18.4
ft
→→
(
∑ V⋅A) = −Vpipe⋅Apipe + n⋅V⋅A = 0
n⋅ A
d
Vpipe = V⋅
= V⋅ n ⋅ ⎛⎜ ⎞
Apipe
⎝ D⎠
2
⎛
⎜
ft
Vpipe = 18.4⋅ × 50 × ⎜
s
⎜
⎝
1
⎞
32
⎟
3
4
⎠
3
(Note that gal = 231 in )
s
CS
Hence
D =
(Number of nozzles)
2
ft
Vpipe = 1.60⋅
s
3
4
⋅ in
Problem 4.20
Given:
Data on flow through nozzles
Find:
Average velocity in head feeder; flow rate
[Difficulty: 1]
Solution:
Basic equation
→→
(
∑ V⋅A) = 0
CS
Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow
Then for the nozzle flow
→→
(
∑ V⋅A) = −Vfeeder⋅Afeeder + 10⋅Vnozzle⋅Anozzle = 0
CS
Hence
⎛ Dnozzle ⎞
Vfeeder = Vnozzle⋅
= Vnozzle⋅ 10⋅ ⎜
Afeeder
⎝ Dfeeder ⎠
10⋅ Anozzle
⎛ 1⎞
⎜ 8
ft
Vfeeder = 10⋅ × 10 × ⎜
s
⎝1⎠
2
2
ft
Vfeeder = 1.56⋅
s
2
The flow rate is
Q = Vfeeder⋅ Afeeder = Vfeeder⋅
Q = 1.56⋅
ft
s
×
π
4
× ⎛⎜ 1 ⋅ in ×
⎝
π⋅ Dfeeder
1 ⋅ ft
4
2
⎞ × 7.48⋅ gal × 60⋅ s
12⋅ in ⎠
1 ⋅ min
3
1 ⋅ ft
Q = 3.82⋅ gpm
Problem 4.21
[Difficulty: 3]
Given:
Data on flow into and out of tank
Find:
Time at which exit pump is switched on; time at which drain is opened; flow rate into drain
Solution:
∂
Basic equation
∂t
M CV +
→→
(
ρ⋅ V⋅ A ) = 0
∑
CS
Assumptions: 1) Uniform flow 2) Incompressible flow
∂
After inlet pump is on
∂t
M CV +
→→
∂
(
ρ⋅ V⋅ A ) = M tank − ρ⋅ Vin⋅ Ain = 0
∑
∂t
CS
⎛ Din ⎞
= Vin⋅
= Vin⋅ ⎜
Atank
dt
⎝ Dtank ⎠
Ain
dh
Hence the time to reach hexit = 0.7 m is
texit =
h exit
dh
∂
dh
M tank = ρ⋅ Atank ⋅
= ρ⋅ Vin⋅ Ain
dt
∂t
2
where h is the level of water in the tank
⎛ Dtank ⎞
=
⋅⎜
Vin
⎝ Din ⎠
h exit
2
texit = 0.7⋅ m ×
1 s
⋅ ×
5 m
⎛ 3⋅ m ⎞
⎜ 0.1⋅ m
⎝
⎠
2
texit = 126 s
dt
∂
After exit pump is on
∂t
→→
dh
∂
(
= Vin⋅ Ain − Vexit ⋅ Aexit
ρ⋅ V⋅ A) = M tank − ρ⋅ Vin⋅ Ain + ρ⋅ Vexit ⋅ Aexit = 0 Atank⋅
∑
dt
∂t
M CV +
CS
2
⎛ Din ⎞
⎛ Dexit ⎞
= Vin⋅
− Vexit ⋅
= Vin⋅ ⎜
− Vexit ⋅ ⎜
Atank
Atank
dt
⎝ Dtank ⎠
⎝ Dtank ⎠
dh
Ain
Hence the time to reach hdrain = 2 m is
Aexit
tdrain = texit +
(hdrain − hexit)
dh
=
(hdrain − hexit)
2
⎛ Din ⎞
⎛ Dexit ⎞
Vin⋅ ⎜
− Vexit ⋅ ⎜
⎝ Dtank ⎠
⎝ Dtank ⎠
dt
tdrain = 126 ⋅ s + ( 2 − 0.7) ⋅ m ×
2
2
1
2
0.1⋅ m ⎞
0.08⋅ m ⎞
m
5 ⋅ × ⎛⎜
− 3 ⋅ × ⎛⎜
s
s
⎝ 3⋅ m ⎠
⎝ 3⋅ m ⎠
m
2
tdrain = 506 s
The flow rate into the drain is equal to the net inflow (the level in the tank is now constant)
2
Qdrain = Vin⋅
π⋅ Din
4
− Vexit ⋅
π⋅ Dexit
4
2
m π
m π
2
2
Qdrain = 5 ⋅ ×
× ( 0.1⋅ m) − 3 ⋅ ×
× ( 0.08⋅ m)
s
4
s
4
3
m
Qdrain = 0.0242
s
Problem 4.22
[Difficulty: 1]
Given:
Data on wind tunnel geometry
Find:
Average speeds in wind tunnel; diameter of section 3
Solution:
Basic equation
Q  V A
Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow
3
Given data:
Q  15
m
D1  1.5 m
s
2
Between sections 1 and 2
Hence
Q  V1 A1  V1
V1 
π
4
For section 3 we can use
V1 
Q
 D1
π D1
4
2
2
 V3 
π D1
2
 V2 A2  V2
4
m
V1  8.49
s
π D2
4
V2 
π
4
π D3
4
m
V3  75
s
D2  1 m
2
or
Q
 D2
2
V1
D3  D1 
V3
m
V2  19.1
s
D3  0.505 m
Problem 4.23
Difficulty: 4]
Given:
Data on flow into and out of cooling tower
Find:
Volume and mass flow rate of cool water; mass flow rate of moist and dry air
Solution:
→→
(
ρ⋅ V⋅ A) = 0
∑
Basic equation
Q = V⋅ A
and at each inlet/exit
CS
Assumptions: 1) Uniform flow 2) Incompressible flow
5 lb
mwarm = 2.5⋅ 10 ⋅
hr
Given data:
At the cool water exit
The mass flow rate is
Qcool = V⋅ A
mcool = ρ⋅ Qcool
D = 6 ⋅ in
V = 5⋅
mcool = 1.94⋅
ft
3
× 0.982 ⋅
ft
lb
ρmoist = 0.065 ⋅
3
ft
s
3
ft π
2
Qcool = 5 ⋅ ×
× ( 0.5⋅ ft)
s
4
slug
ft
3
s
ft
Qcool = 0.982 ⋅
s
Qcool = 441 ⋅ gpm
slug
mcool = 1.91⋅
s
5 lb
mcool = 2.21 × 10 ⋅
hr
NOTE: Software does not allow dots over terms, so m represents mass flow rate, not mass!
For the water flow we need
→→
(
ρ⋅ V⋅ A) = 0
∑
to balance the water flow
CS
We have
−mwarm + mcool + mv = 0
mv = mwarm − mcool
This is the mass flow rate of water vapor. To obtain air flow rates, from psychrometrics
where x is the relative humidity. It is also known (try Googling "density of moist air") that
lb
mv = 29341 ⋅
hr
x=
mv
mair
ρmoist
ρdry
We are given
lb
ρmoist = 0.065 ⋅
3
ft
=
1+x
1 + x⋅
RH2O
Rair
For dry air we could use the ideal gas equation
ρdry = 0.002377⋅
slug
ft
3
p
ρdry =
R⋅ T
but here we use atmospheric air density (Table A.3)
ρdry = 0.002377⋅
slug
ft
3
× 32.2⋅
lb
slug
lb
ρdry = 0.0765⋅
3
ft
Note that moist air is less dense than dry air!
Hence
0.065
0.0765
x =
1+x
=
1 + x⋅
0.065 ⋅
Hence
mv
mair
53.33
0.0765 − 0.065
=x
85.78
53.33
using data from Table A.6
85.78
x = 0.410
− .0765
leads to
Finally, the mass flow rate of moist air is
mv
mair =
x
lb
1
mair = 29341 ⋅ ×
hr 0.41
mmoist = mv + mair
5 lb
mmoist = 1.01 × 10 ⋅
hr
lb
mair = 71563 ⋅
hr
Problem 4.24
Given:
Data on flow through box
Find:
Velocity at station 3
[Difficulty: 1]
Solution:
Basic equation
→→
(
∑ V⋅A) = 0
CS
Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow
Then for the box
∑( )
→→
V⋅ A = −V1⋅ A1 + V2⋅ A2 + V3⋅ A3 = 0
CS
Note that the vectors indicate that flow is in at location 1 and out at location 2; we assume outflow at location 3
Hence
A1
A2
V3 = V1 ⋅
− V2 ⋅
A3
A3
ft 0.5
ft 0.1
V3 = 10⋅ ×
− 20⋅ ×
s
0.6
s
0.6
Based on geometry
Vx = V3 ⋅ sin( 60⋅ deg)
ft
Vx = 4.33⋅
s
Vy = −V3 ⋅ cos( 60⋅ deg)
ft
Vy = −2.5⋅
s
→
⎯
ft
ft
V3 = ⎛⎜ 4.33⋅ , −2.5⋅ ⎞
s
s⎠
⎝
ft
V3 = 5 ⋅
s
Problem 4.25
Given:
Data on flow through device
Find:
Volume flow rate at port 3
[Difficulty: 1]
Solution:
Basic equation
→→
(
∑ V⋅A) = 0
CS
Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow
Then for the box
∑( )
→→
V⋅ A = −V1⋅ A1 + V2⋅ A2 + V3⋅ A3 = −V1⋅ A1 + V2⋅ A2 + Q3
CS
Note we assume outflow at port 3
Hence
Q3 = V1 ⋅ A1 − V2 ⋅ A2
The negative sign indicates the flow at port 3 is inwards.
m
m
2
2
Q3 = 3 ⋅ × 0.1⋅ m − 10⋅ × 0.05⋅ m
s
s
Flow rate at port 3 is 0.2 m3/s inwards
3
m
Q3 = −0.2⋅
s
Problem 4.26
Given:
Water needs of farmer
Find:
Number of supply pipes needed
[Difficulty: 1]
Solution:
Basic equation
Q  V A
Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow
The given data is:
Then
A  150 m 400 m
Q 
If n is the number of pipes
The farmer needs 5 pipes.
A h
t
4 2
A  6  10 m
h  7.5 cm
t  1  hr
or
n 
D  37.5 cm
3
Q  1.25
Q  V
π
4
m
s
2
D n
4 Q
2
π V D
n  4.527
V  2.5
m
s
Problem 4.27
Given:
Data on filling of glass carboy
Find:
Time to fill
[Difficulty: 1]
Solution:
We can treat this as an unsteady problem if we choose the CS as the entire carboy
Basic equation
∂
∂t
→→
(
ρ⋅ V⋅ A) = 0
∑
M CV +
CS
Assumptions: 1) Incompressible flow 2) Uniform flow
Given data:
Q = 3 ⋅ gpm
Hence
∂
∂t
M CV = ρ⋅ A⋅
D = 15⋅ in
dh
dt
= ρ⋅ A⋅
h
τ
h = 2 ⋅ ft
→→
(
ρ⋅ V⋅ A) = ρ⋅ Q
∑
=−
CS
where Q is the fill rate, A is the carboy cross-section area, dh/dt is the rate of rise in the carboy, and τ is the fill time
π
Hence
τ =
4
2
⋅D ⋅h
Q
τ = 6.12⋅ min
Problem 4.28
Given:
Data on filling of a sink
Find:
Time to half fill; rate at which level drops
[Difficulty: 1]
Solution:
This is an unsteady problem if we choose the CS as the entire sink
Basic equation
∂
∂t
→→
(
ρ⋅ V⋅ A) = 0
∑
M CV +
CS
Assumptions: 1) Incompressible flow
Given data:
mrate = 4 ⋅ gpm
Hence
∂
To half fill:
Then, using Eq 1
∂t
w = 18⋅ in
d = 12⋅ in
→→
(
ρ⋅ V⋅ A) = Inflow − Outflow
∑
M CV = −
V
Q = 4 ⋅ gpm
Qdrain = 1 ⋅ gpm
(1)
CS
V =
τ
L = 2 ⋅ ft
1
2
⋅ L⋅ w⋅ d
=Q
After the drain opens, Eq. 1 becomes
V = 1.5 ft
τ =
dV
dt
Qdrain
Vlevel = −
L⋅ w
V
Q
3
V = 11.2 gal
τ = 168 s
τ = 2.81 min
= L⋅ w⋅ Vlevel = −Qdrain
where V level is the speed of water level drop
− 4 ft
Vlevel = −7.43 × 10
s
in
Vlevel = −0.535
min
Problem 4.29
[Difficulty: 1]
Given:
Air flow system
Find:
Flow rate and velocity into each room; narrowest supply duct
Solution:
Basic equation
Q  V A
Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow
The given data is:
L
Qperson  8 
s
n rooms  6
n students  20
h  200  mm
w  500  mm
m
Vmax  1.75
s
Then for each room
Qroom  n students  Qperson
L
Qroom  160
s
and
Vroom 
Qroom
m
Vroom  1.6
s
For the supply duct
Q  n rooms Qroom
and
w h
Q  960
L
s
3
m
Qroom  0.16
s
3
Q  0.96
m
s
Q  Vmax A  Vmax w h where w and h are now the supply duct dimensions
w 
Q
Vmax h
w  1.097 m
h  500  mm
Problem 4.30
Given:
Data on filling of a basement during a storm
Find:
Flow rate of storm into basement
[Difficulty: 1]
Solution:
This is an unsteady problem if we choose the CS as the entire basement
Basic equation
∂
∂t
→→
(
ρ⋅ V⋅ A) = 0
∑
M CV +
CS
Assumptions: 1) Incompressible flow
Given data:
Hence
or
dh
Qpump = 27.5⋅ gpm
∂
∂t
M CV = ρ⋅ A⋅
dh
dt
dt
= 4⋅
in
hr
A = 30⋅ ft⋅ 20⋅ ft
→→
(
ρ⋅ V⋅ A) = ρ⋅ Qstorm − ρ⋅ Qpump
∑
=−
CS
dh
Qstorm = Qpump − A⋅
dt
gal
Qstorm = 27.5⋅
− 30⋅ ft × 20⋅ ft ×
min
Qstorm = 2.57⋅ gpm
⎛ 4 ⋅ ft ⎞ × 7.48⋅ gal × 1 ⋅ hr
⎜ 12 hr
60⋅ min
3
⎝
⎠
ft
where A is the basement area
and dh/dt is the rate at which the
height of water in the basement
changes.
Data on gals from Table G.2
Problem 4.31
Given:
Data on compressible flow
Find:
Downstream density
[Difficulty: 1]
Solution:
Basic equation
→→
(
ρ⋅ V⋅ A ) = 0
∑
CS
Assumptions: 1) Steady flow 2) Uniform flow
Then for the box
→→
(
ρ⋅ V⋅ A ) = −ρu ⋅ Vu⋅ Au + ρd ⋅ Vd ⋅ Ad = 0
∑
CS
Hence
Vd ⋅ Ad
ρu = ρd ⋅
Vu⋅ Au
1000⋅
m
2
kg
s 0.1⋅ m
ρu = 1⋅
⋅
⋅
3
m
2
m 1500⋅
0.25⋅ m
s
kg
ρu = 0.267
3
m
Problem 4.32
[Difficulty: 2]
Given:
Data on flow through device
Find:
Velocity V3; plot V3 against time; find when V3 is zero; total mean flow
Solution:
Governing equation:
−
V3 =
V1⋅ A1 + V2⋅ A2
A3
−
V3 = 6.67⋅ e
→→
V⋅ A = 0
∑
−V1⋅ A1 − V2⋅ A2 + V3⋅ A3 = 0
Applying to the device (assuming V3 is out)
The velocity at A3 is
⌠ →→
⎮
⎮ V dA =
⌡
For incompressible flow (Eq. 4.13) and uniform flow
10⋅ e
t
2 m
⋅
=
s
2
× 0.1⋅ m + 2⋅ cos ( 2⋅ π⋅ t) ⋅
m
s
2
× 0.2⋅ m
2
0.15⋅ m
t
2
+ 2.67⋅ cos( 2 ⋅ π⋅ t)
The total mean volumetric flow at A3 is
∞
⌠
⎮
∞
⎮
⌠
Q = ⎮ V3 ⋅ A3 dt = ⎮
⌡
⌡
0
⎛
−
⎜
⎝ 6.67⋅ e
⎛
−
⎜
Q = lim ⎜ −2 ⋅ e
t→∞⎝
1
⎞
t
2
+ 2.67⋅ cos( 2 ⋅ π⋅ t) ⎠ ⋅ 0.15 dt⋅ ⎜⎛
⎝s
0
2⎞
⋅m
⎠
⎞
t
2
m
+
5⋅ π
⋅ sin( 2 ⋅ π⋅ t)
3
⎠
− ( −2 ) = 2 ⋅ m
The time at which V3 first is zero, and the plot of V3 is shown in the corresponding Excel workbook
3
Q = 2⋅ m
t = 2.39⋅ s
t (s) V 3 (m/s)
9.33
8.50
6.86
4.91
3.30
2.53
2.78
3.87
5.29
6.41
6.71
6.00
4.48
2.66
1.15
0.48
0.84
2.03
3.53
4.74
5.12
4.49
3.04
1.29
-0.15
-0.76
Exit Velocity vs Time
10
8
V 3 (m/s)
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.10
1.20
1.30
1.40
1.50
1.60
1.70
1.80
1.90
2.00
2.10
2.20
2.30
2.40
2.50
6
4
2
0
0.0
0.5
1.0
1.5
2.0
-2
t (s)
The time at which V 3 first becomes zero can be found using Goal Seek
t (s)
V 3 (m/s)
2.39
0.00
2.5
Problem 4.33
Given:
Data on flow down an inclined plane
Find:
Find u max
[Difficulty: 2]
Solution:
Basic equation
⌠
mflow = ⎮ ρu dA
⌡
Assumptions: 1) Steady flow 2) Incompressible flow
Evaluating at 1 and 2
h
h
0
0
⌠
⌠
2
2
⎮
ρ ⋅ g ⋅ sin ( θ) ⋅ w ⎮
y ⎞
ρ⋅ g ⋅ sin ( θ) ⎛
mflow = ⎮ ρ⋅
⋅⎜h ⋅y −
⋅ w dy =
⋅⎮
2 ⎠
μ
μ
⎮
⎝
⎮
⌡
⌡
2
mflow =
ρ ⋅ g ⋅ sin ( θ) ⋅ w
μ
2
Hence
mflow =
⎛ h3
⋅⎜
⎝ 2
ρ ⋅ g ⋅ sin( θ) ⋅ w⋅ h
3⋅ μ
3
−
h
3⎞
6
⎠
2
⎛
y ⎞
⎜h ⋅y −
dy
2 ⎠
⎝
Problem 4.34
[Difficulty: 2]
y
2h

Given:
Data on flow at inlet and outlet of channel
Find:
Find u max
x

CS
Solution:
Basic equation


∫ ρ V ⋅ dA = 0
CS
Assumptions: 1) Steady flow 2) Incompressible flow
Evaluating at 1 and 2
⌠
−ρ⋅ U⋅ 2 ⋅ h ⋅ w + ⎮
⌡
h
ρ⋅ u ( y ) dy = 0
−h
⎡
u max⋅ ⎢[ h − ( −h ) ] −
⎢
⎣
Hence
u max =
3
2
⋅U =
3
2
⌠
⎮
⎮
⎮
⌡
h
−h
⎡ 3 ⎛ 3 ⎞⎤⎤
⎢ h − ⎜ − h ⎥⎥ = 2⋅ h ⋅ U
⎢ 3 ⋅ h2 ⎜ 3 ⋅ h2 ⎥⎥
⎣
⎝
⎠⎦⎦
× 2.5⋅
m
s
⎡
u max⋅ ⎢1 −
⎣
2
⎛ y ⎞ ⎥⎤ dy = 2 ⋅ h⋅ U
⎜h
⎝ ⎠⎦
4
u max⋅ ⋅ h = 2 ⋅ h ⋅ U
3
u max = 3.75⋅
m
s
Problem 4.35
Given:
Data on flow at inlet and outlet of pipe
Find:
Find U
[Difficulty: 2]
Solution:
Basic equation


∫ ρ V ⋅ dA = 0
CS
Assumptions: 1) Steady flow 2) Incompressible flow
Evaluating at inlet and exit
2 ⌠
−ρ⋅ U⋅ π⋅ R + ⎮
⌡
R
ρ⋅ u ( r) ⋅ 2 ⋅ π⋅ r dr = 0
0
u max⋅ ⎜⎛ R −
2
⎝
Hence
U=
1
2
× 3⋅
1
2
m
s
⋅R
⌠
⎮
⎮
⎮
⌡
R
⎡
u max⋅ ⎢1 −
⎣
0
2⎞
⎠
2
= R ⋅U
U=
1
2
⋅ u max
U = 1.5⋅
m
s
2
⎛ r ⎞ ⎤⎥ ⋅ 2⋅ r dr = R2⋅ U
⎜R
⎝ ⎠⎦
Problem 4.36
Given:
Data on flow at inlet and outlet of channel
Find:
Find u max
[Difficulty: 2]
Solution:


∫ ρ V ⋅ dA = 0
Basic equation
CS
Assumptions: 1) Steady flow 2) Incompressible flow
Evaluating at 1 and 2
⌠
−ρ⋅ V1 ⋅ H⋅ w + ⎮
⌡
H
ρ⋅ V2 ( y ) ⋅ w dy = 0
−H
or
⌠
V1 ⋅ H = ⎮
⎮
⌡
H
−H
Hence
π
Vm = ⋅ V1
4
π⋅ y ⎞
dy =
Vm⋅ cos⎛⎜
⎝ 2⋅ H ⎠
⌠
2⋅ ⎮
⎮
⌡
H
0
4 ⋅ H⋅ Vm
2⋅ H ⎛ ⎛ π ⎞
π⋅ y ⎞
dy = 2 ⋅ Vm⋅
⋅ ⎜ sin⎜
− sin( 0 ) ⎞ =
Vm⋅ cos⎛⎜
π ⎝ ⎝2⎠
π
⎝ 2⋅ H ⎠
⎠
Problem 4.37
[Difficulty: 3]
Given:
Velocity distribution in annulus
Find:
Volume flow rate; average velocity; maximum velocity; plot velocity distribution
Solution:
Governing equation
For the flow rate (Eq. 4.14a) and average velocity (Eq.
4.14b)
The given data is
Ro = 5 ⋅ mm
∆p
Ri = 1 ⋅ mm
L
= −10⋅
⌠ →→
⎮
Q = ⎮ V dA
⌡
kPa
μ = 0.1⋅
m
Q
Vav =
A
N⋅ s
(From Fig. A.2)
2
m
2
2
⎛
Ro − Ri
⎛ Ro ⎞ ⎞
−∆p ⎜ 2
2
u ( r) =
⋅ ln⎜
⋅ R −r +
4 ⋅ μ⋅ L ⎜ o
⎝ r ⎠⎟
⎛ Ri ⎞
ln⎜
⎜
⎝
⎝ Ro ⎠
⎠
R
The flow rate is
⌠ o
Q = ⎮ u ( r) ⋅ 2 ⋅ π⋅ r dr
⌡R
⎡ ⎛ R 2 − R 2⎞
⎤
i ⎠ ⎛ 2
2
2⎞ ⎢ ⎝ o
2⎞⎥
⎛
Q =
⋅ R − Ri ⋅
⎠ ⎢ ⎛ R ⎞ − ⎝ Ri + Ro ⎠⎥
8 ⋅ μ⋅ L ⎝ o
o
⎢ ln⎜
⎥
Ri
⎣ ⎝ ⎠
⎦
i
∆p⋅ π
Considerable mathematical manipulation leads to
Substituting values
Q =
3
⋅ ( −10⋅ 10 ) ⋅
8
π
Q = 1.045 × 10
The average velocity is
Q
Vav =
=
A
N
2
2
(
0.1⋅ N⋅ s
2
⋅
m ⋅m
m
3
−5m
Q = 10.45 ⋅
s
2
2
2⎡ 2
2
2
⎞ ⋅ ⎢ 5 − 1 − 5 2 + 12 ⎥⎤ ⋅ ⎛ m ⎞
⎜
⎥ ⎝ 1000 ⎠
⎝ 1000 ⎠ ⎢ ln⎛ 5 ⎞
⎢⎣ ⎜⎝ 1 ⎠
⎥⎦
2⎞
⎠
)
mL
s
1
−5 m
Vav =
× 1.045 × 10 ⋅
×
π
s
1
2
5 −1
2
⋅ ⎛⎜
1000 ⎞
2
⎝ m ⎠
⎛
⎡
⎡
Ro − Ri
⎛ Ro ⎞ ⎞⎥⎤
∆p ⎢
2
d ⎢ −∆p ⎜ 2
⋅ ln⎜
=0=
⋅ Ro − r +
=−
⋅ −2 ⋅ r −
dr
4 ⋅ μ⋅ L ⎢
dx ⎢ 4 ⋅ μ⋅ L ⎜
⎝ r ⎠ ⎟⎥
⎛ Ri ⎞
ln⎜
⎢
⎜
⎥
⎢
⎣
⎝
⎝ Ro ⎠
⎠⎦
⎣
2
The maximum velocity occurs when
(
m
3
Q
π⋅ ⎛ R o − R i
⎝
)
⋅ 5 − 1 ⋅ ⎛⎜
du
2
m
Vav = 0.139
s
⎛ R 2 − R 2⎞ ⎤
i ⎠⎥
⎝ o
⎛ Ri ⎞ ⎥
ln⎜
⋅r ⎥
Ro
⎝ ⎠ ⎦
2
r =
Then
Ri − Ro
2
⎛ Ri ⎞
2 ⋅ ln⎜
⎝ Ro ⎠
r = 2.73⋅ mm
Substituting in u(r)
u max = u ( 2.73⋅ mm) = 0.213 ⋅
The maximum velocity using Solver instead, and the plot, are also shown in an Excel workbook
Ro =
Ri=
¬p /L =
5
1
-10
mm
mm
kPa/m
¬◊ϕ
0.1
N.s/m
2
r (mm) u (m/s)
0.000
0.069
0.120
0.157
0.183
0.201
0.210
0.213
0.210
0.200
0.186
0.166
0.142
0.113
0.079
0.042
0.000
Annular Velocity Distribution
6
5
r (mm)
1.00
1.25
1.50
1.75
2.00
2.25
2.50
2.75
3.00
3.25
3.50
3.75
4.00
4.25
4.50
4.75
5.00
4
3
2
1
0
0.00
The maximum velocity can be found using Solver
r (mm) u (m/s)
2.73
0.213
0.05
0.10
0.15
u (m/s)
0.20
0.25
m
s
Problem 4.38
[Difficulty: 2]
Given:
Data on flow at inlet and outlet of a reducing elbow
Find:
Find the maximum velcoity at section 1
Solution:


∫ ρ V ⋅ dA = 0
Basic equation
CS
Assumptions: 1) Steady flow 2) Incompressible flow
Evaluating at 1, 2 and 3
h
⌠ 1
−⎮ V1 ( y ) ⋅ w dy + V2 ⋅ w⋅ h 2 + V3 ⋅ w⋅ h 3 = 0
⌡
0
or
Hence
h
2
V1max ⌠ 1
V1max h 1
⋅
= V2 ⋅ h 2 + V3 ⋅ h 3
⋅ ⎮ y dy =
2
h 1 ⌡0
h1
(
2
V1max =
⋅ V3 ⋅ h 3 + V2 ⋅ h 2
h1
)
2 ⎛ m
m
V1max =
⋅ ⎜ 5 ⋅ × 0.15⋅ m + 1 ⋅ × 0.2⋅ m⎞
0.5⋅ m ⎝ s
s
⎠
m
V1max = 3.80
s
Problem 4.39
Given:
Data on flow at inlet and outlet of channel
Find:
Find u max
Solution:
Basic equation

[Difficulty: 2]

∫ ρ V ⋅ dA = 0
CS
Assumptions: 1) Steady flow 2) Incompressible flow
h
Evaluating at inlet and exit
⌠
−U⋅ w⋅ h + ⎮ Vexit ( x ) ⋅ w dx = 0
⌡
0
(
)
Here we have
x
Vexit = Vmax − Vmax − Vmin ⋅
h
Hence
x
Vexit = 2 ⋅ Vmin − Vmin⋅
h
h
⌠
⌠
⎮ Vexit ( x ) ⋅ w dx = ⎮
⎮
⌡
⌡
0
h
But we also have
2
⎛ 2 ⋅ V − V ⋅ x ⎞ ⋅ w dx = ⎛⎜ 2⋅ V ⋅ h − V ⋅ h ⎞ ⋅ w = 3 ⋅ V ⋅ h⋅ w
⎜ min
min
min 2 ⋅ h
min h
2 min
⎝
⎠
⎝
⎠
0
3
Hence
2
⋅ Vmin⋅ h ⋅ w = U⋅ w⋅ h
2
m
Vmin =
× 7.5⋅
3
s
Vmax = 2 ⋅ Vmin
2
Vmin = ⋅ U
3
m
Vmin = 5.00⋅
s
Problem 4.30
Problem 4.40
[Difficulty: 2]
Problem 4.27
Problem 4.41
[Difficulty: 2]
Problem 4.31
Problem 4.42
[Difficulty: 2]
Problem 4.28
Problem 4.43
[Difficulty: 2]
Problem 4.33
Problem 4.44
[Difficulty: 2]
Problem 4.45
[Difficulty: 2]
CS
Outflow
Given:
Data on airflow out of tank
Find:
Find rate of change of density of air in tank
Solution:
Basic equation
 
∂
ρ
+
ρ
d
V
V
∫
∫ ⋅ dA = 0
∂t CV
CS
Assumptions: 1) Density in tank is uniform 2) Uniform flow 3) Air is an ideal gas
Hence
Vtank⋅
dρtank
dρtank
dt
dt
+ ρexit ⋅ V⋅ A = 0
3 N
= −300 × 10 ⋅
2
m
× 250 ⋅
dρtank
dt
m
s
=−
ρexit ⋅ V⋅ A
Vtank
2
× 100 ⋅ mm ×
=−
p exit ⋅ V⋅ A
Rair⋅ Texit ⋅ Vtank
2
1
1
⎛ 1⋅ m ⎞ × 1 ⋅ kg⋅ K ×
×
⎜ 1000⋅ mm
( −20 + 273 ) ⋅ K
3
286.9 N⋅ m
⎝
⎠
0.4⋅ m
kg
dρtank
Hence
dt
3
= −0.258 ⋅
m
s
The mass in the tank is decreasing, as expected
Problem 4.32
Problem 4.46
[Difficulty: 2]
Problem 4.35
Problem 4.47
[Difficulty: 2]
Problem 4.48
[Difficulty: 3]
Given:
Data on draining of a tank
Find:
Depth at various times; Plot of depth versus time
Solution:
Basic equation
 
∂
ρ
+
ρ
d
V
V
∫
∫ ⋅ dA = 0
∂t CV
CS
Assumptions: 1) Uniform flow 2) Incompressible flow 3) Neglect air density
Treating the tank as the CV the basic equation becomes
∂ ⌠
⎮
∂t ⌡
y
ρ⋅ Atank dy + ρ⋅ V⋅ Aopening = 0
or
ρ⋅
π
0
V=
Using
Separating variables
dy
1
y
Solving for y
Using the given data
2 dy
⋅D ⋅
dt
+ ρ⋅
π
4
2
⋅d ⋅V = 0
1
2⋅ g⋅ y
and simplifying
dy
dt
2
=
4
⎛ d ⎞ ⋅ 2⋅ g ⋅ dt
⎜D
⎝ ⎠
and integrating
= −⎛⎜
2
⎞ ⋅ 2⋅ g⋅ y 2
⎝ D⎠
d
1⎞
⎛⎜ 1
2
d
2
2
2⋅ ⎜ y − y0
= −⎛⎜ ⎞ ⋅ 2 ⋅ g t
⎝
⎠
⎝ D⎠
2
2
⎡⎢
g ⎛ d ⎞ ⎥⎤
y ( t) = y 0⋅ 1 −
⋅⎜
⋅t
⎢
2⋅ y0 ⎝ D ⎠ ⎥
⎣
⎦
y ( 1 ⋅ min) = 1.73⋅ ft
2
y ( 2 ⋅ min) = 0.804 ⋅ ft
y ( 3 ⋅ min) = 0.229 ⋅ ft
3
Depth (ft)
2.5
2
1.5
1
0.5
0
0.5
1
1.5
t (min)
2
2.5
3
Problem 4.49
[Difficulty: 3]
Given:
Data on draining of a tank
Find:
Times to a depth of 1 foot; Plot of drain timeversus opening size
Solution:
Basic equation
 
∂
ρ
+
ρ
d
V
V
∫
∫ ⋅ dA = 0
∂t CV
CS
Assumptions: 1) Uniform flow 2) Incompressible flow 3) Neglect air density
Treating the tank as the CV the basic equation becomes
y
∂⌠
⎮ ρ⋅ Atank dy + ρ⋅ V⋅ Aopening = 0
∂t ⌡
or
ρ⋅
π
0
Separating variables
dy
1
y
Solving for t
2 dy
⋅D ⋅
dt
+ ρ⋅
π
4
2
⋅d ⋅V = 0
1
V=
Using
4
2⋅ g⋅ y
and simplifying
dt
2
=
dy
⎛ d ⎞ ⋅ 2⋅ g ⋅ dt
⎜D
⎝ ⎠
and integrating
= −⎛⎜
d⎞
⎝ D⎠
2
⋅ 2⋅ g⋅ y
2
1⎞
⎛⎜ 1
2
d
2
2
2⋅ ⎜ y − y0
= −⎛⎜ ⎞ ⋅ 2 ⋅ g t
⎝
⎠
⎝ D⎠
2
t=
2⋅ y0
g
⋅ ⎛⎜
D⎞
2
⎛
⋅⎜1 −
⎝d⎠ ⎝
y
⎞
y0
⎠
Using the given data
t( 2 ⋅ ft) = 45.6 s
Hence for the first drop of 1 foot
∆t = t( 2 ⋅ ft)
∆t = 45.6 s
For the second drop of 1 foot
∆t = t( 1 ⋅ ft) − t( 2 ⋅ ft)
∆t = 59.5 s
t( 1 ⋅ ft) = 105 s
This is because as the level drops the exit speed, hence drain rate, decreases.
Drain Time (min)
15
10
5
0.1
0.2
0.3
d (in)
0.4
0.5
Problem 4.38
Problem 4.50
[Difficulty: 3]
Problem 4.39
Problem 4.51
[Difficulty: 3]
Problem 4.40
Problem 4.52
[Difficulty: 3]
Problem 4.41
Problem 4.53
P4.48.
[Difficulty: 3]
Problem 4.42
Problem 4.54
[Difficulty: 4]
Problem 4.55
[Difficulty: 4]
Given:
Data on draining of a funnel
Find:
Formula for drain time; time to drain from 12 in to 6 in; plot drain time versus hole diameter
Solution:
Basic equation
 
∂
ρ
+
ρ
d
V
V
∫
∫ ⋅ dA = 0
∂t CV
CS
Assumptions: 1) Uniform flow 2) Incompressible flow 3) Neglect air density
Treating the funnel as the CV the basic equation becomes
y
∂⌠
⎮ ρ⋅ Afunnel dy + ρ⋅ V⋅ Aopening = 0
∂t ⌡
0
2
2
For the funnel
Afunnel = π⋅ r = π⋅ ( y ⋅ tan( θ) )
Hence
y
π 2
2∂⌠
2
ρ⋅ π⋅ ( tan( θ) ) ⋅ ⎮ y dy + ρ⋅ V⋅ ⋅ d = 0
⌡
4
∂t 0
Then
2 2 dy
( tan( θ) ) ⋅ y ⋅
dt
= − 2⋅ g⋅ y⋅
d
or
2
⎛ y3 ⎞
d
⎜
= − 2⋅ g⋅ y⋅
4
dt ⎝ 3 ⎠
2d
( tan( θ) ) ⋅
2
4
3
Separating variables
2
y ⋅ dy = −
2⋅ g⋅ d
2
4 ⋅ tan( θ)
2
⋅ dt
0
Hence
⌠
3
⎮
⎮
2⋅ g⋅ d
2
⋅t
⎮ y dy = −
2
⌡y
4 ⋅ tan( θ)
0
5
or
2
5
⋅ y0
2
2⋅ g⋅ d
=
4 ⋅ tan( θ)
5
2
Solving for t
8 tan( θ) ⋅ y 0
t= ⋅
5
2⋅ g⋅ d
2
and using the given data
t = 2.55⋅ min
2
⋅t
To find the time to drain from 12 in to 6 in., we use the time equation with the two depths; this finds the time to drain from 12 in and 6
in, so the difference is the time we want
5
2
8 tan( θ) ⋅ y 0
∆t1 = ⋅
5
2
2⋅ g⋅ d
y 1 = 6 ⋅ in
2
5
2
8 tan( θ) ⋅ y 1
− ⋅
5
2
2⋅ g⋅ d
2
∆t1 = 2.1⋅ min
5
2
8 tan( θ) ⋅ y 1
∆t2 = ⋅
2
5
2⋅ g⋅ d
2
∆t2 = 0.451 ⋅ min
∆t1 + ∆t2 = 2.55⋅ min
Note that
The second time is a bit longer because although the flow rate decreases, the area of the funnel does too.
Drain Time (min)
3
2
1
0.25
0.3
0.35
0.4
d (in)
0.45
0.5
Problem 4.56
[Difficulty: 4]
Given:
Data on draining of a funnel
Find:
Diameter that will drain in 1 min.; plot diamter versus depth y 0
Solution:
 
∂
ρdV + ∫ ρV ⋅ dA = 0
∫
∂t CV
CS
Basic equation
Assumptions: 1) Uniform flow 2) Incompressible flow 3) Neglect air density
Treating the funnel as the CV the basic equation becomes
y
∂⌠
⎮ ρ⋅ Afunnel dy + ρ⋅ V⋅ Aopening = 0
∂t ⌡
0
2
2
For the funnel
Afunnel = π⋅ r = π⋅ ( y ⋅ tan( θ) )
Hence
y
π 2
2∂⌠
2
ρ⋅ π⋅ ( tan( θ) ) ⋅ ⎮ y dy + ρ⋅ V⋅ ⋅ d = 0
4
∂t ⌡0
2
⎛ y3 ⎞
d
⎜
= − 2⋅ g⋅ y⋅
4
dt ⎝ 3 ⎠
2d
( tan( θ) ) ⋅
or
3
2 2 dy
( tan( θ) ) ⋅ y ⋅
Then
dt
= − 2⋅ g⋅ y⋅
d
2
2
y ⋅ dy = −
Separating variables
4
2⋅ g⋅ d
2
4 ⋅ tan( θ)
2
⋅ dt
0
⌠
3
⎮
⎮
2⋅ g⋅ d
2
⋅t
⎮ y dy = −
2
⌡y
4
⋅
tan
(
θ
)
0
Hence
5
2
or
5
⋅ y0
2
=
2⋅ g⋅ d
4 ⋅ tan( θ)
2
⋅t
5
2
d =
Solving for d
8 tan( θ) ⋅ y 0
⋅
5
2⋅ g⋅ t
2
t = 1 min
and using the given data, for
d = 0.399 in
1
d (in)
0.8
0.6
0.4
0.2
0
2
4
6
8
10
12
y0 (in)
14
16
18
20
22
24
Problem 4.57
[Difficulty: 4] Part 1/2
Problem 4.57
For p = 500 kPa, solving Eq. 2 for t we find t = 42.2 days
[Difficulty: 4] Part 2/2
Problem 4.58
Given:
Data on flow through a control surface
Find:
Net rate of momentum flux
Solution:
Basic equation: We need to evaluate
∫
CS
[Difficulty: 3]
 
VρV ⋅ dA
Assumptions: 1) Uniform flow at each section
From Problem 4.24
ft
V1 = 10⋅
s
Then for the control surface
A1 = 0.5⋅ ft
ft
V2 = 20⋅
s
2
A2 = 0.1⋅ ft
2
A3 = 0.6⋅ ft
2
ft
V3 = 5 ⋅
s
It is an outlet
 
     
  
VρV ⋅ dA = V1ρV1 ⋅ A1 + V2 ρV2 ⋅ A2 + V3 ρV3 ⋅ A3
CS
 
 
 
= V1iˆρ V1 ⋅ A1 + V2 ˆjρ V2 ⋅ A2 + V3 sin(60)iˆ − V3 cos(60) ˆj ρ V3 ⋅ A3
= −V1iˆρV1 A1 + V2 ˆjρV2 A2 + V3 sin(60)iˆ − V3 cos(60 ) ˆj ρV3 A3
∫
(
)
(
[
[
]
) [
[
](
]
)
]
= ρ − V12 A1 + V32 A3 sin (60) iˆ + ρ V22 A2 − V32 A3 cos(60 ) ˆj
Hence the x component is
ρ [− V12 A1 + V32 A3 sin (60 )] =
65⋅
lbm
ft
and the y component is
3
(
2
2
⋅s
) ft2 × lbf
= −2406⋅ lbf
lbm⋅ ft
4
2
× −10 × 0.5 + 5 × 0.6 × sin( 60⋅ deg) ⋅
s
ρ [V22 A2 − V32 A3 cos(60 )] =
65⋅
lbm
ft
3
ft
lbf ⋅ s
2
2
× ( 20 × 0.1 − 5 × 0.6 × cos( 60⋅ deg) ) ⋅
×
= 2113⋅ lbf
2
lbm⋅ ft
4
s
2
Problem 4.59
[Difficulty: 3]
y
2h

Given:
Data on flow at inlet and outlet of channel
Find:
Ratio of outlet to inlet momentum flux
Solution:
x

CS

mf x = ∫ uρV ⋅ dA
Basic equation: Momentum flux in x direction at a section
A
Assumptions: 1) Steady flow 2) Incompressible flow
2
Evaluating at 1 and 2
mfx1 = U⋅ ρ⋅ ( −U⋅ 2 ⋅ h ) ⋅ w
mfx1 = 2 ⋅ ρ⋅ w⋅ U ⋅ h
Hence
⌠
⎮
⌠
2
2
mfx2 = ⎮ ρ⋅ u ⋅ w dy = ρ⋅ w⋅ u max ⋅ ⎮
⎮
⌡
−h
⌡
h
h
−h
⎡
⎢1 −
⎣
2⎤
⎛y⎞ ⎥
⎜h
⎝ ⎠⎦
2
⌠
⎮
2
dy = ρ⋅ w⋅ u max ⋅ ⎮
⎮
⌡
h
−h
⎡
⎢1 −
⎣
2 ⋅ ⎛⎜
y⎞
⎝h⎠
2
+
4
⎛ y ⎞ ⎤⎥ dy
⎜
⎝h⎠ ⎦
2
4
2
2 16
mfx2 = ρ⋅ w⋅ u max ⋅ ⎛⎜ 2 ⋅ h − ⋅ h + ⋅ h⎞ = ρ⋅ w⋅ u max ⋅ ⋅ h
5
3
15
⎝
⎠
Then the ratio of momentum fluxes is
16
mfx2
mfx1
But, from Problem 4.34
u max =
=
15
2
⋅ ρ⋅ w⋅ u max ⋅ h
2
2 ⋅ ρ⋅ w⋅ U ⋅ h
3
2
⋅U
⎛ u max ⎞
=
⋅⎜
15 ⎝ U ⎠
2
8
⎛
mfx2
8 ⎜
=
⋅⎜
mfx1
15 ⎝
3
2
⋅U ⎞
U
⎠
2
=
6
5
= 1.2
Hence the momentum increases as it flows in the entrance region of the channel. This appears to contradict common sense, as
friction should reduce flow momentum. What happens is the pressure drops significantly along the channel so the net force on
the CV is to the right.
Problem 4.60
Given:
Data on flow at inlet and outlet of pipe
Find:
Ratio of outlet to inlet momentum flux
Solution:
[Difficulty: 3]

mf x = ∫ uρV ⋅ dA
Basic equation: Momentum flux in x direction at a section
A
Assumptions: 1) Steady flow 2) Incompressible flow
(
)
2
2
Evaluating at 1 and 2
mfx1 = U⋅ ρ⋅ −U⋅ π⋅ R
mfx1 = ρ⋅ π⋅ U ⋅ R
Hence
⌠
⎮
⌠
2⎮
2
mfx2 = ⎮ ρ⋅ u ⋅ 2⋅ π⋅ r dr = 2⋅ ρ⋅ π⋅ u max ⋅
⎮
⌡
0
⌡
2
R
⎡
r ⋅ ⎢1 −
⎣
R
2⎤
⎛r ⎞⎥
⎜R
⎝ ⎠⎦
0
2
⌠
2⎮
dr = 2⋅ ρ⋅ π⋅ u max ⋅ ⎮
⎮
⌡
R
5⎞
3
⎛
⎜ r − 2⋅ r + r dy
4
⎜
2
R ⎠
R
⎝
0
⎛
R
R
2 R
mfx2 = 2⋅ ρ⋅ π⋅ u max ⋅ ⎜
−
+
2
6
2
⎝
2
2
2⎞
⎠
2 R
= ρ⋅ π⋅ u max ⋅
2
3
Then the ratio of momentum fluxes is
1
mfx2
mfx1
But, from Problem 4.35
=
3
u max = 2 ⋅ U
2
⋅ ρ⋅ π⋅ u max ⋅ R
2
ρ⋅ π⋅ U ⋅ R
2
2
⎛ u max ⎞
= ⋅⎜
3 ⎝ U ⎠
2
1
mfx2
mfx1
=
1
3
⋅ ⎛⎜
⎝
2⋅ U ⎞
U
⎠
2
=
4
3
= 1.33
Hence the momentum increases as it flows in the entrance region of the pipe This appears to contradict common sense, as friction
should reduce flow momentum. What happens is the pressure drops significantly along the pipe so the net force on the CV is to
the right.
Problem 4.61
Given:
Data on flow through a bend
Find:
Find net momentum flux
Solution:

Basic equations
[Difficulty: 3]

∫ ρ V ⋅ dA = 0
Momentum fluxes:
mfx =
mfy =
CS
Assumptions: 1) Steady flow 2) Incompressible flow
h
⌠ 1
−⎮ V1 ( y ) ⋅ w dy + V2 ⋅ w⋅ h 2 + V3 ⋅ w⋅ h 3 = 0
⌡
Evaluating mass flux at 1, 2 and 3
0
or
h
h
2
V1max h 1
⌠ 1
⌠ 1
y
dy − V2 ⋅ h 2 =
V3 ⋅ h 3 = ⎮ V1 ( y ) dy − V2 ⋅ h 2 = ⎮ V1max⋅
⋅
− V2 ⋅ h 2
h1
h1
2
⌡
⎮
0
⌡
0
Hence
(
2
V1max =
⋅ V3 ⋅ h 3 + V2 ⋅ h 2
h1
)
Using given data
m
V1max = 3.8
s
For the x momentum, evaluating at 1, 2 and 3
h
⌠ 1
mfx = −⎮ V1 ( y ) ⋅ ρ⋅ V1 ( y ) ⋅ w dy + V3 ⋅ cos( θ) ⋅ ρ⋅ V3 ⋅ h 3 ⋅ w
⌡
0
h
2
3
⌠ 1
2
V1max h 1
⎮ ⎛
y ⎞
2
2
mfx = −⎮ ⎜ V1max⋅
⋅ ρ⋅ w dy + V3 ⋅ ρ⋅ h 3 ⋅ cos( θ) ⋅ w = −
⋅
⋅ ρ⋅ w + V3 ⋅ ρ⋅ h 3 ⋅ w⋅ cos( θ)
2
3
h
1⎠
⎮ ⎝
h1
⌡
0
h
⎛
⎞
2 1
2
mfx = ρ⋅ w⋅ ⎜ −V1max ⋅
+ V3 ⋅ cos( θ) ⋅ h 3
3
⎝
⎠
Using given data
mfx = 841 N
Using given data
mfy = −2075 N
For the y momentum, evaluating at 1, 2 and 3
mfy = −V2 ⋅ ρ⋅ V2 ⋅ h 2 ⋅ w + V3 ⋅ sin( θ) ⋅ ρ⋅ V3 ⋅ h 3 ⋅ w
mfy = ρ⋅ w⋅ ⎛ −V2 ⋅ h 2 − V3 ⋅ sin( θ) ⋅ h 3⎞
⎝
⎠
2
2
Problem 4.49
Problem 4.62
[Difficulty: 2]
Problem 4.63
Given:
Water jet hitting object
Find:
Jet speed; Force generated
[Difficulty: 2]
CS
y
Solution:

x
U
Basic equations: Continuity and Momentum flux in x direction
Rx

Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure throughout 4) Uniform flow
Given data
Q = 1 ⋅ gpm
d = 0.01⋅ in
ρ = 1.94⋅
slug
ft
Using continuity
Q = V⋅ A = U⋅
π
d
⋅d
2
U =
Using data
Q
π
4
Using momentum
⋅d
3
U = 4085
2
ft
s
2
2
2 π⋅ D
Rx = u 1 ⋅ ρ⋅ −u 1 ⋅ A1 = −ρ⋅ U ⋅ A = −ρ⋅ U ⋅
4
(
)
2 π⋅ d
Rx = −ρ⋅ U ⋅
Hence
2
ft
Rx = −1.94⋅
× ⎛⎜ 4085⋅ ⎞ ×
3
s⎠
⎝
ft
slug
π⋅ ⎛⎜
2
4
.01
⋅ ft⎞
2
2
⎝ 12 ⎠ × lbf ⋅ s
4
slug⋅ ft
Rx = −17.7⋅ lbf
U = 2785⋅ mph
FAST!
Problem 4.64
Given:
Fully developed flow in pipe
Find:
Why pressure drops if momentum is constant
[Difficulty: 1]
Solution:
Basic equation: Momentum flux in x direction
Assumptions: 1) Steady flow 2) Fully developed flow
Hence
∆p
Fx =
− τw⋅ As = 0
L
∆p = L⋅ τw⋅ As
where ∆p is the pressure drop over length L, τw is the wall friction and As is the pipe surface area
The sum of forces in the x direction is zero. The friction force on the fluid is in the negative x direction, so the net pressure force
must be in the positive direction. Hence pressure drops in the x direction so that pressure and friction forces balance
Problem 4.65
Given:
Data on flow and system geometry
Find:
Force required to hold plug
[Difficulty: 2]
Solution:
Basic equation:
3
The given data is
Then
D1 = 0.25⋅ m
A1 =
π⋅ D1
D2 = 0.2⋅ m
Q = 1.5⋅
m
p 1 = 3500⋅ kPa
s
ρ = 999 ⋅
kg
3
m
2
2
Q
V1 =
A1
m
V1 = 30.6
s
2
Q
V2 =
A2
m
V2 = 84.9
s
A1 = 0.0491 m
4
π
2
2
A2 = ⋅ ⎛ D1 − D2 ⎞
⎠
4 ⎝
A2 = 0.0177 m
Applying the basic equation
(
)
(
−F + p 1 ⋅ A2 − p 2 ⋅ A2 = 0 + V1 ⋅ −ρ⋅ V1 ⋅ A1 + V2 ⋅ ρ⋅ V2 ⋅ A2
Hence
)
and
p2 = 0
(gage)
F = p 1 ⋅ A1 + ρ⋅ ⎛ V1 ⋅ A1 − V2 ⋅ A2⎞
⎝
⎠
2
F = 3500 ×
kN
2
m
2
2
⋅ 0.0491⋅ m + 999 ⋅
kg
3
m
×
2
2
⎡⎛
⎤
m
⎢⎜ 30.6⋅ ⎞ ⋅ 0.0491⋅ m2 − ⎛⎜ 84.9⋅ m ⎞ ⋅ 0.0177⋅ m2⎥
s⎠
s⎠
⎣⎝
⎝
⎦
F = 90.4⋅ kN
Problem 4.66
Given:
Nozzle hitting stationary cart
Find:
Value of M to hold stationary; plot M versu θ
[Difficulty: 2]
Solution:
Basic equation: Momentum flux in x direction for the tank
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure throughout 4) Uniform flow 5) Exit velocity is V
2
Rx = −M ⋅ g = V⋅ ρ⋅ ( −V⋅ A) + V⋅ cos( θ) ⋅ ( V⋅ A) = ρ⋅ V ⋅ A⋅ ( cos( θ) − 1 )
Hence
2
When θ = 40o
M =
s
9.81⋅ m
× 1000⋅
kg
3
m
× ⎛⎜ 10⋅
⎝
m⎞
s
⎠
2
2
× 0.1⋅ m × ( 1 − cos( 40⋅ deg) )
2
M=
ρ⋅ V ⋅ A
g
⋅ ( 1 − cos( θ) )
M = 238 kg
M (kg)
3000
2000
1000
0
45
90
Angle (deg)
This graph can be plotted in Excel
135
180
Problem 4.67
[Difficulty: 2]
Given:
Large tank with nozzle and wire
Find:
Tension in wire; plot for range of water depths
Solution:
Basic equation: Momentum flux in x direction for the tank
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure throughout 4) Uniform flow
Hence
When y = 0.9 m
2
Rx  T  V ρ ( V A)  ρ V  A  ρ ( 2  g  y ) 
T 
π
2
 1000
kg
3
m
 9.81
m
2
π d
2
T
4
2
 0.9 m  ( 0.015  m) 
s
1
2
 ρ g  y  π d
2
T is linear with y!
2
N s
kg m
T  3.12 N
4
T (N)
3
2
1
0
0.3
0.6
y (m)
This graph can be plotted in Excel
0.9
Problem 4.68
Given:
Water flowing past cylinder
Find:
Horizontal force on cylinder
[Difficulty: 2]
y
V

x
Solution:
CS
Rx
Basic equation: Momentum flux in x direction

V
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure throughout 4) Uniform flow
(
)
(
)
Hence
Rx = u 1 ⋅ ρ⋅ −u 1 ⋅ A1 + u 2 ⋅ ρ⋅ u 2 ⋅ A2 = 0 + ρ⋅ ( −V⋅ sin( θ) ) ⋅ ( V⋅ a⋅ b )
For given data
Rx = −1000⋅
kg
3
m
× ⎛⎜ 3 ⋅
m⎞
⎝ s⎠
2
θ
2
Rx = −ρ⋅ V ⋅ a⋅ b ⋅ sin( θ)
2
× 0.0125⋅ m × 0.0025⋅ m × sin( 20⋅ deg) ×
This is the force on the fluid (it is to the left). Hence the force on the cylinder is
N⋅ s
kg⋅ m
Rx = −Rx
Rx = −0.0962 N
Rx = 0.0962 N
Problem 4.69
[Difficulty: 2]
Given:
Water jet hitting plate with opening
Find:
Force generated on plate; plot force versus diameter d
CS
y
Solution:
x

V
V

Basic equation: Momentum flux in x direction
Rx
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure throughout 4) Uniform flow
2
2 π⋅ D
2 π⋅ d
Rx = u 1 ⋅ ρ⋅ −u 1 ⋅ A1 + u 2 ⋅ ρ⋅ u 2 ⋅ A2 = −ρ⋅ V ⋅
+ ρ⋅ V ⋅
4
4
(
Hence
)
(
)
2
For given data
π
slug ⎛
ft
Rx = − ⋅ 1.94⋅
× ⎜ 15⋅ ⎞ ×
4
3
⎝ s⎠
ft
2
⎛ 1 ⋅ ft⎞ × ⎢⎡1 −
⎜
⎝3 ⎠ ⎣
2
2
2
⎛ 1 ⎞ ⎤⎥ × lbf ⋅ s
⎜
⎝ 4 ⎠ ⎦ slug⋅ ft
2
Rx = −
2
π⋅ ρ⋅ V ⋅ D
4
⎡
⋅ ⎢1 −
⎣
2
⎛ d ⎞ ⎤⎥
⎜D
⎝ ⎠⎦
Rx = −35.7⋅ lbf
From Eq 1 (using the absolute value of Rx)
Force (lbf)
40
30
20
10
0
0.2
0.4
0.6
Diameter Ratio (d/D)
This graph can be plotted in Excel
0.8
1
(1)
Problem 4.70
[Difficulty: 4]
y
V
x
CS
W
Rx
Given:
Water flowing into tank
Find:
Mass flow rates estimated by students. Explain discrepancy
Solution:
Basic equation: Momentum flux in y direction
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure throughout 4) Uniform flow
For the first student
m1 =
ρ⋅ V
where m1 represents mass flow rate (software cannot render a dot above it!)
t
kg
1
3
m1 = 1000⋅
× 3⋅ m ×
3
60⋅ s
m
For the second student
M
m2 =
t
kg
m1 = 50.0
s
where m2 represents mass flow rate
1
m2 = 3150⋅ kg ×
60⋅ s
kg
m2 = 52.5
s
There is a discrepancy because the second student is measuring instantaneous weight PLUS the force generated as the pipe flow
momentum is "killed".
There is a discrepancy because the second student is measuring instantaneous weight PLUS the force generated as the pipe flow
momentum is "killed". To analyse this we first need to find the speed at which the water stream enters the tank, 10 m below the
pipe exit. This would be a good place to use the Bernoulli equation, but this problem is in the set before Bernoulli is covered.
Instead we use the simple concept that the fluid is falling under gravity (a conclusion supported by the Bernoulli equation). From
the equations for falling under gravity:
2
2
Vtank = Vpipe + 2 ⋅ g ⋅ h
where V tank is the speed entering the tank, Vpipe is the speed at the pipe, and h = 10 m is the distance traveled. Vpipe is obtained from
m1
Vpipe =
2
ρ⋅
π⋅ d pipe
4
=
4 ⋅ m1
2
π⋅ ρ⋅ d pipe
3
4
kg
m
Vpipe =
× 50⋅
×
×
π
s
1000⋅ kg
Then
Vtank =
⎛ 1 ⎞
⎜ 0.05⋅ m
⎝
⎠
2
m
Vpipe = 25.5
s
2
2
Vpipe + 2 ⋅ g ⋅ h
⎛ 25.5⋅ m ⎞ + 2 × 9.81⋅ m × 10m
⎜
s⎠
2
⎝
s
Vtank =
m
Vtank = 29.1
s
We can now use the y momentum equation for the CS shown above
(
)
Ry − W = −Vtank⋅ ρ⋅ −Vtank⋅ Atank
Vtank⋅ Atank = Vpipe⋅ Apipe
where A tank is the area of the water flow as it enters the tank. But for the water flow
2
Hence
∆W = Ry − W = ρ⋅ Vtank⋅ Vpipe⋅
π⋅ d pipe
4
This equation indicate the instantaneous difference ΔW between the scale reading (Ry) and the actual weight of water (W) in the tank
∆W = 1000⋅
kg
3
× 29.1⋅
m
m
s
× 25.5⋅
m
s
∆m =
Inducated as a mass, this is
×
π
4
× ( 0.05⋅ m)
2
∆W
g
∆W = 1457 N
∆m = 149 kg
Hence the scale overestimates the weight of water by 1457 N, or a mass of 149 kg
For the second student
M = 3150⋅ kg − 149 ⋅ kg
Hence
M
m2 =
t
M = 3001 kg
where m2 represents mass flow rate
1
kg
m2 = 3001⋅ kg ×
m2 = 50.0
60⋅ s
s
Comparing with the answer obtained from student 1, we see the students now agree! The discrepancy was entirely caused by the
fact that the second student was measuring the weight of tank water PLUS the momentum lost by the water as it entered the tank!
Problem 4.71
[Difficulty: 3]
Given:
Water tank attached to mass
Find:
Whether tank starts moving; Mass to just hold in place
Solution:
Basic equation: Momentum flux in x direction for the tank
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure at exit 4) Uniform flow
Hence
2
2 π⋅ D
Rx = V⋅ cos( θ) ⋅ ρ⋅ ( V⋅ A) = ρ⋅ V ⋅
4
⋅ cos( θ)
V=
We need to find V. We could use the Bernoulli equation, but here it is known that
V =
2 × 9.81⋅
m
2
× 2⋅ m
V = 6.26
s
Hence
Rx = 1000⋅
kg
3
m
× ⎛⎜ 6.26⋅
⎝
m⎞
s
⎠
This force is equal to the tension T in the wire
2
×
π
4
2⋅ g⋅ h
where h = 2 m is the
height of fluid in the tank
m
s
2
× ( 0.05⋅ m) × cos( 60⋅ deg)
Rx = 38.5 N
T = Rx
T = 38.5 N
Fmax = M ⋅ g ⋅ μ
For the block, the maximum friction force a mass of M = 10 kg can generate is
m
where µ is static friction
2
N⋅ s
Fmax = 10⋅ kg × 9.81⋅ × 0.55 ×
2
kg⋅ m
s
Fmax = 54.0 N
Hence the tension T created by the water jet is less than the maximum friction F max; the tank is at rest
The mass that is just sufficient is given by
M=
Rx
g⋅ μ
M ⋅ g ⋅ μ = Rx
M = 38.5⋅ N ×
1
2
⋅
s
9.81 m
×
1
0.55
×
kg⋅ m
2
N⋅ s
M = 7.14 kg
Problem 4.72
Given:
Gate held in place by water jet
Find:
Required jet speed for various water depths
[Difficulty: 4]
Solution:
Basic equation: Momentum flux in x direction for the wall
Note: We use this equation ONLY for the jet impacting the wall. For the hydrostatic force and location we use computing equations
Ixx
FR = p c⋅ A
y' = y c +
A⋅ y c
Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow
Hence
(
2
2 π⋅ D
)
Rx = V⋅ ρ⋅ −V⋅ Ajet = −ρ⋅ V ⋅
4
This force is the force generated by the wall on the jet; the force of the jet hitting the wall is then
2
2 π⋅ D
Fjet = −Rx = ρ⋅ V ⋅
4
For the hydrostatic force
where D is the jet diameter
w⋅ h
h
1
2
FR = p c⋅ A = ρ⋅ g ⋅ ⋅ h ⋅ w = ⋅ ρ⋅ g ⋅ w⋅ h
2
2
3
Ixx
h
2
12
y' = y c +
=
+
= ⋅h
A⋅ y c
2
h
3
w⋅ h ⋅
2
where h is the water depth and w is the gate width
For the gate, we can take moments about the hinge to obtain
h
−Fjet⋅ h jet + FR⋅ ( h − y') = −Fjet⋅ h jet + FR⋅ = 0
3
where h jet is the height of the jet from the ground
2
Hence
For the first case (h = 1 m)
For the second case (h = 0.5 m)
For the first case (h = 0.25 m)
h
1
2 π⋅ D
2 h
Fjet = ρ⋅ V ⋅
⋅ h jet = FR⋅ = ⋅ ρ⋅ g ⋅ w⋅ h ⋅
4
3
2
3
V =
V =
V =
2
3⋅ π
2
3⋅ π
2
3⋅ π
× 9.81⋅
m
2
m
2
m
2
s
V = 28.9
m
V = 10.2
m
V = 3.61
m
2
3
× 1 ⋅ m × ( 0.5⋅ m) ×
3
⎛ 1 ⎞ × 1
⎜ 0.05⋅ m
1⋅ m
⎝
⎠
× 1 ⋅ m × ( 0.25⋅ m) ×
2
3
3 ⋅ π⋅ D ⋅ h j
⎛ 1 ⎞ × 1
⎜
1⋅ m
⎝ 0.05⋅ m ⎠
s
× 9.81⋅
2 ⋅ g ⋅ w⋅ h
2
3
× 1 ⋅ m × ( 1 ⋅ m) ×
s
× 9.81⋅
V=
2
⎛ 1 ⎞ × 1
⎜
1⋅ m
⎝ 0.05⋅ m ⎠
s
s
s
Problem 4.55
Problem 4.73
[Difficulty: 2]
Problem 4.56
Problem 4.74
[Difficulty: 2]
Problem 4.75
[Difficulty: 2]
Problem 4.76
[Difficulty: 3]
Given:
Flow into and out of CV
Find:
Expressions for rate of change of mass, and force
Solution:
Basic equations: Mass and momentum flux
Assumptions: 1) Incompressible flow 2) Uniform flow
dMCV
For the mass equation
dt
dMCV
→→
(
+ ρ⋅ ( −V1 ⋅ A1 − V2 ⋅ A2 + V3 ⋅ A3 + V4 ⋅ A4 ) = 0
ρ⋅ V⋅ A) =
∑
dt
+
CS
dMCV
dt
Fx +
For the x momentum
(
= ρ⋅ V1 ⋅ A1 + V2 ⋅ A2 − V3 ⋅ A3 − V4 ⋅ A4
p 1 ⋅ A1
+
2
5
13
⋅ p 2 ⋅ A2 −
4
5
⋅ p 3 ⋅ A3 −
5
13
)
⋅ p 4 ⋅ A4 = 0 +
V1
(
2
p 1 ⋅ A1
−
2
5
13
⋅ p 2 ⋅ A2 +
Fy +
For the y momentum
4
5
p 1 ⋅ A1
2
⋅ p 3 ⋅ A3 +
−
12
13
5
13
⋅ p 4 ⋅ A4 + ρ⋅ ⎛⎜ −
⎝
⋅ p 2 ⋅ A2 −
3
5
⋅ p 3 ⋅ A3 +
p 1 ⋅ A1
2
+
12
13
⋅ p 2 ⋅ A2 +
3
5
⋅ p 3 ⋅ A3 −
12
13
⋅ p 4 ⋅ A4 + ρ⋅ ⎛⎜ −
⎝
)
)
(
)
5
4
5
2
2
2
2
⋅ V1 ⋅ A1 −
⋅ V2 ⋅ A2 + ⋅ V3 ⋅ A3 +
⋅ V3 ⋅ A3⎞
13
5
13
2
⎠
1
12
13
⋅ p 4 ⋅ A4 = 0 +
V1
(
2
)
(
12
)
⋅ V ⋅ −ρ⋅ V2 ⋅ A2 ...
13 2
12
3
+ ⋅ V3 ⋅ ρ⋅ V3 ⋅ A3 −
⋅ V ⋅ ρ⋅ V3 ⋅ A3
13 3
5
⋅ −ρ⋅ V1 ⋅ A1 −
(
Fy = −
(
5
⋅ V ⋅ −ρ⋅ V2 ⋅ A2 ...
13 2
4
5
+ ⋅ V3 ⋅ ρ⋅ V3 ⋅ A3 +
⋅ V ⋅ ρ⋅ V3 ⋅ A3
5
13 3
(
Fx = −
)
⋅ −ρ⋅ V1 ⋅ A1 +
)
(
)
12
3
12
2
2
2
2
⋅ V1 ⋅ A1 −
⋅ V2 ⋅ A2 + ⋅ V3 ⋅ A3 −
⋅ V3 ⋅ A3⎞
13
5
13
2
⎠
1
Problem 4.77
[Difficulty: 2]
y
x
CS

Rx

Given:
Water flow through elbow
Find:
Force to hold elbow
Solution:
Basic equation: Momentum flux in x direction for the elbow
Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow
Hence
(
From continuity V2 ⋅ A2 = V1 ⋅ A1
Hence
)
(
Rx + p 1g ⋅ A1 + p 2g ⋅ A2 = V1⋅ −ρ⋅ V1⋅ A1 − V2⋅ ρ⋅ V2⋅ A2
⎛ D1 ⎞
V2 = V1⋅
= V1⋅ ⎜
A2
⎝ D2 ⎠
A1
so
3 N
Rx = −350 × 10 ⋅
2
m
×
π⋅ ( 0.2⋅ m)
4
2
3 N
− 75 × 10 ⋅
2
m
×
Rx = −p 1g ⋅ A1 − p 2g ⋅ A2 − ρ⋅ ⎛ V1 ⋅ A1 + V2 ⋅ A2⎞
⎝
⎠
)
2
2
m 0.2 ⎞
V2 = 0.8⋅ ⋅ ⎛⎜
s ⎝ 0.04⎠
π⋅ ( 0.04⋅ m)
4
2
2
m
V2 = 20⋅
s
2
...
2
2
2
2
2
kg ⎡⎛
π⋅ ( 0.2⋅ m)
π⋅ ( .04⋅ m) ⎤ N⋅ s
m
m
⎥×
× ⎢⎜ 0.8⋅ ⎞ ×
+ −1000⋅
+ ⎛⎜ 20⋅ ⎞ ×
3 ⎣⎝
4
4
s⎠
⎝ s⎠
⎦ kg⋅ m
m
Rx = −11.6⋅ kN
The force is to the left: It is needed to hold the elbow on against the high pressures, plus it generates the large change in x momentum
Problem 4.78
[Difficulty: 2]
y
CS
x
Rx
Given:
Water flow through elbow
Find:
Force to hold elbow
Solution:
Basic equation: Momentum flux in x direction for the elbow
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure at exit 4) Uniform flow
Hence
(
)
(
Rx + p 1g ⋅ A1 = V1⋅ −ρ⋅ V1⋅ A1 − V2⋅ ρ⋅ V2⋅ A2
From continuity V2 ⋅ A2 = V1 ⋅ A1
Hence
Rx = −15⋅
lbf
2
in
so
2
× 4⋅ in − 1.94⋅
slug
ft
3
×
)
Rx = −p 1g ⋅ A1 − ρ⋅ ⎛ V1 ⋅ A1 + V2 ⋅ A2⎞
⎝
⎠
2
A1
V2 = V1⋅
A2
2
ft 4
V2 = 10⋅ ⋅
s 1
2
2
⎡⎛ ft ⎞ 2 2 ⎛ ft ⎞ 2
⎤
1⋅ ft ⎞
lbf ⋅ s
2
⎢⎜ 10⋅
×
⋅ 4⋅ in + ⎜ 40⋅
⋅ 1⋅ in ⎥ × ⎛⎜
slug ⋅ ft
⎣⎝ s ⎠
⎝ s⎠
⎦ ⎝ 12⋅ in ⎠
ft
V2 = 40⋅
s
Rx = −86.9⋅ lbf
The force is to the left: It is needed to hold the elbow on against the high pressure, plus it generates the large change in x momentum
Problem 4.79
Given:
Water flow through nozzle
Find:
Force to hold nozzle
[Difficulty: 2]
Solution:
Basic equation: Momentum flux in x direction for the elbow
Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow
Hence
(
From continuity V2 ⋅ A2 = V1 ⋅ A1
Hence
)
(
Rx + p 1g⋅ A1 + p 2g⋅ A2 = V1 ⋅ −ρ⋅ V1 ⋅ A1 + V2 ⋅ cos( θ) ⋅ ρ⋅ V2 ⋅ A2
3 N
Rx = −15 × 10 ⋅
2
m
Rx = −668 ⋅ N
×
⎛ D1 ⎞
= V1 ⋅ ⎜
V2 = V1 ⋅
A2
⎝ D2 ⎠
A1
s
o
π⋅ ( 0.3⋅ m)
4
2
+ 1000⋅
kg
3
m
×
2
)
Rx = −p 1g⋅ A1 + ρ⋅ ⎛ V2 ⋅ A2 ⋅ cos( θ) − V1 ⋅ A1⎞
⎝
⎠
2
m 30
V2 = 1.5⋅ ⋅ ⎛⎜ ⎞
s ⎝ 15 ⎠
2
2
m
V2 = 6 ⋅
s
2
2
2
⎡⎛ m 2 π⋅ ( 0.15⋅ m) 2
m
π⋅ ( .3⋅ m) ⎤ N⋅ s
⎢⎜ 6 ⋅ ⎞ ×
⎥×
⋅ cos( 30⋅ deg) − ⎛⎜ 1.5⋅ ⎞ ×
s⎠
4
4
⎣⎝ s ⎠
⎝
⎦ kg⋅ m
The joint is in tension: It is needed to hold the elbow on against the high pressure, plus it generates the large
change in x momentum
Problem 4.61
Problem 4.80
[Difficulty: 2]
Problem 4.63
Problem 4.81
[Difficulty: 2]
Problem 4.82
[Difficulty: 2]
CS


y
x
Given:
Water flow through orifice plate
Find:
Force to hold plate
Rx
Solution:
Basic equation: Momentum flux in x direction for the elbow
Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow
(
)
(
Hence
Rx + p 1g⋅ A1 − p 2g⋅ A2 = V1 ⋅ −ρ⋅ V1 ⋅ A1 + V2 ⋅ ρ⋅ V2 ⋅ A2
From continuity
Q = V1 ⋅ A1 = V2 ⋅ A2
so
Q
ft
V1 =
= 20⋅
×
A1
s
3
4
1
π⋅ ⎛⎜ ⋅ ft⎞
⎝3 ⎠
2
= 229 ⋅
ft
s
)
Rx = −p 1g⋅ A1 + ρ⋅ ⎛ V2 ⋅ A2 − V1 ⋅ A1⎞
⎝
⎠
2
2
and
2
A1
D
ft
V2 = V1 ⋅
= V1 ⋅ ⎛⎜ ⎞ = 229 ⋅ ×
A2
d
s
⎝ ⎠
2
⎛ 4 ⎞ = 1628⋅ ft
⎜ 1.5
s
⎝ ⎠
NOTE: problem has an error: Flow rate should be 2 ft3/s not 20 ft3/s! We will provide answers to both
Hence
Rx = −200 ⋅
lbf
2
×
π⋅ ( 4 ⋅ in)
2
+ 1.94⋅
4
in
slug
ft
3
×
2
2
2
2⎤
⎡⎛
ft
ft
⎢⎜ 1628⋅ ⎞ × π⋅ ( 1.5⋅ in) − ⎛⎜ 229 ⋅ ⎞ × π⋅ ( 4⋅ in) ⎥ ×
s⎠
s⎠
4
4
⎣⎝
⎝
⎦
×
2
2
2
2⎤
2
2
⎡⎛
ft
ft
⎢⎜ 163 ⋅ ⎞ × π⋅ ( 1.5⋅ in) − ⎛⎜ 22.9⋅ ⎞ × π⋅ ( 4 ⋅ in) ⎥ × ⎛⎜ 1⋅ ft ⎞ × lbf ⋅ s
s⎠
s⎠
4
4
slug⋅ ft
⎣⎝
⎝
⎦ ⎝ 12⋅ in ⎠
2
2
⎛ 1 ⋅ ft ⎞ × lbf ⋅ s
⎜ 12⋅ in
slug⋅ ft
⎝
⎠
Rx = 51707 ⋅ lbf
With more realistic velocities
Hence
Rx = −200 ⋅
lbf
2
in
Rx = −1970⋅ lbf
×
π⋅ ( 4 ⋅ in)
4
2
+ 1.94⋅
slug
ft
3
Problem 4.64
Problem 4.83
[Difficulty: 2]
Problem 4.84
[Difficulty: 2]
CS
Ve
y
x
Rx
Given:
Data on rocket motor
Find:
Thrust produced
Solution:
Basic equation: Momentum flux in x direction for the elbow
Assumptions: 1) Steady flow 2) Neglect change of momentum within CV 3) Uniform flow
Hence


Rx  p eg Ae  Ve ρe Ve Ae  me Ve
Rx  p eg Ae  me Ve
where p eg is the exit pressure (gage), me is the mass flow rate at the exit (software cannot render dot over m!) and V e is the exit velocity
For the mass flow rate
kg
kg
me  mnitricacid  maniline  80
 32
s
s
Hence
Rx  ( 110  101 )  10 
3 N
2
m

π ( 0.6 m)
4
2
 112 
kg
me  112 
s
kg
s
 180 
m
s
2

N s
kg m
Rx  22.7 kN
Problem 4.65
Problem 4.85
[Difficulty: 2]
Problem 4.86
[Difficulty: 3]
Problem 4.87
[Difficulty: 2]
Given:
Data on flow and system geometry
Find:
Deflection angle as a function of speed; jet speed for 10o deflection
Solution:
The given data is
kg
ρ = 999⋅
2
A = 0.01⋅ m
3
L = 2⋅ m
k = 500⋅
m
N
x 0 = 1⋅ m
m
Basic equation (y momentum):
Applying this to the current system in the vertical direction
Fspring = V⋅ sin( θ) ⋅ ( ρ⋅ V⋅ A)
(
)
But
)
2
Hence
k ⋅ x 0 − L⋅ sin( θ) = ρ⋅ V ⋅ A⋅ sin( θ)
Solving for θ
θ = asin⎜
For the speed at which θ =
(
Fspring = k ⋅ x = k ⋅ x 0 − L⋅ sin( θ)
k⋅ x0
⎛
⎞
⎜ k⋅ L + ρ⋅ A⋅ V2
⎝
⎠
10o,
solve
V=
(
k ⋅ x 0 − L⋅ sin( θ)
)
ρ⋅ A⋅ sin( θ)
500 ⋅
V =
999 ⋅
N
m
⋅ ( 1 − 2 ⋅ sin( 5 ⋅ deg) ) ⋅ m
kg
3
⋅
2
kg⋅ m
2
V = 21.8
⋅ 0.01⋅ m ⋅ sin( 5 ⋅ deg) N⋅ s
m
Angle (deg.)
35
30
25
20
15
10
5
0
5
10
15
V (m/s)
20
25
m
s
Problem 4.69
Problem 4.88
[Difficulty: 3]
Problem 4.71
Problem 4.89
[Difficulty: 3]
Problem 4.90
[Difficulty: 2]
y
x
Ry
Rx
CS
Given:
Data on nozzle assembly
Find:
Reaction force
Solution:
Basic equation: Momentum flux in x and y directions
Assumptions: 1) Steady flow 2) Incompressible flow CV 3) Uniform flow
For x momentum
(
)
2
Rx = V2 ⋅ cos( θ) ⋅ ρ⋅ V2 ⋅ A2 = ρ⋅ V2 ⋅
π⋅ D2
4
2
⋅ cos( θ)
⎛ D1 ⎞
V2 = V1 ⋅
= V1 ⋅ ⎜
A2
⎝ D2 ⎠
A1
2
m
V2 = 2 ⋅ ×
s
From continuity
A1 ⋅ V1 = A2 ⋅ V2
Hence
2
2
m⎞
N⋅ s
π
2
⎛
Rx = 1000⋅
×
× ( 0.025 ⋅ m) × cos( 30⋅ deg) ×
× ⎜ 18⋅
3 ⎝
s⎠
kg⋅ m
4
m
For y momentum
Ry − p 1 ⋅ A1 − W − ρ⋅ Vol ⋅ g = −V1 ⋅ −ρ⋅ V1 ⋅ A1 − V2 ⋅ sin( θ) ⋅ ρ⋅ V2 ⋅ A2
(
π⋅ D1
2
+ W + ρ⋅ Vol ⋅ g +
4
W = 4.5⋅ kg × 9.81⋅
m
m
V2 = 18
s
2
3 N
Ry = 125 × 10 ⋅
2
×
×
kg
3
m
Ry = 554 ⋅ N
×
π
4
ρ⋅ π
(
)
⋅ ⎛ V ⋅ D1 − V2 ⋅ D2 ⋅ sin( θ) ⎞
⎠
4 ⎝ 1
N⋅ s
2
2
2
2
kg⋅ m
4
×
3
W = 44.1 N
π⋅ ( 0.075 ⋅ m)
m
+ 1000⋅
)
Rx = 138 ⋅ N
2
s
Hence
2
kg
Ry = p 1⋅
where
⎛ 7.5 ⎞
⎜ 2.5
⎝ ⎠
Vol = 0.002 ⋅ m
2
+ 44.1⋅ N + 1000⋅
⎡⎛ m ⎞ 2
2
⎢⎜ 2 ⋅
× ( 0.075 ⋅ m) −
s
⎣⎝
⎠
kg
3
m
3
× 0.002 ⋅ m × 9.81⋅
m
2
s
2
×
N⋅ s
kg⋅ m
...
2
2
⎛ 18⋅ m ⎞ × ( 0.025 ⋅ m) 2 × sin( 30⋅ deg)⎤⎥ × N⋅ s
⎜ s
⎝
⎠
⎦ kg⋅ m
Problem 4.91
Given:
Data on water jet pump
Find:
Speed at pump exit; pressure rise
[Difficulty: 3]
Solution:
Basic equation: Continuity, and momentum flux in x direction
Assumptions: 1) Steady flow 2) Incompressible flow CV 3) Uniform flow
From continuity
⎛ 0.75 − 0.1 ⎞ + 100 ⋅ ft × 0.1
⎜ 0.75
0.75
s
⎝
⎠
ft
V2 = 10⋅ ×
s
For x momentum
As
Aj
Aj
⎛ A2 − Aj ⎞
V2 = Vs⋅
+ Vj⋅
= Vs⋅ ⎜
+ Vj⋅
A2
A2
A2
⎝ A2 ⎠
−ρ⋅ Vs⋅ As − ρ⋅ Vj⋅ Aj + ρ⋅ V2 ⋅ A2 = 0
(
)
(
ft
V2 = 22⋅
s
)
(
p 1 ⋅ A2 − p 2 ⋅ A2 = Vj⋅ −ρ⋅ Vj⋅ Aj + Vs⋅ −ρ⋅ Vs⋅ As + V2 ⋅ ρ⋅ V2 ⋅ A2
)
A
⎛ 2 Aj
2 s
2⎞
∆p = p 2 − p 1 = ρ⋅ ⎜ Vj ⋅
+ Vs ⋅
− V2
A2
A2
⎝
⎠
∆p = 1.94⋅
slug
ft
Hence
∆p = 1816⋅
3
lbf
ft
2
×
2
2
2
2
⎡⎛
ft
ft
ft ⎤
⎢⎜ 100 ⋅ ⎞ × 0.1 + ⎛⎜ 10⋅ ⎞ × ( 0.75 − 0.1) − ⎛⎜ 22⋅ ⎞ ⎥ × lbf ⋅ s
s⎠
s⎠
0.75 ⎝
0.75
⎣⎝
⎝ s ⎠ ⎦ slug⋅ ft
∆p = 12.6⋅ psi
Problem 4.73
Problem 4.92
[Difficulty: 3]
Problem 4.93
[Difficulty: 3]
V1
V2
CS
p1
p2
Rx
y
x
Given:
Data on adiabatic flow of air
Find:
Force of air on pipe
Solution:
Basic equation: Continuity, and momentum flux in x direction, plus ideal gas equation
p  ρ R T
Assumptions: 1) Steady flow 2) Ideal gas CV 3) Uniform flow
From continuity
ρ1  V1  A1  ρ2  V2  A2  0
ρ1  V1  A  ρ2  V2  A
For x momentum
Rx  p 1  A  p 2  A  V1  ρ1  V1  A  V2  ρ2  V2  A  ρ1  V1  A V2  V1






Rx  p 2  p 1  A  ρ1  V1  A V2  V1
For the air
P1
ρ1 
Rair T1



kg K
1
3 N
ρ1  ( 200  101 )  10 


2
286.9  N m ( 60  273 )  K
m
3 N
Rx  ( 80  200 )  10 
2
m
Hence

ρ1  V1  ρ2  V2
2
 0.05 m  3.15
kg
3
m
 150 
m
s
2
 0.05 m  ( 300  150 ) 
ρ1  3.15
kg
3
m
m
s
2

N s
kg m
Rx  2456 N
This is the force of the pipe on the air; the pipe is opposing flow. Hence the force of the air on the pipe is
Fpipe  2456 N
The air is dragging the pipe to the right
Fpipe  Rx
Problem 4.74
Problem 4.94
[Difficulty: 3]
Problem 4.95
[Difficulty: 3]
V1
V2
CS
p1
p2
ρ1
Rx
y
V3
ρ2
x
Given:
Data on heated flow of gas
Find:
Force of gas on pipe
Solution:
Basic equation: Continuity, and momentum flux in x direction
p = ρ⋅ R⋅ T
Assumptions: 1) Steady flow 2) Uniform flow
From continuity
m3
ρ1
V2 = V1 ⋅
−
ρ2
ρ2 ⋅ A
−ρ1 ⋅ V1 ⋅ A1 + ρ2 ⋅ V2 ⋅ A2 + m3 = 0
3
m
6
kg
m
V2 = 170 ⋅ ×
− 20⋅
×
×
s
2.75
s
2.75⋅ kg
For x momentum
(
m
V2 = 322
s
2
0.15⋅ m
(
)
Rx + p 1 ⋅ A − p 2 ⋅ A = V1 ⋅ −ρ1 ⋅ V1 ⋅ A + V2 ⋅ ρ2 ⋅ V2 ⋅ A
Rx =
⎡( p − p ) + ρ ⋅ V 2 − ρ ⋅ V 2⎤ ⋅ A
2 2
1 1⎦
⎣ 2 1
⎡
3 N
⎢
⎣
m
Rx = ⎢( 300 − 400 ) × 10 ⋅
Hence
)
1
where m3 = 20 kg/s is the mass leaving
through the walls (the software does not
allow a dot)
Rx = 1760 N
2
⎡
kg
⎢
⎣
m
+ ⎢2.75⋅
3
× ⎜⎛ 322 ⋅
⎝
m⎞
s
⎠
2
− 6⋅
kg
3
m
× ⎜⎛ 170 ⋅
⎝
2⎤
⎥ × N⋅ s ⎥ × 0.15⋅ m2
s ⎠ ⎥ kg⋅ m⎥
⎦
⎦
m⎞
2⎤
Problem 4.96
Given:
Data on flow out of pipe device
Find:
Velocities at 1 and 2; force on coupling
[Difficulty: 3]
Solution:
Basic equations (continuity and x and y mom.):
The given data is
ρ = 999⋅
3
kg
D = 20⋅ cm
3
L = 1⋅ m
t = 20⋅ mm
p 3g = 50⋅ kPa Q = 0.3⋅
m
From continuity
Q = A⋅ Vave
Note that at the exit
V( x ) = V1 +
Hence
Q=
1
(
Applying y momentum
V3 =
π
Q
2
⋅D
)
⋅x
(
)
m
V1 = 10
s
⌠
Ry = −⎮
⌡
L
V2 = 2 ⋅ V1
m
V2 = 20
s
m
V3 = 9.549
s
π 2
Rx + p 3g⋅ ⋅ D = −V3 ⋅ ρ⋅ Q
4
0
Expanding and integrating
L
)
4
Applying x momentum
(V2 − V1)
(
1
Vave = ⋅ V1 + V2
2
1
⋅ V1 + V2 ⋅ L⋅ t = ⋅ V1 + 2 ⋅ V1 ⋅ L⋅ t
2
2
2⋅ Q
V1 =
3 ⋅ L⋅ t
At the inlet (location 3)
due to linear velocity distribution
π 2
Rx = −p 3g⋅ ⋅ D − V3 ⋅ ρ⋅ Q
4
⌠
⎮
V( x ) ⋅ ρ⋅ V( x ) ⋅ t dx = −ρ⋅ t⋅ ⎮
⎮
⌡
L
Rx = −4.43⋅ kN
(V2 − V1) ⎤ 2
⎡
⎢V1 +
⋅ x⎥ dx
L
⎣
⎦
0
2
⎡⎢
⎛ V2 − V1 ⎞ L2 ⎛ V2 − V1 ⎞ L3⎥⎤
2
Ry = −ρ⋅ t⋅ ⎢V1 ⋅ L + 2 ⋅ V1 ⋅ ⎜
⋅
+⎜
⋅ ⎥
⎣
⎝ L ⎠ 2 ⎝ L ⎠ 3⎦
Ry = −4.66⋅ kN
m
s
Problem 4.78
Problem 4.97
[Difficulty: 3]
Problem 4.79
Problem 4.98
[Difficulty: 3]
Problem 4.99
[Difficulty: 4]
Given:
Data on flow in wind tunnel
Find:
Mass flow rate in tunnel; Maximum velocity at section 2; Drag on object
Solution:
Basic equations: Continuity, and momentum flux in x direction; ideal gas equation
p = ρ⋅ R⋅ T
Assumptions: 1) Steady flow 2) Uniform density at each section
From continuity
mflow = ρ1 ⋅ V1 ⋅ A1 = ρ1 ⋅ V1 ⋅
π⋅ D1
2
where mflow is the mass flow rate
4
p atm
ρair =
Rair⋅ Tatm
We take ambient conditions for the air density
kg
m π⋅ ( 0.75⋅ m)
mflow = 1.2⋅
× 12.5⋅ ×
3
s
4
m
⌠
⌠
mflow = ⎮ ρ2 ⋅ u 2 dA2 = ρair⋅ ⎮
⎮
⎮
⌡
⌡
Also
R
N
kg⋅ K
1
kg
ρair = 101000⋅
×
×
ρ = 1.2
2
286.9 ⋅ N⋅ m 293 ⋅ K air
3
m
m
2
kg
mflow = 6.63
s
2 ⋅ π⋅ ρair⋅ Vmax ⌠ R 2
2 ⋅ π⋅ ρair⋅ Vmax⋅ R
Vmax⋅ ⋅ 2 ⋅ π⋅ r dr =
⋅ ⎮ r dr =
⌡
R
3
R
0
0
Vmax =
3 ⋅ mflow
2 ⋅ π⋅ ρair⋅ R
3
3
kg
m
Vmax =
× 6.63⋅
×
×
2⋅ π
s
1.2⋅ kg
2
2
r
⎛ 1 ⎞
⎜ 0.375 ⋅ m
⎝
⎠
2
m
Vmax = 18.8
s
⌠
Rx + p 1 ⋅ A − p 2 ⋅ A = V1 ⋅ −ρ1 ⋅ V1 ⋅ A + ⎮ ρ2 ⋅ u 2 ⋅ u 2 dA2
⎮
⌡
R
2
⌠
2
⎮
2 ⋅ π⋅ ρair⋅ Vmax ⌠ R 3
r⎞
⎛
Rx = p 2 − p 1 ⋅ A − V1 ⋅ mflow + ⎮ ρair⋅ ⎜ Vmax⋅
⋅ 2 ⋅ π⋅ r dr = p 2 − p 1 ⋅ A − V1 ⋅ mflow +
⋅ ⎮ r dr
⌡
2
R⎠
⎝
⎮
0
R
⌡
(
For x momentum
(
)
)
(
)
0
(
)
π
2 2
Rx = p 2 − p 1 ⋅ A − V1 ⋅ mflow + ⋅ ρair⋅ Vmax ⋅ R
2
We also have
p 1 = ρ⋅ g ⋅ h 1
p 1 = 1000⋅
kg
3
× 9.81⋅
m
Hence
Rx = ( 147 − 294 ) ⋅
N
2
m
Rx = −54 N
×
m
2
× 0.03⋅ m
p 1 = 294 Pa
p 2 = ρ⋅ g ⋅ h 2
p 2 = 147 ⋅ Pa
s
π⋅ ( 0.75⋅ m)
4
2
⎡
+ ⎢−6.63⋅
⎢
⎣
kg
s
× 12.5⋅
m
s
The drag on the object is equal and opposite
+
π
2
× 1.2⋅
kg
3
m
× ⎛⎜ 18.8⋅
⎝
Fdrag = −Rx
m⎞
s
⎠
2
× ( 0.375 ⋅ m)
⎤
2⎥
⎥
⎦
×
Fdrag = 54.1 N
N
k
Problem 4.100
Given:
Data on wake behind object
Find:
An expression for the drag
[Difficulty: 2]
Solution:
Basic equation:
Momentum
Applying this to the horizontal motion

2
F  U  ρ π 1  U  

1
u ( r)  ρ 2  π r u ( r) dr
0
Integrating and using the limits
1




2
2
F  π ρ U  2   r u ( r) dr


0




2
F  π ρ U  1 




2
 
2 
π r 
2   r  1  cos
dr
 
2
   



2
F  π ρ U   1 




2
4

π

r
  r cos π r  dr
2   r  2  r cos

2
 
 2 


F  π ρ U  1 
 3  2 
8
2 
π 

2


1
0

1
0

F
2
 5 π 2 
 8  π  ρ U


Problem 4.101
Given:
Data on flow in 2D channel
Find:
Maximum velocity; Pressure drop
[Difficulty: 3]
y
2h
x
Solution:

Basic equations: Continuity, and momentum flux in x direction
CS

Assumptions: 1) Steady flow 2) Neglect friction
3
Given data
w = 25⋅ mm
h = 50⋅ mm
Q
U1 =
2⋅ w⋅ h
From continuity
Q = U1 ⋅ 2 ⋅ h ⋅ w
Also
⌠
−ρ⋅ U1⋅ A1 + ⎮ ρ⋅ u 2 dA = 0
⎮
⌡
⌠
⎮
U1 ⋅ 2 ⋅ h ⋅ w = w⋅ ⎮
⎮
⌡
h
⎛
u max⋅ ⎜ 1 −
⎜
⎝
−h
Hence
For x momentum
u max =
3
2
Q = 0.025⋅
y
h
⋅ U1
ρ = 750⋅
s
3
m
⎠
h
4
h
dy = w⋅ u max⋅ ⎡⎢[ h − ( −h ) ] − ⎡⎢ − ⎛⎜ − ⎞⎥⎤⎥⎤ = w⋅ u max⋅ ⋅ h
3
3
3
⎣
⎣
⎝ ⎠⎦⎦
u max = 15
m
s
⌠
Note that there is no Rx (no friction)
p 1 ⋅ A − p 2 ⋅ A = V1 ⋅ −ρ1 ⋅ V1 ⋅ A + ⎮ ρ2 ⋅ u 2 ⋅ u 2 dA2
⎮
⌡
h
⌠
2
2
⎮
2⎞
ρ⋅ u max
1
w ⎮
y
2
2
2
2⎛
⎜
dy = −ρ⋅ U1 +
⋅ ⎡⎢2 ⋅ h − 2 ⋅ ⎛⎜ ⋅ h⎞ + 2 ⋅ ⎛⎜ ⋅ h⎞⎤⎥
p 1 − p 2 = −ρ⋅ U1 + ⋅ ⎮ ρ⋅ u max ⋅ 1 −
h
⎜
A
2
⎣
⎝3 ⎠
⎝ 5 ⎠⎦
h ⎠
⎮
⎝
⌡
(
)
−h
⎡8 3
⎤
8
2
2
∆p = p 1 − p 2 = −ρ⋅ U1 +
⋅ ρ⋅ u max = ρ⋅ U1 ⋅ ⎢ ⋅ ⎛⎜ ⎞ − 1⎥
15
⎣ 15 ⎝ 2 ⎠
⎦
2
Hence
kg
m
U1 = 10.0
s
2⎞
2
m
∆p =
1
5
⋅ ρ⋅ U1
2
∆p = 15.0⋅ kPa
Problem 4.102
Given:
Data on flow in 2D channel
Find:
Maximum velocity; Pressure drop
[Difficulty: 3]
y
2h
x
Solution:

Basic equations: Continuity, and momentum flux in x direction
CS

Assumptions: 1) Steady flow 2) Neglect friction
3
R = 75⋅ mm
Given data
From continuity
Q = 0.1⋅
Q = U1 ⋅ π⋅ R
2
m
ρ = 850⋅
s
π⋅ R
3
m
Q
U1 =
kg
m
U1 = 5.66
s
2
⌠
−ρ⋅ U1⋅ A1 + ⎮ ρ⋅ u 2 dA = 0
⎮
⌡
Also
⌠
2 ⎮
U1 ⋅ π⋅ R = ⎮
⎮
⌡
R
2
2
4
2
⎛
⎛ R2
r ⎞
R
R ⎞
R
⎜
⎜
u max⋅ 1 −
−
= 2 ⋅ π⋅ u max⋅
= π⋅ u max⋅
⋅ 2 ⋅ π⋅ r dr = 2 ⋅ π⋅ u max⋅
2
2
⎜
⎜ 2
2
4
R
4
⋅
R
⎝
⎠
⎝
⎠
0
u max = 2 ⋅ U1
Hence
For x momentum
(
u max = 11.3
m
s
⌠
p 1 ⋅ A − p 2 ⋅ A = V1 ⋅ −ρ1 ⋅ V1 ⋅ A + ⎮ ρ2 ⋅ u 2 ⋅ u 2 dA2
⎮
⌡
(
⌠
⎮
2
2
2 ⎮
p 1 − p 2 ⋅ π⋅ R = −ρ⋅ π⋅ R ⋅ U1 + ⎮
⎮
⌡
)
Note that there is no Rx (no friction)
R
2
2
2
4
6
⎛
r ⎞
R
R ⎞
2
2
2 ⎛R
⋅ 2 ⋅ π⋅ r dr = −ρ⋅ π⋅ R ⋅ U1 + 2 ⋅ π⋅ ρ⋅ u max ⋅ ⎜
ρ⋅ u max ⋅ ⎜ 1 −
− 2⋅
+
⎜
⎜ 2
2
2
4
R ⎠
4⋅ R
6⋅ R ⎠
⎝
⎝
)
2
0
1
1
1
1
2
2
2
2
2
2
∆p = p 1 − p 2 = −ρ⋅ U1 + ⋅ ρ⋅ u max = −ρ⋅ U1 + ⋅ ρ⋅ 2 ⋅ U1 = ρ⋅ U1 ⋅ ⎢⎡ ⋅ ( 2 ) − 1⎥⎤ = ⋅ ρ⋅ U1
3
3
3
3
⎣
⎦
(
Hence
∆p =
1
3
× 850 ⋅
kg
3
m
× ⎛⎜ 5.66⋅
⎝
m⎞
s
⎠
2
)
2
×
N⋅ s
kg⋅ m
∆p = 9.08⋅ kPa
Problem 4.84
Problem 4.103
[Difficulty: 3]
Problem 4.86
Problem 4.104
[Difficulty: 3]
Problem 4.105
[Difficulty: 4]
b
CS
c
y
x
a
d
Ff
Given:
Data on flow of boundary layer
Find:
Plot of velocity profile; force to hold plate
Solution:
Basic equations: Continuity, and momentum flux in x direction
Assumptions: 1) Steady flow 2) Incompressible 3) No net pressure force
ρ = 750 ⋅
Given data
kg
m
U0 = 10⋅
s
w = 1⋅ m
3
m
L = 1⋅ m
δ = 5 ⋅ mm
1
0.8
y
δ
0.6
0.4
0.2
0
0.2
0.4
0.6
0.8
u( y )
U0
δ
From continuity
⌠
−ρ⋅ U0 ⋅ w⋅ δ + mbc + ⎮ ρ⋅ u ⋅ w dy = 0
⌡
0
δ
Hence
where mbc is the mass flow rate across bc (Note:
sotware cannot render a dot!)
⌠
mbc = ⎮ ρ⋅ U0 − u ⋅ w dy
⌡
(
)
0
For x momentum
δ
⌠
⌠
−Ff = U0 ⋅ −ρ⋅ U0 ⋅ w⋅ δ + U0 ⋅ mbc + ⎮ u ⋅ ρ⋅ u ⋅ w dy = ⎮
⌡
⌡
0
(
)
δ
⎡−U 2 + u2 + U ⋅ ( U − u)⎤ ⋅ w dy
0 0
⎣ 0
⎦
0
Then the drag force is
δ
δ
⌠
⌠
u ⎞
2 u ⎛
Ff = ⎮ ρ⋅ u ⋅ U0 − u ⋅ w dy = ⎮ ρ⋅ U0 ⋅
⋅⎜1 −
dy
U
U
⎮
⌡
0
0
⎝
⎠
0
⌡
(
)
0
1
But we have
u
U0
=
3
2
⌠
=⎮
w
⎮
⌡
Ff
⋅η −
η= 1
0
1
2
⋅η
3
where we have used substitution
1
u ⎞
2 ⌠
⋅ ⎛⎜ 1 −
dη = ρ⋅ U0 ⋅ δ⋅ ⎮
ρ⋅ U0 ⋅ δ⋅
⎮
U0
U0
⎝
⎠
⌡
0
u
2
Ff
⎛ 3 ⋅ η − 9 ⋅ η2 − 1 ⋅ η3 + 3 ⋅ η4 − 1 ⋅ η6⎞ dη
⎜2
4
2
2
4
⎝
⎠
3
1
3
1 ⎞
3
2
2
−
+
−
= ρ⋅ U0 ⋅ δ⋅ ⎛⎜ −
= 0.139 ⋅ ρ⋅ U0 ⋅ δ
w
⎝ 4 4 8 10 28 ⎠
Hence
Ff
w
Ff
w
= 0.139 × 750 ⋅
kg
3
m
= 52.1
N
m
× ⎛⎜ 10⋅
⎝
m⎞
s
⎠
2
2
× 0.05⋅ m ×
N⋅ s
kg⋅ m
y = δ⋅ η
Problem 4.106
[Difficulty: 4]
b
CS
c
y
x
a
d
Ff
Given:
Data on flow of boundary layer
Find:
Force on plate per unit width
Solution:
Basic equations: Continuity, and momentum flux in x direction
Assumptions: 1) Steady flow 2) Incompressible 3) No net pressure force
δ
From continuity
⌠
−ρ⋅ U0 ⋅ w⋅ δ + mbc + ⎮ ρ⋅ u ⋅ w dy = 0
⌡
where mbc is the mass flow rate across bc (Note: sotware
cannot render a dot!)
0
δ
Hence
⌠
mbc = ⎮ ρ⋅ U0 − u ⋅ w dy
⌡
(
)
0
For x momentum
δ
⌠
⌠
−Ff = U0 ⋅ −ρ⋅ U0 ⋅ w⋅ δ + U0 ⋅ mbc + ⎮ u ⋅ ρ⋅ u ⋅ w dy = ⎮
⌡
⌡
0
(
)
δ
⎡−U 2 + u2 + U ⋅ ( U − u)⎤ ⋅ w dy
0 0
⎣ 0
⎦
0
Then the drag force is
δ
δ
⌠
⌠
u ⎞
2 u ⎛
Ff = ⎮ ρ⋅ u ⋅ U0 − u ⋅ w dy = ⎮ ρ⋅ U0 ⋅
⋅⎜1 −
dy
U0
U0
⎮
⌡
⎝
⎠
0
⌡
(
)
0
But we have
u
U0
=
y
y = δ⋅ η
where we have used substitution
δ
⌠
=⎮
w
⎮
⌡
Ff
η= 1
1
u ⎞
2 ⌠
⋅ ⎛⎜ 1 −
dη = ρ⋅ U0 ⋅ δ⋅ ⎮ η⋅ ( 1 − η) dη
ρ⋅ U0 ⋅ δ⋅
⌡
U0
U0
0
⎝
⎠
u
2
0
Ff
w
Hence
Ff
w
Ff
w
1
1
1
2
2
= ρ⋅ U0 ⋅ δ⋅ ⎛⎜ − ⎞ = ⋅ ρ⋅ U0 ⋅ δ
3
6
2
⎝
⎠
=
1
6
× 1.225 ⋅
= 0.163 ⋅
kg
3
m
N
m
× ⎛⎜ 20⋅
⎝
m⎞
s
⎠
2
×
2
1000
2
⋅m ×
N⋅ s
kg⋅ m
(using standard atmosphere density)
Problem 4.107
Difficulty: 4] Part 1/2
Problem 4.107
[Difficulty: 4] Part 2/2
Problem 4.108
[Difficulty: 4]
Problem 4.109
[Difficulty: 4]
Problem *4.91
Problem *4.110
[Difficulty: 4]
Problem *4.111
[Difficulty: 4]
CS


Given:
Air jet striking disk
Find:
Manometer deflection; Force to hold disk
Solution:
Basic equations: Hydrostatic pressure, Bernoulli, and momentum flux in x direction
2
p
ρ
+
V
+ g ⋅ z = constant
2
Assumptions: 1) Steady flow 2) Incompressible 3) No friction 4) Flow along streamline 5) Uniform flow 6) Horizontal flow (g x = 0)
Applying Bernoulli between jet exit and stagnation point
p
ρair
2
+
V
2
=
p0
ρair
+0
p0 − p =
1
But from hydrostatics
p 0 − p = SG⋅ ρ⋅ g ⋅ ∆h
∆h =
so
2
1
2
2
⋅ ρair⋅ V
2
⋅ ρair⋅ V
SG ⋅ ρ⋅ g
3
2
2
=
ρair⋅ V
2 ⋅ SG ⋅ ρ⋅ g
2
ft
ft
s
1
∆h = 0.002377⋅
×
×
× ⎜⎛ 225 ⋅ ⎞ ×
3
s⎠
2 ⋅ 1.75 1.94⋅ slug 32.2⋅ ft
⎝
ft
slug
For x momentum
(
)
∆h = 0.55⋅ ft
2
2 π⋅ D
Rx = V⋅ −ρair⋅ A⋅ V = −ρair⋅ V ⋅
4
2
ft
Rx = −0.002377⋅
× ⎛⎜ 225 ⋅ ⎞ ×
3
s⎠
⎝
ft
slug
The force of the jet on the plate is then
F = −Rx
π⋅ ⎛⎜
0.5
⋅ ft⎞
2
2
⎝ 12 ⎠ × lbf ⋅ s
4
slug⋅ ft
Rx = −0.164 ⋅ lbf
F = 0.164 ⋅ lbf
∆h = 6.6⋅ in
Problem *4.112
[Difficulty: 3]

CS
y
x
V, A
Rx

Given:
Water jet shooting upwards; striking surface
Find:
Flow rate; maximum pressure; Force on hand
Solution:
Basic equations: Bernoulli and momentum flux in x direction
p
ρ
2

V
2
 g  z  constant
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure throughout 4) Uniform flow
Given data
h  10 m
ρ  1000
kg
D  1  cm
3
m
p atm
Using Bernoulli between the jet exit and its maximum height h
ρ
or
Then
V 
Q 
2 g h
π
4
V  14.0
2
D V
Q  66.0
2

V

V
2

p atm
ρ
 g h
m
s
L
min
For Dr. Pritchard the maximum pressure is obtained from Bernoulli
p atm
ρ
2
2


p max
ρ

p 
1
2
2
 ρ V
p  98.1 kPa
(gage)
2
For Dr. Pritchard blocking the jet, from x momentum applied to the CV Rx  u 1  ρ u 1  A1  ρ V  A
Hence
Repeating for Dr. Fox
2 π
F  ρ V 
4
2
D
h  15 m
p 
1
2
V 
2
 ρ V
2 π
F  ρ V 
F  15.4 N
4
2 g h
p  147.1  kPa
2
D
F  23.1 N
V  17.2
(gage)
m
s
Q 
π
4
2
D V
Q  80.8
L
min
Problem *4.113
[Difficulty: 3]
Problem *4.114
[Difficulty: 3]
CS


Given:
Water jet striking disk
Find:
Expression for speed of jet as function of height; Height for stationary disk
Solution:
Basic equations: Bernoulli; Momentum flux in z direction
p
ρ
2
+
V
2
+ g ⋅ z = constant
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow
The Bernoulli equation becomes
V0
2
2
2
V
+ g⋅ 0 =
2
+ g⋅ h
2
(
2
V = V0 − 2 ⋅ g ⋅ h
)
V=
2
V0 − 2 ⋅ g ⋅ h
2
Hence
−M ⋅ g = w1 ⋅ −ρ⋅ w1 ⋅ A1 = −ρ⋅ V ⋅ A
But from continuity
ρ⋅ V0 ⋅ A0 = ρ⋅ V⋅ A
Hence we get
M ⋅ g = ρ⋅ V⋅ V⋅ A = ρ⋅ V0 ⋅ A0 ⋅ V0 − 2 ⋅ g ⋅ h
Solving for h
h=
V⋅ A = V0 ⋅ A0
so
2
1
⎢⎡
2
⋅ V −
2⋅ g ⎢ 0
⎣
⎛ M⋅ g ⎞
⎜ ρ⋅ V ⋅ A
⎝ 0 0⎠
2⎤
⎥
⎥
⎦
⎡⎢
2
m
h =
×
× ⎢⎛⎜ 10⋅ ⎞ −
2 9.81⋅ m ⎢⎝
s⎠
⎢⎣
1
h = 4.28 m
2
s
3
⎡
⎤
s
4
⎢2⋅ kg × 9.81⋅ m × m
⎥
×
×
1000⋅ kg 10⋅ m
2⎥
2
⎢
25
s
⋅ m⎞ ⎥
π⋅ ⎜⎛
⎢
⎣
⎝ 1000 ⎠ ⎦
2⎤
⎥
⎥
⎥
⎥⎦
Problem *4.96
Problem *4.115
[Difficulty: 4] Part 1/2
Problem *4.96 cont'd
Problem *4.115
[Difficulty: 4] Part 2/2
Problem *4.116
Given:
Stream of water striking a vane
Find:
Water speed; horizontal force on vane
[Difficulty: 3]
Solution:
Basic equations: Bernoulli; Momentum flux in x direction
p
ρ
2
+
V
+ g ⋅ z = constant
2
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow
Given or available data
From Bernoulli
Combining
D = 50⋅ mm
p0 = p +
1
2
2
⋅ ρwater⋅ V
1
kg
ρwater = 1000⋅
3
m
ρHg = 13.6⋅ ρwater
and for the manometer
p 0 − p = ρHg⋅ g ⋅ ∆h
2
⋅ρ
⋅ V = ρHg⋅ g ⋅ ∆h
2 water
Applying x momentum to the vane
V =
or
2 ⋅ ρHg⋅ g ⋅ ∆h
ρwater
θ = 30⋅ deg
V = 14.1
∆h = 0.75⋅ m
m
s
π 2
π 2
Rx = ρwater⋅ V⋅ ⎛⎜ −V⋅ ⋅ D ⎞ + ρwater⋅ ( −V⋅ cos( θ) ) ⋅ ⎛⎜ V⋅ ⋅ D ⎞
4
4
⎝
⎠
2 π 2
Rx = −ρwater⋅ V ⋅ ⋅ D ⋅ ( 1 + cos( θ) )
4
⎝
⎠
Rx = −733 N
Assuming frictionless, incompressible flow with no net pressure force is realistic, except along the vane where friction will
reduce flow momentum at the exit.
Problem *4.117
[Difficulty: 2]
Given:
Data on flow and venturi geometry
Find:
Force on convergent section; water pressure
Solution:
Basic equations:
2
p
Bernoulli equation and x momentum
+
ρ
ρ = 999 ⋅
The given data is
kg
V
2
+ g ⋅ z = const
D = 100 ⋅ mm
3
d = 50⋅ mm
p 1 = 200 ⋅ kPa
Q = 1000⋅
A1 = 0.00785 m
π 2
A2 = ⋅ d
4
A2 = 0.00196 m
m
V1 = 2.12
s
Q
V2 =
A2
m
V2 = 8.49
s
m
L
min
For pressure we first need the velocities
2
A1 =
π⋅ D
V1 =
π
Then
2
4
4
Q
2
⋅D
p1
Applying Bernoulli between inlet and throat
+
ρ
V1
2
=
2
2
p2
ρ
+
V2
2
2
ρ
2
2
p 2 = p 1 + ⋅ ⎛ V1 − V2 ⎞
⎠
2 ⎝
Solving for p 2
p 2 = 200 ⋅ kPa +
1
2
⋅ 999 ⋅
kg
3
(
)
2
2 m
2
× 2.12 − 8.49 ⋅
m
2
s
2
×
N⋅ s
kg⋅ m
×
kN
1000⋅ N
p 2 = 166 ⋅ kPa
Applying the horizontal component of momentum
(
)
(
−F + p 1 ⋅ A2 − p 2 ⋅ A2 = V1 ⋅ −ρ⋅ V1 ⋅ A1 + V2 ⋅ ρ⋅ V2 ⋅ A2
)
F = p 1 ⋅ A1 − p 2 ⋅ A2 + ρ⋅ ⎛ V1 ⋅ A1 − V2 ⋅ A2⎞
⎝
⎠
2
or
F = 200 ⋅
kN
2
m
F = 1.14 kN
2
× 0.00785 ⋅ m − 166 ⋅
kN
2
m
2
2
× 0.00196 ⋅ m + 999 ⋅
kg
3
m
×
2
⎡⎛
⎢⎜ 2.12⋅ m ⎞ ⋅ 0.00785 ⋅ m2 −
s⎠
⎣⎝
2
2
⎛ 8.49⋅ m ⎞ ⋅ 0.00196 ⋅ m2⎤⎥ ⋅ N⋅ s
⎜
s⎠
⎝
⎦ kg × m
Problem *4.118
Given:
Nozzle flow striking inclined plate
Find:
Mimimum gage pressure
[Difficulty: 3]
Solution:
Basic equations: Bernoulli and y momentum
p
ρ
The given data is
2

V
 g  z  const
2
ρ  999 
kg
3
L
q  1200
s m
m
q
V2 
W
For the exit velocity and nozzle velocity
Then from Bernoulli
p1 
ρ
ρ
2
2
 V1  p atm   V2
2
2
W  80 mm h  0.25 m w  20 mm
m
V2  15.0
s
w
V1  V2 
W
or
p1 
θ  30 deg
m
V1  3.75
s
  V  V1
2  2
ρ
H  7.5 m
2
2
  ρ g h
p 1  103  kPa
(gage)
Applying Bernoulli between 2 and the plate (state 3)
p atm 
ρ
2
ρ
2
2
 V2  p atm   V3  ρ g  H
2
V3 
2
V2  2  g  H
m
V3  19.3
s
For the plate there is no force along the plate (x momentum) as there is no friction. For the force normal to the plate
(y momentum) we have


Ry  V3  cos( θ)  ρ V3  A3  V3  cos( θ)  ( ρ q )
Ry  V3  cos( θ)  ρ q
Ry  20.0
kN
m
Problem *4.119
Given:
Water faucet flow
Find:
Expressions for stream speed and diameter; plot
[Difficulty: 3]
Solution:
p
Basic equation: Bernoulli
ρ
2
+
V
+ g ⋅ z = const
2
Assumptions: Laminar, frictionless, uniform flow
D0 = 5 ⋅ mm
The given data is
V0 =
π
The initial velocity is
4
h = 50⋅ mm
Q
⋅ D0
Q =
p atm
ρ
V( z) =
Evaluating at h
2
V0 + 2 ⋅ g ⋅ z
V( h ) = 1.03
Q = 0.333⋅
L
min
m
V0 = 0.283
s
2
Then applying Bernoulli between the exit and any other location
Then
1⋅ L
3⋅ min
Also
m
π
π 2
2
V0⋅ ⋅ D0 = V⋅ ⋅ D
4
4
2
+
V0
2
=
ρ
2
+
V
2
D( z) =
so
− g⋅ z
(z downwards)
D0
1
2⋅ g ⋅ z ⎞
⎜⎛ 1 +
2
⎜
V0
⎝
⎠
D( h ) = 2.62⋅ mm
s
p atm
4
1.25
10
Height (mm)
V (m/s)
1
0.75
0.5
20
30
40
0.25
0
10
20
30
z (mm)
40
50
− 2.5 − 1.5 − 0.5
0.5
Diameter (mm)
1.5
2.5
Problem *4.99
Problem *4.120
[Difficulty: 4]
Problem *4.100
Problem *4.121
[Difficulty: 4]
Problem *4.102
Problem *4.122
[Difficulty: 4]
Problem *4.123
[Difficulty: 4] Part 1/2
Problem *4.123
[Difficulty: 4] Part 2/2
Problem *4.124
[Difficulty: 5]
Given:
Plates coming together
Find:
Expression for velcoity field; exit velocity; plot
Solution:
Apply continuity using deformable CV as shown
Basic equation:
=0
Assumptions: Incompressible, uniform flow
m
V0  0.01
s
Given data:
Continuity becomes
or
2 dh
π r 
R  100  mm
or

t
2
 V 2  π r h  π r  V0  V 2  π r h  0
dt
If V0 is constant
h  h 0  V0  t
Evaluating
V( R 0 )  0.250
Exit Velocity (m/s)
h 0  2  mm
so
m
s
V( r t) 

V0  r
2  h 0  V0  t
V( R 0.1 s)  0.500

πr2h  V2πrh  0
Hence
r
V( r)  V0 
2 h
Note that
tmax 
h0
V0
tmax  0.200 s
m
s
6
4
2
0
0.05
0.1
0.15
t (s)
The velocity greatly increases as the constant flow rate exits through a gap that becomes narrower with time.
0.2
Problem *4.104
Problem *4.125
[Difficulty: 5] Part 1/2
Problem *4.104 cont'd
Problem *4.125
[Difficulty: 5] Part 2/2
Problem *4.126
[Difficulty: 4] Part 1/2
Problem *4.126
[Difficulty: 4] Part 2/2
Problem *4.127
[Difficulty: 3]

CS (moves
at speed U)
y

x
Rx
Ry
Given:
Water jet striking moving vane
Find:
Force needed to hold vane to speed U = 5 m/s
Solution:
Basic equations: Momentum flux in x and y directions
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Jet relative velocity is
constant
Then
(
)
(
)
Rx = u 1 ⋅ −ρ⋅ V1 ⋅ A1 + u 2 ⋅ ρ⋅ V2 ⋅ A2 = −( V − U) ⋅ [ ρ⋅ ( V − U) ⋅ A] + ( V − U) ⋅ cos( θ) ⋅ [ ρ⋅ ( V − U) ⋅ A]
2
Rx = ρ( V − U) ⋅ A⋅ ( cos( θ) − 1 )
A =
π
4
⋅ ⎜⎛
40
⎝ 1000
⋅ m⎞
2
−3
A = 1.26 × 10
⎠
2
m
Using given data
Rx = 1000⋅
kg
3
× ⎡⎢( 25 − 5 ) ⋅
⎣
m
Then
(
)
(
m⎤
2
2
N⋅ s
−3 2
× 1.26 × 10 ⋅ m × ( cos( 150 ⋅ deg) − 1 ) ×
⎥
s⎦
kg⋅ m
Rx = −940 N
)
Ry = v 1 ⋅ −ρ⋅ V1 ⋅ A1 + v 2 ⋅ ρ⋅ V2 ⋅ A2 = −0 + ( V − U) ⋅ sin( θ) ⋅ [ ρ⋅ ( V − U) ⋅ A]
2
Ry = ρ( V − U) ⋅ A⋅ sin( θ)
Ry = 1000⋅
kg
3
m
× ⎡⎢( 25 − 5 ) ⋅
⎣
m⎤
2
2
N⋅ s
−3 2
R = 252 N
× 1.26 × 10 ⋅ m × sin( 150 ⋅ deg) ×
⎥
s⎦
kg⋅ m y
Hence the force required is 940 N to the left and 252 N upwards to maintain motion at 5 m/s
Problem 4.128
[Difficulty: 3]

CS (moves
at speed U)

y
Rx
Ry
Given:
Water jet striking moving vane
Find:
Force needed to hold vane to speed U = 10 m/s
x
Solution:
Basic equations: Momentum flux in x and y directions
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Jet relative velocity is
constant
Then
(
)
(
)
Rx = u 1 ⋅ −ρ⋅ V1 ⋅ A1 + u 2 ⋅ ρ⋅ V2 ⋅ A2 = −( V − U) ⋅ [ ρ⋅ ( V − U) ⋅ A] + ( V − U) ⋅ cos( θ) ⋅ [ ρ⋅ ( V − U) ⋅ A]
2
Rx = ρ( V − U) ⋅ A⋅ ( cos( θ) − 1 )
Using given data
Rx = 1000⋅
kg
3
× ⎡⎢( 30 − 10) ⋅
⎣
m
Then
(
)
(
m⎤
2
2
N⋅ s
2
× 0.004 ⋅ m × ( cos( 120 ⋅ deg) − 1 ) ×
⎥
s⎦
kg⋅ m
Rx = −2400 N
)
Ry = v 1 ⋅ −ρ⋅ V1 ⋅ A1 + v 2 ⋅ ρ⋅ V2 ⋅ A2 = −0 + ( V − U) ⋅ sin( θ) ⋅ [ ρ⋅ ( V − U) ⋅ A]
2
Ry = ρ( V − U) ⋅ A⋅ sin( θ)
Ry = 1000⋅
kg
3
m
× ⎡⎢( 30 − 10) ⋅
⎣
m⎤
2
2
N⋅ s
2
× 0.004 ⋅ m × sin( 120 ⋅ deg) ×
⎥
s⎦
kg⋅ m
Hence the force required is 2400 N to the left and 1390 N upwards to maintain motion at 10 m/s
Ry = 1386 N
Problem 4.129
[Difficulty: 2]
Problem 4.130
[Difficulty: 3]
Given:
Data on jet boat
Find:
Formula for boat speed; flow rate; value of k; new speed and flow rate
Solution:
CV in boat
coordinates
Basic equation:
Momentum
Given data
m
D = 75⋅ mm Vj = 15⋅
s
V = 10⋅
m
kg
ρ = 1000⋅
s
3
m
Applying the horizontal component of momentum
Fdrag = V⋅ ( −ρ⋅ Q) + Vj⋅ ( ρ⋅ Q)
2
2
Fdrag = k ⋅ V
or, with
k ⋅ V = ρ⋅ Q⋅ Vj − ρ⋅ Q⋅ V
2
k ⋅ V + ρ⋅ Q⋅ V − ρ⋅ Q⋅ Vj = 0
Solving for V
For the flow rate
To find k from Eq 1, let
V= −
ρ⋅ Q
2⋅ k
2
⎛ ρ⋅ Q ⎞ + ρ⋅ Q⋅ Vj
⎜ 2⋅ k
k
⎝
⎠
+
π
2
Q = Vj⋅ ⋅ D
4
α=
ρ⋅ Q
3
Q = 0.0663
m
s
2
2
2
For
k =
ρ⋅ Q
2⋅ α
m
Vj = 25⋅
s
k = 3.31
α + 2 ⋅ α⋅ Vj
2
( V + α) = V + 2 ⋅ α⋅ V + α = α + 2 ⋅ α⋅ Vj
Hence
2
V = −α +
then
2⋅ k
2
(1)
α =
or
(
V
)
2 ⋅ Vj − V
α = 10
m
s
N
⎛ m⎞
⎜
⎝s⎠
π 2
Q = Vj⋅ ⋅ D
4
2
3
Q = 0.11
m
s
2
⎡ ρ⋅ Q
ρ⋅ Q⋅ Vj⎤
ρ⋅ Q ⎞
⎥ V = 16.7 m
+ ⎛⎜
+
k ⎦
s
⎣ 2⋅ k
⎝ 2⋅ k ⎠
V = ⎢−
Problem 4.110
Problem 4.131
[Difficulty: 2]
Problem 4.112
Problem 4.132
[Difficulty: 2]
Problem 4.133
[Difficulty: 3]

CS (moves
at speed U)

y
Ry
Rx
Given:
Water jet striking moving vane
Find:
Expressions for force and power; Show that maximum power is when U = V/3
Solution:
Basic equation: Momentum flux for inertial CV
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow
5) Jet relative velocity is constant
Then




Rx  u 1 ρ V1 A1  u 2 ρ V2 A2  ( V  U)  [ ρ ( V  U)  A ]  ( V  U)  cos ( θ)  [ ρ ( V  U)  A ]
2
Rx  ρ( V  U)  A  ( cos ( θ)  1)
This is force on vane; Force exerted by vane is equal and opposite
2
Fx  ρ ( V  U)  A ( 1  cos( θ) )
The power produced is then
2
P  U Fx  ρ U ( V  U)  A ( 1  cos( θ) )
To maximize power wrt to U
dP
dU
Hence
2
 ρ ( V  U)  A ( 1  cos( θ) )  ρ ( 2 )  ( 1 )  ( V  U)  U A ( 1  cos( θ) )  0
V  U  2 U  V  3 U  0
Note that there is a vertical force, but it generates no power
U
V
3
for maximum power
x
Problem 4.134
[Difficulty: 3]
CS (moves to
left at speed Vc) 
Vj + Vc
Vj + Vc

y
R
Rx
x
t
Given:
Water jet striking moving cone
Find:
Thickness of jet sheet; Force needed to move cone
Solution:
Basic equations: Mass conservation; Momentum flux in x direction
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Jet relative velocity is
constant
Then
Hence
(
−ρ⋅ V1 ⋅ A1 + ρ⋅ V2 ⋅ A2 = 0
t=
Dj
)
−ρ⋅ Vj + Vc ⋅
π⋅ Dj
4
2
(
)
+ ρ⋅ Vj + Vc ⋅ 2 ⋅ π⋅ R⋅ t = 0
(Refer to sketch)
2
t =
8⋅ R
1
8
2
× ( 4 ⋅ in) ×
1
t = 0.222 ⋅ in
9 ⋅ in
Using relative velocities, x momentum is
(
)
(
)
(
) (
)
(
)
(
)
Rx = u 1 ⋅ −ρ⋅ V1 ⋅ A1 + u 2 ⋅ ρ⋅ V2 ⋅ A2 = − Vj + Vc ⋅ ⎡ρ⋅ Vj + Vc ⋅ Aj⎤ + Vj + Vc ⋅ cos( θ) ⋅ ⎡ρ⋅ Vj + Vc ⋅ Aj⎤
⎣
⎦
⎣
⎦
(
)2
Rx = ρ Vj + Vc ⋅ Aj⋅ ( cos( θ) − 1 )
Using given data
2
ft
Rx = 1.94⋅
× ⎡⎢( 100 + 45) ⋅ ⎥⎤ ×
3
s⎦
⎣
ft
slug
π⋅ ⎛⎜
4
⋅ ft⎞
2
2
⎝ 12 ⎠ × ( cos( 60⋅ deg) − 1 ) × lbf ⋅ s
4
Hence the force is 1780 lbf to the left; the upwards equals the weight
slug⋅ ft
Rx = −1780⋅ lbf
Problem 4.114
Problem 4.135
[Difficulty: 3]
Problem 4.116
Problem 4.136
[Difficulty: 3]
Problem 4.117
Problem 4.137
[Difficulty: 3]
Problem 4.138
[Difficulty: 2]
Problem 4.139
Given:
Jet impacting a splitter vane
Find:
Mass flow rate ratio; new speed U
Solution:
Apply momentum equation to inertial CV
[Difficulty: 4]
Assumptions: No pressure force; neglect water mass on vane; steady flow wrt vane; uniform flow; no change of speed wrt the vane
Basic equation
Given data
V = 25⋅
−5
m
A = 7.85⋅ 10
s
2
⋅m
U = 10⋅
)
Hence
0 = 0 + ( V − U) ⋅ m2 − ( V − U) ⋅ sin( θ) ⋅ m3
Note that
m1 = ρ⋅ A⋅ ( V − U)
m1 = m2 + m3 so
s
ρ = 999⋅
kg
3
m
0 = v 1⋅ −m1 + v 2⋅ m2 + v 3⋅ m3
For no vertical force, y momentum becomes
and
θ = 30⋅ deg
V− U
For constant speed wrt the vane, the jet velocity at each location is
(
m
where v i and mi are the vertical components
of velocity and mass flow rates, respectively,
at the inlet and exits, wrt the vane
coordinates
m2
1
m2 = m3 ⋅ sin( θ)
= sin( θ) =
m3
2
or
kg
m1 = 1.18
s
m3
m1 = m3 ⋅ sin( θ) + m3
(
m1
=
1
m3
1 + sin( θ)
m1
)
(
=
kg
m3 = 0.784
s
2
3
)
and using x momentum
Rx = u 1 ⋅ −m1 + u 2 ⋅ m2 + u 3 ⋅ m3 = ( V − U) ⋅ −m1 + 0 + ( V − U) ⋅ cos( θ) ⋅ m3
Writing in terms of m1
Rx = ( V − U) ⋅ m1 ⋅ ⎛⎜
Instead, the force is now
Rx = −16⋅ N
Hence
Rx = ( V − U) ⋅ ρ⋅ A⋅ ⎛⎜
Solving for U
U = V−
cos( θ)
⎝ 1 + sin( θ)
Rx = −7.46 N
⎠
Rx = ( V − U) ⋅ m1 ⋅ ⎛⎜
but
2
− 1⎞
cos( θ)
⎝ 1 + sin( θ)
cos( θ)
⎝ 1 + sin( θ)
⎡ρ⋅ A⋅ ⎛
− 1⎞⎥⎤
⎢ ⎜
⎣ ⎝ 1 + sin( θ)
⎠⎦
cos( θ)
− 1⎞
⎠
− 1⎞
Rx
kg
m2 = 0.392
s
⎠
U = 3.03
m
s
and
m1 = ρ⋅ A⋅ ( V − U)
Problem 4.120
Problem 4.133
Problem 4.140
[Difficulty: 3]
Problem 4.141
[Difficulty: 2]
Problem 4.142
[Difficulty: 3]
Problem 4.143
[Difficulty: 4]
Given:
Data on vane/slider
Find:
Formula for acceleration and speed; plot
Solution:
The given data is
ρ = 999 ⋅
kg
2
M = 30⋅ kg
3
A = 0.005 ⋅ m
m
V = 20⋅
dU
The equation of motion, from Problem 4.141, is
dt
2
ρ⋅ ( V − U) ⋅ A
=
M
− g ⋅ μk
2
The acceleration is thus
a=
ρ⋅ ( V − U) ⋅ A
− g ⋅ μk
M
μk = 0.3
s
m
dU
Separating variables
ρ⋅ ( V − U) ⋅ A
M
Substitute
u= V− U
du
dU = −du
ρ⋅ A⋅ u
and u = V - U so
Using initial conditions
⌠
⎮
⎮
⎮
⎮
⌡
−
−
1
⎛ ρ⋅ A⋅ u2
⎞
⎜
− g ⋅ μk
⎝ M
⎠
M
g ⋅ μk ⋅ ρ⋅ A
M
g ⋅ μk ⋅ ρ⋅ A
V− U=
U= V−
Note that
⎛
ρ⋅ A
⎝
g ⋅ μk ⋅ M
⎡
ρ⋅ A
⋅ atanh⎜
⋅ atanh⎢
⎣ g ⋅ μk⋅ M
g ⋅ μk ⋅ M
ρ⋅ A
g ⋅ μk ⋅ M
ρ⋅ A
⎛
ρ⋅ A
⎝
g ⋅ μk ⋅ M
atanh⎜
du = −
− g ⋅ μk
⎛
M
g ⋅ μk ⋅ ρ⋅ A
ρ⋅ A
⋅ atanh⎜
M
⎠
g ⋅ μk ⋅ ρ⋅ A
⎞
⎡
ρ⋅ A
⎣
g ⋅ μk ⋅ M
⋅ atanh⎢
⎤
M
⎦
g ⋅ μk ⋅ ρ⋅ A
⋅ ( V − U)⎥ +
⋅u
⎝ g⋅ μk ⋅ M ⎠
⎞
⋅u = −
− g ⋅ μk
= −dt
2
M
But
= dt
2
⎛
⋅ atanh⎜
⎤
⋅ ( V − U)⎥
⎦
ρ⋅ A
⎞
⋅ V = −t
⎝ g ⋅ μk⋅ M ⎠
⎛ g⋅ μk ⋅ ρ⋅ A
⎛ ρ⋅ A ⋅ V⎞ ⎞
⋅ t + atanh⎜
⎜⎝
M
⎝ g⋅ μk ⋅ M ⎠ ⎠
⋅ tanh⎜
⎛ g⋅ μk ⋅ ρ⋅ A
⎛ ρ⋅ A ⋅ V⎞ ⎞
⋅ t + atanh⎜
⎜⎝
M
⎝ g⋅ μk ⋅ M ⎠ ⎠
⋅ tanh⎜
⎞
π
⎠
2
⋅ V = 0.213 −
⋅i
which is complex and difficult to handle in Excel, so we use the identity
atanh( x ) = atanh⎛⎜
1⎞
⎝x⎠
−
π
2
⋅i
for x > 1
so
U= V−
and finally the identity
tanh⎛⎜ x −
⎝
g ⋅ μk ⋅ M
ρ⋅ A
π
2
⋅ i⎞ =
⎠
⎛ g⋅ μk ⋅ ρ⋅ A
1
⎞ − π ⋅ i⎞
⋅ t + atanh⎛
⎜
M
2 ⎟
ρ⋅ A
⎜
⋅V
⎜
⎜
⎝
⎝ g⋅ μk ⋅ M ⎠
⎠
⋅ tanh⎜
1
tanh( x )
g ⋅ μk ⋅ M
to obtain
ρ⋅ A
U( t) = V −
⎛ g⋅ μk ⋅ M 1 ⎞ ⎞
⎛ g⋅ μk ⋅ ρ⋅ A
⋅ t + atanh⎜
⋅
M
⎝
⎝ ρ⋅ A V ⎠ ⎠
tanh⎜
g ⋅ μk ⋅ M
2
a=
Note that
ρ⋅ ( V − U) ⋅ A
M
− g ⋅ μk
⎛ g ⋅ μk⋅ ρ⋅ A
⎛ g ⋅ μk⋅ M 1 ⎞ ⎞
⋅ t + atanh⎜
⋅
M
⎝
⎝ ρ⋅ A V ⎠ ⎠
tanh⎜
g ⋅ μk
a( t ) =
Hence
ρ⋅ A
V− U=
and
⎛ g⋅ μk ⋅ ρ⋅ A
⎛ g⋅ μk ⋅ M 1 ⎞ ⎞
⋅ t + atanh⎜
⋅
M
⎝
⎝ ρ⋅ A V ⎠ ⎠
2
− g ⋅ μk
tanh⎜
The plots are presented below
20
U (m/s)
15
10
5
0
0.5
1
1.5
2
2.5
3
2
2.5
3
t (s)
a (m/s2)
60
40
20
0
0.5
1
1.5
t (s)
Problem 1.24
Problem 4.133
Problem 4.144
[Difficulty: 3]
Problem 4.145
[Difficulty: 4]

CS (moves at
speed
instantaneous
speed U)

y
x
Given:
Water jet striking moving vane/cart assembly
Find:
Angle θ at t = 5 s; Plot θ(t)
Solution:
Basic equation: Momentum flux in x direction for accelerating CV
Assumptions: 1) No changes in CV 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Constant jet relative
velocity
Then
(
)
(
)
−M ⋅ arfx = u 1 ⋅ −ρ⋅ V1 ⋅ A1 + u 2 ⋅ ρ⋅ V2 ⋅ A2 = −( V − U) ⋅ [ ρ⋅ ( V − U) ⋅ A] + ( V − U) ⋅ cos( θ) ⋅ [ ρ⋅ ( V − U) ⋅ A]
2
Since
−M ⋅ arfx = ρ( V − U) ⋅ A⋅ ( cos( θ) − 1 )
or
arfx = constant
U = arfx⋅ t
⎡
θ = acos⎢1 −
⎢
⎣
then
M ⋅ arfx
cos( θ) = 1 −
cos( θ) = 1 −
2
ρ⋅ ( V − U) ⋅ A
M ⋅ arfx
(
)
2
ρ⋅ V − arfx⋅ t ⋅ A
⎤
⎥
2
ρ⋅ ( V − arfx⋅ t) ⋅ A⎥
⎦
M ⋅ arfx
Using given data
⎡⎢
⎢
⎢
⎢⎣
θ = acos 1 − 55⋅ kg × 1.5⋅
m
2
3
×
s
m
1000⋅ kg
1
×
⎛ 15⋅ m − 1.5⋅ m × 5 ⋅ s⎞
⎜ s
2
s
⎝
⎠
2
×
⎥⎤
2⎥
0.025 ⋅ m ⎥
⎥⎦
1
20
Angle
Speed
135
15
90
10
45
5
0
at t = 5 s
0
2.5
5
7.5
Time t (s)
The solution is only valid for θ up to 180 o (when t = 9.14 s). This graph can be plotted in Excel
0
10
Speed U (m/s)
Angle (deg)
180
θ = 19.7⋅ deg
Problem 4.146
[Difficulty: 3]
Given:
Vaned cart with negligible resistance
Find:
Initial jet speed; jet and cart speeds at 2.5 s and 5 s; what happens to V - U?
Solution:
Apply x momentum
Assumptions: 1) All changes wrt CV 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Constant jet area
Given data
ρ = 999 ⋅
kg
2
M = 5 ⋅ kg
3
A = 50⋅ mm
a = 2.5⋅
m
Then
2
Hence
a⋅ M = ρ⋅ ( V − U) ⋅ ( 1 − cos( θ) ) ⋅ A
Solving for V
V( t) = a⋅ t +
Also, for constant acceleration
θ = 120 ⋅ deg
2
s
−a⋅ M = u 1 ⋅ [ −ρ⋅ ( V − U) ⋅ A] + u 1 ⋅ [ ρ⋅ ( V − U) ⋅ A]
Hence, evaluating
m
where
u1 = V − U
and
u 2 = ( V − U) ⋅ cos( θ)
From this equation we can see that for constant acceleration V and U must
increase at the same rate!
M⋅ a
ρ⋅ ( 1 − cos( θ) ) ⋅ A
V( 0 ) = 12.9
U( t) = a⋅ t
m
s
V( 2.5⋅ s) = 19.2
so
m
s
V( 5 ⋅ s) = 25.4
V− U=
m
s
M⋅ a
ρ⋅ ( 1 − cos( θ) ) ⋅ A
= const!
Problem 4.147
[Difficulty: 3] Part 1/2
Problem 4.147
[Difficulty: 3] Part 2/2
Problem 4.148
[Difficulty: 3]
Problem 4.149
[Difficulty: 3] Part 1/2
Problem 4.149
[Difficulty: 3] Part 2/2
Problem 4.130
Problem 4.150
[Difficulty: 3]
Problem 4.151
[Difficulty: 4]
Given:
Vaned cart being hit by jet
Find:
Jet speed to stop cart in 1s and 2 s; distance traveled
Solution:
Apply x momentum
Assumptions: 1) All changes wrt CV 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Constant jet area
Given data
ρ = 999 ⋅
kg
M = 5 ⋅ kg
3
D = 35⋅ mm
θ = 60⋅ deg
m
π
A =
4
2
2
⋅D
A = 962 ⋅ mm
Then
−arf ⋅ M = u 1 ⋅ [ −ρ⋅ ( V + U) ⋅ A] + u 2 ⋅ [ ρ⋅ ( V + U) ⋅ A]
where
arf =
Hence
or
−
dU
−
dU
dt
dt
dU
u 1 = −( V + U)
dt
2
2
2
⋅ M = ρ⋅ ( V + U) ⋅ A − ρ⋅ ( V + U) ⋅ A⋅ cos( θ) = ρ⋅ ( V + U) ⋅ A⋅ ( 1 − cos( θ) )
2
⋅ M = ρ⋅ ( V + U) ⋅ A⋅ ( 1 − cos( θ) )
(1)
−
Hence
U0
V⋅ V + U0
V= −
U0
2
)
+
=
1
V
ρ⋅ ( 1 − cos( θ) ) ⋅ A⋅ t
M
U0
4
2
+
d ( V + U)
( V + U)
Integrating from U0 at t = 0 to U = 0 at t
(
u 2 = −( V + U) ⋅ cos( θ)
and
Note that V is constant, so dU = d(V+U), separating variables
Solving for V
m
U0 = 5 ⋅
s
U0 ⋅ M
ρ⋅ A⋅ ( 1 − cos( θ) ) ⋅ t
or
−
2
=
1
V + U0
ρ⋅ ( 1 − cos( θ) ) ⋅ A
M
=
⋅ dt
ρ⋅ ( 1 − cos( θ) ) ⋅ A
M
M ⋅ U0
2
V + V⋅ U0 −
ρ⋅ ( 1 − cos( θ) ) ⋅ A⋅ t
⋅t
To find distances note that
dU
dt
so Eq. 1 can be rewritten as
Separating variables
−U⋅
=
dU
dU dx
⋅
= U⋅
dx
dx dt
dU
⋅ M = ρ⋅ ( V + U) ⋅ A⋅ ( 1 − cos( θ) )
dx
2
U⋅ dU
( V + U)
2
=−
ρ⋅ A⋅ ( 1 − cos( θ) )
M
⋅ dx
0
It can be shown that
⌠
V ⎞ V
V
U
⎮
dU = ln⎛⎜
+
−
⎮
2
⎝ V + U0 ⎠ V V + U0
⎮ ( V + U)
⌡U
(Remember that V is constant)
0
ln⎛⎜
V
⎞+1−
⎝ V + U0 ⎠
x=−
Solving for x
V
V + U0
M
ρ⋅ A⋅ ( 1 − cos( θ) )
⋅ ⎜⎛ ln⎛⎜
⎝
=−
ρ⋅ A⋅ ( 1 − cos( θ) )
V
M
⎞+1−
⎝ V + U0 ⎠
⋅x
V
⎞
V + U0
⎠
Substituting values:
To stop in
To stop in
t = 1⋅ s
V = −
and
x = −
U0
2
+
U0
2
+
4
U0 ⋅ M
ρ⋅ A⋅ ( 1 − cos( θ) ) ⋅ t
⋅ ⎛⎜ ln⎛⎜
⎞+1− V ⎞
ρ⋅ A⋅ ( 1 − cos( θ) )
V + U0
⎝ ⎝ V + U0 ⎠
⎠
M
t = 2⋅ s
V = −
and
x = −
U0
2
+
U0
4
2
+
V
U0 ⋅ M
ρ⋅ A⋅ ( 1 − cos( θ) ) ⋅ t
⎞+1− V ⎞
ρ⋅ A⋅ ( 1 − cos( θ) )
V + U0
⎝ ⎝ V + U0 ⎠
⎠
M
⋅ ⎛⎜ ln⎛⎜
V
V = 5.13
m
s
x = 1.94 m
V = 3.18
m
s
x = 3.47 m
Problem 4.132
Problem 4.152
[Difficulty: 3]
Problem 4.153
[Difficulty: 4]
Given:
Data on vane/slider
Find:
Formula for acceleration, speed, and position; plot
Solution:
Apply x momentum
Assumptions: 1) All changes wrt CV 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Constant jet area
The given data is
ρ  999 
kg
2
M  30 kg
3
A  0.005  m
m
Then
k U  M  arf  u 1 [ ρ ( V  U)  A ]  u 2 m2  u 3 m3
where
arf 
Hence
or
dU
u1  V  U
dt
k  U  M 
dU
dt
dU
dt
ρ ( V  U)  A
M
a
s
k  7.5
Ns
m
u3  0
2

k U
M
2
The acceleration is thus
m
 ρ ( V  U)  A
2

u2  0
V  20
ρ ( V  U)  A
M

k U
M
The differential equation for U can be solved analytically, but is quite messy. Instead we use a simple numerical method - Euler's
method
 ρ ( V  U( n) ) 2 A k U( n )
  ∆t
U( n  1 )  U( n )  

M
M 

For the position x
dx
dt
so
U
x ( n  1 )  x ( n )  U( n )  ∆t
The final set of equations is
U( n  1 )  U( n ) 
 ρ ( V  U( n) ) 2 A k U( n )

  ∆t

M 
M

2
a( n ) 
ρ ( V  U( n ) )  A
M

x ( n  1 )  x ( n )  U( n )  ∆t
k  U( n )
M
where Δt is the time step
The results can be plotted in Excel
Position x vs Time
45
x (m)
40
35
30
25
20
a (m/s 2)
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
0.0
0.0
0.7
1.6
2.7
3.9
5.2
6.6
7.9
9.3
10.8
12.2
13.7
15.2
16.6
18.1
19.6
21.1
22.6
24.1
25.7
27.2
28.7
30.2
31.7
33.2
34.8
36.3
37.8
39.3
40.8
0.0
6.7
9.5
11.1
12.1
12.9
13.4
13.8
14.1
14.3
14.5
14.6
14.7
14.8
14.9
15.0
15.0
15.1
15.1
15.1
15.1
15.1
15.2
15.2
15.2
15.2
15.2
15.2
15.2
15.2
15.2
66.6
28.0
16.1
10.5
7.30
5.29
3.95
3.01
2.32
1.82
1.43
1.14
0.907
0.727
0.585
0.472
0.381
0.309
0.250
0.203
0.165
0.134
0.109
0.0889
0.0724
0.0590
0.0481
0.0392
0.0319
0.0260
0.0212
5
0
-5 0.0
0.5
1.0
1.5
2.0
2.5
3.0
2.5
3.0
t (s)
Velocity U vs Time
16
14
U (m/s)
U (m/s)
12
10
8
6
4
2
0
0.0
0.5
1.0
1.5
2.0
t (s)
70
Acceleration a vs Time
60
2
x (m)
a (m/s )
t (s)
15
10
50
40
30
20
10
0
0
1
1
2
t (s)
2
3
3
Problem 4.134
Problem 4.154
[Difficulty: 3]
Problem 4.136
Problem 4.155
[Difficulty: 3]
Problem 4.156
[Difficulty: 3]
Given:
Data on system
Find:
Jet speed to stop cart after 1 s; plot speed & position; maximum x; time to return to origin
Solution:
Apply x momentum
Assumptions: 1) All changes wrt CV 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Constant jet area
The given data is
kg
ρ = 999 ⋅
2
M = 100⋅ kg
3
A = 0.01⋅ m
m
Then
−arf ⋅ M = u 1⋅ [ −ρ⋅ ( V + U) ⋅ A ] + u 2⋅ m2 + u 3⋅ m3
where
arf =
Hence
−
dU
dt
dU
u 1 = −( V + U)
dt
2
⋅ M = ρ⋅ ( V + U) ⋅ A
or
dU
dt
u2 = u3 = 0
and
2
=−
ρ⋅ ( V + U) ⋅ A
d ( V + U)
which leads to
M
( V + U)
V + U0
U = −V +
Integrating and using the IC U = U0 at t = 0
m
U0 = 5⋅
s
1+
(
ρ⋅ A⋅ V + U0
2
= −⎛⎜
ρ⋅ A
⎝ M
⋅ dt⎞
⎠
) ⋅t
M
To find the jet speed V to stop the cart after 1 s, solve the above equation for V, with U = 0 and t = 1 s. (The equation becomes a
quadratic in V). Instead we use Excel's Goal Seek in the associated workbook
From Excel
V = 5⋅
m
s
dx
For the position x we need to integrate
dt
The result is
x = −V⋅ t +
⎡
ρ⋅ A ⎣
M
⋅ ln⎢1 +
V + U0
= U = −V +
1+
(
ρ⋅ A⋅ V + U0
M
)
(
ρ⋅ A⋅ V + U0
M
) ⋅t
⎤
⎦
⋅ t⎥
This equation (or the one for U with U = 0) can be easily used to find the maximum value of x by differentiating, as well as the time for x
to be zero again. Instead we use Excel's Goal Seek and Solver in the associated workbook
From Excel
x max = 1.93⋅ m
The complete set of equations is
t( x = 0 ) = 2.51⋅ s
V + U0
U = −V +
ρ⋅ A⋅ V + U0
1+
⋅t
M
(
)
x = −V⋅ t +
⎡
ρ⋅ A ⎣
M
⋅ ln⎢1 +
(
ρ⋅ A⋅ V + U0
M
)
⎤
⎦
⋅ t⎥
The plots are presented in the Excel workbook:
t (s)
x (m)
U (m/s)
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
2.6
2.8
3.0
0.00
0.82
1.36
1.70
1.88
1.93
1.88
1.75
1.56
1.30
0.99
0.63
0.24
-0.19
-0.65
-1.14
5.00
3.33
2.14
1.25
0.56
0.00
-0.45
-0.83
-1.15
-1.43
-1.67
-1.88
-2.06
-2.22
-2.37
-2.50
To find V for U = 0 in 1 s, use Goal Seek
t (s)
U (m/s)
V (m/s)
1.0
0.00
5.00
To find the maximum x , use Solver
t (s)
x (m)
1.0
1.93
To find the time at which x = 0 use Goal Seek
t (s)
x (m)
2.51
0.00
Cart Position x vs Time
2.5
2.0
x (m)
1.5
1.0
0.5
0.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
2.5
3.0
-1.0
-1.5
t (s)
Cart Speed U vs Time
6
5
U (m/s)
4
3
2
1
0
-1
0.0
0.5
1.0
1.5
-2
-3
t (s)
2.0
Problem 4.157
[Difficulty: 2]
Given:
Mass moving betweem two jets
Find:
Time st slow to 2.5 m/s; plot position; rest position; explain
Solution:
Apply x momentum
Assumptions: 1) All changes wrt CV 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow 5) Constant jet area
The given data is
kg
ρ = 999 ⋅
M = 5⋅ kg
3
V = 20⋅
m
m
m
U0 = 10⋅
s
s
Then
−arf ⋅ M = u 1⋅ [ −ρ⋅ ( V − U) ⋅ A ] + u 2⋅ [ −ρ⋅ ( V + U) ⋅ A ] + u 3⋅ m3
where
arf =
−
Hence
dU
dt
dU
u1 = V − U
dt
⋅ M = ρ⋅ A ⋅ ⎡⎣−( V − U) + ( V + U)
2
dU
Separating and integrating
U
Solving for t
For position x
t = −
=−
dx
⋅ ln⎛⎜
4 ⋅ ρ⋅ A⋅ V
M
⎞
4 ⋅ ρ⋅ V⋅ A
U0
⎝ ⎠
M
−
U
m
2
A = 100⋅ mm
s
u3 = 0
2⎤
⎦ = 4⋅ ρ⋅ A ⋅ V⋅ U
or
( )
ln( U) − ln U0 = −
4 ⋅ ρ⋅ A⋅ V
M
−
⋅t
4⋅ ρ ⋅ A ⋅ V
t = 0.867 s for
⋅t
M
U = U0 ⋅ e
U = 2.5
(1)
m
s
⋅t
M
= U = U0 ⋅ e
dt
x final =
⋅ dt
and
and using given data
4⋅ ρ ⋅ A ⋅ V
and a straightforward integration leads to
For large time
u 2 = −( V + U)
U = 2.5⋅
M ⋅ U0
4 ⋅ ρ⋅ V⋅ A
x ( t) =
M ⋅ U0
4 ⋅ ρ⋅ V⋅ A
⎛
−
⎜
⋅⎝1 − e
4⋅ ρ ⋅ V ⋅ A
M
⎞
⋅t
⎠
For
t = 0.867 s
x ( t) = 4.69 m
x final = 6.26 m
8
x (m)
6
4
2
0
1
2
t (s)
3
4
Problem *4.158
[Difficulty: 3]


Given:
Water jet striking moving disk
Find:
Acceleration of disk when at a height of 3 m
CS moving
at speed U
Solution:
Basic equations: Bernoulli; Momentum flux in z direction (treated as upwards) for linear accelerating CV
p
ρ
2
+
V
2
+ g ⋅ z = constant
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow
The Bernoulli equation becomes
V0
2
2
+ g⋅ 0 =
V1
2
2
(
+ g ⋅ z − z0
)
V1 =
2
⎛ 15⋅ m ⎞ + 2 × 9.81⋅ m ⋅ ( 0 − 3) ⋅ m
⎜
2
⎝ s⎠
s
V1 =
(All in jet)
(
2
V0 + 2 ⋅ g ⋅ z0 − z
)
m
V1 = 12.9
s
The momentum equation becomes
(
)
(
) (
)
(
)
−W − M ⋅ arfz = w1 ⋅ −ρ⋅ V1 ⋅ A1 + w2 ⋅ ρ⋅ V2 ⋅ A2 = V1 − U ⋅ ⎡−ρ⋅ V1 − U ⋅ A1⎤ + 0
⎣
⎦
Hence
arfz =
(
)2
ρ⋅ V1 − U ⋅ A1 − W
arfz = 1000⋅
M
kg
3
m
=
× ⎡⎢( 12.9 − 5 ) ⋅
⎣
(
)2
ρ⋅ V1 − U ⋅ A1
M
m⎤
2
V0
2
ρ⋅ V1 − U ⋅ A0 ⋅
V1
(
−g=
15
)
M
1
m
2
×
− 9.81⋅
× 0.005 ⋅ m ×
⎥
s⎦
2
12.9 30⋅ kg
s
−g
arfz = 2.28
using
m
2
s
V1 ⋅ A1 = V0 ⋅ A0
Problem 4.159
[Difficulty: 4]
M = 35 kg


CS moving
at speed U
D = 75 mm
Given:
Water jet striking disk
Find:
Plot mass versus flow rate to find flow rate for a steady height of 3 m
Solution:
Basic equations: Bernoulli; Momentum flux in z direction (treated as upwards)
p
ρ
2
V

2
 g  z  constant
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow
V0
The Bernoulli equation becomes
2
2
 g 0 
V1
(All in jet)
2
2
 g h
V1 


2
V0  2  g  h
The momentum equation becomes




M  g  w1  ρ V1  A1  w2  ρ V2  A2  V1  ρ V1  A1  0
2
M
Hence
M
ρ V1  A1
V1  A1  V0  A0
but from continuity
g
ρ V1  V0  A0
g
2
π ρ V0  D0
2
 
 V0  2  g  h
g
4
and also
Q  V0  A0
This equation is difficult to solve for V 0 for a given M. Instead we plot first:
M (kg)
150
100
50
0.02
0.04
0.06
0.08
Q (cubic meter/s)
3
Goal Seek or Solver in Excel feature can be used to find Q when M = 35 kg
Q  0.0469
m
s
Problem 4.160
[Difficulty: 3]
Problem 4.161
[Difficulty: 3]
Problem 4.142
Problem 4.162
[Difficulty: 3] Part 1/2
Problem 4.142 cont'd
Problem 4.162
Difficulty: [3] Part 2/2
Problem 4.163
[Difficulty: 4]
Given:
Rocket sled on track
Find:
Plot speed versus time; maximum speed; effect of reducing k
Solution:
Basic equation: Momentum flux in x direction
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure in jet 4) Uniform flow
Given data
M 0 = 5000⋅ kg
k = 50⋅
N⋅ s
m
Ve = 1750⋅
s
m
−FR − arf ⋅ M = u e⋅ mrate = −Ve⋅ mrate
The momentum equation becomes
From continuity
M = M 0 − mrate⋅ t
Hence, combining
dU
−k ⋅ U − M 0 − mrate⋅ t ⋅
= −Ve⋅ mrate
dt
Separating variables
Integrating
Simplifying
Solving for U
Using given data
(
)
dU
Ve⋅ mrate − k ⋅ U
1
=
(
dt
M 0 − mrate⋅ t
(( (
)
k
⎛ Ve⋅ mrate − k ⋅ U ⎞
⋅ ln⎜
⎝
Ve⋅ mrate
⎠
⎡
⎢
Ve⋅ mrate ⎢
U( t) =
⋅ ⎢1 −
k
⎣
U( 10⋅ s) = 175
m
s
1
k
and
FR = k ⋅ U
dU
dt
))) =
⋅ ln⎛⎜ 1 −
⎝
kg
mrate = 50⋅
s
=
Ve⋅ mrate − k ⋅ U
M 0 − mrate⋅ t
)
(
=
M fuel = 1000⋅ kg
or
⋅ ln Ve⋅ mrate − k ⋅ U − ln Ve⋅ mrate
k
1
(All in jet)
1
mrate
( (
)
( ))
⋅ ln M 0 − mrate⋅ t − ln M 0
⎞ = 1 ⋅ ln⎛ M0 − mrate⋅ t ⎞ = 1 ⋅ ln⎛ 1 − mrate⋅ t ⎞
⎜
⎜
M0
M0
Ve⋅ mrate
⎠ mrate ⎝
⎠ mrate ⎝
⎠
k⋅ U
k
mrate⋅ t ⎞
⎛
⎜1 − M
0 ⎠
⎝
⎤
mrate⎥
⎥
⎥
⎦
and fuel is used up when
tfuel =
M fuel
mrate
tfuel = 20 s
This is when the speed is maximum
With 10% reduction in k
The percent improvement is
(
)
Umax = U tfuel
k 2 = 0.9⋅ k
Umax2 − Umax
Umax
m
Umax = 350
s
k2 ⎤
⎡
⎢
⎥
mrate⎥
⎢
mrate⋅ tfuel ⎞
Ve⋅ mrate ⎢
⎛
⎥
Umax2 =
⋅ 1 − ⎜1 −
⎢
⎥
M0
k2
⎣ ⎝
⎠
⎦
= 1.08⋅ %
When the fuel runs out the momentum equation simplifies from
(
)
dU
−k ⋅ U − M 0 − mrate⋅ t ⋅
= −Ve⋅ mrate
dt
−
The solution to this (with U = Umax when t = tfuel)
m
Umax2 = 354
s
Uempty( t) = Umax⋅ e
(
k⋅ t− tfuel
to
−k ⋅ U −
dU
dt
)
M0− Mfuel
400
U (m/s)
300
200
100
0
0
20
40
t (s)
60
=0
Problem 4.164
[Difficulty: 3]
CS at speed U
y
x
Ve
Y
X
Given:
Data on rocket sled
Find:
Minimum fuel to get to 265 m/s
Solution:
Basic equation: Momentum flux in x direction
Assumptions: 1) No resistance 2) p e = p atm 3) Uniform flow 4) Use relative velocities
From continuity
dM
dt
Hence from momentum
= mrate = constant
−arfx⋅ M = −
M = M 0 − mrate⋅ t
so
dU
dt
(
)
(
(Note: Software cannot render a dot!)
)
⋅ M 0 − mrate⋅ t = u e⋅ ρe⋅ Ve⋅ Ae = −Ve⋅ mrate
Ve⋅ mrate
Separating variables
dU =
Integrating
M0
mrate⋅ t ⎞
⎞
⎛
⎛
U = Ve⋅ ln⎜
= −Ve⋅ ln⎜ 1 −
or
M0
⎝ M0 − mrate⋅ t ⎠
⎝
⎠
M 0 − mrate⋅ t
⋅ dt
⎛⎜
−
⎜
The mass of fuel consumed is mf = mrate⋅ t = M 0 ⋅ ⎝ 1 − e
Hence
⎛
−
⎜
mf = 900 ⋅ kg × ⎝ 1 − e
U
Ve
265
⎞
2750
⎠
⎛⎜
−
M0
⎜
t=
⋅⎝1 − e
mrate
⎞
⎠
mf = 82.7 kg
U
Ve
⎞
⎠
Problem 4.165
[Difficulty: 3]
CS at speed U
y
x
Ve
Y
X
Given:
Data on rocket weapon
Find:
Expression for speed of weapon; minimum fraction of mass that must be fuel
Solution:
Basic equation: Momentum flux in x direction
Assumptions: 1) No resistance 2) p e = p atm 3) Uniform flow 4) Use relative velocities 5) Constant mass flow rate
From continuity
dM
dt
= mrate = constant
M = M 0 − mrate⋅ t
so
(
)
(
(Note: Software cannot render a dot!)
)
dU
Hence from momentum −arfx⋅ M = −
⋅ M 0 − mrate⋅ t = u e⋅ ρe⋅ Ve⋅ Ae = −Ve⋅ mrate
dt
Separating variables
dU =
Ve⋅ mrate
M 0 − mrate⋅ t
⋅ dt
Integrating from U = U0 at t = 0 to U = U at t = t
( (
)
⎛
( )) = −Ve⋅ ln⎜ 1 −
U − U0 = −Ve⋅ ln M 0 − mrate⋅ t − ln M 0
⎛
U = U0 − Ve⋅ ln⎜ 1 −
⎝
Rearranging
mrate⋅ t ⎞
⎝
M0
⎠
mrate⋅ t ⎞
M0
MassFractionConsumed =
⎠
mrate⋅ t
M0
−
=1−e
( U−U0)
Ve
−
=1−e
( 3500− 600)
6000
= 0.383
Hence 38.3% of the mass must be fuel to accomplish the task. In reality, a much higher percentage would be needed due to drag
effects
Problem 4.166
[Difficulty: 3] Part 1/2
Problem 4.166
[Difficulty: 3] Part 2/2
Problem 4.147
Problem 4.167
[Difficulty: 3]
Problem 4.168
[Difficulty: 3] Part 1/2
Problem 4.168
[Difficulty: 3] Part 2/2
Problem 4.148
Problem 4.169
[Difficulty: 3]
Problem 4.170
[Difficulty: 3]
CS at speed V
y
x
Y
Ve
X
Given:
Data on rocket
Find:
Speed after 5 s; Maximum velocity; Plot of speed versus time
Solution:
Basic equation: Momentum flux in y direction
Assumptions: 1) No resistance 2) p e = p atm 3) Uniform flow 4) Use relative velocities 5) Constant mass flow rate
From continuity
dM
dt
Hence from momentum
Separating variables
= mrate = constant
M = M 0 − mrate⋅ t
so
(
)
−M ⋅ g − arfy⋅ M = u e⋅ ρe⋅ Ve⋅ Ae = −Ve⋅ mrate
dV =
or
(Note: Software cannot render a dot!)
arfy =
dV
dt
=
Ve⋅ mrate
M
−g=
Ve⋅ mrate
M 0 − mrate⋅ t
⎞
⎛ Ve⋅ mrate
⎜ M − m ⋅ t − g ⋅ dt
rate
⎝ 0
⎠
Integrating from V = at t = 0 to V = V at t = t
( (
)
⎛
( )) − g⋅ t = −Ve⋅ ln⎜ 1 −
V = −Ve⋅ ln M 0 − mrate⋅ t − ln M 0
At t = 5 s
⎝
mrate⋅ t ⎞
M0
⎠
⎛
V = −Ve⋅ ln⎜ 1 −
− g⋅ t
⎝
m
Vmax = −2500⋅ ⋅ ln⎛⎜ 1 − 10⋅
×
× 5 ⋅ s⎞ − 9.81⋅ × 5 ⋅ s
s ⎝
s
350 ⋅ kg
2
⎠
s
m
kg
mrate⋅ t ⎞
1
For the motion after 5 s, assuming the fuel is used up, the equation of motion becomes
M0
⎠
− g⋅ t
m
Vmax = 336
s
a = −M ⋅ g
500
V (m/s)
300
100
− 100 0
20
40
− 300
− 500
Time (s)
60
−g
Problem 4.151
Problem 4.171
[Difficulty: 3]
Problem 4.172
[Difficulty: 4]
y
x

CS (moves
at speed U)

Ry
Ff
Given:
Water jet striking moving vane
Find:
Plot of terminal speed versus turning angle; angle to overcome static friction
Solution:
Basic equations: Momentum flux in x and y directions
Assumptions: 1) Incompressible flow 2) Atmospheric pressure in jet 3) Uniform flow 4) Jet relative velocity is constant
(
)
(
)
−Ff − M ⋅ arfx = u 1 ⋅ −ρ⋅ V1 ⋅ A1 + u 2 ⋅ ρ⋅ V2 ⋅ A2 = −( V − U) ⋅ [ ρ⋅ ( V − U) ⋅ A] + ( V − U) ⋅ cos( θ) ⋅ [ ρ⋅ ( V − U) ⋅ A]
Then
2
arfx =
ρ( V − U) ⋅ A⋅ ( 1 − cos( θ) ) − Ff
(1)
M
(
)
Ry − M ⋅ g = v 1 ⋅ −ρ⋅ V1 ⋅ A1 + v 2 ⋅ ρ⋅ V2 ⋅ A2 = 0 + ( V − U) ⋅ sin( θ) ⋅ [ ρ⋅ ( V − U) ⋅ A]
Also
2
Ry = M ⋅ g + ρ( V − U) ⋅ A⋅ sin( θ)
At terminal speed arfx = 0 and Ff = µkRy. Hence in Eq 1
0=
or
ρ⋅ V − Ut ⋅ A⋅ ( 1 − cos( θ) ) − μk ⋅ ⎡M ⋅ g + ρ⋅ V − Ut ⋅ A⋅ sin( θ)⎤
⎣
⎦
(
)2
(
)2
M
V − Ut =
(
μk ⋅ M ⋅ g
ρ⋅ A⋅ 1 − cos( θ) − μk ⋅ sin( θ)
)
=
(
)2 (
ρ⋅ V − Ut ⋅ A⋅ 1 − cos( θ) − μk ⋅ sin( θ)
Ut = V −
M
(
μk ⋅ M ⋅ g
ρ⋅ A⋅ 1 − cos( θ) − μk ⋅ sin( θ)
The terminal speed as a function of angle is plotted below; it can be generated in Excel
)
)
− μk ⋅ g
Terminal Speed (m/s)
20
15
10
5
0
10
20
30
40
50
60
70
80
Angle (deg)
For the static case
Ff = μs⋅ Ry
and
arfx = 0
(the cart is about to move, but hasn't)
Substituting in Eq 1, with U = 0
2
0=
or
(
ρ⋅ V ⋅ A⋅ ⎡1 − cos( θ) − μs⋅ ρ⋅ V ⋅ A⋅ sin( θ) + M ⋅ g
⎣
cos( θ) + μs⋅ sin( θ) = 1 −
2
)
M
μs⋅ M ⋅ g
2
ρ⋅ V ⋅ A
We need to solve this for θ! This can be done by hand or by using Excel's Goal Seek or Solver
Note that we need θ = 19o, but once started we can throttle back to about θ = 12.5 o and still keep moving!
θ = 19.0⋅ deg
90
Problem 4.173
[Difficulty: 3]
Problem 4.174
[Difficulty: 3]
Problem 4.175
[Difficulty: 3]
Problem 4.176
[Difficulty: 4]
CS at speed V
y
x
Y
Ve
X
Given:
Data on rocket
Find:
Maximum speed and height; Plot of speed and distance versus time
Solution:
Basic equation: Momentum flux in y direction
Assumptions: 1) No resistance 2) p e = p atm 3) Uniform flow 4) Use relative velocities 5) Constant mass flow rate
From continuity
dM
dt
= mrate = constant
M = M 0 − mrate⋅ t
so
(
)
Hence from momentum
−M ⋅ g − arfy⋅ M = u e⋅ ρe⋅ Ve⋅ Ae = −Ve⋅ mrate
Hence
arfy =
Separating variables
dV =
dV
dt
=
Ve⋅ mrate
M
(Note: Software cannot render a dot!)
−g=
Ve⋅ mrate
M 0 − mrate⋅ t
−g
⎞
⎛ Ve⋅ mrate
⎜ M − m ⋅ t − g ⋅ dt
rate
⎝ 0
⎠
Integrating from V = at t = 0 to V = V at t = t
( (
)
⎛
( )) − g⋅ t = −Ve⋅ ln⎜ 1 −
V = −Ve⋅ ln M 0 − mrate⋅ t − ln M 0
mrate⋅ t ⎞
⎛
V = −Ve⋅ ln⎜ 1 −
− g⋅ t
M0
⎝
⎠
⎝
for
t ≤ tb
To evaluate at tb = 1.7 s, we need V e and mrate
mf
mrate =
tb
mrate =
Also note that the thrust Ft is due to
momentum flux from the rocket
Ft = mrate⋅ Ve
Ft
Ve =
mrate
Hence
⎛
Vmax = −Ve⋅ ln⎜ 1 −
⎝
mrate⋅ tb ⎞
M0
⎠
mrate⋅ t ⎞
M0
⎠
− g⋅ t
(burn time)
12.5⋅ gm
1.7⋅ s
(1)
− 3 kg
mrate = 7.35 × 10
Ve =
5.75⋅ N
7.35 × 10
− 3 kg
⋅
s
×
kg⋅ m
2
s ⋅N
m
Ve = 782
s
s
− g ⋅ tb
m
m
1
− 3 kg
Vmax = −782 ⋅ ⋅ ln⎜⎛ 1 − 7.35 × 10 ⋅
×
× 1.7⋅ s⎞ − 9.81⋅ × 1.7⋅ s
2
s ⎝
s
0.0696⋅ kg
⎠
s
m
Vmax = 138
s
To obtain Y(t) we set V = dY/dt in Eq 1, and integrate to find
Y=
Ve⋅ M 0
mrate
mrate⋅ t ⎞
⎡⎛
⋅ ⎢⎜ 1 −
⎣⎝
M0
⎠⎝ ⎝
m
Yb = 782 ⋅ × 0.0696⋅ kg ×
s
At t = tb
+−
1
2
× 9.81⋅
m
2
mrate⋅ t ⎞
⎛ ⎛
⋅ ⎜ ln⎜ 1 −
M0
s
7.35 × 10
× ( 1.7⋅ s)
−3
⋅ kg
⎠
⎞
⎤
1
⎠
⎦
2
− 1 + 1⎥ −
⋅ ⎡⎢⎛⎜ 1 −
⎣⎝
⋅ g⋅ t
2
t ≤ tb
tb = 1.7⋅ s
(2)
0.00735 ⋅ 1.7 ⎞ ⎛
.00735⋅ 1.7 ⎞
⎛
− 1⎞ + 1⎥⎤ ...
⎜ ln⎜ 1 − .0696
⎠⎝ ⎝
⎠ ⎠ ⎦
0.0696
2
s
Yb = 113 m
After burnout the rocket is in free assent. Ignoring drag
(
V( t) = Vmax − g ⋅ t − tb
)
(
(3)
)
(
)
1
2
Y( t) = Yb + Vmax⋅ t − tb − ⋅ g ⋅ t − tb
2
t > tb
The speed and position as functions of time are plotted below. These are obtained from Eqs 1 through 4, and can be plotted in
Excel
150
V (m/s)
100
50
0
5
10
15
20
− 50
Time (s)
Y (m)
1500
1000
500
0
5
10
15
20
Time (s)
Using Solver, or by differentiating y(t) and setting to zero, or by setting V(t) = 0, we find for the maximum y
t = 15.8 s
y max = 1085 m
(4)
Problem 4.177
[Difficulty: 3]
Given:
Data on "jet pack" rocket
Find:
Initial exhaust mass flow rate; mass flow rate at end; maximum time of flight
Solution:
Basic equation: Momentum flux in y direction
Assumptions: 1) Jet pack just hovers 2) Steady flow 3) Uniform flow 4) Use relative velocities
Given data
m
Ve = 3000⋅
s
M 0 = 200⋅ kg
mrateinit =
M 0 ⋅ g moon
Ve
−M ⋅ g moon = −v 1 ⋅ mrate = −Ve⋅ mrate
M f ⋅ g moon
mratefinal =
dM
dt
=
Ve
M ⋅ g moon
so
Ve
Integrating
Solving for t
t=−
g moon
M f = M 0 − M fuel
dM
= mrate
dt
dM
M
⎛ M0 ⎞ gmoon
ln⎜
=
⋅t
Ve
⎝M⎠
Ve
m
⋅ ln⎛⎜
⎞
M0
⎝ ⎠
M
mrate =
or
2
M ⋅ g moon
Ve
M f = 100 kg
kg
mratefinal = 0.0556
s
Flight ends as fuel is used up. To find this, from continuity
Hence
g moon = 1.67
kg
mrateinit = 0.111
s
Finally, when all the fuel is just used up, the mass is
Then
g moon = 0.17⋅ g
s
At all instants, the momentum becomes
Hence, initially
M fuel = 100⋅ kg
=
g moon
Ve
−
or
M = M0⋅ e
so when
M = Mf
mrate =
but
M ⋅ g moon
Ve
⋅ dt
gmoon
⋅t
Ve
tfinal = −
Ve
g moon
⎛ Mf ⎞
⋅ ln⎜
⎝
M0
⎠
tfinal = 20.8 min
Problem 4.178
[Difficulty: 5] Part 1/3
Problem 4.178
[Difficulty: 5] Part 2/3
Problem 4.178
[Difficulty: 5] Part 3/3
Problem 4.179
[Difficulty: 5] Part 1/2
Problem 4.179
[Difficulty: 5] Part 2/2
Problem 4.180
[Difficulty: 5]


Given:
Water jet striking moving disk
Find:
Motion of disk; steady state height
CS moving
at speed U
Solution:
Basic equations: Bernoulli; Momentum flux in z direction (treated as upwards) for linear accelerating CV
p
ρ
2
+
V
2
+ g ⋅ z = constant
Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure 4) Uniform flow 5) velocities wrt CV
V0
The Bernoulli equation becomes
2
+ g⋅ 0 =
2
V1
2
+ g⋅ h
2
2
(
2
arfz =
d h
U=
and
2
dt
(1)
m
V1 = 12.9
s
(
) (
)
2
dh
−M ⋅ g − M ⋅
we get
dt
d h
(
2
= −ρ⋅ ⎛⎜ V1 −
⎝
dt
2
d h
V1 ⋅ A1 = V0 ⋅ A0
Using Eq 1, and from continuity
)
V0 − 2 ⋅ g ⋅ h
)
−M ⋅ g − M ⋅ arfz = w1 ⋅ −ρ⋅ V1 ⋅ A1 + w2 ⋅ ρ⋅ V2 ⋅ A2 = V1 − U ⋅ ⎡−ρ⋅ V1 − U ⋅ A1⎤ + 0
⎣
⎦
The momentum equation becomes
With
2
V1 =
⎛ 15⋅ m ⎞ + 2 × 9.81⋅ m ⋅ ( 0 − 3) ⋅ m
⎜ s
2
⎝
⎠
s
V1 =
(All in jet)
2
dt
=
dh ⎞
dt ⎠
2
⋅ A1
2
ρ⋅ A0 ⋅ V0
⎛ V 2 − 2 ⋅ g⋅ h − dh ⎞ ⋅
−g
⎜ 0
dt ⎠
⎝
2
M ⋅ V0 − 2 ⋅ g ⋅ h
(2)
This must be solved numerically! One approach is to use Euler's method (see the Excel solution)
dh
At equilibrium h = h 0
dt
2
d h
=0
2
=0
so
dt
⎛ V 2 − 2 ⋅ g⋅ h ⎞ ⋅ ρ⋅ A ⋅ V − M⋅ g = 0
0⎠
0 0
⎝ 0
Hence
⎡
⎢
⎛
h0 =
× ⎜ 15⋅
×
× 1−
2 ⎝
s⎠
9.81⋅ m ⎢
⎣
1
m⎞
2
2
s
and
h0 =
V0
2
2⋅ g
⎡
⋅ ⎢1 −
⎢
⎣
⎛⎜ M⋅ g ⎞
⎜ ρ⋅ V02⋅ A0
⎝
⎠
2
3
⎡
s ⎞
1 ⎥⎤
⎢30⋅ kg × 9.81⋅ m × m
× ⎛⎜
×
⎢
2
1000⋅ kg ⎝ 15⋅ m ⎠
2⎥
s
.005⋅ m ⎦
⎣
2⎤
2⎤
⎥
⎥
⎦
⎥
⎥ h 0 = 10.7 m
⎦
In Excel:
Ξt = 0.05 s
2
A 0 = 0.005 m
2
9.81 m/s
15 m/s
30 kg
Ξ = 1000 kg/m3
2
2
h (m)
2.000
2.000
2.061
2.167
2.310
2.481
2.673
2.883
3.107
3.340
3.582
3.829
4.080
4.333
4.587
4.840
dh/dt (m/s)
0.000
1.213
2.137
2.852
3.412
3.853
4.199
4.468
4.675
4.830
4.942
5.016
5.059
5.074
5.066
5.038
d h/dt (m/s )
24.263
18.468
14.311
11.206
8.811
6.917
5.391
4.140
3.100
2.227
1.486
0.854
0.309
-0.161
-0.570
-0.926
0.800
0.850
0.900
0.950
1.000
1.050
1.100
1.150
1.200
1.250
1.300
1.350
1.400
1.450
1.500
1.550
1.600
1.650
1.700
1.750
1.800
1.850
1.900
1.950
2.000
2.050
2.100
2.150
2.200
2.250
2.300
2.350
2.400
2.450
5.092
5.341
5.588
5.830
6.069
6.302
6.530
6.753
6.969
7.179
7.383
7.579
7.769
7.952
8.127
8.296
8.457
8.611
8.757
8.896
9.029
9.154
9.272
9.384
9.488
9.587
9.679
9.765
9.845
9.919
9.989
10.052
10.111
10.166
4.991
4.930
4.854
4.767
4.669
4.563
4.449
4.328
4.201
4.069
3.934
3.795
3.654
3.510
3.366
3.221
3.076
2.931
2.787
2.645
2.504
2.365
2.230
2.097
1.967
1.842
1.720
1.602
1.489
1.381
1.278
1.179
1.085
0.997
-1.236
-1.507
-1.744
-1.951
-2.130
-2.286
-2.420
-2.535
-2.631
-2.711
-2.776
-2.826
-2.864
-2.889
-2.902
-2.904
-2.896
-2.878
-2.850
-2.814
-2.769
-2.716
-2.655
-2.588
-2.514
-2.435
-2.350
-2.261
-2.167
-2.071
-1.972
-1.871
-1.769
-1.666
12
6
10
5
8
4
Position
6
3
Speed
4
2
2
1
0
0
0
1
2
3
4
Time t (s)
t (s)
2.950
3.000
3.050
3.100
3.150
3.200
3.250
3.300
3.350
3.400
3.450
3.500
3.550
3.600
3.650
3.700
3.750
3.800
3.850
3.900
3.950
4.000
4.050
4.100
4.150
4.200
4.250
4.300
4.350
4.400
4.450
4.500
4.550
h (m) dh/dt (m/s)
10.506
0.380
10.525
0.341
10.542
0.307
10.558
0.275
10.571
0.246
10.584
0.220
10.595
0.197
10.604
0.176
10.613
0.157
10.621
0.140
10.628
0.124
10.634
0.111
10.640
0.098
10.645
0.087
10.649
0.078
10.653
0.069
10.656
0.061
10.659
0.054
10.662
0.048
10.665
0.043
10.667
0.038
10.669
0.033
10.670
0.030
10.672
0.026
10.673
0.023
10.674
0.021
10.675
0.018
10.676
0.016
10.677
0.014
10.678
0.013
10.678
0.011
10.679
0.010
10.679
0.009
2
2
2
d h/dt (m/s )
-0.766
-0.698
-0.634
-0.574
-0.519
-0.469
-0.422
-0.380
-0.341
-0.306
-0.274
-0.245
-0.219
-0.195
-0.174
-0.155
-0.138
-0.123
-0.109
-0.097
-0.086
-0.077
-0.068
-0.060
-0.053
-0.047
-0.042
-0.037
-0.033
-0.029
-0.026
-0.023
-0.020
5
Speed (m/s)
2
t (s)
0.000
0.050
0.100
0.150
0.200
0.250
0.300
0.350
0.400
0.450
0.500
0.550
0.600
0.650
0.700
0.750
Position (m)
g=
V=
M=
Problem 4.181
[Difficulty: 5] Part 1/2
Problem 4.181
[Difficulty: 5] Part 2/2
Problem 4.182
[Difficulty: 5] Part 1/3
Problem 4.133
Problem 4.182
[Difficulty: 5] Part 2/3
Problem 4.182
[Difficulty: 5] Part 3/3
Problem 4.183
[Difficulty: 5] Part 1/2
4.184
4.184
4.184
Problem 4.183
[Difficulty: 5] Part 2/2
Problem 4.184
4.137
[Difficulty: 5] Part 1/4
Problem 4.184
[Difficulty: 5] Part 2/4
Problem 4.184
[Difficulty: 5] Part 3/4
Problem 4.184
[Difficulty: 5] Part 4/4
Problem 4.185
[Difficulty: 3] Part 1/2
Problem 4.185
[Difficulty: 3] Part 2/2
Problem *4.165
Problem 4.186
[Difficulty: 2]
Example 4.6
4.6
Problem 4.187
[Difficulty: 3]
Problem 4.188
[Difficulty: 3]
Given:
Data on rotating spray system
Find:
Torque required to hold stationary; steady-state speed
Solution:
Basic equation: Rotating CV
Assumptions: 1) No surface force; 2) Body torques cancel; 3) Sprinkler stationary; 4) Steady flow; 5) Uniform flow; 6) L<<r
Q = 15⋅
The given data is
V=
For each branch
L
R = 225⋅ mm
min
1
⋅
2 π
4
ρ = 999⋅
kg
3
m
Q
⋅d
d = 5⋅ mm
V = 6.37
2
m
s
The basic equation reduces to a single scalar equation (FOR EACH BRANCH)
(
)
⌠→ → →
⎮
Tshaft − ⎮ r × α × r ⋅ ρ dV =
⌡
But
(
)
→ → →
2
r × α × r = r ⋅α
⌠ → ⎯⎯
→ ⎯⎯
→→
⎮ r × V ⋅ ρ⋅ V
xyz
xyz dA
⎮
⌡
(r and α perpendicular); the volume integral is
Tshaft −
R
3
3
⋅ α⋅ ρ⋅
π
4
2
⋅ d = R⋅ V⋅ ρ⋅
(
)
⌠→ → →
⎮
⎮ r × α × r ⋅ ρ dV =
⌡
3
⌠
R
π 2
⎮ 2
d
r
⋅
α
⋅
ρ
V
=
⋅ α⋅ ⋅ d
⎮
3
4
⌡
⌠ → ⎯⎯
→ ⎯⎯
→→
Q
⎮ r × V ⋅ ρ⋅ V
xyz
xyz dA = R⋅ V⋅ ρ⋅ 2
⎮
⌡
For the surface integral (FOR EACH BRANCH)
Combining
where α is the angular acceleration
Q
(1)
2
Q
When the sprayer is at rest, α = 0, so
Tshaft = R⋅ V⋅ ρ⋅
The total torque is then
Ttotal = 2 ⋅ Tshaft
2
Tshaft = 0.179 N⋅ m
Ttotal = 0.358 N⋅ m
When the device is released is released (Tshaft = 0 in Eq 1), we can solve for α
α =
6 ⋅ ρ⋅ Q⋅ V
2
ρ⋅ π⋅ d ⋅ R
2
3 1
α = 2.402 × 10
2
s
Problem 4.189
[Difficulty: 3]
Given:
Data on rotating spray system
Find:
Differential equation for motion; steady speed
Solution:
Basic equation: Rotating CV
Assumptions: 1) No surface force; 2) Body torques cancel; 3) Steady flow; 5) Uniform flow; 6) L<<r
Q = 15⋅
The given data is
V =
For each branch
1
L
R = 225 ⋅ mm
min
2 π
4
V = 6.37
⋅d
ρ = 999 ⋅
kg
3
m
Q
⋅
d = 5 ⋅ mm
m
2
A =
s
π
4
⋅d
2
2
A = 19.6 mm
The basic equation reduces to a single scalar equation (FOR EACH BRANCH)
(
)
⌠→
→ → → → →
⎮
−⎮ r × 2 ⋅ ω × V × r + α × r ⋅ ρ dV =
⌡
But
(
⌠ → ⎯⎯
→ ⎯⎯
→→
⎮ r × V ⋅ ρ⋅ V
dA
xyz
xyz
⎮
⌡
)
→
→ → → → →
2
r × 2 ⋅ ω × V × r + α × r = 2 ⋅ ω⋅ r⋅ V + α⋅ r
The volume integral is then
(r and α perpendicular)
3
⌠→
⎛ 2
→ → → → →
R ⎞
⎮
−⎮ r × 2 ⋅ ω × V × r + α × r ⋅ ρ dV = −⎜ ω⋅ R ⋅ V + α⋅
⋅ ρ⋅ A
3 ⎠
⌡
⎝
(
)
⌠ → ⎯⎯
→ ⎯⎯
→→
Q
⎮ r × V ⋅ ρ⋅ V
dA = R⋅ V⋅ ρ⋅
xyz
xyz
⎮
2
⌡
For the surface integral (FOR EACH BRANCH)
⎛
Combining
−⎜ ω⋅ R ⋅ V + α⋅
⎝
2
where α is the angular acceleration
R
3⎞
3
⎠
⋅ ρ⋅ A = R⋅ V⋅ ρ⋅
The steady state speed (α = 0 in Eq 1) is then when
Q
2
or
α=
3
⋅ ⎛⎜ −ω⋅ V⋅ A⋅ R −
A⋅ R ⎝
−ωmax⋅ V⋅ A⋅ R −
1
ωmax = −28.3
s
2
Q⋅ V
2
=0
Q⋅ V ⎞
2
or
⎠
(1)
Q
ωmax = −
2 ⋅ A⋅ R
ωmax = −270 rpm
Problem 4.190
[Difficulty: 3]
NOTE ERROR: Retarding torque is 0.05 N.m!
Given:
Data on rotating spray system
Find:
Differential equation for motion; steady speed; troque to stop
Solution:
Basic equation: Rotating CV
Assumptions: 1) No surface force; 2) Body torques cancel; 3) Steady flow; 5) Uniform flow; 6) L<<r
Q = 15⋅
The given data is
V =
For each branch
L
R = 225 ⋅ mm
min
d = 5 ⋅ mm
ρ = 999 ⋅
kg
T = 0.05⋅ N⋅ m
3
m
1
Q
⋅
2 π 2
⋅d
4
V = 6.37
m
A =
s
π
4
⋅d
2
2
A = 19.6⋅ mm
The basic equation reduces to a single scalar equation (FOR EACH BRANCH)
(
)
⌠→
→ → → → →
⎮
− ⎮ r × 2⋅ ω × V × r + α × r ⋅ ρ dV =
2 ⌡
T
But
(
⌠ → ⎯⎯
→ ⎯⎯
→ →
⎮ r × V ⋅ ρ⋅ V dA
xyz
xyz
⎮
⌡
)
→
→ → → → →
2
r × 2 ⋅ ω × V × r + α × r = 2 ⋅ ω⋅ r⋅ V + α⋅ r
(r and α perpendicular)
3
⌠→
⎛ 2
→ → → → →
R ⎞
⎮
−⎮ r × 2⋅ ω × V × r + α × r ⋅ ρ dV = −⎜ ω⋅ R ⋅ V + α ⋅
⋅ ρ⋅ A
3 ⎠
⌡
⎝
(
The volume integral is then
)
⌠ → ⎯⎯
→ ⎯⎯
→ →
⎮ r × V ⋅ ρ⋅ V dA = R⋅ V⋅ ρ⋅ Q
xyz
xyz
⎮
2
⌡
For the surface integral (FOR EACH BRANCH)
Combining
T
2
⎛
− ⎜ ω⋅ R ⋅ V + α⋅
⎝
where T is the retarding torque α is the angular
acceleration
2
R
3⎞
3
⎠
⋅ ρ⋅ A = R⋅ V⋅ ρ⋅
The steady state speed (α = 0 in Eq 1) is then when
For no rotation use α = ω = 0 in Eq 1, and solve for Tmax
Q
2
or
3
α=
2 ⋅ ρ⋅ A⋅ R
3
(
)
2
⋅ T − 2 ⋅ ρ⋅ A⋅ ω⋅ R ⋅ V − ρ⋅ R⋅ Q⋅ V
2
T − 2 ⋅ ρ⋅ A⋅ ωmax⋅ R ⋅ V − ρ⋅ R⋅ Q⋅ V = 0 or
1
ωmax = −24.3
s
ωmax = −232 ⋅ rpm
Tmax = ρ⋅ Q⋅ R⋅ V
Tmax = 0.358 ⋅ N⋅ m
ωmax =
(1)
T − ρ⋅ R⋅ Q⋅ V
2
2 ⋅ ρ⋅ A⋅ R ⋅ V
Problem 4.191
[Difficulty: 4]
Given:
Data on rotating spray system
Find:
Torque required to hold stationary; steady-state speed
Solution:
Basic equation: Rotating CV
The given data is
ρ = 999 ⋅
kg
3
δ = 2.5⋅ mm
ro = 300 ⋅ mm
m
L
Qin = 3 ⋅
s
ri = ( 300 − 250 ) ⋅ mm
For no rotation (ω = 0) the basic equation reduces to a single scalar equation
⌠ → ⎯⎯
→ ⎯⎯
→→
Tshaft = ⎮ r × Vxyz⋅ ρ⋅ Vxyz dA
⎮
⌡
r
r
i
i
o
⌠o
2
2
2
2 ⌠
Tshaft = 2 ⋅ δ⋅ ⎮ r⋅ V⋅ ρ⋅ V dr = 2 ⋅ ρ⋅ V ⋅ δ⋅ ⎮ r dr = ρ⋅ V ⋅ δ⋅ ⎛ ro − ri ⎞
⎝
⎠
⌡r
⌡r
or
V =
where V is the exit velocity with respect to the CV
Qin
(
2 ⋅ δ⋅ ro − ri
V = 2.40
)
2
Hence
⎡ Qin ⎤ ⎛ 2
2
Tshaft = ρ⋅ ⎢
⎥ ⋅ δ⋅ ⎝ ro − ri ⎞⎠
2 ⋅ δ⋅ ( ro − ri)
⎣
⎦
2
Tshaft =
ρ⋅ Qin
4⋅ δ
⋅
m
s
(ro + ri)
(ro − ri)
2
⎛ L 10− 3⋅ m3 ⎞
1
( 0.3 + 0.05)
999 ⋅ kg
Tshaft =
×
×
×
× ⎜ 3⋅ ×
4 ⎝ s
L
0.0025⋅ m ( 0.3 − 0.05)
3
⎠
m
1
For the steady rotation speed the equation becomes
⌠→
→
→ ⎯⎯
−⎮ r × ⎛ 2 ⋅ ω × Vxyz⎞ ⋅ ρ dV =
⎝
⎠
⎮
⌡
Tshaft = 1.26⋅ N⋅ m
⌠ → ⎯⎯
→ ⎯⎯
→→
⎮ r × V ⋅ ρ⋅ V
dA
xyz
xyz
⎮
⌡
⌠→
→
→ ⎯⎯
The volume integral term −⎮ r × ⎛ 2 ⋅ ω × Vxyz⎞ ⋅ ρ dV must be evaluated for the CV. The velocity in the CV varies with r. This
⎝
⎠
⎮
⌡
variation can be found from mass conservation
For an infinitesmal CV of length dr and cross-section A at radial position r, if the flow in is Q, the flow out is Q + dQ, and the loss
through the slot is Vδdr. Hence mass conservation leads to
( Q + dQ) + V⋅ δ⋅ dr − Q = 0
dQ = −V⋅ δ⋅ dr
Q( r) = −V⋅ δ⋅ r + const
At the inlet (r = ri)
Qin
Q = Qi =
2
Hence
Qin
Qin
Q = Qi + V⋅ δ⋅ ri − r =
+
⋅ δ⋅ ri − r
2
2 ⋅ δ⋅ ro − ri
(
)
(
v ( r) =
and along each rotor the water speed is
(
)
Q
A
=
Qin
2⋅ A
)
Q=
Qin
2
⎞
⎛
ri − r
⎝
ro − ri
⎠
⋅⎜1 +
=
Qin
2
⎛ ro − r ⎞
⋅⎜
⎝ ro − ri ⎠
⌠→
→ ⎯⎯
→
Hence the term - ⎮ r × ⎛ 2 ⋅ ω × Vxyz⎞ ⋅ ρ dV becomes
⎝
⎠
⎮
⌡
r
r
⌠ o Q ⎛ r − r⎞
⌠→
⌠o
→
→ ⎯⎯
in
o
⎮
⎮
⎛
⎞
−
r × 2 ⋅ ω × Vxyz ⋅ ρ dV = 4 ⋅ ρ⋅ A⋅ ω⋅ ⎮ r⋅ v ( r) dr = 4 ⋅ ρ⋅ ω⋅ ⎮ r⋅
⋅⎜
dr
⎝
⎠
⎮
2
r
− ri
⌡r
⌡
o
⎝
⎠
⎮
i
⌡r
i
ro
or
(
i
Recall that
⌠ → ⎯⎯
→ ⎯⎯
→→
2
2
2
⎮ r × V ⋅ ρ⋅ V
dA = ρ⋅ V ⋅ δ⋅ ⎛ ro − ri ⎞
xyz
xyz
⎝
⎠
⎮
⌡
Hence equation
⌠→
→
→ ⎯⎯
−⎮ r × ⎛ 2 ⋅ ω × Vxyz⎞ ⋅ ρ dV =
⎝
⎠
⎮
⌡
3
ρ⋅ Qin⋅ ω⋅
Solving for ω
(
3
2
⌠
⌠→
ro + ri ⋅ 2 ⋅ ri − 3 ⋅ ro
→
→ ⎯⎯
⎛ ro − r ⎞
⎮
⎮
⎛
⎞
−
r × 2 ⋅ ω × Vxyz ⋅ ρ dV = 2 ⋅ ρ⋅ Qin⋅ ω⋅ ⎮ r⋅ ⎜
dr = ρ⋅ Qin⋅ ω⋅
⎝
⎠
⎮
ro − ri
3 ⋅ ro − ri
⌡
⎝
⎠
⎮
⌡r
ω =
2
(
ro + ri ⋅ 2 ⋅ ri − 3 ⋅ ro
(
3 ⋅ ro − ri
)
)
3 ⋅ ro − ri ⋅ V ⋅ δ⋅ ⎛ ro − ri
⎝
(
)
2
2
⌠ → ⎯⎯
→ ⎯⎯
→→
⎮ r × V ⋅ ρ⋅ V
xyz
xyz dA
⎮
⌡
= ρ⋅ V ⋅ δ⋅ ⎛ ro − ri
⎝
2
2
becomes
2⎞
⎠
2⎞
⎠
3
2
Qin⋅ ⎡ro + ri ⋅ ( 2 ⋅ ri − 3 ⋅ ro )⎤
⎣
⎦
ω = 120 ⋅ rpm
)
)
⎛ ro − r ⎞
⋅⎜
⎝ ro − ri ⎠
Problem 4.192
[Difficulty: 4]
Given:
Data on rotating spray system
Find:
Torque required to hold stationary; steady-state speed
Solution:
Governing equation: Rotating CV
The given data is
kg
ρ = 999 ⋅
δ = 2.5⋅ mm
3
ro = 300 ⋅ mm
ri = ( 300 − 250 ) ⋅ mm
m
L
Qin = 3 ⋅
s
For no rotation (ω = 0) this equation reduces to a single scalar equation
r
⌠ → ⎯⎯
→ ⎯⎯
→→
Tshaft = ⎮ r × Vxyz⋅ ρ⋅ Vxyz dA
⎮
⌡
⌠o
Tshaft = 2 ⋅ δ⋅ ⎮ r⋅ V⋅ ρ⋅ V dr
⌡r
or
i
where V is the exit velocity with respect to the CV. We need to find V(r). To do this we use mass conservation, and the fact that the
distribution is linear
( )
(
)
(r − ri)
r − ri
V( r) = Vmax⋅
ro − ri
so
V( r) =
Qin
δ
⋅
(ro − ri)
ro
Hence
⌠
Tshaft = 2 ⋅ ρ⋅ δ⋅ ⎮
⌡r
(
)
1
2 ⋅ ⋅ Vmax⋅ ro − ri ⋅ δ = Qin
2
and
2
ro
2 ⌠
(
)
2
ρ⋅ Qin ⎮
⎡ r − ri ⎤
⎥ dr
⋅ ⎮ r⋅ ⎢
r⋅ V dr = 2 ⋅
δ
2⎥
⎮
⎢
i
⎣ ro − ri ⎦
⎮
⌡r
2
(
)
(
)
6 ⋅ δ⋅ ( ro − ri)
2
Tshaft =
ρ⋅ Qin ⋅ ri + 3 ⋅ ro
i
2
Tshaft =
1
6
⎛ L 10− 3⋅ m3 ⎞
1
( 0.05 + 3 ⋅ 0.3)
999 ⋅ kg
×
×
×
×
L
0.0025⋅ m
( 0.3 − 0.05)
3
⎝ s
⎠
m
× ⎜ 3⋅
For the steady rotation speed the equation becomes
⌠→
→ ⎯⎯
→
−⎮ r × ⎛ 2 ⋅ ω × Vxyz⎞ ⋅ ρ dV =
⎝
⎠
⎮
⌡
⌠ → ⎯⎯
→ ⎯⎯
→→
⎮ r × V ⋅ ρ⋅ V
xyz
xyz dA
⎮
⌡
Tshaft = 2.28⋅ N⋅ m
⌠→
→ ⎯⎯
→
The volume integral term −⎮ r × ⎛ 2 ⋅ ω × Vxyz⎞ ⋅ ρ dV must be evaluated for the CV. The velocity in the CV varies with r. This
⎝
⎠
⎮
⌡
variation can be found from mass conservation
For an infinitesmal CV of length dr and cross-section A at radial position r, if the flow in is Q, the flow out is Q + dQ, and the loss
through the slot is Vδdr Hence mass conservation leads to
r
( Q + dQ) + V⋅ δ⋅ dr − Q = 0
(
⌠ Q
in r −
Q( r) = Qi −δ⋅ ⎮
⋅
⎮ δ
ro −
⎮
⌡r
dQ = −V⋅ δ⋅ dr
(
i
r
)
(
⌠
r−
dr = Qi − ⎮ Qin⋅
⎮
2
ri
ro −
⎮
⌡r
ri
)
(
ri
)
ri
)
2
dr
i
Qin
At the inlet (r = ri)
Q = Qi =
2
Hence
⎡
(r − ri)2 ⎥⎤
⎢
Q( r) =
⋅ 1−
2 ⎢
2⎥
⎣ ( ro − ri) ⎦
Qin
and along each rotor the water speed is
⎡
(r − ri)2 ⎥⎤
⎢
v ( r) =
=
⋅ 1−
A
2⋅ A ⎢
2⎥
⎣ ( ro − ri) ⎦
⌠→
→ ⎯⎯
→
Hence the term - ⎮ r × ⎛ 2 ⋅ ω × Vxyz⎞ ⋅ ρ dV becomes
⎝
⎠
⎮
⌡
⌠
⎞
⎛ ⌠ ro
⎮
⎜
4 ⋅ ρ⋅ A⋅ ω⋅ ⎮ r⋅ v ( r) dr = 4 ⋅ ρ⋅ ω⋅ ⎮
⎜ ⌡r
⎮
⎝ i
⎠
⎮
Qin
Q
ro
⌡r
ro
or
⎡
(r − ri)2 ⎤⎥
⎢
dr
⋅ r⋅ 1 −
2
⎢
2⎥
r
−
r
⎣ ( o i) ⎦
Qin
i
⌠
2
⎡
⎮
ro − r ⎤
⎢
⎥ dr = ρ⋅ Q ⋅ ω⋅ ⎛⎜ 1 ⋅ r 2 + 1 ⋅ r ⋅ r − 1 ⋅ r 2⎞
2 ⋅ ρ⋅ Qin⋅ ω⋅ ⎮ r⋅ 1 ⋅ −
in
o
3 i o 2 i ⎠
⎢ r − r 2⎥
⎮
⎝6
o
i ⎦
⎣
⎮
⌡r
(
)
(
)
i
2
(
)
Recall that
⌠ → ⎯⎯
→ ⎯⎯
→ → ρ⋅ Qin ⋅ ri + 3 ⋅ ro
⎮ r × V ⋅ ρ⋅ V
xyz
xyz dA =
⎮
6 ⋅ ro − ri ⋅ δ
⌡
Hence equation
⌠→
→
→ ⎯⎯
−⎮ r × ⎛ 2 ⋅ ω × Vxyz⎞ ⋅ ρ dV =
⎝
⎠
⎮
⌡
becomes
2
1 2 1
1 2⎞ ρ⋅ Qin ⋅ ( ri + 3 ⋅ ro )
⎛
ρ⋅ Qin⋅ ω⋅ ⎜ ⋅ ro + ⋅ ri⋅ ro − ⋅ ri =
6 ⋅ ( ro − ri) ⋅ δ
3
2
⎝6
⎠
Solving for ω
ω =
(
(
ρ⋅ Qin⋅ ri + 3 ⋅ ro
)
⌠ → ⎯⎯
→ ⎯⎯
→→
⎮ r × V ⋅ ρ⋅ V
dA
xyz
xyz
⎮
⌡
)
⎛ r 2 + 2 ⋅ r ⋅ r − 3 ⋅ r 2⎞ ⋅ ( r − r ) ⋅ ρ⋅ δ
i o
i ⎠ o
i
⎝o
ω = 387 ⋅ rpm
Problem 4.193
[Difficulty: 3]
Problem 4.194
[Difficulty: 3]
Problem *4.175
Problem 4.195
[Difficulty: 3]
Problem *4.176
Problem 4.196
[Difficulty: 3]
Problem 4.197
[Difficulty: 4]
Problem *4.178
Problem 4.198
[Difficulty: 4]
Problem *4.179
Problem 4.199
[Difficulty: 4] Part 1/2
Problem *4.179 cont'd
Problem 4.199
Difficulty: [4] Part 2/2
Problem *4.180
Problem 4.200
[Difficulty: 4] Part 1/3
Problem *4.180 cont'd
Problem 4.200
[Difficulty: 4] Part 2/3
Problem *4.180 cont'd
Problem 4.200
[Difficulty: 4] Part 3/3
Problem *4.181
Problem 4.201
[Difficulty: 5] Part 1/2
Problem *4.181 cont'd
Problem 4.201
[Difficulty: 5] Part 2/2
Problem 4.202
[Difficulty: 5] Part 1/2
Problem 4.202
[Difficulty: 5] Part 2/2
Problem 4.183
Problem 4.203
[Difficulty: 2]
Problem 4.204
Given:
Compressed air bottle
Find:
Rate of temperature change
[Difficulty: 3]
Solution:
Basic equations: Continuity; First Law of Thermodynamics for a CV
Assumptions: 1) Adiabatic 2) No work 3) Neglect KE 4) Uniform properties at exit 5) Ideal gas
Given data
p = 500⋅ kPa
Also
Rair =
From continuity
∂
∂t
∂
∂t
T = 20°C
286.9⋅ N ⋅ m
N⋅m
kg⋅ K
where mexit is the mass flow rate at the exit (Note: Software does not allow a dot!)
M CV = −mexit
p
⎛∂ ⎞
⎛∂ ⎞
p
∂⌠
⎮ u dM + ⎛⎜ u + ⎞ ⋅ mexit = u ⋅ ⎜ M + M ⋅ ⎜ u + ⎛⎜ u + ⎞ ⋅ mexit
ρ⎠
ρ⎠
∂t ⌡
⎝
⎝ ∂t ⎠
⎝ ∂t ⎠ ⎝
From the 1st law
0=
Hence
dT
p
u ⋅ −mexit + M ⋅ cv ⋅
+ u ⋅ mexit + ⋅ mexit = 0
dt
ρ
dT
M = ρ⋅ V
dT
But
For air
kg
mexit = 0.01⋅
s
V = 100⋅ L
cv = 717.4⋅
kg⋅ K
M CV + mexit = 0
T = 293K
(
ρ =
)
dt
(where V is volume) so
dt
p
3 N
ρ = 500 × 10 ⋅
Rair⋅ T
2
m
×
kg⋅ K
286.9 ⋅ N⋅ m
×
=−
=−
mexit ⋅ p
M ⋅ cv ⋅ ρ
mexit ⋅ p
2
V⋅ cv ⋅ ρ
1
( 20 + 273 ) ⋅ K
ρ = 5.95
3
m
2
Hence
kg
⎛ m3 ⎞
C
K
× 500 × 10 ⋅
×
×
×
= −1.97⋅ = −1.97⋅
= −0.01⋅
×⎜
dt
s
2
100 ⋅ L
−3 3
717.4 ⋅ N⋅ m ⎝ 5.95⋅ kg ⎠
s
s
m
10 ⋅ m
dT
kg
3 N
1
L
kg⋅ K
Problem 4.205
Given:
Data on centrifugal water pump
Find:
Pump efficiency
[Difficulty: 3]
Solution:
Basic equations:
(4.56)
Ws
∆p = SG Hg⋅ ρ⋅ g ⋅ ∆h
η=
D1 = 0.1⋅ m
D2 = 0.1⋅ m
Q = 0.02⋅
SG Hg = 13.6
h 1 = −0.2⋅ m
Pin
3
Available data:
ρ = 1000
kg
3
m
s
Pin = 6.75⋅ kW
p 2 = 240 ⋅ kPa
m
Assumptions: 1) Adiabatic 2) Only shaft work 3) Steady 4) Neglect Δu 5) Δz = 0 6) Incompressible 7) Uniform flow
Then
2
⎛⎜
V1 ⎞
−Ws = ⎜ p 1 ⋅ v 1 +
⋅ ( −mrate) +
2 ⎠
⎝
Since
mrate = ρ⋅ Q
V1 = V2
and
(
2
⎛⎜
V2 ⎞
⎜ p2 ⋅ v2 + 2 ⋅ ( mrate)
⎝
⎠
)
(
−Ws = ρ⋅ Q⋅ p 2 ⋅ v 2 − p 1 ⋅ v 1 = Q⋅ p 2 − p 1
p 1 = ρHg⋅ g ⋅ h
(
or
Ws = Q⋅ p 1 − p 2
η =
Ws
Pin
)
(from continuity)
)
p 1 = SGHg⋅ ρ⋅ g ⋅ h 1
p 1 = −26.7⋅ kPa
Ws = −5.33⋅ kW
The negative sign indicates work in
η = 79.0⋅ %
Problem 4.187
Problem 4.206
[Difficulty: 2]
Problem 4.186
Problem 4.207
[Difficulty: 2]
Problem 4.188
Problem 4.208
[Difficulty: 2]
Problem 4.209
[Difficulty: 3]
e
zmax
CV (b)
d
V2
CV (a)
z
x
c
Given:
Data on fire boat hose system
Find:
Volume flow rate of nozzle; Maximum water height; Force on boat
Solution:
Basic equation: First Law of Thermodynamics for a CV
Assumptions: 1) Neglect losses 2) No work 3) Neglect KE at 1 4) Uniform properties at exit 5) Incompressible 6) p atm at 1 and 2
⎛⎜ V 2
⎞
2
−Ws = ⎜
+ g ⋅ z2 ⋅ mexit
⎝ 2
⎠
Hence for CV (a)
mexit = ρ⋅ V2 ⋅ A2
where mexit is mass flow rate (Note:
Software cannot render a dot!)
⎛ 1 ⋅ V 2 + g ⋅ z ⎞ ⋅ ρ⋅ V ⋅ A = −W which is a cubic for V 2!
⎜2 2
2
2 2
s
⎝
⎠
Hence, for V 2 (to get the flow rate) we need to solve
To solve this we could ignore the gravity term, solve for velocity, and then check that the gravity term is in fact
minor. Alternatively we could manually iterate, or use a calculator or Excel, to solve. The answer is
Hence the flow rate is
Q = V2 ⋅ A2 = V2 ⋅
π⋅ D2
2
Q = 114 ⋅
4
ft
×
s
π
4
×
⎛ 1 ⋅ ft⎞
⎜ 12
⎝
⎠
2
Q = 0.622 ⋅
ft
V2 = 114 ⋅
s
ft
3
s
Q = 279 ⋅ gpm
−Ws = g ⋅ zmax⋅ mexit
To find zmax, use the first law again to (to CV (b)) to get
550⋅ ft⋅ lbf
zmax = −
Ws
g ⋅ mexit
=−
Ws
zmax = 15⋅ hp ×
g ⋅ ρ⋅ Q
s
1 ⋅ hp
2
×
s
32.2⋅ ft
×
ft
3
1.94⋅ slug
×
s
0.622 ⋅ ft
3
×
slug⋅ ft
2
s ⋅ lbf
zmax = 212 ⋅ ft
For the force in the x direction when jet is horizontal we need x momentum
Then
(
)
(
)
Rx = u 1 ⋅ −ρ⋅ V1 ⋅ A1 + u 2 ⋅ ρ⋅ V2 ⋅ A2 = 0 + V2 ⋅ ρ⋅ Q
Rx = 1.94⋅
slug
ft
3
× 0.622 ⋅
ft
3
s
× 114 ⋅
ft
s
Rx = ρ⋅ Q⋅ V2
2
×
lbf ⋅ s
slug⋅ ft
Rx = 138 ⋅ lbf
Problem 4.189
Problem 4.210
[Difficulty: 3]
Problem 4.211
Given:
Data on helicopter-type craft
Find:
Air speed; Minimum power needed
[Difficulty: 4]
Solution:
Basic equation: Contunity, z momentum; First Law of Thermodynamics for a CV; Bernoulli; Ideal gas
=0
p
ρ
2
+
V
2
+ g ⋅ z = const
p = ρ⋅ Rair⋅ T
∆h = cp ⋅ ∆T
Assumptions: 1) Atmospheric at exit 2) Standard air 3) Uniform properties at exit 4) Incompressible
Given data
M = 1000⋅ kg
p = 101 ⋅ kPa
Then
π
2
A1 = ⋅ Do
4
A1 = 15.9 m
(
T = 15 °C
π
2
2
A2 = ⋅ ⎛ Do − Di ⎞
⎠
4 ⎝
2
) (
Do = 4.5⋅ m
)
2
A2 = 1.72 m
0 = −ρ⋅ V1 ⋅ A1 + ρ⋅ V2 ⋅ A2
From momentum
−p 1g⋅ A1 − M ⋅ g = w1 ⋅ −ρ⋅ V1 ⋅ A1 + w2 ⋅ ρ⋅ V2 ⋅ A2
Then
−p 1g⋅ A1 − M ⋅ g = V1 ⋅ ρ⋅ V1 ⋅ A1 − V2 ⋅ ρ⋅ V2 ⋅ A2 = −ρ⋅ V2 ⋅ A2 ⋅ V2 − V1
or
)
(
)
w1 = −V1
(
For this flow Bernoulli also applies between the atmosphere and location 1
Using continuity
ρ =
Rair =
p
Rair⋅ T
286.9 ⋅ N⋅ m
kg⋅ K
ρ = 1.222
kg
3
m
A2
V1 =
⋅V
A1 1
From continuity
(
Di = 4.25⋅ m
w2 = −V1
ρ⋅ V1 ⋅ A1 = ρ⋅ V2 ⋅ A2
)
1
2
p atm = p 1 + ⋅ ρ⋅ V1
2
A2
1
1
1
2
2
p 1g⋅ A1 = − ⋅ ρ⋅ V1 ⋅ A1 = − ⋅ ρ⋅ V2 ⋅ A2 ⋅ V1 = − ⋅ ρ⋅ V2 ⋅ A2 ⋅
2
2
2
A1
and
1
2
p 1g = − ⋅ ρ⋅ V1
2
Substituting into the momentum equation and using continuity
A2
V1 ⎞
A2 ⎞
⎛
⎛
2
2
2
or
⋅ ρ⋅ V2 ⋅ A2 ⋅
− M ⋅ g = −ρ⋅ V2 ⋅ A2 ⋅ ⎜ 1 −
= −ρ⋅ V2 ⋅ A2 ⋅ ⎜ 1 −
A1
V2
A1
2
⎝
⎠
⎝
⎠
1
Hence
V2 =
M⋅ g
⎛
1 A2 ⎞
ρ⋅ A2 ⋅ ⎜ 1 − ⋅
2 A1
⎝
⎠
Substituting values
⎛
1 A2 ⎞
2
M ⋅ g = ρ⋅ V2 ⋅ A2 ⋅ ⎜ 1 − ⋅
2 A1
⎝
m
V2 = 70.3
s
For power we use the First Law
We have additional assumptions 5) pv = const 6) Neglect Δz
Then
⎛⎜ V 2 − V 2 ⎞
2
1
dQ ⎞
−Ws = mrate⋅ ⎜
+ mrate⋅ ⎛⎜ u 2 − u 1 −
2
dm ⎠
⎝
⎠
⎝
The last term is non-mechanical energy; the minimum possible work is when this is zero. Hence
2
⎛⎜ V 2 − V 2 ⎞
V2 ⎡⎢
2
1
−Ws = −Wmin = mrate⋅ ⎜
= mrate⋅
⋅ 1−
2
2 ⎢
⎝
⎠
⎣
Wmin =
ρ⋅ A2 ⋅ V2
2
3
⎡
⎢
⋅ 1−
⎢
⎣
⎛ A2 ⎞
⎜A
⎝ 1⎠
2⎤
2 ⎡
⎛ V1 ⎞ ⎥ ρ⋅ A2 ⋅ V2 ⎢
⋅ 1−
⎜V ⎥ =
⎢
2
2
⎝ ⎠⎦
⎣
2⎤
⎥
⎥
⎦
Using given data
Wmin = 360 ⋅ kW
2⎤
⎛ A2 ⎞ ⎥
⎜A ⎥
⎝ 1⎠ ⎦
⎠
Problem 4.192
Problem 4.212
[Difficulty: 4] Part 1/2
Problem 4.192 cont'd
Problem 4.212
[Difficulty: 4] Part 2/2
Problem 5.1
Given:
Find:
Solution:
[Difficulty: 1]
The list of velocity fields provided above
Which of these fields possibly represent two-dimensional, incompressible flow
We will check these flow fields against the continuity equation
Governing
Equations:

u    v    w    0 (Continuity equation)
y
z
t
x
Assumptions:
(1) Incompressible flow (ρ is constant)
(2) Two dimensional flow (velocity is not a function of z)
Based on the two assumptions listed above, the continuity equation reduces to:

x
u 

y
v 0
This is the criterion against which we will check all of the flow fields.
a)
b)
c)
d)
2
2
2
2
3
u ( xy t)  2 x  y  x  y
v ( xy t)  x  x y  4 y
Hence

x
2
u 

y
2
2
Hence

x
2
y
2
v ( xy t)  x t  y  t
Hence

x

y

x
v 0
u ( xy t)  x  t  2 y
u 
x

x
v 0
v ( xy t)  3 ( x  y )  y  t


x
Hence

x
y
u 
u ( xy t)  2 y  2 x y
u ( xy t)  2 t x
NOT INCOMPRESSIBLE
v ( xy t)  t ( 3 x  3 y )  3 t y

y
v 0

y
v ( xy t)  x ( 2 y  4)

y
v ( xy t)  2 x  2 y
NOT INCOMPRESSIBLE
u ( xy t)  ( 2 x  4 y )  x t
u ( xy t)  t ( 2 x  4 y )  2 t x
u ( xy t)  4 x  2 x y
INCOMPRESSIBLE
v ( xy t)  2 x y  y  x


v 0
u ( xy t)  2 x y  x  y
u 

NOT INCOMPRESSIBLE

y
v ( xy t)  t
Problem 5.2
[Difficulty: 2]
Given:
Velocity fields
Find:
Which are 3D incompressible
Solution:
We will check these flow fields against the continuity equation
Governing
Equation:

u    v    w    0 (Continuity equation)
y
z
t
x
Assumption:
Incompressible flow (ρ is constant)

Based on the assumption, the continuity equation reduces to:
x
u 

y
v 

z
w0
This is the criterion against which we will check all of the flow fields.
a)
2
2
w( x y z t)  3  x  z  x  y



u ( x y z t)  2  z
y

Hence
x
2
v ( x y z t)  6  x  z  2  z
u 

y
v 

z
w0
u ( x y z t)  x  y  z t
v ( x y z t)  x  y  z t


x
u ( x y z t)  t y  z
y

Hence
c)
3 4
v ( x y z t)  2  y  z  6  x  y  z
x
b)
2 2
u ( x y z t)  2  y  2  x  z
x
2
u ( x y z t)  x  2  y  z

x
u ( x y z t)  2  x
Hence
2
2
2
v ( x y z t)  t  x  z
u 

y
v 

z
w0
z
2
w( x y z t)  6  x  z
NOT INCOMPRESSIBLE
2

2
w( x y z t)  z  x  t  y  t

z
2

w( x y z t)  2  z t  x  t y

NOT INCOMPRESSIBLE
2
v ( x y z t)  x  2  y  z
w( x y z t)  2  x  z  y  2  z


y

x
v ( x y z t)  2
u 

y
v 

z
w0
z
w( x y z t)  2  2  x
INCOMPRESSIBLE
Problem 5.3
[Difficulty: 2]
Given:
x component of velocity
Find:
y component for incompressible flow; Valid for unsteady?; How many y components?
Solution:
Basic equation:

x
( ρ u ) 

y
( ρ v ) 

z
( ρ w) 

t
ρ0
Assumption: Incompressible flow; flow in x-y plane
Hence

x
u 

y
v 0
or



v   u   [ A x  ( y  B) ]  A ( y  B)
y
x
x
 y2


v ( x y )   A ( y  B) dy  A 
 B y   f ( x )

2

This basic equation is valid for steady and unsteady flow (t is not explicit)
Integrating
There are an infinite number of solutions, since f(x) can be any function of x. The simplest is f(x) = 0
 y2
v ( x y )  A 
2

 B y 

v ( x y )  6  y 
y
2
2
Problem 5.4
Given:
Find:
Solution:
[Difficulty: 1]
The velocity field provided above
The conditions under which this fields could represent incompressible flow
We will check this flow field against the continuity equation
Governing
Equations:

u    v    w    0 (Continuity equation)
y
z
t
x
Assumptions:
(1) Incompressible flow (ρ is constant)
Based on the assumption listed, the continuity equation reduces to:
u v w


0
x y z
Calculating the partial derivatives of the velocity components:
u
A
x
v
E
y
w
J
z
Applying this information to the continuity equation we get the necessary condition for incompressible flow:
A E J 0
(B, C, D, F, G, and H are arbitrary)
Problem 5.5
[Difficulty: 2]
Given:
x component of velocity
Find:
y component for incompressible flow; Valid for unsteady? How many y components?
Solution:
Basic
Equation:
∂
∂x
Assumptions:
Hence
∂
∂x
( ρ⋅ u ) +
∂
∂y
( ρ⋅ v ) +
∂
∂z
( ρ⋅ w) +
∂
∂t
ρ=0
Incompressible flow (ρ is constant)
Flow is only in the x-y plane
u +
∂
∂y
v =0
or
∂
(
)
2
3
∂
∂
v = − u = − 3 ⋅ x ⋅ y − y = −6 ⋅ x ⋅ y
∂y
∂x
∂x
⌠
2
v ( x , y ) = −⎮ 6 ⋅ x ⋅ y dy = −3 ⋅ x ⋅ y + f ( x )
⌡
This basic equation is valid for steady and unsteady flow (t is not explicit)
Integrating
There are an infinite number of solutions, since f(x) can be any function of x. The simplest is f(x) = 0
v ( x , y ) = −3 ⋅ x ⋅ y
2
Problem 5.6
Given:
Find:
Solution:
[Difficulty: 2]
The x-component of velocity in a steady, incompressible flow field
The simplest y-component of velocity for this flow field
We will check this flow field against the continuity equation
Governing
Equations:

u    v    w    0 (Continuity equation)
y
z
t
x
Assumptions:
(1) Incompressible flow (ρ is constant)
(2) Two dimensional flow (velocity is not a function of z)
u v

0
x y
Based on the two assumptions listed above, the continuity equation reduces to:
The partial of u with respect to x is:
u
A
v
u
A
  2 Therefore from continuity, we have
 2

x
x
x x
y
Integrating this expression will yield the y-component of velocity:
v





The simplest version of this velocity component would result when f(x) = 0:
A
2
x
dy  f ( x) 
Ay
2
 f ( x)
x
v
Ay
2
x
Problem 5.7
[Difficulty: 2]
Given:
y component of velocity
Find:
x component for incompressible flow; Simplest x components?
Solution:
Basic
equation:

x
Assumptions:
Hence
Integrating

x
( ρ u ) 

y
( ρ v ) 

z
( ρ w) 

t
ρ0
Incompressible flow (ρ is constant)
Flow is only in the x-y plane
u 

y
v 0




2
2
2
2


u   v   A x  y  x  y   A x  x  y  A x  y  2  y
x
y
y

or



1
3

3
2
4
2 2
u ( x y )   A x  3  x  y dx    A x   A x  y  f ( y )
4
2

This basic equation is valid for steady and unsteady flow (t is not explicit)
There are an infinite number of solutions, since f(y) can be any function of y. The simplest is f(y) = 0
u ( x y ) 
3
2
2 2
 A x  y 
1
4
 A x
4
u ( x y ) 
9
2
2 2
x y 
3
4
x
4
Problem 5.8
[Difficulty: 3]
Given:
y component of velocity
Find:
x component for incompressible flow; Simplest x component
Solution:
Basic equation:

x
( ρ u ) 

y
( ρ v ) 

( ρ w) 
z

t
ρ0
Assumption: Incompressible flow; flow in x-y plane
Hence
Integrating

x
u 

y
v 0


u ( x y )  



u ( x y ) 
u ( x y ) 
 2 x x  3 y
2 x y 


u   v   
 

2
3
x
y
y  2

2 
2
2
x

y
x

y










1
2
2
1
2
x y
2



2 y
x



x2  y2
2
y
2 y


2
2
2

2
 f (y)
2
Note: Instead of this approach we could have verified that u and v satisfy continuity
2
 1
2 y


x  x 2  y 2
2
2
x y





    2  x y
2 y
 x2  y2





  0 However, this does not verify
the solution is the simplest.


2

2
2
2
2
2
2
 2 x x2  3 y2 
x y
x  y  2 y

 dx 
 f (y) 
 f ( y)
3 
2
2

2
2
2
2
2
2
x y
x y
 x y

x y
The simplest form is
2
or


Problem 5.9
[Difficulty: 2]
Given:
x component of velocity
Find:
y component for incompressible flow; Valid for unsteady? How many y components?
Solution:
Basic equation:

x
( ρ u ) 

y
( ρ v ) 

z
( ρ w) 
Assumption: Incompressible flow; flow in x-y plane
Hence
Integrating

x
u 

y
v 0



v ( x y )  


or
x
A
b
t
ρ0
x
 x







y
y
A
b
b



v   u    A e  cos      e  cos  
y
x
x 
 b 
b
 b 
x
 e  cos
b

y
b
dy  A e  sin   f ( x )

b
b
y
This basic equation is valid for steady and unsteady flow (t is not explicit)
There are an infinite number of solutions, since f(x) can be any function of x. The simplest is f(x) = 0
x
x
v ( x y )  A e  sin
b
y

b
v ( x y )  10 e  sin
5
y

5
Problem 5.10
Given:
Approximate profile for a laminar boundary layer:
U y
u
δ  c x (c is constant)
δ
Find:
(a) Show that the simplest form of v is
v
[Difficulty: 2]
u y

4 x
(b) Evaluate maximum value of v/u where δ = 5 mm and x = 0.5 m
Solution:
We will check this flow field using the continuity equation
Governing
Equations:

u    v    w    0 (Continuity equation)
x
y
z
t
Assumptions:
(1) Incompressible flow (ρ is constant)
(2) Two dimensional flow (velocity is not a function of z)
u v

0
x y
Based on the two assumptions listed above, the continuity equation reduces to:
u u d
Uy 1 
Uy
v
u
Uy

  2  cx 2  


3 Therefore from continuity:
3
x  dx
2

y
x
2cx 2
2cx 2
1
The partial of u with respect to x is:
Integrating this expression will yield the y-component of velocity:

v




U y
3
2  c x
2
U y
2
Now due to the no-slip condition at the wall (y = 0) we get f(x) = 0. Thus: v 
δ
v
The maximum value of v/U is where y = δ: v ratmax 

4 x
u
v ratmax 
dy  f ( x ) 
3
4  c x
U y
2
5  10
2
3
4  c x
U y


y
1 4 x
c x
3
m
4  0.5 m

 f ( x)
2
u y
4 x
(Q.E.D.)
v
u y

4 x
2
v ratmax  0.0025
Problem 5.11
Given:
Approximate (parabolic) profile for a laminar boundary layer:
u
U
Find:
[Difficulty: 3]
 2  
y
  

δ δ
y
2
δ  c x
(c is constant)
(a) Show that the simplest form of v for incompressible flow is
v
U

1 y
1 y
        
x 2  δ 
3 δ
δ
2
3


(b) Plot v/U versus y/δ
(c) Evaluate maximum value of v/U where δ = 5 mm and x = 0.5 m
Solution:
We will check this flow field using the continuity equation
Governing
Equations:

u    v    w    0 (Continuity equation)
y
z
t
x
Assumptions:
(1) Incompressible flow (ρ is constant)
(2) Two dimensional flow (velocity is not a function of z)
u v

0
x y
Based on the two assumptions listed above, the continuity equation reduces to:
1

 2 y 2 y 2  1 1
u u d

 U  2  3   cx 2 Now since δ  c x 2
The partial of u with respect to x is:
x  dx
  2
 
u Uc 2  y
y 2  Uc 2




x
   2  3   2
  y   y 2 
       Therefore from continuity:
       
Integrating this expression will yield the y-component of velocity:


v


u Uc 2
v


x  2
y
2
2

c
and thus
δ
 y   y  2 
     
      
2
 y
 y   dy  f ( x )

  
δ  δ   δ  
U c
 
1
3
2
3
2
 y2
y 
U c  1  y 
1 y 
2

δ
           f ( x)
v


 f ( x) 
2 2
Since δ  c x c 
2
2  2 δ
δ 2  δ 
3 δ 
3 δ 
δ 
x
U c
x
1
2
Thus:
Evaluating:
δ 1 y
1 y
v  U         
x 2  δ 
3 δ
2
3
  f (x)

Now due to the no-slip condition at the wall (y = 0) we get f(x) = 0. Therefore:
v
U

1 y
1 y
        
x 2  δ 
3 δ
2
δ
3


v
(Q.E.D.)
U

1 y
1 y
        
x 2  δ 
3 δ
2
δ
3


Plotting this relationship shows:
1
Dimensionless height (y/delta)
Assuming x = 0.5 m and δ = 5 mm
0.5
0
0
5 10
4
0.001
0.0015
0.002
Dimensionless Velocity (v/U)
v
δ 1
1
δ
The maximum value of v/U is where y = δ: v ratmax 
     
U
x 2
3  6 x
v ratmax 
5  10
3
m
6  0.5 m
v ratmax  0.00167
Problem 5.12
Given:
Approximate (sinusoidal) profile for a laminar boundary layer:
u
U
Find:
[Difficulty: 3]
 sin
π y 

 2 δ 
δ  c x
(c is constant)
(a) Show that the simplest form of v for incompressible flow is
v
U

1 δ  π y
  cos   
π x   2 δ
 π  y   sin π  y   1
 2 δ  2 δ 

 
 
(b) Plot v/U versus y/δ
(c) Evaluate maximum value of v/U where δ = 5 mm and x = 0.5 m
Solution:
We will check this flow field using the continuity equation
Governing
Equations:

u    v    w    0 (Continuity equation)
y
z
t
x
Assumptions:
(1) Incompressible flow (ρ is constant)
(2) Two dimensional flow (velocity is not a function of z)
Based on the two assumptions listed above, the continuity equation reduces to:
u v

0
x y
 y
u u d
Ucy  2
 y 
 y  1 

 U  2 cos   cx 2  
x cos 
2
The partial of u with respect to x is: x
 dx
4
 2 
 2  2
 2
1
1
Now since δ  c x
2

x
1
2

c
δ
and thus
Uc 2 y  y 
u
cos  Therefore from continuity:

x
4 3
 2 
Integrating this expression will yield the y-component of velocity:
v
1


v


2
π U c  y
3
4 δ
 cos
π y 
v Uc 2 y
 y 
cos 

3
y
4
 2 
 dy  f ( x )
 2 δ 
2
2
2 
π U c  2  δ y
π U c 
π y 
 π  y   4  δ  cos π  y    f ( x )

d

f
(
x
)

y  cos
y


sin

 δ
 δ 
3  π
3 
 2 δ 
 2  π2
 2 
4 δ
4 δ


Evaluating:
Simplifying this expression:
1
v
U c
2
2 δ 
2
π y  2 δ
π y
 
 cos     f ( x )
 2 δ π
 2 δ 
  y  sin
v
U δ y
π y
π y
2
    sin     cos     f ( x )
2 x δ
2
δ
π


 2 δ 
0
U δ 2
    cos( 0 )   f ( x )
2 x π

v
U δ π y
π y
π y
     sin    cos    1
π x 2 δ
 2 δ
 2 δ 
f ( x)  
U δ
π x
Since
δ  c x
2
2
c 
2
δ
Thus:
x
Now due to the no-slip condition at the wall (y = 0) we get:
Therefore: v 
v
Thus:
U

U δ y
π y
π y
2
U δ
    sin     cos    
2 x δ
 2 δ  π  2 δ   π x
π y
π y
π y
  sin    cos    1
π x  2 δ
 2 δ
 2 δ 
δ
 
v
U
δ

π x
Simplifying:
(Q.E.D.)
π y
π y
π y
  sin    cos    1
2
2
δ
δ



 2 δ 
 
Plotting this relationship shows:
Assuming x = 0.5 m and δ = 5 mm
1
Dimensionless height (y/delta)
Dimensionless height (y/delta)
1
0.5
0
0
0.5
1
0.5
0
0
0.0005
Dimensionless Velocity (u/U)
0.001
0.0015
0.002
Dimensionless Velocity (v/U)
v
δ π
π
π
δ π
The maximum value of v/U is where y = δ: v ratmax 

   sin   cos   1 
   1
U
π x  2
2
 2   π x  2

v ratmax 
5  10
3
m
π  0.5 m

 π  1
2



v ratmax  0.00182
Problem 5.13
[Difficulty: 3]
Given:
Data on boundary layer
Find:
y component of velocity ratio; location of maximum value; plot velocity profiles; evaluate at particular point
Solution:
3 y  1  y 
u ( x y )  U   
  

 2  δ( x )  2  δ( x ) 
so
For incompressible flow
Hence
so
3


3
y  1  y 
u ( x y )  U   
  

2
  c x  2  c x 

x
u 

y
and
δ( x )  c x
3


v 0

d
v ( x y )  
u ( x y ) dy
 dx



v ( x y )  


and
du
dx

3
4
 y3 x5 y x3 
 U  
   dy
4
 c3 2 c 2 


3
4 
 y2
y

v ( x y )   U 

8 
3
5
 2
3 2
2 c  x 
 c x
3
The maximum occurs at
yδ
v max 
 y3
y 

5
3

 3 2
2
c x 
 c x
 U 
v ( x y ) 
as seen in the Excel work shown below.
δ
1
 U   1   1
8
x 
2 
3
At δ  5  mm and x  0.5 m, the maximum vertical velocity is
v max
U
 0.00188
δ  y
1 y
 U       
8
x  δ 
2 δ
3
2
4


To find when v /U is maximum, use Solver in Excel
y /δ
0.00188
1.0
v /U
y /δ
0.000000
0.000037
0.000147
0.000322
0.000552
0.00082
0.00111
0.00139
0.00163
0.00181
0.00188
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Vertical Velocity Distribution In Boundary layer
1.0
0.8
y /δ
v /U
0.6
0.4
0.2
0.0
0.0000
0.0005
0.0010
v /U
0.0015
0.0020
Problem 5.14
Given:
[Difficulty: 3]
Steady, incompressible flow in x-y plane:
2 2
u  A x  y
3 1
A  0.3 m
s
Find:
(a) a possible y component of velocity for this flow field
(b) if the result is valid for unsteady, incompressible flow
(c) number of possible y components for velocity
(d) equation of the streamlines for the flow
(e) plot streamlines through points (1,4) and (2,4)
Solution:
We will check this flow field using the continuity equation
Governing
Equations:

u    v    w    0 (Continuity equation)
y
z
t
x
Assumptions:
(1) Incompressible flow (ρ is constant)
(2) Two dimensional flow (velocity is not a function of z)
Based on the two assumptions listed above, the continuity equation reduces to:
The partial of u with respect to x is:
u
 2Axy 2
x
Therefore from continuity:
u v

0
x y
v
u

 2Axy 2
y
x


2
Integrating this expression will yield the y-component of velocity: v 
 2  A x  y dx  f ( x )

The basic equation reduces for the same form for unsteady flow. Hence
Since f(x) is arbitrary:
2
2 3
v    A x  y  f ( x )
3
The result is valid for unsteady, incompressible flow.
There are an infinite number of possible y-components of velocity.
The simplest version of v is when f(x) = 0. Therefore, the equation of the corresponding streamline is:
dy
dx

v
u

2
2 3
 A x  y
3
2 2
A x  y
3
2 y
   Separating variables and integrating:
3 x
2
2 dx
2
 
ln( y )    ln( x ) Thus: x  y  constant
3
y
3 x
dy
10
are the equations of the streamlines of this flow field.
Plotting streamline for point (1, 4): 1  4
Plotting streamline for point (2, 4): 2  4
2
8
3
8
x y
2
3
3
2
2
 16 x  y
8
y (m)
3
6
4
2
 16
The two streamlines are plotted here in red (1,4) and blue (2,4):
0
0
2
4
x (m)
6
8
10
Problem 5.15
Given:
[Difficulty: 3]
Steady, incompressible flow in x-y plane:
v  B x  y
3 1
3
B  0.2 m
s
Find:
(a) the simplest x component of velocity for this flow field
(b) equation of the streamlines for the flow
(c) plot streamlines through points (1,4) and (2,4)
Solution:
We will check this flow field using the continuity equation
Governing
Equations:

u    v    w    0 (Continuity equation)
y
z
t
x
Assumptions:
(1) Incompressible flow (ρ is constant)
(2) Two dimensional flow (velocity is not a function of z)
Based on the two assumptions listed above, the continuity equation reduces to:
The partial of v with respect to y is:
v
 3Bxy 2 Therefore from continuity:
y
u v

0
x y
u
v

 3Bxy 2
x
y


2
Evaluating the integral:
Integrating this expression will yield the x-component of velocity: u 
 3  B  x  y dx  f ( y )

3
3
2 2
2 2
u   B x  y  f ( y ) The simplest version of this equation is obtained when f(y) = 0:
u   B x  y
2
2
dy
The equation of a streamline is:
dx

v
u

B x  y
3
2 y
 
3
3 x
2 2
 B x  y
2
2 dx
 
y
3 x
dy
Separating variables and integrating:
3
3
2
2
ln( y )    ln( x ) Thus: x  y  constant
3
3
Plotting streamline for point (1, 4): 1  4
2
x y
are the equations of the streamlines of this flow field.
2
 constant
3
8
x y
2
10
8
8
Plotting streamline for point (2, 4): 2  4
2
3
 16 x  y
2
 16
The two streamlines are plotted here in red (1,4) and blue (2,4):
y (m)
3
6
4
2
0
0
2
4
6
x (m)
8
10
Problem 5.16
[Difficulty: 5]
Discussion: Refer back to the discussion of streamlines, pathlines, and streaklines in
Section 2-2.
Because the sprinkler jet oscillates, this is an unsteady flow. Therefore pathlines and
streaklines need not coincide.
A pathline is a line tracing the path of an individual fluid particle. The path of each
particle is determined by the jet angle and the speed at which the particle leaves the jet.
Once a particle leaves the jet it is subject to gravity and drag forces. If aerodynamic drag
were negligible, the path of each particle would be parabolic. The horizontal speed of the
particle would remain constant throughout its trajectory. The vertical speed would be
slowed by gravity until reaching peak height, and then it would become increasingly
negative until the particle strikes the ground. The effect of aerodynamic drag is to reduce
the particle speed. With drag the particle will not rise as high vertically nor travel as far
horizontally. At each instant the particle trajectory will be lower and closer to the jet
compared to the no-friction case. The trajectory after the particle reaches its peak height
will be steeper than in the no-friction case.
A streamline is a line drawn in the flow that is tangent everywhere to the velocity vectors
of the fluid motion. It is difficult to visualize the streamlines for an unsteady flow field
because they move laterally. However, the streamline pattern may be drawn at an instant.
A streakline is the locus of the present locations of fluid particles that passed a reference
point at previous times. As an example, choose the exit of a jet as the reference point.
Imagine marking particles that pass the jet exit at a given instant and at uniform time
intervals later. The first particle will travel farthest from the jet exit and on the lowest
trajectory; the last particle will be located right at the jet exit. The curve joining the
present positions of the particles will resemble a spiral whose radius increases with
distance from the jet opening.
Problem 5.17
Given:
Find:
Solution:
[Difficulty: 4]
Conservation of mass in rectangular coordinates
Identical result to Equation 5.1a by expanding products of density and velocity in a Taylor Series.
We will use the diagram in Figure 5.1 (shown here). We will apply the conservation of mass evaluating the
derivatives at point O:
Governing
Equations:

u    v    w    0 (Continuity equation - Eqn 5.1a)
y
z
t
x
Assumptions:
Expansion of density and velocity via Taylor series is valid
around point O.
In the x-direction, the mass flux is:
At the right face:
At the left face:
m x  udA  udydz
 u  dx 

m x  dx 2   u 
dydz (out of the volume)


x
2



 u   dx 
m x  dx 2   u 
   dydz (into the volume)
x  2 

The net mass flux out of the volume in the x-direction would then be:

 u  dx 
 u   dx 
  u 

m x ( net )  m x  dx 2  m x  dx 2   u 
dydz   u 
dxdydz
   dydz 

x 2 
x  2 
x


Similarly, the net mass fluxes in the y-direction and z-direction are:
The rate of mass accumulation in the volume is:
m ( net ) 
m y ( net ) 

dm 
dxdydz
 
dt  vol t
 v 
dxdydz
x
m z ( net ) 
 w
dxdydz
x
Now the net outflux must balance the accumulation:

  u 
 v 
  w 
dm 
dxdydz 
dxdydz 
dxdydz 
dxdydz  0
  0 Therefore we may write:
x
x
x
t
dt  vol
We may divide the volume out of all terms:
 u   v   w 



 0 (Q.E.D.)
x
x
x
t
 u   v   w 



0
x
x
x
t
Problem 5.18
Given:
Find:
Solution:
[Difficulty: 2]
The list of velocity fields provided above
Which of these fields possibly represent incompressible flow
We will check these flow fields against the continuity equation
Governing
Equations:
1 
rVr   1  V    V z     0
z
t
r r
r 
Assumptions:
(1) Incompressible flow (ρ is constant)
(2) Two dimensional flow (velocity is not a function of z)
Based on the two assumptions listed above, the continuity equation reduces to:
(Continuity equation)
 rVr  V

0
r

This is the criterion against which we will check all of the flow fields.
 rVr  V
 U cos     U cos    0


r
(a) Vr  U cos( θ)
Vθ  U sin( θ)
This could be an incompressible flow field.
q
(b) Vr  
2  π r
 rVr  V
00 0


r
K
Vθ 
2  π r
This could be an incompressible flow field.

(c) Vr  U cos( θ)  1 


 a 
 r
 
Vθ  U sin( θ)  1 

2
2
 a  
 r
 
  a 2 
  a 2 
 rVr  V

U
U
cos  1      0
 cos  1    


r
r


  r  


This could be an incompressible flow field.
Problem 5.19
Given:
Find:
Solution:
[Difficulty: 2]
The list of velocity fields provided above
Which of these fields possibly represent incompressible flow
We will check these flow fields against the continuity equation
Governing
Equations:
1 
rVr   1  V    V z     0
z
t
r r
r 
Assumptions:
(1) Incompressible flow (ρ is constant)
(2) Two dimensional flow (velocity is not a function of z)
Based on the two assumptions listed above, the continuity equation reduces to:
(Continuity equation)


r Vr  Vθ  0

r
θ
This is the criterion against which we will check all of the flow fields.
a)
b)
b)

r Vr(r θ t)
r
K
Vr( r θ t)  
r
Vθ( r θ t)  0
Hence


r Vr  Vθ  0

r
θ
Vr( r θ t)  0
K
Vθ( r θ t) 
r
Hence

Vr( r θ t)  
Vθ( r θ t)  

r Vr(r θ t)
r
2

r Vr(r θ t)
r
Hence

 0
INCOMPRESSIBLE
K sin( θ)
r
2
r
K cos( θ)

2
θ
r

Vθ( r θ t)  

r Vr  Vθ  0

r
θ

θ
Vθ( r θ t)  0
INCOMPRESSIBLE

r Vr  Vθ  0

r
θ
K cos( θ)
 0
K cos( θ)
2
r
INCOMPRESSIBLE

θ
Vθ( r θ t)  0
Problem 5.20
[Difficulty: 3]
Given:
r component of velocity
Find:
θ component for incompressible flow; How many θ components
Solution:
Basic equation:
Assumptions:
1 ∂
1 ∂
∂
∂
⋅ ρ⋅ r⋅ Vr + ⋅
ρ⋅ Vθ +
ρ⋅ Vz + ρ = 0
r ∂r
r ∂θ
∂z
∂t
(
)
(
)
(
)
Incompressible flow
Flow in r-θ plane
Hence
1 ∂
1 ∂
⋅ r⋅ Vr + ⋅
V =0
r ∂r
r ∂θ θ
Integrating
⌠
Vθ( r , θ) = −⎮ U⋅ cos( θ) dθ = −U⋅ sin( θ) + f ( r)
⌡
( )
( )
or
Vθ( r , θ) = −U⋅ sin( θ) + f ( r)
There are an infinite number of solutions as f(r) can be any function of r
The simplest form is
Vθ( r , θ) = −U⋅ sin( θ)
∂
∂
∂
Vθ = − r⋅ Vr = − ( r⋅ U⋅ cos( θ) ) = −U⋅ cos( θ)
∂θ
∂r
∂r
( )
Problem 5.21
[Difficulty: 3]
Given:
r component of velocity
Find:
θ component for incompressible flow; How many θ components
Solution:
Basic equation:
1 
1 


 ρ r Vr  
ρ Vθ 
ρ Vz  ρ  0
r r
r θ
z
t






Assumption: Incompressible flow; flow in r-θ plane
Hence
1 
1 
 r Vr  
V 0
r r
r θ θ
Integrating


Vθ( r θ)  



 
Vθ( r θ)  
 
Λ  cos( θ)
2
dθ  
r
Λ  sin( θ)
2
Λ  cos( θ) 
Λ  cos( θ)



Vθ   r Vr    

r
2
θ
r
r 

r

o
r
Λ  sin( θ)
2
 f ( r)
r
 f ( r)
r
There are an infinite number of solutions as f(r) can be any function of r
The simplest form is
Vθ( r θ)  
Λ  sin( θ)
2
r
 
Problem 5.22
[Difficulty: 2]
Given:
Flow between parallel disks as shown. Velocity is purely tangential. No-slip
condition is satisfied, so velocity varies linearly with z.
Find:
Solution:
An expression for the velocity field
We will apply the continuity equation to this system.
Governing
Equations:
1 
rVr   1  V    V z     0
z
t
r r
r 

V  Vr eˆr  V eˆ  V z kˆ
Assumptions:
Since the velocity is linear with z, we may write: Vθ( r z)  z f ( r)  C
Vθ( r 0 )  0
(Velocity flow field)
(1) Incompressible flow (ρ is constant)
(2) Purely tangential flow
(3) Linear velocity variation with z
Based on the first two assumptions, the continuity equation reduces to:
1:
(Continuity equation)
0  f ( r)  C  0
Therefore the tangential velocity is:
C0
z
Vθ  ω r
h
2:
Vθ( r h )  r ω
V
 0 thus:

Vθ  Vθ( r z)
Now we apply known boundary conditions:
h  f ( r)  r ω
f ( r) 
r ω
h
Thus, the velocity field is:

z
V  r eˆ
h
Problem 5.23
[Difficulty: 4]
Given:
Find:
Definition of "del" operator in cylindrical coordinates, velocity vector
Solution:
We will apply the velocity field to the del operator and simplify.
 


(a) An expression for   V in cylindrical coordinates.
(b) Show result is identical to Equation 5.2c.
Governing
Equations:


1 

  eˆr
 eˆ
 kˆ
r
z
r 

V  Vr eˆr  V eˆ  V z kˆ
(Definition of "del" operator)
(Velocity flow field)
1 
rVr   1  V    V z   0
r r
r 
z
ˆ
ˆ
e
er
 eˆr
 eˆ


 
(Equation 5.2c)
(Hints from footnote)


  V using the governing equations yields:

  

1 
 kˆ    Vr eˆr  V eˆ  V z kˆ
 eˆ
  V   eˆr
z 
r 
 r


1 
  Vr eˆr  V eˆ  V z kˆ  kˆ   Vr eˆr  V eˆ  V z kˆ
  Vr eˆr  V eˆ  V z kˆ  eˆ
 eˆr
r 
z
r
ˆ

Vr   eˆ  eˆr 1  Vr   eˆ  1 er Vr 
 eˆr  eˆr
r
r 
r 
ˆ
1 
V   eˆ  1 e V   kˆ  kˆ  Vz 
 eˆ  eˆ
r 
r 
z
Substituting
 








eˆ  eˆr  eˆr  eˆ  0
eˆ  eˆ  eˆr  eˆr  kˆ  kˆ  1


ˆ
1 eˆr

Vr   1  V   eˆ  1 e V    V z 
  V  Vr   eˆ 
r
r 
r 
r 
z
1
1 

V   eˆ  eˆr 1 V    Vz 
 Vr   eˆ  eˆ Vr  
r
r 
r
r
z
1
1 

V    Vz 
  V r    V r  
r
r
r 
z

1
Combining the first two terms:
Vr   Vr   1  rVr  which can be verified through differentiation. Thus:
r
r r
r
Using the hints listed above, and knowing that:
 
 

 1 
rVr   1  V    V z  (Q.E.D.)
  V 
r r
r 
z
Problem 5.24
[Difficulty: 3]
Given:
The velocity field
Find:
Whether or not it is a incompressible flow; sketch various streamlines
Solution:
A
Vr 
r
B
Vθ 
r
 
For incompressible flow
1 d
1 d
 r Vr   Vθ  0
r dr
r dθ
Hence
1 d
1 d
 r Vr   Vθ  0
r dr
r dθ
For the streamlines
 
dr
Vr
so





 
1 d
 r Vr  0
r dr
Flow is incompressible
r dθ
r dr
Vθ
A

dr  

r

1
A
Equation of streamlines is r  C e
A
B
dθ
1 d
 V 0
r dθ θ
2

r  dθ
B
ln( r) 
Integrating
θ
A
B
 θ  const
4
B
(a) For A = B = 1 m2/s, passing through point (1m, /2)
θ
2
π
2
r e
(b) For A = 1 m2/s, B = 0 m2/s, passing through point (1m, /2)
θ
π
4
2
0
2
(c) For A = 0 m2/s, B = 1 m2/s, passing through point (1m, /2)
2
r  1 m
4
(a)
(b)
(c)
2
4
Problem *5.25
[Difficulty: 2]

y
V  U iˆ
h
Given:
Velocity field for viscometric flow of Example 5.7:
Find:
(a) Stream function
(b) Locate streamline that divides flow rate equally
Solution:
The flow is incompressible, so the stream function may be derived
Governing
Equations:
u

y
v

x
(Definition of stream function)
2


y
U y
ψ   u dy  f ( x )   U dy  f ( x ) 
 f ( x)

h
2 h


Integrating the velocity will result in the stream function:
Let ψ = 0 at y = 0, so f(x) = 0:
ψ
2
U h
U h
The flow rate is:
The stream function is a maximum value at y = h: ψmax 

2 h
2
So the streamline which splits the flow rate into two equal parts is:
Therefore, the equation of this streamline would be:
U y
2
2 h

U h
4
ψhalfQ 
1
2
 ψmax 
U y
2
2 h
U h
U h
 ψmax  ψmin 
0
2
2
w
Q
1 U h
U h


2 2
4
2
Simplifying this equation: y 
h
2
2
or: y 
h
2
y
h
2
Problem *5.26
Given:
Velocity field
Find:
Stream function ψ
[Difficulty: 3]
Solution:
Basic equations:  ( ρ u )   ( ρ v)   ( ρ w)   ρ  0
x
Assumptions:
Hence
y
z
t

x
u 

y
v 0

2

v  x ( x  1)  2 y   ψ
x
Checking
y

v ψ
x
ψ
x
3
3

2
x
y
x
[ 2  y  ( 2x  1 ) ] 
ψ


ψ( x y )  

2
g(y)  y
and
2
2
ψ( x y )  y  2  x  y 

y
x ( x  1 )  2  y2  0

2
2
ψ( x y )   2  y  ( 2  x  1 ) dy  2  x  y  y  f ( x )

and
f ( x)  

or
u  2 y ( 2 x  1) 
The stream function is

Incompressible flow
Flow in x-y plane
Hence
Comparing these
u
x
2
2

x
2
3
x ( x  1 )  2  y2 dx   x  x  2  x y 2  g( y
2
3
3
3
 2
x
x 
2
  u( x y )  2  y  4  x y
u ( x y )   y  2  x  y 

2
3 
y 
2
3

x
x 
2
  2
  v ( x y)  x2  x  2 y 2
v ( x y )  
y  2 x y 

2
3 
x 

2
3
2
Problem *5.27
[Difficulty: 3]
Given:
The velocity field
Find:
Whether or not it is a incompressible flow; sketch stream function
Solution:
A
Vr 
r
 
For incompressible flow
1 d
1 d
 r Vr   Vθ  0
r dr
r dθ
Hence
1 d
1 d
 r Vr   Vθ  0
r dr
r dθ
For the stream function

 
θ
ψ  r Vr  A

Integrating
B
ψ  Vθ  
r
r
Comparing, stream function is
ψ  A θ  B ln( r)
ψ
B
Vθ 
r
 
1 d
 r Vr  0
r dr
Flow is incompressible
ψ  A θ  f ( r)
ψ  B ln( r)  g ( θ)
1 d
 V 0
r dθ θ
Problem *5.28
Given:
[Difficulty: 2]
Stream function for an incompressible flow field:
ψ  U r sin( θ) 
Find:
q
2 π
θ
(a) Expression for the velocity field
(b) Location of stagnation points
(c) Show that the stream function is equal to zero at the stagnation points.
We will generate the velocity field from the stream function.
Solution:
Governing
1 

Vr 
V  
Equations:
r
r 
Taking the derivatives of the stream function:
q
Vr  U cos ( θ) 
2 π r
(Definition of stream function)
Vθ  U sin ( θ)
 
q 
V    U cos  
eˆr  U sin  eˆ
2R 

So the velocity field is:
To find the stagnation points we must find the places where both velocity components are zero. When
When Vθ  0
For θ = 0: r 
sin( θ)  0 therefore:
q
2  π U cos( 0 )

q
2  π U
θ  0 π
For θ = π:
q
Vr  0 r 
2  π U cos( θ)
Now we can apply these values of θ to the above relation to find r:
r
q
2  π cos( π)

q
2  π U
These represent the same point:
Stagnation point at:
( r θ) 
At the stagnation point:
ψstagnation  U
q
2  π U
 sin( 0 ) 
q
2 π
 q

 2  π U 0


0  0
ψstagnation  0
Problem *5.29
[Difficulty: 3]
Given:
Velocity field
Find:
Whether it's 1D, 2D or 3D flow; Incompressible or not; Stream function ψ
Solution:
Basic equation:

x

( ρ u ) 
y
( ρ v ) 

z
( ρ w) 

t
ρ0
u

z

w  ψ
x
ψ
Assumption: Incompressible flow; flow in x-z plane (v = 0)
Velocity field is a function of x and z only, so is 2D
Check for incompressible

x

x
Hence

x

u 
z
w0


z 3  x2  z2   6  x z

u 
z


3 2 2 1 4

2
2
ψ( x z)   z 3  x  z dz   x  z   z  f ( x )
2
4

   ψ


ψ( x z)  

and
w  x x  3 z
2
2
Checking
x

z
x
4
and
4
ψ( x z)  
x
4

4

 ψ
2
The stream function is
2

x  x 2  3 z2   6 x  z
Flow is INCOMPRESSIBLE
u  z 3  x  z
f ( x)  
z
w0
Hence
Comparing these

3
2
2 2
x z 
z
g ( z)  
z
4
4
4
 x 4 3 2 2 z4 
  u( x z)  3 x 2 z  z3
u ( x z)   
 x z 
2
4 
z  4


3
2
3
3

w( x z)   z y  z  y  w( x z)  z  3  y  z
y


4
x  x 2  3 z2  dx   x  3  x2 z2  g ( z)
4


4
2
Problem *5.30
Given:
[Difficulty: 3]
Stream function for an incompressible flow field:
ψ  5  A x  2  A y
A  2
m
s
Find:
(a) Sketch streamlines ψ = 0 and ψ = 5
(b) Velocity vector at (0, 0)
(c) Flow rate between streamlines passing through points (2, 2) and (4, 1)
Solution:
We will generate the velocity field from the stream function.

y

x
Governing
Equations:
u
Assumptions:
Incompressible flow (ρ is constant)
Flow is only in the x-y plane
v
(Definition of stream function)
For ψ = 0: 0  5  A x  2  A y Solving for y:
5
y   x
2
For ψ = 5: 5  5  A x  2  A y Solving for y:
5
5 m
s
5
5
y   x  

  x  m
2
2 s
2 m
2
2
2
Here is the plot of the two streamlines:
ψ =0 is in red; ψ = 5 is in blue
10
y (m)
5
4
2
0
2
4
5
Generating the velocity components from the stream function derivatives:
u  2  A
v  5 A
 10
Therefore, the velocity vector at (0, 0) is:

V  4iˆ  10 ˆj
At the point (2, 2) the stream function value is:
At the point (4, 1) the stream function value is:
The flow rate between these two streamlines is:
ψa  5  2 
m
ψb  5  2 
m
s
s
Q  ψb  ψa
x (m)
 2 m  2  2
m
 4 m  2  2
m

2
Q   44

s
s
2
 2  m ψa  28
m
 1  mψb  44
m
2
2

   28 m 
s 
s  
m
s
s
3
Q  16
m
s m
Flow rate is 16 m3/s per meter of depth
Problem *5.31
[Difficulty: 3]
Given:
Approximate profile for a laminar boundary layer:
U y
u
δ  c x (c is constant)
δ
Find:
(a) Stream function for the flow field
(b) Location of streamlines at one-quarter and one-half the total flow rate in the boundary layer.
Solution:
We will generate the stream function from the velocity field.
Governing
Equations:
u

y
v

x
(Definition of stream function)
Integrating the x-component of velocity yields the stream function:

ψ 


U y
δ
dy  f ( x) 
U y
2
2 δ
 f ( x)
If we set ψ  0 at y  0 then the stream function would be:
2
The total flow rate per unit depth within the boundary layer is:
At one-quarter of the flow rate in the boundary layer:
1
8
 U δ 
U y
2
2 δ
Solving for y:
2
y 
4
 U δ 
U y
2
2 δ
Solving for y:
2
y 
Q  ψ ( δ)  ψ ( 0) 
U δ
2 δ
 0
1
2
Q
2
2 δ
 U δ
1 1
1
  U δ   U δ Therefore, the streamline would be located at:
4 2
8
1 2
δ So at one-quarter of the flow rate:
4
At one-half of the flow rate in the boundary layer:
1
Q
U y
ψ 
y
δ

1
2
1 1
1
  U δ   U δ Therefore, the streamline would be located at:
2 2
4
1 2
δ So at one-quarter of the flow rate:
2
y
δ

1
2
Problem *5.32
Given:
[3]
Approximate profile for a laminar boundary layer:
u
U
 2  
y

δ

y
δ
 
2
δ  c x
(c is constant)
Find:
(a) Stream function for the flow field
(b) Location of streamlines at one-quarter and one-half the total flow rate in the boundary layer.
Solution:
We will generate the stream function from the velocity field.
Governing
Equations:
u

y
v

x
(Definition of stream function)
Integrating the x-component of velocity yields the stream function:


 y
ψ   U 2    

 δ

2
2
3
 y   dy  f ( x )  U δ  y   1   y    f ( x) If we set ψ  0 at y  0 the stream function would be:
δ
 
 
3 δ 
 
 δ 
 y 2 1 y 3
ψ  U δ       
3 δ 
 δ 
The total flow rate per unit depth within the boundary layer is:
Q
At one-quarter of the flow rate in the boundary layer:
1
6
3
 y  2 1  y  3
y 
or

2




 
δ
3 δ 
 δ 
 
 U δ  U δ 
6  
y
 δ  2 1  δ  3
2
       0   U δ
3
3  δ 
 δ 
Q  ψ( δ)  ψ( 0 )  U δ 
1 2
1
  U δ   U δ Therefore, the streamline would be located at:
4 3
6
2
  1  0 We may solve this cubic for y/δ using several methods,
δ
including Goal Seek in Excel or polyroots in Mathcad. Once the roots are determined, only one root would make physical sense.
So at one-quarter of the flow rate:
y
δ
At one-half of the flow rate in the boundary layer:
1
3
Q
1 2
1
  U δ   U δ
2 3
3
3
2
 y  2 1 y  3

y
y
  3   δ   or  δ   3    1  0
 δ 
 
 
δ
 U δ  U δ 
 0.442
Therefore, the streamline would be located at:
We solve this cubic as we solved the previous one.
So at one-half of the flow rate:
y
δ
 0.653
Problem *5.33
Given:
[Difficulty: 3]
Approximate profile for a laminar boundary layer:
u
U
 sin
π y 

 2 δ 
δ  c x
(c is constant)
Find:
(a) Stream function for the flow field
(b) Location of streamlines at one-quarter and one-half the total flow rate in the boundary layer.
Solution:
We will generate the stream function from the velocity field.
Governing
Equations:
u

y
v

x
(Definition of stream function)
Integrating the x-component of velocity yields the stream function:

π y 
2  U δ
 π y 
ψ   U sin
 dy  f ( x )   π  cos 2  δ   f ( x)

2

δ





If we set ψ  0 at y  0 the stream function would be:
ψ
Q  ψ( δ)  ψ( 0 )  
The total flow rate per unit depth within the boundary layer is:
Q
At one-quarter of the flow rate in the boundary layer:
U δ
2 π

2  U δ
π
  1  cos
π y  
  or
 2 δ  

1
4
 1  cos
1 2  U δ
U δ


4 π
2 π
π y 
 solving for y/δ:
 2 δ 
2  U δ
π
2  U δ
π
 cos
π y 

 2 δ 
2  U δ
 cos( 0 )  

π
2

  cos

π
Therefore, the streamline would be located at:
y
δ

2
π
 acos
3

4
So at one-quarter of the flow rate:
y
δ
Q
At one-quarter of the flow rate in the boundary layer:
U δ
π

2  U δ
π
  1  cos

π y  
  or
 2 δ  
1
2
 1  cos
π y 
1 2  U δ
U δ


2 π
π
 solving for y/δ:
 2 δ 
 0.460
Therefore, the streamline would be located at:
y
δ

2
π
 acos
1

2
So at one-half of the flow rate:
y
δ
 0.667
Problem *5.34
[Difficulty: 3]
Given:
Data on boundary layer
Find:
Stream function; locate streamlines at 1/4 and 1/2 of total flow rate
Solution:
3 y
1 y
u ( x y )  U        
2
δ
   2 δ
3


3 y
1 y
For the stream function u   ψ  U        
y
2  δ  2  δ 
Hence
δ( x )  c x
and
3



3

3 y
1 y 
ψ   U          dy

2  δ  2  δ  

3 y
1 y 
ψ  U  
    f ( x)
 4 δ 8 δ3 


2
4
3 y
1 y
ψ  U δ        
4
δ
8 δ
  
2
Let ψ = 0 = 0 along y = 0, so f(x) = 0, so
4


The total flow rate in the boundary layer is
Q
At 1/4 of the total
 ψ( δ)  ψ( 0 )  U δ 
1
y
2

δ
 4  
y
δ

y

4
 5
δ
X 
The solution to the
quadratic is
24 
X  0.216
2 4
y
2
X 
y
δ

Note that the other root is
2
X 
y
δ
2
24 
24  4  4  5
2 4
 5.784
X  0.465
4
y
  2    5
δ
δ
The solution to the
quadratic is
where
2
24  4  4  5
2
12 
2
4  X  24 X  5  0
or
3 y
1 y
At 1/2 of the total flow ψ  ψ0  U δ        
8 δ
4  δ 
Hence
5
   U δ
W
4 8 8
 3 y 2 1 y 4 1 5
ψ  ψ0  U δ              U δ
8 δ  4 8
4  δ 

24 
Hence
3
12 
X  0.671
or
4
  1   5  U δ
 2 8

2
2  X  12 X  5  0
where
2
12  4  2  5.
2 2
X  0.450
Note that the other root is
2
X 
12 
y
δ
2
12  4  2  5
2 2
 5.55
Problem *5.35
Given:
[Difficulty: 3]
Rigid body motion in Example Problem 5.6

V  r eˆ
ω  0.5
rad
s
Find:
(a) Stream function for the flow field
(b) Volume flow rate per unit depth between r = 0.10 m and 0.12 m
(c) Sketch velocity profiles along a line of constant θ
(d) Check the volume flow rate calculated from the stream function by integrating the velocity profile
along this line.
Solution:
We will generate the stream function from the velocity field.
Governing
Equations:
Vr 
1 
r 
V  

r
(Definition of stream function)
2

ω r
ψ   r ω dr  f ( θ)  
 f ( θ)
2

Integrating the θ-component of velocity yields the stream function:
2
1 df
Vr  
 0 Therefore, f ( θ)  C
r dθ
Now take the derivative of the stream function:
ψ
ω r
2
C
2
 ω r 2

  ω r1
2
ω
2
2
The volume flow rate per unit depth is: Q  ψ r2   ψ r1    
 C   
 C    r1  r2 


2
2
2

 

Substituting in known values:
Q 
1
2
 0.5
rad
s

2
2

2
 0.10  0.12  m
3
Q  0.001100
m
s m
Because Q<0, the flow is in the direction of eθ
Along a line of constant θ, the velocity varies linearly:
From the linear velocity variation, Vθ  ω r Thus the flow rate is:
r
r
2
2
ω
2
2

Q
Vθ dr  ω  r dr    r2  r1 



2
r
r
1
1
Substituting known values: Q 
1
2
 0.5
rad
s

2
2

2
 0.12  0.10  m
3
Q  0.001100
m
s m
These two expressions are the same
with the exception of the sign.
Problem *5.36
[Difficulty: 3]
U
h
y
x
Given:
Linear velocity profile
Find:
Stream function ψ; y coordinate for half of flow
Solution:
Basic equations:
u

y

v ψ
x
ψ
and we
have
u  U 
y

h
v0
Assumption: Incompressible flow; flow in x-y plane
Check for incompressible

x

u 
y
v 0
 U y   0


x  h 
y

Flow is INCOMPRESSIBLE

Hence
Hence
x

u 
y
u  U
y
h
v 0


y
and

v0 ψ
x
Comparing these
f ( x)  0
The stream function is
ψ( x y ) 

ψ( x y )   0 dx  g ( y )

and
U y
For half the flow rate
Hence
0 0
2

y
U y
ψ( x y )   U dy 
 f (x)

h
2 h

ψ
g(y) 
U y
2
2 h
2
2 h
h
For the flow (0 < y < h)

h

U 
U h
Q   u dy    y dy 

h 0
2
0
h
hhalf
2
U h half
 half
1 U h  U h
U 

u dy   
y dy 
  
 4

2  2 
2 h
h 0
2
0
Q
2
h half 
1
2
h
2
h half 
1
2
h 
1.5 m
2
 1.06 m
Problem *5.37
Given:
[Difficulty: 3]
Rigid body motion in Example Problem 5.6
 C
V  eˆ
r
2
C  0.3
m
s
Find:
(a) Stream function for the flow field
(b) Volume flow rate per unit depth between r = 0.20 m and 0.24 m
(c) Sketch velocity profiles along a line of constant θ
(d) Check the volume flow rate calculated from the stream function by integrating the velocity profile
along this line.
Solution:
We will generate the stream function from the velocity field.
Governing
Equations:
Vr 
1 
r 
V  

r
(Definition of stream function)
Assumptions: Incompressible flow
Flow is in the r-θ plane only
Integrating the θ-component of velocity yields the stream function:
Now take the derivative of the stream function:
C
r
dr  f ( θ)  C ln( r)  f ( θ)
1 df
Vr  
 0 Therefore, f ( θ)  C1
r dθ
ψ  C ln( r)  C1
 r1 
Q  ψ r2  ψ r1  C ln r2  C1  C ln r1  C1  C ln 
 r2 
 
The volume flow rate per unit depth is:
2
Substituting in known values:

ψ  


Q  0.3
m
s
 ln
  
 
 
 
0.20 

 0.24 

3
Q  0.0547
m
s m
Because Q<0, the flow is in the direction of eθ
Along a line of constant θ, the velocity varies inversely with r:
From the velocity profile,
r
2
Q   Vθ dr 

r
1
C
Vθ 
r
Thus the flow rate is:
r
2
2 C
 r2 
m
0.24 

Substituting
known
values:
Q  0.3
 ln
dr  C ln 

r1
s
 r
 0.20 


r
3
Q  0.0547
m
s m
1
These two expressions are the same
with the exception of the sign.
Problem 5.38
[Difficulty: 2]
Given:
Find:
The velocity field provided above
Solution:
We will check this flow field against the continuity equation, and then apply the definition of acceleration
(a) the number of dimensions of the flow
(b) if this describes a possible incompressible flow
(c) the acceleration of a fluid particle at point (1,2,3)
Governing
Equations:

u    v    w    0 (Continuity equation)
y
z
t
x






V V (Particle acceleration)
V
DV
V

w
v
ap 
u
t
z
y
x
Dt
Assumptions:
(1) Incompressible flow (ρ is constant)
(2) Two dimensional flow (velocity is not a function of z)
(3) Steady flow (velocity is not a function of t)
Based on assumption (2), we may state that:
The flow is two dimensional.
Based on assumptions (1) and (3), the continuity equation reduces to:
u v

0
x y
This is the criterion against which we will check the flow field.
u  x y
1
2
v   y
3
u v
 y2  y2  0

x y
3
Based on assumptions (2) and (3), the acceleration reduces to:
This could be an incompressible flow field.



V and the partial derivatives of velocity are:
V
v
ap  u
y
x


V
V

2ˆ
ˆ
 y i  yk and
 2 xyiˆ  y 2 ˆj  xkˆ Therefore the acceleration vector is equal to:
x
y

1
1
1
2
a p  xy 2 y 2 iˆ  ykˆ  y 3 2 xyiˆ  y 2 ˆj  xkˆ  xy 4 iˆ  y 5 ˆj  xy 3 kˆ At point (1,2,3), the acceleration is:
3
3
3
3





32 ˆ 16 ˆ
16
1
 1
 2

a p    1  2 4 iˆ    2 5  ˆj    1  2 3 kˆ  iˆ 
j k
3
3
3
3
 3
 3


16
32 ˆ 16 ˆ
a p  iˆ 
j k
3
3
3
Problem 5.39
[Difficulty: 3]
Given:
Velocity field
Find:
Whether flow is incompressible; Acceleration of particle at (2,1)
Solution:

Basic equations:
x
u 

y
v 0
4
2 2
u ( x y )  A x  6  x  y  y

For incompressible flow
x

Checking
x

Hence
x
For this flow
ax  u 

x
u 

y

u 

4

y

y
2
x
4

y

4 
2


ay  4  A  y  x  y
ax  4 
x

4 
2









y

x






A 4  x y 3  4 x 3 y   A 4 x  y3  4  x3 y   A 4 x  y3  4  x3 y 
y
3
 1  1   2 m  ( 2 m) 2  ( 1  m) 2
4 3 
 m s 
2
ay  4 
y
A 4 x  y3  4  x3 y   A 4  x3  12 x y 2
A x 4  6 x 2 y2  y 4   A 4  x y 3  4 x 3 y   A x 4  6 x 2 y2  y 4 
2
Hence at (2,1)

v
2 2
2

3
ay  A x  6  x  y  y 
2

v 0
2 2
v  v
3
u
ax  4  A  x  x  y

3
v 0
ax  A x  6  x  y  y 
ay  u 

v ( x y )  A 4  x  y  4  x  y
A x4  6  x2 y 2  y4   A 4  x3  12 x y 2
u  v
2

4
 1  1   1 m  ( 2 m) 2  ( 1  m) 2
4 3 
 m s 
3
ax  62.5
m
2
s
3
ay  31.3
m
2
s
a 
2
ax  ay
2
a  69.9
m
2
s

Problem 5.40
[Difficulty: 2]
Given:
Find:
The velocity field provided above
Solution:
We will check this flow field against the continuity equation, and then apply the definition of acceleration
(a) the number of dimensions of the flow
(b) if this describes a possible incompressible flow
(c) the acceleration of a fluid particle at point (2,1,3)
Governing
Equations:

u    v    w    0 (Continuity equation)
y
z
t
x






V V (Particle acceleration)
V
DV
V

w
v
ap 
u
t
z
y
x
Dt
Assumptions:
(1) Incompressible flow (ρ is constant)
(2) Steady flow (velocity is not a function of t)
Since the velocity is a function of x, y, and z, we may state that:
Based on assumptions (1) and (2), the continuity equation reduces to:
The flow is three dimensional.
u v w


0
x y z
This is the criterion against which we will check the flow field.
2
u  a x  y
u v w


 2axy  b  2cz  0
x y z
v  b  y
w  c z
This can not be incompressible.
2
Based on assumption (2), the acceleration reduces to:




V and the partial derivatives of velocity are:
V
V
w
ap  u
v
z
y
x



V
V

V

2
 2axyiˆ
 2czkˆ Therefore the acceleration vector is equal to:
 ax iˆ  bˆj and
x
z
y

2
2
a p  ax y 2axyiˆ  by ax iˆ  bˆj  cz 2 2czkˆ  2a 2 x 3 y 2  abx 2 y iˆ  b 2 y ˆj  2c 2 z 3 kˆ




  
   

At point (2,1,3):
  2 2
  2  2

  1 2


2
2
3
2
2
3
ˆ
ˆ
a p  2   2   2 m   1 m   2   2 m   1 m i     1 m j  2  
  3 m   kˆ
m s s
  m  s 
  s 

  m  s 

m
 48iˆ  4 ˆj  54kˆ 2
s

m
a p  48iˆ  4 ˆj  54kˆ 2
s
Problem 5.41
[Difficulty: 3]
Given:
x component of velocity field
Find:
Simplest y component for incompressible flow; Acceleration of particle at (1,3)
Solution:


v ψ
x
Basic equations
u
We are given
u ( x y )  A x  10 x  y  5  x  y
y
ψ
5
3 2
4





10 3 3

5
3 2
4
5
 5
Hence for incompressible flow ψ( x y )   u dy 
 A x  10 x  y  5  x  y dy  A  x  y  3  x  y  x  y   f ( x )






10 3 3
5
5
4
2 3
5


v ( x y )   ψ x y   A  x  y 
 x  y  x  y   f ( x )  A 5  x  y  10 x  y  y  F( x )
3
x
x  


 
2 3
5
 4
  F(x)
4
2 3
5
v ( x y )  A  5  x  y  10 x  y  y 
v ( x y )  A 5  x  y  10 x  y  y
Hence
The simplest is
For this flow
5
ax  u 

x
u  v

y

2
2
ax  5  A  x  x  y
ay  u 
5

x
u
4 
3 2
ax  A x  10 x  y  5  x  y 
v  v

y


2


ay  5  A  y  x  y
ax  5 
x
2







y

4
 1  1   1 m  ( 1 m) 2  ( 3  m) 2
2 4 
 m s 
2
ay  5 

y
A 5 x 4 y  10 x2 y 3  y5   A 5 x 4 y  10 x2 y 3  y5   A 5 x 4 y  10 x2 y 3  y5 
2
Hence at (1,3)

4
4 
3 2

A x 5  10 x 3 y2  5  x y 4   A 5  x4 y  10 x 2 y3  y 5   A x5  10 x3 y 2  5 x  y4 
v
ay  A x  10 x  y  5  x  y 
2
x
2
where F(x) is an arbitrary function of x
 1  1   3 m  ( 1 m) 2  ( 3  m) 2


2 4 
 m s 
4
4m
ax  1.25  10
2
s
4
4m
ay  3.75  10
2
s
a 
2
ax  ay
2
a  3.95  10
4m
2
s
Problem 5.42
[Difficulty: 2]
Given:
Find:
The velocity field provided above
Solution:
We will check this flow field against the continuity equation, and then apply the definition of acceleration
(a) if this describes a possible incompressible flow
(b) the acceleration of a fluid particle at point (x,y) = (0.5 m, 5 mm)
(c) the slope of the streamline through that point
Governing
Equations:

u    v    w    0 (Continuity equation)
y
z
t
x






V V (Particle acceleration)
V
DV
V

w
v
ap 
u
t
z
y
x
Dt
Assumptions:
(1) Incompressible flow (ρ is constant)
(2) Two-dimensional flow (velocity is not a function of z)
(3) Steady flow (velocity is not a function of t)
Based on the assumptions above, the continuity equation reduces to:
u
A U y
1
x
2
v
A U y
2
3
4 x
2
u v
1 AUy
AUy


2 3 0
3
x y
2 2
x
4x 2
Based on assumptions (2) and (3), the acceleration reduces to:

V
3 AUy 2 ˆ
AUy
j
  3 2 iˆ 
x
2x
8x 5 2
u v

0
x y
This is the criterion against which we
will check the flow field.
This represents a possible incompressible flow field.



V and the partial derivatives of velocity are:
V
v
ap  u
y
x

V
AU
AUy

and
 1 2 iˆ  3 2 ˆj Therefore the acceleration vector is equal to:
y
2x
x

AUy  AUy ˆ 3 AUy 2 ˆ  AUy 2  AU ˆ AUy ˆ 
A 2U 2 y 2 ˆ A 2U 2 y 3 ˆ
At (5 m, 5 mm):


ap  1 2  3 2 i 
j 
i  3 2 j  
i
j
52
32  12
2
3
8x
2x
4x
4x
x  2x

 4x  x
2
2
2
2
2
3
 1  141  

m   0.005   ˆ  1  141  
m   0.005   ˆ
a p      1 2    0.240   
 i    
   0.240   
 j
s   0.5    4  m1 2  
s   0.5  
 4  m  

The slope of the streamline is given by: slope 
v
u

A U y
3
4 x
2

x
2
A U y



m
a p  2.86 10  2 iˆ  10  4 ˆj 2
s
1
y
4 x
2
Therefore, slope 
0.005
4  0.5
slope  2.50  10
3
Problem 5.43
[Difficulty: 2]
Given:
X component of a 2-dimensional transient flow.
Find:
Y component of flow and total acceleration.
Solution:
Governing
Equations:
u v w
 
0
(Continuity Equation for an Incompressible Fluid)
x y z





V
V
DV
V V

u
v
ap 
w

(Material Derivative)
x
Dt
y
z
t
Incompressible fluid
Assumptions: No motion along the wall (x = 0) limited to two dimensions (w = 0).
 2t 
u  Ax  sin 

 T 
The given or available data is:
w0
v
u
 2t 

  A  sin 

y
x
 T 
 2t 
v  Ay  sin 
C
 T 
Simplify the continuity equation to find v:
Integrate:
Use the boundary condition of no flow at the origin to solve for the constant of integration
Give the velocity in vector form:

 2t 
V  A  sin 
  xiˆ  yˆj
 T 


 2t 
v  Ay  sin 

 T 


V
V

v
Use the material derivative to find the acceleration. Start with the convective terms.
a p ,conv  u
x
y


V
V

 2t 
 2t  ˆ
 2t 
 2t  ˆ
v
 Ax  sin 
a p ,conv  u
  A  sin 
i  Ay  sin 
  A  sin 
j
x
y
 T 
 T 
 T 
 T 
 2t 
 A 2  sin 2 
  xiˆ  yˆj
 T 

Finish the local term:


 2t 
a p ,conv  A 2  sin 2 
  xiˆ  yˆj
T




V
2

 2t 
 2t 
a p ,local 
A  cos
 A  sin 
  xiˆ  yˆj 

T
t
 T 
 T 



2A  2t 

a p ,local 
cos

T
 T 
Problem 5.44
[Difficulty: 2]
Given:
Find:
The 2-dimensional, incompressible velocity field provided above
Solution:
We will check the dimensions against the function definition, check the flow field against the continuity equation,
and then apply the definition of acceleration.
(a) dimensions of the constant A
(b) simplest x-component of the velocity
(c) acceleration of a particle at (1,2)
Governing
Equations:

u    v    w    0 (Continuity equation)
y
z
t
x






V V (Particle acceleration)
V
DV
V

w
v
ap 
u
t
z
y
x
Dt
Assumptions:
(1) Incompressible flow (ρ is constant)
(2) Two-dimensional flow (velocity is not a function of z)
(3) Steady flow (velocity is not a function of t)
Since
v  A x  y it follows that A  
v
x y
and the dimensions of A are given by:
A   v   L  1  1
 xy 
Based on the assumptions above, the continuity equation reduces to:
t L L
1
Lt
v
u v
u

 0 Therefore:
  Ax  
x
x y
y

1
2
Integrating with respect to x will yield the x-component of velocity: u   A  x dx  f ( y )   A  x  f ( y )
2

The simplest x-component of velocity is obtained for f(y) = 0:
Based on assumptions (2) and (3), the acceleration reduces to:
A 
u
1
2
 A x



V and the partial derivatives of velocity are:
V
v
ap  u
y
x


V
V

 Axiˆ  Ayˆj and
  Axˆj Therefore the acceleration vector is equal to:
x
y

1
1
1
a p  Ax 2 Axiˆ  Ayˆj  Axy  Axˆj  A 2 x 3 iˆ  A 2 x 2 yˆj At (1 , 2):
2
2
2





1
 1

a p    A 2  13 iˆ    A 2  12  2  ˆj
2
 2


1
a p  A 2  iˆ 
2
ˆj 

2
Problem 5.45
[Difficulty: 2]
Given:
Velocity field
Find:
Whether flow is incompressible; expression for acceleration; evaluate acceleration along axes and along y = x
Solution:
2
The given data is
For incompressible flow
Hence, checking
A  10

x

x
m

u 
y

u 
u ( x y ) 
s
y
A x
2
x y
A y
v ( x y ) 
2
2
x y
2
v 0
v  A
x2  y2  A x2  y2  0
2
2
 x 2  y 2
x2  y2
Incompressible flow
The acceleration is given by
du
du
For the present steady, 2D flow ax  u 
 v

dx
dy


2
 A x2  y 2 
A x
A y  2  A x  y 

ax  
 

2
2
2
2
2
2
x y 
2
2
x y 
2
2 
2
2 
x y
 x y 
 x y 
A x
 


2
2
dv
dv
A x  2  A x  y 
A y  A  x  y 


ay  u 
 v

 



dx
dy
2
2
2
2
2
2 
x y   2
2 
2
2
x y
 x y  
 x  y  
2
Along the x axis
ax  
A
x
3



100
x


2
 x 2  y 2
2
2
ay  
A y
ay  0
3
2
Along the y axis
ax  0
ay  
A
y
2
Along the line x = y
ax  
A x
4

r
where
r
2
x y
3

100
y
2
100  x
ay  
4
r
A y
4

3
100  y
r
2
2
For this last case the acceleration along the line x = y is
a
2
4
r
2
A
A
100
A
100
2
2
2
2
ax  ay    x  y  

a

4
3
3
3
3
r
r
r
r
r
In each case the acceleration vector points towards the origin, proportional to 1/distance 3, so the flow field is a radial
decelerating flow.
Problem 5.46
Given:
Duct flow with incompressible, inviscid liquid
U  5
Find:
[Difficulty: 2]
m
s
L  0.3 m
u ( x )  U  1 

x


2 L 
Expression for acceleration along the centerline of the duct
We will apply the definition of acceleration to the velocity.
Solution:





Governing

V V (Particle acceleration)
V
DV
V

w
v
ap 
u
Equation:
t
z
y
x
Dt
Assumptions:
(1) Incompressible flow (ρ is constant)
(2) One-dimensional flow along centerline (u = u(x) only)
(3) Steady flow (velocity is not a function of t)
Based on assumptions (2) and (3), the acceleration reduces to:
apx  u 

x
u  U  1 
 
2
x 
U 
  U 
    2 L    2 L   1  2 L 
2 L  



x
2
apx  
U
2 L
  1 

x


2 L 
Problem 5.47
[Difficulty: 4]
6 ft
y
4 in
1 in
x
Given:
Flow in a pipe with variable diameter
Find:
Solution:
Expression for particle acceleration; Plot of velocity and acceleration along centerline
Basic equations:
Assumptions:
1) Incompressible flow
2) Uniform flow
2
Continuity reduces to
But
Hence
and for the flow rate
D  Di 
Do  Di
x
π D
4
where D i and Do are the inlet and exit diameters, and x is
distance along the pipe of length L: D(0) = D i, D(L) = Do.
2
Do  Di 


2
π Di 
 x
π Di
L


Vi
 V
4
4
V  Vi
Di
2
2
Do  Di 


Di 
 x
L


Some representative values are V( 0  ft)  3 
The acceleration is given by
L
Q  V A  V
ft
s
Vi



1 

V
 Do
 
 D  1 
 i
  x

L

L
ft
  7.68 s
2
2
Vi
V( x ) 


1 

V( L)  48
ft
s
 Do
 
 D  1 
 i
  x

L

2
2
For this flow
ax  V

x
ax 
V
2
ax ( x )  
Vi


1 

 Do
 Do
 
 D  1 
 i
  x

L


x 
 

 1 


 1
 Di

  
2
5
  Do
 Do
  
 

 1
x 
 1

  Di
 Di
  x 
  1
L 



L
L
 


Vi
  Do
x 

  Di
L 
L

 
1
  1



5
ax 
ft
Some representative values are ax ( 0  m)  2.25
2
s
L
  23.6
2
ft
2
3 ft
ax ( L)  2.30  10 
V (ft/s)
40
30
20
10
2
4
6
4
6
x (ft)
a (ft/s2)
2000
1500
1000
500
0
2
x (ft)
2
s
s
The following plots can be done in Excel
0

 1

2  Vi  
 Di
2

 Do
2  Vi  
Problem 5.48
Given:
Find:
[Difficulty: 2]
Incompressible, inviscid flow of air between parallel disks
(a) simplified version of continuity equation valid
 in this flow field
(b) show that the velocity is described by: V  V R r eˆ
r
(c) acceleration of a particle at r = ri, r = R

Solution:

We will apply the conservation of mass and the definition of acceleration to the velocity.
Governing
Equations:

1 
1 V
rVr  
V    Vz     0 (Continuity Equation)
t
r r 
r 
 z

 V

DV
 V  V 
ap 
(Particle acceleration)
Dt
t

Assumptions:

(1) Incompressible flow (ρ is constant)
(2) One-dimensional flow (velocity not a function of θ or z)
(3) Flow is only in the r-direction
(4) Steady flow (velocity is not a function of t)
Based on the above assumptions, the continuity equation reduces to:
C
should be the form of the solution. Now since at r = R:
Thus: Vr 
r
1 
 r Vr  0 or
r r
 
r Vr  C
R V  C it follows that:
R
Vr   V or:
r

V  V R r eˆr
(Q.E.D.)
Based on assumptions (2) - (4), acceleration is radial only, and that acceleration is equal to:
apr   V

2
R
R
V
  V      



r
2
R r
 r 
R
3
Therefore, at r = ri:
Therefore, at r = R:
apr   15

apr   15


apr  Vr Vr
r
2
75
1

  

s
0.075  m  25 
m
m
2
1
75
  
 
0.075  m  75 
s
3
4m
apr  8.1  10
2
s
3
apr  3  10
3m
2
s
Problem 5.49
Given:
Find:
[Difficulty: 2]
Incompressible flow between parallel plates as shown
(a) Show that the radial component of velocity is:
(b) Acceleration in the gap
Q
Vr 
2  π r h
We will apply the conservation of mass and the definition of acceleration to the velocity.
Solution:

Governing
1 
1 V
rVr  
V    Vz     0 (Continuity Equation)
Equation:
t
r r 
r 
 z

 V

DV
 V  V 
ap 
(Particle acceleration)
Dt
t

Assumptions:

(1) Incompressible flow (ρ is constant)
(2) One-dimensional flow (velocity not a function of θ or z)
(3) Flow is only in the r-direction
(4) Steady flow (velocity is not a function of t)
Based on the above assumptions, the continuity equation reduces to:
1 
 r Vr  0 or
r r
 
r Vr  C
C
should be the form of the solution. Now since the volumetric flow rate is: Q  2  π r h  Vr it follows that:
Thus: Vr 
r
Q
Vr 
2  π r h
(Q.E.D.)
Based on assumptions (2) - (4), acceleration is radial only, and that acceleration is equal to:
apr 
Q
2  π r h

Q
2
2  π r  h
 
2
  1 Therefore, the particle acceleration is:

 2 π h  r3
Q

apr  Vr Vr
r
2

 Q  1
a p  
 3 eˆr
 2h  r
Problem 5.50
[Difficulty: 4]
Given:
Data on pollution concentration
Find:
Plot of concentration; Plot of concentration over time for moving vehicle; Location and value of maximum rate
change
Solution:
D
 


u v w 
Dt
z t
y
x
Basic equation:
Assumption:
(Material Derivative)
Concentration of pollution is a function of x only
Sensor travels in x-direction only
For this case we have
uU
Hence
  
dc
Dc
d  
 u
 U A  e
Dt
v0
dx
 

c( x )  A  e
w 0
x
2 a
dx
 


e
Dt
a 
Dc
U A
a

1
2

e
2 a
 
U A 
a 
e
e
 
a 

We need to convert this to a function of time. For this motion u = U so
U t
x 
x

x
a

e
x
a

1
2

e
x
2 a




x  U t
U t 
2 a



The following plots can be done in Excel
1 10
c (ppm)
8 10
6 10
5
6
6
4 10
6
2 10
6
0
10
20
x (ft)
30
Dc/Dt (ppm/s)
4 10
4
3 10
4
2 10
4
1 10
4
 1 10
4
0
0.1
0.2
0.3
0.4
0.5
t (s)
The magnitude of the rate of change is maximized when

 
d  Dc 
d  U A 
e
   
dx  Dt  dx  a 
 
U A  1
 e
2
4
a 
x
2 a
x
a

1
2

e
x
2 a


  0

x
x

a
e
0


or
e
x max  2  a ln( 4 )  2  3  ft  ln( 4 )
tmax 
x max
U
 8.32 ft 



Dcmax
U A

e
Dt
a 
Dcmax
Dt
s

1
2
4
x max  8.32 ft
tmax  0.119  s
70 ft
xmax
a
2 a

xmax 
e
2 a



 

ft
1
5
 70  3  10  ppm 
 e
s
3  ft 
8.32
3

1
2

e
8.32 
2 3



Dcmax
Dt
 4.38  10
 5 ppm

s
Note that there is another maximum rate, at t = 0 (x = 0)
Dcmax
Dt
 70
ft
s
5
 3  10
 ppm 
1
3  ft
  1 

1

2
Dcmax
Dt
 4 ppm
 3.50  10

s
Problem 5.51
Given:
Sediment concentration fates in a river after a rainfall are:

t
Find:
[Difficulty: 2]
c  100
ppm

hr
x
c  50
ppm
mi
Stream speed is 0.5 mph, where a boat is used to survey the concentration.
The boat speed is 2.5 mph.
(a) rates of change of sediment concentration observed when boat travels
upstream, drifts with the current, or travels downstream.
(b) explain why the observed rates differ
We will apply the concept of substantial derivative
Solution:
Dc
c
c
c c
Governing
u v w 
Equation:
Dt
y
z t
x
Assumptions:
(Substantial Derivative)
(1) One-dimensional motion (velocity not a function of y or z)
(2) Steady flow (velocity is not a function of t)
Dc
c c
u 
Dt
x t
Based on the above assumptions, the substantial derivative reduces to:
u  uB
u B  0.5 mph  2.5 mph
To obtain the rates of change from the boat, we set
(i) For travel upstream, u B  u s  Vb
mi
6
10
u B  2 mph
6
6
10
Dcup  2.0
 50
 100
hr
mi
hr
(ii) For drifting, u B  u s
mi
10
Dcup  0.00
hr
u B  0.5 mph
6
10
6
6
10
Dcdrift  0.5
 50
 100
hr
mi
hr
10
Dcdrift  125.0
hr
(iii) For travel downstream, u B  u s  Vb
6
mi
10
Dcdown  3.0
 50
hr
mi
 100 
u B  0.5 mph  2.5 mph
10
6
hr
u B  3 mph
6
10
Dcdown  250 
hr
Physically the observed rates of change differ because the
observer is convected through the flow. The convective
change may add to or subtract from the local rate of change.
Problem 5.52
Given:
[Difficulty: 2]
Instruments on board an aircraft flying through a cold front show ambient temperature
dropping at 0.7 oF/min, air speed of 400 knots and 2500 ft/min rate of climb.
Rate of temperature change with respect to horizontal distance through cold front.
Find:
We will apply the concept of substantial derivative
Solution:
T
T
T T
DT
Governing
(Substantial Derivative)
w

v
u
Equation:
y
z t
x
Dt
Assumptions:
(1) Two-dimensional motion (velocity not a function of z)
(2) Steady flow (velocity is not a function of t)
(3) Temperature is constant in y direction
DT
T
u
Dt
x
Based on the above assumptions, the substantial derivative reduces to:
Finding the velocity components:
V  400 
nmi
hr

6080 ft
nmi

hr
3600 s
V  675.56
2
Therefore:
u 
ft 
ft 


 675.56 s    41.67  
s

 
So the rate of change of temperature through the cold front is:
δTx 
0.7 ∆°F
min

ft
v  2500
s
2
u  674.27
s
674.27 ft

ft
min

min
60 s
v  41.67 
ft
s
ft
s
min
60 s

5280 ft
mi
∆°F
δTx  0.0914
mi
Problem 5.53
Given:
[Difficulty: 2]
Aircraft flying north with speed of 300 mph with respect to ground, 3000 ft/min
vertical. Rate of temperature change is -3 deg F/1000 ft altitude. Ground temperature
varied 1 deg F/mile.
Rate of temperature change shown by on-board flight recorder
Find:
We will apply the concept of substantial derivative
Solution:
T
T
T T
DT
Governing
w

v
u
Equation:
y
z t
x
Dt
Assumptions:
(Substantial Derivative)
(1) Two-dimensional motion (velocity not a function of z)
(2) Steady flow (velocity is not a function of t)
DT
T
T
u
v
Dt
y
x
Based on the above assumptions, the substantial derivative reduces to:
Substituting numerical values: δT   300 

mi
hr

1  ∆°F
mi

hr
3  ∆°F 
ft
 
   3000 min  1000 ft 


60 min 
δT  14
∆°F
min
Problem 5.54
Given:
Z component of an axisymmetric transient flow.
Find:
Radial component of flow and total acceleration.
Solution:
Governing
Equations:
[Difficulty: 4]
1  rVr  1 V Vz


 0 (Continuity Equation for an Incompressible Fluid)
r r
r 
z
V V Vr V2
V V

 Vz r  r
a r , p  Vr r  
r
z
t (Particle acceleration)
r
r 
V V Vz
V V
 Vz z  z
a z , p  Vr z  
r
z
t
r 
Incompressible fluid
Assumptions: No motion along the wall (z = 0) limited to two dimensions (Vθ = 0 and all partials with respect to θ are zero).
The given or available data is:
 2t 
VZ  Az  sin 

 T 
V  0
 
0

Simplify the continuity equation to find Vr:
rVr 
V
1 rVr 
 2t 
 r   A  sin 
 z 

r
r r
z
 T 
Solve using separation of variables:
rVr  
r2A
 2t 
 sin 
C
2
 T 
Use the boundary condition of no flow at the origin to solve for the constant of integration
Find the convective terms of acceleration.
ar ,conv  Vr
Vr  
Vr
V
rA  2t 
A  2t 
 2t 
 Vz r   sin 
   sin 
  Az sin 
0
r
z
2
2  T 
 T 
 T 
ar ,conv 
a z ,conv  Vr
rA
 2t 
 sin 

2
 T 
Vz
V
rA  2t 
 2t 
 2t 
 Vz z   sin 
  0  Az  sin 
  A  sin 

r
z
2
 T 
 T 
 T 
rA 2
 2t 
sin 2 

4
 T 
 2t 
a z ,conv  zA2 sin 2 

 T 
Find the local terms:
ar ,local
V
2 rA  2t 
 r 
 cos

T
t
2
 T 
a z ,local 
Vz
2
 2t 

 Az  cos

T
t
 T 
ar ,local 
a z ,local 
 rA  2t 
cos

T
 T 
2zA
 2t 
cos

T
 T 
Problem 5.55
Given:
Find:
Solution:
[Difficulty: 3]
Definition of "del" operator
an expression for the convective acceleration for a fluid particle.
We will directly substitute the velocity vector into the expression.
Governing
Equation:
Assumptions:
 ˆ  ˆ  ˆ
i
j k
z
y
x

V  uiˆ  vˆj  wkˆ

("del" operator in rectangular coordinates)
(velocity vector)
None.
Directly substituting we get:
V   V  uiˆ  vˆj  wkˆ   x iˆ  y ˆj  z kˆ uiˆ  vˆj  wkˆ 






 


  u  v  w  uiˆ  vˆj  wkˆ
y
z 
 x
 u
w
w 
u
u   v
v
v   w
v
 w kˆ
  u
v
 w iˆ   u  v  w  ˆj   u
y
z 
y
z   x
y
z   x
 x
The components of this vector are the x-, y-, and z-components of the convective acceleration:
 u
u
u  u
v
 w  
a xp   u
y
z  t
 x
 v
v
v  v
a yp   u  v  w  
y
z  t
 x
 w
w
w  w
v
 w  
a zp   u
y
z  t
 x
Problem 5.56
Given:
Find:
[Difficulty: 3]
Steady, two-dimensional velocity field represented above
(a) proof that streamlines are hyperbolas (xy = C)
(b) acceleration of a particle in this field
(c) acceleration of particles at (x,y) = (1/2m, 2m), (1m,1m), and (2m, 1/2m)
(d) plot streamlines corresponding to C = 0, 1, and 2 m 2 and show accelerations
We will apply the acceleration definition, and determine the streamline slope.
Solution:





Governing

V V (Particle acceleration)
V
DV
V

w
v
ap 
u
Equations:
t
z
y
x
Dt
Assumptions:
(1) Two-dimensional flow (velocity is not a function of z)
(2) Incompressible flow
Streamlines along the x-y plane are defined by
dy
dx

v
u

A  y
dx
Thus:
Ax
x

dy
y
0
After integrating: ln( x)  ln( y )  ln( C) which yields:
x y  C (Q.E.D.)



V

V

Based on the above assumptions the particle acceleration reduces to:
Substituting in the field:
v
ap  u
y
x


a p   Ax Aiˆ   Ay  A ˆj  A 2 xiˆ  yˆj which simplifies to a p  A 2 xiˆ  yˆj




m
a p  0.5iˆ  2 ˆj 2
s

At (x,y) = (1m, 1m) a p


m
 iˆ  ˆj 2
s


At (x,y) = (2m, 0.5m) a p


5
Here is the plot of the streamlines:
4
(When C = 0 the streamline is on
the x- and y-axes.)
3
2
1
0
0
1
2
3
X (m)

m
 2iˆ  0.5 ˆj 2
s
Y (m)
At (x,y) = (0.5m, 2m)

4
5
Problem 5.57
Given:
Find:
Solution:
[Difficulty: 3]
Steady, two-dimensional velocity field represented above
(a) general acceleration of a particle in this field
(b) acceleration of particles at (x,y) = (0m, 4/3m), (1m,2m), and (2m, 4m)
(c) plot streamlines with acceleration vectors
We will apply the acceleration definition, and determine the streamline slope.






V V (Particle acceleration)
V
V
DV

w
v
ap 
u
t
z
y
x
Dt
Governing
Equations:
Assumptions:
(1) Two-dimensional flow (velocity is not a function of z)
(2) Incompressible flow
Based on the above assumptions the particle acceleration reduces to:

 


a p   Ax  B Aiˆ   Ay  A ˆj  A 2 x  AB iˆ  A 2 y ˆj


At (x,y) = (0m, 4/3m)

m
a p   0.12iˆ  0.0533 ˆj 2
s
At (x,y) = (2m, 0.5m)

m
a p   0.04iˆ  0.160 ˆj 2
s

dy
dx
1
A
 ln( y ) 
1
A

v
u

A y
A x  B
dx
Thus:
A x  B
 ln( C) which yields:



m
a p   0.08iˆ  0.0800 ˆj 2
s

dy
A y
0
After integrating:
6
5
Here is the plot of the streamlines:
4
( A x  B)  y  C
3
2
Y (m)
A
 ln( A x  B) 
At (x,y) = (1m, 1m)

Streamlines along the x-y plane are defined by
1



V Substituting in the field:
V
v
ap  u
y
x

a p  A 2 x  AB iˆ  A 2 y  ˆj
1
0
-1
-2
-3
-4
-5
-6
-3 -2 -1
0
1
2
3
X (m)
4
5
6
7
8
9
Problem 5.58
Given:
Find:
[Difficulty: 3]
Velocity field represented above
(a) the proper value for C if the flow field is incompressible
(b) acceleration of a particle at (x,y) = (3m,2m)
(c) sketch the streamlines in the x-y plane
Solution:
We will check the velocity field against the continuity equation, apply the acceleration definition, and
determine the streamline slope.

u    v    w    0 (Continuity equation)
z
t
y
x






V V (Particle acceleration)
V
DV
V

w
v
ap 
u
t
z
y
x
Dt
Governing
Equations:
Assumptions:
(1) Two-dimensional flow (velocity is not a function of z)
(2) Incompressible flow
Based on the above assumptions the continuity equation reduces to:
The partial derivatives are:

x
u A
and

y
v C
x
u 

y
Thus from continuity:
Based on the above assumptions the particle acceleration reduces to:




a p   Ax  B Aiˆ  Cy Cˆj  Dkˆ  A 2 x  AB iˆ  C 2 yˆj  Dkˆ
v  0 This is the criterion to check the velocity.
A  C  0 or




V V
V

ap  u
v
y t
x
dy
dx

v
u

C y
A x  B
At (x,y) = (3m, 2m)

Thus:
dx
1 dy
 

or
A x  B
A y

B
  constant
A
Here is a plot of the streamlines
passing through (3, 2):
dx
x
B

dy
y
0
A
3
Y (m)
y   x 


m
a p  4iˆ  8 ˆj  5kˆ 2
s
B
Solving this ODE by integrating: ln x    ln( y )  const
A

Therefore:
C  2  s
Substituting in the field:
2
 2  2

m
mˆ  2 
2

4





a p   3 m
  2 mˆj  5 2 kˆ
i 
s   s 
s
s
 s 
Streamlines along the x-y plane are defined by
1
C  A
2
1
0
0
1
2
X (m)
3
4
Problem 5.59
Given:
Find:
[Difficulty: 3]
Linear approximate profile for two-dimensional boundary layer
(a) x- and y-components of acceleration of a fluid particle
(b) locate the maximum values of acceleration
(c) compute ratio of maximum acceleration components
We will apply the acceleration definition.
Solution:





Governing

V V (Particle acceleration)
V
V
DV

w
v
ap 
u
Equation:
t
z
y
x
Dt
(1) Two-dimensional flow (velocity is not a function of z)
Assumptions:
(2) Incompressible flow
(3) Steady flow
Based on the above assumptions the particle acceleration reduces to:



V The velocities and derivatives are:
V
v
ap  u
y
x
1
u
U y
v
δ
u y
4 x

U y
2
4  δ x
δ  c x

2
x
u 
 U y    d δ    U y  δ   U y

 
2 2 x
2  δ x
δ  δ   dx 
δ

2
2
2
 U y2    U y2   d 
U y
U y
3  U y
δ
 
   δ  



v  
2
2
2
2 x
x
x  4  δ x  δ  4  δ x   dx 
4  δ x
4 δ  x
8  δ x


apx  u 
apy  u 

x

x
u  v
v  v

y

y
2
y
v 
U y
2  δ x
y
u 
U
δ
So the accelerations are:
2 2
2
u 
U y U
U y
U y U y


 
4  δ x δ
4  δ x
δ 2  δ x
v 
U y
U y
U y 3  U y
U y




2
4  δ x 2  δ x
2 2
δ
8  δ x
4 δ  x
2


2
apx  
3
y
y
2
4 x
2
apy  
 
U
U
4 x
 
When x = 0.5 m and δ = 5 mm: ratio 
0.5 m
0.005  m
apxmax  
U
4 x

δ
 
δ
2
The maximum values are when y = δ:
2
apymax
U 4 x δ
δ
The ratio of the accelerations is:


 
4 x 2 x
apxmax
x
U
y
x
2
apymax  
2
U

δ
4 x x
ratio  100
Problem 5.60
[Difficulty: 4]
U
y
x
Given:
Flow in boundary layer
Find:
Expression for particle acceleration a x; Plot acceleration and find maximum at x = 0.8 m
Solution:
Basic equations
u
U
 2  
y
δ

We need to evaluate
ax  u 
First, substitute
λ( x y ) 
Then

x

x

x
Collecting terms
To find the maximum
u 
x
y
δ
 
u  v

y
2
v
U

1 y
1 y
        
x 2  δ  3  δ 
δ
3
δ  c x


u
y
u
so
δ( x )
U
 2 λ  λ
v
2
U
y dδ
du dλ

 U ( 2  2  λ)     

2  dx
dλ dx
 δ 
u  U ( 2  2  λ)   

u  U ( 2  2  λ) 

δ

dδ
dx
x

1
1 3
1
   λ   λ 
3
2


1
2

2
 c x

1
1
λ 1
λ  1
2
2
  c x
 U ( 2  2  λ)   
   c x

δ 2
1 2

λ
2 x


U λ  λ
2
 c x 2 

x


2
2
  2 U   y   y    2 U λ  λ
 
δ
2
y
δ δ  δ  
y
δ 

2
2 


1
1
2 U λ  λ 


  U δ    λ   λ3   2 U λ  λ 
ax  u  u  v  u  U 2  λ  λ 
3
x
x 2
y
x
y




2
2
3
4
2
4 y
1 y 
U  2 4 3 1 4 U   y 
  λ   λ   λ  
           
ax 
3 δ
3 δ 
x 
3
3
 x  δ

Hence


u  U 
2
 2
y
 

dax
dλ
2
0
The solution of this quadratic (λ < 1) is
U
x
  2  λ  4  λ 
2

λ 
3
2
3
4
3
λ
3



or
λ  0.634

1  2  λ 
y
δ
 0.634
2
3
2
λ  0

2
At λ = 0.634
ax 
U
x
  0.634 
2

ax  0.116   6 

4
3
m
s
2
3
 0.634 
1
3
 0.634
4


2
 0.116 
1
 
0.8 m

U
x
ax  5.22
m
2
s
The following plot can be done in Excel
1
0.9
0.8
0.7
y/d
0.6
0.5
0.4
0.3
0.2
0.1
6
5
4
3
a (m/s2)
2
1
0
Problem 5.61
Given:
Find:
Solution:
[Difficulty: 3]
Steady, two-dimensional velocity field represented above
(a) prove velocity field represents a possible incompressible flow field
(b) expression for the streamline at t = 1.5 s
(c) plot of the streamline through (x,y) = (2m,4m) at that instant
(d) local velocity vector
(e) vectors representing local, convective, and total accelerations
We will apply the acceleration definition, and determine the streamline slope.






V V (Particle acceleration)
V
DV
V

w
v
ap 
u
t
z
y
x
Dt

u    v    w    0 (Continuity equation)
y
x
z
t
Governing
Equations:
Assumptions:
(1) Two-dimensional flow (velocity is not a function of z)
(2) Incompressible flow
Based on the two assumptions listed above, the continuity equation reduces to:

x
u 

y
v 0
This is the criterion against which we will check all of the flow fields.
u  ( 2  cos ( ω t) )  ax

v  ( 2  cos ( ω t) )  ay
x
u 

y
v  ( 2  cos ( ω t) )  ( a  a )  0
Streamlines along the x-y plane are defined by
dy
dx
ln( x)  ln( y )  ln( C)
which yields: x y  C
At (x,y,t) = (2m,4m,1.5s)

v
u

a  y  ( 2  cos ( ω t) )
a  x ( 2  cos ( ω t) )
At (x,y) = (2m,4m)

y
Thus:
dx
x
C  2 m  4 m
x
2


dy
y
0
After integrating:
2
C  8m
x y  8 m (plot shown below)



m
V  10iˆ  20 ˆj
s
 5

 2

V  2 miˆ  4 mˆj  2  cos
 1.5 s  
s
 s



Based on the above assumptions the particle acceleration reduces to:
The local acceleration is:
This could be an incompressible flow field.




V V
V

ap  u
v
y t
x



V
5 2
 2

 a xiˆ  yˆj  sin t   
 1.5 s  
2 miˆ  4 mˆj   sin 
a p ,local 
t
s s
 s









m
a p ,local  0iˆ  0 ˆj 2
s
The convective acceleration is:



V
V
m
5  m
 a uiˆ  vˆj 2  cos t    10 iˆ  20
v
a p ,conv  u
y
x
s
s  s

ˆj  2  cos 2  1.5 s  



 s





m
a p ,conv  50iˆ  100 ˆj 2
s



m
a p  50iˆ  100 ˆj 2
s
The total acceleration is the sum of the two acceleration terms:
10
Here is the plot of the streamline
and the vectors:
8
6
Y (m)
Acceleration
4
2
Velocity
0
0
2
4
6
X (m)
8
10
Problem 5.62
Given:
Find:
[Difficulty: 3]
Sinusoidal profile for two-dimensional boundary layer
(a) x- and y-components of acceleration of a fluid particle
(b) plot components as functions of y/δ for U = 20 ft/s, x = 3 ft, δ = 0.04 in
(c) maximum values of acceleration at this x location
We will apply the acceleration definition.
Solution:
Governing
Equation:
Assumptions:
(Particle acceleration)
(1) Two-dimensional flow (velocity is not a function of z)
(2) Incompressible flow
(3) Steady flow



V
V
v
ap  u
y
x
Based on the above assumptions the particle acceleration reduces to:
1
π y
η    η( x y )
2 δ

y

x
η 
η 
The velocities and derivatives are:
π
δ  c x
2 δ
u  U sin( η)

x
y

x
U δ
  ( cos( η)  η sin( η)  1 )
π x
v 
1
1
δ
2
d
δ   c x

2
2 x
dx
2
π y δ
π y
  d 



 η    δ  
2 2 x
4 x δ
δ   dx 
2 δ

v

To make this easier, define η:
u 
u 

u

η x

u

η y
η  U cos( η)  
η  U cos( η) 
π

y
4 x δ
π
2 δ


U π
2 δ
U
2 x
 η cos( η)
 cos( η)
(Eqn. 1)
(Eqn. 2)
We find the derivatives of v using product and chain rules:
1 δ
δ 
π y
δ
Simplifying this expression:


  
 ( cos( η)  η sin( η)  1 )   ( sin( η)  sin( η)  η cos( η) )  



x 2 x
2
4 x δ
π
x
x 


U

x
v 
U δ
2  π x
 ( cos( η)  η sin( η)  1 )  η  cos( η)
2
2
(Eqn. 3)

y
v 

v

η y
η 
π
U
U δ
  ( sin( η)  sin( η)  η cos( η) ) 

 η cos( η)
2 δ
2 x
π x
(Eqn. 4)
So the accelerations are:
apx  u 

x
u  v

y
u  U sin( η)  
U
2 x
 η cos( η) 
U δ
U π
  ( cos( η)  η sin( η)  1 ) 
 cos( η)
π x
2 δ
Simplifying this expression:
2
apx 
apy  u 

x
v  v

y
v  U sin( η)  
U δ
2  π x
2

2
apy 
Simplifying this expression:

2
 cos( η)  η sin( η)  1  η  cos( η) 
U δ
2  π x
2
U
2 x
 cos( η)  ( cos( η)  1 )
U δ
U
  ( cos( η)  η sin( η)  1 ) 
 η cos( η)
π x
2 x


 η cos( η)  ( cos( η)  η sin( η)  1 )  sin( η)   1  η  cos( η)  η sin( η)  1
2
Here are the plots of the acceleration components:
X-component of Acceleration in Boundary Layer
Y-component of Acceleration in Boundary Layer
1
Non Dimensional Height (y/delta)
Non Dimensional Height (y/delta)
1
0.5
0
 20
 15
 10
5
0
0.5
0
 0.02
 0.01
X-acceleration (ft/s^2)
The maximum values and their locations may be found using Excel or Mathcad:
0
Y-acceleration (ft/s^2)
apxmax  16.7
ft
2
apymax  0.0178
s
y
δ
 0.667
ft
2
s
y
δ
 0.839
Problem 5.63
Given:
Find:
[Difficulty: 3]
Flow between parallel disks through porous surface
(a) show that V r = vor/2h
(b) expression for the z-component of velocity (v o<<V)
(c) expression for acceleration of fluid particle in the gap
We will apply the continuity equation to the control volume shown:
Solution:
 

Governing
0


d
V

V

  dA
(Continuity)
Equations:
t CV
CS



 V

DV
 V  V 
ap 
(Particle Accleration)
Dt
t

Assumptions:

(1) Steady flow
(2) Incompressible flow
(3) Uniform flow at every section
(4) Velocity in θ-direction is zero
Based on the above assumptions the continuity equation reduces to:
1 

 r Vr  Vz  0
r r
z
 
We apply the differential form of continuity to find Vz :


Vz  


r
2
0  ρ v 0  π r  ρ Vr 2  π r h Solving for Vr: Vr  v 0 
2 h
v0
z
dz  f ( r)  v 0   f ( r) Now at z = 0: Vz  v 0
h
h
v0
1 

 r Vr 
  Vz
r r
h
z
 
Therefore we can solve for f(r):
0
v 0  v 0   f ( r) f ( r)  v 0
h
So we find that the z-component of velocity is:
Based on the above assumptions the particle acceleration reduces to:
v0
Vr 
2 h
r


z
Vr  0

r
Vz  0
v0
Vz  
h
z



V
V
a p  Vr
 Vz
z
r
Therefore:
z
Vz  v 0   1  
h


So the accelerations are:
v0
apr  Vr Vr  Vz Vr  v 0 

 v 0   1 
2

h
2

h
r
z



r
apz  Vr Vz  Vz Vz  v 0 
 0  v 0   1 
2 h
r
z



r
2
z
 0
apr 
h
z
v0
 
h h
apz 
v0
h
2
 
z
h
v0  r
4 h
2
 1

Problem 5.64
Given:
Find:
[Difficulty: 3]
Steady inviscid flow over a circular cylinder of radius R
Solution:
Governing
Equation:
(a) Expression for acceleration of particle moving along θ = π
(b) Expression for accleeration of particle moving along r = R
(c) Locations at which accelerations in r- and θ- directions reach maximum and minimum values
(d) Plot a r as a function of R/r for θ = π and as a function of θ for r = R
(e) Plot aθ as a function of θ for r = R
We will apply the particle acceleration definition to the velocity field



 V

DV
 V  V 
ap 
Dt
t

Assumptions:

(Particle Accleration)
(1) Steady flow
(2) Inviscid flow
(3) No flow in z-direction, velocity is not a function of z
Based on the above assumptions the particle acceleration reduces to:



V V V

a p  Vr
r
r 
2
Vθ
Vr Vθ


apθ  Vr Vθ 
 Vθ 
r
r θ
r
Vθ
Vθ

apr  Vr Vr 
 Vr 
r
r θ
r


Vr  U 1 

When θ = π:
2
 R   V  0
r
θ
 

So the accelerations are:
apr  U 1 


R
2
R
2
 2  U
Vr  U 2  
3
3
r
r
r
2
2
2
3
 R    2  U R  2 U   R   1 
r
 
R r 
3
 
r

θ
and the components are:
Vr  0

r
2
 R  
 
r

Vθ  0
θ
2
apr 
2 U
R
Vθ  0
 
R
3

 1 

r 
2
 R  
 
r
apθ  0
To find the maximum acceleration, we take the derivative of the accleration and set it to zero: Let
2


2


2 U
2 U  2
2
3
4
2
d
 3  η  1  η  η  2  η 
5  η  3η  0 Therefore:
apr 
R
R
dη
3
2
  1   1  



R  1.291  
 1.291  
2
The maximum acceleration would then be:
aprmax 
η 
2 U
 
1
3
5
η
or
R
r
r  1.291  R
2
aprmax  0.372 
U
R
When r = R:

Vr  0 Vθ  2  U sin( θ)
So the accelerations are: apr  
apθ 
r
( 2  U cos( θ) )
2
R

θ
Vr  0
2

R
2  U sin( θ)
Vr  0
4 U
R
 ( sin( θ) )

r

Vθ  0
θ
Vθ  2  U cos( θ)
2
2
apr  
2
 2  U cos( θ) 
4 U
R
4 U
R
 ( sin( θ) )
2
 sin( θ)  cos( θ)
4 U
apθ 
R
 sin( θ)  cos( θ)
Radial acceleration is minimum at θ  180  deg
2
armin  4 
U
R
aθ  0
Acceleration along Stagnation Streamline
0.4
Azimuthal acceleration is maximum at θ  45 deg
2
Accelerations at this angle are:
ar  2 
U
R
2
aθmax  2 
U
R
Azimuthal acceleration is minimum at θ  135  deg
2
Accelerations at this angle are:
ar  2 
U
R
Radial Acceleration (ar*R/U^2)
Accelerations at this angle are:
0.3
0.2
0.1
2
aθmin  2 
U
0
R
0
1
2
3
4
Radius Ratio (r/R)
Acceleration along Cylinder Surface
Radial and Azimuthal Accelerations (a*R/U^2)
The plots of acceleration along the stagnation streamline and
the cylinder surface are shown here. In all cases the
accelerations have been normalized by U2/R
2
2
Radial
Azimuthal
0
2
4
0
50
100
150
Azimuthal Position along Surface (deg)
5
Problem 5.65
Given:
Find:
[Difficulty: 3]
Air flow through porous surface into narrow gap
(a) show that u(x) = v ox/h
(b) expression for the y-component of velocity
(c) expression for acceleration of fluid particle in the gap
We will apply the continuity equation to the control volume shown:
Solution:
 

Governing
0


d
V

V

  dA
(Continuity)
Equations:
t CV
CS






V V
V
DV
V

w
v
ap 
u
(Particle Accleration)
t
z
y
x
Dt
Assumptions:
(1) Steady flow
(2) Incompressible flow
(3) Uniform flow at every section
Based on the above assumptions the continuity equation reduces to:
We apply the differential form of continuity to find v:


v  


v0
y
dy  f ( x )  v 0   f ( x ) Now at y = 0:
h
h

x
u 

y
v 0

x
u 
v0

  v Therefore the y-velocity v is:
h
y
0
v  v 0 Therefore we can solve for f(x): v 0  v 0   f ( x ) f ( x )  v 0
h
So we find that the y-component of velocity is:
Based on the above assumptions the particle acceleration reduces to:

x
u 
v0

h
y
u 0

x
v 0

y
v 
v0
x
u( x)  v0
h
0  x  w v 0  h  w u ( x ) Solving for u:
y
v  v 0   1  
h




V
V
v
ap  u
y
x
h
So the accelerations are:
2
x v0
y
 v 0   1    0
apx  u  u  v  u  v 0  
h
h
h
x
y



apy  u  v  v  v  v 0   0  v 0   1 
h
x
y



x
The acceleration vector would be:
y
v0
 
h h
apx 
v0  x
h
apy 
v0
h
2
 
y
h
2
 1

v2  x

y  
a p  0  iˆ    1 ˆj 
h h
h  
Problem 5.66
Given:
Velocity field and nozzle geometry
Find:
Acceleration along centerline; plot
[Difficulty: 3]
Solution:
Assumption: Incompressible flow
A0  5  ft
The given data is
2
L  20 ft
b  0.2 ft
u ( xt) 
A0
A ( x)
ft
rad
U0  20
ω  0.16
s
s
A ( x)  A0  ( 1  b  x)
u ( x)  A ( x)  U0 Ao
The velocity on the centerline is obtained from continuity
so
1
 U0 ( 0.5  0.5 cos ( ω t) ) 
U0
( 1  b  x)
 ( 0.5  0.5 cos ( ω t) )
The acceleration is given by
For the present 1D flow

t
u  u

x
u 
0.5 U0  ω sin( ω t)
1  b x

U0
( 1  b x)
 U0  b ( 0.5 cos( ω t)  0.5)
2


( 1  b x)


 ( 0.5  0.5 cos( ω t) )  
 U0 b  ( 0.5 cos( ω t)  0.5) 
2


( 1  b x)



( 1  b x) 

U0
 ( 0.5 ω sin( ω t) )  ( 0.5  0.5 cos( ω t) )  
The plot is shown here:
Acceleration in a Nozzle
40
t=0s
35
t = 10 s
2
Acceleration a x (ft/s )
ax 
ax 
30
t = 20 s
25
t = 30 s
20
15
10
5
0
0
5
10
x (ft)
15
20
Problem 5.67
[Difficulty: 4]
Given:
Find:
Steady, two-dimensional velocity field of Problem 5.56
Solution:
We will apply the particle acceleration definition to the velocity field
(a) expressions for particle coordinates, x p = f1(t) and yp = f2(t)
(b) Time requires for particle to travel from (0.5, 2) to (1, 1) and (2, 0.5)
(c) compare acceleration determined from f1(t) and f2(t) to those found in Problem 5.56



 V

DV
 V  V 
ap 
Dt
t

Governing
Equation:
Assumptions:

(Particle Accleration)
(1) Incompressible flow
(2) Two-dimensional flow
(3) Steady flow
For the given flow, u  A x and v  A y Thus
up 
d
f1  A x p  A f1 or
dt
f
1 1
 f1 

df1  ln   A t
 f1
 x0 
x
Solving for f1 yields:
f1 ( t)  x 0  e
df1
f1
 A dt Integrating from x 0 to f1 yields:
A t
0
Similarly, we can find: v p  d f2  A y p  A f2 or
dt
df2
f2
 A dt Integrating from y 0 to f2 yields:
f2 ( t)  y 0  e
 A t
1
In this problem, x 0   mand y 0  2  m Knowing the final position, we can solve for the time required.
2
To reach (1, 1):
x  1.0 m t  ln( 2 )  1 s
t  0.693 s
y  1.0 m t  ln
1
t  0.693 s
t  1.386 s
y  0.5 m t  ln
1
t  1.386 s
  1 s t  0.693 s
2
To reach (2, 0.5):
x  2.0 m t  ln( 4 )  1 s
  1 s t  1.386 s
4
The acceleration components are:
ax 
d
2
dt
2 A t
f  x0 A  e
2 1
2
 A  f1 ( t) ay 
d
2
dt
2  A t
f  x 0  ( A)  e
2 2
2
 A  f2 ( t)
At (x, y) = (1, 1):
At (x, y) = (2, 0.5):



m
a p  iˆ  ˆj 2
s



m
a p  2iˆ  0.5 ˆj 2
s
These results are identical to the accelerations calculated in Problem 5.56.
Problem 5.68
[Difficulty: 3]
Given:
Find:
One-dimensional, incompressible flow through circular channel.
Solution:
We will apply the particle acceleration definition to the velocity field
(a) the acceleration of a particle at the channel exit
(b) plot as a function of time for a compleye cycle.
(c) plot acceleration if channel is constant area
(d) explain difference between the two acceleration cases



 V

DV
 V  V 
ap 
Dt
t

Governing
Equations:
0
Assumptions:

(Particle Accleration)
 

dV   V  dA

t CV
CS
(Continuity equation)
(1) Incompressible flow
(2) One-dimensional flow
Based on the above assumptions the continuity equation can provide the velocity at any location:
Now based on the geometry of the channel we can write
R1
u  U
2
x

 R1  ∆R L 


2

U0  U1 sin( ω t) 
 ∆R  x 
1  R   L 
1



 R1 
u  U


A
 r 
A1
2

x
x
r  R1  R1  R2   R1  ∆R Therefore the flow speed is:
L
L
Based on the above assumptions the particle acceleration reduces to:
2

 u u  ˆ Substituting the velocity and derivatives into this expression we can get the acceleration in the x-direction:
 i
a p  u
 x t 
ax 
ax 
U0  U1 sin(ω t) U0  U1 sin(ω t)
 ∆R  x 
1  R   L 
1


2  ∆R
R1 L

2

1  ∆R   x 
 

R1  L 


U0  U1 sin( ω t) 2
1  ∆R   x 
 

R1  L 


5

3
  ω U1  cos( ω t)

 R1  L   ∆R  x  2
1  R   L 
1


 ( 2 )   
ω U1  cos( ω t)
1  ∆R   x 
 

R1  L 


∆R
When we simplify this expression we get:
Now we substitute the given values into this expression we get:
2
1
0.2 m


ax  32  20  2  sin 0.3
 

2
2 m
1
 ( 20  2  sin( ω t) ) 
1 m
rad
s
2
s
2
 t   2.4 cos 0.3


1

 1  0.1 m  1


0.2 m


 m
s  2
s
rad
 t 
Here is a plot of the acceleration versus time.
For a constant area channel, ΔR = 0 and the acceleration becomes:
ax   0.6 cos 0.3


rad
s
 t  
m
5
 0.3
2 10
Acceleration (m/s^2)
ax  2  0.1 m 
rad
s
 2
m
s
 cos( ω t) 
1
2
 1  0.1 m  1


0.2 m


Acceleration in Converging Channel
4
4
1.5 10
1 10
4
5 10
3
  s2
0
0
10
20
Time (s)
The plot of that acceleration is shown below.
The acceleration is so much larger for the converging channel
than in the constant area channel because the convective
acceleration is generated by the converging channel - the
constant area channel has only local acceleration.
Acceleration in Constant-Area Channel
Acceleration (m/s^2)
1
0.5
0
 0.5
1
0
10
Time (s)
20
Problem 5.69
Given:
Velocity components
Find:
Which flow fields are irrotational
[Difficulty: 2]
Solution:

For a 2D field, the irrotationality the test is
a)
2
u ( x y t)  2  x  y
Hence
b)

y

x
v 

y
x
v 

y

x
v 

y


x
2
2
v ( x y t)  3  x  y  2  y

y
u ( x y t)  2  y
Not irrotational
2
v ( x y t)  2  x  y  y  x
2

x
v ( x y t)  2  x  2  y

y
2

x
v ( x y t)  t
2

y
u 0
u 0
u ( x y t)  2  x  1
Not irrotational
v ( x y t)  x  t  y  t
u ( x y t)  ( x  2  y )  x  t
Hence
2
u 0
u 0
u ( x y t)  x  t  2  y

y
u 0
2
Hence
d)
x
v 
3

v ( x y t)  x  x  y  2  y
u ( x y t)  2  x  y  x  y
Hence
c)

2
x
v 
u ( x y t)  2
Not irrotational
v ( x y t)  ( 2  x  y )  y  t

x
v ( x y t)  2  t y

y
u ( x y t)  2  t x
Not irrotational
Problem 5.70
[Difficulty: 4]
Given:
Find:
Definition of "del" operator in cylindrical coordinates, velocity vector
Solution:
We will apply the velocity field to the del operator and simplify.


  
(a) An expression for V   V in cylindrical coordinates.
(b) Show result is identical to Equations 5.12.
Governing
Equations:


1 

  eˆr
 eˆ
 kˆ
r
z
r 

V  Vr eˆr  V eˆ  V z kˆ
V   V
  
V  V  V eˆ
  
r r

(Velocity flow field)
eˆ
 eˆr

eˆr
 eˆ

Substituting
(Definition of "del" operator)
(Hints from footnote)
using the governing equations yields:


1 
 
 
 kˆ  Vr eˆr  V eˆ  V z kˆ
 eˆ
 V eˆ  V z kˆ   eˆr
r 
z 
 r




  V 
  Vr

 V z  Vr eˆr  V eˆ  V z kˆ
z 
 r r 
V 


Vr eˆr  V eˆ  V z kˆ  V z
Vr eˆr  V eˆ  V z kˆ
Vr eˆr  V eˆ  V z kˆ  
 Vr
r 
r
z
V
V




Vr eˆr     V eˆ   V  V z kˆ  Vz  Vr eˆr
 Vr Vr eˆr  Vr V eˆ  Vr V z kˆ  
r
r
r
z
r 
r 
r 


 V z V eˆ  V z V z kˆ
z
z






Applying the product rule to isolate derivatives of the unit vectors:
V   V  V
  
V eˆ
V V
V eˆ
V V z ˆ
V
V z ˆ V Vr
Vr
k
eˆr  Vr  eˆ  Vr
eˆr   r Vr  
eˆ    V  
k
r 
r 
r 
r 
r 
r
r
r
V
V
V z ˆ
 V z r eˆr  V z  eˆ  V z
k
z
z
z
r
Collecting terms:
V   V  V
V
V V VrV
Vr V Vr V2
V 
 V


 V z r eˆr  Vr   

 Vz 
r
r
z 
r
z
r
r 
r
r 


V 
 V z V V z
  Vr

 V z z kˆ
r
z 
r 


eˆ

The three terms in parentheses are the three components of convective acceleration given in Equations 5.12.
Problem 5.71
[Difficulty: 4]
Given:
Sinusoidal approximation to boundary-layer velocity profile:
π y
u  U sin   where δ  5  mm at x  0.5 m
 2 δ
m
Neglect the vertical component of velocity. U  0.5
s
Find:
(a) Circulation about a contour bounded by x = 0.4 m, x = 0.6 m, y = 0, and y = 8 mm.
(b) Result if evaluated Δx = 0.2 m further downstream
Solution:
We will apply the definition of circulation to the given velocity field.
 
   V  ds
Governing
Equation:
(Definition of circulation)
From the definition of circulation we break up the integral:
 
 
 
 
   V  ds   V  ds   V  ds   V  ds
ab
x
bc
cd
 d
Γ   U dx  U x c  x d
x

da

Γ  0.5
c
At the downstream location, since
m
s
Since the velocity is zero over ab, and since
the velocity and path are perpendicular over bc
and da:
2
 ( 0.6 m  0.4 m)
1
1
δ  c x
2
Γ  0.1
δ'  δ 
x

 x' 
2
1
δ'  5  mm 
0.8 m 

 0.5 m 
m
s
2
δ'  6.325  mm
Now since the boundary layer is less than 8 mm thick at point c', the integral along c'c will be the same as that along cd.
Γbb'c'c  Γabcd
Problem 5.72
Given:
Find:
Solution:
[Difficulty: 2]
Velocity field for flow in a rectangular corner as in Example 5.8.
Circulation about the unit square shown above.
We will apply the definition of circulation to the given velocity field.
 
   V  ds
Governing
Equation:
(Definition of circulation)
From the definition of circulation we break up the integral:
 
 
 
 
   V  ds   V  ds   V  ds   V  ds
ab


bc
cd
da

 
The integrand is equal to: V  ds  Axiˆ  Ayˆj  dxiˆ  dyˆj  Axdx  Aydy Therefore, the circulation is equal to:
x
y
x
y
a
d
c
b
 a
 b
 c
 d
A
2
2
2
2
2
2
2
2
Γ   A x dx   A y dy   A x dx   A y dy    x d  x a    y c  y d    x b  x c    y a  y b 









2
y
x
y
x
Γ 
1
2
 0.3
1
s


 
 
 

 22  1 2  22  1 2  12  2 2  12  2 2   m2
2
Γ 0
m
s
This result is to be expected since the flow is irrotational
and by Stokes' theorem, the circulation is equal to the curl
of the velocity over the bounded area (Eqn. 5.18).
Problem 5.73
Given:
Flow field
Find:
If the flow is incompressible and irrotational
[Difficulty: 3]
Solution:

Basic equations: Incompressibility
a)
7
x
b)

x
y
3 4
6
4 2
u 

y

x
6
2 4
v 0
5 2
3 4
5
3 3
v ( x y )  42 x  y  140  x  y  42 x  y
v 

y
6
5
u 0
Note that if we define

Irrotationality
x
6
v 

y
u 0
4 3
2 5
v ( x y )  7  x  y  35 x  y  21 x  y  y
6

y
6
4 2
7
2 4
v ( x y )  7  x  105  x  y  105  x  y  7  y
6
COMPRESSIBLE
7
x
v 0
u ( x y )  7  x  105  x  y  105  x  y  7  y
u ( x y )  x  21 x  y  35 x  y  7  x  y

Hence
5 2

u ( x y )  x  21 x  y  35 x  y  7  x  y

Hence
x
u 
6
4 3
2 5
v ( x y )  7  x  y  35 x  y  21 x  y  y
7
5
3 3
5

 u ( x y )  42 x  y  140  x  y  42 x  y
y
ROTATIONAL

6
4 3
2 5
v ( x y )   7  x  y  35 x  y  21 x  y  y
7

then the flow is incompressible and irrotational!
Problem 5.74
Given:
Find:
[Difficulty: 2]
Two-dimensional flow field
(a) show that the velocity field represents a possible incompressible flow
(b) Rotation at (x, y) = (1, 1)
(c) Circulation about the unit square shown above
We will apply the definition of circulation to the given velocity field.
Solution:




Governing

u   v   w 
 0 (Continuity equation)
Equations:
y
z
t
x

Assumptions:

1
2
   V
(Definition of rotation)
 
   V  ds
(Definition of circulation)
(1) Steady flow
(2) Incompressible flow
(3) Two dimensional flow (velocity is not a function of z)
Based on the assumptions listed above, the continuity equation reduces to:

x
u 

y
v 0
This is the criterion against which we will check the flow field.

x

u 
y
v  2A x  B x  2 
1
2  ft s
x 
1
ft s
x  0
This could be an incompressible flow field.
kˆ

1
From the definition of rotation:
At (x, y) = (1, 1)
 Bykˆ
z 2
0
 
 
 
 
From the definition of circulation we break up the integral:   V  ds  V  ds  V  ds  V  ds




iˆ
 1 

2 x
Ax 2
ˆj

y
Bxy
ab


bc
cd

  0.5kˆ
rad
s
da

 
The integrand is equal to: V  ds  Ax 2 iˆ  Bxyˆj  dxiˆ  dyˆj  Ax 2 dx  Bxydy Therefore, the circulation is equal to:
y
x
x
y
 a
 d
 c
 b
2
2
Γ   A x dx   B x  y dy   A x dx   B x  y dy
y
x
y
x
b
a
Γ
c
  x  xa  xd  xc
3  b
A
Γ
3
3
 x   y  yb
2 c c
B
2
3
2

3


Evaluating the integrals:
d
B  2
2
2
2
x  y  y b   x a  y a  y d  Since x a  x d  0 and



2 c c
Substituting given values:
Γ 
1
2
  


  1  ft  12  0 2  ft2

 ft s 
1
x b  x c we can simplify:
Γ  0.500 
ft
2
s
Problem 5.75
Given:
Find:
[Difficulty: 2]
Two-dimensional flow field
(a) show that the velocity field represents a possible incompressible flow
(b) Rotation at (x, y) = (1, 1)
(c) Circulation about the unit square shown above
We will apply the definition of circulation to the given velocity field.
Solution:




Governing

u   v   w 
 0 (Continuity equation)
Equations:
y
z
t
x

Assumptions:

1
2
   V
(Definition of rotation)
 
   V  ds
(Definition of circulation)
(1) Steady flow
(2) Incompressible flow
(3) Two dimensional flow (velocity is not a function of z)
Based on the assumptions listed above, the continuity equation reduces to:

x
u 

y
v 0
This is the criterion against which we will check the flow field.

x
u 

y
v  A y  2  B y 
1
m s
y  2 
1
2  m s
y  0
This could be an incompressible flow field.
kˆ

1
From the definition of rotation:
  Axkˆ At (x, y) = (1, 1)
z
2
0
 
 
 
 
From the definition of circulation we break up the integral:   V  ds  V  ds  V  ds  V  ds




ˆj
iˆ
 1 


y
2 x
Axy By 2
ab


bc
cd

  0.5kˆ
rad
s
da

 
The integrand is equal to: V  ds  Axyiˆ  By 2 ˆj  dxiˆ  dyˆj  Axydx  By 2 dy Therefore, the circulation is equal to:
x
y
x
y
 a
 d
 c
 b
A
B
2
2
2
2
3
3
3
3
Γ   A x  y dx   B y dy   A x  y dx   B y dy    x b  x a  y a  y c   y c  y b  y a  y d 




2
3
y
x
y
x

a
Since y a  y d  0 and
b
c

d
A
2
2
y b  y c we can simplify: Γ     x b  x a   y c Substituting given values:


2
Γ 
1
2

1
m s
2
2

2
 1  0  m  1  m Γ  0.5
2
m
s
Problem 5.76
Given:
Stream function
Find:
If the flow is incompressible and irrotational
[Difficulty: 3]
Solution:
Basic equations:

Incompressibility
u 

v 0
Irrotationality

x
y
x
Note: The fact that ψ exists means the flow is incompressible, but we check anyway
5
3 3
ψ( x y )  3  x  y  10 x  y  3  x  y
Hence
u ( x y ) 

y
v 

y
u 0
5
5
3 2
2 2
4
ψ( x y )  3  x  30 x  y  15 x  y
4
2 3
4
5

v ( x y )   ψ( x y )  30 x  y  15 x  y  3  y
x
For incompressibility

x
Hence

x
4
u ( x y )  15 x  90 x  y  15 y
u 

y
v 0

y
2 2
4
v ( x y )  90 x  y  15 x  15 y
INCOMPRESSIBLE
For irrotationality

x
Hence

x
3
3
v ( x y )  60 x  y  60 x  y
v 

y
u 0
3
3

 u ( x y )  60 x  y  60 x  y
y
IRROTATIONAL
4
Problem 5.77
Given:
Stream function
Find:
If the flow is incompressible and irrotational
[Difficulty: 3]
Solution:
Basic equations:

Incompressibility
u 

v 0
Irrotationality

x
y
x
Note: The fact that ψ exists means the flow is incompressible, but we check anyway
6
4 2
2 4
ψ( x y )  x  15 x  y  15 x  y  y
Hence
u ( x y ) 

y
2 3
v 

y
u 0
6
4
ψ( x y )  60 x  y  30 x  y  6  y
5
3 2
5
4

v ( x y )   ψ( x y )  60 x  y  6  x  30 x  y
x
For incompressibility

x
Hence

x
3

3
u ( x y )  120  x  y  120  x  y
u 

y
y
v 0
3
v ( x y )  120  x  y  120  x  y
3
INCOMPRESSIBLE
For irrotationality

x
Hence

x
2 2
4
v ( x y )  180  x  y  30 x  30 y
v 

y
u 0
4
4
2 2
4

 u ( x y )  30 x  180  x  y  30 y
y
IRROTATIONAL
Problem 5.78
[Difficulty: 2]
Given:
Find:
Velocity field for motion in the x-direction with constant shear
Solution:
We will apply the definition of circulation to the given velocity field.
(a) Expression for the velocity field
(b) Rate of rotation
(c) Stream function
Governing
Equations:

u    v    w    0 (Continuity equation)
y
z
t
x

 1
(Definition of rotation)
   V
2
Assumptions:
(1) Steady flow
(2) Incompressible flow

The x-component of velocity is: u   A dy  f ( x )  Ay  f ( x ) Since flow is parallel to the x-axis:

From the definition of rotation:
ˆj
iˆ
 1



x
y
2
Ay  f  x  0
From the definition of the stream function

V  Ay  f  x iˆ
kˆ

1
  Akˆ
z
2
0

  0.05kˆ
rad
s


1
2
ψ   u dy  g ( x )   ( A y  f ( x ) ) dy  g ( x )   A y  f ( x )  y  g ( x )
2



d
d
v   ψ   f ( x)  y 
g ( x )  0 Therefore, the derivatives of both f and g are zero, and thus f and g are constants:
x
d
dx
x
ψ
1
2
2
 A y  c1  y  c2
Problem 5.79
[Difficulty: 2]
Given:
The stream function
Find:
Whether or not the flow is incompressible; whether or not the flow is irrotational
Solution:
A
ψ( x y )  
The stream function is
2
2  π x  y
u ( x y ) 
The velocity components are

y

2
A y
ψ( x y ) 
2
π x  y
2


v ( x y )   ψ( x y )  
x
2
Because a stream function exists, the flow is:

Alternatively, we can check with
x

x
For a 2D field, the irrotionality the test is

x
v ( x y ) 

y

x
A x
2
π x  y
2

2
Incompressible
u 
u 
v 
u ( x y ) 

y

y

y
v 0
4  A x  y
v 
2
π x y
2

3

4  A x  y
2
π x y

2
3
0
Incompressible
u 0
4  A x
2
2
2
π  x  y 
3

2 A
2
2
π  x  y 
2

4  A y
2
2
2
π  x  y 
3

2 A
2
2
π  x  y 
2 Not irrotational
Problem 5.80
[Difficulty: 2]
Given:
Find:
Flow field represented by a stream function.
Solution:
We will apply the definition of rotation to the given velocity field.
(a) Show that this represents an incompressible velocity field
(b) the rotation of the flow
(c) Plot several streamlines in the upper half plane


1
2
Governing
Equation:
   V
Assumptions:
(1) Steady flow
(2) Incompressible flow
(Definition of rotation)
From the definition of the stream function: u   ψ  A x  2  A y
y

v   ψ  A y
x

x
ˆj
iˆ
 1


From the definition of rotation:   2
x
y
A x  2 y   Ay
u 

y
Applying the continuity equation:
v  A A 0
This could be an incompressible
flow field
kˆ

1
  2 Akˆ   Akˆ
z 2
0
The streamlines are curves where the stream function is constant, i.e.,
ψ  constant

   Akˆ
Here is a plot of streamlines:
Streamline Plot
5
psi = 0
psi = -2
psi = 6
4
Y (m)
3
2
1
0
4
2
0
X (m)
2
4
Problem 5.81
[Difficulty: 3]
Given:
Find:
Flow field represented by a stream function.
Solution:
We will apply the definition of circulation to the given velocity field.
(a) Expression for the velocity field
(b) Show that flow field is irrotational
(c) Plot several streamlines and illustrate velocity field
1
2

   V
Assumptions:
(1) Steady flow
(2) Incompressible flow
(Definition of rotation)
From the definition of the stream function: u   ψ  2  y
y
From the definition of rotation:
ˆj
iˆ
 1 


2 x
y
 2 y  2x

V  2 yiˆ  2 xˆj

v   ψ  2  x In vector notation:
x
kˆ

1

  2  2kˆ  0
z 2
0
The streamlines are curves where the stream function is constant, i.e.,
ψ  constant


0
Flow is irrotational
Here is a plot of streamlines:
Streamline Plot
5
psi = 0
psi = 4
psi = 8
4
3
Y (m)

Governing
Equation:
2
1
0
0
1
2
3
X (m)
4
5
Problem 5.82
[Difficulty: 2]
Given:
Find:
Flow field represented by a velocity function.
Solution:
We will apply the definition of rotation and circulation to the given velocity field.
(a) Fluid rotation
(b) Circulation about the curve shown
(c) Stream function
(d) Plot several streamlines in first quadrant

Governing
Equation:
Assumption:

1
2
   V
(Definition of rotation)
 
   V  ds
(Definition of circulation)
Steady flow
kˆ

1
From the definition of rotation:
 By kˆ
z 2
0
 
 
 
 
From the definition of circulation we break up the integral:   V  ds  V  ds  V  ds  V  ds




iˆ
 1 

2 x
Ax 2
ˆj

y
Bxy
ab


bc
cd

   ykˆ
rad
ft  s
da

 
The integrand is equal to: V  ds  Ax 2 iˆ  Bxyˆj  dxiˆ  dyˆj  Ax 2 dx  Bxydy Therefore, the circulation is equal to:
x
y
x
y
 a
 d
 c
 b
2
2
Γ   A x dx   B x  y dy   A x dx   B x  y dy
y
x
y
x
Γ
c
b
a
  x  xa  xd  xc
3  b
A
Γ
3
3
3
 x   y  yb
2 c c
B
2
2

1
ft s
B  2
2
2
 2
Since x a  x d  0 and x b  x c we can simplify:
  2 xc  yc  y b   xa  ya  y d 
Substituting given values:
2
1
ft s
Γ 
1
2

2
ft s
2
2

 1  ft  1  0  ft
2
Γ  1.000 
ft
2
s



2
2
ψ   u dy  f ( x )   A x dy  f ( x )  A x  y  f ( x )




B 2
ψ   v dx  g ( y )   B x  y dx  g ( y )    x  y  g ( y ) Comparing the two stream functions:
2



v ψ
x
 x  y  f (x) 
d
3
From the definition of the stream function: u   ψ
y
In addition,
Evaluating the integral:
2
 x  y  g ( y ) Thus, f  g  constant
Taking f(x) = 0:
2
ψ  A x  y
ψ  constant
Here is a plot of streamlines:
Streamline Plot
5
ψ=1
ψ=4
ψ=8
ψ = 16
4
3
Y (ft)
The streamlines are curves where the stream function is constant, i.e.,
2
1
0
0
1
2
3
X (ft)
4
5
Problem 5.83
[Difficulty: 2]
Given:
Find:
Flow field represented by a velocity function.
Solution:
We will apply the definition of circulation to the given velocity field.
(a) An expression for the stream function
(b) Circulation about the curve shown
(c) Plot several streamlines (including the stagnation streamline) in first quadrant
Governing
Equation:
 
   V  ds
Assumptions:
Steady flow
(Definition of circulation)
From the definition of the stream function: u   ψ
y
In addition,


A 2
ψ   u dy  f ( x )   ( A y  B) dy  f ( x )   y  B y  f ( x )
2




A 2
ψ   v dx  g ( y )   A x dx  g ( y )    x  g ( y )
2



v ψ
x
A 2
A 2
 y  B y  f ( x )    x  g ( y ) Thus,
2
2
A 2
f ( x )    x  C Taking C = 0:
2
From the definition of circulation we break up the integral:


bc
cd
 y x
2
  B y
da

y
x
y
x
 a
 c
 d
 b
Γ   ( A y  B) dx   A x dy   ( A y  B) dx   A x dy
y
y
x
x


Evaluating the integral:
d
c
b
a
 



Γ  A y a  B  x b  x a  A x b  y c  y b  A y c  B  x d  x c  A x d  y a  y d
Γ 
2

2
A
 
V  ds   Ay  B iˆ  Axˆj  dxiˆ  dyˆj   Ay  B dx  Axdy Therefore, the circulation is:
The integrand is equal to:

ψ
 
 
 
 
   V  ds   V  ds   V  ds   V  ds
ab

Comparing the two stream functions:
10
ft 
 10
  0  ft  10   ( 1  0 )  ft  s  1  ft  ( 1  0 )  ft 
s
 s

Substituting known values:
 10  1 ft  10 ft   ( 0  1)  ft  10  0 ft  ( 1  0)  ft


s
s
 s
Γ  0
ft
2
s
The streamlines are curves where the stream function is constant, i.e.,
ψ  constant
The stagnation streamline is the one running through the point
where the velocity vanishes:
A y stag  B  0
y stag  
B
A
Here is a plot of streamlines:
Streamline Plot
5
ψ = -5
ψ=0
ψ=5
ψ = 10
 1  ft
4
x stag  0
Plugging this information in to find the stream function at
the stagnation point yields:
ψstag 
10
2 s
ψstag  5 
ft
2
2
 ( 1  ft)  ( 0  ft)   10  1  ft
3
Y (ft)
A x stag  0
2
s
ft
2
1
s
0
0
1
2
3
X (ft)
4
5
Problem 5.84
[Difficulty: 3]
Given:
Find:
Viscometric flow of Example 5.7, V = U(y/h)i, where U = 4 mm/s and h = 4 mm
Solution:
We will apply the definition of rotation to the given velocity field.
(a) Average rate of rotation of two line segments at +/- 45 degrees
(b) Show that this is the same as in the example


1
2
Governing
Equation:
   V
Assumptions:
(1) Steady flow
(2) Incompressible flow
Considering the lines shown:
uc  ua sinθ1
ωac 
l

ωac 
y
ub  ud 
ωbd 
ω

y
u
y

 



 
u  l sin θ1
y
  2  h  sinθ12


y
U
u  sin θ1
ωbd 
l sinθ2 sinθ2
l
y
 
l
u  l sin θ2

since the component normal to l is u  sin θ1
l sinθ1 sinθ1

u
uc  ua 
(Definition of rotation)
ud  ub sinθ2
l
U
1 U
2
2
 ωac  ωbd      sin θ1
 sin θ2 


2
2 h
1


  
1 U  1 
ω     
 
2 h  2 
2
Substituting for U and h:
ω  
1
2
 
since the component normal to l is u  sin θ2
  2  h  sinθ22
u  sin θ2
  
2
 1  
 
 2 
 4
mm
s
Now consider this sketch:
Now we sum these terms:
When θ1  45 deg and
θ2  135  deg
1 U
ω 
2 h

1
4  mm
ω  0.5
1
s
Problem 5.85
[Difficulty: 3]
Given:
Find:
Velocity field for pressure-driven flow between stationary parallel plates
Solution:
We will apply the definition of circulation to the given velocity field.
(a) Expression for circulation about a closed contour of height h and length L
(b) Evaluate part (a) for h = b/2 and h = b
(c) Show that the same result is obtained from area integral of Stokes Theorem (Eq. 5.14)
 
   V  ds
Governing
Equations:
(Definition of circulation)

   V dA   V  ds


(Stokes Theorem)
A
Assumptions:
(1) Steady flow
From the definition of circulation we break up the integral:
 
 
 
 
   V  ds   V  ds   V  ds   V  ds
1
The integrand is equal to:

2
3
4

 
y
y
y
y
V  ds  U 1  iˆ  dxiˆ  dyˆj  U 1  dx Therefore, the circulation is equal to:
b b
b b
L
0
0
L


h
0
0
h
h
h
Γ   U   1   dx   U   1   dx  U L   1  


b 
b 
b
b
b
b 


1 b
1 b
For h = b/2: Γ  U L    1   
b 2 
b 2
From Stokes Theorem:

Γ  U L 


h
0

Γ 
U L
For h = b:
4

h
h
Γ  U L   1  
b 
b
Γ  U L 1  ( 1  1 )

 v u 
 1 2y 
     V dA     dA  U   
dA
b b 
x y 
A
A
A
2
 1  2  y dy  U L  h  h   U L h   1 



b 
b b 
b b 
h

b
Γ 0
We define dA = L dy:
h
h
Γ  U L   1  
b 
b
Problem 5.86
[Difficulty: 3]
Given:
Find:
Velocity field approximation for the core of a tornado
Solution:
We will apply the definition of rotation to the given velocity field.
(a) Whether or not this is an irrotational flow
(b) Stream function for the flow

Governing
Equation:

1
2
   V
(Definition of rotation)


1 

  eˆr
 eˆ
 kˆ
r
r 
z
(Definition of "del" operator)
eˆ
 eˆr

eˆr
 eˆ

Assumptions:
(Hints from text)
(1) Steady flow
(2) Two-dimensional flow (no z velocity, velocity is not a function of θ or z)
1
2

   eˆr
From the definition of rotation:
1 
 

 kˆ   Vr eˆr  V eˆ 
 eˆ
r 
z 
r
Employing assumption (2) yields:

V 
1 
1  
1

 V
product
Vr eˆr  V eˆ  From
 eˆ
  Vr eˆr  V eˆ   eˆr   eˆr r  eˆ    eˆ 
rule:
r
r  


r
r
2

r





1
2

ˆ
ˆ
1
eˆr  eˆr  Vr  eˆr  eˆ  V  eˆ 1   eˆr Vr  Vr er  eˆ V  V e 

r
r


 
2
r  

1
eˆr  eˆr  Vr  eˆr  eˆ  V  1 Vr  V   eˆ  eˆ  1 V  Vr

r
2
r 
r
 r r 
 r 
   eˆr



Since V is only a function of r:



ψ  



q θ
2 π
 1  V 1 Vr V  ˆ

 k
  
r 
 2  r r 
Flow is
irrotational.


q
q θ
dθ  f ( r)  
ψ   r Vr dθ  f ( r)  
 f ( r)

 2 π
2 π



K
K
Vθ dr  g ( θ)  
dr  g ( θ)  
 ln( r)  g ( θ) Comparing these two expressions:
 2  π r
2 π

To build the stream function:Vr 

Vθ   ψ
r
K
K ˆ 
1  V V  ˆ 1 
 k   


k  0
r 
2  2r 2 2r 2 
2  r
Using the hints from the
text:
1 
 ψ
r θ
 f ( r)  
K
2 π
 ln( r)  g ( θ) f ( r)  
K
2 π
 ln( r)
ψ
K
2 π
 ln( r) 
q θ
2 π
Problem 5.87
Given:
Find:
Velocity field for fully-developed flow in a circular tube
Solution:
We will apply the definition of vorticity to the given velocity field.
[Difficulty: 2]
(a) Rates of linear and angjular deformation for this flow
(b) Expression for the vorticity vector


Governing
Equation:
   V
Assumptions:
(1) Steady flow
(Definition of vorticity)
 1 rVr  1 V V z

0
 V 

z
r r
r 
1  Vθ  
r-θ plane:   
 V 0
r r  r  θ r
The volume dilation rate of the flow is:
The angular deformations are:
Rates of linear deformation in all three
directions is zero.
1
θ-z plane:  Vθ    Vz  0
r θ
z
2 r
z-r plane:  Vr   Vz  Vmax
2
z
r
R
The vorticity in cylindrical coordinates is:


2 r
angdef  Vmax
2
R
V 
 1 rV 1 Vr
 1 V z V 
 V


eˆr   r  z eˆ  
r 
z 
r 
 z
 r r
 r 
   V  

  Vmax
ˆ
k

2r
eˆ
R2
Problem 5.88
[Difficulty: 3]
Given:
Find:
Velocity field for pressure-driven flow between stationary parallel plates
Solution:
We will apply the definition of vorticity to the given velocity field.
(a) Rates of linear and angjular deformation for this flow
(b) Expression for the vorticity vector
(c) Location of maximum vorticity


Governing
Equation:
   V
Assumptions:
(1) Steady flow
The volume dilation rate of the flow is:
The angular deformations are:
(Definition of vorticity)
 u v w


0
 V 
x y z
Rates of linear deformation in all three
directions is zero.
2 y
x-y plane:  v   u  u max
2
x
y
b
y-z plane:  w   v  0
y
z
2 y
angdef  u max
2
b
z-x plane:  u   w  0
z
x

The vorticity is:

   V 
iˆ

x
ˆj

y
kˆ

2y
 u max 2 kˆ
z
b
  y 2 
u max 1    
  b  
0
0

  u max
2y ˆ
k
b2
The vorticity is a maximum at y=b
and y=-b
Problem 5.89
Given:
Flow between infinite plates
Find:
Prove that u = 0, dP/dy = constant
[Difficulty: 2]
Solution:
Governing
Equations:
u v w
 
0
x y z
(Continuity Equation)
  2u  2u  2u 
 u
u
u
u 
P
u
v
 w   g x 
   2  2  2 
x
y
z 
x
y
z 
 t
 x
 
  2v  2v  2v 
 v
v 
v
v
P
   2  2  2 
   u  v  w   g y 
y
x
t
y
z





y
z 


 x
(Navier-Stokes Equations)
 2w 2w 2w 
 w
P
w
w 
w
   2  2  2 
v
 w   g z 
u
z
x
y
z 
y
z 
 t
 x
 
Incompressible fluid
Assumptions: No motion along the wall (x = 0) limited to two dimensions (w = 0).
Prove that u = 0:
Given that
 
u v w
V  V (z ) this means that

0

z z z


Also given that the flow is fully developed which means that V  V ( y ) so that


u v w


0
y y y
And steady flow implies that V  V (t )
u
 0 , but because u  u ( y, z , t ) then u  u (x) meaning that the partial derivative here
x
du
becomes an ordinary derivative:
0
dx
The continuity equation becomes
Integrating the ordinary derivative gives:
u  constant
By the no-slip boundary condition u = 0 at the surface of either plate meaning the constant must be zero.
Hence:
u0
Prove that
P
 constant :
y
Due to the fact that u = 0, and gravity is in the negative y-direction the x-component of the Navier-Stokes Equation becomes:
P
 0 hence P  P(x)
x
Due to the fact that w = 0, and gravity is in the negative y-direction the z-component of the Navier-Stokes Equation becomes:
P
 0 hence P  P(z )
z
The y-component of the Navier-Stokes Equation reduces to:
  2v 
P
0
 g    2 
y
 x 
So then
  2v 
P
  g    2  [1]
y
 x 
It has been shown that P  P ( x, z ) and because the flow is steady P  P (t ) meaning that P  P ( y ). This means that the left
hand side of [1] can only be a function of y or a constant. Additionally, by the fully developed, steady flow, and
 
V  V (z ) conditions it is shown that v  v(x). For this reason the right hand side of [1] can only be a function or x or a constant.
Mathematically speaking it is impossible for:
Hence,
P
 constant
y
f ( y )  g ( x) so each side of [1] must be a constant.
Problem 5.90
Given:
temperature profile and temperature-dependent viscosity expression
Find:
Velocity Profile
[Difficulty: 3]
Solution:
Governing
Equations:
u v w
 
0
x y z
(Continuity Equation)
  2u  2u  2u 
 u
P
u
u
u 
   2  2  2 
u
v
 w   g x 
x
x
y
z 
y
z 
 t
 x
 
  2v  2v  2v 
 v
P
v 
v
v
   2  2  2 
 u  v  w   g y 
z 
y
y
x
y
z 
 t
 x
 
 2w 2w 2w 
 w
P
w 
w
w
   2  2  2 
 w   g z 
 
v
u
z 
y
x
z
y
z 
 t
 x
Assumptions: Incompressible fluid
Similar to the Example 5.9, the x-component momentum equation can be simplified to
d yx
dy
  g sin 
(1)
Integrating once, one has
 yx   gy sin   C1
Using the boundary condition:  yx ( y
(2)
 h)  0
c1  gh sin 
(3)
Substituting c1 into eq. (2),
 yx  
du
 g (h  y ) sin 
dy
Here, the fluid viscosity depends on the temperature,
(4)
(Navier-Stokes Equations)

0
1  a(Tw  T0 )(1  y / h)
(5)
Substituting equation (5) into equation (4), we have
du gh(1  y / h) sin 

(1  a(Tw  T0 )(1  y / h))
dy
0
(6)
Integrating equation (6) once
u
y y2
y
gh sin 
( y (1  )  a(Tw  T0 ) y (1   2 ))  C2
0
2h
h 3h
(7)
At y=0, u=0: c2=0.
Substituting c2=0 into eq. (7), one obtains
u
y
gh sin 
y y2
( y (1  )  a(Tw  T0 ) y (1   2 ))
h 3h
0
2h
When a=0, eq. (8) can be simplified to
u
(8)
gh sin 
y
y (1  ) , and it is exactly the same velocity profile in Example 5.9.
0
2h
Problem 5.91
Given:
Find:
[Difficulty: 2]
Sinusoidal approximation for velocity profile in laminar boundary layer
(a) Express shear force per unit volume in the x-direction
(b) Maximum value at these conditions
We will evaluate a differential volume of fluid in this flow field
Solution:
(1) Steady flow
Assumptions:
The differential of shear force would be: dFshear  ( τ  dτ)  dx dz  τ dx dz  dτ dx dz and
π y 
π U
From the given profile: d u 
 cos

2 δ
dy
 2 δ 
d
and
2
dy
2
π
The maximum magnitue for this shear force is when y = δ:
For water: μ  0.001 
N s
2
m
U  3
m
s
δ  2  mm
dV
2
 π y 

  sin

2

δ
 
 2 δ 
u  U 
dFsx
Thus,

dτ
dy
dFsx
dV
Fvmax 

2
d  du 
d
 μ dy   μ 2 u
dy 

dy
2
 π y 

  sin

2

δ
 
 2 δ 
 μ U 
dFsxmax
dV
π
 μ U 


 2 δ 
π
2
Substituting these values:
Fvmax  0.001 
N s
2
m
 3
m
s

1
π

 2  0.002  m 


2
kN
Fvmax  1.851 
3
m
Problem 5.92
[Difficulty: 3]
Given:
Find:
Linear approximation for velocity profile in laminar boundary layer
Solution:
We will apply the definition of rotation to the given velocity field.
(a) Express rotation, find maximum
(b) Express angular deformation, locate maximum
(c) Express linear deformation, locate maximum
(d) Express shear force per unit volume, locate maximum

1
2

Governing
Equation:
   V
Assumptions:
(1) Steady flow
iˆ
 1 
The rotation is:  
2 x
y
U
kˆ

1    U y y    y  ˆ
  
  U  k
z 2  x  4 x   y   
ˆj

y
U y y
4 x

(Definition of rotation)
0
1 1 3 U y
1 U
1  3 U y
ωz     
 

   
2 4 2
5
2
2 8
1
5
2
c x
2


   U  1  3   y  
1
8 x 
1 
2
2
c x 
2  c x
2
2
c x
2
Computing the partial derivatives:


c x
U 
3 y
ωz  
 1    
2 δ 
8 x
U
2
2
2
c x
2


 c x 2 


angdef 
y 
1 U y
Linear deformation:  u    U
 
1
2
3
x
x 

 c x 2 

y
v 
c x
x
u 
x
U
δ
2
x

3

8
 1 
2

y
v 
U y
Maximum value at y = δ

2δ x
U y

2δ x


 
2
 U y2  2 U y
 
 
3
4
3
y  4


2
2
c x
 c x 




Maximum value at y = δ
U y   


   U y   U   1  3  y  1  
The angular deformation is:  v   u    
1
3  y
1
x
y
x  4

 c 4 2 5


2
Maximum value at y = δ
y
U
1
c x

3

8
 1 
 
y
2

x 
2
2

x 
Maximum value at y = 0
The shear stress is
 3 y 2

  μ U 
v  u 
 1     
8 x 
δ 
y 
 x
τyx  μ 
The net shear force on a fluid element is dτ dx dz:
Therefore the shear stress per unit volume is:
dτ 

y
τ  dy 
3  μ U y
d
F

4  δ x x
dV
μ U
δ
3 2 y 
3  μ U y
 dy
 
 8 2   dy  
2
x 
4  δ x

Maximum value at y = δ
Problem 5.93
Given:
Find:
[Difficulty: 2]
Velocity profile for fully developed laminar flow in a tube
(a) Express shear force per unit volume in the x-direction
(b) Maximum value at these conditions
We will evaluate a differential volume of fluid in this flow field
Solution:
(1) Steady flow
Assumptions:
The differential of shear force would be: dFshear  ( τ  dτ)  2  π r dz dr  τ 2  π r dz dr  2  π r dτ dz dr
dFsz
and in cylindrical coordinates:

dV
2 r
Therefore:
From the given profile: d u  u max
2
dr
R
dFsz
dV
 2 
1
1d 
d
d 
 ( r τ)  2  π dr dz 
 r μ u 
2  π r dr dz dr
r dr 
dr 
μ u max
r
 
μ u max
μ u max 2  r
d r 
 2 

 4 
2
2
r
dr  2 
R
R
R 
2

Fvmax 
For water: μ  2.1  10
 5 lbf  s

ft
2
u max  10
ft
s
R  3  in
dFszmax
dV
 4 
μ u max
R
2
Substituting these values:
 5 lbf  s
Fvmax  4  2.1  10

ft
2
 10
ft
s

1
( 3  in)
2

 12 in 
 ft 


2
lbf
Fvmax  0.0134
3
ft
Problem 5.94
Given:
Horizontal, fully developed flow
Find:
Velocity Profile and pressure gradient
[Difficulty: 3]
Solution:
u v w
 
0
x y z
Governing
Equations:
(Continuity Equation)
  2u  2u  2u 
 u
u
u
u 
P
u
v
 w   g x 
   2  2  2 
x
y
z 
x
y
z 
 t
 x
 
  2v  2v  2v 
 v
P
v 
v
v
   2  2  2 
 u  v  w   g y 
y
z 
y
x
y
z 
 t
 x
 
 2w 2w 2w 
 w
P
w
w
w 
   2  2  2 
 w   g z 
v
u
z
x
z 
y
y
z 
 t
 x
 
(1) Incompressible fluid
Assumptions: (2) Zero net flow rate
For fully developed flow
d 2u 1 dp

dy 2  dx
(1)
The general solution for equation (1) is
u
y 2 dp
 C1 y  C2
2  dx
(2)
where C1 and C2 are constants.
Apply the boundary conditions
u  0 at y  0
du
 C at y  h

dy
Then, we can get C1 
u
1

(C  h
2
dp
) and C2  0
dx
y
h dp ' 1 ' 2 Ch '
(y  y ) 
y , where y ' 
2
 dx

h
The net flow or flow rate is zero:
(Navier-Stokes Equations)
h 2 dp 1 ' 1 '2
Ch 1 '
(
y
y
dy
y dy


)
 dx 0
 0
2
dp 3 C
Thus,

dx 2 h
0
Problem 5.95
Given:
N-S equations and simplification assumptions
Find:
Fluid Velocity
[Difficulty: 3]
Solution:
Governing
Equations:
 u  0
(Continuity Equation)
ρu  u  -p   2 u  J  B
(Momentum Equation)
(1) Incompressible fluid
Assumptions: (2) Two dimensional, fully developed flow driven by Lorentz force
(2) Zero pressure gradient
Write the 2D continuity and momentum equations in Cartesian coordinates:
u u
0

x y
ρ(u
 2u  2u
p
u
u
)    ( 2  2 )  JB
v
x
y
x
x
y
(1)
(2)
ρ(u
 2v  2v
p
v
v
)    ( 2  2 )
v
y
y
x
x
y
(3)
Simplify the above equations:
u
v0
 0  u  u( y)
x
Using the assumption of zero pressure gradient, equation (3) vanishes, and equation (2) can be
simplified as
d 2u
(4)
0   2  JB
dy
General solution for equation (4) is
1 JB 2
u
y  C1 y  C2
(5)
2 
Apply the no slip boundary conditions into equation (5), we get
2

h
1 JB h
h
u
(
)
0




 C1  C 2


2
2  4
2

2
 u ( h )  0   1 JB h  C h  C
1
2
 2
2  4
2
JB 2
h
Therefore, C1=0 and C 2 
8
The fluid velocity is given as
JB 2
u( y) 
(h  4 y 2 )
8
Problem 5.96
[Difficulty: 2]
Given:
Temperature-dependent fluid density and the Navier-Stokes equations
Find:
Explanation for the buoyancy-driven flow; effect of angle on fluid velocity
Solution:
Governing
Equations:
u v w
 
0
x y z
(Continuity Equation)
  2u  2u  2u 
 u
u
u
u 
P
   2  2  2 
   u  v  w   g x 
x
y
z 
x
y
z 
 t
 x
  2v  2v  2v 
 v
v
v
P
v 
   2  2  2 
 u  v  w   g y 
y
z 
y
x
y
z 
 t
 x
 
(Navier-Stokes Equations)
 2w 2w 2w 
 w
P
w 
w
w
   2  2  2 
 w   g z 
v
u
z
z 
y
x
y
z 
 t
 x
 
Assumption: Incompressible fluid
(1) The first term in the right-hand-side of the momentum equations (5.27a)-(5.27c) represents the
gravitational body force, which is proportional to the local fluid density. The fluid density in the region
at temperature 72oC is higher than that in the region at temperature 90-94 oC, and meanwhile is lower
than that in the region at temperature 50-55 oC. Thus, the net gravitational force induces counterclockwise fluid circulation within the loop.
(2) Since the fluid circulation is driven by buoyancy force which is proportional to gcos where g is the
gravitational acceleration, one can control the flow rate in the loop by adjusting the inclination angle .
When the angle =90o, there is no fluid motion. When =0, the flow rate is the maximum.
Problem 5.97
Given:
N-S equations
Find:
Fluid velocity
Solution:
  u  0 (Continuity equation)
ρu  u  -p   2u (Momentum equation)
(1) Two-dimensional fully developed flow
Assumptions: (2) Zero pressure gradient
Governing
Equations:
(1) Write the continuity and momentum equations in Cartesian form:
 u v
 x  y  0

 2u  2u
p
u
u

  (u  v )     ( 2  2 )
y
x
x
y
x

2
 v  2v
p
v
v






(
)
(
)


v
u

x 2 y 2
y
y
x

(1)
(2)
(3)
[Difficulty: 3]
Simplify the above equations:
u
v0
 0  u  u( y)
x
Using the assumption of zero pressure gradient, equation (3) vanishes, and equation (2) can be simplified as
d 2u
0 2
(4)
dy
General solution for equation (4) is given as
u  C1 y  C2
(5)
Apply the boundary condition into equation (5), we get
h
h


u ( 2 )    E  C1 2  C2


h
h
 u( )  
E  C1  C2


2
2
Therefore, C1=0 and C2  

E

The fluid velocity is given as
u( y)  

E

(6)
(2) Pressure-driven flow has a parabolic flow velocity profile; while EOF has a plug velocity profile and it is
independent of the channel size.
(3) Substituting =7.0810-10 CV-1m-1, =0.1V, Pa.s, and E=1000 V/m into equation (6), one obtains
7.08  10 -10 C  V 1  m 1   0.1V
 1000 V/m
10 3 Pa  s
C V
N
-6
70.8
10
 70.8  10 -6 m / s


 70.8  10 -6
2
Pa  s  m
Pa  s  m
u( y)  
Problem 5.98
ρ =
3
250
1
4
µ =
999
0.001
h =
12
mm
mm
m
m
Draining a Tank
1
h =
y (0 ) = y 0
y n +1 = y n + hkyn
k=−
k = 0.000099 s
t n (min)
0
12
24
36
48
60
72
84
96
108
120
Error:
1
min
dy
d 4 ρg
y
=−
dt
32 D 2 µL
n
0
1
2
3
4
5
6
7
8
9
10
Exact
Euler (10 pt)
Euler (20 pt)
3
kg/m
N·s/m 2
d 4 ρg
32 D 2 µL
y Exact (t ) = y0 e
6
−
d 4 ρg
32 D 2 µL
min
Depth y (m)
d =
D =
y0 =
L =
[Difficulty: 3]
t
t n+1 = t n + h
1
0
-1
y n (m)
1
0.929
0.862
0.801
0.743
0.690
0.641
0.595
0.553
0.513
0.477
3%
n
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
t n (min)
0.0
6.0
12.0
18.0
24.0
30.0
36.0
42.0
48.0
54.0
60.0
66.0
72.0
78.0
84.0
90.0
96.0
102.0
108.0
114.0
120.0
Error:
y n (m)
1
0.964
0.930
0.897
0.865
0.834
0.804
0.775
0.748
0.721
0.695
0.670
0.646
0.623
0.601
0.579
0.559
0.539
0.520
0.501
0.483
1%
y Exact(m)
1
0.965
0.931
0.898
0.867
0.836
0.807
0.779
0.751
0.725
0.700
0.675
0.651
0.629
0.606
0.585
0.565
0.545
0.526
0.507
0.489
0
0
30
60
Time t (min)
90
120
Problem 5.99
[Difficulty: 3]
y n +1 = y n + ∆x cos( xn )
∆x
0.06545
n
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
x
0.000
0.065
0.131
0.196
0.262
0.327
0.393
0.458
0.524
0.589
0.654
0.720
0.785
0.851
0.916
0.982
1.047
1.113
1.178
1.244
1.309
1.374
1.440
1.505
1.571
Error
∆x
0.032725
y
0.000
0.065
0.131
0.196
0.260
0.323
0.385
0.446
0.504
0.561
0.615
0.667
0.716
0.763
0.806
0.846
0.882
0.915
0.944
0.969
0.990
1.007
1.020
1.028
1.032
3.24%
1.5
Euler (Large steps)
Euler (Medium steps)
Euler (Small steps)
Exact
1.0
0.5
n
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
x
0.000
0.033
0.065
0.098
0.131
0.164
0.196
0.229
0.262
0.295
0.327
0.360
0.393
0.425
0.458
0.491
0.524
0.556
0.589
0.622
0.654
0.687
0.720
0.753
0.785
0.818
0.851
0.884
0.916
0.949
0.982
1.014
1.047
1.080
1.113
1.145
1.178
1.211
1.244
1.276
1.309
1.342
1.374
1.407
1.440
1.473
1.505
1.538
1.571
∆x
0.021817
y
0.000
0.033
0.065
0.098
0.131
0.163
0.195
0.227
0.259
0.291
0.322
0.353
0.384
0.414
0.444
0.473
0.502
0.530
0.558
0.585
0.612
0.638
0.663
0.688
0.712
0.735
0.757
0.779
0.800
0.820
0.839
0.857
0.874
0.890
0.906
0.920
0.934
0.946
0.958
0.968
0.978
0.986
0.994
1.000
1.006
1.010
1.013
1.015
1.016
0.0
0.0
0.2
xn +1 = xn + ∆x
0.4
0.6
0.8
Error
1.0
1.63%
n
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
1.2
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
x
0.000
0.022
0.044
0.065
0.087
0.109
0.131
0.153
0.175
0.196
0.218
0.240
0.262
0.284
0.305
0.327
0.349
0.371
0.393
0.415
0.436
0.458
0.480
0.502
0.524
0.545
0.567
0.589
0.611
0.633
0.654
0.676
0.698
0.720
0.742
0.764
0.785
0.807
0.829
0.851
0.873
0.894
0.916
0.938
0.960
0.982
1.004
1.025
1.047
1.069
1.091
1.113
1.134
1.4
1.156
1.178
1.200
1.222
1.244
1.265
1.287
1.309
1.331
1.353
1.374
1.396
1.418
1.440
1.462
1.484
1.505
1.527
1.549
1.571
y
0.000
0.022
0.044
0.065
0.087
0.109
0.131
0.152
0.174
0.195
0.217
0.238
0.259
0.280
0.301
0.322
0.343
0.363
0.383
0.404
0.424
0.443
0.463
0.482
0.501
0.520
0.539
0.557
0.576
0.593
0.611
0.628
0.645
0.662
0.678
0.695
0.710
0.726
0.741
0.756
0.770
0.784
0.798
0.811
0.824
0.836
0.848
0.860
0.871
0.882
0.893
0.903
0.913
0.922
0.931
0.939
0.947
0.954
0.961
0.968
0.974
0.980
0.985
0.990
0.994
0.998
1.001
1.004
1.006
1.008
1.009
1.010
1.011
Error
1.09%
y Exact
0.000
0.022
0.044
0.065
0.087
0.109
0.131
0.152
0.174
0.195
0.216
0.238
0.259
0.280
0.301
0.321
0.342
0.362
0.383
0.403
0.423
0.442
0.462
0.481
0.500
0.519
0.537
0.556
0.574
0.591
0.609
0.626
0.643
0.659
0.676
0.692
0.707
0.722
0.737
0.752
0.766
0.780
0.793
0.806
0.819
0.831
0.843
0.855
0.866
0.877
0.887
0.897
0.906
1.6 0.915
0.924
0.932
0.940
0.947
0.954
0.960
0.966
0.971
0.976
0.981
0.985
0.988
0.991
0.994
0.996
0.998
0.999
1.000
1.000
1.8
Problem 5.100
[Difficulty: 3]
N =4
x = 0.333
x
0.000
0.333
0.667
1.000
Eq. 5.34 (LHS)
1.000
-1.000
0.000
0.000
0.000
1.333
-1.000
0.000
0.000
0.000
1.333
-1.000
0.000
0.000
0.000
1.333
(RHS)
1
0
0
0
Inverse Matrix
1.000
0.750
0.563
0.422
0.000
0.750
0.563
0.422
0.000
0.000
0.750
0.563
0.000
0.000
0.000
0.750
Result
1.000
0.750
0.563
0.422
Exact
1.000
0.717
0.513
0.368
Error
0.000
0.000
0.001
0.001
0.040
N =8
x = 0.143
x
0.000
0.143
0.286
0.429
0.571
0.714
0.857
1.000
N = 16
x = 0.067
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Eq. 5.34 (LHS)
1.000
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.143
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.143
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.143
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.143
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.143
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.143
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.143
(RHS)
1
0
0
0
0
0
0
0
Inverse Matrix
1
1.000
0.875
0.766
0.670
0.586
0.513
0.449
0.393
2
0.000
0.875
0.766
0.670
0.586
0.513
0.449
0.393
3
0.000
0.000
0.875
0.766
0.670
0.586
0.513
0.449
4
0.000
0.000
0.000
0.875
0.766
0.670
0.586
0.513
5
0.000
0.000
0.000
0.000
0.875
0.766
0.670
0.586
6
0.000
0.000
0.000
0.000
0.000
0.875
0.766
0.670
7
0.000
0.000
0.000
0.000
0.000
0.000
0.875
0.766
8
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.875
Result
1.000
0.875
0.766
0.670
0.586
0.513
0.449
0.393
Eq. 5.34 (LHS)
1
1.000
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
2
0.000
1.067
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
3
0.000
0.000
1.067
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
4
0.000
0.000
0.000
1.067
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
5
0.000
0.000
0.000
0.000
1.067
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
6
0.000
0.000
0.000
0.000
0.000
1.067
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
7
8
9
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
1.067 0.000 0.000
-1.000 1.067 0.000
0.000 -1.000 1.067
0.000 0.000 -1.000
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
10
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.067
-1.000
0.000
0.000
0.000
0.000
0.000
Exact
1.000
0.867
0.751
0.651
0.565
0.490
0.424
0.368
Error
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.019
11
12
13
14
15
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
1.067 0.000 0.000 0.000 0.000
-1.000 1.067 0.000 0.000 0.000
0.000 -1.000 1.067 0.000 0.000
0.000 0.000 -1.000 1.067 0.000
0.000 0.000 0.000 -1.000 1.067
0.000 0.000 0.000 0.000 -1.000
16
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.067
(RHS)
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Inverse Matrix
1.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.597
0.559
0.524
0.492
0.461
0.432
0.405
0.380
x
0.000
0.067
0.133
0.200
0.267
0.333
0.400
0.467
0.533
0.600
0.667
0.733
0.800
0.867
0.933
1.000
N
4
8
16
x
0.333
0.143
0.067
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.597
0.559
0.524
0.492
0.461
0.432
0.405
0.380
Error
0.040
0.019
0.009
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.597
0.559
0.524
0.492
0.461
0.432
0.405
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.597
0.559
0.524
0.492
0.461
0.432
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.597
0.559
0.524
0.492
0.461
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.597
0.559
0.524
0.492
0.000
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.597
0.559
0.524
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.597
0.559
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.597
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.938
Result
1.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.597
0.559
0.524
0.492
0.461
0.432
0.405
0.380
Exact
1.000
0.936
0.875
0.819
0.766
0.717
0.670
0.627
0.587
0.549
0.513
0.480
0.449
0.420
0.393
0.368
Error
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.009
Problem 5.101
New Eq. 5.37:
[Difficulty: 3]
 u i 1  1   x u i  2  x  cos 2 x i 
N =4
x = 0.333
x
0.000
0.333
0.667
1.000
Eq. 5.34 (LHS)
1.000
-1.000
0.000
0.000
0.000
1.333
-1.000
0.000
0.000
0.000
1.333
-1.000
0.000
0.000
0.000
1.333
(RHS)
0
0.52392
0.15683
-0.2774
Inverse Matrix
1.000
0.750
0.563
0.422
0.000
0.750
0.563
0.422
0.000
0.000
0.750
0.563
0.000
0.000
0.000
0.750
Result
0.000
0.393
0.412
0.101
Eq. 5.34 (LHS)
1.000
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.143
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.143
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.143
-1.000
0.000
0.000
0.000
Exact
0.000
0.522
0.666
0.414
Error
0.000
0.004
0.016
0.024
0.212
N =8
x = 0.143
0.000
0.000
0.000
0.000
1.143
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.143
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.143
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.143
(RHS)
0
0.27413
0.24032
0.18703
0.11857
0.0405
-0.0409
-0.1189
x
0.000
0.143
0.286
0.429
0.571
0.714
0.857
1.000
N = 16
x = 0.067
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Inverse Matrix
1
1.000
0.875
0.766
0.670
0.586
0.513
0.449
0.393
2
0.000
0.875
0.766
0.670
0.586
0.513
0.449
0.393
3
0.000
0.000
0.875
0.766
0.670
0.586
0.513
0.449
4
0.000
0.000
0.000
0.875
0.766
0.670
0.586
0.513
5
0.000
0.000
0.000
0.000
0.875
0.766
0.670
0.586
6
0.000
0.000
0.000
0.000
0.000
0.875
0.766
0.670
7
0.000
0.000
0.000
0.000
0.000
0.000
0.875
0.766
8
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.875
Result
0.000
0.240
0.420
0.531
0.569
0.533
0.431
0.273
Eq. 5.34 (LHS)
1
1.000
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
2
0.000
1.067
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
3
0.000
0.000
1.067
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
4
0.000
0.000
0.000
1.067
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
5
0.000
0.000
0.000
0.000
1.067
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
6
0.000
0.000
0.000
0.000
0.000
1.067
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
7
8
9
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
1.067 0.000 0.000
-1.000 1.067 0.000
0.000 -1.000 1.067
0.000 0.000 -1.000
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
10
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.067
-1.000
0.000
0.000
0.000
0.000
0.000
Exact
0.000
0.263
0.469
0.606
0.668
0.653
0.565
0.414
Error
0.000
0.000
0.000
0.001
0.001
0.002
0.002
0.002
0.094
11
12
13
14
15
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
1.067 0.000 0.000 0.000 0.000
-1.000 1.067 0.000 0.000 0.000
0.000 -1.000 1.067 0.000 0.000
0.000 0.000 -1.000 1.067 0.000
0.000 0.000 0.000 -1.000 1.067
0.000 0.000 0.000 0.000 -1.000
16
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.067
(RHS)
0
0.13215
0.12862
0.12281
0.11482
0.10478
0.09289
0.07935
0.06441
0.04831
0.03137
0.01386
-0.0039
-0.0216
-0.0389
-0.0555
Inverse Matrix
1.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.597
0.559
0.524
0.492
0.461
0.432
0.405
0.380
x
0.000
0.067
0.133
0.200
0.267
0.333
0.400
0.467
0.533
0.600
0.667
0.733
0.800
0.867
0.933
1.000
N
4
8
16
x
0.333
0.143
0.067
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.597
0.559
0.524
0.492
0.461
0.432
0.405
0.380
Error
0.212
0.094
0.044
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.597
0.559
0.524
0.492
0.461
0.432
0.405
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.597
0.559
0.524
0.492
0.461
0.432
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.597
0.559
0.524
0.492
0.461
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.597
0.559
0.524
0.492
0.000
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.597
0.559
0.524
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.597
0.559
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.597
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.938
Result
0.000
0.124
0.237
0.337
0.424
0.495
0.552
0.591
0.615
0.622
0.612
0.587
0.547
0.492
0.425
0.346
Exact
0.000
0.129
0.247
0.352
0.445
0.522
0.584
0.630
0.659
0.671
0.666
0.645
0.608
0.557
0.491
0.414
Error
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.044
Problem 5.102
[Difficulty: 3]
New Eq. 5.37:  ui1 
1 xui  x  2xi2  xi 
N =4
x = 0.333
x
0.000
0.333
0.667
1.000
Eq. 5.34 (LHS)
1.000
-1.000
0.000
0.000
0.000
1.333
-1.000
0.000
0.000
0.000
1.333
-1.000
0.000
0.000
0.000
1.333
(RHS)
3
0.18519
0.51852
1
Inverse Matrix
1.000
0.750
0.563
0.422
0.000
0.750
0.563
0.422
0.000
0.000
0.750
0.563
0.000
0.000
0.000
0.750
Result
3.000
2.389
2.181
2.385
Eq. 5.34 (LHS)
1.000
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.143
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.143
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.143
-1.000
0.000
0.000
0.000
Exact
3.000
2.222
1.889
2.000
Error
0.000
0.007
0.021
0.037
0.256
N =8
x = 0.143
0.000
0.000
0.000
0.000
1.143
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.143
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.143
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.143
(RHS)
3
0.02624
0.06414
0.1137
0.17493
0.24781
0.33236
0.42857
x
0.000
0.143
0.286
0.429
0.571
0.714
0.857
1.000
N = 16
x = 0.067
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Inverse Matrix
1
1.000
0.875
0.766
0.670
0.586
0.513
0.449
0.393
2
0.000
0.875
0.766
0.670
0.586
0.513
0.449
0.393
3
0.000
0.000
0.875
0.766
0.670
0.586
0.513
0.449
4
0.000
0.000
0.000
0.875
0.766
0.670
0.586
0.513
5
0.000
0.000
0.000
0.000
0.875
0.766
0.670
0.586
6
0.000
0.000
0.000
0.000
0.000
0.875
0.766
0.670
7
0.000
0.000
0.000
0.000
0.000
0.000
0.875
0.766
8
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.875
Result
3.000
2.648
2.373
2.176
2.057
2.017
2.055
2.174
Eq. 5.34 (LHS)
1
1.000
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
2
0.000
1.067
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
3
0.000
0.000
1.067
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
4
0.000
0.000
0.000
1.067
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
5
0.000
0.000
0.000
0.000
1.067
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
6
0.000
0.000
0.000
0.000
0.000
1.067
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
7
8
9
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
1.067 0.000 0.000
-1.000 1.067 0.000
0.000 -1.000 1.067
0.000 0.000 -1.000
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
10
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.067
-1.000
0.000
0.000
0.000
0.000
0.000
Exact
3.000
2.612
2.306
2.082
1.939
1.878
1.898
2.000
Error
0.000
0.000
0.001
0.001
0.002
0.002
0.003
0.004
0.113
11
12
13
14
15
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
1.067 0.000 0.000 0.000 0.000
-1.000 1.067 0.000 0.000 0.000
0.000 -1.000 1.067 0.000 0.000
0.000 0.000 -1.000 1.067 0.000
0.000 0.000 0.000 -1.000 1.067
0.000 0.000 0.000 0.000 -1.000
16
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.067
(RHS)
3
0.00504
0.01126
0.01867
0.02726
0.03704
0.048
0.06015
0.07348
0.088
0.1037
0.12059
0.13867
0.15793
0.17837
0.2
Inverse Matrix
1.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.597
0.559
0.524
0.492
0.461
0.432
0.405
0.380
x
0.000
0.067
0.133
0.200
0.267
0.333
0.400
0.467
0.533
0.600
0.667
0.733
0.800
0.867
0.933
1.000
N
4
8
16
x
0.333
0.143
0.067
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.597
0.559
0.524
0.492
0.461
0.432
0.405
0.380
Error
0.256
0.113
0.054
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.597
0.559
0.524
0.492
0.461
0.432
0.405
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.597
0.559
0.524
0.492
0.461
0.432
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.597
0.559
0.524
0.492
0.461
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.597
0.559
0.524
0.492
0.000
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.597
0.559
0.524
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.597
0.559
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.597
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.637
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.679
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.724
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.772
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.824
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.938
0.879
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.938
Result
3.000
2.817
2.652
2.503
2.373
2.259
2.163
2.084
2.023
1.979
1.952
1.943
1.952
1.978
2.022
2.083
Exact
3.000
2.809
2.636
2.480
2.342
2.222
2.120
2.036
1.969
1.920
1.889
1.876
1.880
1.902
1.942
2.000
Error
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.054
Problem 5.103
[Difficulty: 3]
Equation of motion:
M
u
du
du
A   A
 
dt
dy

du
 A 
 
u  0
dt
 M 
du
 k u  0
dt
New Eq. 5.37:
 u i  1  1  k  t u i  0
2
N =4
t  0.333
A = 0.0025 m
 = 0.5 mm
Eq. 5.34 (LHS)
1.000
-1.000
0.000
0.000
t
0.000
0.333
0.667
1.000
0.000
0.000
0.000
1.250 0.000 0.000
-1.000 1.250 0.000
0.000 -1.000 1.250
Inverse Matrix
1.000
0.800
0.640
0.512
0.000
0.800
0.640
0.512
0.000
0.000
0.800
0.640
0.000
0.000
0.000
0.800
Eq. 5.34 (LHS)
1.000
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.107
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.107
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.107
-1.000
0.000
0.000
0.000
(RHS)
1
=
M =
0
0
0
k =
Result
1.000
0.800
0.640
0.512
Exact
1.000
0.779
0.607
0.472
Error
0.0E+00
1.1E-04
2.8E-04
3.9E-04
0.028
N =8
t  0.143
0.000
0.000
0.000
0.000
1.107
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.107
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.107
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.107
(RHS)
1
0
0
0
0
0
0
0
0.45 N.s/m2
3
kg
0.75 s
-1
t
0.000
0.143
0.286
0.429
0.571
0.714
0.857
1.000
N = 16
t  0.067
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Inverse Matrix
1
1.000
0.903
0.816
0.737
0.666
0.601
0.543
0.490
2
0.000
0.903
0.816
0.737
0.666
0.601
0.543
0.490
3
0.000
0.000
0.903
0.816
0.737
0.666
0.601
0.543
4
0.000
0.000
0.000
0.903
0.816
0.737
0.666
0.601
5
0.000
0.000
0.000
0.000
0.903
0.816
0.737
0.666
6
0.000
0.000
0.000
0.000
0.000
0.903
0.816
0.737
7
0.000
0.000
0.000
0.000
0.000
0.000
0.903
0.816
8
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.903
Result
1.000
0.903
0.816
0.737
0.666
0.601
0.543
0.490
Eq. 5.34 (LHS)
1
1.000
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
2
0.000
1.050
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
3
0.000
0.000
1.050
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
4
0.000
0.000
0.000
1.050
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
5
0.000
0.000
0.000
0.000
1.050
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
6
0.000
0.000
0.000
0.000
0.000
1.050
-1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
7
8
9
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
1.050 0.000 0.000
-1.000 1.050 0.000
0.000 -1.000 1.050
0.000 0.000 -1.000
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
0.000 0.000 0.000
10
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.050
-1.000
0.000
0.000
0.000
0.000
0.000
Exact
1.000
0.898
0.807
0.725
0.651
0.585
0.526
0.472
Error
0.0E+00
2.9E-06
9.5E-06
1.7E-05
2.5E-05
3.2E-05
3.7E-05
4.1E-05
0.013
11
12
13
14
15
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
1.050 0.000 0.000 0.000 0.000
-1.000 1.050 0.000 0.000 0.000
0.000 -1.000 1.050 0.000 0.000
0.000 0.000 -1.000 1.050 0.000
0.000 0.000 0.000 -1.000 1.050
0.000 0.000 0.000 0.000 -1.000
16
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.050
(RHS)
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Inverse Matrix
1.000
0.952
0.907
0.864
0.823
0.784
0.746
0.711
0.677
0.645
0.614
0.585
0.557
0.530
0.505
0.481
t
0.000
0.067
0.133
0.200
0.267
0.333
0.400
0.467
0.533
0.600
0.667
0.733
0.800
0.867
0.933
1.000
N
4
8
16
t
0.333
0.143
0.067
0.000
0.952
0.907
0.864
0.823
0.784
0.746
0.711
0.677
0.645
0.614
0.585
0.557
0.530
0.505
0.481
Error
0.028
0.013
0.006
0.000
0.000
0.952
0.907
0.864
0.823
0.784
0.746
0.711
0.677
0.645
0.614
0.585
0.557
0.530
0.505
0.000
0.000
0.000
0.952
0.907
0.864
0.823
0.784
0.746
0.711
0.677
0.645
0.614
0.585
0.557
0.530
0.000
0.000
0.000
0.000
0.952
0.907
0.864
0.823
0.784
0.746
0.711
0.677
0.645
0.614
0.585
0.557
0.000
0.000
0.000
0.000
0.000
0.952
0.907
0.864
0.823
0.784
0.746
0.711
0.677
0.645
0.614
0.585
0.000
0.000
0.000
0.000
0.000
0.000
0.952
0.907
0.864
0.823
0.784
0.746
0.711
0.677
0.645
0.614
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.952
0.907
0.864
0.823
0.784
0.746
0.711
0.677
0.645
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.952
0.907
0.864
0.823
0.784
0.746
0.711
0.677
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.952
0.907
0.864
0.823
0.784
0.746
0.711
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.952
0.907
0.864
0.823
0.784
0.746
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.952
0.907
0.864
0.823
0.784
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.952
0.907
0.864
0.823
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.952
0.907
0.864
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.952
0.907
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.952
Result
1.000
0.952
0.907
0.864
0.823
0.784
0.746
0.711
0.677
0.645
0.614
0.585
0.557
0.530
0.505
0.481
Exact
1.000
0.951
0.905
0.861
0.819
0.779
0.741
0.705
0.670
0.638
0.607
0.577
0.549
0.522
0.497
0.472
Error
0.0E+00
8.3E-08
3.0E-07
6.1E-07
9.9E-07
1.4E-06
1.8E-06
2.2E-06
2.7E-06
3.0E-06
3.4E-06
3.7E-06
4.0E-06
4.3E-06
4.5E-06
4.7E-06
0.006
Problem 5.104
ui 
x 
[Difficulty: 3]
ug i 1  x ug2i
1  2x ug i
0.333
x
0.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
Iteration
0
1
2
3
4
5
6
Exact
0.333
1.000
0.800
0.791
0.791
0.791
0.791
0.791
0.750
0.667
1.000
0.800
0.661
0.650
0.650
0.650
0.650
0.600
1.000
1.000
0.800
0.661
0.560
0.550
0.550
0.550
0.500
Residuals
0.204
0.127
0.068
0.007
0.000
0.000
1E+00
1.0
1E-01
1E-02
1E-03
Residual R
Iterations = 2
Iterations = 4
Iterations = 6
Exact Solution
0.9
1E-04
0.8
1E-05
u
1E-06
0.7
1E-07
1E-08
0.6
1E-09
0.5
1E-10
0
1
2
3
Iteration N
4
5
6
0.0
0.2
0.4
0.6
x
0.8
1.0
Problem 5.105
ui 
x 
[Difficulty: 3]
ug i 1  x ug2i
1  2x ug i
0.0667
Iteration
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
0.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
0.067
1.000
0.941
0.941
0.941
0.941
0.941
0.941
0.941
0.941
0.941
0.941
0.941
0.941
0.941
0.941
0.941
0.941
0.941
0.941
0.941
0.941
0.941
0.941
0.941
0.941
0.941
0.941
0.941
0.941
0.941
0.941
0.133
1.000
0.941
0.889
0.888
0.888
0.888
0.888
0.888
0.888
0.888
0.888
0.888
0.888
0.888
0.888
0.888
0.888
0.888
0.888
0.888
0.888
0.888
0.888
0.888
0.888
0.888
0.888
0.888
0.888
0.888
0.888
0.200
1.000
0.941
0.889
0.842
0.841
0.841
0.841
0.841
0.841
0.841
0.841
0.841
0.841
0.841
0.841
0.841
0.841
0.841
0.841
0.841
0.841
0.841
0.841
0.841
0.841
0.841
0.841
0.841
0.841
0.841
0.841
0.267
1.000
0.941
0.889
0.842
0.799
0.799
0.799
0.799
0.799
0.799
0.799
0.799
0.799
0.799
0.799
0.799
0.799
0.799
0.799
0.799
0.799
0.799
0.799
0.799
0.799
0.799
0.799
0.799
0.799
0.799
0.799
0.333
1.000
0.941
0.889
0.842
0.799
0.761
0.760
0.760
0.760
0.760
0.760
0.760
0.760
0.760
0.760
0.760
0.760
0.760
0.760
0.760
0.760
0.760
0.760
0.760
0.760
0.760
0.760
0.760
0.760
0.760
0.760
0.400
1.000
0.941
0.889
0.842
0.799
0.761
0.726
0.725
0.725
0.725
0.725
0.725
0.725
0.725
0.725
0.725
0.725
0.725
0.725
0.725
0.725
0.725
0.725
0.725
0.725
0.725
0.725
0.725
0.725
0.725
0.725
x
0.467
1.000
0.941
0.889
0.842
0.799
0.761
0.726
0.694
0.693
0.693
0.693
0.693
0.693
0.693
0.693
0.693
0.693
0.693
0.693
0.693
0.693
0.693
0.693
0.693
0.693
0.693
0.693
0.693
0.693
0.693
0.693
0.533
1.000
0.941
0.889
0.842
0.799
0.761
0.726
0.694
0.664
0.664
0.664
0.664
0.664
0.664
0.664
0.664
0.664
0.664
0.664
0.664
0.664
0.664
0.664
0.664
0.664
0.664
0.664
0.664
0.664
0.664
0.664
0.600
1.000
0.941
0.889
0.842
0.799
0.761
0.726
0.694
0.664
0.637
0.637
0.637
0.637
0.637
0.637
0.637
0.637
0.637
0.637
0.637
0.637
0.637
0.637
0.637
0.637
0.637
0.637
0.637
0.637
0.637
0.637
0.667
1.000
0.941
0.889
0.842
0.799
0.761
0.726
0.694
0.664
0.637
0.612
0.612
0.612
0.612
0.612
0.612
0.612
0.612
0.612
0.612
0.612
0.612
0.612
0.612
0.612
0.612
0.612
0.612
0.612
0.612
0.612
0.733
1.000
0.941
0.889
0.842
0.799
0.761
0.726
0.694
0.664
0.637
0.612
0.589
0.589
0.589
0.589
0.589
0.589
0.589
0.589
0.589
0.589
0.589
0.589
0.589
0.589
0.589
0.589
0.589
0.589
0.589
0.589
0.800
1.000
0.941
0.889
0.842
0.799
0.761
0.726
0.694
0.664
0.637
0.612
0.589
0.568
0.567
0.567
0.567
0.567
0.567
0.567
0.567
0.567
0.567
0.567
0.567
0.567
0.567
0.567
0.567
0.567
0.567
0.567
0.867
1.000
0.941
0.889
0.842
0.799
0.761
0.726
0.694
0.664
0.637
0.612
0.589
0.568
0.548
0.547
0.547
0.547
0.547
0.547
0.547
0.547
0.547
0.547
0.547
0.547
0.547
0.547
0.547
0.547
0.547
0.547
0.933
1.000
0.941
0.889
0.842
0.799
0.761
0.726
0.694
0.664
0.637
0.612
0.589
0.568
0.548
0.529
0.529
0.529
0.529
0.529
0.529
0.529
0.529
0.529
0.529
0.529
0.529
0.529
0.529
0.529
0.529
0.529
1.000
1.000
0.941
0.889
0.842
0.799
0.761
0.726
0.694
0.664
0.637
0.612
0.589
0.568
0.548
0.529
0.512
0.511
0.511
0.511
0.511
0.511
0.511
0.511
0.511
0.511
0.511
0.511
0.511
0.511
0.511
0.511
Exact
1.000
0.938
0.882
0.833
0.789
0.750
0.714
0.682
0.652
0.625
0.600
0.577
0.556
0.536
0.517
0.500
1.0
Iterations = 10
Iterations = 20
Iterations = 30
Exact Solution
0.9
0.8
u
0.7
0.6
0.5
0.0
0.2
0.4
0.6
x
0.8
1.0
Problem 5.106
[Difficulty: 3]
ui  ui 1 1
 0
ui
x
ui  ui  ug i
1
1
1  ui 


1
ui u g i  ui u g i 
u g i 
ui  ui 1 1  ui  ug i
1

ug i 
ug i
x
ui  ui 1 1 
u

2 i
ug i 
ug i
x
x 
 x 
2 x
ui 1  2   ui 1 
 ug 
ug i
i 

2 x
ui 1 
ug i
ui 
x
1 2
ug i

0



0


1.500
x
Iteration
0
1
2
3
4
5
6
Exact
x 
Iteration
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
0.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
1.500
3.000
2.400
2.366
2.366
2.366
2.366
2.366
2.449
3.000
3.000
2.400
1.555
1.151
1.816
1.310
0.601
1.732
4.500
3.000
2.400
1.555
-0.986
-7.737
2.260
-0.025
0.000
0.300
3.000
2.897
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
0.600
3.000
2.897
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
0.900
3.000
2.897
2.789
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
0.300
0.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
1.200
3.000
2.897
2.789
2.677
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
1.500
3.000
2.897
2.789
2.677
2.560
2.438
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
1.800
3.000
2.897
2.789
2.677
2.560
2.438
2.308
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
x
2.100
3.000
2.897
2.789
2.677
2.560
2.438
2.308
2.170
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.400
3.000
2.897
2.789
2.677
2.560
2.438
2.308
2.170
2.023
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.700
3.000
2.897
2.789
2.677
2.560
2.438
2.308
2.170
2.023
1.862
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
3.000
3.000
2.897
2.789
2.677
2.560
2.438
2.308
2.170
2.023
1.862
1.686
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
3.300
3.000
2.897
2.789
2.677
2.560
2.438
2.308
2.170
2.023
1.862
1.686
1.487
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
3.600
3.000
2.897
2.789
2.677
2.560
2.438
2.308
2.170
2.023
1.862
1.686
1.487
1.254
1.233
1.233
1.233
1.233
1.233
1.233
1.233
3.900
3.000
2.897
2.789
2.677
2.560
2.438
2.308
2.170
2.023
1.862
1.686
1.487
1.254
0.958
0.901
0.899
0.899
0.899
0.899
0.899
4.200
3.000
2.897
2.789
2.677
2.560
2.438
2.308
2.170
2.023
1.862
1.686
1.487
1.254
0.958
0.493
1.349
0.544
14.403
0.859
0.338
4.500
3.000
2.897
2.789
2.677
2.560
2.438
2.308
2.170
2.023
1.862
1.686
1.487
1.254
0.958
0.493
3.091
1.192
0.051
-0.024
-0.051
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
3.000
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.896
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.789
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.677
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.560
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.436
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.306
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.168
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
2.019
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.858
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.679
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.476
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
1.233
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.899
0.538
5.953
0.805
0.286
0.450
0.900
0.369
0.605
-0.517
-17.059
0.935
0.392
0.663
-0.020
-0.041
-0.088
-0.204
-0.621
8.435
0.831
0.313
0.494
1.379
0.551
-16.722
0.936
0.392
0.664
-0.014
-0.029
-0.061
-0.135
-0.347
-1.765
1.371
0.549
-40.363
0.914
0.379
0.627
-0.243
-0.105
-0.239
-1.998
1.195
-0.273
-0.876
2.601
0.145
0.266
0.858
-29.971
0.955
-0.352
-1.662
0.383
1.534
-0.549
198.629
-0.624
41.087
0.817
-0.765
2.623
1.203
0.066
0.377
0.591
-4.391
0.813
-1.376
0.483
4.578
-0.270
-0.603
-4.389
1.532
0.180
5.316
0.810
-0.668
4.652
Exact
3.000
2.898
2.793
2.683
2.569
2.449
2.324
2.191
2.049
1.897
1.732
1.549
1.342
1.095
0.775
0.000
Here are graphs comparing the numerical and exact solutions.
3.5
3.5
Iterations = 2
Iterations = 4
Iterations = 6
Exact Solution
3.0
2.5
Iterations = 20
Iterations = 40
Iterations = 60
Exact Solution
3.0
2.5
2.0
2.0
u
u
1.5
1.5
1.0
1.0
0.5
0.5
0.0
0.0
0
1
2
3
x
4
5
0
1
2
3
x
4
5
Problem 5.107
[Difficulty: 3]
du
2
 k U  u 
dt
v U u
M
dv   du
dv
 kv 2
dt
dv
k 2

v 0
dt M
M
t 
k =
M =
1.000
0.02
0.3
vi2  2 v g i vi  v g2 i
vi  vi 1
k

2 v g i vi  v g2 i  0
t
M
k
 t v g2 i
v g i 1 
M
vi 
k
1 2
t v g i
M


lbf.s2/ft2
slug
t
Iteration
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
0.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
1.000
25.000
15.385
13.365
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
2.000
25.000
15.385
10.213
8.603
8.477
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
3.000
25.000
15.385
10.213
7.269
6.158
6.043
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
4.000
25.000
15.385
10.213
7.269
5.480
4.715
4.621
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
5.000
25.000
15.385
10.213
7.269
5.480
4.323
3.781
3.706
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
6.000
25.000
15.385
10.213
7.269
5.480
4.323
3.533
3.136
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
7.000
25.000
15.385
10.213
7.269
5.480
4.323
3.533
2.967
2.668
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
8.000
25.000
15.385
10.213
7.269
5.480
4.323
3.533
2.967
2.547
2.314
2.274
2.273
2.273
2.273
2.273
2.273
2.273
2.273
2.273
2.273
2.273
2.273
2.273
2.273
2.273
2.273
2.273
2.273
2.273
9.000
25.000
15.385
10.213
7.269
5.480
4.323
3.533
2.967
2.547
2.224
2.039
2.006
2.005
2.005
2.005
2.005
2.005
2.005
2.005
2.005
2.005
2.005
2.005
2.005
2.005
2.005
2.005
2.005
2.005
10.000
25.000
15.385
10.213
7.269
5.480
4.323
3.533
2.967
2.547
2.224
1.970
1.820
1.792
1.791
1.791
1.791
1.791
1.791
1.791
1.791
1.791
1.791
1.791
1.791
1.791
1.791
1.791
1.791
1.791
11.000
25.000
15.385
10.213
7.269
5.480
4.323
3.533
2.967
2.547
2.224
1.970
1.765
1.641
1.617
1.617
1.617
1.617
1.617
1.617
1.617
1.617
1.617
1.617
1.617
1.617
1.617
1.617
1.617
1.617
12.000
25.000
15.385
10.213
7.269
5.480
4.323
3.533
2.967
2.547
2.224
1.970
1.765
1.597
1.493
1.473
1.472
1.472
1.472
1.472
1.472
1.472
1.472
1.472
1.472
1.472
1.472
1.472
1.472
1.472
13.000
25.000
15.385
10.213
7.269
5.480
4.323
3.533
2.967
2.547
2.224
1.970
1.765
1.597
1.457
1.369
1.351
1.351
1.351
1.351
1.351
1.351
1.351
1.351
1.351
1.351
1.351
1.351
1.351
1.351
14.000
25.000
15.385
10.213
7.269
5.480
4.323
3.533
2.967
2.547
2.224
1.970
1.765
1.597
1.457
1.338
1.263
1.247
1.247
1.247
1.247
1.247
1.247
1.247
1.247
1.247
1.247
1.247
1.247
1.247
15.000
25.000
15.385
10.213
7.269
5.480
4.323
3.533
2.967
2.547
2.224
1.970
1.765
1.597
1.457
1.338
1.237
1.172
1.158
1.158
1.158
1.158
1.158
1.158
1.158
1.158
1.158
1.158
1.158
1.158
29
30
31
32
33
34
35
36
37
38
39
40
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
25.000
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
13.267
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
8.476
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
6.042
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
4.620
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.705
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
3.075
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.618
2.273
2.273
2.273
2.273
2.273
2.273
2.273
2.273
2.273
2.273
2.273
2.273
2.005
2.005
2.005
2.005
2.005
2.005
2.005
2.005
2.005
2.005
2.005
2.005
1.791
1.791
1.791
1.791
1.791
1.791
1.791
1.791
1.791
1.791
1.791
1.791
1.617
1.617
1.617
1.617
1.617
1.617
1.617
1.617
1.617
1.617
1.617
1.617
1.472
1.472
1.472
1.472
1.472
1.472
1.472
1.472
1.472
1.472
1.472
1.472
1.351
1.351
1.351
1.351
1.351
1.351
1.351
1.351
1.351
1.351
1.351
1.351
1.247
1.247
1.247
1.247
1.247
1.247
1.247
1.247
1.247
1.247
1.247
1.247
1.158
1.158
1.158
1.158
1.158
1.158
1.158
1.158
1.158
1.158
1.158
1.158
Above values are for v! To get u we compute u = U - v
Iteration
10
20
40
0.000
0.000
0.000
11.733
11.733
11.733
16.524
16.524
16.524
18.958
18.958
18.958
20.380
20.380
20.380
21.295
21.295
21.295
21.925
21.925
21.925
22.382
22.382
22.382
22.726
22.727
22.727
22.961
22.995
22.995
23.030
23.209
23.209
23.030
23.383
23.383
23.030
23.528
23.528
23.030
23.649
23.649
23.030
23.753
23.753
23.030
23.842
23.842
Exact
0.000
15.625
19.231
20.833
21.739
22.321
22.727
23.026
23.256
23.438
23.585
23.707
23.810
23.897
23.973
24.038
30
25
u (ft/s)
20
Iterations = 10
Iterations = 20
Iterations = 40
Exact Solution
15
10
5
0
0
2
4
6
8
t (s)
10
12
14
16
Problem 6.1
[Difficulty: 2]
Given:
Velocity field
Find:
Acceleration of particle and pressure gradient at (1,1)
Solution:
NOTE: Units of B are s-1 not ft-1s-1
Basic equations
(2
u ( x , y ) = A⋅ y − x
For this flow
ax = u ⋅
∂
∂x
u + v⋅
∂
∂y
2
) − B⋅ x
v ( x , y ) = 2 ⋅ A⋅ x ⋅ y + B⋅ y
(2
u = ⎡⎣A⋅ y − x
(
2
) − B⋅x⎤⎦ ⋅∂ ⎡⎣A⋅(y2 − x2) − B⋅x⎤⎦ + (2⋅A⋅x⋅y + B⋅y)⋅∂ ⎡⎣A⋅(y2 − x2) − B⋅x⎤⎦
∂x
2
ax = ( B + 2 ⋅ A⋅ x ) ⋅ A⋅ x + B⋅ x + A⋅ y
ay = u ⋅
∂
∂x
v + v⋅
∂
∂y
(2
v = ⎡⎣A⋅ y − x
2
2
∂y
)
) − B⋅x⎤⎦ ⋅∂ (2⋅A⋅x⋅y + B⋅y) + (2⋅A⋅x⋅y + B⋅y)⋅∂ (2⋅A⋅x⋅y + B⋅y)
∂x
∂y
(2
ay = ( B + 2 ⋅ A⋅ x ) ⋅ ( B⋅ y + 2 ⋅ A⋅ x ⋅ y ) − 2 ⋅ A⋅ y ⋅ ⎡⎣B⋅ x + A⋅ x − y
ax = ( 1 + 2 ⋅ 1 ⋅ 1 ) ⋅
Hence at (1,1)
ay = ( 1 + 2 ⋅ 1 ⋅ 1 ) ⋅
a =
2
ax + ay
1
s
1
s
(
)s
)⎦
2⎤
2 ft
2
× 1⋅ 1 + 1⋅ 1 + 1⋅ 1 ⋅
× ( 1⋅ 1 + 2⋅ 1⋅ 1⋅ 1) ⋅
2
ft
s
ft
ax = 9 ⋅
− 2⋅ 1⋅ 1⋅
1
s
(2
× ⎡⎣1 ⋅ 1 + 1 ⋅ 1 − 1
)⎦ ⋅ s
2 ⎤ ft
⎛ ay ⎞
θ = atan⎜
ay = 7 ⋅
2
s
a = 11.4⋅
⎝ ax ⎠
2
s
ft
ft
θ = 37.9⋅ deg
2
s
For the pressure gradient
lbf
∂
∂x
p = ρ⋅ g x − ρ⋅ ax = −2 ⋅
slug
ft
3
× 9⋅
ft
2
s
2
×
lbf ⋅ s
∂
slug⋅ ft
∂x
2
p = −18⋅
ft
= −0.125 ⋅
ft
psi
ft
lbf
∂
∂y
p = ρ⋅ g y − ρ⋅ ay = 2 ⋅
slug
ft
3
× ( −32.2 − 7 ) ⋅
ft
2
s
2
×
lbf ⋅ s
∂
slug⋅ ft
∂y
2
p = −78.4⋅
ft
ft
= −0.544 ⋅
psi
ft
Problem 6.2
[Difficulty: 2]
Given:
Velocity field
Find:
Acceleration of particle and pressure gradient at (2,2)
Solution:
Basic equations
Given data
For this flow
A = 1⋅
B = 3⋅
s
1
s
x = 2⋅ m
y = 2⋅ m
∂
∂x
ay = u ⋅
u + v⋅
∂
∂x
∂y
v + v⋅
ax = ( 1 + 9)
a =
∂
2
1
s
∂
∂y
kg
3
v ( x , y ) = B⋅ x − A⋅ y
u = ( A ⋅ x + B⋅ y ) ⋅
∂
∂x
v = ( A ⋅ x + B⋅ y ) ⋅
( A ⋅ x + B⋅ y ) + ( B⋅ x − A ⋅ y ) ⋅
∂
∂x
∂
∂y
( B⋅ x − A ⋅ y ) + ( B⋅ x − A ⋅ y ) ⋅
m
× 2⋅ m
ax = 20
2
θ = atan ⎜
ax + ay
ρ = 999 ⋅
m
u ( x , y ) = A⋅ x + B⋅ y
ax = u ⋅
Hence at (2,2)
1
ay = ( 1 + 9)
s
⎛ ay ⎞
a = 28.28
⎝ ax ⎠
∂
∂y
1
s
(
2
2
)
(
2
2
( A ⋅ x + B⋅ y )
ax = A + B ⋅ x
( B⋅ x − A ⋅ y )
ay = A + B ⋅ y
× 2⋅ m
ay = 20
m
)
m
s
θ = 45⋅ deg
s
For the pressure gradient
∂
kg
m
N⋅s
2
∂
× 20⋅
×
p = ρ⋅ g x − ρ⋅ ax = −999⋅
2
kg⋅ m
3
∂x
s
m
∂x
2
kg
m
N⋅s
p = −ρ⋅ g y − ρ⋅ ay = 999⋅
× ( −9.81 − 20) ⋅
×
3
2
kg⋅ m
∂y
m
s
∂
∂
∂y
p = −20000⋅
Pa
p = −29800⋅
Pa
m
m
= −20.0⋅
kPa
= −29.8⋅
kPa
m
m
Problem 6.3
[Difficulty: 2]
Given:
Velocity field
Find:
Acceleration of particle and pressure gradient at (1,2)
Solution:
Basic equations
Given data
A = 1⋅
1
B = 2⋅
s
m
2
x = 1⋅ m
y = 2⋅ m
t = 5⋅ s
ρ = 999 ⋅
s
u ( x , y , t) = −A⋅ x + B⋅ t
kg
3
m
v ( x , y , t) = A⋅ y + B⋅ t
The acceleration components and values are
axt( x , y , t) =
∂
∂t
u ( x , y , t) = B
∂
∂t
m
axt( x , y , t) = 2
2
s
axc( x , y , t) = u ( x , y , t) ⋅
ayt( x , y , t) =
axt( x , y , t) = B
∂
∂x
u ( x , y , t) + v ( x , y , t) ⋅
v ( x , y , t)
ayc( x , y , t) = u ( x , y , t) ⋅
∂
∂y
2
u ( x , y , t) axc( x , y , t) = A ⋅ x − A⋅ B⋅ t
axc( x , y , t) = −9
m
2
s
ayt( x , y , t) = B
m
ayt( x , y , t) = 2
2
s
∂
∂x
v ( x , y , t) + v ( x , y , t) ⋅
ax ( x , y , t) = axt( x , y , t) + axc( x , y , t)
∂
∂y
2
ayc( x , y , t) = y ⋅ A + B⋅ t⋅ A
v( x , y , t)
ayc( x , y , t) = 12
2
s
2
ax ( x , y , t ) = x ⋅ A − B ⋅ t ⋅ A + B
ax ( x , y , t) = −7
m
2
s
ay ( x , y , t) = ayt( x , y , t) + ayc( x , y , t)
m
2
ay ( x , y , t ) = y ⋅ A + B ⋅ t ⋅ A + B
ay ( x , y , t) = 14
m
2
s
For overall acceleration
a( x , y , t ) =
2
ax ( x , y , t ) + ay ( x , y , t )
2
(x⋅A2 − B⋅t⋅A + B) + (y⋅A2 + B⋅t⋅A + B)
2
a( x , y , t ) =
2
a( x , y , t) = 15.7
m
2
s
⎛ ay ( x , y , t ) ⎞
θ = atan⎜
⎝
ax ( x , y , t )
⎠
For the pressure gradient we need
θ = −63.4⋅ deg
−ρ⋅ ax ( x , y , t) = 6.99⋅
kPa
m
−ρ⋅ ay ( x , y , t) = −13.99 ⋅
kPa
−ρ⋅ g = −9.80⋅
m
Hence for the pressure gradient
2
m
N⋅ s
kg
× 7⋅ ×
p = ρ⋅ g x − ρ⋅ ax = 999 ⋅
2
kg⋅ m
3
∂x
s
m
∂
∂
∂x
2
m
N⋅ s
kg
× ( −9.81 − 14) ⋅ ×
p = −ρ⋅ g y − ρ⋅ ay = 999 ⋅
2
kg
⋅m
3
∂y
s
m
∂
∂
∂y
p = 6990⋅
Pa
m
p = −23800 ⋅
= 6.99⋅
Pa
m
kPa
m
= −23.8⋅
kPa
m
kPa
m
Problem 6.4
Given:
Velocity field
Find:
Pressure gradient at (1,1) at 1 s
[Difficulty: 2]
Solution:
Basic equations
Given data
A = 2⋅
1
B = 1⋅
2
s
1
x = 1⋅ m
2
y = 1⋅ m
t = 1⋅ s
ρ = 1000⋅
s
kg
3
m
u ( x , y , t) = ( −A⋅ x + B⋅ y ) ⋅ t
v ( x , y , t) = ( A⋅ y + B⋅ x ) ⋅ t
The acceleration components and values are
axt( x , y , t) =
∂
∂t
u ( x , y , t) = B⋅ y − A⋅ x
∂
∂t
m
axt( x , y , t) = −1
2
s
axc( x , y , t) = u ( x , y , t) ⋅
ayt( x , y , t) =
axt( x , y , t) = B⋅ y − A⋅ x
∂
∂x
u ( x , y , t) + v ( x , y , t) ⋅
v ( x , y , t)
∂
∂y
(
2
2
u ( x , y , t) axc( x , y , t) = t ⋅ x ⋅ A + B
)
2
axc( x , y , t) = 5
m
2
s
ayt( x , y , t) = A⋅ y + B⋅ x
m
ayt( x , y , t) = 3
2
s
ayc( x , y , t) = u ( x , y , t) ⋅
∂
∂x
v ( x , y , t) + v ( x , y , t) ⋅
∂
∂y
v( x , y , t)
2
2
)
2
ayc( x , y , t) = 5
m
2
s
2 2
ax ( x , y , t) = axt( x , y , t) + axc( x , y , t)
(
ayc( x , y , t) = t ⋅ y ⋅ A + B
2 2
ax ( x , y , t ) = x ⋅ A ⋅ t − x ⋅ A + x ⋅ B ⋅ t + y ⋅ B
ax ( x , y , t ) = 4
m
2
s
2 2
ay ( x , y , t) = ayt( x , y , t) + ayc( x , y , t)
2 2
ay ( x , y , t ) = y ⋅ A ⋅ t + y ⋅ A + y ⋅ B ⋅ t + x ⋅ B
ay ( x , y , t ) = 8
m
2
s
Hence for the pressure gradient
∂
∂x
∂
∂y
p = −ρ⋅ ax = −1000⋅
kg
3
× 4⋅
kg
3
m
2
2
×
s
m
p = −ρ⋅ ay = −1000⋅
m
× 8⋅
m
2
s
N⋅ s
∂
kg⋅ m
∂x
2
×
N⋅ s
∂
kg⋅ m
∂y
p = −4000⋅
Pa
p = −8000⋅
Pa
m
m
= −4 ⋅
kPa
= −8 ⋅
kPa
m
m
Problem 6.5
[Difficulty: 2]
Given:
Velocity field
Find:
Acceleration of particle and pressure gradient at (1,1)
Solution:
Basic equations
ax = u ⋅
∂
∂x
(2
2
(2
2
u ( x , y ) = A⋅ x − y
For this flow
u + v⋅
∂
∂y
u = ⎡⎣A⋅ x − y
) − 3 ⋅ B⋅ x
v ( x , y ) = −2 ⋅ A⋅ x ⋅ y + 3 ⋅ B⋅ y
) − 3⋅B⋅x⎤⎦ ⋅∂ ⎡⎣A⋅(x2 − y2) − 3⋅B⋅x⎤⎦ + (−2⋅A⋅x⋅y + 3⋅B⋅y)⋅∂ ⎡⎣A⋅(x2 − y2) − 3⋅B⋅x⎤⎦
∂x
(
∂y
2
ax = ( 2 ⋅ A⋅ x − 3 ⋅ B) ⋅ A⋅ x − 3 ⋅ B⋅ x + A⋅ y
ay = u ⋅
∂
∂x
v + v⋅
∂
∂y
(2
v = ⎡⎣A⋅ x − y
2
2
)
) − 3⋅B⋅x⎤⎦ ⋅∂ (−2⋅A⋅x⋅y + 3⋅B⋅y) + (−2⋅A⋅x⋅y + 3⋅B⋅y)⋅∂ (−2⋅A⋅x⋅y + 3⋅B⋅y)
∂x
∂y
(2
ay = ( 3 ⋅ B⋅ y − 2 ⋅ A⋅ x ⋅ y ) ⋅ ( 3 ⋅ B − 2 ⋅ A⋅ x ) − 2 ⋅ A⋅ y ⋅ ⎡⎣A⋅ x − y
Hence at (1,1)
ax = ( 2 ⋅ 1 ⋅ 1 − 3 ⋅ 1 ) ⋅
1
s
2
ax + ay
) − 3⋅B⋅x⎤⎦
)s
2 ft
2
× 1⋅ 1 − 3⋅ 1⋅ 1 + 1⋅ 1 ⋅
ay = ( 3 ⋅ 1 ⋅ 1 − 2 ⋅ 1 ⋅ 1 ⋅ 1 ) ⋅
a =
(
2
1
× ( 3⋅ 1 − 2⋅ 1⋅ 1) ⋅
s
ft
ax = 1 ⋅
1
− 2⋅ 1⋅ 1⋅
s
s
⎛ ay ⎞
2
θ = atan⎜
(2
× ⎡⎣1 ⋅ 1 − 1
2
⎝ ax ⎠
ay = 7 ⋅
∂x
ft
ft
θ = 81.9⋅ deg
2
s
lbf
p = ρ⋅ g x − ρ⋅ ax = −2 ⋅
slug
ft
3
× 1⋅
ft
2
s
2
×
lbf ⋅ s
∂
slug⋅ ft
∂x
2
p = −2 ⋅
ft
ft
= −0.0139⋅
psi
ft
lbf
∂
∂y
p = ρ⋅ g y − ρ⋅ ay = 2 ⋅
slug
ft
3
× ( −32.2 − 7 ) ⋅
ft
2
s
2
×
lbf ⋅ s
∂
slug⋅ ft
∂y
2
s
For the pressure gradient
∂
2
s
) − 3⋅1⋅1⎤⎦ ⋅ fts
a = 7.1⋅
ft
2
p = −78.4⋅
ft
ft
= −0.544 ⋅
psi
ft
Problem 6.6
[Difficulty: 2]
Given:
Velocity field
Find:
Simplest y component of velocity; Acceleration of particle and pressure gradient at (2,1); pressure on x axis
Solution:
Basic equations
For this flow
u ( x , y ) = A⋅ x
Hence
v ( x , y ) = −A⋅ y
For acceleration
ax = u ⋅
∂
∂x
ay = u ⋅
⌠
⌠
⎮ ∂
so
u +
v ( x, y ) = −⎮
u dy = −⎮ A dy = −A ⋅ y + c
v =0
⌡
∂x
∂y
⎮ ∂x
⌡
is the simplest y component of velocity
∂
u + v⋅
∂
∂x
∂
u = A ⋅ x⋅
∂y
v + v⋅
∂
∂y
ax =
a =
⎛ 2 ⎞ × 2⋅ m
⎜s
⎝ ⎠
2
ax + ay
∂
∂x
v = A⋅ x ⋅
2
Hence at (2,1)
∂
2
( A ⋅ x) + ( −A ⋅ y ) ⋅
∂
∂x
∂
∂y
( −A⋅ y ) + ( −A⋅ y ) ⋅
2
2
ax = A ⋅ x
( A ⋅ x) = A ⋅ x
∂
∂y
2
ay = A ⋅ y
( −A⋅ y )
2
⎛ 2 ⎞ × 1⋅ m
⎜s
⎝ ⎠
⎛ ay ⎞
θ = atan⎜
⎝ ax ⎠
ax = 8
ay =
m
ay = 4
2
s
a = 8.94
m
θ = 26.6⋅ deg
2
s
2
N⋅ s
m
kg
∂
× 8⋅ ×
p = ρ⋅ g x − ρ⋅ ax = −1.50⋅
2
kg⋅ m
3
∂x
s
m
∂x
2
m
N⋅ s
kg
× 4⋅ ×
p = ρ⋅ g y − ρ⋅ ay = −1.50⋅
2
kg⋅ m
3
∂y
s
m
∂
∂
∂z
kg
p = ρ⋅ g z − ρ⋅ az = 1.50 ×
For the pressure on the x axis
1
2 2
p ( x ) = p 0 − ⋅ ρ⋅ A ⋅ x
2
3
× ( −9.81) ⋅
m
dp =
∂
∂x
p
p ( x ) = 190 ⋅ kPa −
x
0
2
⋅ 1.5⋅
2
s
⌠
p − p0 = ⎮
⌡
1
m
kg
3
m
(
∂y
N⋅ s
∂
kg⋅ m
∂y
⌠
ρ⋅ g x − ρ⋅ ax dx = ⎮
⌡
)
x
0
2
×
∂
2
×
2
s
For the pressure gradient
∂
m
p = −6 ⋅
Pa
m
Pa
m
p = −14.7⋅
Pa
m
(−ρ⋅A2⋅x) dx = − 12 ⋅ρ⋅A2⋅x2
2
⎛ 2 ⎞ × N⋅ s × x 2
⎜
kg⋅ m
⎝ s⎠
p = −12⋅
p ( x ) = 190 −
3
1000
⋅x
2
(p in kPa, x in m)
Problem 6.7
[Difficulty: 3]
Given:
Velocity field
Find:
Expressions for local, convective and total acceleration; evaluate at several points; evaluate pressure gradient
Solution:
A  2
The given data is
1
ω  1
s

Check for incompressible flow
x

Hence
x
u 
u 
1
ρ  2
s

y

y
kg
u  A x  sin( 2  π ω t)
3
v  A y  sin( 2  π ω t)
m
v 0
v  A sin( 2  π ω t)  A sin( 2  π ω t)  0
Incompressible flow
The governing equation for acceleration is
The local acceleration is then

x - component
t

y - component
t
u  2  π A ω x  cos( 2  π ω t)
v  2  π A ω y  cos( 2  π ω t)
For the present steady, 2D flow, the convective acceleration is
x - component
u
y - component
u

x

x
u  v
v  v
The total acceleration is then

2
y

u  A x  sin( 2  π ω t)  ( A sin( 2  π ω t) )  ( A y  sin( 2  π ω t) )  0  A  x  sin( 2  π ω t)
2
y
2
v  A x  sin( 2  π ω t)  0  ( A y  sin( 2  π ω t) )  ( A sin( 2  π ω t) )  A  y  sin( 2  π ω t)
x - component
y - component

t

t
u  u
v  u

x

x
u  v
v  v

y

y
2
2
u  2  π A ω x  cos( 2  π ω t)  A  x  sin( 2  π ω t)
2
2
v  2  π A ω y  cos( 2  π ω t)  A  y  sin( 2  π ω t)
2
Evaluating at point (1,1) at
t  0 s
12.6
Local
m
m
12.6
and
2
s
12.6
Total
m
m
12.6
and
2
12.6
and
2
s
t  1 s
12.6
Local
12.6
Convective
2
0
m
0
and
2
s
y
Evaluated at (1,1) and time
2
m
2
m
12.6
and
2
m
Convective
2
s
m
12.6
and
2
0
m
0
and
2
s
m
2
s
p  ρ
Du

Dt
x
p  ρ
Dv

Dt
x


2
2
p  ρ 2  π A ω y  cos( 2  π ω t)  A  y  sin( 2  π ω t)
t  0 s
x comp.
t  0.5 s
x comp.
t  1 s
2
p  ρ 2  π A ω x  cos( 2  π ω t)  A  x  sin( 2  π ω t)

x comp.
25.1
25.1
Pa
y comp.
m
Pa
y comp.
m
25.1
Pa
m
m
2
s
Hence, the components of pressure gradient (neglecting gravity) are

m
s
The governing equation (assuming inviscid flow) for computing the pressure gradient is
x
2
s
s
s

m
2
m
s
Total
s
s
m
12.6
0
and
2
s
s
Total
m
m
12.6
and
2
12.6
Local
0
s
s
t  0.5 s
Convective
2
y comp.
25.1
Pa
m
25.1
25.1

2
Pa
m
Pa
m
Problem 6.8
[Difficulty: 3]
Given:
Velocity field
Find:
Expressions for velocity and acceleration along wall; plot; verify vertical components are zero; plot pressure
gradient
Solution:
3
m
q = 2⋅
The given data is
u=
s
h = 1⋅ m
m
ρ = 1000⋅
kg
3
m
q⋅ x
2 ⋅ π⎡⎣x + ( y − h )
2
2⎤
+
⎦
q⋅ x
2 ⋅ π⎡⎣x + ( y + h )
2
v=
2⎤
⎦
q⋅ ( y − h)
2 ⋅ π⎡⎣x + ( y − h )
2
2⎤
+
⎦
q⋅ ( y + h)
2 ⋅ π⎡⎣x + ( y + h )
2
2⎤
⎦
The governing equation for acceleration is
For steady, 2D flow this reduces to (after considerable math!)
x - component
y - component
(
π⋅ x + h
∂y
2
u =−
⎡
(2
q ⋅ x⋅ ⎣ x + y
2
)
)
2
2
2
(2
− h ⋅ h − 4⋅ y
2
2⎤
)⎦
2
⎡⎣x2 + ( y + h ) 2⎤⎦ ⋅ ⎡⎣x 2 + ( y − h) 2⎤⎦ ⋅ π2
2
2 ⎡ 2
2
2 2
2⎤
q ⋅ y⋅ ⎣ x + y
− h ⋅ h + 4⋅ x ⎦
∂
∂
ay = u ⋅ v + v ⋅ v = −
2
2
∂x
∂y
2 2
2
2
2
π ⋅ ⎡⎣x + ( y + h ) ⎤⎦ ⋅ ⎡⎣x + ( y − h ) ⎤⎦
(
)
2
q⋅ x
2
∂x
u + v⋅
∂
(
)
y = 0⋅ m
For motion along the wall
u=
ax = u ⋅
∂
v=0
(No normal velocity)
ax = −
(2
q ⋅ x⋅ x − h
2
(2
π ⋅ x +h
)
2
)
2
3
ay = 0
(No normal acceleration)
The governing equation (assuming inviscid flow) for computing the pressure gradient is
Hence, the component of pressure gradient (neglecting gravity) along the wall is
∂
∂x
p = −ρ⋅
Du
∂
Dt
∂x
(2
2
p =
ρ⋅ q ⋅ x ⋅ x − h
2
(2
π ⋅ x +h
2
)
)
2
3
The plots of velocity, acceleration, and pressure gradient are shown below, done in Excel. From the plots it is clear that the fluid
experiences an adverse pressure gradient from the origin to x = 1 m, then a negative one promoting fluid acceleration. If flow
separates, it will likely be in the region x = 0 to x = h.
q =
h =
2
1
m 3/s/m
m
0.35
∠=
1000
kg/m 3
0.30
0.00000
0.00000
0.01945
0.00973
0.00495
0.00277
0.00168
0.00109
0.00074
0.00053
0.00039
0.00
0.00
-19.45
-9.73
-4.95
-2.77
-1.68
-1.09
-0.74
-0.53
-0.39
u (m/s)
0.00
0.32
0.25
0.19
0.15
0.12
0.10
0.09
0.08
0.07
0.06
dp /dx (Pa/m)
0.25
0.20
0.15
0.10
0.05
0.00
0
1
2
3
4
5
6
7
8
9
10
8
9
10
9
10
x (m)
Acceleration Along Wall Near A Source
0.025
0.020
a (m/s 2)
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
a (m/s2)
0.015
0.010
0.005
0.000
0
1
2
3
4
5
6
7
-0.005
x (m)
Pressure Gradient Along Wall
5
dp /dx (Pa/m)
x (m) u (m/s)
Velocity Along Wall Near A Source
0
0
1
2
3
4
5
-5
-10
-15
-20
-25
x (m)
6
7
8
Problem 6.9
[Difficulty: 2]
Problem 6.10
Given:
Velocity field
Find:
Expression for pressure field; evaluate at (2,2)
[Difficulty: 2]
Solution:
Basic equations
Given data
A = 4⋅
1
B = 2⋅
s
1
x = 2⋅ m
s
y = 2⋅ m
u ( x , y ) = A⋅ x + B⋅ y
v ( x , y ) = B⋅ x − A⋅ y
Note that
∂
∂
Then
∂x
∂
∂y
v( x , y) = 0
ax ( x , y ) = u ( x , y ) ⋅
ay ( x , y ) = u ( x , y ) ⋅
∂
∂x
∂
∂x
∂x
u( x , y) + v( x , y) ⋅
v( x , y) + v( x , y) ⋅
∂
The momentum equation becomes
∂x
p = −ρ⋅ ax
v( x , y) −
∂
∂y
∂
∂y
∂
∂y
∂
∂y
(
x
y
0
0
p ( x , y ) = 80⋅ kPa
(
2
2
)
2
2
−
(
2
)
2
ax ( x , y ) = 40
m
2
s
(
2
ay ( x , y ) = y ⋅ A + B
v( x , y)
)
2
ay ( x , y ) = 40
m
2
s
p = −ρ⋅ ay
⌠
⌠
p ( x , y ) = p 0 − ρ⋅ ⎮ ax ( x , y ) dx − ρ⋅ ⎮ ay ( x , y ) dy
⌡
⌡
ρ⋅ A + B ⋅ y
p 0 = 200 ⋅ kPa
3
u( x , y) = 0
ax ( x , y ) = x ⋅ A + B
u( x , y)
Integrating
p( x , y) = p0 −
kg
m
For this flow
u( x , y) +
ρ = 1500⋅
2
2
)
ρ⋅ A + B ⋅ x
2
2
and
p = dx⋅
∂
∂x
p + dy⋅
∂
∂y
p
Problem 6.11
[Difficulty: 2]
Given:
Velocity field
Find:
Expression for pressure gradient; plot; evaluate pressure at outlet
Solution:
Basic equations
Given data
U = 15⋅
m
L = 5⋅ m
s
ρ = 1250⋅
kg
3
m
u ( x ) = U⋅ ⎛⎜ 1 −
Here
p in = 100 ⋅ kPa
⎝
x⎞
u ( 0 ) = 15
L⎠
ρ⋅ u ⋅
The x momentum becomes
dp
The pressure gradient is then
dx
s
u ( L) = 0
d
d
u = ρ⋅ aa = − p
dx
dx
ax ( x ) = u ( x ) ⋅
Hence
m
2
= −ρ⋅
U
∂
∂x
⋅ ⎛⎜
x
⎝L
L
m
s
U ⋅ ⎛⎜
2
ax ( x ) =
u( x)
⎝L
⎠
L
− 1⎞
⎠
x
⌠
p ( x ) = p in − ρ⋅ ⎮ ax ( x ) dx
⌡
Integrating momentum
− 1⎞
x
2
p ( x ) = p in −
U ⋅ ρ⋅ x ⋅ ( x − 2 ⋅ L)
0
2⋅ L
2
2
p ( L) =
Hence
ρ⋅ U
2
+ p in
p ( L) = 241 ⋅ kPa
dp/dx (kPa/m)
60
40
20
0
1
2
3
x (m)
4
5
Problem 6.12
[Difficulty: 2]
Given:
Velocity field
Find:
Expression for acceleration and pressure gradient; plot; evaluate pressure at outlet
Solution:
Basic equations
Given data
U = 20⋅
m
L = 2⋅ m
s
u ( x ) = U⋅ e
ρ = 900 ⋅
kg
3
m
−
Here
p in = 50⋅ kPa
x
L
u ( 0 ) = 20
m
u ( L) = 7.36
s
m
s
−
The x component of acceleration is then
The x momentum becomes
The pressure gradient is then
ax ( x ) = u ( x ) ⋅
ρ⋅ u ⋅
dp
dx
2
∂
∂x
ax ( x ) = −
u( x)
=
ρ
L
−
L
2
⋅U ⋅e
2⋅ x
L
⌠
p ( x ) = p in − ρ⋅ ⎮ ax ( x ) dx
⌡
0
2
Hence
L
d
d
u = ρ⋅ aa = − p
dx
dx
x
Integrating momentum
U ⋅e
2⋅ x
p ( L) = p in −
( − 2 − 1)
U ⋅ ρ⋅ e
2
⎛ − 2⋅ x
⎞
2 ⎜
L
− 1⎠
U ⋅ ρ⋅ ⎝ e
p ( x ) = p in −
2
p ( L) = 206 ⋅ kPa
dp/dx (kPa/m)
200
150
100
50
0
0.5
1
1.5
2
x (m)
ax (m/s2)
0
0.5
1
− 50
− 100
− 150
− 200
x (m)
1.5
2
Problem 6.13
[Difficulty: 2]
Problem 6.14
[Difficulty: 3]
Given:
Velocity field
Find:
The acceleration at several points; evaluate pressure gradient
Solution:
3
3
m
The given data is
q  2
m
s
K  1
m
s
m
ρ  1000
kg
q
Vr  
2  π r
3
m
K
Vθ 
2  π r
The governing equations for this 2D flow are
The total acceleration for this steady flow is then
2
2
Vθ
Vθ

ar  Vr Vr 
 Vr 
r
r θ
r
ar  
θ - component
Vθ
Vr Vθ


aθ  Vr Vθ 
 Vθ 
r
r θ
r
aθ  0
Evaluating at point (1,0)
ar  0.127
r - component

m
2
q K
2 3
4 π  r
aθ  0
2
s
Evaluating at point (1,π/2)
ar  0.127
m
aθ  0
2
s
Evaluating at point (2,0)
ar  0.0158
m
2
aθ  0
s
From Eq. 6.3, pressure gradient is
Evaluating at point (1,0)
Evaluating at point (1,π/2)
Evaluating at point (2,0)

r

p  ρ ar
r
p 
2
1 
 p 0
r θ

1 
 p 0
r θ
r

r

r
Pa
p  127 
Pa
m
p  15.8
m
Pa
m
2 3
4 π  r
1 
 p  ρ aθ
r θ
p  127 
2
ρ q  K
1 
 p 0
r θ
1 
 p 0
r θ

Problem 6.15
[Difficulty: 3]
Problem 6.16
[Difficulty: 3]
Given:
Flow in a pipe with variable area
Find:
Expression for pressure gradient and pressure; Plot them; exit pressure
Solution:
Assumptions: 1) Incompressible flow 2) Flow profile remains unchanged so centerline velocity can represent average velocity
Basic equations
Q = V⋅ A
Given data
ρ = 1.75⋅
slug
ft
For this 1D flow
2
p i = 35⋅ psi
3
Q = u i⋅ Ai = u ⋅ A
Ai = 15⋅ in
(Ai − Ae)
A = Ai −
L
⋅x
so
2
Ae = 2.5⋅ in
L = 10⋅ ft
Ai
u ( x ) = u i⋅
= u i⋅
A
ui = 5⋅
Ai
Ai −
⎡ ( Ai − Ae) ⎤
⎢
⋅ x⎥
⎣ L
⎦
Ai
⎤ A ⋅ L ⋅ u i ⋅ ( Ae − Ai)
⎡
⎥= i
ax = u ⋅ u + v ⋅ u = u i ⋅
⋅ ⎢u i⋅
∂x
∂y
⎡ ( Ai − Ae) ⎤ ∂x ⎢
⎡ ( Ai − Ae) ⎤ ⎥ ( A ⋅ L + A ⋅ x − A ⋅ x) 3
i
e
i
Ai − ⎢
⋅ x⎥
⋅ x⎥ ⎥
Ai − ⎢
⎢
⎣ L
⎦ ⎣
⎣ L
⎦⎦
∂
For the pressure
∂
∂x
Ai
∂
2
p = −ρ⋅ ax − ρ⋅ g x = −
2
2
2
∂
(
ρ⋅ Ai ⋅ L ⋅ u i ⋅ Ae − Ai
(Ai⋅ L + Ae⋅ x − Ai⋅ x)
dp =
∂
∂x
3
⌠
x
2 2 2
⎮
⌠
ρ⋅ Ai ⋅ L ⋅ u i ⋅ Ae − Ai
∂
⎮
⎮
dx
p − pi =
p dx =
−
⎮ ∂x
⎮
3
Ai⋅ L + Ae⋅ x − Ai⋅ x
⌡
⎮
0
⌡
p ⋅ dx
(
(
This is a tricky integral, so instead consider the following:
x
x
0
0
∂
∂x
p = −ρ⋅ ax = −ρ⋅ u ⋅
( )
∂
1 ∂
2
u
u = − ⋅ ρ⋅
2 ∂x
∂x
⌠
⌠
ρ
ρ
2
2
2
∂
∂
p − pi = ⎮
p dx = − ⋅ ⎮
u dx = ⋅ u ( x = 0 ) − u ( x )
⎮ ∂x
⎮
2
2
∂x
⌡
⌡
( )
(
)
)
)
0
Hence
2
)
x
and
2
ft
s
ρ
2
2
p( x) = pi + ⋅ ⎛ ui − u( x) ⎞
⎠
2 ⎝
p( x) = pi +
Hence
⎡
⎢
⋅ 1−
2 ⎢
⎢
⎣
ρ⋅ u i
2
which we recognise as the Bernoulli equation!
Ai
⎤
⎡⎢
⎥
⎢
⎡ ( Ai − Ae) ⎤ ⎥
⋅ x⎥ ⎥
⎢ Ai − ⎢ L
⎣
⎣
⎦⎦
2⎤
⎥
⎥
⎥
⎦
p ( L) = 29.7 psi
At the exit
Pressure Gradient (psi/ft)
The following plots can be done in Excel
6
4
2
0
2
4
6
8
10
6
8
10
Pressure (psi)
x (ft)
35
30
25
0
2
4
x (ft)
Problem 6.17
[Difficulty: 3]
Given:
Flow in a pipe with variable area
Find:
Expression for pressure gradient and pressure; Plot them
Solution:
Assumptions: 1) Incompressible flow 2) Flow profile remains unchanged so centerline velocity can represent average velocity
Basic equations
Given data
Q = V⋅ A
ρ = 1250⋅
kg
2
A0 = 0.25⋅ m
3
a = 1.5⋅ m
L = 5⋅ m
m
For this 1D flow
Q = u 0 ⋅ A0 = u ⋅ A
so
A0
u( x) = u0⋅
=
A
⎛
−
⎜
A( x ) = A0 ⋅ ⎝ 1 + e
x
a
−
−e
x
⎞
2⋅ a
⎠
u 0 = 10⋅
m
u0
⎛
−
⎜
⎝1 + e
x
a
−
−e
x
⎞
2⋅ a
⎠
⎛ −
2
2⋅ a ⎜
⋅ ⎝ 2⋅ e
u0 ⋅ e
−
ax = u ⋅
∂
∂x
u + v⋅
∂
∂y
u =
∂
∂x
u0
⎡
⎢
x ⎞ ∂x ⎢ ⎛
x
−
−
−
⎜
⎢
2⋅ a
a
−e
+
−
e
⎠ ⎣⎝1 e
u0
⎛
−
⎜
+
⎝1 e
x
a
p = −ρ⋅ ax − ρ⋅ g x = −
⋅
⎛ −
2
2⋅ a ⎜
⋅ ⎝ 2⋅ e
ρ⋅ u 0 ⋅ e
−
For the pressure
p 0 = 300 ⋅ kPa
s
⎛ −
⎜
2 ⋅ a⋅ ⎝ e
a
−
−e
⎞
x
x
x
∂
2⋅ a
x
2⋅ a
− 1⎠
⎞
+ 1⎠
3
⎤
⎥=
x ⎞⎥
⎛ −
2⋅ a ⎥
⎜
⎠ ⎦ 2 ⋅ a⋅ ⎝ e
x
x
a
−
−e
⎞
x
2⋅ a
x
2⋅ a
− 1⎠
⎞
+ 1⎠
3
x
dp =
and
∂
∂x
⌠
⎮
−
⎮
x
2
⌠
ρ
u
⋅
⋅
e
⎮
0
∂
p − pi = ⎮
p dx = ⎮ −
⎮ ∂x
⎮
⌡
⎛ −
0
⎮
⎜
⎮
2 ⋅ a⋅ ⎝ e
⌡
p ⋅ dx
⎛ −
2⋅ a ⎜
⋅ ⎝ 2⋅ e
x
x
a
−
−e
⎞
x
2⋅ a
− 1⎠
⎞
x
2⋅ a
3
dx
+ 1⎠
0
∂
This is a tricky integral, so instead consider the following:
x
x
0
0
∂x
p = −ρ⋅ ax = −ρ⋅ u ⋅
( )
∂
1 ∂
2
u
u = − ⋅ ρ⋅
2
∂x
∂x
⌠
⌠
ρ
ρ
2
2
2
∂
∂
p − pi = ⎮
p dx = − ⋅ ⎮
u dx = ⋅ u ( x = 0 ) − u ( x )
⎮ ∂x
⎮
2
2
∂x
⌡
⌡
Hence
( )
ρ
2
2
p(x) = p0 + ⋅ ⎛ u0 − u(x) ⎞
⎝
⎠
2
p( x) = p0 +
ρ⋅ u 0
2
2
⎡
⋅ ⎢1 −
⎢
⎢
⎣
(
)
which we recognise as the Bernoulli equation!
1
⎡⎢
x
⎢⎛
−
−
⎜
⎢⎣ ⎝ 1 + e a − e
⎥⎤
x ⎞
⎥
2⋅ a ⎥
⎠⎦
2⎤
⎥
⎥
⎥
⎦
The following plots can be done in Excel
0.26
Area (m2)
0.24
0.22
0.2
0.18
0
1
2
3
x (m)
4
5
Pressure Gradient (kPa/m)
20
0
1
2
3
4
5
− 20
− 40
− 60
x (m)
300
Pressure (kPa)
290
280
270
260
250
0
1
2
3
x (m)
4
5
Problem 6.18
[Difficulty: 3]
Given:
Nozzle geometry
Find:
Acceleration of fluid particle; Plot; Plot pressure gradient; find L such that pressure gradient < 5 MPa/m in
absolute value
Solution:
The given data is
Di = 0.1⋅ m
Do = 0.02⋅ m
D( x ) = Di +
For a linear decrease in diameter
From continuity
Hence
or
Q = V⋅ A = V⋅
V( x ) ⋅
π
4
V( x ) =
m
Vi = 1 ⋅
s
L = 0.5⋅ m
Do − Di
L
kg
ρ = 1000⋅
3
m
⋅x
3
π
π 2
2
⋅ D = Vi⋅ ⋅ Di
4
4
Q = 0.00785
2
⋅ D( x ) = Q
V( x ) =
Vi
m
s
4⋅ Q
2
Do − Di ⎞
⎛
π⋅ ⎜ Di +
⋅x
L
⎝
⎠
2
Do − Di ⎞
⎛
⋅x
⎜1 +
L⋅ Di
⎝
⎠
The governing equation for this flow is
or, for steady 1D flow, in the notation of the problem
d
ax = V⋅ V =
dx
Vi
Do − Di ⎞
⎛
⋅x
⎜1 +
L⋅ Di
⎝
⎠
d
⋅
2 dx
2
Vi
2
Do − Di ⎞
⎛
⋅x
⎜1 +
L⋅ Di
⎝
⎠
This is plotted in the associated Excel workbook
From Eq. 6.2a, pressure gradient is
∂
∂x
p = −ρ⋅ ax
∂
∂x
2
p =
(
2 ⋅ ρ⋅ Vi ⋅ Do − Di
)
5
⎡ ( Do − Di) ⎤
Di⋅ L⋅ ⎢1 +
⋅ x⎥
Di⋅ L
⎣
⎦
ax ( x ) = −
(
2 ⋅ Vi ⋅ Do − Di
)
5
⎡ ( Do − Di) ⎤
Di⋅ L⋅ ⎢1 +
⋅ x⎥
Di⋅ L
⎣
⎦
This is also plotted in the associated Excel workbook. Note that the pressure gradient is always negative: separation is unlikely to
occur in the nozzle
∂
At the inlet
∂x
p = −3.2⋅
kPa
∂
At the exit
∂x
m
p = −10⋅
To find the length L for which the absolute pressure gradient is no more than 5 MPa/m, we need to solve
∂
∂x
2
p ≤ 5⋅
MPa
m
=
(
2 ⋅ ρ⋅ Vi ⋅ Do − Di
)
5
⎡ ( Do − Di) ⎤
Di⋅ L⋅ ⎢1 +
⋅ x⎥
Di⋅ L
⎣
⎦
with x = L m (the largest pressure gradient is at the outlet)
2
L≥
Hence
(
2 ⋅ ρ⋅ Vi ⋅ Do − Di
)
5
L ≥ 1⋅ m
⎛ Do ⎞ ∂
Di⋅ ⎜
⋅
p
Di
∂
x
⎝ ⎠
This result is also obtained using Goal Seek in the Excel workbook
From Excel
x (m) a (m/s 2)
0.000
0.050
0.100
0.150
0.200
0.250
0.300
0.350
0.400
0.420
0.440
0.460
0.470
0.480
0.490
0.500
3.20
4.86
7.65
12.6
22.0
41.2
84.2
194
529
843
1408
2495
3411
4761
6806
10000
dp /dx (kPa/m)
-3.20
-4.86
-7.65
-12.6
-22.0
-41.2
-84.2
-194
-529
-843
-1408
-2495
-3411
-4761
-6806
-10000
For the length L required
for the pressure gradient
to be less than 5 MPa/m (abs)
use Goal Seek
L =
1.00
x (m)
dp /dx (kPa/m)
1.00
-5000
m
MPa
m
Acceleration Through A Nozzle
12000
2
a (m/s )
10000
8000
6000
4000
2000
0
0.0
0.1
0.1
0.2
0.2
0.3
0.3
0.4
0.4
0.5
0.5
0.4
0.5
0.5
x (m)
Pressure Gradient Along A Nozzle
0
dp/dx (kPa/m)
0.0
0.1
0.1
0.2
0.2
0.3
-2000
-4000
-6000
-8000
-10000
-12000
x (m)
0.3
0.4
Problem 6.19
[Difficulty: 3]
Given:
Diffuser geometry
Find:
Acceleration of a fluid particle; plot it; plot pressure gradient; find L such that pressure gradient is less than
25 kPa/m
Solution:
The given data is
Di = 0.25⋅ m
Do = 0.75⋅ m
D( x) = Di +
For a linear increase in diameter
From continuity
Hence
Q = V⋅ A = V⋅
V( x) ⋅
π
4
Do − Di
L
π 2
2
⋅ D = Vi⋅ ⋅ Di
4
4
2
V( x) =
ρ = 1000⋅
⋅x
Q = 0.245
4⋅ Q
m
s
Vi
V( x) =
or
2
Do − Di ⎞
⎛
π⋅ ⎜ Di +
⋅x
L
⎝
⎠
ax = V⋅
or, for steady 1D flow, in the notation of the problem
2
Hence
ax ( x ) = −
(
Do − Di ⎞
⎛
⋅x
⎜1 +
L⋅ Di
⎝
⎠
d
V=
dx
Vi
d
2 dx
⋅
Do − Di ⎞
⎛
⋅x
⎜1 +
L⋅ Di
⎝
⎠
Vi
2
Do − Di ⎞
⎛
⋅x
⎜1 +
L⋅ Di
⎝
⎠
)
5
⎡ ( Do − Di) ⎤
Di⋅ L⋅ ⎢1 +
⋅ x⎥
Di⋅ L
⎣
⎦
This can be plotted in Excel (see below)
From Eq. 6.2a, pressure gradient is
∂
∂x
3
m
The governing equation for this flow is
2 ⋅ Vi ⋅ Do − Di
kg
3
π
⋅ D( x) = Q
m
Vi = 5⋅
s
L = 1⋅ m
p = −ρ⋅ ax
∂
∂x
2
p =
(
2 ⋅ ρ⋅ Vi ⋅ Do − Di
)
5
⎡ ( Do − Di) ⎤
Di⋅ L⋅ ⎢1 +
⋅ x⎥
Di⋅ L
⎣
⎦
2
This can also plotted in Excel. Note that the pressure gradient is adverse: separation is likely to occur in the diffuser, and occur
near the entrance
∂
At the inlet
∂x
p = 100 ⋅
kPa
∂
At the
exit
m
∂x
p = 412 ⋅
Pa
m
To find the length L for which the pressure gradient is no more than 25 kPa/m, we need to solve
2
∂
∂x
p ≤ 25⋅
kPa
m
=
(
2 ⋅ ρ⋅ Vi ⋅ Do − Di
)
5
⎡ ( Do − Di) ⎤
Di⋅ L⋅ ⎢1 +
⋅ x⎥
Di⋅ L
⎣
⎦
with x = 0 m (the largest pressure gradient is at the inlet)
2
L≥
Hence
(
2 ⋅ ρ⋅ Vi ⋅ Do − Di
Di⋅
∂
∂x
)
L ≥ 4⋅ m
p
This result is also obtained using Goal Seek in Excel.
In Excel:
Di
Do
L
Vi
=
=
=
=
0.25
0.75
1
5
m
m
m
m/s
( =
1000
kg/m 3
x (m) a (m/s 2)
dp /dx (kPa/m)
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.40
0.50
0.60
-100
-62.1
-40.2
-26.9
-18.59
-13.17
-9.54
-5.29
-3.125
-1.940
100
62.1
40.2
26.93
18.59
13.17
9.54
5.29
3.125
1.940
0.70
0.80
0.90
1.00
-1.256
-0.842
-0.581
-0.412
1.256
0.842
0.581
0.412
For the length L required
for the pressure gradient
to be less than 25 kPa/m
use Goal Seek
L =
4.00
x (m)
dp /dx (kPa/m)
0.0
25.0
m
Acceleration Through a Diffuser
0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
0.9
1.0
2
a (m/s )
-20
-40
-60
-80
-100
-120
x (m)
Pressure Gradient Along A Diffuser
dp /dx (kPa/m)
120
100
80
60
40
20
0
0.0
0.1
0.2
0.3
0.4
0.5
x (m)
0.6
0.7
0.8
Problem 6.20
[Difficulty: 4]
Problem 6.21
[Difficulty: 4]
Problem 6.20
Problem 6.22
[Difficulty: 3]
Problem 6.23
[Difficulty: 4]
Given:
Rectangular chip flow
Find:
Velocity field; acceleration; pressure gradient; net force; required flow rate; plot pressure
Solution:
Basic equations
→→
(
∑ V⋅A) = 0
∂
∂x
CS
The given data is
ρ = 1.23⋅
kg
u +
∂
∂y
v =0
p atm = 101 ⋅ kPa
3
h = 0.5⋅ mm
b = 40⋅ mm
M length = 0.005 ⋅
m
Assuming a CV that is from the centerline to any point x, and noting that q is inflow per unit area, continuity leads to
q ⋅ x ⋅ L = U⋅ h ⋅ L
u ( x ) = U( x ) = q ⋅
or
x
h
For acceleration we will need the vertical velocity v; we can use
∂
∂x
Hence
u +
∂
∂y
v =0
x
du
q
∂
d
v =− u =−
= − ⎛⎜ q ⋅ ⎞ = −
h
dx
h
x
d
∂y
∂x
⎝ ⎠
∂
or
⌠
v ( y = y ) − v ( y = 0 ) = −⎮
⎮
⌡
y
0
But
v( y = 0) = q
For the x acceleration
ax = u ⋅
∂
∂x
u + v⋅
∂
∂y
q
h
dy = −q ⋅
y
h
so
v ( y ) = q ⋅ ⎛⎜ 1 −
u
y
x q
ax = q ⋅ ⋅ ⎛⎜ ⎞ + q ⋅ ⎛⎜ 1 − ⎞ ⋅ ( 0 )
h⎠
h ⎝h⎠
⎝
⎝
y⎞
h⎠
ax =
q
h
2
2
⋅x
kg
m
For the y acceleration
ay = u ⋅
∂
∂x
v + v⋅
∂
x
y
q
ay = q ⋅ ⋅ ( 0 ) + q ⋅ ⎛⎜ 1 − ⎞ ⋅ ⎜⎛ − ⎞
h
h⎠ ⎝ h⎠
⎝
v
∂y
∂
Hence
∂x
ρ⋅
Also
p = −ρ⋅
q
h
Dv
Dx
Du
ρ⋅
For the pressure gradient we use x and y momentum (Euler equation)
Dx
ax =
q
2
h
⋅ ⎛⎜
y
⎝h
− 1⎞
⎠
⎛ ∂
∂ ⎞
∂
u + v ⋅ u = ρ⋅ ax = − p
∂y ⎠
∂x
⎝ ∂x
= ρ⋅ ⎜ u ⋅
2
2
⋅x
⎛ ∂
∂ ⎞
∂
v + v ⋅ v = ρ⋅ ay = − p
∂y ⎠
∂y
⎝ ∂x
∂
= ρ⋅ ⎜ u ⋅
p = ρ⋅
∂y
q
2
h
⋅ ⎛⎜ 1 −
⎝
y⎞
h⎠
For the pressure distribution, integrating from the outside edge (x = b/2) to any point x
x
p ( x = x ) − p ⎛⎜ x =
⎝
b⎞
2⎠
x
⌠
2
2
⌠
2
⎮
q
q
q
2
2
⎮ ∂
p dx = ⎮ −ρ⋅ ⋅ x dx = −ρ⋅
⋅ x + ρ⋅
⋅b
= p ( x ) − p atm =
⎮ ∂x
2
2
2
⎮
8⋅ h
2⋅ h
h
⎮
⎮b
⌡b
⌡
2
2 2
p ( x ) = p atm + ρ⋅
q ⋅b
⎡
⋅ ⎢1 − 4 ⋅ ⎛⎜
x⎞
⎥
⎝b⎠ ⎦
8⋅ h ⎣
2
2
2⎤
For the net force we need to integrate this ... actually the gage pressure, as this pressure is opposed on the outer surface by p atm
2 2
ρ⋅ q ⋅ b
pg( x) =
b
8⋅ h
2
⎡
⋅ ⎢1 − 4 ⋅ ⎛⎜
⎣
x⎞
2⎤
⎥
⎝b⎠ ⎦
b
⌠2
⌠2
2 2
2
⎮
⎮ ρ⋅ q 2⋅ b 2 ⎡
x⎞ ⎤
ρ⋅ q ⋅ b ⋅ L ⎛ b
1 b
⎛
⎥ dx =
Fnet = 2 ⋅ L⋅ ⎮ p g ( x ) dx = 2 ⋅ L⋅ ⎮
⋅ ⎢1 − 4 ⋅ ⎜
⋅⎜ − ⋅ ⎞
2 ⎣
2
⌡
⎝b⎠ ⎦
⎝2 3 2⎠
⎮
8⋅ h
4⋅ h
0
⌡
2 3
Fnet =
ρ⋅ q ⋅ b ⋅ L
12⋅ h
2
0
2 3
The weight of the chip must balance this force
M ⋅ g = M length ⋅ L⋅ g = Fnet =
ρ⋅ q ⋅ b ⋅ L
12⋅ h
2
2 3
or
M length ⋅ g =
ρ⋅ q ⋅ b
12⋅ h
3
m
2
Solving for q for the given mass/length
q =
12⋅ h ⋅ g ⋅ M length
ρ⋅ b
q = 0.0432⋅
3
s
2
m
b
The maximum speed
b⋅ q
Umax =
2⋅ h
2
b
Umax = u ⎛⎜ x = ⎞ = q ⋅
h
2⎠
⎝
2 2
The following plot can be done in Excel
pg( x) =
ρ⋅ q ⋅ b
8⋅ h
2
⎡
⋅ ⎢1 − 4 ⋅ ⎛⎜
⎣
x⎞
2⎤
⎥
⎝b⎠ ⎦
m
Umax = 1.73
s
2
2
Pressure (Pa)
1.5
1
0.5
− 0.02
− 0.01
0
0.01
0.02
x (m)
The net force is such that the chip is floating on air due to a Bernoulli effect: the speed is maximum at the edges and zero at the
center; pressure has the opposite trend - pressure is minimum (p atm) at the edges and maximum at the center.
Problem 6.24
[Difficulty: 3]
Problem 6.25
[Difficulty: 4]
Given:
Velocity field
Find:
Constant B for incompressible flow; Acceleration of particle at (2,1); acceleration normal to velocity at (2,1)
Solution:
Basic equations
For this flow
3
u ( x , y ) = A⋅ x + B⋅ x ⋅ y
∂
∂x
∂
∂x
u( x , y) +
u( x , y) +
∂
∂y
∂
∂y
2
v( x , y) =
3
∂
∂x
Hence for ax
ax = u ⋅
u + v⋅
∂
∂y
(
ay = u ⋅
∂
(2
∂x
v + v⋅
2
∂
∂y
(2
3
3
ay = 3 ⋅ ⎛
0.2
a =
2
ax + ay
1
2
)
2
)
∂x
3
) (A⋅y3 − 3⋅A⋅x2⋅y) + (A⋅y3 − 3⋅A⋅x2⋅y)⋅∂ (A⋅y3 − 3⋅A⋅x2⋅y)
2 ∂
∂x
2
∂y
2
2
2
∂y
2
(
2
2
) (A⋅x3 − 3⋅A⋅x⋅y2) + (A⋅y3 − 3⋅A⋅x2⋅y)⋅∂ (A⋅x3 − 3⋅A⋅x⋅y2)
⎞ × 1⋅ m × ⎡⎣( 2⋅ m) 2 + ( 1 ⋅ m) 2⎤⎦
⎜ 2
⎝ m ⋅s ⎠
0.2
B = −0.6
2 ∂
⎞ × 2⋅ m × ⎡( 2⋅ m) 2 + ( 1 ⋅ m) 2⎤
⎣
⎦
⎜ 2
m
⋅
s
⎝
⎠
ax = 3 ⋅ ⎛
B = −3 ⋅ A
Hence
v ( x , y ) = A⋅ y − 3 ⋅ A⋅ x ⋅ y
v = A⋅ x − 3 ⋅ A⋅ x ⋅ y ⋅
ay = 3 ⋅ A ⋅ y ⋅ x + y
Hence at (2,1)
2
u = A⋅ x − 3 ⋅ A⋅ x ⋅ y ⋅
ax = 3 ⋅ A ⋅ x ⋅ x + y
For ay
)=0
2
m ⋅s
3
2
∂y
(2
u ( x , y ) = A⋅ x − 3 ⋅ A⋅ x ⋅ y
∂x
(A⋅x3 + B⋅x⋅y2) + ∂ (A⋅y3 + B⋅x2⋅y) = 0
v ( x , y ) = ( 3 ⋅ A + B) ⋅ x + y
We can write
∂
2
v ( x , y ) = A⋅ y + B⋅ x ⋅ y
2
ax = 6.00⋅
m
2
s
2
ay = 3.00⋅
m
2
s
a = 6.71
m
2
s
We need to find the component of acceleration normal to the velocity vector
At (2,1) the velocity vector is at angle

a
⎛ 1 − 3⋅ 2 ⋅ 1 ⎞
θvel = atan⎜
⎜ 23 − 3⋅ 2⋅ 12
⎝
⎠
θvel = −79.7⋅ deg
⎛ ay ⎞
θaccel = atan⎜
⎝ ax ⎠
1
θaccel = atan⎛⎜ ⎞
⎝2⎠
θaccel = 26.6⋅ deg
∆θ = θaccel − θvel
∆θ = 106 ⋅ deg
3
At (1,2) the acceleration vector is at angle

V
⎛ A⋅ y 3 − 3⋅ A⋅ x2⋅ y ⎞
v
θvel = atan⎛⎜ ⎞ = atan⎜
⎜ A⋅ x 3 − 3⋅ A⋅ x⋅ y 2
⎝u⎠
⎝
⎠
Hence the angle between the acceleration and velocity vectors is
The component of acceleration normal to the velocity is then
2
an = a⋅ sin( ∆θ) = 6.71⋅
∆θ
m
2
s
⋅ sin( 106 ⋅ deg)
an = 6.45⋅
m
2
s
Problem 6.26
[Difficulty: 4] Part 1/2
Problem 6.26
[Difficulty: 4] Part 2/2
Problem 6.27
[Difficulty: 5]
Given:
Velocity field
Find:
Constant B for incompressible flow; Equation for streamline through (1,2); Acceleration of particle; streamline
curvature
Solution:
Basic equations
(4
2 2
u ( x, y ) = A ⋅ x − 6⋅ x ⋅ y + y
For this flow
∂
∂x
∂
∂x
u ( x, y ) +
u( x , y) +
∂
∂y
v ( x, y ) =
∂
∂y
∂
∂x
4
)
v ( x, y ) = B⋅ x ⋅ y − x⋅ y
(3
(
)
)
3
(
)
⎡⎣A ⋅ x4 − 6⋅ x2⋅ y2 + y 4 ⎤⎦ + ∂ ⎡⎣B⋅ x3⋅ y − x⋅ y 3 ⎤⎦ = 0
(3
v ( x , y) = B⋅ x − 3⋅ x ⋅ y
∂y
) + A⋅(4⋅x3 − 12⋅x⋅y2) = (4⋅A + B)⋅x⋅(x2 − 3⋅y2) = 0
2
B = −4 ⋅ A
Hence
1
B = −8
3
m ⋅s
Hence for ax
ax = u ⋅
∂
∂x
u + v⋅
∂
∂y
(4
)
4 ∂
2 2
u = A⋅ x − 6 ⋅ x ⋅ y + y ⋅
2
(2
ax = 4 ⋅ A ⋅ x ⋅ x + y
)
2
∂x
(
)
(
)
(
)
⎡⎣A⋅ x 4 − 6⋅ x 2⋅ y2 + y 4 ⎤⎦ + ⎡⎣−4 ⋅ A⋅ x3⋅ y − x⋅ y 3 ⎤⎦ ⋅ ∂ ⎡⎣A⋅ x 4 − 6⋅ x 2⋅ y2 + y 4 ⎤⎦
∂y
3
For ay
ay = u ⋅
∂
∂x
v + v⋅
∂
∂y
(4
2
(2
ay = 4 ⋅ A ⋅ y ⋅ x + y
For a streamline
dy
dx
Let
)
4 ∂
2 2
v = A⋅ x − 6 ⋅ x ⋅ y + y ⋅
u=
=
v
u
so
)
2
∂x
(
)
(
)
(
)
⎡⎣−4⋅ A⋅ x 3⋅ y − x ⋅ y3 ⎤⎦ + ⎡⎣−4⋅ A⋅ x 3⋅ y − x ⋅ y3 ⎤⎦ ⋅ ∂ ⎡⎣−4 ⋅ A⋅ x3⋅ y − x⋅ y 3 ⎤⎦
∂y
3
dy
dx
y
du
x
dx
=
=
(3
−4 ⋅ A⋅ x ⋅ y − x ⋅ y
3
)
A⋅ x − 6 ⋅ x ⋅ y + y
4
)
=−
d ⎛⎜
d ⎛⎜
1⎞
(4
2 2
y⎞
(3
4⋅ x ⋅ y − x⋅ y
(x4 − 6⋅x2⋅y2 + y4)
⎝ x ⎠ = 1 ⋅ dy + y⋅ ⎝ x ⎠ = 1 ⋅ dy − y
dx
x dx
dx
)
3
x dx
x
2
so
dy
dx
= x⋅
du
dx
+u
dy
Hence
dx
x⋅
du
dx
dx
Separating variables
x
= x⋅
du
+u=−
dx
(3
4⋅ x ⋅ y − x⋅ y
4
2
u⋅ (u
(
− 10⋅ u
2
+ 5)
3 2
)u +
(
2
3⎞
⎛1
⎜ u − 6⋅ u + u
⎝
⎠
)
4⋅ 1 − u
)
2
3⎞
⎛1
⎜ u − 6⋅ u + u
⎝
⎠
)
1
5
3
ln( x ) = − ⋅ ln u − 10⋅ u + 5 ⋅ u + C
5
⋅ du
(u5 − 10⋅u3 + 5⋅u)⋅x5 = c
5
(
4⋅ 1 − u
(
2
u − 6⋅ u + 1
4
)
=−
4
2
⎥⎤ = − u⋅ u − 10⋅ u + 5
4
2
⎛ 1 − 6⋅ u + u3⎞ ⎥
u − 6⋅ u + 1
⎜u
⎥
⎝
⎠⎦
4⋅ 1 − u
=−u+
=−
)
( x 4 − 6 ⋅ x 2⋅ y 2 + y 4 )
(
⎡⎢
⎢
⎢⎣
3
5
3 2
4
y − 10⋅ y ⋅ x + 5 ⋅ y ⋅ x = const
4
y − 10⋅ y ⋅ x + 5 ⋅ y ⋅ x = −38
For the streamline through (1,2)
Note that it would be MUCH easier to use the stream function method here!
2
an = −
To find the radius of curvature we use
2
V
V
R =
or
R
an

V
We need to find the component of acceleration normal to the velocity vector
⎡
v
θvel = atan⎛⎜ ⎞ = atan⎢−
⎢
⎝u⎠
(3
)
⎤
⎥
4
2 2
4 ⎥
⎣ x − 6⋅ x ⋅ y + y ⎦
At (1,2) the velocity vector is at angle
4⋅ x ⋅ y − x⋅ y
(
4⋅ ( 2 − 8) ⎤
θvel = atan⎡⎢−
⎥
⎣ 1 − 24 + 16⎦
3

a
)
∆θ
θvel = −73.7⋅ deg
At (1,2) the acceleration vector is at angle
⎡ 2
⎢ 4⋅ A ⋅ y⋅ x2 + y2
⎛ ay ⎞
θaccel = atan⎜
= atan⎢
⎝ ax ⎠
⎢ 4 ⋅ A2⋅ x ⋅ x 2 + y2
⎣
) ⎥⎤ = atan⎛ y ⎞
⎥
⎜
3
x
⎥
)⎦ ⎝ ⎠
(
(
3
∆θ = θaccel − θvel
Hence the angle between the acceleration and velocity vectors is
an = a⋅ sin( ∆θ)
The component of acceleration normal to the velocity is then
At (1,2)
(2
2
ax = 4 ⋅ A ⋅ x ⋅ x + y
a =
2
2
)
3
7
2
a = 4472
2
s
(4
2 2
u = A⋅ x − 6 ⋅ x ⋅ y + y
2
Then
R =
V
an
θaccel = 63.4⋅ deg
∆θ = 137 ⋅ deg
2
m
an = a⋅ sin( ∆θ)
2
) = −14⋅ ms
(3
⎝
m⎞
s
2
2
(2
3
)
2
⎠
2
×
3
) = 48⋅ ms
1
an = 3040
m
2
V=
2
s
3040 m
2
u + v = 50⋅
2
⋅
= 4000⋅
m
2
s
s
v = B⋅ x ⋅ y − x ⋅ y
R = ⎛⎜ 50⋅
2
ay = 4 ⋅ A ⋅ y ⋅ x + y
s
4
ax + ay
a=
where
⎛ 2 ⎞ = 2000⋅ m
⎜ 3
2
s
⎝ m ⋅s ⎠
7
= 500 ⋅ m × A = 500 ⋅ m ×
2 m
2000 + 4000 ⋅
2
θaccel = atan⎛⎜ ⎞
⎝1⎠
R = 0.822 m
m
s
Problem 6.28
Given:
Velocity field for doublet
Find:
Expression for pressure gradient
[Difficulty: 2]
Solution:
Basic equations
For this flow
Hence for r momentum
Λ
Vr( r , θ) = − ⋅ cos( θ)
2
r
Λ
Vθ( r , θ) = − ⋅ sin( θ)
2
r
Vz = 0
2
⎛
Vθ
Vθ ⎞
⎜ ∂
∂
ρ⋅ g r − p = ρ⋅ ⎜ Vr⋅ Vr +
⋅ V −
r ⎠
r ∂θ r
∂r
⎝ ∂r
∂
Ignoring gravity
⎡⎢
⎢
Λ
Λ
∂
∂
p = −ρ⋅ ⎢⎛ − ⋅ cos( θ) ⎞ ⋅ ⎛ − ⋅ cos( θ) ⎞ +
⎜
⎜
2
2
∂r
⎢⎣⎝ r
⎠ ∂r ⎝ r
⎠
For θ momentum
ρ⋅ g θ −
⎛ − Λ ⋅ sin( θ)⎞
⎜ 2
⎝ r
⎠ ⋅ ∂ ⎛ − Λ ⋅ cos( θ) ⎞ −
r
∂θ ⎜ r2
⎝
⎠
⎛ Λ
⎞
⎜ − 2 ⋅ sin( θ)
⎝ r
⎠
r
2⎤
⎥
⎥
⎥
⎥⎦
∂
∂r
2
p =
2⋅ Λ ⋅ ρ
5
r
Vθ
Vr⋅ Vθ ⎞
⎛ ∂
1 ∂
∂
⋅ p = ρ⋅ ⎜ Vr⋅ Vθ +
⋅ Vθ +
r ∂θ
r ⎠
r ∂θ
⎝ ∂r
Ignoring gravity
⎡
⎢
Λ
∂
⎢ Λ
∂
p = −r⋅ ρ⋅ ⎛ − ⋅ cos( θ) ⎞ ⋅ ⎛ − ⋅ sin( θ) ⎞ +
⎜
⎜
⎢
2
2
∂θ
⎣⎝ r
⎠ ∂r ⎝ r
⎠
The pressure gradient is purely radial
⎛ Λ
⎞
⎜ − 2 ⋅ sin( θ)
⎝ r
⎠ ⋅ ∂ ⎛ − Λ ⋅ sin( θ)⎞ +
r
∂θ ⎜ r2
⎝
⎠
⎛ − Λ ⋅ sin( θ)⎞ ⋅ ⎛ − Λ ⋅ cos( θ) ⎞ ⎤
⎥
⎜ 2
⎜ 2
⎝ r
⎠⎝ r
⎠⎥
⎥
r
⎦
∂
∂θ
p =0
Problem 6.29
[Diffculty: 4]
Given:
Velocity field for flow over a cylinder
Find:
Expression for pressure gradient; pressure variation; minimum pressure; plot velocity
Solution:
Basic equations
Given data
ρ = 1.23⋅
kg
a = 150 ⋅ mm
3
U = 75⋅
m
For this flow
⎡ a
⎤
Vr = U⋅ ⎢⎛⎜ ⎞ − 1⎥ ⋅ cos( θ)
r
On the surface r = a
Vr = 0
2
Hence on the surface:
For r momentum
For θ momentum
⎣⎝ ⎠
⎦
m
s
⎡ a
⎤
Vθ = U⋅ ⎢⎛⎜ ⎞ + 1⎥ ⋅ sin( θ)
r
2
⎣⎝ ⎠
⎦
Vθ = 2 ⋅ U⋅ sin( θ)
2
⎛
⎛⎜ V 2 ⎞
Vθ
Vθ ⎞
⎜ ∂
θ
∂
∂
ρ⋅ ⎜ Vr⋅ Vr +
⋅ Vr −
= ρ⋅ ⎜ −
=− p
r ⎠
r ∂θ
⎝ a ⎠ ∂r
⎝ ∂r
∂
∂r
2
p = ρ⋅
Vθ
a
2
= ρ⋅ 4 ⋅ U ⋅ sin( θ)
Vθ
Vr⋅ Vθ ⎞
⎛ ∂
⎛ Vθ ∂
⎞
1 ∂
2 ⋅ U⋅ sin( θ)
∂
ρ⋅ ⎜ Vr⋅ Vθ +
⋅ 2 ⋅ U⋅ cos( θ) = − ⋅ p
⋅ Vθ +
⋅ Vθ = ρ⋅
= ρ⋅ ⎜
r ∂θ
r ⎠
r ∂θ
a
⎝ ∂r
⎝ a ∂θ ⎠
2
∂
∂θ
p =−
4 ⋅ ρ⋅ U
a
2
⋅ sin( θ) ⋅ cos( θ) = −
2 ⋅ ρ⋅ U
a
⋅ sin( 2 ⋅ θ)
For the pressure distribution we integrate from θ = 0 to θ = θ, assuming p(0) = p atm (a stagnation point)
θ
⌠
∂
p ( θ) − p atm = ⎮
p dθ =
⎮ ∂θ
⌡
0
θ
⌠
2
⎮
4 ⋅ ρ⋅ U
⎮ −
⋅ sin( θ) ⋅ cos( θ) dθ
a
⎮
⌡
0
2
2⌠
θ
2
p ( θ) = −4 ⋅ ρ⋅ U ⎮ sin( θ) ⋅ cos( θ) dθ
⌡
p ( θ) = −2 ⋅ U ⋅ ρ⋅ sin( θ)
2
Minimum p:
0
Pressure (kPa)
0
50
100
p ⎛⎜
π⎞
⎝2⎠
= −13.8⋅ kPa
150
−5
− 10
− 15
x (m)
⎡ a
⎤
Vθ( r) = U⋅ ⎢⎛⎜ ⎞ + 1⎥
r
2
For the velocity as a function of radial position at θ = π/2
Vr = 0
so
V = Vθ
⎣⎝ ⎠
⎦
5
r/a
4
3
2
1
75
100
125
150
V(r) (m/s)
The velocity falls off to V = U as directly above the cylinder we have uniform horizontal as the effect of the cylinder decreases
m
Vθ( 100 ⋅ a) = 75
s
Problem 6.30
[Difficulty: 2]
Given:
Flow in a curved section
Find:
Expression for pressure distribution; plot; V for wall static pressure of 35 kPa
Solution:
Basic equation
∂
∂n
2
p = ρ⋅
V
R
Assumptions: Steady; frictionless; no body force; constant speed along streamline
ρ = 999 ⋅
Given data
kg
3
V = 10⋅
m
At the inlet section
p = p( y)
m
L = 75⋅ mm
s
∂
hence
∂n
2
p( y) = pc −
Integrating from y = 0 to y = y
ρ⋅ V
R0⋅ L
⋅y
p =−
2
dp
dy
R0 = 0.2⋅ m
2
= ρ⋅
V
R
p c = 50⋅ kPa
2 2⋅ y
= ρ⋅ V ⋅
2 2⋅ y
dp = −ρ⋅ V ⋅
L⋅ R 0
p ⎛⎜
p ( 0 ) = 50⋅ kPa
(1)
L⎞
⎝2⎠
L⋅ R 0
⋅ dy
= 40.6⋅ kPa
40
y (mm)
30
20
10
40
42
44
46
48
50
p (kPa)
For a new wall pressure
p wall = 35⋅ kPa
solving Eq 1 for V gives
V =
(
)
4 ⋅ R0 ⋅ p c − p wall
ρ⋅ L
V = 12.7
m
s
Problem 6.31
[Difficulty: 2]
Problem 6.32
Given:
Velocity field for free vortex flow in elbow
Find:
Similar solution to Example 6.1; find k (above)
[Difficulty: 3]
Solution:
Basic equation
∂
∂r
2
p =
ρ⋅ V
c
V = Vθ =
r
with
r
Assumptions: 1) Frictionless 2) Incompressible 3) free vortex
2
∂
ρ⋅ c
ρ⋅ V
d
=
p =
3
r
dr
r
2
For this flow
p ≠ p ( θ)
Hence
2⎛ 2
2⎞
⌠ 2
2
2
ρ⋅ c ⎛ 1
1 ⎞ ρ⋅ c ⋅ ⎝ r2 − r1 ⎠
⎮ ρ⋅ c
dr =
⋅⎜
∆p = p 2 − p 1 = ⎮
−
=
2 2
3
2 ⎜ 2
2
2⋅ r1 ⋅ r2
r
r1
r2
⎮
⎝
⎠
⌡r
so
∂r
p =
r
(1)
1
Next we obtain c in terms of Q
⌠ →→
⎮
Q = ⎮ V dA =
⌡
r
r
⌠2
⌠ 2 w⋅ c
⎛ r2 ⎞
⎮ V⋅ w dr = ⎮
dr = w⋅ c⋅ ln⎜
r
⎮
⌡r
⎝ r1 ⎠
1
⌡r
1
Hence
c=
Q
⎛ r2 ⎞
w⋅ ln⎜
⎝ r1 ⎠
ρ⋅ c ⋅ ⎛ r2 − r1
⎝
2
Using this in Eq 1
∆p = p 2 − p 1 =
2
2
2 ⋅ r1 ⋅ r2
2
Solving for Q
2
2
2⎞
⎠ =
2 ⋅ r1 ⋅ r2
⎛ r2 ⎞
Q = w⋅ ln⎜
⋅
⋅ ∆p
⎝ r1 ⎠ ρ⋅ ⎛⎝ r2 2 − r12⎞⎠
ρ⋅ Q ⋅ ⎛ r2 − r1
⎝
2
2
2⎞
⎠
2
⎛ r2 ⎞ 2 2
2 ⋅ w ⋅ ln⎜
⋅ r1 ⋅ r2
⎝ r1 ⎠
2
2
2
2 ⋅ r1 ⋅ r2
⎛ r2 ⎞
k = w⋅ ln⎜
⋅
⎝ r1 ⎠ ρ⋅ ⎛⎝ r2 2 − r12⎞⎠
Problem 6.33
[Difficulty: 4] Part 1/2
Problem 6.33
[Difficulty: 4] Part 2/2
Problem 6.34
[Difficulty: 2]
Given:
Flow rates in elbow for uniform flow and free vortes
Find:
Plot discrepancy
Solution:
(
)
For Example 6.1 QUniform = V⋅ A = w⋅ r2 − r1 ⋅
⎛ r2 ⎞
For Problem 6.32 Q = w⋅ ln⎜
⎝ r1 ⎠
2
⋅
2 ⋅ r1 ⋅ r2
1
⎛ r2 ⎞
ρ⋅ ln⎜
⎝ r1 ⎠
⋅ ∆p
or
⎛ r2
⎞
−1
⎜
QUniform⋅ ρ
⎝ r1
⎠
=
w⋅ r1 ⋅ ∆p
⎛ r2 ⎞
ln⎜
⎝ r1 ⎠
2
2
2
ρ⋅ ⎛ r2 − r1 ⎞
⎝
⎠
⋅ ∆p
or
Q⋅ ρ
w⋅ r1 ⋅ ∆p
=
(1)
⎛ r2 ⎞ ⎛ r2 ⎞
2
⎜ r ⋅ ln⎜ r ⋅
⎝ 1 ⎠ ⎝ 1 ⎠ ⎡⎢⎛ r2 ⎞ 2
−
⎢⎜ r
⎣⎝ 1 ⎠
⎤
⎥
1
⎥
⎦
(2)
It is convenient to plot these as functions of r2/r1
Eq. 1
Eq. 2
Error
1.01
1.05
1.10
1.15
1.20
1.25
1.30
1.35
1.40
1.45
1.50
1.55
1.60
1.65
1.70
1.75
1.80
1.85
1.90
1.95
2.00
2.05
2.10
2.15
2.20
2.25
2.30
2.35
2.40
2.45
2.50
0.100
0.226
0.324
0.401
0.468
0.529
0.586
0.639
0.690
0.738
0.785
0.831
0.875
0.919
0.961
1.003
1.043
1.084
1.123
1.162
1.201
1.239
1.277
1.314
1.351
1.388
1.424
1.460
1.496
1.532
1.567
0.100
0.226
0.324
0.400
0.466
0.526
0.581
0.632
0.680
0.726
0.769
0.811
0.851
0.890
0.928
0.964
1.000
1.034
1.068
1.100
1.132
1.163
1.193
1.223
1.252
1.280
1.308
1.335
1.362
1.388
1.414
0.0%
0.0%
0.1%
0.2%
0.4%
0.6%
0.9%
1.1%
1.4%
1.7%
2.1%
2.4%
2.8%
3.2%
3.6%
4.0%
4.4%
4.8%
5.2%
5.7%
6.1%
6.6%
7.0%
7.5%
8.0%
8.4%
8.9%
9.4%
9.9%
10.3%
10.8%
10.0%
7.5%
Error
r2/r1
5.0%
2.5%
0.0%
1.0
1.2
1.4
1 .6
1 .8
r 2 /r 1
2.0
2.2
2.4
2.6
Problem 6.35
[Difficulty: 4] Part 1/2
Problem 6.35
[Difficulty: 4] Part 2/2
Problem 6.36
[Difficulty: 4]
Given:
x component of velocity field
Find:
y component of velocity field; acceleration at several points; estimate radius of curvature; plot streamlines
Solution:
3
Λ = 2⋅
The given data is
The basic equation (continuity) is
∂
∂x
u +
m
u=−
s
∂
∂y
(2
Λ⋅ x − y
)
2
( x 2 + y 2)
2
v =0
The basic equation for acceleration is
⌠
v = −⎮
⎮
⌡
Hence
Integrating (using an integrating factor)
v=−
⌠
⎮
dy = −⎮
dx
⎮
⎮
⌡
du
(2
2⋅ Λ⋅ x⋅ x − 3⋅ y
( x 2 + y 2)
3
) dy
2
2⋅ Λ⋅ x⋅ y
(x2 + y2)
2
Alternatively, we could check that the given velocities u and v satisfy continuity
u=−
so
∂
∂x
(2
Λ⋅ x − y
u +
2
( x 2 + y 2)
∂
∂y
v =0
)
2
∂
∂x
u =
(2
2⋅ Λ⋅ x⋅ x − 3⋅ y
( x 2 + y 2)
3
2
)
v=−
2⋅ Λ⋅ x⋅ y
(x2 + y2)
∂
2
∂y
v =−
(2
2⋅ Λ⋅ x⋅ x − 3⋅ y
( x 2 + y 2)
3
2
)
For steady, 2D flow the acceleration components reduce to (after considerable math!):
x - component
ax = u ⋅
∂
∂x
u + v⋅
∂
∂y
(
u
)
(
)
(
⎡ Λ⋅ x2 − y2 ⎤ ⎡ 2⋅ Λ⋅ x⋅ x2 − 3⋅ y2 ⎤ ⎡ 2⋅ Λ⋅ x⋅ y ⎡ 2⋅ Λ⋅ y⋅ 3⋅ x2 − y2
⎤
⎥⋅⎢
⎥ + ⎢−
⎥ ⋅⎢
2
3
2
3
⎢ 2 2 ⎥⎢
⎥ ⎢ 2 2 ⎥⎢
2
2
2
2
x +y
x +y
⎣ x +y ⎦⎣
⎦ ⎣ x +y ⎦⎣
ax = ⎢−
y - component
ay = u ⋅
(
∂
∂x
)
v + v⋅
∂
∂y
(
(
)
(
)
(
)
Evaluating at point (0,1)
Evaluating at point (0,2)
Evaluating at point (0,3)
u = 2⋅
(
)
)
m
(
)
(
(
)
v = 0⋅
s
u = 0.5⋅
m
v = 0⋅
s
u = 0.222 ⋅
m
s
v = 0⋅
(
m
)
ax = 0 ⋅
s
(
m
2
)
y = 1m
)⎥⎤a
y=−
⎥
⎦
2
2⋅ Λ ⋅ x
( x 2 + y 2)
⎥
⎦
s
m
ax = 0 ⋅
s
m
2
ax = 0 ⋅
s
m
2
⎛ 2⋅ m ⎞
⎜ s
⎝
⎠
r =
or
( x 2 + y 2)
m
2
ay = −0.25⋅
m
2
s
ay = −0.0333⋅
s
u
aradial = −ay = −
r
2⋅ Λ ⋅ y
s
s
m
8⋅
m
2
s
r= −
u
2
ay
2
r = 0.5 m
m
2
s
y = 2m
⎛ m⎞
⎜ 0.5⋅ s
⎝
⎠
r =
0.25⋅
2
r = 1m
m
2
s
y = 3m
3
2
ay = −8 ⋅
2
The instantaneous radius of curvature is obtained from
For the three points
x=−
v
⎡ Λ⋅ x2 − y2 ⎤ ⎡ 2⋅ Λ⋅ y⋅ 3⋅ x2 − y2 ⎤ ⎡ 2⋅ Λ⋅ x⋅ y ⎡ 2⋅ Λ⋅ y⋅ 3⋅ y2 − x2
⎤
⎥⋅⎢
⎥ + ⎢−
⎥ ⋅⎢
2
3
2
3
⎢ 2 2 ⎥⎢
⎥ ⎢ 2 2 ⎥⎢
2
2
2
2
x +y
x +y
⎣ x +y ⎦⎣
⎦ ⎣ x +y ⎦⎣
ay = ⎢−
)⎥⎤a
m⎞
⎛
⎜ 0.2222⋅ s
⎝
⎠
r =
0.03333 ⋅
m
2
r = 1.5⋅ m
2
s
The radius of curvature in each case is 1/2 of the vertical distance from the origin. The streamlines form circles tangent to the x axis
3
2⋅ Λ⋅ x⋅ y
−
( x 2 + y 2) = 2 ⋅ x ⋅ y
=
=
2
2
dx
u
(x2 − y2)
Λ⋅ (x − y )
−
2
( x 2 + y 2)
dy
The streamlines are given by
2
v
(2
)
2
−2 ⋅ x ⋅ y ⋅ dx + x − y ⋅ dy = 0
so
This is an inexact integral, so an integrating factor is needed
R=
First we try
F=e
Then the integrating factor is
(
)
⎡d 2 2 d ( −2⋅ x ⋅ y)⎤ = − 2
⋅⎢ x − y −
⎥
−2 ⋅ x ⋅ y ⎣dx
y
dy
⎦
1
⌠
⎮
2
⎮ − dy
y
⎮
⌡
=
1
y
(2
2
2
) ⋅dy = 0
The equation becomes an exact integral
x
x −y
−2 ⋅ ⋅ dx +
y
2
y
So
2
⌠
x
x
u = ⎮ −2 ⋅ dx = −
+ f ( y)
⎮
y
y
⌡
ψ=
Comparing solutions
x
and
2
y
+y
(1)
(x2 − y2) dy = − x2 − y + g(x)
⌠
⎮
u=⎮
⎮
⌡
y
2
2
3.50
5.29
4.95
4.64
4.38
4.14
3.95
3.79
3.66
3.57
3.52
3.50
3.75
5.42
5.10
4.82
4.57
4.35
4.17
4.02
3.90
3.82
3.77
3.75
y
2
x + y = ψ⋅ y = const ⋅ y
or
These form circles that are tangential to the x axis, as can be shown in Excel:
x values
The stream function can be evaluated using Eq 1
2.50
2.25
2.00
1.75
1.50
1.25
1.00
0.75
0.50
0.25
0.00
0.10
62.6
50.7
40.1
30.7
22.6
15.7
10.1
5.73
2.60
0.73
0.10
0.25
25.3
20.5
16.3
12.5
9.25
6.50
4.25
2.50
1.25
0.50
0.25
See next page for plot:
0.50
13.0
10.6
8.50
6.63
5.00
3.63
2.50
1.63
1.00
0.63
0.50
0.75
9.08
7.50
6.08
4.83
3.75
2.83
2.08
1.50
1.08
0.83
0.75
1.00
7.25
6.06
5.00
4.06
3.25
2.56
2.00
1.56
1.25
1.06
1.00
1.25
6.25
5.30
4.45
3.70
3.05
2.50
2.05
1.70
1.45
1.30
1.25
1.50
5.67
4.88
4.17
3.54
3.00
2.54
2.17
1.88
1.67
1.54
1.50
1.75
5.32
4.64
4.04
3.50
3.04
2.64
2.32
2.07
1.89
1.79
1.75
2.00
5.13
4.53
4.00
3.53
3.13
2.78
2.50
2.28
2.13
2.03
2.00
2.25
5.03
4.50
4.03
3.61
3.25
2.94
2.69
2.50
2.36
2.28
2.25
y values
2.50
5.00
4.53
4.10
3.73
3.40
3.13
2.90
2.73
2.60
2.53
2.50
2.75
5.02
4.59
4.20
3.86
3.57
3.32
3.11
2.95
2.84
2.77
2.75
3.00
5.08
4.69
4.33
4.02
3.75
3.52
3.33
3.19
3.08
3.02
3.00
3.25
5.17
4.81
4.48
4.19
3.94
3.73
3.56
3.42
3.33
3.27
3.25
4.00
5.56
5.27
5.00
4.77
4.56
4.39
4.25
4.14
4.06
4.02
4.00
4.25
5.72
5.44
5.19
4.97
4.78
4.62
4.49
4.38
4.31
4.26
4.25
4.50
5.89
5.63
5.39
5.18
5.00
4.85
4.72
4.63
4.56
4.51
4.50
4.75
6.07
5.82
5.59
5.39
5.22
5.08
4.96
4.87
4.80
4.76
4.75
5.00
6.25
6.01
5.80
5.61
5.45
5.31
5.20
5.11
5.05
5.01
5.00
Problem 6.37
[Difficulty: 4] Part 1/2
Problem 6.37
[Difficulty: 4 ] Part 2/2
Problem 6.38
Given:
Water at speed 25 ft/s
Find:
Dynamic pressure in in. Hg
[Difficulty: 1]
Solution:
Basic equations
p dynamic =
1
2
2
⋅ ρ⋅ V
p = ρHg⋅ g ⋅ ∆h = SGHg⋅ ρ⋅ g ⋅ ∆h
2
Hence
∆h =
∆h =
2
ρ⋅ V
2⋅ SGHg⋅ ρ⋅ g
1
2
× ⎜⎛ 25⋅
⎝
ft ⎞
s
⎠
=
V
2⋅ SG Hg⋅ g
2
×
1
13.6
×
s
2
32.2⋅ ft
×
12⋅ in
1⋅ ft
∆h = 8.56⋅ in
Problem 6.39
Given:
Air speed of 100 km/hr
Find:
Dynamic pressure in mm water
[Difficulty: 1]
Solution:
Basic equations
Hence
p dynamic =
∆h =
1
2
⋅ ρair⋅ V
2
p = ρw⋅ g ⋅ ∆h
ρair V2
⋅
ρw 2 ⋅ g
1.23⋅
kg
3
m
∆h =
999 ⋅
kg
3
m
×
1
2
× ⎜⎛ 100 ⋅
⎝
km ⎞
hr
⎠
2
2
×
2
2
⎛ 1000⋅ m ⎞ × ⎛ 1 ⋅ hr ⎞ × s
⎜
⎜
9.81⋅ m
⎝ 1⋅ km ⎠ ⎝ 3600⋅ s ⎠
∆h = 48.4⋅ mm
Problem 6.40
Given:
Air speed
Find:
Plot dynamic pressure in mm Hg
[Difficulty: 2]
Solution:
p dynamic 
Basic equations
1
2
2
 ρair V
kg
ρw  999 
3
m
Available data
1
Hence
2
kg
ρair  1.23
3
m
SG Hg  13.6
2
 ρair V  SGHg ρw g  ∆h
V( ∆h) 
Solving for V
p  ρHg g  ∆h  SGHg ρw g  ∆h
2  SG Hg ρw g  ∆h
ρair
250
V (m/s
200
150
100
50
0
50
100
150
Dh (mm)
200
250
Problem 6.41
Given:
Velocity of automobile
Find:
Estimates of aerodynamic force on hand
[Difficulty: 2]
Solution:
The basic equation is the Bernoulli equation (in coordinates attached to the vehicle)
p atm +
1
2
2
⋅ ρ⋅ V = p stag
where V is the free stream velocity
For air
ρ = 0.00238 ⋅
slug
ft
3
We need an estimate of the area of a typical hand. Personal inspection indicates that a good approximation is a square of sides 9
cm and 17 cm
A = 9 ⋅ cm × 17⋅ cm
A = 153 ⋅ cm
2
Hence, for p stag on the front side of the hand, and p atm on the rear, by assumption,
(
)
1
2
F = p stag − p atm ⋅ A = ⋅ ρ⋅ V ⋅ A
2
(a)
V = 30⋅ mph
2
ft ⎞
⎛
22⋅
⎜
1
1
slug
s
2
2
F = ⋅ ρ⋅ V ⋅ A = × 0.00238 ⋅
× ⎜ 30⋅ mph⋅
× 153 ⋅ cm ×
15⋅ mph ⎠
2
2
3
⎝
ft
(b)
⎛ 1 ⋅ ft ⎞
⎜ 12
⎜
⎝ 2.54⋅ cm ⎠
2
⎛ 1 ⋅ ft ⎞
⎜ 12
⎜
⎝ 2.54⋅ cm ⎠
2
F = 0.379 ⋅ lbf
V = 60⋅ mph
2
ft ⎞
⎛
22⋅
⎜
1
1
slug
s
2
2
F = ⋅ ρ⋅ V ⋅ A = × 0.00238 ⋅
× ⎜ 60⋅ mph⋅
× 153 ⋅ cm ×
15
⋅
mph
2
2
3
⎝
⎠
ft
F = 1.52⋅ lbf
These values pretty much agree with experience. However, they overestimate a bit as the entire front of the hand is not at
stagnation pressure - there is flow around the had - so the pressure is less than stagnation over most of the surface.
Problem 6.42
Given:
Air jet hitting wall generating pressures
Find:
Speed of air at two locations
[Difficulty: 2]
Solution:
Basic equations
p
ρair
2
V
+
p
ρair =
Rair⋅ T
+ g ⋅ z = const
2
∆p = ρHg⋅ g ⋅ ∆h = SG Hg⋅ ρ⋅ g ⋅ ∆h
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Available data
For the air
R = 287 ⋅
J
T = −10 °C
kg⋅ K
kg
ρ = 999 ⋅
p = 200 ⋅ kPa
3
SG Hg = 13.6
m
p
ρair =
R⋅ T
ρair = 2.65
kg
3
m
Hence, applying Bernoulli between the jet and where it hits the wall directly
p atm
ρair
Hence
∆h = 25⋅ mm
+
Vj
2
=
2
p wall
p wall =
ρair
p wall = SGHg⋅ ρ⋅ g ⋅ ∆h =
Vj =
hence
ρair⋅ Vj
ρair⋅ Vj
2
(working in gage pressures)
2
2
Vj =
so
2
2 × 13.6 × 999 ⋅
kg
3
m
×
1
2 ⋅ SGHg⋅ ρ⋅ g ⋅ ∆h
ρair
3
⋅
m
2.65 kg
× 9.81⋅
m
2
× 25⋅ mm ×
s
1⋅ m
m
Vj = 50.1
s
1000⋅ mm
Repeating the analysis for the second point
∆h = 5 ⋅ mm
p atm
ρair
+
Vj
2
2
=
p wall
ρair
2
Hence
V =
2
+
V
2
V=
2
Vj −
2 ⋅ p wall
ρair
=
2
Vj −
2 ⋅ SG Hg⋅ ρ⋅ g ⋅ ∆h
ρair
3
⎛ 50.1⋅ m ⎞ − 2 × 13.6 × 999 ⋅ kg × 1 ⋅ m × 9.81⋅ m × 5 ⋅ mm × 1 ⋅ m
⎜
s⎠
3
2.65 kg
2
1000⋅ mm
⎝
m
s
V = 44.8
m
s
Problem 6.43
[Difficulty: 2]
Problem 6.44
[Difficulty: 2]
Given:
Wind tunnel with inlet section
Find:
Dynamic and static pressures on centerline; compare with Speed of air at two locations
Solution:
Basic equations
p dyn =
1
2
⋅ρ ⋅U
2 air
p 0 = p s + p dyn
p
ρair =
Rair⋅ T
∆p = ρw⋅ g ⋅ ∆h
p atm = 101⋅ kPa
kg
h 0 = −10⋅ mm ρw = 999⋅
3
m
p s = −1.738 kPa
hs =
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Available data
T = −5 °C
m
J
U = 50R
⋅ = 287⋅
s
kg⋅ K
For air
p atm
ρair =
R⋅ T
ρair = 1.31
p dyn =
1
2
kg
3
m
2
⋅ ρair⋅ U
Also
p 0 = ρw⋅ g ⋅ h 0
and
p 0 = p s + p dyn
p dyn = 1.64⋅ kPa
p 0 = −98.0 Pa
so
(gage)
p s = p 0 − p dyn
(gage)
∂
Streamlines in the test section are straight so
In the curved section
∂
∂n
p =0
and
2
V
p = ρair⋅
R
∂n
so
p w < p centerline
p w = p centerline
ps
ρw⋅ g
h s = −177 mm
Problem 6.45
[Difficulty: 2]
Problem 6.46
[Difficulty: 2]
Problem 6.47
4.123
[Difficulty: 4]
Problem 6.48
[Difficulty: 2]
Given:
Flow in pipe/nozzle device
Find:
Gage pressure needed for flow rate; repeat for inverted
Solution:
Basic equations
p
Q = V⋅ A
ρ
2
+
V
2
+ g ⋅ z = const
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Available data
From continuity
D1 = 1 ⋅ in
D2 = 0.5⋅ in
Q = V1 ⋅ A1 = V2 ⋅ A2
ft
V2 = 30⋅
s
z2 = 10⋅ ft
A2
⎛ D2 ⎞
V1 = V2 ⋅ ⎜
⎝ D1 ⎠
V1 = V2 ⋅
A1
ρ = 1.94⋅
ft
or
2
ft
V1 = 7.50
s
Hence, applying Bernoulli between locations 1 and 2
p1
ρ
+
V1
2
2
+0=
p2
ρ
+
V2
2
2
Solving for p 1 (gage)
⎛⎜ V 2 − V 2
⎞
2
1
p 1 = ρ⋅ ⎜
+ g ⋅ z2
2
⎝
⎠
When it is inverted
z2 = −10⋅ ft
⎞
⎛⎜ V 2 − V 2
2
1
p 2 = ρ⋅ ⎜
+ g ⋅ z2
2
⎝
⎠
+ g ⋅ z2 = 0 +
V2
2
slug
2
+ g ⋅ z2working in gage pressures
p 1 = 10.0⋅ psi
p 2 = 1.35⋅ psi
3
Problem 6.49
[Difficulty: 2]
Problem 6.50
Given:
Siphoning of gasoline
Find:
Flow rate
[Difficulty: 2]
Solution:
Basic equation
p
ρgas
2

V
2
 g  z  const
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Hence, applying Bernoulli between the gas tank free surface and the siphon exit
p atm
ρgas

p atm
ρgas
Hence
V
The flow rate is then
Q  V A 
2
V

2
 g h
where we assume the tank free surface is slowly changing so V tank <<,
and h is the difference in levels
2 g h
2
Q 
π
4
π D
4
 2 g  h
2
 ( .5 in) 
1  ft
2
2
144  in

2  32.2
ft
2
s
 1  ft
Q  0.0109
ft
3
s
Q  4.91
gal
min
Problem 6.51
Given:
Siphoning of wort
Find:
Flow rate; plot; height for a flow of 2 L/min
[Difficulty: 2]
Solution:
Basic equation
p
ρwort
2
V

 g  z  const
2
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Hence, applying Bernoulli between the open surface of the full tank and tube exit to atmosphere
p atm
ρgas

2
p atm

ρgas
Hence
V
The flow rate is then
Q  V A 
V
2
 g h
where we assume the tank free surface is slowly changing so V tank <<,
and h is the difference in levels
2 g h
2
π D
4
 2 g h
D  5  mm
For
Q (L/min)
3
2
1
0
50
100
150
200
250
h (mm)
For a flow rate of
Q  2
L
min
3
5m
Q  3.33  10
s
2
solving for h
h 
8 Q
2
4
π D g
h  147  mm
Problem 6.52
Given:
Ruptured pipe
Find:
Height benzene rises from tank
[Difficulty: 2]
Solution:
Basic equation
p
ρben
2

V
 g  z  const
2
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Available data
ρ  999 
kg
p ben  50 kPa
3
(gage)
From Table A.2
m
Hence, applying Bernoulli between the pipe and the rise height of the benzene
p ben
ρben
Hence
h 

p atm
ρben
 g h
p ben
SG ben ρ g
h  5.81 m
where we assume Vpipe <<, and h is the rise height
where p ben is now the gage pressure
SG ben  0.879
Problem 6.53
Given:
Ruptured Coke can
Find:
Pressure in can
[Difficulty: 2]
Solution:
Basic equation
p
ρCoke
2

V
2
 g  z  const
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Available data
kg
ρw  999 
3
m
h  0.5 m
From a web search
SG DietCoke  1
SG RegularCoke  1.11
Hence, applying Bernoulli between the coke can and the rise height of the coke
p can
ρCoke

p atm
ρCoke
 g h
where we assume VCoke <<, and h is the rise height
Hence
p Coke  ρCoke  g  h  SG Coke  ρw g  h where p Coke is now the gage pressure
Hence
p Diet  SG DietCoke ρw g  h
p Diet  4.90 kPa
(gage)
and
p Regular  SGRegularCoke  ρw g  h
p Regular  5.44 kPa
(gage)
Problem 6.54
Given:
Flow rate through siphon
Find:
Maximum height h to avoid cavitation
[Difficulty: 3]
Solution:
Basic equation
p
ρ
2
V
+
+ g ⋅ z = const
2
Q = V⋅ A
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Available data
From continuity
Q = 5⋅
V=
3
−3m
L
Q
A
Q = 5 × 10
s
D = 25⋅ mm
s
p atm = 101⋅ kPa
3
m
4⋅ Q
=
kg
ρ = 999⋅
V=
2
π⋅ D
4
π
3
× 0.005⋅
m
s
×
⎛ 1 ⎞
⎜ .025⋅ m
⎝
⎠
2
V = 10.2
m
s
Hence, applying Bernoulli between the free surface and point A
p atm
ρ
=
pA
ρ
2
+ g⋅ h +
V
where we assume VSurface <<
2
2
Hence
V
p A = p atm − ρ⋅ g ⋅ h − ρ⋅
2
p v = 2.358 ⋅ kPa
From the steam tables, at 20oC the vapor pressure is
This is the lowest permissible value of pA
2
Hence
V
p A = p v = p atm − ρ⋅ g ⋅ h − ρ⋅
2
Hence
h = ( 101 − 2.358 ) × 10 ⋅
3 N
2
m
×
h=
or
1
3
⋅
m
999 kg
p atm − p v
ρ⋅ g
2
×
s
9.81⋅ m
×
kg⋅ m
2
N⋅ s
2
−
−
V
2⋅ g
1
2
× ⎛⎜ 10.2
⎝
m⎞
s
⎠
2
2
×
s
9.81⋅ m
h = 4.76 m
Problem 6.55
[Difficulty: 2]
Problem 6.56
Given:
Flow through tank-pipe system
Find:
Velocity in pipe; Rate of discharge
[Difficulty: 2]
Solution:
Basic equations
p
ρ
2

V
 g  z  const
2
∆p  ρ g  ∆h
Q  V A
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Hence, applying Bernoulli between the free surface and the manometer location
p atm
ρ

p
ρ
2
 g H 
V
where we assume VSurface <<, and H = 4 m
2
2
Hence
V
p  p atm  ρ g  H  ρ
2
For the manometer
p  p atm  SGHg ρ g  h 2  ρ g  h 1
Note that we have water on one side and mercury on
the other of the manometer
2
Combining equations
Hence
ρ g  H  ρ
V 
V
2
 SGHg ρ g  h 2  ρ g  h 1
2  9.81
m
2
or
V

2  g  H  SG Hg h 2  h 2
 ( 4  13.6  0.15  0.75)  m
V  7.29
s
2
The flow rate is
Q  V
π D
4

Q 
π
4
 7.29
m
s
 ( 0.05 m)
2
m
s
3
Q  0.0143
m
s
Problem 6.57
[Difficulty: 2]
Problem 6.58
[Difficulty: 3]
Given:
Air flow over a wing
Find:
Air speed relative to wing at a point; absolute air speed
Solution:
Basic equation
p
ρ
2
V
+
+ g ⋅ z = const
2
p = ρ⋅ R⋅ T
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Available data
T = −10 °C
For air
ρ=
km
V1 = 200 ⋅
hr
p 1 = 65⋅ kPa
p1
3 N
ρ = ( 65) × 10 ⋅
R⋅ T
2
m
×
kg⋅ K
286.9 ⋅ N⋅ m
×
p 2 = 60⋅ kPa
1
( −10 + 273 ) ⋅ K
R = 286.9 ⋅
ρ = 0.861
N⋅ m
kg⋅ K
kg
3
m
Hence, applying Bernoulli between the upstream point (1) and the point on the wing (2)
p1
ρ
Hence
+
V1
2
=
2
2
p2
ρ
+
2
where we ignore gravity effects
2
( p1 − p2)
V2 =
V1 + 2 ⋅
V2 =
kg⋅ m
m
3 N
⎛ 200 ⋅ km ⎞ × ⎛ 1000⋅ m ⎞ × ⎛ 1⋅ hr ⎞ + 2 × m
× ( 65 − 60) × 10 ⋅
×
V2 = 121
⎜
⎜
⎜
hr ⎠
2
2
s
0.861 ⋅ kg
⎝
⎝ 1 ⋅ km ⎠ ⎝ 3600⋅ s ⎠
m
N⋅ s
ρ
2
Then
V2
2
2
3
NOTE: At this speed, significant density changes will occur, so this result is not very realistic
The absolute velocity is
V2abs = V2 − V1
m
V2abs = 65.7
s
Problem 6.59
[Difficulty: 3]
Given:
Water flow over a hydrofoil
Find:
Stagnation pressure; water speed relative to airfoil at a point; absolute value
Solution:
Basic equations
p
ρ
2

V
 g  z  const
2
∆p  ρ g  h
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Available data
m
V1  20
s
ρ  999 
kg
3
h  3 m
p 2  75 kPa
p 1  ρ g  h
p 1  29.4 kPa
(gage)
m
Using coordinates fixed to the hydrofoil, the pressure at depth h is
Applying Bernoulli between the upstream (1) and the stagnation point (at the front of the hydrofoil)
p1
ρ

V1
2

2
p0
or
ρ
1
2
p 0  p 1   ρ V1
2
p 0  229  kPa
Applying Bernoulli between the upstream point (1) and the point on the hydrofoil (2)
p1
ρ
Hence

V2 
V1
2
2

2
p2
ρ
V1  2 

V2
2
2
p1  p2
ρ
m
V2  24.7
s
This is the speed of the water relative to the hydrofoil; in absolute coordinates
V2abs  V2  V1
m
V2abs  44.7
s
Problem 6.60
[Difficulty: 3]
Problem 6.61
Given:
Flow through fire nozzle
Find:
Maximum flow rate
[Difficulty: 2]
Solution:
Basic equation
p
ρ
2
+
V
+ g ⋅ z = const
2
Q = V⋅ A
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Hence, applying Bernoulli between the inlet (1) and exit (2)
p1
ρ
+
V1
2
=
2
p2
ρ
+
V2
2
where we ignore gravity effects
2
2
But we have
π⋅ D
π⋅ d
Q = V1 ⋅ A1 = V1 ⋅
= V2 ⋅ A2 =
4
4
(
4
2⋅ p2 − p1
2
2 d
V2 − V2 ⋅ ⎛⎜ ⎞ =
ρ
⎝ D⎠
Hence
V2 =
(
2⋅ p1 − p2
⎡
ρ⋅ ⎢1 −
⎣
Then
V2 =
2×
π⋅ d
Q = V2 ⋅
4
d
V1 = V2 ⋅ ⎛⎜ ⎞
D
so
2
⎝ ⎠
)
)
4
⎛ d ⎞ ⎥⎤
⎜D
⎝ ⎠⎦
ft
3
1.94⋅ slug
2
2
× ( 100 − 0 ) ⋅
lbf
2
in
2
×
⎛ 12⋅ in ⎞ ×
⎜ 1⋅ ft
⎝
⎠
1
Q =
× 124 ⋅ × ⎛⎜ ⋅ ft⎞
4
s ⎝ 12 ⎠
π
ft
1
1−
⎛1⎞
⎜
⎝3⎠
3
×
slug⋅ ft
ft
V2 = 124 ⋅
s
2
lbf ⋅ s
2
Q = 0.676 ⋅
ft
3
s
Q = 304 ⋅
gal
min
Problem 6.62
[Difficulty: 2]
Given:
Race car on straightaway
Find:
Air inlet where speed is 60 mph; static pressure; pressure rise
Solution:
Basic equation
p
ρ
2
+
V
+ g ⋅ z = const
2
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline 5) Standard atmosphere
Available data
p atm = 101 ⋅ kPa
slug
ρ = 0.002377⋅
ft
3
V1 = 235⋅ mph
V2 = 60⋅ mph
Between location 1 (the upstream flow at 235 mph with respect to the car), and point 2 (on the car where V = 60 mph),
Bernoulli becomes
p1
ρ
Hence
2
+
V1
2
=
ρ
2
+
V1
2
=
⎡
⎢
p 2 = p atm + ⋅ ρ⋅ V1 ⋅ 1 −
⎢
2
⎣
Note that the pressure rise is
1
2
1
p2
ρ
2
+
⎛ V2 ⎞
⎜V
⎝ 1⎠
⎡
⎢
∆p = ⋅ ρ⋅ V1 ⋅ 1 −
⎢
2
⎣
The freestream dynamic pressure is
Then
p atm
2
V2
2
2⎤
⎥
⎥
⎦
p 2 = 15.6 psi
2⎤
⎛ V2 ⎞ ⎥
⎜V ⎥
⎝ 1⎠ ⎦
q =
1
∆p
= 93.5⋅ %
q
2
⋅ ρ⋅ V1
Note that at this speed the flow is borderline compressible!
2
∆p = 0.917 ⋅ psi
q = 0.980 ⋅ psi
Problem 6.63
[Difficulty: 2]
Problem 6.64
[Difficulty: 3]
Given:
Velocity field
Find:
Pressure distribution along wall; plot distribution; net force on wall
Solution:
3
m
q = 2⋅
The given data is
u=
s
h = 1⋅ m
m
kg
ρ = 1000⋅
3
m
q⋅ x
2 ⋅ π⎡⎣x + ( y − h )
2
+
2⎤
⎦
q⋅ x
2 ⋅ π⎡⎣x + ( y + h )
2
v=
2⎤
⎦
q⋅ ( y − h)
2 ⋅ π⎡⎣x + ( y − h )
2
+
2⎤
⎦
q⋅ ( y + h)
2 ⋅ π⎡⎣x + ( y + h )
2
2⎤
⎦
The governing equation is the Bernoulli equation
p
ρ
+
1
2
2
⋅ V + g ⋅ z = const
V=
where
2
u +v
2
Apply this to point arbitrary point (x,0) on the wall and at infinity (neglecting gravity)
x →0
At
u=
At point (x,0)
u→0
q⋅ x
(2
π⋅ x + h
2
v→0
V→0
v=0
)
V=
q⋅ x
(2
π⋅ x + h
p atm
Hence the Bernoulli equation becomes
ρ
=
p
ρ
+
⋅⎡
q⋅ x
⎤
2 ⎢
2
2⎥
⎣ π⋅ x + h ⎦
1
(
2
)
2
)
ρ
p(x) = − ⋅ ⎡
2 ⎢
q⋅ x
⎤
2
2⎥
⎣ π⋅ x + h ⎦
or (with pressure expressed as gage pressure)
(
2
)
(Alternatively, the pressure distribution could have been obtained from Problem 6.8, where the momentum equation was used to find
the pressure gradient
∂
∂x
2
p =
(2
ρ⋅ q ⋅ x ⋅ x − h
2
(2
π ⋅ x +h
2
)
) along the wall. Integration of this with respect to x leads to the same result for p(x))
2
3
The plot of pressure can be done in Excel (see below). From the plot it is clear that the wall experiences a negative gage pressure
on the upper surface (and zero gage pressure on the lower), so the net force on the wall is upwards, towards the source
10⋅ h
⌠
F=⎮
⌡
The force per width on the wall is given by
(pupper − plower) dx
F=−
− 10⋅ h
ρ⋅ q
2 ⌠
⎮
⋅⎮
2
2⋅ π ⎮
⎮
⌡
10⋅ h
− 10⋅ h
⌠
⎮
⎮
⎮
⎮
⌡
The integral is
x
( x 2 + h 2)
F=−
so
atan⎛⎜
2
ρ⋅ q
2
2
⋅ ⎜⎛ −
10
101
2⋅ π ⋅ h ⎝
2
(x
2
2
+h
2
)
2
x⎞
⎝h⎠ −
dx =
x
2⋅ h
x
2
2⋅ h + 2⋅ x
2
+ atan( 10) ⎞
⎠
2
2
⎛ m2 ⎞
10
N⋅ s
1
F = −
× 1000⋅
×
+ atan( 10) ⎞ ×
× ⎜⎛ −
× ⎜ 2⋅
2
3 ⎝
s ⎠
1 ⋅ m ⎝ 101
⎠ kg⋅ m
2⋅ π
m
1
kg
F = −278 ⋅
N
m
In Excel:
q =
h =
2
1
m3/s/m
m
ℵ= 1000 kg/m 3
x (m) p (Pa)
0.00
-50.66
-32.42
-18.24
-11.22
-7.49
-5.33
-3.97
-3.07
-2.44
-1.99
Pressure Distribution Along Wall
0
0
1
2
3
4
5
-10
p (Pa)
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
-20
-30
-40
-50
-60
x (m)
6
7
8
9
10
dx
Problem 6.65
[Difficulty: 3]
Given:
Velocity field for plane doublet
Find:
Pressure distribution along x axis; plot distribution
Solution:
p
The governing equation is the Bernoulli equation
ρ
+
1
2
2
⋅ V + g ⋅ z = const
3
Λ = 3⋅
The given data is
m
s
2
u +v
2
p 0 = 100 ⋅ kPa
3
m
Λ
Vr = − ⋅ cos( θ)
2
r
From Table 6.1
kg
ρ = 1000⋅
V=
where
Λ
Vθ = − ⋅ sin( θ)
2
r
where Vr and Vθ are the velocity components in cylindrical coordinates (r,θ). For points along the x axis, r = x, θ = 0, Vr = u and Vθ
=v=0
Λ
u=−
v = 0
2
x
p
so (neglecting gravity)
ρ
+
1
2
2
⋅ u = const
Apply this to point arbitrary point (x,0) on the x axis and at infinity
x →0
At
u→0
p → p0
u=−
At point (x,0)
Λ
x
Hence the Bernoulli equation becomes
p0
ρ
=
p
ρ
+
Λ
2
2⋅ x
or
4
p(x) = p0 −
ρ⋅ Λ
2⋅ x
2
2
4
The plot of pressure can be done in Excel:
x (m) p (Pa)
99.892
99.948
99.972
99.984
99.990
99.993
99.995
99.997
99.998
99.998
99.999
99.999
99.999
99.999
99.999
100.000
Pressure Distribution Along x axis
100.0
p (kPa)
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
100.0
99.9
99.9
99.8
0.0
0.2
0.4
0.6
0.8
1.0
x (m)
1.2
1.4
1.6
1.8
2.0
Problem 6.66
[Difficulty: 3]
Problem 6.67
[Difficulty: 3]


Rx
Given:
Flow through fire nozzle
Find:
Maximum flow rate
Solution:
Basic equation
p
ρ
2
+
V
+ g ⋅ z = const
2
Q = V⋅ A
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Hence, applying Bernoulli between the inlet (1) and exit (2)
p1
+
ρ
V1
2
=
2
p2
ρ
V2
+
2
where we ignore gravity effects
2
2
But we have
π⋅ D
π⋅ d
Q = V1 ⋅ A1 = V1 ⋅
= V2 ⋅
4
4
Hence in Bernoulli
4
2⋅ p2 − p1
2
2 d
V2 − V2 ⋅ ⎛⎜ ⎞ =
ρ
⎝ D⎠
(
2
)
so
d
V1 = V2 ⋅ ⎛⎜ ⎞
D
or
V2 =
⎝ ⎠
(
2⋅ p1 − p2
⎡
ρ⋅ ⎢1 −
⎣
3
V2 =
2×
π⋅ d
m
3 N
1000⋅ kg
× ( 700 − 0 ) × 10 ⋅
2
⎛ 25 ⎞
⎜ 75
⎝ ⎠
4
× ( 0.025 ⋅ m)
2
m
2
Q = V2 ⋅
4
Then
1
×
Q =
(
π
× 37.6⋅
4
)
1−
m
s
(
)
×
2
3
Q = 0.0185⋅
Hence
⎡
π⋅ D
π⋅ D
+ ρ⋅ Q⋅ V2 ⋅ ⎢1 −
Rx = −p 1 ⋅
+ ρ⋅ Q⋅ V2 − V1 = −p 1 ⋅
4
4
⎣
3 N
Rx = −700 × 10 ⋅
2
m
×
π
4
2
⋅ ( 0.075 ⋅ m) + 1000⋅
kg
3
m
3
× 0.0185⋅
m
s
m
s
Q = 18.5⋅
L
s
using gage pressures
2
)
m
V2 = 37.6
s
N⋅ s
Rx + p 1 ⋅ A1 = u 1 ⋅ −ρ⋅ V1 ⋅ A1 + u 2 ⋅ ρ⋅ V2 ⋅ A2
(
)
4
⎛ d ⎞ ⎥⎤
⎜D
⎝ ⎠⎦
kg⋅ m
From x momentum
2
2
× 37.6⋅
m
s
⎡
2⎤
⎛d⎞ ⎥
⎜
⎝ D⎠ ⎦
× ⎢1 −
⎣
3
2
⎛ 25 ⎞ ⎥⎤ × N⋅ s
⎜ 75
⎝ ⎠ ⎦ kg⋅ m
Rx = −2423 N
This is the force of the nozzle on the fluid; hence the force of the fluid on the nozzle is 2400 N to the right; the nozzle is in tension
Problem 6.68
[Difficulty: 3]
Problem 6.69
[Difficulty: 3]
Given:
Flow through reducing elbow
Find:
Gage pressure at location 1; x component of force
Solution:
Basic equations:
p
ρ
2
+
V
+ g ⋅ z = const
2
Q = V⋅ A
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline 5) Ignore elevation change 6) p 2 = p atm
Available data:
From contnuity
Q = 2.5⋅
V1 =
L
Q = 2.5 × 10
s
Q
3
−3m
D = 45⋅ mm
s
V2 =
p1
Hence, applying Bernoulli between the inlet (1) and exit (2)
ρ
or, in gage pressures
From x-momentum
p 1g =
ρ
2
⋅ ⎛ V2 − V1
⎝
2
(
ρ = 999 ⋅
kg
3
m
m
V1 = 1.57
s
⎛ π⋅ D2 ⎞
⎜
⎝ 4 ⎠
d = 25⋅ mm
2⎞
+
Q
m
V2 = 5.09
s
⎛ π⋅ d 2 ⎞
⎜
⎝ 4 ⎠
V1
2
2
=
p2
ρ
+
V2
2
2
p 1g = 11.7⋅ kPa
⎠
)
(
)
Rx + p 1g⋅ A1 = u 1 ⋅ −mrate + u 2 ⋅ mrate = −mrate⋅ V1 = −ρ⋅ Q⋅ V1
because
u 1 = V1
u2 = 0
2
π⋅ D
Rx = −p 1g⋅
− ρ⋅ Q⋅ V1
4
The force on the supply pipe is then
Rx = −22.6 N
Kx = −Rx
Kx = 22.6 N
on the pipe to the right
Problem 6.70
[Difficulty: 3]
Given:
Flow nozzle
Find:
Mass flow rate in terms of Δp, T1 and D 1 and D 2
Solution:
Basic equation
p
ρ
2
+
V
+ g ⋅ z = const
2
Q = V⋅ A
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Hence, applying Bernoulli between the inlet (1) and exit (2)
p1
+
ρ
But we have
V1
2
=
2
p2
ρ
Q = V1 ⋅ A1 = V1 ⋅
+
V2
2
where we ignore gravity effects
2
π⋅ D1
4
2
= V2 ⋅
π⋅ D2
2
⎛ D2 ⎞
V1 = V2 ⋅ ⎜
⎝ D1 ⎠
so
4
2
Note that we assume the flow at D 2 is at the same pressure as the entire section 2; this will be true if there is turbulent mixing
Hence
Then the mass flow rate is
Using
For a flow nozzle
4
2⋅ (p2 − p1)
⎛ D2 ⎞
V2 − V2 ⋅ ⎜
=
ρ
⎝ D1 ⎠
2
2
mflow = ρ⋅ V2 ⋅ A2 = ρ⋅
p = ρ⋅ R⋅ T
mflow = k ⋅ ∆p where
π⋅ D2
4
2
⋅
(
2⋅ p1 − p2
⎡
⎢
ρ⋅ 1 −
⎢
⎣
k=
)
⎛ D2 ⎞
⎜D
⎝ 1⎠
mflow =
4⎤
2⋅ 2
⋅
π⋅ D2
⎡
⎢
ρ⋅ 1 −
⎢
⎣
2
2⋅ 2
2
2⋅ 2
2
=
⎥
⎥
⎦
π⋅ D2
π⋅ D2
(
2⋅ p1 − p2
V2 =
or
⋅
⋅
)
⎛ D2 ⎞
⎜D
⎝ 1⎠
4⎤
⎥
⎥
⎦
∆p⋅ ρ
⎡
⎢
⎢1 −
⎣
⎛ D2 ⎞
⎜
⎝ D1 ⎠
4⎤
⎥
⎥
⎦
∆p⋅ p 1
⎡
⎢
R ⋅ T1 ⋅ 1 −
⎢
⎣
⎛ D2 ⎞
⎜D
⎝ 1⎠
4⎤
⎥
⎥
⎦
p1
⎡
⎢
R ⋅ T1 ⋅ 1 −
⎢
⎣
⎛ D2 ⎞
⎜D
⎝ 1⎠
4⎤
⎥
⎥
⎦
We can expect the actual flow will be less because there is actually significant loss in the device. Also the flow will experience a
vena contracta so that the minimum diameter is actually smaller than D 2. We will discuss this device in Chapter 8.
Problem 6.71
[Difficulty: 4]
Given:
Flow through branching blood vessel
Find:
Blood pressure in each branch; force at branch
Solution:
Basic equations
p
ρ
2
+
V
∑ Q= 0
+ g ⋅ z = const
2
Q = V⋅ A
∆p = ρ⋅ g ⋅ ∆h
CV
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Given data
L
Q1 = 4.5⋅
min
L
Q2 = 2 ⋅
min
SG Hg = 13.6
ρ = 999 ⋅
kg
3
m
For Q 3 we have
∑ Q = −Q1 + Q2 + Q3 = 0
D1 = 10⋅ mm
D2 = 5 ⋅ mm
D3 = 3 ⋅ mm
kg
ρb = 1060⋅
3
m
h 1 = 140 ⋅ mm
(pressure in in. Hg)
so
Q3 = Q1 − Q2
L
Q3 = 2.50⋅
min
CV
Q1
V1 =
A1
We will need each velocity
Similarly
V2 =
V1 =
4 ⋅ Q2
π⋅ D2
4 ⋅ Q1
π⋅ D1
2
m
V2 = 1.70
s
2
m
V1 = 0.955
s
V3 =
Hence, applying Bernoulli between the inlet (1) and exit (2)
p1
ρ
+
V1
2
2
=
p2
ρ
+
V2
2
2
where we ignore gravity effects
4 ⋅ Q3
π⋅ D3
2
m
V3 = 5.89
s
ρ
2
2
p 2 = p 1 + ⋅ ⎛ V1 − V2 ⎞
⎝
⎠
2
and
Hence
ρb
2
2
p2 = p1 +
⋅ ⎛ V − V2 ⎞
⎠
2 ⎝ 1
p 2 = 17.6⋅ kPa
In mm Hg
h2 =
Similarly for exit (3)
ρ
2
2
p 3 = p 1 + ⋅ ⎛ V1 − V3 ⎞
⎝
⎠
2
In mm Hg
h3 =
p2
p 1 = SGHg⋅ ρ⋅ g ⋅ h 1
p 1 = 18.7⋅ kPa
h 2 = 132 ⋅ mm
SGHg⋅ ρ⋅ g
p3
p 3 = 1.75⋅ kPa
h 3 = 13.2⋅ mm
SGHg⋅ ρ⋅ g
Note that all pressures are gage.
(
)
(
Rx + p 3 ⋅ A3 ⋅ cos( 60⋅ deg) − p 2 ⋅ A2 ⋅ cos( 45⋅ deg) = u 3 ⋅ ρ⋅ Q3 + u 2 ⋅ ρ⋅ Q2
For x momentum
)
(
Rx = p 2 ⋅ A2 ⋅ cos( 45⋅ deg) − p 3 ⋅ A3 ⋅ cos( 60⋅ deg) + ρ⋅ Q2 ⋅ V2 ⋅ cos( 45⋅ deg) − Q3 ⋅ V3 ⋅ cos( 60⋅ deg)
Rx = p2 ⋅
π⋅ D2
2
4
⋅ cos( 45⋅ deg) − p 3 ⋅
π⋅ D3
2
4
(
⋅ cos( 60⋅ deg) + ρ⋅ Q2 ⋅ V2 ⋅ cos( 45⋅ deg) − Q3 ⋅ V3 ⋅ cos( 60⋅ deg)
(
)
(
Ry − p 3 ⋅ A3 ⋅ sin( 60⋅ deg) − p 2 ⋅ A2 ⋅ sin( 45⋅ deg) + p 1 ⋅ A1 = v 3 ⋅ ρ⋅ Q3 + v 2 ⋅ ρ⋅ Q2
For y momentum
)
)
Rx = 0.156 N
)
(
Ry = p 2 ⋅ A2 ⋅ sin( 45⋅ deg) + p 3 ⋅ A3 ⋅ sin( 60⋅ deg) − p 1 ⋅ A1 + ρ⋅ Q2 ⋅ V2 ⋅ sin( 45⋅ deg) + Q3 ⋅ V3 ⋅ sin( 60⋅ deg)
Ry = p2 ⋅
π⋅ D2
4
2
⋅ sin( 45⋅ deg) + p 3 ⋅
π⋅ D3
4
2
⋅ sin( 60⋅ deg) − p 1 ⋅
π⋅ D1
4
)
2
(
+ ρ⋅ Q2 ⋅ V2 ⋅ sin( 45⋅ deg) + Q3 ⋅ V3 ⋅ sin( 60⋅ deg)
)
Ry = −0.957 N
Problem 6.72
[Difficulty: 3] Part 1/2
Problem 6.72
[Difficulty: 3] Part 2/2
Problem 6.73
[Difficulty: 3]
Problem 6.74
[Difficulty: 3]
c
V
H
CS
W
y
x
Ry
Given:
Flow through kitchen faucet
Find:
Area variation with height; force to hold plate as function of height
Solution:
2
p
Basic equation
ρ

V
 g  z  const
2
Q  V A
Assumptions: 1) Incompressible flow 2) Inviscid 3) Steady 4) Along a streamline
Hence, applying Bernoulli between the faucet (1) and any height y
V1
2
2
2
 g H 
2
 g y
where we assume the water is at patm
2
V( y ) 
Hence
V
V1  2  g  ( H  y )
m
V1  0.815
s
The problem doesn't require a plot, but it looks like
V( 0  m)  3.08
m
s
5
V (m/s)
4
3
2
1
0
5
10
15
20
25
30
35
40
45
y (cm)
The speed increases as y decreases because the fluid particles "trade" potential energy for kinetic, just as a falling solid particle does!
2
But we have
Hence
π D
Q  V1  A1  V1 
 V A
4
A
V1  A1
V
2
A( y ) 
π D1  V1
2
4  V1  2  g  ( H  y )
45
A( H)  1.23 cm
y (cm)
The problem doesn't require a plot, but it looks like
2
A( 0 )  0.325  cm
30
15
2
0
0.5
1
A (cm2)
The area decreases as the speed increases. If the stream falls far enough the flow will change to turbulent.
For the CV above


Ry  W  u in ρ Vin Ain  V ( ρ Q)
2
2
Ry  W  ρ V  A  W  ρ Q V1  2  g  ( H  y )
Hence Ry increases in the same way as V as the height y varies; the maximum force is when y = H
2
Rymax  W  ρ Q V1  2  g  H
1.5
Problem 6.75
[Difficulty: 4]
Open-Ended Problem Statement: An old magic trick uses an empty thread spool and a
playing card. The playing card is placed against the bottom of the spool. Contrary to
intuition, when one blows downward through the central hole in the spool, the card is not
blown away. Instead it is ‘‘sucked’’ up against the spool. Explain.
Discussion: The secret to this “parlor trick” lies in the velocity distribution, and hence
the pressure distribution, that exists between the spool and the playing cards.
Neglect viscous effects for the purposes of discussion. Consider the space between the
end of the spool and the playing card as a pair of parallel disks. Air from the hole in the
spool enters the annular space surrounding the hole, and then flows radially outward
between the parallel disks. For a given flow rate of air the edge of the hole is the crosssection of minimum flow area and therefore the location of maximum air speed.
After entering the space between the parallel disks, air flows radially outward. The flow
area becomes larger as the radius increases. Thus the air slows and its pressure increases.
The largest flow area, slowest air speed, and highest pressure between the disks occur at
the outer periphery of the spool where the air is discharged from an annular area.
The air leaving the annular space between the disk and card must be at atmospheric
pressure. This is the location of the highest pressure in the space between the parallel
disks. Therefore pressure at smaller radii between the disks must be lower, and hence the
pressure between the disks is sub-atmospheric. Pressure above the card is less than
atmospheric pressure; pressure beneath the card is atmospheric. Each portion of the card
experiences a pressure difference acting upward. This causes a net pressure force to act
upward on the whole card. The upward pressure force acting on the card tends to keep it
from blowing off the spool when air is introduced through the central hole in the spool.
Viscous effects are present in the narrow space between the disk and card. However, they
only reduce the pressure rise as the air flows outward, they do not dominate the flow
behavior.
Problem 6.76
[Difficulty: 4]
Given:
Air jet striking disk
Find:
Manometer deflection; Force to hold disk; Force assuming p 0 on entire disk; plot pressure distribution
Solution:
Basic equations: Hydrostatic pressure, Bernoulli, and momentum flux in x direction
p
∆p = SG ⋅ ρ⋅ g ⋅ ∆h
ρ
2
V
+
2
+ g ⋅ z = constant
Assumptions: 1) Steady flow 2) Incompressible 3) No friction 4) Flow along streamline 5) Uniform flow 6) Horizontal flow (g x = 0)
Applying Bernoulli between jet exit and stagnation point
p atm
ρair
2
+
V
2
=
p0
1
2
p 0 − p atm = ⋅ ρair⋅ V
2
+0
ρair
1
But from hydrostatics
p 0 − p atm = SG ⋅ ρ⋅ g ⋅ ∆h
∆h = 0.002377⋅
slug
ft
For x momentum
(
3
× ⎜⎛ 225 ⋅
⎝
ft ⎞
2
s⎠
2 π⋅ d
)
1
×
2 ⋅ 1.75
×
2
SG ⋅ ρ⋅ g
ft
3
2
=
ρair⋅ V
2 ⋅ SG ⋅ ρ⋅ g
2
1.94⋅ slug
×
s
32.2⋅ ft
∆h = 0.55⋅ ft
2
Rx = V⋅ −ρair⋅ A⋅ V = −ρair⋅ V ⋅
4
Rx = −0.002377⋅
slug
ft
3
× ⎛⎜ 225 ⋅
The force of the jet on the plate is then F = −Rx
The stagnation pressure is
∆h =
so
2
⋅ ρair⋅ V
1
2
p 0 = p atm + ⋅ ρair⋅ V
2
⎝
ft ⎞
s⎠
π⋅ ⎛⎜
2
×
0.4
⋅ ft⎞
2
2
⎝ 12 ⎠ × lbf ⋅ s
4
slug⋅ ft
Rx = −0.105 ⋅ lbf
F = 0.105 ⋅ lbf
∆h = 6.60⋅ in
The force on the plate, assuming stagnation pressure on the front face, is
2
1
2 π⋅ D
F = p 0 − p ⋅ A = ⋅ ρair⋅ V ⋅
2
4
(
F =
)
π
8
× 0.002377⋅
slug
ft
3
× ⎛⎜ 225 ⋅
⎝
ft ⎞
2
s⎠
2
×
2
lbf ⋅ s
⎛ 7.5 ⎞
F = 18.5⋅ lbf
⎜ ⋅ ft ×
slug⋅ ft
⎝ 12 ⎠
Obviously this is a huge overestimate!
For the pressure distribution on the disk, we use Bernoulli between the disk outside edge any radius r for radial flow
p atm
ρair
+
1
2
p
2
⋅ v edge =
ρair
+
1
2
⋅v
2
We need to obtain the speed v as a function of radius. If we assume the flow remains constant thickness h, then
Q = v ⋅ 2 ⋅ π⋅ r⋅ h = V⋅
π⋅ d
2
v ( r) = V⋅
4
d
2
8⋅ h⋅ r
We need an estimate for h. As an approximation, we assume that h = d (this assumption will change the scale of p(r) but not the
basic shape)
d
Hence
v ( r) = V⋅
Using this in Bernoulli
ρair⋅ V ⋅ d
4
1
2
2
p ( r) = p atm + ⋅ ρair⋅ ⎛ v edge − v ( r) ⎞ = p atm +
− ⎞
⋅⎛
⎝
⎠
⎜
2
128
2
2
r ⎠
⎝D
8⋅ r
2 2
1
2 2
Expressed as a gage pressure
0
p ( r) =
ρair⋅ V ⋅ d
128
1
⋅⎛
4
1⎞
⎜ 2− 2
r ⎠
⎝D
2
p (psi)
− 0.1
− 0.2
− 0.3
r (in)
3
4
Problem 6.77
[Difficulty: 4] Part 1/2
Problem 6.77
[Difficulty: 4] Part 2/2
Problem 6.78
[Difficulty: 4] Part 1/2
Problem 6.78
[Difficulty: 4] Part 2/2
Problem 6.79
[Difficulty: 4]
Problem 6.80
Given:
Air flow over "bubble" structure
Find:
Net vertical force
Solution:
The net force is given by
L = 50⋅ ft
Available data
[Difficulty: 3]
→ ⌠
→
F = ⎮ p dA
⌡
R = 25ft
∆p = ρ⋅ g ⋅ ∆h
also
V = 35⋅ mph
∆h = 0.75⋅ in
ρ = 1.94⋅
slug
ft
The internal pressure is
∆p = ρ⋅ g ⋅ ∆h
⌠
FV = ⎮
⌡
π
where pi is the internal pressure and p the external
π
slug
ft
(pi − p)⋅ sin(θ)⋅ R⋅ L dθ
0
⌠
FV = ⎮
⎮
⌡
ρair = 0.00238 ⋅
∆p = 187 Pa
Through symmetry only the vertical component of force is no-zero
Hence
3
(
(
1
2
2
p = p atm − ⋅ ρair⋅ V ⋅ 1 − 4 ⋅ sin( θ)
2
p i = p atm + ∆p
)
)
⎡∆p − 1 ⋅ ρ ⋅ V2⋅ 1 − 4 ⋅ sin( θ) 2 ⎤ ⋅ sin( θ) ⋅ R⋅ L dθ
⎢
⎥
2 air
⎣
⎦
0
⌠
FV = R⋅ L⋅ ∆p⋅ ⎮
⌡
π
1
2⌠
π
sin( θ) dθ − R⋅ L⋅ ⋅ ρair⋅ V ⋅ ⎮
⌡
2
0
0
But
⌠
⎮
⎮
⌡
(sin(θ) − 4⋅sin(θ)3) dθ = −cos(θ) + 4⋅⎛⎜ cos(θ) − 13 ⋅cos(θ)3⎞
⎝
⌠
⎮ sin( θ) dθ = −cos( θ)
⌡
Combining results
5
2
FV = R⋅ L⋅ ⎛⎜ 2 ⋅ ∆p + ⋅ ρair⋅ V ⎞
3
⎝
⎠
⎠
(1 − 4⋅sin(θ)2)⋅sin(θ) dθ
so
⌠
⎮
⌡
π
0
so
⌠
⎮
⌡
(sin(θ) − 4⋅sin(θ)3) dθ = − 103
π
sin( θ) dθ = 2
0
4
FV = 2.28 × 10 ⋅ lbf
FV = 22.8⋅ kip
3
Problem 6.81
[Difficulty: 4] Part 1/2
Problem 6.81
[Difficulty: 4] Part 2/2
Problem 6.82
[Difficulty: 4]
Given:
Water flow out of tube
Find:
Pressure indicated by gage; force to hold body in place
Solution:
Basic equations: Bernoulli, and momentum flux in x direction
p
ρ
2
+
V
+ g ⋅ z = constant
2
Q = V⋅ A
Assumptions: 1) Steady flow 2) Incompressible 3) No friction 4) Flow along streamline 5) Uniform flow 6) Horizontal flow (g x = 0)
Applying Bernoulli between jet exit and stagnation point
p1
ρ
+
p1 =
V1
2
2
=
p2
+
ρ
⋅ ⎛ V − V1
2 ⎝ 2
ρ
V2
2
2
2
V2
=
2
where we work in gage pressure
2
2⎞
⎠
2
A1
D
V2 = V1 ⋅
= V1 ⋅
A2
2
2
D −d
But from continuity Q = V1 ⋅ A1 = V2 ⋅ A2
⎞
ft ⎛
2
V2 = 20⋅ ⋅ ⎜
s ⎜ 2
2
⎝ 2 − 1.5 ⎠
2
p1 =
Hence
The x mometum is
1
2
× 1.94⋅
slug
ft
3
ft
V2 = 45.7⋅
s
(
)
× 45.7 − 20 ⋅ ⎛⎜
2
2
(
ft ⎞
2
⎝s⎠
)
F = p 1 ⋅ A1 + ρ⋅ ⎛ V1 ⋅ A1 − V2 ⋅ A2⎞
⎝
⎠
F = 11.4⋅
lbf
2
in
×
2
×
(
lbf ⋅ s
π⋅ ( 2 ⋅ in)
4
2
+ 1.94⋅
F = 14.1⋅ lbf
slug
ft
3
×
2
p 1 = 1638⋅
slug⋅ ft
−F + p 1 ⋅ A1 − p 2 ⋅ A2 = u 1 ⋅ −ρ⋅ V1 ⋅ A1 + u 2 ⋅ ρ⋅ V2 ⋅ A2
2
where D = 2 in and d = 1.5 in
lbf
ft
2
p 1 = 11.4⋅ psi
(gage)
)
using gage pressures
2
2
2
2
2
⎡⎛ ft ⎞ 2 π⋅ ( 2⋅ in) 2 ⎛
π⋅ ⎡⎣( 2 ⋅ in) − ( 1.5⋅ in) ⎤⎦ ⎤ ⎛ 1 ⋅ ft ⎞
lbf ⋅ s
ft
⎢⎜ 20⋅
⎥×⎜
×
− ⎜ 45.7⋅ ⎞ ×
×
4
4
slug⋅ ft
s⎠
⎣⎝ s ⎠
⎝
⎦ ⎝ 12⋅ in ⎠
in the direction shown
Problem 6.83
[Difficulty: 4] Part 1/2
Problem 6.83
[Difficulty: 4] Part 2/2
Problem 6.84
[Difficulty: 5]
Open-Ended Problem Statement: Describe the pressure distribution on the exterior of a
multistory building in a steady wind. Identify the locations of the maximum and
minimum pressures on the outside of the building. Discuss the effect of these pressures
on infiltration of outside air into the building.
Discussion: A multi-story building acts as a bluff-body obstruction in a thick
atmospheric boundary layer. The boundary-layer velocity profile causes the air speed
near the top of the building to be highest and that toward the ground to be lower.
Obstruction of air flow by the building causes regions of stagnation pressure on upwind
surfaces. The stagnation pressure is highest where the air speed is highest. Therefore the
maximum surface pressure occurs near the roof on the upwind side of the building.
Minimum pressure on the upwind surface of the building occurs near the ground where
the air speed is lowest.
The minimum pressure on the entire building will likely be in the low-speed, lowpressure wake region on the downwind side of the building.
Static pressure inside the building will tend to be an average of all the surface pressures
that act on the outside of the building. It is never possible to seal all openings completely.
Therefore air will tend to infiltrate into the building in regions where the outside surface
pressure is above the interior pressure, and will tend to pass out of the building in regions
where the outside surface pressure is below the interior pressure. Thus generally air will
tend to move through the building from the upper floors toward the lower floors, and
from the upwind side to the downwind side.
Problem 6.85
[Difficulty: 5]
Open-Ended Problem Statement: Imagine a garden hose with a stream of water
flowing out through a nozzle. Explain why the end of the hose may be unstable when
held a half meter or so from the nozzle end.
Discussion: Water flowing out of the nozzle tends to exert a thrust force on the end of the
hose. The thrust force is aligned with the flow from the nozzle and is directed toward the
hose.
Any misalignment of the hose will lead to a tendency for the thrust force to bend the hose
further. This will quickly become unstable, with the result that the free end of the hose
will “flail” about, spraying water from the nozzle in all directions.
This instability phenomenon can be demonstrated easily in the backyard. However, it will
tend to do least damage when the person demonstrating it is wearing a bathing suit!
Problem 6.86
[Difficulty: 5]
Open-Ended Problem Statement: An aspirator provides suction by using a stream of
water flowing through a venturi. Analyze the shape and dimensions of such a device.
Comment on any limitations on its use.
Discussion: The basic shape of the aspirator channel should be a converging nozzle
section to reduce pressure followed by a diverging diffuser section to promote pressure
recovery. The basic shape is that of a venturi flow meter.
If the diffuser exhausts to atmosphere, the exit pressure will be atmospheric. The pressure
rise in the diffuser will cause the pressure at the diffuser inlet (venturi throat) to be below
atmospheric.
A small tube can be brought in from the side of the throat to aspirate another liquid or gas
into the throat as a result of the reduced pressure there.
The following comments can be made about limitations on the aspirator:
1. It is desirable to minimize the area of the aspirator tube compared to the flow area
of the venturi throat. This minimizes the disturbance of the main flow through the
venturi and promotes the best possible pressure recovery in the diffuser.
2. It is desirable to avoid cavitation in the throat of the venturi. Cavitation alters the
effective shape of the flow channel and destroys the pressure recovery in the
diffuser. To avoid cavitation, the reduced pressure must always be above the
vapor pressure of the driver liquid.
3. It is desirable to limit the flow rate of gas into the venturi throat. A large amount
of gas can alter the flow pattern and adversely affect pressure recovery in the
diffuser.
The best combination of specific dimensions could be determined experimentally by a
systematic study of aspirator performance. A good starting point probably would be to
use dimensions similar to those of a commercially available venturi flow meter.
Problem 6.87
[Difficulty: 5]
Problem 6.88
[Difficulty: 3]
Problem 6.89
Given:
Unsteady water flow out of tube
Find:
Pressure in the tank
[Difficulty: 3]
Solution:
Basic equation: Unsteady Bernoulli
Assumptions: 1) Unsteady flow 2) Incompressible 3) No friction 4) Flow along streamline 5) Uniform flow 6) Horizontal flow (g x = 0
Applying unsteady Bernoulli between reservoir and tube exit
2
2
⌠
2
dV ⌠
V
∂
+ g⋅ h =
V ds =
+⎮
+
⋅ ⎮ 1 ds
⎮ ∂t
2
ρ
dt ⌡1
2
⌡
2
V
p
where we work in gage pressure
1
Hence
Hence
⎛ V2
p = ρ⋅ ⎜
⎝ 2
p = 1.94⋅
− g⋅ h +
slug
ft
3
×
dV
dt
⎞
⋅L
⎠
2
2
⎛ 62
⎞
⎜ − 32.2 × 4.5 + 7.5 × 35 ⋅ ⎛⎜ ft ⎞ × lbf ⋅ s
slug⋅ ft
⎝2
⎠ ⎝s⎠
p = 263 ⋅
lbf
ft
2
p = 1.83⋅ psi
(gage)
Problem 6.90
Given:
Unsteady water flow out of tube
Find:
Initial acceleration
[Difficulty: 3]
Solution:
Basic equation: Unsteady Bernoulli
Assumptions: 1) Unsteady flow 2) Incompressible 3) No friction 4) Flow along streamline 5) Uniform flow 6) Horizontal flow (g x = 0
Applying unsteady Bernoulli between reservoir and tube exit
2
2
⌠
dV ⌠
∂
+ g⋅ h = ⎮
V ds =
⋅ ⎮ 1 ds = a x ⋅ L
⎮ ∂t
ρ
dt ⌡1
⌡
p
where we work in gage pressure
1
Hence
Hence
ax =
ax =
1
L
⋅ ⎜⎛
p
⎝ρ
1
35⋅ ft
+ g⋅ h⎞
⎠
3
⎡ lbf ⎛ 12⋅ in 2
⎤
slug⋅ ft
ft
⎞ × ft
×
+ 32.2⋅ × 4.5⋅ ft⎥
×⎜
2
2
⎢ in2 ⎝ 1 ⋅ ft ⎠
⎥
1.94⋅ slug
s ⋅ lbf
s
⎣
⎦
× ⎢3 ⋅
ax = 10.5⋅
ft
2
s
Note that we obtain the same result if we treat the water in the pipe as a single body at rest with gage pressure p + ρgh at the left end!
Problem 6.91
[Difficulty: 4]
Problem 6.92
[Difficulty: 4]
Given:
Unsteady water flow out of tube
Find:
Differential equation for velocity; Integrate; Plot v versus time
Solution:
Basic equation: Unsteady Bernoulli
Assumptions: 1) Unsteady flow 2) Incompressible 3) No friction 4) Flow along streamline 5) Uniform flow 6) Horizontal flow (g x = 0
Applying unsteady Bernoulli between reservoir and tube exit
2
⌠
2
2
2
⎮ ∂
V
p
dV ⌠
dV
V
V
+ g⋅ h =
+
⋅L
+⎮
V ds =
+
⋅ ⎮ 1 ds =
⌡
2
ρ
dt
dt
2
2
⎮ ∂t
1
⌡
2
where we work in gage pressure
1
dV
Hence
dt
2
+
V
2⋅ L
=
1
L
L⋅ dV
Separating variables
p
ρ
2
+ g⋅ h −
⋅ ⎜⎛
p
⎝ρ
+ g⋅ h ⎞
is the differential equation for the flow
⎠
= dt
V
2
Integrating and using limits V(0) = 0 and V(t) = V
p
⎞
⎜⎛
+ g⋅ h
ρ
V( t) = 2 ⋅ ⎜ + g ⋅ h⎞ ⋅ tanh⎜
⋅ t⎟
2
ρ
⎜
⎝
⎠
⎝ 2⋅ L
⎠
⎛p
25
V (ft/s)
20
15
10
5
0
1
2
3
t (s)
This graph is suitable for plotting in Excel
For large times
V =
2 ⋅ ⎛⎜
p
⎝ρ
+ g ⋅ h⎞
⎠
V = 22.6⋅
ft
s
4
5
Problem 6.93
[Difficulty: 5] Part 1/2
Problem 6.93
4.48
[Difficulty: 5] Part 2/2
.04
Problem 6.94
[Difficulty: 5]
Problem 6.95
(a)
[Difficulty: 2]
Note that the effect of friction would be that the EGL would tend to drop:
suddenly at the contraction, gradually in the large pipe, more steeply in the
small pipe. The HGL would then “hang” below the HGL in a manner similar
to that shown.
EGL
Turbine
HGL
(b)
Note that the effect of friction would be that the EGL would tend to drop:
suddenly at the contraction, gradually in the large pipe, more steeply in the
small pipe. The HGL would then “hang” below the HGL in a manner similar
to that shown.
EGL
Turbine
HGL
Problem 6.96
(a)
[Difficulty: 2]
Note that the effect of friction would be that the EGL would tend to drop from
right to left: steeply in the small pipe, gradually in the large pipe, and
suddenly at the expansion. The HGL would then “hang” below the HGL in a
manner similar to that shown.
EGL
Flow
Pump
HGL
(b)
Note that the effect of friction would be that the EGL would tend to drop from
right to left: steeply in the small pipe, gradually in the large pipe, and
suddenly at the expansion. The HGL would then “hang” below the HGL in a
manner similar to that shown.
EGL
Flow
HGL
Pump
Problem 6.97
[Difficulty: 2]
Problem 6.98
[Difficulty: 3]
Given:
2D incompressible, inviscid flow field
Find:
Relationships among constants; stream function; velocity potential
Solution:
Basic equations
u=
Incompressible
∂
∂
v=− ψ
∂x
ψ
∂y
u ( x , y ) = a⋅ x + b ⋅ y
Check incompressibility
∂
∂x
∂
u( x , y) +
v( x , y) −
∂
∂y
∂
Irrotational
∂
v=− ϕ
∂y
v ( x , y ) = c⋅ x + d ⋅ y
v( x , y) = a + d
Hence must have
d = −a
u( x , y) = c − b
Hence must have
c = b
Check irrotational
∂x
Hence for the streamfunction
⌠
1
2
ψ ( x , y ) = ⎮ u ( x , y ) dy = a ⋅ x ⋅ y + ⋅ b ⋅ y + f ( x )
2
⌡
∂y
∂
u=− ϕ
∂x
⌠
1
2
ψ( x , y ) = −⎮ v ( x , y ) dx = − ⋅ c⋅ x − d ⋅ x ⋅ y + g ( y )
2
⌡
Comparing
Check
Hence for the velocity potential
ψ( x , y ) =
u( x , y) =
1
2
2
⋅ b⋅ y −
∂
∂y
1
2
2
⋅ c⋅ x + a⋅ x ⋅ y
ψ( x , y ) = a ⋅ x + b ⋅ y
ψ( x , y ) = a ⋅ x ⋅ y +
1
2
(2
⋅ b⋅ y − x
2
)
∂
v ( x , y ) = − ψ( x , y ) = b ⋅ x − a ⋅ y
∂x
⌠
1
2
ϕ( x , y ) = −⎮ u ( x , y ) dx = − ⋅ a⋅ x − b ⋅ x ⋅ y + f ( y )
2
⌡
⌠
1
2
ψ( x , y ) = −⎮ v ( x , y ) dy = −c⋅ x ⋅ y − ⋅ d ⋅ y + g ( x )
2
⌡
Comparing
Check
1
(2
)
1
1
2
2
ϕ( x , y ) = − ⋅ a⋅ x − ⋅ d ⋅ y − b ⋅ x ⋅ y
2
2
ϕ( x , y ) = −b ⋅ x ⋅ y −
∂
u ( x , y ) = − ϕ( x , y ) = a⋅ x + b ⋅ y
∂x
∂
v ( x , y ) = − ϕ( x , y ) = b ⋅ x − a⋅ y
∂y
2
⋅ a⋅ x − y
2
Problem 6.99
[Difficulty: 3]
Given:
Stream function
Find:
If the flow is irrotational; Pressure difference between points (1,4) and (2,1)
Solution:
Basic equations: Incompressibility because ψ exists
u=
2
∂
∂y
∂
v=− ψ
∂x
ψ
∂
Irrotationality
∂x
v −
∂
∂y
u =0
ψ( x , y ) = A⋅ x ⋅ y
u( x , y) =
∂
∂y
ψ( x , y ) =
∂
∂y
(A⋅x2⋅y)
(
2
∂
∂
v ( x , y ) = − ψ( x , y ) = − A⋅ x ⋅ y
∂x
∂x
Hence
∂
∂x
v( x , y) −
∂
∂y
u ( x , y ) = A⋅ x
)
2
v ( x , y ) = −2 ⋅ A⋅ x ⋅ y
∂
u ( x , y ) = −2 ⋅ A⋅ y
∂x
v −
∂
∂y
u ≠0
so flow is NOT IRROTATIONAL
Since flow is rotational, we must be on same streamline to be able to use Bernoulli
At point (1,4)
ψ( 1 , 4 ) = 4 A
ψ( 2 , 1 ) = 4 A
and at point (2,1)
Hence these points are on same streamline so Bernoulli can be used. The velocity at a point is
2
2
2
2
Hence at (1,4)
V1 =
⎡ 2.5 × ( 1 ⋅ m) 2⎤ + ⎛ −2 × 2.5 × 1⋅ m × 4⋅ m⎞
⎢ m⋅ s
⎥ ⎜
m⋅ s
⎣
⎦ ⎝
⎠
Hence at (2,1)
V2 =
⎡ 2.5 × ( 2 ⋅ m) 2⎤ + ⎛ −2 × 2.5 × 2⋅ m × 1⋅ m⎞
⎢ m⋅ s
⎥ ⎜
m⋅ s
⎣
⎦ ⎝
⎠
Using Bernoulli
p1
+
1
∆p =
1
ρ
2
2
⋅ V1 =
2
p2
+
ρ
× 1200⋅
kg
3
m
1
2
⋅ V2
(
2
2
)
m⎞
⎝s⎠
2
2
u( x , y) + v( x , y)
m
V1 = 20.2
s
m
V2 = 14.1
s
∆p =
× 14.1 − 20.2 ⋅ ⎛⎜
2
V( x , y ) =
⋅ ⎛ V − V1
2 ⎝ 2
ρ
2
2
×
N⋅ s
kg⋅ m
∆p = −126 ⋅ kPa
2⎞
⎠
2
Problem 6.100
[Difficulty: 2]
Problem 6.101
[Difficulty: 3]
Given:
Data from Table 6.2
Find:
Stream function and velocity potential for a source in a corner; plot; velocity along one plane
Solution:
From Table 6.2, for a source at the origin
ψ( r , θ) =
Expressed in Cartesian coordinates
ψ( x , y ) =
q
2⋅ π
⋅θ
q
2⋅ π
ϕ( r , θ) = −
⋅ atan⎛⎜
y⎞
q
2⋅ π
ϕ( x , y ) = −
⎝x⎠
⋅ ln( r)
q
4⋅ π
(2
⋅ ln x + y
2
)
To build flow in a corner, we need image sources at three locations so that there is symmetry about both axes. We need sources at
(h,h), (h,- h), (- h,h), and (- h,- h)
Hence the composite stream function and velocity potential are
ψ( x , y ) =
q
2⋅ π
ϕ( x , y ) = −
⋅ ⎛⎜ atan⎛⎜
⎝
q
4⋅ π
y − h⎞
⎝x − h⎠
+ atan⎛⎜
y + h⎞
+ atan⎛⎜
⎝x − h⎠
y + h⎞
⋅ ln⎡⎣⎡⎣( x − h ) + ( y − h ) ⎤⎦ ⋅ ⎡⎣( x − h ) + ( y + h )
2
+ atan⎛⎜
⎝x + h⎠
2
2
y − h ⎞⎞
⎝ x + h ⎠⎠
2⎤⎤
⎦⎦ −
q
⋅ ⎡⎣( x + h ) + ( y + h ) ⎤⎦ ⋅ ⎡⎣( x + h ) + ( y − h )
2
4⋅ π
2
2
2⎤
⎦
By a similar reasoning the horizontal velocity is given by
u=
q⋅ ( x − h)
2 ⋅ π⎡⎣( x − h ) + ( y − h )
2
2⎤
⎦
+
q⋅ ( x − h)
2 ⋅ π⎡⎣( x − h ) + ( y + h )
2
2⎤
⎦
+
q⋅ ( x + h)
2 ⋅ π⎡⎣( x + h ) + ( y + h )
2
2⎤
+
⎦
q⋅ ( x + h)
2 ⋅ π⎡⎣( x + h ) + ( y + h )
2
Along the horizontal wall (y = 0)
u=
or
q⋅ ( x − h)
2 ⋅ π⎡⎣( x − h ) + h
u(x) =
2
q
π
2⎤
⎦
+
q⋅ ( x − h)
2 ⋅ π⎡⎣( x − h ) + h
2
x−h
x+h
⎤
+
⎢
2
2
2
2⎥
( x + h) + h ⎦
⎣ ( x − h) + h
⋅⎡
2⎤
⎦
+
q⋅ ( x + h)
2 ⋅ π⎡⎣( x + h ) + h
2
2⎤
⎦
+
q⋅ ( x + h)
2 ⋅ π⎡⎣( x + h ) + h
2
2⎤
⎦
2⎤
⎦
The results in Excel are:
Velocity Potential
y
x
Stream Function
y
x
Problem 6.102
[Difficulty: 3]
Given:
Velocity field of irrotational and incompressible flow
Find:
Stream function and velocity potential; plot
Solution:
The velocity field is
u=
q⋅ x
2 ⋅ π⎡⎣x + ( y − h )
2
∂
2⎤
+
⎦
q⋅ x
2 ⋅ π⎡⎣x + ( y + h )
2
∂
v=− ψ
∂x
2⎤
v=
⎦
∂
u=− ϕ
∂x
q⋅ ( y − h)
2 ⋅ π⎡⎣x + ( y − h )
2
∂
v=− ϕ
∂y
The basic equations are
u=
Hence for the stream function
⌠
q ⎛
y − h⎞
y + h ⎞⎞
ψ = ⎮ u ( x , y ) dy =
⋅ ⎜ atan⎛⎜
+ atan⎛⎜
+ f ( x)
2⋅ π ⎝
⌡
⎝ x ⎠
⎝ x ⎠⎠
∂y
ψ
⌠
q ⎛
y − h⎞
y + h ⎞⎞
ψ = −⎮ v ( x , y ) dx =
⋅ ⎜ atan⎛⎜
+ atan⎛⎜
+ g( y)
2⋅ π ⎝
⌡
⎝ x ⎠
⎝ x ⎠⎠
q
⋅ ⎜⎛ atan⎛⎜
y − h⎞
+ atan⎛⎜
y + h ⎞⎞
The simplest expression for ψ is
ψ( x , y ) =
For the stream function
⌠
q
2
2
2
2
ϕ = −⎮ u ( x , y ) dx = −
⋅ ln⎡⎣⎡⎣x + ( y − h ) ⎤⎦ ⋅ ⎡⎣x + ( y + h ) ⎤⎦⎤⎦ + f ( y )
4⋅ π
⌡
2⋅ π
⎝
⎝ x ⎠
⎝ x ⎠⎠
⌠
q
2
2
2
2
ϕ = −⎮ v ( x , y ) dy = −
⋅ ln⎡⎣⎡⎣x + ( y − h ) ⎤⎦ ⋅ ⎡⎣x + ( y + h ) ⎤⎦⎤⎦ + g ( x )
4⋅ π
⌡
The simplest expression for φ is
ϕ( x , y ) = −
q
4⋅ π
⋅ ln⎡⎣⎡⎣x + ( y − h ) ⎤⎦ ⋅ ⎡⎣x + ( y + h )
2
2
2
2⎤⎤
⎦⎦
2⎤
⎦
+
q⋅ ( y + h)
2 ⋅ π⎡⎣x + ( y + h )
2
2
In Excel:
Stream Function
Velocity Potential
Problem 6.103
[Difficulty: 3]
Given:
Data from Table 6.2
Find:
Stream function and velocity potential for a vortex in a corner; plot; velocity along one plane
Solution:
From Table 6.2, for a vortex at the origin
ϕ( r , θ) =
Expressed in Cartesian coordinates
ϕ( x , y ) =
K
2⋅ π
⋅θ
q
2⋅ π
ψ( r , θ) = −
⋅ atan⎛⎜
y⎞
K
2⋅ π
ψ( x , y ) = −
⎝x⎠
⋅ ln( r)
q
4⋅ π
(2
⋅ ln x + y
2
)
To build flow in a corner, we need image vortices at three locations so that there is symmetry about both axes. We need vortices
at (h,h), (h,- h), (- h,h), and (- h,- h). Note that some of them must have strengths of - K!
Hence the composite velocity potential and stream function are
ϕ( x , y ) =
K
2⋅ π
y − h⎞
⋅ ⎛⎜ atan⎛⎜
⎝x − h⎠
⎝
− atan⎛⎜
y + h⎞
⎝x − h⎠
+ atan⎛⎜
y + h⎞
⎝x + h⎠
− atan⎛⎜
y − h ⎞⎞
⎝ x + h ⎠⎠
⎡ ( x − h) 2 + ( y − h) 2 ( x + h) 2 + ( y + h) 2⎤
⎥
ψ( x , y ) = −
⋅
⋅ ln⎢
4⋅ π ⎢
2
2
2
2⎥
⎣ ( x − h) + ( y + h) ( x + h) + ( y − h) ⎦
K
By a similar reasoning the horizontal velocity is given by
u=−
K⋅ ( y − h )
2 ⋅ π⎡⎣( x − h ) + ( y − h )
2
2⎤
+
⎦
K⋅ ( y + h )
2 ⋅ π⎡⎣( x − h ) + ( y + h )
2
2⎤
⎦
−
K⋅ ( y + h )
2 ⋅ π⎡⎣( x + h ) + ( y + h )
2
2⎤
⎦
+
K⋅ ( y − h )
2 ⋅ π⎡⎣( x + h ) + ( y − h )
2
Along the horizontal wall (y = 0)
u=
or
K⋅ h
2 ⋅ π⎡⎣( x − h ) + h
u(x) =
2
K⋅ h
⋅⎡
2⎤
⎦
+
K⋅ h
2 ⋅ π⎡⎣( x − h ) + h
2
2⎤
⎦
1
1
⎤
−
⎢
π
2
2
2
2⎥
( x + h) + h ⎦
⎣ ( x − h) + h
−
K⋅ h
2 ⋅ π⎡⎣( x + h ) + h
2
2⎤
⎦
−
K⋅ h
2 ⋅ π⎡⎣( x + h ) + h
2
2⎤
⎦
2⎤
⎦
In Excel:
y
Stream Function
x
Velocity Potential
y
x
Problem 6.104
Given:
Stream function
Find:
Velocity potential
[Difficulty: 2]
Solution:
Basic equations: Incompressibility because ψ exists
∂
Irrotationality
2
ψ ( x, y ) = A ⋅ x ⋅ y − B⋅ y
We have
Then
∂x
u ( x, y ) =
∂
∂y
Hence
∂
∂x
v ( x, y ) −
∂
∂y
∂
∂y
∂
∂y
∂
v=− ψ
∂x
ψ
∂
u=− φ
∂x
∂
v=− φ
∂y
u =0
3
2
ψ ( x, y )
u ( x, y ) = A ⋅ x − 3⋅ B⋅ y
∂
v ( x, y ) = − ψ ( x, y )
∂x
Then
v −
u=
2
v ( x, y ) = −2⋅ A ⋅ x⋅ y
u ( x, y ) = 6⋅ B⋅ y − 2⋅ A ⋅ y
but
6⋅ B − 2⋅ A = 0
1
m⋅ s
hence flow is IRROTATIONAL
∂
u=− φ
∂x
so
3
⌠
A⋅ x
2
φ( x , y ) = −⎮ u ( x , y ) dx + f ( y ) → φ( x , y ) = f ( y ) −
+ 3⋅ B⋅ x⋅ y
3
⌡
∂
v=− φ
∂y
so
⌠
2
φ( x , y ) = −⎮ v ( x , y ) dy + g ( x ) → φ( x , y ) = A⋅ x ⋅ y + g ( x )
⌡
Comparing, the simplest velocity potential is then
2
φ( x , y ) = A⋅ x ⋅ y −
A⋅ x
3
3
Problem 6.105
[Difficulty: 2]
Given:
Stream function
Find:
Velocity field; Show flow is irrotational; Velocity potential
Solution:
Basic equations: Incompressibility because ψ exists
∂
Irrotationality
∂x
5
v −
∂
∂y
3 2
∂
∂y
ψ( x , y )
∂x
Hence
v( x , y) −
∂
∂y
∂y
∂
v=− ψ
∂x
ψ
∂
u=− φ
∂x
∂
v=− φ
∂y
4
3
3
2 2
4
u ( x , y ) = 20⋅ x ⋅ y − 20⋅ x ⋅ y
∂
v ( x , y ) = − ψ( x , y )
∂x
∂
∂
u =0
ψ( x , y ) = x − 10⋅ x ⋅ y + 5 ⋅ x ⋅ y
u( x , y) =
u=
v ( x , y ) = 30⋅ x ⋅ y − 5 ⋅ x − 5 ⋅ y
u( x , y) = 0
4
Hence flow is IRROTATIONAL
∂
u=− φ
∂x
so
⌠
4
2 3
φ( x , y ) = −⎮ u ( x , y ) dx + f ( y ) = 5 ⋅ x ⋅ y − 10⋅ x ⋅ y + f ( y )
⌡
∂
v=− φ
∂y
so
⌠
4
2 3
5
φ( x , y ) = −⎮ v ( x , y ) dy + g ( x ) = 5 ⋅ x ⋅ y − 10⋅ x ⋅ y + y + g ( x )
⌡
Comparing, the simplest velocity potential is then
4
2 3
φ( x , y ) = 5 ⋅ x ⋅ y − 10⋅ x ⋅ y + y
5
Problem 6.106
Given:
Stream function
Find:
Velocity potential
[Difficulty: 2]
Solution:
Basic equations: Incompressibility because ψ exists
∂
Irrotationality
3
ψ( x , y ) = A⋅ x − B⋅ x ⋅ y
We have
Then
∂x
u( x , y) =
∂
∂y
ψ( x , y )
Hence
∂
∂x
v( x , y) −
∂
∂y
∂
∂y
∂
∂y
∂
v=− ψ
∂x
ψ
∂
u=− φ
∂x
∂
v=− φ
∂y
u =0
2
u ( x , y ) = −2 ⋅ B⋅ x ⋅ y
∂
v ( x , y ) = − ψ( x , y )
∂x
Then
v −
u=
2
v ( x , y ) = B⋅ y − 3 ⋅ A⋅ x
u ( x , y ) = 2 ⋅ B⋅ x − 6 ⋅ A⋅ x
but
2
2⋅ B − 6⋅ A = 0
1
m⋅ s
hence flow is IRROTATIONAL
∂
u=− φ
∂x
so
⌠
2
φ( x , y ) = −⎮ u ( x , y ) dx + f ( y ) → φ( x , y ) = B⋅ y ⋅ x + f ( y )
⌡
∂
v=− φ
∂y
so
3
⌠
B⋅ y
2
φ( x , y ) = −⎮ v ( x , y ) dy + g ( x ) → φ( x , y ) = g ( x ) −
+ 3 ⋅ A⋅ x ⋅ y
3
⌡
Comparing, the simplest velocity potential is then
2
φ( x , y ) = 3 ⋅ A⋅ x ⋅ y −
B⋅ y
3
3
Problem 6.107
[Difficulty: 2]
Given:
Stream function
Find:
Find A vs B if flow is irrotational; Velocity potential
Solution:
Basic equations: Incompressibility because ψ exists
∂
Irrotationality
We have
∂x
3
(
v −
∂
∂y
2
2
∂
∂y
2
∂x
Hence
v( x , y) −
ψ( x , y )
∂
∂y
∂y
ψ
∂
v=− ψ
∂x
∂
u=− φ
∂x
∂
v=− φ
∂y
)
u ( x , y ) = −B⋅ ( 2 ⋅ y − 2 ⋅ x ⋅ y )
∂
v ( x , y ) = − ψ( x , y )
∂x
∂
∂
u =0
ψ( x , y ) = A⋅ x + B⋅ x ⋅ y + x − y
u( x , y) =
u=
2
(2
v ( x , y ) = −3 ⋅ A⋅ x − B⋅ y + 2 ⋅ x
u ( x , y ) = −2 ⋅ x ⋅ ( 3 ⋅ A + B)
Hence flow is IRROTATIONAL if
)
B = −3 ⋅ A
∂
u=− φ
∂x
so
⌠
2
φ( x , y ) = −⎮ u ( x , y ) dx + f ( y ) = 2 ⋅ B⋅ y ⋅ x − B⋅ y ⋅ x + f ( y )
⌡
∂
v=− φ
∂y
so
3
⌠
B⋅ y
2
φ( x , y ) = −⎮ v ( x , y ) dy + g ( x ) = 3 ⋅ A⋅ x ⋅ y + 2 ⋅ B⋅ x ⋅ y +
+ g( x)
3
⌡
Comparing, the simplest velocity potential is then
2
φ( x , y ) = 2 ⋅ B⋅ y ⋅ x − B⋅ y ⋅ x + B⋅
y
3
3
Problem 6.108
[Difficulty: 2]
Given:
Stream function
Find:
Velocity field; Show flow is irrotational; Velocity potential
Solution:
Basic equations: Incompressibility because ψ exists
∂
Irrotationality
We have
∂x
6
v −
∂
∂y
4 2
∂
∂y
∂x
v( x , y) −
∂
∂y
∂y
∂
v=− ψ
∂x
ψ
2 4
∂
u=− φ
∂x
4
u ( x , y ) = 60⋅ x ⋅ y − 30⋅ x ⋅ y − 6 ⋅ y
3 2
5
v ( x , y ) = 60⋅ x ⋅ y − 6 ⋅ x − 30⋅ x ⋅ y
u( x , y) = 0
∂
v=− φ
∂y
6
2 3
ψ( x , y )
∂
v ( x , y ) = − ψ( x , y )
∂x
∂
∂
u =0
ψ( x , y ) = x − 15⋅ x ⋅ y + 15⋅ x ⋅ y − y
u( x , y) =
Hence
u=
5
4
Hence flow is IRROTATIONAL
∂
u=− φ
∂x
so
⌠
5
3 3
5
φ( x , y ) = −⎮ u ( x , y ) dx + f ( y ) = 6 ⋅ x ⋅ y − 20⋅ x ⋅ y + 6 ⋅ x ⋅ y + f ( y )
⌡
∂
v=− φ
∂y
so
⌠
5
3 3
5
φ( x , y ) = −⎮ v ( x , y ) dy + g ( x ) = 6 ⋅ x ⋅ y − 20⋅ x ⋅ y + 6 ⋅ x ⋅ y + g ( x )
⌡
Comparing, the simplest velocity potential is then
5
3 3
φ( x , y ) = 6 ⋅ x ⋅ y − 20⋅ x ⋅ y + 6 ⋅ x ⋅ y
5
Problem 6.109
Given:
Velocity potential
Find:
Show flow is incompressible; Stream function
[Difficulty: 2]
Solution:
Basic equations: Irrotationality because φ exists
Incompressibility
We have
Hence
Hence
∂
∂x
u +
2
u=
∂
∂y
φ( x , y ) = A⋅ x + B⋅ x ⋅ y − A⋅ y
∂
∂y
ψ
∂
v=− ψ
∂x
2
u ( x , y ) = −2 ⋅ A⋅ x − B⋅ y
∂
v ( x , y ) = − φ( x , y )
∂y
v ( x , y ) = 2 ⋅ A⋅ y − B⋅ x
∂x
u( x , y) +
u=
∂
∂y
ψ
∂
v=− ψ
∂x
∂
∂y
v( x , y) = 0
∂
v=− φ
∂y
v =0
∂
u ( x , y ) = − φ( x , y )
∂x
∂
∂
u=− φ
∂x
Hence flow is INCOMPRESSIBLE
so
⌠
1
2
ψ( x , y ) = ⎮ u ( x , y ) dy + f ( x ) = −2 ⋅ A⋅ x ⋅ y − ⋅ B⋅ y + f ( x )
2
⌡
so
⌠
1
2
ψ( x , y ) = −⎮ v ( x , y ) dx + g ( y ) = −2 ⋅ A⋅ x ⋅ y + ⋅ B⋅ x + g ( y )
2
⌡
Comparing, the simplest stream function is then
ψ( x , y ) = −2 ⋅ A⋅ x ⋅ y +
1
2
2
⋅ B⋅ x −
1
2
⋅ B⋅ y
2
Problem 6.110
Given:
Velocity potential
Find:
Show flow is incompressible; Stream function
[Difficulty: 2]
Solution:
Basic equations: Irrotationality because φ exists
∂
Incompressibility
We have
Hence
Hence
5
∂x
u +
3 2
∂
∂y
u=
∂
∂y
∂
v=− ψ
∂x
ψ
∂
u=− φ
∂x
v =0
4
2
φ( x , y ) = x − 10⋅ x ⋅ y + 5 ⋅ x ⋅ y − x + y
2
∂
u ( x , y ) = − φ( x , y )
∂x
u ( x , y ) = 30⋅ x ⋅ y − 5 ⋅ x + 2 ⋅ x − 5 ⋅ y
∂
v ( x , y ) = − φ( x , y )
∂y
v ( x , y ) = 20⋅ x ⋅ y − 20⋅ x ⋅ y − 2 ⋅ y
∂
∂x
u( x , y) +
u=
∂
∂y
ψ
∂
v=− ψ
∂x
∂
∂y
2 2
4
3
v( x , y) = 0
∂
v=− φ
∂y
4
3
Hence flow is INCOMPRESSIBLE
so
⌠
2 3
4
5
ψ( x , y ) = ⎮ u ( x , y ) dy + f ( x ) = 10⋅ x ⋅ y − 5 ⋅ x ⋅ y + 2 ⋅ x ⋅ y − y + f ( x )
⌡
so
⌠
2 3
4
ψ( x , y ) = −⎮ v ( x , y ) dx + g ( y ) = 10⋅ x ⋅ y − 5 ⋅ x ⋅ y + 2 ⋅ x ⋅ y + g ( y )
⌡
Comparing, the simplest stream function is then
2 3
4
ψ( x , y ) = 10⋅ x ⋅ y − 5 ⋅ x ⋅ y + 2 ⋅ x ⋅ y − y
5
Problem 6.111
Given:
Velocity potential
Find:
Show flow is incompressible; Stream function
[Difficulty: 2]
Solution:
Basic equations: Irrotationality because φ exists
∂
Incompressibility
6
∂x
u +
4 2
u=
∂
∂y
Hence
∂y
∂
v=− ψ
∂x
ψ
∂
u=− φ
∂x
2 4
6
∂
u ( x , y ) = − φ( x , y )
∂x
u ( x, y ) = 60⋅ x ⋅ y − 6⋅ x − 30⋅ x⋅ y
∂
v ( x , y ) = − φ( x , y )
∂y
v ( x, y ) = 30⋅ x ⋅ y − 60⋅ x ⋅ y + 6⋅ y
∂
∂x
u ( x, y ) +
u=
∂
∂y
ψ
∂
v=− ψ
∂x
∂
∂y
v ( x, y ) = 0
∂
v=− φ
∂y
v =0
φ( x, y ) = x − 15⋅ x ⋅ y + 15⋅ x ⋅ y − y
Hence
∂
3 2
5
4
4
2 3
5
Hence flow is INCOMPRESSIBLE
so
⌠
3 3
5
5
ψ( x , y ) = ⎮ u ( x , y ) dy + f ( x ) = 20⋅ x ⋅ y − 6 ⋅ x ⋅ y − 6 ⋅ x ⋅ y + f ( x )
⌡
so
⌠
3 3
5
5
ψ( x , y ) = −⎮ v ( x , y ) dx + g ( y ) = 20⋅ x ⋅ y − 6 ⋅ x ⋅ y − 6 ⋅ x ⋅ y + g ( y )
⌡
Comparing, the simplest stream function is then
3 3
5
ψ( x , y ) = 20⋅ x ⋅ y − 6 ⋅ x ⋅ y − 6 ⋅ x ⋅ y
5
Problem 6.112
[Difficulty: 4]
Given:
Complex function
Find:
Show it leads to velocity potential and stream function of irrotational incompressible flow; Show that df/dz leads
to u and v
Solution:
Basic equations: Irrotationality because φ exists
∂
Incompressibility
6
f ( z) = z = ( x + i ⋅ y )
6
∂x
u +
∂
∂y
u=
v =0
∂
∂y
∂
v=− ψ
∂x
ψ
Irrotationality
6
∂
∂x
4 2
v −
∂
∂y
2 4
∂
u=− φ
∂x
∂
v=− φ
∂y
u =0
(
6
5
5
We are thus to check the following
6
4 2
2 4
φ( x , y ) = x − 15⋅ x ⋅ y + 15⋅ x ⋅ y − y
∂
u ( x , y ) = − φ( x , y )
∂x
so
∂
v ( x , y ) = − φ( x , y )
∂y
so
6
(
)
5
5
3 3
3 2
5
4
4
2 3
5
3 2
5
4
4
2 3
5
ψ( x , y ) = − 6 ⋅ x ⋅ y + 6 ⋅ x ⋅ y − 20⋅ x ⋅ y
u ( x , y ) = 60⋅ x ⋅ y − 6 ⋅ x − 30⋅ x ⋅ y
v ( x , y ) = 30⋅ x ⋅ y − 60⋅ x ⋅ y + 6 ⋅ y
An alternative derivation of u and v is
u( x , y) =
∂
∂y
ψ( x , y )
u ( x , y ) = 60⋅ x ⋅ y − 6 ⋅ x − 30⋅ x ⋅ y
∂
v ( x , y ) = − ψ( x , y )
∂x
Hence
Hence
Next we find
∂
∂x
∂
∂x
v( x , y) −
u( x , y) +
df
dz
Hence we see
df
dz
=
∂
∂y
∂
∂y
v ( x , y ) = 30⋅ x ⋅ y − 60⋅ x ⋅ y + 6 ⋅ y
u( x , y) = 0
Hence flow is IRROTATIONAL
v( x , y) = 0
Hence flow is INCOMPRESSIBLE
( 6) = 6⋅z5 = 6⋅(x + i⋅y)5 = (6⋅x5 − 60⋅x3⋅y2 + 30⋅x⋅y4) + i⋅(30⋅x4⋅y + 6⋅y5 − 60⋅x2⋅y3)
dz
d z
= −u + i⋅ v
Hence the results are verified;
These interesting results are explained in Problem 6.113!
u = −Re⎛⎜
df
⎞
⎝ dz ⎠
and
)
3 3
f ( z) = x − 15⋅ x ⋅ y + 15⋅ x ⋅ y − y + i⋅ 6 ⋅ x ⋅ y + 6 ⋅ x ⋅ y − 20⋅ x ⋅ y
Expanding
v = Im⎛⎜
df
⎞
⎝ dz ⎠
Problem 6.113
[Difficulty: 4]
Given:
Complex function
Find:
Show it leads to velocity potential and stream function of irrotational incompressible flow; Show that df/dz
leads to u and v
Solution:
Basic equations: u = ∂ ψ
∂y
First consider
∂
∂x
f =
2
Hence
∂
∂x
2
2
Combining
∂
∂x
2
∂
∂
v=− ψ
∂x
∂
u=− φ
∂x
d
d
d
f = 1⋅ f = f
dz
dz
∂x dz
f =
f +
z⋅
(1)
and also
⎛ ∂ ⎞ d ⎛ d ⎞ d2
⎜ f = ⎜ f = 2f
∂x ⎝ ∂x ⎠ dz ⎝ dz ⎠
dz
2
∂y
2
f =
d
2
dz
2
f −
d
2
dz
2
f =0
and
∂y
f =
∂
∂y
2
∂
d
d
d
f = i⋅ f = i⋅ f
dz
dz
∂y dz
f =
z⋅
(2)
2
⎛∂ ⎞
d ⎛ d ⎞
d
⎜ f = i⋅ ⎜ i⋅ f = − 2 f
dz ⎝ dz ⎠
∂y ⎝ ∂y ⎠
dz
∂
Any differentiable function f(z) automatically satisfies the Laplace
Equation; so do its real and imaginary parts!
We demonstrate derivation of velocities u and v
From Eq 1
d
∂
∂
∂
∂
f = f = ( φ − i⋅ ψ) = φ − i⋅ ψ = −u + i⋅ v
dz
∂x
∂x
∂x
∂x
From Eq 2
1 ∂
1 ∂
d
∂
∂
f = ⋅ f = ⋅ ( φ − i⋅ ψ) = −i⋅ φ − ψ = i⋅ v − u
i ∂y
i ∂y
dz
∂y
∂y
Hence we have demonstrated that
∂
2
∂
∂
∂
v=− φ
∂y
df
dz
= −u + i⋅ v
Problem 6.114
[Difficulty: 2]
Problem 6.115
[Difficulty: 2] Part 1/2
Problem 6.115
[Difficulty: 2] Part 2/2
Problem 6.116
[Difficulty: 3]
Problem 6.117
[Difficulty: 3]
Problem 6.118
[Difficulty: 3] Part 1/2
Problem 6.118
[Difficulty: 3] Part 2/2
Problem 6.119
[Difficulty: 3]
Open-Ended Problem Statement: Consider flow around a circular cylinder with
freestream velocity from right to left and a counterclockwise free vortex. Show that the
lift force on the cylinder can be expressed as FL = −ρUΓ, as illustrated in Example 6.12.
Discussion: The only change in this flow from the flow of Example 6.12 is that the
directions of the freestream velocity and the vortex are changed. This changes the sign of
the freestream velocity from U to −U and the sign of the vortex strength from K to −K.
Consequently the signs of both terms in the equation for lift are changed. Therefore the
direction of the lift force remains unchanged.
The analysis of Example 6.12 shows that only the term involving the vortex strength
contributes to the lift force. Therefore the expression for lift obtained with the changed
freestream velocity and vortex strength is identical to that derived in Example 6.12. Thus
the general solution of Example 6.12 holds for any orientation of the freestream and
vortex velocities. For the present case, FL = −ρUΓ, as shown for the general case in
Example 6.12.
Problem 6.120
[Difficulty: 2]
Problem 6.121
[Difficulty: 3]
Problem 6.122
[Difficulty: 3]
Problem 6.123
[Difficulty: 3]
Problem 6.124
[Difficulty: 3] Part 1/2
Problem 6.124
[Difficulty: 3] Part 2/2
Problem 6.125
[Difficulty: 4]
Problem 6.126
[Difficulty: 3] Part 1/2
Problem 6.126
[Difficulty: 3] Part 2/2
Problem 6.127
[Difficulty: 4] Part 1/2
Problem 6.127
[Difficulty: 4] Part 2/2
Problem 7.1
[Difficulty: 2]
Given:
Find:
Equation describing the propagation speed of surface waves in a region of uniform depth
Solution:
To nondimensionalize the equation all lengths are divided by the reference length and all velocities are divided by
the reference velocity. Denoting the nondimensional quantities by an asterisk:
Nondimensionalization for the equation using length scale L and velocity scale Vo. Obtain the dimensionless
groups that characterize the flow.
* 
Substituting into the governing equation:

h* 
L
c V 
2
*
0
h
L
c* 
c
V0
  2 g* L 
2h* L

 

tanh
*
* L
2 
 L
Simplifying this expression:
  2 gL *
2
 2
c*  
2
*
 LV0  V0 2
The dimensionless group is
g L
V0
which is the reciprocal of the square of the Froude number, and
2
which is the inverse of the Weber number.
σ
ρ L V0
2

2h*
 tanh *


Problem 7.2
Given:
Equation for beam
Find:
Dimensionless groups
[Difficulty: 2]
Solution:
Denoting nondimensional quantities by an asterisk
A* 
Hence
A
L2
A  L2 A *
Substituting into the governing equation
The final dimensionless equation is
The dimensionless group is
y* 
y
L
t*  t
y  Ly*
t
I* 
t*

I
L4
I  L4 I *
x* 
x
L
x  Lx*
2 y *
4 y *
4 1
*
0
 EL 4 LI
t *2
x *4
L
2 y *  E  4 y *
I *
A*

0
t *2   L2 2  x *4
L L A *
2
2
 E 
 2 2 
 L 
Problem 7.3
Given:
Find:
Solution:
[Difficulty: 2]
Equation describing the slope of a steady wave in a shallow liquid layer
Nondimensionalization for the equation using length scale L and velocity scale V o. Obtain the dimensionless
groups that characterize the flow.
To nondimensionalize the equation all lengths are divided by the reference length and all velocities are divided by
the reference velocity. Denoting the nondimensional quantities by an asterisk:
h* 
Substituting into the governing equation:
The dimensionless group is
V0
h
L
 
 
x* 
x
L
u* 
u
V0
 
 
 h* L
u*V0  u *V0


 x* L
g  x* L
2
g L
which is the square of the Froude number.
h*
V02 * u *


u
x*
gL x*
Problem 7.4
[Difficulty: 2]
Given:
Find:
Equation describing one-dimensional unsteady flow in a thin liquid layer
Solution:
To nondimensionalize the equation all lengths are divided by the reference length and all velocities are divided by
the reference velocity. Denoting the nondimensional quantities by an asterisk:
Nondimensionalization for the equation using length scale L and velocity scale Vo. Obtain the dimensionless
groups that characterize the flow.
x* 
x
L

h* 

h
L
u* 
u
V0
 
 
t* 
t
L V0
 
 
 u *V0
 u *V0
 h* L
*



u
V
g
Substituting into the governing equation:
0
 t * L V0
 x* L
 x* L


V02 u * V02 * u *
h *

u


g
L t *
L
x *
x *
The dimensionless group is
g L
V0
2
Simplifying this expression:
Thus:
which is the reciprocal of the square of the Froude number.
*
u *
gL h *
* u



u
t *
x *
V02 x *
Problem 7.5
Given:
Find:
Solution:
[Difficulty: 2]
Equation describing two-dimensional steady flow in a liquid
Nondimensionalization for the equation using length scale L and velocity scale V 0. Obtain the dimensionless
groups that characterize the flow.
To nondimensionalize the equation all lengths are divided by the reference length and all velocities are divided by
the reference velocity. Denoting the nondimensional quantities by an asterisk:
x* 
x
L
h* 
h
L
u* 
u
V0
y* 
y
L
Substituting into the governing equation:
 
 
 
 




 u *V0
 2 u *V0
 h * L    2 u *V0

u V0


 g
 x* L
 x * L    x * L  x * L  y * L  y * L
*
     



Simplifying this expression:
V02 * u *
h * V0
u
 g *  2
L
L
x *
x
  2u *  2u * 



 x * 2 y * 2 


Thus:
The dimensionless groups are
g L
V0
reciprocal of the Reynolds number.
2
u*
gL h *
u *
   2 u *  2 u * 




V02 x * V0 L  x * 2 y * 2 
x *
which is the reciprocal of the square of the Froude number, and
μ
V0  ρ L
which is the
Problem 7.6
Given:
Equations for modeling atmospheric motion
Find:
Non-dimensionalized equation; Dimensionless groups
[Difficulty: 2]
Solution:
Recall that the total acceleration is



DV V 
 V  V

t
Dt
Nondimensionalizing the velocity vector, pressure, angular velocity, spatial measure, and time, (using a typical velocity magnitude V
and angular velocity magnitude ):


V
V* 
V
Hence


V VV *



* 

p
p* 
p
x* 


  *
p  p p *
x
L
t*  t
x  Lx*
V
L
t
L
t*
V
Substituting into the governing equation




V V *
V 
1 p
V
 V V *  * V * 2V  * V *  
p *
 L
L t *
L
The final dimensionless equation is
The dimensionless groups are



V * 
p
 L  
 V *  * V * 2 
p *
  * V  
V
t *
V2



p
V
2
L
V
The second term on the left of the governing equation is the Coriolis force due to a rotating coordinate system. This is a very
significant term in atmospheric studies, leading to such phenomena as geostrophic flow.
Problem 7.7
Given:
Find:
Solution:
[Difficulty: 4]
The Prandtl boundary-layer equations for steady, incompressible, two-dimensional flow neglecting gravity
Nondimensionalization for the equation using length scale L and velocity scale V 0. Obtain the dimensionless
groups that characterize the flow.
To nondimensionalize the equation all lengths are divided by the reference length and all velocities are divided by
the reference velocity. Denoting the nondimensional quantities by an asterisk:
x* 
Substituting into the continuity equation:
x
L
y* 
y
L
u* 
u
V0
v* 
v
V0
   
   
*
*
 u *V0  v *V0
Simplifying this expression: V0 u  V0 v  0


0
L x * L y *
 x* L
 y*L
u * v *

0
x * y *
We expand out the second derivative in the momentum equation by writing it as the derivative of the derivative. Upon substitution:
 
 
 
 
 
   
 u *V0
 u *V0
1 p
  u *V0
*
 v V0


u V0
  x* L
 x* L
 y*L
 y*L  y*L
*
u*
 
Simplifying this expression yields:
*
u *
1 p
  2 u * Now every term in this equation has been non-dimensionalized except the
* u
v




V02 x * V0 L y * 2 pressure gradient. We define a dimensionless pressure as:
x *
y *
p* 
p
V02
Substituting this into the momentum equation:
u*
Simplifying this expression yields:
The dimensionless group is
ν
V0  L


*
1  p * V02
  2u *
u *
* u
u
v




V0 L y * 2
V02
x *
y *
x *
*
which is the reciprocal of the Reynolds number.
*
  2u *
u *
p *
* u



v

x *
y *
x * V0 L y * 2
Problem 7.8
Given:
Equation for unsteady, 2D compressible, inviscid flow
Find:
Dimensionless groups
[Difficulty: 2]
Solution:
Denoting nondimensional quantities by an asterisk
x* 
x
L
y* 
y
L
u* 
u
c0
v* 
v
c0
c* 
c
c0
t* 
t c0
L
* 

L c0
Note that the stream function indicates volume flow rate/unit depth!
Hence
x  Lx*
y  Ly*
u  c0 u *
v  c0 v *
c  c0 c *
t
Lt *
c0
  L c0  *
Substituting into the governing equation
2
2
 c03   2 *  c03   u *2  v *2   c03  2
 c03  2
 c03 
 2 *
2   *
2   *
 












u
c
v
c
u
v






0
*
*
*
*
2
*
*
2
L
L
t
x *2  L 
y *2  L 
x * y *
 L  t *
 
 
The final dimensionless equation is
2
2
 2 *
 2 *  u *2  v *2 
2
2  *
2
2  *




0





*
*
2
*
*

u
c
v
c
u
v
*
*
x * y *
x *2
y *2
t *2
t
No dimensionless group is needed for this equation!
Problem 7.9
Given:
Equations Describing pipe flow
Find:
Non-dimensionalized equation; Dimensionless groups
[Difficulty: 2]
Solution:
Nondimensionalizing the velocity, pressure, spatial measures, and time:
u* 
u
V
p* 
p
p
x* 
x
L
r* 
r
L
t*  t
V
L
Hence
u V u*
p  p p *
x  Lx*
r  Dr*
t
L
t*
V
Substituting into the governing equation
V u *
u
1
1 p *
1   2 u * 1 u * 
  p
 V 2 

V

L t *
L x *
t
D  r *2 r * r * 
The final dimensionless equation is
u *
p p *    L   2 u * 1 u * 



 
t *
V 2 x *  DV  D  r *2 r * r * 
The dimensionless groups are
p
V
2

DV
L
D
Problem 7.10
Given:
Find:
Solution:
Functional relationship between pressure drop through orifice plate and physical parameters
Appropriate dimensionless parameters
We will use the Buckingham pi-theorem.


1
p
2
Select primary dimensions M, L, t:
3
p


V
D
d
M
Lt 2
M
L3
M
Lt
L
t
L
L

V
D
4
5
[Difficulty: 2]
V
D
d
n = 6 parameters
r = 3 dimensions
m = r = 3 repeating parameters
We have n - m = 3 dimensionless groups. Setting up dimensional equations:
a
b
c
Π1  Δp ρ  V  D
a
Thus:
b
 M    M    L   L c  M 0 L 0 t 0
 2   3   t 
 L t   L 
Summing exponents:
M: 1  a  0
The solution to this system is:
L:
1  3  a  b  c  0
t:
2  b  0
a  1 b  2 c  0
Check using F, L, t primary dimensions:
F
L
a
b
c
Π2  μ ρ  V  D
2

L
4
F t
a
Thus:
2

t
Π1 
Δp
2
ρ V
2
 1 Checks
out.
L
2
b
 M    M    L   L c  M 0 L 0 t 0
 L t   3   
  L   t 
Summing exponents:
M: 1  a  0
L:
1  3  a  b  c  0
t:
1  b  0
The solution to this system is:
a  1 b  1 c  1
μ
Π2 
ρ V D
(This is the Reynolds number, so it checks out)
a
b
c
Π3  d  ρ  V  D
Thus:
L 
M
a
 
L
b
c
0 0 0
 3   t   L  M  L  t
L 
Summing exponents:
M: a  0
L:
1c0
t:
b0
The solution to this system is:
a0
b0
c  1
d
Π3 
D
(This checks out)
Problem 7.11
Given:
That drag depends on speed, air density and frontal area
Find:
How drag force depend on speed
Solution:
Apply the Buckingham  procedure
 F

V
A
n = 4 parameters
 Select primary dimensions M, L, t
F
V

ML
L
t
M
A

r = 3 primary dimensions
t2
 V
L3

L2
A
m = r = 3 repeat parameters
 Then n – m = 1 dimensionless groups will result. Setting up a dimensional equation,
1  V a  b Ac F
b
a
 
L M 
    3  L2
 t  L 
c
ML
t2
 M 0 L0 t 0
Summing exponents,
M:
b 1  0
b  1
L:
a  3b  2c  1  0
c  1
t:
a20
a  2
Hence
1 
F
V 2 A
 Check using F, L, t as primary dimensions
1 
 1
F
Ft
2
L2
L4 t 2
L2
The relation between drag force F and speed V must then be
F  V 2 A  V 2
The drag is proportional to the square of the speed.
[Difficulty: 2]
Problem 7.12
[Difficulty: 2]
At low speeds, drag F on a sphere is only dependent upon speed V, viscosity μ, and diameter D
Given:
Find:
Solution:
Appropriate dimensionless parameters
We will use the Buckingham pi-theorem.

1
F
2
Select primary dimensions M, L, t:
3
F
V

ML
t2
L
t
M
Lt
V

D
n = 4 parameters
D
n = 4 parameters
L
r = 3 dimensions
D
4
V
5
We have n - m = 1 dimensionless group. Setting up a dimensional equation:
a b
c
Π1  F V  μ  D
m = r = 3 repeating parameters
a
Thus:
b
 M  L    L    M   L c  M 0 L 0 t 0
 2   t   L t 
 t 
Summing exponents:
M: 1  b  0
L:
1abc0
t:
2  a  b  0
The solution to this system is:
a  1 b  1 c  1
F
Π1 
μ V D
2
Check using F, L, t primary dimensions:
t L 1
F 
  1 Checks out.
L F t L
Since the procedure produces only one dimensionless group, it must be a constant. Therefore:
F
μ V D
 constant
Problem 7.13
[Difficulty: 2]
Given:
Find:
Functional relationship between the drag on a satellite and other physical parameters
Solution:
We will use the Buckingham pi-theorem.
1
Expression for FD in terms of the other variables
2
FD
λ
ρ
L
c
Select primary dimensions M, L, t:
3
FD
λ
M L
t
4
5
L
2
ρ
ρ
M
L
L
L
c
L
L
3
n = 5 parameters
r = 3 dimensions
t
m = r = 3 repeating parameters
c
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
a b d
Π1  D ρ  L  c
Thus:
M L
t
Summing exponents:
2
a
 L  
 3
L 
b
a  1
L:
1  3 a  b  d  0
t:
2  d  0
Check using F, L, t dimensions:
F
L
4
F t
a b d
M
L
d
0 0 0
  M L t
 
t
The solution to this system is:
M: 1  a  0
Π2  λ ρ  L  c

Thus:
L 
t:
d0
a
a0
Check using F, L, t dimensions:
The functional relationship is:
M
2

t
L
1
L
FD
2 2
ρ L  c
2
L
2
 L  
b
d  2
1
L
d
0 0 0
  M L t
t
λ
Π2 
L
The solution to this system is:
M: a  0
1  3 a  b  d  0
L
 3
L 
Summing exponents:
L:
2
1

b  2
Π1 
b  1
d0
(Π2 is sometimes referred to
as the Knudsen number.)
1
 
Π1  f Π2
FD
2 2
ρ L  c
 f 
λ

 L
λ
2 2
FD  ρ L  c  f  
L
 
Problem 7.14
Given:
Find:
Solution:
Functional relationship between buoyant force of a fluid and physical parameters
Buoyant force is proportional to the specific weight as demonstrated in Chapter 3.
We will use the Buckingham pi-theorem.
1
FB
2
Select primary dimensions F, L, t:
3
FB
V
F
L
5
V
n = 3 parameters
γ
V
γ
3
F
L
4
[Difficulty: 2]
r = 2 dimensions
3
m = r = 2 repeating parameters
γ
We have n - m = 1 dimensionless group. Setting up dimensional equations:
a b
Π1  FB V  γ
 3  F3 
a
Thus:
b
F L
0
 F L
0
L 
Summing exponents:
F:
1b0
The solution to this system is:
L:
3 a  3 b  0
a  1 b  1
2
FB
Π1 
V γ
2
Check using M, L, t dimensions:
M L 1 t  L
 
1
2
3 M
t
L
The functional relationship is:
Π1  C
FB
V γ
C
Solving for the buoyant force:
FB  C V γ
Buoyant force is
proportional to γ
(Q.E.D.)
Problem 7.15
[Difficulty: 2]
Given:
Find:
Functional relationship between drag on an object in a supersonic flow and physical parameters
Solution:
We will use the Buckingham pi-theorem.
Functional relationship for this problem using dimensionless parameters
1
FD
2
Select primary dimensions M, L, t:
3
FD
V
ρ
M L
L
M
2
t
V
ρ
t
4
5
ρ
V
A
A
L
3
L
n = 5 parameters
c
c
L
2
r = 3 dimensions
t
m = r = 3 repeating parameters
A
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
a b
c
Π1  FD V  ρ  A
Thus:
M L
t
Summing exponents:
 
2
M: 1  b  0
1  a  3 b  2 c  0
t:
2  a  0
t
2
L
c
 
c
Thus:
Π1 
a  2 b  1 c  1
Check using F, L, t dimensions: F
a b
b
The solution to this system is:
L:
Π2  c V  ρ  A
a
M
0 0 0
2
 M L t
   L



 t   L3 
L
L
t
2
 
L

4
F t
L
2
a


1
L
2
M

   L3 
t
FD
2
V  ρ A
1
b
 2
 L
c
0
0 0
 M L t
Summing exponents:
M: b  0
The solution to this system is:
L:
1  a  3 b  2 c  0
t:
1  a  0
Check using F, L, t dimensions:
The functional relationship is:
a  1 b  0
c0
c
Π2 
V
(The reciprocal of Π 2 is also referred
to as the Mach number.)
L t
 1
t L
 
Π1  g Π2
FD
2
V  ρ A
 f 


 V
c
Problem 7.16
[Difficulty: 2]
Given:
That speed of shallow waves depends on depth, density, gravity and surface tension
Find:
Dimensionless groups; Simplest form of V
Solution:
Apply the Buckingham  procedure
 V

D

g
n = 5 parameters
 Select primary dimensions M, L, t

V


L
 t


g

D

g
L
M
L3
L
t2

 


M
t 2 
D
r = 3 primary dimensions
m = r = 3 repeat parameters
 Then n – m = 2 dimensionless groups will result. Setting up a dimensional equation,
a
b
L M 
c L
1  g a  b D cV   2   3  L   M 0 L0t 0
t
t   L 
M:
b0
b0
1
L : a  3b  c  1  0 c  
Hence
Summing exponents,
2
1
t:
a
 2a  1  0
2
a
b
 L M 
c M
 2  g a  b D c   2   3  L  2  M 0 L0t 0
t
t   L 
M:
b 1  0
b  1
L : a  3b  c  0 c  2
Hence
Summing exponents,
t:
 2 a  2  0 a  1
L
t
 Check using F, L, t as primary dimensions
 
 1
1
The relation between drag force speed V is
L
 2
t
1
2
L

1  f  2 
  
V
 f 

2 
gD
 gD 
1 
V
gD
2 

g D 2
F
L
2 
 1
L Ft 2 2
L
t 2 L4
  
V  gD f 

2 
 gD 
Problem 7.17
Given:
Find:
Solution:
Functional relationship between wall shear stress in a boundary layer and physical parameters
Appropriate dimensionless parameters
We will use the Buckingham pi-theorem.


1
w
2
Select primary dimensions M, L, t:
3
w
x


U
M
Lt 2
L
M
L3
M
Lt
L
t

x
U
4
5
[Difficulty: 2]
x
U
n = 5 parameters
r = 3 dimensions
m = r = 3 repeating parameters
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
a b
c
Π1  τw ρ  x  U
a
Thus:
c
 M    M   L b   L   M 0 L 0 t 0
 
 2  3
t
 L t   L 
Summing exponents:
M: 1  a  0
The solution to this system is:
L:
1  3  a  b  c  0
t:
2  c  0
Check using F, L, t dimensions:
a b
c
Π2  μ ρ  x  U
a  1 b  0
τw
2
ρ U
4
2
 F    L    t   1
 2  2  L 
 L   F t 
a
Thus:
c  2
Π1 
c
 M    M   L b   L   M 0 L 0 t 0
 L t   3 
 
  L 
t
Summing exponents:
M: 1  a  0
L:
1  3  a  b  c  0
t:
1  c  0
Check using F, L, t dimensions:
The solution to this system is:
a  1 b  1 c  1
4
2
 F t    L    1    t   1


 2   2  L  L 
 L   F t 
μ
Π2 
ρ x  U
The functional relationship is:
 
Π1  f Π2
Problem 7.18
Given:
Find:
Solution:
Functional relationship between boundary layer thickness and physical parameters
Appropriate dimensionless parameters
We will use the Buckingham pi-theorem.


1

2
Select primary dimensions M, L, t:
3

x


U
L
L
M
L3
M
Lt
L
t
x
U
4
5
x

[Difficulty: 2]
U
n = 5 parameters
r = 3 dimensions
m = r = 3 repeating parameters
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
a b
c
Π1  δ ρ  x  U
Thus:
L 
M
a
 3
L 
 L  
b
L
c
0 0 0
  M L t
t
Summing exponents:
M: 0  a  0
The solution to this system is:
L:
1  3 a  b  c  0
t:
0c0
a0
Check using F, L, t dimensions: ( L)  
b  1 c  0
δ
Π1 
x
1
1
 L
a b
c
Π2  μ ρ  x  U
a
Thus:
c
 M    M   L b   L   M 0 L 0 t 0
 L t   3 
 
  L 
t
Summing exponents:
M: 1  a  0
L:
1  3  a  b  c  0
t:
1  c  0
Check using F, L, t dimensions:
The solution to this system is:
a  1 b  1 c  1
4
2
 F t    L    1    t   1
 2   2   L   L 
 L   F t 
μ
Π2 
ρ x  U
The functional relationship is:
 
Π1  f Π2
Problem 7.19
Given:
That light objects can be supported by surface tension
Find:
Dimensionless groups
[Difficulty: 2]
Solution:
Apply the Buckingham  procedure
 W
p


g
n = 5 parameters
 Select primary dimensions M, L, t

W


 ML
 t 2


g

p

g
L
M
L3
L
t2
p

 


M
t 2 
r = 3 primary dimensions
m = r = 3 repeat parameters
 Then n – m = 2 dimensionless groups will result. Setting up a dimensional equation,
L
1  g  p W   2 
t 
M:
b 1  0
L : a  3b  c  1  0
t:
 2a  2  0
a
Summing exponents,
b
a
c
a
b
1 
W
gp 3
2 

gp 2
b
L M 
c M
 2  g  p    2   3  L  2  M 0 L0t 0
t
t   L 
M:
b 1  0
b  1
L : a  3b  c  0 c  2
Hence
t:
 2 a  2  0 a  1
a
Summing exponents,
b
M 
c ML
0 0 0
 3  L  2  M L t
t
L 
b  1
c  3
Hence
a  1
c
F
F
L
 Check using F, L, t as primary dimensions
1 
 1
2 
 1
L Ft 2 2
L Ft 2 3
L
L
t 2 L4
t 2 L4
1 Wp
Note: Any combination of 1 and 2 is a  group, e.g.,

, so 1 and 2 are not unique!
2 
Problem 7.20
Given:
Find:
Solution:
Functional relationship between the speed of a free-surface gravity wave in deep water and physical parameters
The dependence of the speed on the other variables
We will use the Buckingham pi-theorem.
1
V
2
Select primary dimensions M, L, t:
3
V
λ
L
t
4
5
D
D
λ
D
L
L
ρ
n = 5 parameters
g
ρ
g
M
L
3
2
L
ρ
[Difficulty: 2]
t
r = 3 dimensions
m = r = 3 repeating parameters
g
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
a b c
Π1  V D  ρ  g
L
Thus:
t
L 
a
M
b

L
c
0 0 0
 3   2   M L t
L  t 
Summing exponents:
M: b  0
L:
1  a  3 b  c  0
t:
1  2  c  0
The solution to this system is:
1
1
a b0 c
2
2
Check using F, L, t dimensions:
a b c
Π2  λ D  ρ  g
Thus:
Π1 
V
g D
 L   t   1
 t   L
  
L L  
a
M
b

L
c
0 0 0
 3   2   M L t
L  t 
Summing exponents:
M: b  0
L:
1  a  3 b  c  0
t:
2  c  0
The solution to this system is:
a  1 b  0
Check using F, L, t dimensions: L
The functional relationship is:
1
L
c0
λ
Π2 
D
1
 
Π1  f Π2
V
g D
 f 
λ

 D
Therefore the velocity is:
V
g  D f 
λ

 D
Problem 7.21
[Difficulty: 2]
Given:
Functional relationship between mean velocity for turbulent flow in a pipe or boundary layer and physical
parameters
Find:
(a) Appropriate dimensionless parameters containing mean velocity and one containing the distance from the
wall that are suitable for organizing experimental data.
(b) Show that the result may be written as:
u

 yu 
 f  *  where u*  w is the friction velocity
u*

  
Solution:
We will use the Buckingham pi-theorem.
w


1
u
2
Select primary dimensions M, L, t:
3
u
w
L
t
M
Lt 2
4
5

y


M
L3
M
Lt
y
L
w
y
n = 5 parameters
r = 3 dimensions
m = r = 3 repeating parameters
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
1  u y 
a
b c
w
Thus:
L
t

M
 3
L 
a
L 
b
c
0 0 0

 2   M L t
 L t 
M
Summing exponents:
M: a  c  0
L:
1  3 a  b  c  0
t:
1  2  c  0
Check using F, L, t dimensions:
 2   a y b wc
Thus:
The solution to this system is:
1
1
a
b0 c
2
2
 L   t   1
t 
   L
M
L t

M
 3
L 
a
L 
b
c
0 0 0

 2   M L t
 L t 
M
1  u
u


 w u*
Summing exponents:
M: 1  a  c  0
L:
1  3  a  b  c  0
t:
1  2  c  0
The solution to this system is:
1
1
a
b  1 c  
2
2
2 

 





y  w y  w yu* yu*
Π2 is the reciprocal of the Reynolds number, so we know that it checks out.
The functional relationship
is:
 
Π1  g Π2
  
u

 g 
u*
 yu* 
which may be rewritten as:
u
 yu 
 f *
u*
  
Problem 7.22
[Difficulty: 2]
(The solution to this problem was first devised by G.I. Taylor
in the paper "The formation of a blast wave by a very intense
explosion. I. Theoretical discussion," Proceedings of the Royal
Society of London. Series A, Mathematical and Physical
Sciences, Vol. 201, No. 1065, pages 159 - 174 (22 March 1950).)
Given:
Find:
Functional relationship between the energy released by an explosion and other physical parameters
Solution:
We will use the Buckingham pi-theorem.
Expression for E in terms of the other variables
1
E
2
Select primary dimensions M, L, t:
3
E
t
M L
t
4
5
t
R
t
L
ρ
p
M
L t
t
n = 5 parameters
ρ
p
2
2
ρ
R
2
M
L
r = 3 dimensions
3
m = r = 3 repeating parameters
R
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
a b
Π1  E ρ  t  R
c
Thus:
M L
t
Summing exponents:
2  3 a  c  0
t:
2  b  0
a b
c
Thus:
F L
M
t:
2  b  0
b c
0 0 0
 3  t L  M L t
L 
4
2 1
t 
F t
2

M
L
5
c  5
E t
2
ρ R
5
1
a
a  1
Check using F, L, t dimensions:
F
L
The functional relationship is:
L
b2
Π1 
The solution to this system is:
M: 1  a  0
1  3  a  c  0
a
b c
0 0 0
 3  t L  M L t
L t  L 
2
Summing exponents:
L:
M
a  1
Check using F, L, t dimensions:
Π2  p  ρ  t  R
2

The solution to this system is:
M: 1  a  0
L:
2
2

L
4
F t
 
Π1  f Π2
2
b2
2 1
t 
L
2
Π2 
c  2
p t
2
ρ R
2
1
E t
2
ρ R
5
 p t2 

 2
 ρ R 
 f
E
ρ R
t
2
5
 p t2 

 2
 ρ R 
f 
Problem 7.23
Given:
Find:
Solution:
Functional relationship between the speed of a capillary wave and other physical parameters
An expression for V based on the other variables
We will use the Buckingham pi-theorem.
1
V
2
Select primary dimensions M, L, t:
3
V
σ
L
M
t
2
4
5
σ
σ
t
λ
[Difficulty: 2]
λ
λ
L
n = 4 parameters
ρ
ρ
M
L
r = 3 dimensions
3
m = r = 3 repeating parameters
ρ
We have n - m = 1 dimensionless group. Setting up dimensional equations:
a b c
Π1  V σ  λ  ρ
Thus:
t
Summing exponents:
1  b  3 c  0
t:
1  2  a  0
Check using F, L, t dimensions:
The functional relationship is:

M
 2
t 
a
L 
b
M
c
0 0
 3   L t
L 
The solution to this system is:
1
1
1
a
b
c
2
2
2
M: a  c  0
L:
L
Π1  C
Π1  V
λ ρ
σ
2
 L   L F t  L  1
t
4 F
 
L
V
λ ρ
σ
C
Therefore the velocity is:
V  C
σ
λ ρ
Problem 7.24
Given:
Find:
Solution:
Functional relationship between the flow rate over a weir and physical parameters
An expression for Q based on the other variables
We will use the Buckingham pi-theorem.
1
Q
2
Select primary dimensions L, t:
3
Q
L
h
5
g
h
g
3
h
L
L
t
4
[Difficulty: 2]
t
2
n = 5 parameters
b
b
r = 2 dimensions
L
m = r = 2 repeating parameters
g
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
a b
Π1  Q h  g
Thus:
L
3
t
L 
a
L
0 0
 2   L t
t 
Summing exponents:
L:
3ab0
t:
1  2  b  0
Check:
The solution to this system is:
5
1
a
b
2
2
Π1 
Q
2
h  g h
 L3   1 2  t
         1
 t   L  L
a b
Π2  b  h  g
Thus:
L L  
a
L
L:
1ab0
t:
2  b  0
L
1
L
b
0 0
 2   L t
t 
Summing exponents:
Check:
b
The solution to this system is:
a  1
b0
b
Π2 
h
1
The functional relationship is:
 
Π1  f Π2
Q
2
h  g D
 f 
b

h
Therefore the flow rate is:
Q  h  g  h  f 
2
b

h
Problem 7.25
Given:
That automobile buffer depends on several parameters
Find:
Dimensionless groups
[Difficulty: 2]
Solution:
Apply the Buckingham  procedure
 T

F
e


n = 6 parameters
 Select primary dimensions M, L, t

 T


 ML2
 2
 t


F
e

1
t
F
ML
t2
e
L

M
Lt


 


M

t2 
r = 3 primary dimensions
m = r = 3 repeat parameters
 Then n – m = 3 dimensionless groups will result. Setting up a dimensional equation,
a
c
2
 ML 
b  1  ML
1  F e  T   2  L    2  M 0 L0t 0
t t
 t 
a b
c
Summing exponents,
M:
L:
a 1  0
ab20
a  1
b  1
t:
 2a  c  2  0
c0
a
Summing exponents,
a
T
Fe
2 
e2
F
c
 ML 
b1 M
 3  F e     2  L    2  M 0 L0t 0
 t 
t t
M:
a 1  0
a  1
L:
ab0
b 1
Hence
t :  2a  c  2  0 c  0
 Check using F, L, t as primary dimensions
1 
c
 ML 
b1 M
 2  F a eb c    2  L   
 M 0 L0t 0
 t 
 t  Lt
M:
a 1  0
a  1
L:
a  b 1  0
b2
Hence
t :  2a  c  1  0 c  1
a b
Summing exponents,
Hence
c
3 
e
F
Ft 2 1
F
L
L
2
FL
L
L
t
1 
 1
3 
 1
2 
 1
F
FL
F
T
1
Note: Any combination of 1, 2 and 3 is a  group, e.g.,

, so 1, 2 and 3 are not unique!
 2 e3
Problem 7.26
Given:
That the power of a vacuum depends on various parameters
Find:
Dimensionless groups
[Difficulty: 2]
Solution:
Apply the Buckingham  procedure
 P
p
D
d


di
do
n = 8 parameters
 Select primary dimensions M, L, t

 P


 ML2
 3
 t



D
p
M
Lt 2
D d 
L
L
1
t



do 



L

di
M
L3
L
r = 3 primary dimensions
m = r = 3 repeat parameters
 Then n – m = 5 dimensionless groups will result. Setting up a dimensional equation,
M
1   D  P   3
L
M:
a 1  0
L :  3a  b  2  0
c30
t:
a
Summing exponents,
b
c
a
c
2

b  1  ML
0 0 0


 L   3  M Lt

t t
a  1
b  5
Hence
c  3
a
1 
P
D 5 3
c
M 
b 1 M
 2   D  Δp   3  L   
 M 0 L0t 0
2
L 
 t  Lt
M:
a 1  0
a  1
p
Summing exponents,
2 
L :  3a  b  1  0 b  2
Hence
D 2 2
c20
t:
c  2
d
d
d
The other  groups can be found by inspection:
4  i
5  o
3 
D
D
D
a
b
c
 Check using F, L, t as primary dimensions
1 
FL
t
2
 1
2 
F
L2
2
 1
3   4  5 
Ft 2 1
L 2
L4
t
1
P
Note: Any combination of 1, 2 and 3 is a  group, e.g.,
, so the ’s are not unique!

 2 pD 3
Ft 5 1
L 3
L4
t
L
 1
L
Problem 7.27
Given:
Find:
Solution:
Functional relationship between the load bearing capacity of a journal bearing and other physical parameters
Dimensionless parameters that characterize the problem.
We will use the Buckingham pi-theorem.
1
W
2
Select primary dimensions F, L, t:
3
W
D
l
c
F
L
L
L
4
5
D
[Difficulty: 2]
D
ω
l
c
n = 6 parameters
ω
μ
ω
μ
1
F t
t
2
L
r = 3 dimensions
m = r = 3 repeating parameters
μ
We have n - m = 3 dimensionless groups. Setting up dimensional equations:
a
b
c
Π1  W D  ω  μ
Thus:
a
Summing exponents:
F:
1c0
L:
a  2 c  0
t:
b  c  0
1
c
The solution to this system is:
a  2
Check using M, L, t dimensions:
The functional relationship is:
b
F t 
0 0 0

 F L t



 t   L2 
F L  
b  1
L t
M L 1
  t
1
M
2
2
t
L

Π1  f Π2 Π3

c  1
Π1 
W
2
D  ω μ
By inspection, we can see that:
l
Π2 
D
W
2
D  ω μ
c
Π3 
D
 f 


 D D
l

c
Problem 7.28 (In Excel)
[Difficulty: 2]
Given: That drain time depends on fluid viscosity and density, orifice diameter, and gravity
Find: Functional dependence of t on other variables
Solution:
We will use the workbook of Example 7.1, modified for the current problem
n
r
m =r
n -m
The number of parameters is:
The number of primary dimensions is:
The number of repeat parameters is:
The number of  groups is:
=5
=3
=3
=2
Enter the dimensions (M, L, t) of
the repeating parameters, and of up to
four other parameters (for up to four  groups).
The spreadsheet will compute the exponents a , b , and c for each.
REPEATING PARAMETERS: Choose , g , d
M
1

g
d
L
-3
1
1
t
-2
 GROUPS:
t
 1:
M
0
L
0
t
1
a =
b =
c =
0
0.5
-0.5

 2:
M
1
L
-1
a =
b =
c =
-1
-0.5
-1.5
M
0
L
0
a =
b =
c =
0
0
0
t
-1
The following  groups from Example 7.1 are not used:
 3:
Hence
1  t
The final result is
g
d
t
and
M
0
L
0
a =
b =
c =
0
0
0

2 
g
d
g
 2 
f 2 3
  gd 


1 3
2d 2

t
0
 4:
2
 gd 3
2
with  1  f  2 
t
0
Problem 7.29
[Difficulty: 2]
Given:
Find:
Functional relationship between the power transmited by a sound wave and other physical parameters
Solution:
We will use the Buckingham pi-theorem.
Expression for E in terms of the other variables
1
E
2
Select primary dimensions M, L, t:
3
E
V
ρ
M
L
M
3
t
3
t
4
5
ρ
V
ρ
V
L
r
n = 5 parameters
n
r
n
1
L
r = 3 dimensions
t
m = r = 3 repeating parameters
r
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
a
b c
Π1  E ρ  V  r
Thus:

M
t:
3  b  0
a  1
Check using F, L, t dimensions:
a
b c
Π2  n  ρ  V  r
Thus:
F
L t
1
t
Summing
exponents:
M: a  0
L:
3  a  b  c  0
t:
1  b  0
Check using F, L, t
dimensions:
The functional relationship is:
L
b
The solution to this system is:
M: 1  a  0
3  a  b  c  0
 
c
0 0 0
 3   t   L  M  L  t
t L 
3
Summing exponents:
L:
a
M
L

4
F t

2

b  3
t
c0
3
a
 
L
b
c
0 0 0
 3   t   L  M  L  t
L 
a0
t
3
ρ V
1
n r
Π2 
V
The solution to this system is:
1
E
3
L
M
Π1 
 L
 
Π1  f Π2
t
L
b  1
c1
1
E
3
ρ V
 f 
n r 

 V
E  ρ V  f 
3
n r 

 V
Problem 7.30
[Difficulty: 2]
Given:
Functional relationship between the time needed to drain a tank through an orifice plate and other physical
parameters
Find:
(a) the number of dimensionless parameters
(b) the number of repeating variables
(c) the Π term which contains the viscosity
Solution:
We will use the Buckingham pi-theorem.
1
τ
2
Select primary dimensions M, L, t:
3
τ
h0
D
d
T
L
L
L
h0
D
d
μ
g
ρ
μ
L
M
M
2
3
L t
t
4
5
ρ
d
n = 7 parameters
ρ
g
L
r = 3 dimensions
m = r = 3 repeating parameters
g
We have n - m = 4 dimensionless groups.
Setting up dimensional equation including the viscosity:
a b c
Π1  μ ρ  d  g
Thus:
Summing exponents:
M: 1  a  0
L:
1  3  a  b  c  0
t:
1  2  c  0
M

M
L t  3 
L 
a
L 
b
L
0 0 0
 2   M L t
t 
The solution to this system is:
3
1
a  1
b
c
2
2
4
Check using F, L, t dimensions:
c
1
t
F t L



1
1
2
2 3
L F t
L
2
L
2
Π1 
μ
3
1
2
2
ρ d  g
Problem 7.31
Given:
[Difficulty: 3]
Find:
Functional relationship between the flow rate of viscous liquid dragged out of a bath and other physical
parameters
Expression for Q in terms of the other variables
Solution:
We will use the Buckingham pi-theorem.
1
Q
2
Select primary dimensions M, L, t:
3
Q
L
μ
μ
3
t
4
5
ρ
ρ
V
g
ρ
g
M
M
L
L t
3
2
L
t
h
n = 6 parameters
V
h
V
L
L
r = 3 dimensions
t
m = r = 3 repeating parameters
h
We have n - m = 3 dimensionless groups. Setting up dimensional equations:
a
b c
Π1  Q ρ  V  h
Thus:
L
3
M
a
 
a0
L:
3  3 a  b  c  0
t:
1  b  0
Check using F, L, t dimensions:
b c
b
The solution to this system is:
M: a  0
Π2  μ ρ  V  h
L
c
0 0 0
L  M L t



t
3
t
L   
Summing exponents:
a

Thus:
Summing exponents:
M: 1  a  0
L:
1  3  a  b  c  0
t:
1  b  0
c  2
Q
V h
2
3
t 1
 
1
t L 2
L
L
M

M
a
 
L
b
c
0 0 0
L  M L t



L t
3
t
L   
The solution to this system is:
a  1
4
Check using F, L, t dimensions:
b  1
Π1 
b  1
t 1
F t L

  1
2
2 L L
L F t
c  1
μ
Π2 
ρ V h
a
b c
Π3  g  ρ  V  h
Thus:
L
t:
2  b  0
Check using F, L, t dimensions:
 
L
a0
L
t
The functional relationship is:
a
b
The solution to this system is:
M: a  0
1  3 a  b  c  0
M
c
0 0 0
 3   t   L  M  L  t
t L 
2
Summing exponents:
L:

2
 L
t
c1
g h
2
V
2
L

b  2
Π3 
2
1
Π1  f Π2 Π3

Q
V h
2
 ρ V h V2 


 μ g h 
 f
 ρ V h V2 


 μ g h 
Q  V h  f 
2
Problem 7.32
[Difficulty: 2]
Given:
Find:
Functional relationship between the power required to drive a fan and other physical parameters
Solution:
We will use the Buckingham pi-theorem.
Expression for P in terms of the other variables
1
P
2
Select primary dimensions M, L, t:
3
P
ρ
M L
t
4
5
Q
M
L
3
t
2
L
D
n = 5 parameters
ω
D
ρ
3
ρ
Q
ω
D
3
1
L
r = 3 dimensions
t
m = r = 3 repeating parameters
ω
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
a
b
c
Π1  P ρ  D  ω
Thus:
M L
t
Summing exponents:
2  3 a  b  0
t:
3  c  0

M
 3
L 
3
a
 L  
b
1
t
c
0 0 0
  M L t

The solution to this system is:
a  1
M: 1  a  0
L:
2
b  5
Π1 
c  3
P
5
3
ρ D  ω
4
Check using F, L, t dimensions:
a
b
c
Π2  Q ρ  D  ω
Thus:
1 3
F L L

 t  1
t
2 5
F t L
L
3
3  3 a  b  0
t:
1  c  0
Check using F, L, t
dimensions:
The functional relationship is:
M
 3
L 
t
Summing
exponents:
M: a  0
L:

a
 L  
b
1
t
c
0 0 0
  M L t

The solution to this system is:
a0
1
t
 L
 
Π1  f Π2
t
L
b  3
Π2 
c  1
Q
3
D ω
1
P
5
3
ρ D  ω
 f
Q 
 3 
 D ω 
P  ρ D  ω  f 
5
3
Q 
 3 
 D ω 
Problem 7.33
[Difficulty: 2]
Given:
Functional relationship between the mass flow rate exiting a tank through a rounded drain hole and other
physical parameters
Find:
(a) Number of dimensionless parameters that will result
(b) Number of repeating parameters
(c) The Π term that contains the viscosity
Solution:
We will use the Buckingham pi-theorem.
1
m
2
Select primary dimensions M, L, t:
3
m
h0
M
t
4
5
ρ
D
d
h0
D
d
L
L
L
μ
g
ρ
μ
L
M
M
3
L t
t
d
n = 7 parameters
ρ
g
2
L
r = 3 dimensions
We have n - r = 4 dimensionless groups.
m = r = 3 repeating parameters
g
Setting up dimensional equation involving the viscosity:
a b c
Π1  μ ρ  d  g
Thus:
Summing exponents:
M: 1  a  0
L:
1  3  a  b  c  0
t:
1  2  c  0
M

M
L t  3 
L 
a
L 
b
L
0 0 0
 2   M L t
t 
The solution to this system is:
3
1
a  1
b
c
2
2
4
Check using F, L, t dimensions:
c
1
t
F t L



1
1
2
2 3
L F t
L
2
L
2
Π1 
μ
3
ρ d  g
Problem 7.34
[Difficulty: 3]
Given:
Functional relationship between the deflection of the bottom of a cylindrical tank and other physical parameters
Find:
Functional relationship between these parameters using dimensionless groups.
Solution:
We will use the Buckingham pi-theorem.
1
δ
2
Select primary dimensions F, L, t:
3
δ
D
h
d
L
L
L
L
D
h
d
γ
E
γ
E
F
L
4
5
D
n = 6 parameters
F
3
L
r = 2 dimensions
2
m = r = 2 repeating parameters
γ
We have n - m = 4 dimensionless groups.
Setting up dimensional equations:
a b
Π1  δ D  γ
Thus:
L L  
a
Summing exponents:
F:
b0
L:
1  a  3 b  0
b
0 0

 3   F L
L 
F
a  1
b0
Check using M, L, t dimensions:
Now since h and d have the same dimensions as δ, it
would follow that the the next two pi terms would be:
a b
Π4  E D  γ
Thus:
F
L
Summing exponents:
F:
1b0
L:
2  a  3  b  0
δ
Π1 
D
The solution to this system is:
L 
a
2
h
Π2 
D
L
1
L
d
Π3 
D
b
0 0

 3   F L
L 
F
The solution to this system is:
a  1
E
Π4 
D γ
b  1
Check using M, L, t dimensions:

Π1  f Π2 Π3 Π4

2 2
1 L t
 
1
2 L M
M
L t
The functional relationship is:
1
δ
D
 f 
h d E 
 

 D D D γ 
(For further reading, one should consult an appropriate text, such as Advanced Strength of Materials by Cook and Young)
Problem 7.35
Given:
Functional relationship between the diameter of droplets formed during jet breakup and other physical
parameters
(a) The number of dimensionless parameters needed to characterize the process
(b) The ratios (Π-terms)
Find:
Solution:
We will use the Buckingham pi-theorem.
1
d
2
Select primary dimensions M, L, t:
3
d
ρ
5
V
μ
σ
V
σ
V
M
M
M
L
3
L t
2
t
L
ρ
μ
ρ
L
4
[Difficulty: 3]
t
n = 6 parameters
D
D
r = 3 dimensions
L
m = r = 3 repeating parameters
D
We have n - m = 3 dimensionless groups.
Setting up dimensional equations:
a
b
c
Π1  d  ρ  V  D
Thus:
L 
L
a0
L:
1  3 a  b  c  0
t:
b  0
b
b
b0
c
Thus:
M

M
a
 
L
a  1
b  1
c  1
1  3  a  b  c  0
c
Summing exponents:
M: 1  a  0
L:
3  a  b  c  0
t:
2  b  0
Thus:
L
1
μ
Π2 
ρ V D
4
Check using F, L, t dimensions:
1  b  0
b
1
b
The solution to this system is:
M: 1  a  0
a
L
c
0 0 0
L  M L t



L t
3
t
L   
Summing exponents:
Π3  σ ρ  V  D
d
Π1 
D
c  1
Check using F, L, t dimensions:
Π2  μ ρ  V  D
t:
 
The solution to this system is:
M: a  0
L:
a
c
0 0 0
 3   t   L  M  L  t
L 
Summing exponents:
a
M
M

M
a
 
L
t 1
F t L

  1
2
2 L L
L F t
b
c
0 0 0
 3   t   L  M  L  t
t L 
2
The solution to this system is:
a  1
b  2
c  1
Π3 
4
Check using F, L, t dimensions:
2
σ
2
ρ V  D
F L
t 1

  1
L
2 2 L
F t L
Problem 7.36
Given:
Find:
Solution:
Functional relationship between the height of a ball suported by a vertical air jet and other physical parameters
The Π terms that characterize this phenomenon
We will use the Buckingham pi-theorem.
1
h
2
Select primary dimensions M, L, t:
3
h
D
d
L
L
L
4
5
ρ
[Difficulty: 3]
D
V
d
V
ρ
μ
W
μ
W
n = 7 parameters
V
ρ
L
M
M
M L
t
3
L t
2
L
t
r = 3 dimensions
m = r = 3 repeating parameters
d
We have n - m = 4 dimensionless groups. Setting up dimensional equations:
a
b c
Π1  h  ρ  V  d
Thus:
Summing exponents:
M
a
 
L:
1  3 a  b  c  0
t:
b  0
b c
b
c
0 0 0
 3   t   L  M  L  t
L 
a0
Π2  D ρ  V  d
L
The solution to this system is:
M: a  0
a
L 
b0
c  1
Check using F, L, t dimensions:
Thus:
Summing exponents:
M: a  0
L:
1  3 a  b  c  0
t:
b  0
L 
M
a
 
L
h
Π1 
d
L
1
L
1
b
c
0 0 0
 3   t   L  M  L  t
L 
The solution to this system is:
a0
b0
D
Π2 
d
c  1
Check using F, L, t dimensions:
L
1
L
1
a
b c
Π3  μ ρ  V  d
Thus:
M
L t
Summing exponents:
t:
 
L
b
c
0 0 0
 3   t   L  M  L  t
L 
a  1
b  1
b c
μ
Π3 
ρ V d
c  1
4
Check using F, L, t dimensions:
1  b  0
a
Thus:
M L
t
Summing exponents:
M: 1  a  0
t:
a
1  3  a  b  c  0
Π4  W ρ  V  d
L:
M
The solution to this system is:
M: 1  a  0
L:

2

M
a
 
L
t 1
F t L

  1
2
2 L L
L F t
b
c
0 0 0
 3   t   L  M  L  t
L 
The solution to this system is:
a  1
b  2
Π4 
c  2
1  3 a  b  c  0
2  b  0
Check using F, L, t dimensions:
F
L
4
F t
2

t
2
L
2

W
2 2
ρ V  d
1
L
2
1
Problem 7.37 (In Excel)
[Difficulty: 3]
Given: That dot size depends on ink viscosity, density, and surface tension, and geometry
Find:  groups
Solution:
We will use the workbook of Example 7.1, modified for the current problem
n
r
m =r
n -m
The number of parameters is:
The number of primary dimensions is:
The number of repeat parameters is:
The number of  groups is:
=7
=3
=3
=4
Enter the dimensions (M, L, t) of
the repeating parameters, and of up to
four other parameters (for up to four  groups).
The spreadsheet will compute the exponents a , b , and c for each.
REPEATING PARAMETERS: Choose , V , D
M
1

V
D
L
-3
1
1
t
-1
 GROUPS:
d
1:

3:
Hence
1 
d
D
2 
M
0
L
1
a =
b =
c =
0
0
-1
M
1
L
0
a =
b =
c =
-1
-2
-1

VD

VD

Note that groups 1 and 4 can be obtained by inspection
t
0

2:
t
-2
L
4:
3 

V 2 D
4 
L
D
M
1
L
-1
a =
b =
c =
-1
-1
-1
M
0
L
1
a =
b =
c =
0
0
-1
t
-1
t
0
Problem 7.38 (In Excel)
[Difficulty: 3]
Given: Bubble size depends on viscosity, density, surface tension, geometry and pressure
Find:  groups
Solution:
We will use the workbook of Example 7.1, modified for the current problem
n
r
m =r
n -m
The number of parameters is:
The number of primary dimensions is:
The number of repeat parameters is:
The number of  groups is:
=6
=3
=3
=3
Enter the dimensions (M, L, t) of
the repeating parameters, and of up to
four other parameters (for up to four  groups).
The spreadsheet will compute the exponents a , b , and c for each.
REPEATING PARAMETERS: Choose , p , D

p
D
M
1
1
L
-3
-1
1
M
0
L
1
a =
b =
c =
0
0
-1
M
1
L
0
a =
b =
c =
0
-1
-1
t
-2
 GROUPS:
d
1:

3:
Hence
1 
d
D

2 

1
1
2 p 2 D

2
pD 2
Note that the 1 group can be obtained by inspection
t
0

2:
t
-2
4:
3 

Dp
M
1
L
-1
a =
b =
c =
-0.5
-0.5
-1
M
0
L
0
a =
b =
c =
0
0
0
t
-1
t
0
Problem 7.39 (In Excel)
[Difficulty: 3]
Given: Speed depends on mass, area, gravity, slope, and air viscosity and thickness
Find:  groups
Solution:
We will use the workbook of Example 7.1, modified for the current problem
n
r
m =r
n -m
The number of parameters is:
The number of primary dimensions is:
The number of repeat parameters is:
The number of  groups is:
=7
=3
=3
=4
Enter the dimensions (M, L, t) of
the repeating parameters, and of up to
four other parameters (for up to four  groups).
The spreadsheet will compute the exponents a , b , and c for each.
REPEATING PARAMETERS: Choose g , , m
M
L
1
1
t
-2
t
-1
g

m
1
V
M
0
L
1
a =
b =
c =
-0.5
-0.5
0
M
0
L
0
a =
b =
c =
0
0
0
 GROUPS:
1 :

3 :
Hence
V
1 
g
1 1
2 2
V2

g
2 
g
2 :
t
0
3
2
1
2m

 2 3
2
m g
Note that the 1 , 3 and 4 groups can be obtained by inspection
L
-1
a =
b =
c =
-0.5
1.5
-1
M
0
L
2
a =
b =
c =
0
-2
0
A
4 :

M
1

3  
4 
A
2
t
-1
t
0
Problem 7.40
Given:
Find:
Solution:
Functional relationship between the length of a wake behind an airfoil and other physical parameters
The Π terms that characterize this phenomenon
We will use the Buckingham pi-theorem.
1
w
2
Select primary dimensions M, L, t:
3
w
V
5
ρ
L
V
L
L
4
[Difficulty: 2]
t
V
t
L
t
L
L
ρ
μ
ρ
n = 6 parameters
μ
M
M
3
L t
L
r = 3 dimensions
m = r = 3 repeating parameters
L
We have n - m = 3 dimensionless groups. Setting up dimensional equations:
a
b
Π1  w ρ  V  L
c
Thus:
Summing exponents:
L 
t:
b  0
b c
Thus:
Summing exponents:
L 
b0
M
a
 
L
w
Π1 
L
c  1
L
b0
t
Π2 
L
c  1
Check using F, L, t dimensions:
b  0
b
c
Thus:
Summing exponents:
M: 1  a  0
M

M
a
 
L
L
1
L
1
b
c
0 0 0
L  M L t



L t
3
t
L   
μ
Π3 
ρ V L
The solution to this system is:
a  1
b  1
c  1
1  3  a  b  c  0
1  b  0
1
b
1  3 a  b  c  0
a
L
The solution to this system is:
a0
Π3  μ ρ  V  L
1
c
0 0 0
 3   t   L  M  L  t
L 
M: a  0
t:
b
Check using F, L, t dimensions:
Π2  t ρ  V  L
L:
L
a0
1  3 a  b  c  0
t:
 
The solution to this system is:
L:
L:
a
c
0 0 0
 3   t   L  M  L  t
L 
M: a  0
a
M
4
Check using F, L, t dimensions:
F t L
t 1

  1
2
2 L L
L F t
Problem 7.41
Given:
That the power of a washing machine agitator depends on various parameters
Find:
Dimensionless groups
[Difficulty: 2]
Solution:
Apply the Buckingham  procedure
 P
H
D
max
h

f

n = 8 parameters
 Select primary dimensions M, L, t

 P


 ML2
 3
 t



D
H
D
h max
L
L
L
1
t
max
f

1
t
M
L3



 r = 3 primary dimensions
M

Lt 
m = r = 3 repeat parameters
 Then n – m = 5 dimensionless groups will result. Setting up a dimensional equation,
a
c
2
M 
b  1  ML


1   D  P   3  L   3  M 0 L0t 0
L 
t t
M:
a 1  0
a  1
L :  3a  b  2  0 b  5
Hence
t:
c30
c  3
a
Summing exponents,
b
c
max
a
1 
P
3
D max
5
c
M 
b1 M
 M 0 L0t 0
 2   D     3  L   
L
t
Lt
 
 
M:
a 1  0
a  1

Summing exponents,
2 
L :  3a  b  1  0 b  2
Hence
2
D max
 c 1  0
t:
c  1
f
h
H
5 
3 
4 
The other  groups can be found by inspection:
D
D
max
a
b
c
max
 Check using F, L, t as primary dimensions
1 
FL
t
Ft 2 5 1
L 3
L4
t
 1
Note: Any combination of ’s is a  group, e.g.,
Ft
L2
 3   4   5  1
 1
Ft 2 2 1
L
L4
t
1
P

, so the ’s are not unique!
2
 2 D 3max

2 
Problem 7.42
Given:
[Difficulty: 3]
Find:
Functional relationship between the mass flow rate of gas through a choked-flow nozzle and other physical
parameters
(a) How many independent Π terms that characterize this phenomenon
(b) Find the Π terms
(c) State the functional relationship for the mass flow rate in terms of the Π terms
Solution:
We will use the Buckingham pi-theorem.
1
m
2
Select primary dimensions M, L, t:
3
m
A
A
M
L
t
4
5
p
p
p
T
n = 5 parameters
R
L
T
2
(Mathcad can't render dots!)
R
T
M
2
L t
A
T
2
r = 4 dimensions
2
t T
m = r = 4 repeating parameters
R
We have n - m = 1 dimensionless group.
Setting up dimensional equations:
a
b
c
Π1  m p  A  T  R
d
Thus:
Summing exponents:
 
The solution to this system is:
1
a  1
b  1 c 
2
M: 1  a  0
L:
d
a
b
 2
M 
c L 
2
0 0 0 0
 L T 

 M L t T
2 
t  2

 L t 
 t T 
M
d
m
Π1 
 R T
p A
1
2
a  2  b  2  d  0
t:
1  2  a  2  d  0
T:
cd0
1
2
Check using F, L, t dimensions:
L
F t L 1
2
  
T  1
2
1
L F
L
t T
The functional relationship is:
Π1  C
m
p A
 R T  C
So the mass flow rate is:
2
m  C
p A
R T
Problem 7.43 (In Excel)
[Difficulty: 3]
Given: Time to speed up depends on inertia, speed, torque, oil viscosity and geometry
Find:  groups
Solution:
We will use the workbook of Example 7.1, modified for the current problem
n
r
m =r
n -m
The number of parameters is:
The number of primary dimensions is:
The number of repeat parameters is:
The number of  groups is:
=8
=3
=3
=5
Enter the dimensions (M, L, t) of
the repeating parameters, and of up to
four other parameters (for up to four  groups).
The spreadsheet will compute the exponents a , b , and c for each.
REPEATING PARAMETERS: Choose , D , T

D
T
M
L
1
1
2
t
-1
-2
 GROUPS:
Two  groups can be obtained by inspection: /D and L /D . The others are obtained below
t
1:
I
3:
M
0
L
0
a =
b =
c =
1
0
0
M
1
L
2
a =
b =
c =
2
0
-1
Hence the  groups are

 D 3
I 2
D
T
T
Note that the 1 group can also be easily obtained by inspection
t
L
D
t
1

2:
t
0
4:
M
1
L
-1
a =
b =
c =
1
3
-1
M
0
L
0
a =
b =
c =
0
0
0
t
-1
t
0
Problem 7.44
Given:
[Difficulty: 3]
Find:
Functional relationship between the mass flow rate of liquid from a pressurized tank through a contoured nozzle
and other physical parameters
(a) How many independent Π terms that characterize this phenomenon
(b) Find the Π terms
(c) State the functional relationship for the mass flow rate in terms of the Π terms
Solution:
We will use the Buckingham pi-theorem.
1
m
2
Select primary dimensions M, L, t:
3
m
L
t
5
ρ
h
M
2
L
A
h
ρ
A
M
4
ρ
A
L
3
∆p
∆p
g
M
L
L t
n = 6 parameters
g
2
t
r = 3 dimensions
2
m = r = 3 repeating parameters
g
We have n - m = 3 dimensionless groups.
Setting up dimensional equations:
a
b c
Π1  m ρ  A  g
Thus:
M
t
Summing exponents:

M
 3
L 
a
 2  L2 
b
 L
c
0
0 0
 M L t
t 
The solution to this system is:
5
1
a  1
b
c
4
2
M: 1  a  0
L:
3  a  2  b  c  0
t:
1  2  c  0
a
b c
5
1
4
2
ρ A  g
4
Check using F, L, t dimensions:
1
t
F t L



1
1
L
2 5
F t
L
Π2  h  ρ  A  g
m
Π1 
Thus:
Summing exponents:
M: a  0
L:
1  3 a  2 b  c  0
t:
2  c  0
L 
M
 3
L 
a
 2  L2 
b
 L
c
0
2
L
0 0
 M L t
t 
The solution to this system is:
1
a0
b
c0
2
Check using F, L, t dimensions:
Π2 
L
1
L
1
h
A
2
a
b c
Π3  ∆p ρ  A  g
Thus:
M

M
2  3
L t  L 
Summing exponents:
a
 2  L2 
b
 L
c
0 0 0
 M L t
t 
The solution to this system is:
1
a  1
b
c  1
2
M: 1  a  0
L:
1  3  a  2  b  c  0
t:
2  2  c  0
Π3 
F
Check using F, L, t dimensions:
L
The functional relationship is:

Π1  f Π2 Π3

m
5
1
4
2
ρ A  g
 f 

h

∆p
2


A ρ g  A 

L
4

F t
t
∆p
ρ g  A
2

1
2 L L
1
So the mass flow rate is:
5
1
m  ρ A  g  f 
4
2

h

∆p


A ρ g  A 
Problem 7.45
Given:
Find:
Solution:
Functional relationship between the aerodynamic torque on a spinning ball and other physical parameters
The Π terms that characterize this phenomenon
We will use the Buckingham pi-theorem.
1
T
2
Select primary dimensions M, L, t:
3
T
t
4
5
ρ
V
M L
ρ
L
M
M
t
3
L t
2
V
μ
V
2
ρ
[Difficulty: 3]
L
μ
D
D
ω
d
ω
d
1
L
n = 7 parameters
r = 3 dimensions
L
t
m = r = 3 repeating parameters
D
We have n - m = 4 dimensionless groups. Setting up dimensional equations:
a
b
c
Π1  T ρ  V  D
Thus:
M L
t
Summing exponents:
2
2

M
a
 
L
b
c
0 0 0
 3   t   L  M  L  t
L 
The solution to this system is:
a  1
M: 1  a  0
L:
2  3 a  b  c  0
t:
2  b  0
b  2
c  3
Check using F, L, t dimensions:
F L
L
4
F t
a
b
c
Π2  μ ρ  V  D
Thus:
Summing exponents:
M: 1  a  0
L:
1  3  a  b  c  0
t:
1  b  0
M

M
a
 
L
T
Π1 
2

t
2
2
L
3
ρ V  D
2

1
L
3
1
b
c
0 0 0
L  M L t



L t
3
t
L   
The solution to this system is:
a  1
b  1
c  1
μ
Π2 
ρ V D
4
Check using F, L, t dimensions:
F t L
t 1

  1
2
2 L L
L F t
a
b
c
Π3  ω ρ  V  D
Thus:
Summing exponents:
1

M
a
 
L
a0
3  a  b  c  0
t:
1  b  0
ω D
Π3 
V
The solution to this system is:
M: a  0
L:
b
c
0 0 0
L  M L t



t
3
t
L   
b  1
c1
Check using F, L, t dimensions:
1
t
a
b
c
Π4  d  ρ  V  D
Thus:
L 
t:
b  0
 
L
a0
b0
d
Π4 
D
c  1
Check using F, L, t dimensions:
1
L
The functional relationship is:
1
b
The solution to this system is:
M: a  0
1  3 a  b  c  0
a
t
L
c
0 0 0
 3   t   L  M  L  t
L 
Summing exponents:
L:
M
 L

Π1  f Π2 Π3 Π4

L  1
T
2
3
ρ V  D
ω D d 
 
 ρ V D V D 
 f 
μ

Problem 7.46
Given:
Ventilation system of cruise ship clubhouse
Find:
Dimensionless groups
[Difficulty: 2]
Solution:
Apply the Buckingham  procedure
 c
N
p
D

 Select primary dimensions M, L, t

c


1
 L3



D
p

N
p
D 
p

g
1
M
Lt 2
1
t
M
L3
M
L3
L
t2
L


g
n = 9 parameters




M
Lt 
r = 3 primary dimensions
m = r = 3 repeat parameters
 Then n – m = 6 dimensionless groups will result. Setting up a dimensional equation,
a
c
M 
b1 M
 M 0 L0t 0
1   D  Δp   3  L   
2
L
t
Lt
 
 
M:
a 1  0
a  1
L :  3a  b  1  0 b  2
Hence
t:
c20
c  2
a
Summing exponents,
b
c
M
2   D     3
L
M:
a 1  0
L :  3a  b  1  0
 c 1  0
t:
a
Summing exponents,
b
The other  groups can be found by inspection:
c
a
p
D 2 2
c

b1 M
 M 0 L0t 0
 L   
t
Lt
 

a  1
b  2
Hence
c  1
 3  cD 3
1 
2 
p


D 2
6 
g
D 2
4  N
5 
 1
 3   4   5   6  1
 Check using F, L, t as primary dimensions
1 
F
L2
Ft 2 2 1
L 2
L4
t
 1
Note: Any combination of ’s is a  group, e.g.,
2 
Ft
L2
Ft 2 2 1
L
L4
t
p
1

, so the ’s are not unique!
 2 
Problem 7.47
[Difficulty: 3]
Given:
Find:
Functional relationship between the mass burning rate of a combustible mixture and other physical parameters
Solution:
We will use the Buckingham pi-theorem.
The dependence of mass burning rate
1
m
2
Select primary dimensions M, L, t:
3
m
δ
δ
M
5
δ
(Mathcad can't render dots!)
D
α
M
L
3
t
L
ρ
α
ρ
L
t
4
ρ
n = 5 parameters
D
2
L
2
r = 3 dimensions
t
m = r = 3 repeating parameters
α
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
c
a b
c
Π1  m δ  ρ  α
Thus:
b  2
M
L
0 0 0
L      M L t
 3  t 
t
L 
M
Summing exponents:
a
The solution to this system is:
M: 1  b  0
L:
a  3 b  2 c  0
t:
1  c  0
a b
c
Π2  D δ  ρ  α
a  1
Thus:
Summing exponents:
M: b  0
L:
2  a  3 b  2 c  0
t:
1  c  0
b  1
Check using F, L, t dimensions:
The functional relationship is:
c  1
c
b  2
M
L
0 0 0
L      M L t
 3  t 
t
L 
L
2
a
The solution to this system is:
a0
b0
4
6
m
Π1 
δ ρ α
F t 1 L
t
 

1
L L
2 2
F t L
 
Π1  f Π2
D
Π2 
α
c  1
L
2
t

t
L
2
1
m
δ ρ α
 f 
D

α
Problem 7.48
Given:
Find:
Functional relationship between the power loss in a journal bearing and other physical parameters
The Π terms that characterize this phenomenon and the function form of the dependence of P on these
parameters
We will use the Buckingham pi-theorem.
Solution:
1
P
2
Select primary dimensions F, L, t:
3
P
l
F L
t
4
5
[Difficulty: 3]
D
ω
D
c
l
D
c
L
L
L
ω
μ
p
ω
μ
p
1
F t
t
L
n = 7 parameters
F
2
L
2
r = 3 dimensions
m = r = 3 repeating parameters
p
We have n - m = 4 dimensionless groups. Setting up dimensional equations:
a
b c
Π1  P D  ω  p
Thus:
t
Summing exponents:
F:
1c0
L:
1  a  2 c  0
t:
1  b  0
a
b c
Π2  l D  ω  p
c0
L:
1  a  2 c  0
t:
b  0
a
 L  
a
a  3
b c
Summing exponents:
F:
c0
L:
1  a  2 c  0
t:
b  0
c
Thus:
L L  
a
b  1
b
c  1
L L  
a
P
3
D  ω p
c
The solution to this system is:
Thus:
Π1 
F
0 0 0
   F L t



 t   L2 
1
a  1
Π3  c D  ω  p
b
F
0 0 0
   F L t



 t   L2 
1
The solution to this system is:
Summing exponents:
F:
F L
b0
b
c0
l
Π2 
D
c
F
0 0 0
   F L t



 t   L2 
1
The solution to this system is:
a  1
b0
c0
c
Π3 
D
a
b c
Π4  μ D  ω  p
Thus:
F t
L
2
Summing exponents:
6
F:
1c0
L:
2  a  2  c  0
t:
1b0
a
b
c
F
0 0 0
   F L t



 t   L2 
1
The solution to this system is:
a0
Check using M, L, t dimensions:
M L
t
The functional relationship is:
 L  
3

2

1
L
3
b1
 t
L t
Π1  f Π2 Π3 Π4
c  1
2
M

Π4 
 1 L
1
L
P
3
ω p  D
1
L
1
L
l
p
2
1
c μ ω 
 

D D p 
 f 
μ ω
M 1 L t
 
1
L t t M
c μ ω 
 

D D p 
P  ω p  D  f 
3
l
Problem 7.49
Given:
Find:
Solution:
Functional relationship between the heat transfer rate in a convection oven and other physical parameters
The number of Π terms that characterize this phenomenon and the Π terms
We will use the Buckingham pi-theorem.
1
Q
2
Select primary dimensions F, L, t, T (temperature):
3
Q
4
5
Θ
cp
cp
L
t
t T
2
V
L
Θ
L
T
L
2
F L
ρ
[Difficulty: 4]
μ
ρ
F t
L
2
4
μ
V
F t
L
2
t
L
Setting up dimensional equations:
a
b
c
Π1  Q ρ  V  L  Θ
d
Thus:
1a0
a
 F t2   L b c d 0 0 0 0
     L  T  F  L  t  T

t  4  t
L 
a  1
L:
1  4 a  b  c  0
t:
1  2  a  b  0
T:
d0
c
Π2  cp  ρ  V  L  Θ
We have n - m = 3 dimensionless
groups.
The solution to this system is:
F:
b
r = 4 dimensions
F L
Summing exponents:
a
n = 7 parameters
V
m = r = 4 repeating parameters
Θ
L
ρ
d
Thus:
Summing exponents:
b  3
c  2 d  0
a0
L:
2  4 a  b  c  0
t:
2  2  a  b  0
T:
1  d  0
Q
3
ρ V  L
a
 F t2   L b c d 0 0 0 0
     L  T  F  L  t  T

2

4   t
t T  L 
L
2
The solution to this system is:
F:
Π1 
a0
b  2
c0
d1
Π2 
cp  Θ
2
V
a
a
b
c
Π3  μ ρ  V  L  Θ
d
Thus:
Summing exponents:
F:
1a0
L:
2  4  a  b  c  0
t:
1  2 a  b  0
T:
d0
 F t2   L  b c d 0 0 0 0
   L T  F L t T

2  4   t
L  L 
F t
The solution to this system is:
a  1
b  1
c  1 d  0
μ
Π3 
ρ V L
2
6
Check using M, L, t, T dimensions:
M L
t
The functional relationship is:
3

2

L
3
M

Π1  f Π2 Π3
t
2
L

2

1
L
3
1
L
2

2
t
2
t T L
Q
3
ρ V  L
2
2
3
T  1
M L t 1
   1
L t M L L
 cp Θ μ 

 V2 ρ V L 


 f
 cp  Θ μ 

 V2 ρ V L 


Q  ρ V  L  f 
3
2
Problem 7.50
Given:
Find:
Solution:
Functional relationship between the thrust of a marine propeller and other physical parameters
The Π terms that characterize this phenomenon
We will use the Buckingham pi-theorem.
1
FT
2
Select primary dimensions F, L, t:
3
FT
ρ
M L
M
t
4
5
ρ
2
ρ
[Difficulty: 3]
L
V
D
D
L
3
g
ω
p
μ
V
g
ω
p
μ
L
L
1
M
2
t
L t
V
t
t
n = 8 parameters
M
L t
2
r = 3 dimensions
m = r = 3 repeating parameters
D
We have n - m = 5 dimensionless groups. Setting up dimensional equations:
a
b
c
Π1  FT ρ  V  D
Thus:
M L
t
Summing exponents:
1  3 a  b  c  0
t:
2  b  0
b
c
Thus:
L
L:
1  3 a  b  c  0
t:
2  b  0
c
Summing exponents:
M: a  0
L:
3  a  b  c  0
t:
1  b  0
b
c
0 0 0
 3   t   L  M  L  t
L 

M
b  2
a
 
a0
b
L
L
c  2
Thus:
Π1 
FT
2
1

M
b  2
a
 
L
c1
Π2 
g D
2
V
b
c
0 0 0
L  M L t



t
3
t
L   
The solution to this system is:
a0
b  1
c1
2
ρ V  D
b
The solution to this system is:
M: a  0
a
 
c
0 0 0
L  M L t



2
3
t
t L   
Summing exponents:
Π3  ω ρ  V  D
a
a  1
L:
a
M
The solution to this system is:
M: 1  a  0
Π2  g  ρ  V  D
2

ω D
Π3 
V
a
b
c
Π4  p  ρ  V  D
M
2
Summing exponents:
1  3  a  b  c  0
t:
2  b  0
b
c
M
L
b
1  3  a  b  c  0
t:
1  b  0
Check using F, L, t dimensions: F
M
a
 
L
c0
p
2
ρ V
b
μ
Π5 
ρ V D
The solution to this system is:
M: 1  a  0
L:

b  2
Π4 
c
0 0 0
L  M L t



L t
3
t
L   
Thus:
Summing exponents:
6
 
a  1
L:
a
a
M
The solution to this system is:
M: 1  a  0
Π5  μ ρ  V  D

c
0 0 0
 3   t   L  M  L  t
L t  L 
Thus:
a  1
L
4
F t
2

t
2
L
2

1
L
2
b  1
1
L
t
2
c  1
 L
t
2
L
2
1
1
t
 L
t
L
1
F
L
2

L
4
F t
2

t
2
L
2
4
1
t 1
F t L

  1
2
2 L L
L F t
Problem 7.51
[Difficulty: 3]
Given:
That the cooling rate depends on rice properties and air properties
Find:
The  groups
Solution:
Apply the Buckingham  procedure
 dT/dt
c
k
L

cp

V
n = 8 parameters
 Select primary dimensions M, L, t and T (temperature)
dT dt
c
k
L
T
t
L2
ML
t 2T
t 2T
cp


V
L2
M
t 2T
L3
M
Lt
L
t

r = 4 primary dimensions
 V

L
L
cp
m = r = 4 repeat parameters
Then n – m = 4 dimensionless groups will result. By inspection, one  group is c/cp. Setting up a dimensional equation,
d
2
dT  L   M 
c L  T

1  V  L c
    3  L   2 
 T 0 M 0 L0t 0
dt  t   L 
t T  t
a
a
b
b c d
p
Summing exponents,
T:
d  1  0
d 1
M:
b0
b0
L:
t:
Hence
1 
a  3b  c  2d  0 a  c  2  c  1
 a  2d  1  0
a  3
dT Lc p
dt V 3
By a similar process, we find
2 
k
L2 c p
and
3 

LV
Hence
 c
dT Lc p
 , k , 
f

 c p L2 c LV
dt V 3
p





Problem 7.52
Given:
Find:
Functional relationship between the power to drive a marine propeller and other physical parameters
(a) The number of Π terms that characterize this phenomenon
(b) The Π terms
Solution:
We will use the Buckingham pi-theorem.
1
P
2
Select primary dimensions F, L, t:
3
P
ρ
t
5
D
ρ
M L
4
[Difficulty: 3]
2
M
3
ρ
D
L
V
L
3
c
ω
μ
V
c
ω
μ
L
L
1
M
t
t
t
L t
V
n = 7 parameters
r = 3 dimensions
m = r = 3 repeating parameters
D
We have n - m = 4 dimensionless groups. Setting up dimensional equations:
a
b
c
Π1  P ρ  V  D
Thus:
M L
t
Summing exponents:
2  3 a  b  c  0
t:
3  b  0
b
c
Thus:
L
L:
1  3 a  b  c  0
t:
1  b  0
c
Summing exponents:
M: a  0
L:
3  a  b  c  0
t:
1  b  0
L
b
c
0 0 0
 3   t   L  M  L  t
L 

M
b  3
a
 
a0
b
 
L
c  2
Thus:
Π1 
P
3
1

M
b  1
a
 
L
c0
c
Π2 
V
b
c
0 0 0
L  M L t



t
3
t
L   
The solution to this system is:
a0
b  1
c1
2
ρ V  D
b
The solution to this system is:
M: a  0
Π3  ω ρ  V  D
a
c
0 0 0
L  M L t



t
3
t
L   
Summing exponents:
a
M
a  1
L:
a

The solution to this system is:
M: 1  a  0
Π2  c ρ  V  D
3
2
ω D
Π3 
V
a
b
c
Π4  μ ρ  V  D
Thus:
Summing exponents:
M: 1  a  0
L:
1  3  a  b  c  0
t:
1  b  0
M
L t

M
Check using F, L, t dimensions:
 
L
b
c
0 0 0
 3   t   L  M  L  t
L 
μ
Π4 
ρ V D
The solution to this system is:
a  1
4
6
a
b  1
3
F L L
t 1

 
1
t
2 3 2
F t L L
c  1
L t
 1
t L
1
t
 L
t
L
4
1
F t L
t 1

  1
2
2 L L
L F t
Problem 7.53
[Difficulty: 2]
Given:
Boundary layer profile
Find:
Two  groups by inspection; One  that is a standard fluid mechanics group; Dimensionless groups
Solution:
Two obvious  groups are u/U and y/. A dimensionless group common in fluid mechanics is U (Reynolds number)
Apply the Buckingham  procedure
 u
y
U
dU/dx




n = 6 parameters
 Select primary dimensions M, L, t

 U

u


L

t
y U
L
L
t
dU dx
2
1
t
L
t




L


m = r = 3 primary dimensions
m = r = 2 repeat parameters
 Then n – m = 4 dimensionless groups will result. We can easily do these by inspection
1 
u
U
2 
y

3 
dU
dy 
U
 Check using F, L, t as primary dimensions, is not really needed here
Note: Any combination of ’s can be used; they are not unique!
4 

U
Problem 7.54
Given:
Functional relationship between the maximum pressure experienced in a water hammer wave and other physical
parameters
(a) The number of Π terms that characterize this phenomenon
(b) The functional relationship between the Π terms
Find:
Solution:
1
We will use the Buckingham pi-theorem.
p max
ρ
2
U0
EV
Select primary dimensions M, L, t:
3
p max
ρ
U0
M
L
M
3
t
L t
M
L t
4
5
ρ
2
L
n = 4 parameters
EV
2
r = 3 dimensions
m = 2 repeating parameters because p max and Ev have the same dimensions.
We have n - m = 2 dimensionless groups.
U0
Setting up dimensional equations:
a
Π1  p max ρ  U0
b
M
2
1  3  a  b  0
t:
2  b  0
a
Π2  Ev  ρ  U0
b
M
t:
2  b  0
 
L
b
Check using F, L, t dimensions:
2
M
a
 
L
p max
ρ U0
2
b
The solution to this system is:
a  1
F
L
The functional relationship is:

b  2
Π1 
0 0 0
 3   t   M  L  t
L t  L 
Thus:
M: 1  a  0
1  3  a  b  0
a
a  1
Summing exponents:
L:
M
The solution to this system is:
M: 1  a  0
L:

0 0 0
 3   t   M  L  t
L t  L 
Thus:
Summing exponents:
6
[Difficulty: 4]
2

L
4
F t
2

 
Π1  f Π2
t
b  2
2
L
2
1
Thus:
F
L
2

L
4
F t
2

t
Π2 
Ev
ρ U0
2
2
L
2
1
p max
ρ U0
2
 Ev 

 ρ U 2 
 0 
 f
Problem 7.55
[Difficulty: 3]
Given: Model scale for on balloon
Find: Required water model water speed; drag on protype based on model drag
Solution:
From Appendix A (inc. Fig. A.2)
The given data is
For dynamic similarity we assume
μair  1.8  10
m
Vair  5 
s
Lratio  20
ρw Vw Lw
μw
Then
 5 Ns
kg
ρair  1.24
3
m


2
m
 3 Ns
kg
ρw  999
3
m
μw  10
ρair Vair Lair
μair
μw ρair Lair
μw ρair
m
Vw  Vair


 Vair

 Lratio  5  
μair ρw Lw
μair ρw
s
 10 3 


 1.8  10 5 


 1.24   20


 999 
m
Vw  6.90
s
Fair
1
2

 ρair Aair Vair
2
Fw
1
2
2
 ρw Aw Vw
2
 Lair 
2

 Lratio

Aw
 Lw 
Aair
Hence the prototype drag is
2
 Vair 
Fair  Fw
 Lratio  
  2000 N 
ρw
 Vw 
ρair
2
 1.24   202   5 
 999 
 6.9 


 
2
m
Fw  2 kN
For the same Reynolds numbers, the drag coefficients will be the same so we have
where

2
Fair  522 N
Problem 7.56
Given:
[Difficulty: 3]
Find:
Airship is to operate at 20 m/s in air at standard conditions. A 1/20 scale model is to be tested in a wind tunnel at
the same temperature to determine drag.
(a) Criterion needed to obtain dynamic similarity
(b) Air pressure required if air speed in wind tunnel is 75 m/s
(c) Prototype drag if the drag on the model is 250 N
Solution:
Dimensional analysis predicts:
F
2
ρ V  L
2
 f 
ρ V L 
 Therefore, for dynamic similarity, it would follow that:
 μ 
ρm Vm Lm
μm

ρp  Vp  Lp
μp
Since the tests are performed at the same temperature, the viscosities are the same. Solving for the ratio of densities:
ρm
ρp

Vp Lp μm
20
p
Thus:



 20  1  5.333 Now from the ideal gas equation of state: ρ 
Vm Lm μp
75
R T
ρm Tp
pm  pp

ρp Tm
5
p m  101  kPa  5.333  1
Fp
From the force ratios:
2
ρp  Vp  Lp
Substituting known values:
2

p m  5.39  10 Pa
Fm
2
Thus:
2
ρm Vm  Lm
1
Fp  250  N 

5.333
2
 Vp   Lp 
Fp  Fm

  
ρm Vm
   Lm 
ρp
2
2
 20   ( 20) 2
 75 
 
Fp  1.333  kN
Problem 7.57
[Difficulty: 2]
Given:
Find:
A model is to be subjected to the same Reynolds number in air flow and water flow
Solution:
For dynamic similarity:
(a) Which flow will require the higher flow speed
(b) How much higher the flow speed needs to be
ρw Vw Lw
μw

From Tables A.8 and A.10 at 20 deg C: νw  1.00  10
Va
Vw

1.51  10
1.00  10
ρa  Va La
μa
2
6 m

s
and
We know that Lw  La
2
5 m
νa  1.51  10

Thus:
Va
Vw

ρw μa
νa


ρa μw
νw
Therefore:
s
5
6
 15.1
Air speed must be higher than
water speed.
To match Reynolds number:
Va  15.1 Vw
Problem 7.58
Given:
[Difficulty: 5]
Vessel to be powered by a rotating circular cylinder. Model tests are planned to determine the required power
for the prototype.
(a) List of parameters that should be included in the analysis
(b) Perform dimensional analysis to identify the important dimensionless groups
Find:
From an inspection of the physical problem: P  f ( ρ μ V ω D H)
Solution:
We will now use the Buckingham pi-theorem to find the dimensionless groups.
1
P
2
Select primary dimensions M, L, t:
3
P
ρ
ρ
M L
t
4
5
μ
2
3
ρ
V
ω
μ
V
ω
M
M
L
1
3
L t
t
t
L
ω
D
H
D
H
L
L
n = 7 parameters
r = 3 dimensions
m = r = 3 repeating parameters
D
We have n - m = 4 dimensionless groups. Setting up dimensional equations:
a
b
c
Π1  P ρ  ω  D
Thus:
M L
t
Summing exponents:
2  3 a  c  0
t:
3  b  0
a
b
c
Π2  μ ρ  ω  D
Thus:
M
L t
t:
1  b  0
a
b
c
Π3  V ρ  ω  D
Summing exponents:
M: a  0
L:
1  3 a  c  0
t:
1  b  0
a
 
1
b
c
0 0 0
 3   t   L  M  L  t
L 

b  3
M
a
 
1
c  5
L

P
3
5
ρ ω  D
c
0 0 0
 3   t   L  M  L  t
L 
a  1
Thus:
Π1 
b
The solution to this system is:
M: 1  a  0
1  3  a  c  0
M
a  1
Summing exponents:
L:
3

The solution to this system is:
M: 1  a  0
L:
2
M
b  1
a
 
1
c  2
Π2 
μ
b
c
0 0 0
L  M L t



t
3
t
L   
The solution to this system is:
a0
b  1
c  1
2
ρ ω D
V
Π3 
ω D
a
b
c
Π4  H ρ  ω  D
Thus:
L 
t:
b  0
 
1
a0
b0
4
6
Check using F, L, t dimensions:
The functional relationship is:
b
The solution to this system is:
M: a  0
1  3 a  c  0
a
c
0 0 0
 3   t   L  M  L  t
L 
Summing exponents:
L:
M
c  1
4
F L L
F t L
1
3 1

t 
1

 t
1
t
2
2
2
5
2
F t
L F t
L
L

Π1  f Π2 Π3 Π4

H
Π4 
D
L
t
 t
1
L
1
L
1
1
L
P
3
5
ρ ω  D
 f
μ
V H



2 ω D D 
 ρ ω D

Problem 7.59
[Difficulty: 3]
Given:
Find:
Measurements of drag are made on a model car in a fresh water tank. The model is 1/5-scale.
Solution:
The flows must be geometrically and kinematically similar, and have equal Reynolds numbers to be dynamically
similar:
(a) Conditions requred to ensure dynamic similarity between the model and the prototype.
(b) Required fraction of speed in air at which the model needs to be tested in water to ensure dynamically similar
conditions.
(c) Drag force on the prototype model traveling at 90 kph in air if the model drag is 182 N traveling at 4 m/s in
water.
Geometric similarity requires a true model in all respects.
Kinematic similarity requires the same flow pattern, i.e., no free-surface or cavitation effects.
The problem may be stated as F = f(ρ,V,L,μ)
F
Dimensional analysis gives this relation:
2
ρ V  L
2
Vm Lm
Matching Reynolds numbers between the model and prototype
flows:
νw  1.00  10
From Tables A.8 and A.10 at 20 deg C:
Vm
Vp
6

1.00  10
5
1.51  10

5
1
 g ( Re)
νm
2
6 m


Vp  Lp
μ
Thus:
νp

Vm
Vp
νa  1.51  10

V L
ν

νm Lp

νp Lm
Therefore:
s
Vm
 0.331
If the conditions are dynamically similar:
Vp
Fm
2
2
ρm Vm  Lm

Fp
2
ρp  Vp  Lp
Thus:
2
2
 Vp   Lp 
Fp  Fm

  
ρm Vm
   Lm 
ρp
2
Substituting in known values:
ρ V L
Re 
2
5 m
and
s
where
1.20 
km 1000 m
hr
s 
Fp  182  N 



  90
 
999
hr
km
3600

s
4

m

5
 
1
 0.331
2
2
Fp  213 N
Problem 7.60
[Difficulty: 2]
Given:
Flow around ship's propeller
Find:
Model propeller speed using Froude number and Reynolds number
Solution:
Basic equations:
Fr 
V
Re 
g L
V L
ν
Assumptions: (a) The model and the actual propeller are geometrically similar
(b) The flows about the propellers are kinematically and dynamically similar
Using the Froude number
But the angular velocity is given by
Comparing Eqs. 1 and 2
The model rotation speed is then
Using the Reynolds number
Vm
Fr m 
g  Lm
 Fr p 
Vp
g  Lp
V  L ω
or
Vp
so
Vm
Vp
Lm ωm


Lp ωp
Lm
ωm
Lp
ωp
Lp
ωm  ωp 
Lm
Rem 
Vm
Vm Lm
νm



Lm
(1)
Lp
Lm ωm

Lp ωp
Lp
Lm
ωm  100  rpm 
 Rep 
Vp  Lp
or
νp
Vm
Vp

(2)
9
ωm  300  rpm
1
Lp νm
Lp


Lm νp
Lm
(3)
(We have assumed the viscosities of the sea water and model water are comparable)
Comparing Eqs. 2 and 3
Lm ωm
Lp


Lp ωp
Lm
The model rotation speed is then
 Lp 
ωm  ωp  

 Lm 
 Lp 


ωp
 Lm 
ωm
2
2
ωm  100  rpm 
9
 
1
2
Of the two models, the Froude number appears most realistic; at 8100 rpm serious cavitation will occur, which would
invalidate the similarity assumptions. Both flows will likely have high Reynolds numbers so that the flow becomes
independent of Reynolds number; the Froude number is likely to be a good indicator of static pressure to dynamic
pressure for this (although cavitation number would be better).
ωm  8100 rpm
Problem 7.61
[Difficulty: 3]
Given:
A torpedo with D = 533 mm and L = 6.7 m is to travel at 28 m/s in water. A 1/5 scale model of the torpedo is to be
tested in a wind tunnel. The maximum speed in the tunnel is fixed at 110 m/s, but the pressure can be varied at a
constant temperature of 20 deg C.
Find:
(a) Minimum pressure required in the wind tunnel for dynamically similar testing.
(b) The expected drag on the prototype if the model drag is 618 N.
Solution:
The problem may be stated as:
F
2
2
 g ( Re)
F  f ( ρ V D μ) From the Buckingham pi theorem, we expect 2 Π terms:
ρ V D
Re 
where
μ
ρ V  D
ρm Vm Dm
Matching Reynolds numbers between the model and prototype flows:
μm
At 20 deg C:
 3 N s
μp  1.00  10

and
2
μm  1.81  10
 5 N s

m

ρp  Vp  Dp
μp
Vp Dp μm
Thus: ρm  ρp 


Vm Dm μp
So substituting in values yields:
2
m
5
kg
28
5
1.81  10
kg
ρm  998 

 
ρm  23.0
3
110 1
3
3
m
1.00  10
m
Substituting in values:
p m  23.0
kg
3
m
If the conditions are dynamically similar:
 287 
N m
kg K
p m  ρm R Tm
2
 293  K 
Fm
2
2
ρm Vm  Dm
Substituting in known values:
From the ideal gas equation of state:
998
Fp  618  N 

23.0
Pa m
p m  1.934  MPa
N
Fp

Thus:
2
ρp  Vp  Dp
2
 28    5 
 110   

 1
2
 Vp 
Fp  Fm


ρm Vm
 
ρp
2
 Dp 


 Dm 
2
2
Fp  43.4 kN
Problem 7.62
[Difficulty: 3]
Given:
A 1/10 scale airfoil was tested in a wind tunnel at known test conditions. Prototype airfoil has a chord length of
6 ft and is to be flown at standard conditions.
Find:
(a) Reynolds number at which the model was tested
(b) Corresponding prototype speed
Solution:
Assumptions: (a) The viscosity of air does not vary appreciably between 1 and 5 atmospheres
(b) Geometric, kinematic, and dynamic similarity applies
F  f ( ρ V L μ)
The problem may be stated as:
F
2
ρ V  L
2
 g ( Re)
where
Re 
ρ V L
From the Buckingham pi theorem, we expect 2 Π terms:
Lm 
The model chord length is
μ

At 59 deg F:
2116 lbf 
atm ft
2



lbm R
53.33  ft lbf
μm  3.74  10
 7 lbf  s

ft

1
519  R

slug
32.2 lbm
ρm  0.0119
2
slug
ft
Rem  0.0119
Therefore:
 1.20 ft
5
pm
Substituting known values:
ρm 
R  Tm
We can calculate the model flow density from the ideal gas equation of state:
ρm   5  atm 

6  ft
slug
ft
3
 130 
3
ft
s
 1.2 ft 
ft
2
3.74  10
7
2

 lbf  s
lbf  s
slug ft
6
Rem  5.0  10
Matching Reynolds numbers between the model and prototype flows:
ρm Vm Lm
μm
From the ideal gas equation of state:
ρm
ρp


ρp  Vp  Lp
μp
ρm Lm μp
Thus: Vp  Vm


ρp Lp μm
p m Tp
p m Tp Lm μp
Therefore: Vp  Vm
So substituting in values yields:




p p Tm Lp μm
p p Tm
7
ft 5
519
1 3.74  10
Vp  130   
 
s
1
519
5
7
3.74  10
ft
Vp  130.0 
s
Problem 7.63
[Difficulty: 2]
Given:
Model of weather balloon
Find:
Model test speed; drag force expected on full-scale balloon
Solution:
From Buckingham Π
F
2
2
ρ V  D
 f 
ν
V

 V D c 
For similarity
Rep  Rem
Hence
Rep 

 F( Re M )
Mp  Mm
and
Vp  Dp
νp
 Rem 
(Mach number criterion
satisified because M<<1)
Vm Dm
νm
νm Dp
Vm  Vp 

νp Dm
From Table A.7 at 68 oF
νm  1.08  10
 5 ft

2
From Table A.9 at 68 oF
s


 1.08  10 5 ft 
ft
s 
Vm  5   

s 
2
ft
 1.62  10 4 
s 

 4 ft
νp  1.62  10

2
s
2
Then
Fm
2
2
ρm Vm  Dm

 10 ft 
1 
  ft 
6 
ft
Vm  20.0
s
2
2
ρp  Vp  Dp
2
 0.00234  slug 

3 
ft 

Fp  0.85 lbf 

slug 

1.94

3 
ft


2
ρp Vp Dp
Fp  Fm


ρm.
2
2
Vm Dm
Fp
2
 5 ft 
 s   10 ft  2
 ft   

 20   1  ft 
 s 6 
Fp  0.231  lbf
Problem 7.64
[Difficulty: 2]
Given:
Model of wing
Find:
Model test speed for dynamic similarity; ratio of model to prototype forces
Solution:
We would expect
From Buckingham Π
F  F( l s V ρ μ)
F
2
ρ V  l s
For dynamic similarity
where F is the force (lift or drag), l is the chord and s the span
ρ V l l 
 
 μ s
 f 
ρm Vm lm
μm

ρp  Vp  lp
μp
Hence
ρp lp μm
Vm  Vp 
 
ρm lm μp
From Table A.8 at 20 oC
μm  1.01  10
 3 N s

From Table A.10 at 20oC
2
μp  1.81  10
m
Fm
2
ρm Vm  lm sm


2
m
 1.21 kg 

3
m 
m   10 
Vm  7.5 
 
s
kg   1 

 998  3 
m 

Then
 5 N s
Fp
2
ρp  Vp  lp  sp
 1.01  10 3 N s 

2 
m 


 5 N s 
 1.81  10  2 
m 

2
m
Vm  5.07
s
ρm Vm lm sm
998





2 l p  sp
Fp
ρp
1.21
Vp
Fm
2
 5.07   1  1  3.77
 7.5 
10 10


Problem 7.65
Given:
[Difficulty: 3]
The fluid dynamic charachteristics of a gold ball are the be tested using a model in a wind tunnel. The
dependent variables are the drag and lift forces. Independent variables include the angular speed and dimple
depth. A pro golfer can hit a ball at a speed of 75 m/s and 8100 rpm. Wind tunnel maximum speed is 25 m/s.
(a) Suitable dimensionless parameters and express the functional dependence between them.
(b) Required diameter of model
(c) Required rotational speed of model
Find:
Solution:
Assumption:
Wind tunnel is at standard conditions
FD  FD( D V ω d ρ μ) FL  FL( D V ω d ρ μ) n = 7 and m = r = 3, so
from the Buckingham pi theorem, we expect two sets of four Π terms. The application of the Buckingham pi theorem will not be
shown here, but the functional dependences would be:
FD
FL
ρ V D ω D d 
ρ V D ω D d 

 

 
 f 
 g 
2 2
μ
D
2
2
μ
V
V D



ρ V  D
ρ V  D
The problem may be stated as:
To determine the required model diameter, we match Reynolds numbers between the model and prototype flows:
ρm Vm Dm
μm

ρp  Vp  Dp
μp
ρp Vp μm
Thus: Dm  Dp 
Substituting known values:


ρm Vm μp
75
Dm  4.27 cm  1 
1
25
Dm  12.81  cm
To determine the required angular speed of the model, we match the dimensionless rotational speed between the flows:
ωm Dm
Vm

ωp  Dp
Vp
Dp Vm
Thus: ωm  ωp 

Dm Vp
4.27
25
Substituting known values: ωm  8100 rpm 

12.81
75
ωm  900  rpm
Problem 7.66
[Difficulty: 3]
Given:
Model of water pump
Find:
Model flow rate for dynamic similarity (ignoring Re); Power of prototype
Solution:
From Buckingham Π
Q
3
ω D
Hence
3
Qp

ωm Dm
ωp  Dp
 Dm 
Qm  Qp 


ωp Dp
 
3
ρp  1.94
3
1
4
 
5
ft
Qm  0.750 
s
From Table A.9 at 68 oF
ρm  0.00234 
slug
ft
3
Pp

3
ρp  ωp  Dp
3
5
 ωp   Dp 
Pp  Pm

  
ρm ωm
   Dm 
ρp
3
3
3
Pm
ρm ωm  Dm
3
 2400  
 750 


slug
ft
Then
3
ωm
ft
Qm  15

s
From Table A.8 at 68 oF
where Q is flow rate, ω is angular speed, d
is diameter, and ρ is density (these Π
groups will be discussed in Chapter 10)
5
ρ ω  D
Qm
For dynamic similarity
P
and
3
1.94
Pp  0.1 hp 

0.00234
5
3
 750    4 
 2400   

 1
5
3
Pp  2.59  10  hp
Note that if we had used water instead of air as the working fluid for the model pump, it would have drawn 83 hp. Water would have
been an acceptable working fluid for the model, and there would have been less discrepancy in the Reynolds number.
Problem 7.67
Given:
Model of Frisbee
Find:
Dimensionless parameters; Prototype speed and angular speed
[Difficulty: 2]
Solution:
Assumption: Geometric, kinematic, and dynamic similarity between model and prototype.
F  F( D V ω h ρ μ)
The functional dependence is
From Buckingham Π
F
2
2
ρ V  D
For dynamic similarity
where F represents lift or drag
ρ V D ω D h 

 
V D
 μ
 f 
ρm Vm Dm
μm

ρp  Vp  Dp
μp
ρm Dm μp
Vp  Vm


ρp Dp μm
ft
Vp  140   ( 1 ) 
s
 1   ( 1)
7
 
ft
Vp  20
s
Also
ωm Dm
Vm

ωp  Dp
Vp
Dm Vp
ωp  ωm

Dp Vm
ωp  5000 rpm 
1 
7
 
 20 
 140 


ωp  102  rpm
Problem 7.68
[Difficulty: 3]
Given:
A 1:20 model of a hydrofoil is to be tested in water at 130 deg F. The prototype operates at a speed of 60 knots
in water at 45 deg F. To model the cavitation, the cavitation number must be duplicated.
Find:
Ambient pressure at which the test must be run
Solution:
To duplicate the Froude number between the model and the prototype requires:
Lm
Vm  Vp 
Lp
1
Vm  60 knot
20
p m  p vm
1
2
 Vm 

p m  p vm   p p  p vp 

ρp Vp
 
2
g  Lm
Vp

g  Lp
Thus:
Vm  13.42  knot
To match the cavitation number between the model and the prototype:
ρm
Vm
2
 ρm Vm

p p  p vp
1
2
 ρp  Vp
Therefore:
2
 Vm 
Assuming that the densities are equal: p m  p vm  p p  p vp  

 Vp 

From table A.7: at 130 deg F p vm  2.23 psi
p m  2.23 psi  ( 14.7 psi  0.15 psi)  
13.42 

 60 
at 45 deg F
p vp  0.15 psi

Thus the model pressure is:
2
p m  2.96 psi
2
Problem 7.69
[Difficulty: 2]
Given:
Oil flow in pipe and dynamically similar water flow
Find:
Average water speed and pressure drop
Solution:
From Example 7.2
Δp
2
ρ V
 f 
l e
  
 ρ V D D D 
μ
μH2O
ρH2O VH2O DH2O
From Fig. A.3 at 77 oF
νOil  10.8  8  10
From Table A.8 at 60 oF
μOil

For dynamic similarity
 5 ft


Hence

VH2O 
Then
ΔpOil
ρOil VOil
From Table A.2
2

2
s
2
2
 3
ft
s
ft
VH2O  0.0420
s
s
2
ΔpH2O


2
s
 4 ft
8.64  10
 4 ft
 8.64  10
νH2O
μH2O ρOil
VH2O 

 Voil 
V
ρH2O μOil
νOil Oil
s
 5 ft
1.21  10
2
s
 5 ft
νH2O  1.21  10
so
ρOil VOil DOil
2
ρH2O VH2O
ΔpH2O 
ρH2O VH2O
ρOil  VOil
2
 ΔpOil
SG Oil  0.92
ΔpH2O 
1
0.92
2

 0.0420   7 psi
 3 


3
ΔpH2O  1.49  10
 psi
Problem 7.70
[Difficulty: 3]
Given:
The frequency of vortex shedding from the rear of a bluff cylinder is a function of ρ, d, V, and μ. Vortex shedding
occurs in standard air on two cylinders with a diameter ratio of 2.
Find:
(a) Functional relationship for f using dimensional analysis
(b) Velocity ratio for vortex shedding
(c) Frequency ratio for vortex shedding
Solution:
We will use the Buckingham pi-theorem.
1
f
2
Select primary dimensions F, L, t:
3
f
ρ
1
M
ρ
t
4
5
ρ
L
V
d
d
3
L
V
n = 5 parameters
μ
V
μ
L
M
t
L t
r = 3 dimensions
m = r = 3 repeating parameters
d
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
a
b c
Π1  f  ρ  V  d
Thus:
Summing exponents:
1

a
 
L
The solution to this system is:
a0
L:
3  a  b  c  0
t:
1  b  0
b c
Π2  μ ρ  V  d
b
c
0 0 0
L  M L t



t
3
t
L   
M: a  0
a
M
Thus:
Summing exponents:
M: 1  a  0
L:
1  3  a  b  c  0
t:
1  b  0
M

M
b  1
a
 
L
b
c
0 0 0
L  M L t



L t
3
t
L   
The solution to this system is:
a  1
b  1
4
6
Check using F, L, t dimensions:
c1
f d
Π1 
V
c  1
1 t
F t L
t 1
 L  1

  1
t L
2
2 L L
L F t
μ
Π2 
ρ V d
The functional relationship is:
f d
 
Π1  f Π2
V
 f 
ρ V d 

 μ 
To achieve dynamic similarity between geometrically similar flows, we must duplicate all but one of the dimensionless groups:
ρ1  V1  d 1
μ1
Now if

ρ2  V2  d 2
V1
μ2
V2
ρ1  V1  d 1
μ1

ρ2  V2  d 2
μ2

ρ2 d 2 μ1
1
 
 1  1
ρ1 d 1 μ2
2
it follows that:
f1  d 1
V1

f2  d 2
V2
V1
V2
and
f1
f2

d 2 V1
1
1

 
2
d 1 V2
2
f1
f2


1
2
1
4
Problem 7.71
[Difficulty: 3]
Given:
Find:
1/8-scale model of a tractor-trailer rig was tested in a pressurized wind tunnel.
Solution:
We will use definitions of the drag coefficient and Reynolds number.
(a) Aerodynamic drag coefficient for the model
(b) Compare the Reynolds numbers for the model and the prototype vehicle at 55 mph
(c) Calculate aerodynamic drag on the prototype at a speed of 55 mph into a headwind of 10 mph
Governing
Equations:
CD 
FD
1
2
Re 
(Drag Coefficient)
2
 ρ V  A
ρ V L
(Reynolds Number)
μ
Am  Wm Hm
Assume that the frontal area for the model is:
3
The drag coefficient would then be:
From the definition of Re:
Rem
Rep
Rem
Rep

3.23
1.23
  75

m
s

hr
55 mi


CDm  2  128  N 
ρm Vm Lm μp



ρp Vp Lp μm
mi
5280 ft

ft
0.3048 m
2
Am  0.305  m  0.476  m
Am  0.1452 m
2
m
3.23 kg

kg m
1
 s  

 75.0 m 
2
2


N s
0.1452 m
CDm  0.0970
Assuming standard conditions and equal viscosities:

3600 s 
hr
1
 8 11

Rem  Rep
Since the Reynolds numbers match, assuming geometric and kinetic similarity we can say that the drag coefficients are equal:
1
2
FDp   CD ρp  Vp  Ap
2
Susbstituting known values yields:
2
2
1
kg 
mi 5280 ft 0.3048 m
hr 
N s
2
2
FDp 
 0.0970  1.23



 ( 55  10)
  0.1452 m  8 
2
3 
hr
ft
3600 s
mi
kg m
m
FDp  468 N
Problem 7.72
Given:
Flow around cruise ship smoke stack
Find:
Range of wind tunnel speeds
[Difficulty: 2]
Solution:
For dynamic similarity
Vm Dm
νm
Since
1  knot  1 
Hence for
nmi
hr
and

Vp  Dp
or
νp
Dp
15
Vm 
 Vp 
 V  15 Vp
Dm
1 p
1  nmi  6076.1 ft
nmi 6076.1 ft
hr
Vp  12


hr
nmi
3600 s
ft
Vp  20.254
s
ft
Vm  15  20.254
s
ft
Vm  304 
s
nmi 6076.1 ft
hr
Vp  24


hr
nmi
3600 s
ft
Vp  40.507
s
ft
Vm  15  40.507
s
ft
Vm  608 
s
Note that these speeds are very high - compressibility effects may become important, since the Mach number is
no longer much less than 1!
Problem 7.73
[Difficulty: 2]
Given:
Model of flying insect
Find:
Wind tunnel speed and wing frequency; select a better model fluid
Solution:
For dynamic similarity the following dimensionless groups must be the same in the insect and model (these are
Reynolds number and Strouhal number, and can be obtained from a Buckingham Π analysis)
Vinsect Linsect
νair
From Table A.9 (68oF)
The given data is

Vm Lm
ωinsect Linsect
νm
Vinsect

kg
ρair  1.21
3
m
νair  1.50  10
ωinsect  60 Hz
m
Vinsect  1.5
s
ωm Lm
Vm
2
5 m

s
Linsect
Lm

1
8
Linsect νm
Linsect
m 1
m
Hence in the wind tunnel Vm  Vinsect

 Vinsect
Vm  1.5 
Vm  0.1875
Lm νair
Lm
s
8
s
Vm Linsect
0.1875 1
Also
ωm  ωinsect

ωm  60 Hz 

ωm  0.9375 Hz
Vinsect Lm
1.5
8
It is unlikely measurable wing lift can be measured at such a low wing frequency (unless the measured lift was averaged, using an
integrator circuit, or perhaps a load cell and data acquisition system). Maybe try hot air (100 oC) for the model
2
5 m
νhot  2.29  10
For hot air try
Hence
Also
Vinsect Linsect
νair

Vm Lm

s
Linsect νhot
Vm  Vinsect

Lm νair
νhot
Vm Linsect
ωm  ωinsect

Vinsect Lm
ωm  60 Hz 
Hot air does not improve things much. Try modeling in water
Hence
Vinsect Linsect
νair

Vm Lm
νw
νair  1.50  10
instead of
0.286
1.5

m

s
1
Vm  1.5  
s
8
5
2.29  10
5
1.50  10
1
2
6 m

s
6
m 1 1.01  10
m
Vm  1.5  
Vm  0.01262
s
8
5
s
1.50  10
Vm Linsect
Vm
0.01262
1
ωm  ωinsect

 ωinsect
L
ωm  60 Hz 

Vinsect Lm
Vinsect ratio
1.5
8
This is even worse! It seems the best bet is hot (very hot) air for the wind tunnel. Alternatively, choose a much
smaller wind tunnel model, e.g., a 2.5 X model would lead to V m = 0.6 m/s and ωm = 9.6 Hz
Also
m
Vm  0.286
s
ωm  1.43 Hz
8
νw  1.01  10
Linsect νw
Vm  Vinsect

Lm νair
2
5 m
ωm  0.0631 Hz
Problem 7.74
A model test of a 1:4 scale tractor-trailer rig is performed in standard air. The drag force is a function of A, V, ρ,
and μ.
(a) Dimensionless parameters to characterize the model test results
(b) Conditions for dynamic similarity
(c) Drag force on the prototype vehicle based on test results
(d) Power needed to overcome the drag force
We will use the Buckingham pi-theorem.
Given:
Find:
Solution:
1
FD
2
Select primary dimensions F, L, t:
3
FD
A
4
5
L
2
ρ
V
A
M L
t
V
2
ρ
n = 5 parameters
μ
V
ρ
L
M
M
3
L t
t
L
μ
r = 3 dimensions
m = r = 3 repeating parameters
A
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
a
b
c
Π1  F ρ  V  A
Thus:
M L
t
2
Summing exponents:
1  3  a  b  2c  0
t:
2  b  0
a
b
c
M
a
 
L
 
b
c
0 0 0
2
 3   t   L  M  L  t
L 
a  1
L:
Π2  μ ρ  V  A

The solution to this system is:
M: 1  a  0
Thus:
M
L t
Summing exponents:
L:
1  3  a  b  2  c  0
t:
1  b  0
Check using F, L, t dimensions: F

M
b  2
a
 
L
 3   t 
L 
b
 2
 L
Π1 
c  1
c
0
a  1
L
4
F t
2

t
2
L
2

b  1
1
L
2
c
4
1
FD
2
ρ V  A
0 0
 M L t
The solution to this system is:
M: 1  a  0
6
[Difficulty: 3]
1
2
F t L
t 1

  1
2
2 L L
L F t
Π2 
μ
ρ V A
For dynamic similarity:
We must have geometric and kinematic similarity, and
The Reynolds numbers must match.
FDm
Once dynamic similarity is insured, the drag coefficients must be equal:
1
2
ρ V A
2 m m m

FDp
1
2
2
 Vp  Ap
0.00237  75 
2
So for the prototype: FDp  FDm


F
550

lbf



 A
 300   4
Dp
ρm Vm
0.00237


  m
ρp
The power requirement would be:
P  FDp Vp
P  550  lbf  75
ft
s

hp s
550  ft lbf
2
 ρ  V  Ap
2 p p
FDp  550  lbf
P  75.0 hp
P  55.9 kW
Problem 7.75
[Difficulty: 2]
Given:
Model of boat
Find:
Model kinematic viscosity for dynamic similarity
Solution:
For dynamic similarity
Vm Lm
νm
Hence from Eq 2
Vm
Vp


Vp  Lp
Vm
(1)
g  Lm
νp
g  Lm
Vp
(from Buckingham Π; the first
is the Reynolds number, the
second the Froude number)
(2)
g  Lp
Lm

g  Lp

Lp
3
Using this in Eq 1
Vm Lm
Lm Lm
 Lm 
νm  νp 

 νp 

 νp  

Vp Lp
Lp Lp
 Lp 
2
3
From Table A.8 at 50 oF
 5 ft
νp  1.41  10

2
s
νm  1.41  10
 5 ft

2
s

1
 10 
 
2
 7 ft
νm  4.46  10

2
s
Note that there aren't any fluids in Figure A.3 with viscosities that low!
Problem 7.76
[Difficulty: 4]
Given:
Model the motion of a glacier using glycerine. Assume ice as Newtonian fluid with density of glycerine but one
million times as viscous. In laboratory test the professor reappears in 9.6 hours.
Find:
(a) Dimensionless parameters to characterize the model test results
(b) Time needed for professor to reappear
Solution:
We will use the Buckingham pi-theorem.
1
V
2
Select primary dimensions F, L, t:
3
V
ρ
g
L
M
L
M
2
L t
ρ
t
4
5
ρ
g
L
g
3
t
μ
μ
D
H
L
D
H
L
L
L
L
n = 7 parameters
r = 3 dimensions
m = r = 3 repeating parameters
D
We have n - m = 4 dimensionless groups. Setting up dimensional equations:
a b
c
Π1  V ρ  g  D
Thus:
Summing exponents:
M: a  0
L:
1  3 a  b  c  0
t:
1  2  b  0
a b
c
Π2  μ ρ  g  D
Thus:
Summing exponents:
M: 1  a  0
L:
1  3  a  b  c  0
t:
1  2  b  0
L

M
a

L
b
c
0 0 0
L  M L t




t
3
2
L  t 
The solution to this system is:
1
1
a0
b
c
2
2
M

M
a

L
Π1 
V
g D
b
c
0 0 0
L  M L t




L t
3
2
L  t 
The solution to this system is:
1
3
a  1
b
c
2
2
Π2 
μ
3
ρ g D
(This is a gravity-driven
version of Reynolds #)
a b
c
Π3  H ρ  g  D
L 
Summing exponents:
a

a0
L:
1  3  a  b  c  0
t:
2  b  0
L
Check using F, L, t dimensions:
t
t

1
L
For dynamic similarity:

2
b0
1
1
L
4
μm
2
L
μp

ρm g m Dm
Lm
Lp

2
L
L
1
L
1
2
Matching the last two terms insures geometric similarity.
SG ice  0.92
From Tables A.1 and A.2:
ρp  g p  Dp
2
So
t
1
F t L



1
3
2
2 1
L F t
1

3
Dp
H
Π3 
D
c  1
Π1  f Π2 Π3 Π4
The functional relationship would be:
Dm
b
L
Π4 
D
By inspection we can see that
Therefore:
L
The solution to this system is:
M: a  0
6
M
c
0 0 0
 3   2  L  M L t
L  t 
Thus:
3
SG glycerine  1.26
2
3

3
 μm ρp 
 1  0.92   8.11  10 5 Since we have geometric similarity, the last two terms


μ ρ 
 6 1.26 
must match for model and prototype:
 p m
 10

 8.11  10
5
Lm  1850 m  8.11  10
5
Matching the first Π term:
Vm
Vp

Dm
Dp
 0.00900
Lm  0.1500 m
The time needed to reappear would be:
τ
L
V
Thus:
Lm
Lm
τm 
Vm 
Vm
τm
Lm Lp Vm
Lp
Lp Vm
1
day
τp 



 τm

τp  9.6 hr 
 0.00900 
Vp
Lm Vp
Vm Lm Vp
5
24 hr
8.11 10
Solving for the actual time:
τp  44.4 day
Your professor will be back before
the end of the semester!
Problem 7.77
[Difficulty: 3]
Given:
Model of automobile
Find:
Factors for kinematic similarity; Model speed; ratio of protype and model drags; minimum pressure for no cavitation
Solution:
For dynamic similarity
ρm Vm Lm
μm
ρp  Vp  Lp

ρp Lp μm
Vm  Vp 


ρm Lm μp
μp
For air (Table A.9) and water (Table A.7) at 68 oF
ρp  0.00234 
ρm  1.94
slug
ft
slug
ft
μp  3.79  10
3
3
ft
s
Vm  60 mph 

60 mph
Fm
2
2
Hence
Fp
Fm
For Ca = 0.5

ρp  Vp  Lp
2
2
2
2

ρm Vm  Lm
p min  p v
 0.5
1
2
 ρ V
2
From steam tables, for water at 68oF
 0.00234  
 1.94 


ρp  Vp  Lp
2
2
5 
1
 
 2.10  10 5 



 7
 3.79  10 
ft
Vm  29.4
s
Fp

ρm Vm  Lm

ft
 5 lbf  s
μm  2.10  10 
2
ft
88
Then
 7 lbf  s
2
 0.00234  
 1.94 


1
Fp
Fm
slug
for the water tank
p tank  3.25 psi  0.7  1.94
slug
2
2
lbf  s
1  ft 
 
 

slug ft  12 in 
s
  29.4
ft 
  29.4
ft 
2

ft
This is the minimum allowable pressure in the water tank; we can use it to find the required tank pressure
p min  p tank
1.4
2
2
Cp  1.4 
p tank  p min 
 ρ V  p min  0.7 ρ V
1
2
2
 ρ V
2
4
 0.270
so
 1.94
p min  0.339  psi 
2
1
2
p min  p v   ρ V
4
so we get
p v  0.339  psi
2
 88    5 
 29.4   

 1
3
ft
3

2
2
lbf  s
1  ft 
 
 

s
slug ft  12 in 
p min  3.25 psi
2
p tank  11.4 psi
Problem 7.78
[Difficulty: 3]
Given:
A scale model of a submarine is to be tested in fresh water under two conditions:
1 - on the surface
2 - far below the surface
Find:
(a) Speed for the model test on the surface
(b) Speed for the model test submerged
(c) Ratio of full-scale drag to model drag
Solution:
On the surface, we need to match Froude numbers:
g  Lm

Vp
Lm
or: Vm  Vp 
Lp
g  Lp
1
Vm  24 knot 
Thus for 1:50 scale:
Vm
50
ρm Vm Lm
When submerged, we need to match Reynolds numbers:
μm
From Table A.2, SG seawater  1.025 and μseawater  1.08  10
1.025
50
Vm  0.35 knot 


0.998
1
1.08  10
1.00  10


For submerged travel:
0.998
FDm
m
Vm  9.99
s
Vm  19.41  knot or
3
FDm

2
 ρm Vm  Am
FDp
1
2
2
 ρp  Vp  Ap
2
2
 Vp  Ap ρp  Vp Lp 


  A  ρ   V  L  Substituting in known values:
FDm ρm Vm
m  m m
  m
1.025
FDp
at 20oC. Thus for 1:50 scale:
ρp
FDp
Solving for the ratio of forces:
FDm
μp
m
3
2

ρp Lp μm
or: Vm  Vp 


ρm Lm μp
2
1
FDp
ρp  Vp  Lp
 3 N s
Under dynamically similar conditions, the drag coefficients will match:
For surface travel:
m
Vm  1.75
s
Vm  3.39 knot or

2

5
 24 50 
 3.39  1   1.29  10


1.025
0.998
2

 0.35  50   0.835
 19.41 1 


FDp
FDm
 1.29  10
FDp
FDm
5
 0.835
(on surface)
(submerged)
Problem 7.79
[Difficulty: 2]
Given:
Model size, model speed, and air temperatures.
Find:
Equivalent speed of the full scale vehicle corresponding to the different air temperatures.
Solution:
Re L 
Governing
Equation:
VL

(Reynolds Number)
where V is the air velocity, L is the length of the rocket or model, and , ν is the kinematic viscosity of air.
Subscript m corresponds to the model and r is the rocket.
Assumption: Modeling follows the Reynolds equivalency.
The given or available data is:
Lm  LR  12in
VT  100mph
VR  120mph
ft 2
ft 2
(Table A.9)
 68 F  1.62 10  4
(Table A.9)
s
s
2
4 ft
 150 F  2.09 10
(Table A.9)
s
m2
 CO2  8.3 10 6
(Figure A.3 or other source)
s
 40 F  1.47 10 4
Determine
Re L 
VR LR
R
the

120
Reynolds
Number
for
expected
maximum
speed
ft
mile 5280ft
hr


 12in 
12in
hr
mile 3600s
2
ft
1.62  10-4 
s
at
ambient
Re L  1.09  106
Re-arrange the Reynolds Number Equation for speed equivalents:
Re L 
VL

 VR  VT 
LM  68 F

LR  T
In this problem, the only term that changes is νT
Solve for speed at the low temperature: VR  VT 
LM  68 F

LR  40 F
ft 2
12in
s  110mph
 100mph 

ft 2
12in
1.47  10  4
s
1.62  10  4
temperature:
Solve for speed at the high temperature: VR  VT 
Solve for CO2:
VR  VT 
VR @ 40 F  110mph
LM  68 F

LR  CO2
LM  68 F

LR  150 F
ft 2
1.62 10
12in
s  77.5mph
 100mph 

2
ft
12in
2.09 10  4
s
4
ft 2
12in
s
 100mph 

 181mph
2
2
12in
m
ft



8.3 10 6

s  0.305m 
1.62  10 4
VR @150 F  77.5mph
VR @ CO2  181mph
Chilling the air to 40°F increases the model speed, but not enough to achieve the target. Heating the air works against the desired
outcome.
This shows that the equivalent speed can be increased by decreasing the kinematic viscosity. An inspection of figure A.3 shows that
cooling air decreases the kinematic viscosity. It also shows that CO2 has a lower kinematic viscosity than air resulting in much higher
model speeds.
Problem 7.80
Given:
Find:
Solution:
[Difficulty: 3]
The drag force on a circular cylinder immersed in a water flow can be expressed as a function of D, l, V, ρ, and μ.
Static pressure distribution can be expressed in terms of the pressure coefficient. At the minimum static
pressure, the pressure coefficient is equal to -2.4. Cavitation onset occurs at a cavitation number of 0.5.
(a) Drag force in dimensionless form as a function of all relevant variables
(b) Maximum speed at which a cylinder could be towed in water at atmospheric pressure without cavitation
The functional relationship for drag force is: FD  FD( D l V ρ μ) From the Buckingham Π-theorem, we have
6 variables and 3 repeating parameters. Therefore, we will have 3 dimensionless groups. The functional form of
these groups is:
FD
2
2
ρ V  D
The pressure coefficient is:
CP 
p  p inf
1
2
At the minimum pressure point
At the onset of cavitation


ρ  Ca  CPmin
2 p inf  p v
2
 ρ V
l
D

ρ V D 
μ
p  pv
1
2
2
 ρ V
1
2
p min  p inf   ρ Vmax  CPmin where CPmin  2.4
2
1
2
p min  p v   ρ Vmax  Ca
2
p inf 
Equating these two expressions:
Vmax 
and the cavitation number is: Ca 
 g 
where Ca  0.5
1
1
2
2
 ρ Vmax  CPmin  p v   ρ Vmax  Ca and if we solve for Vmax:
2
2
At room temperature (68 deg F): p v  0.339  psi
ρ  1.94
slug
ft
3
Substituting values we get:
Vmax 
2  ( 14.7  0.339 ) 
lbf
2
in

ft
3
1.94 slug

1
[ 0.5  ( 2.4) ]

slug ft
2
lbf  s
2

144  in
ft
2
ft
Vmax  27.1
s


Problem 7.81
Given:
Find:
Solution:
[Difficulty: 4]
A circular container partiall filled with water is rotate about its axis at constant angular velocity ω. Velocity in the
θ direction is a function of r, τ, ω, ρ, and μ.
(a) Dimensionless parameters that characterize this problem
(b) If honey would attain steady motion as quickly as water if rotated at the same angular speed
(c) Why Reynolds number is not an important parameter in scaling the steady-state motion of liquid in the
container.
The functional relationship for drag force is: Vθ  Vθ( ω r τ ρ μ) From the Buckingham Π-theorem, we have
6 variables and 3 repeating parameters. Therefore, we will have 3 dimensionless groups. The functional form of
these groups is:
Vθ
ω r
μ
From the above result Π2 
Π2  Π3 
μ
2
ρ ω r
νh  τh  νw τw
 ω τ 
containing the properties μ and ρ, and
2
ρ ω r
μ τ
2
ρ r

ν τ
2
r
νw
τh  τw
νh
Now for steady flow:
νh  τh
2
r

 g
μ
ω τ


2
 ρ ω r

Π3  ω τ containing the time τ
νw τw
and at the same radius:
2
r
Now since honey is more viscous than water, it follows that:
τh  τw
At steady state, solid body rotation exists. There are no viscous forces, and therefore, the Reynolds number would
not be important.
Problem 7.82
[Difficulty: 3]
Given:
Model of tractor-trailer truck
Find:
Drag coefficient; Drag on prototype; Model speed for dynamic similarity
Solution
:For kinematic similarity we need to ensure the geometries of model and prototype are similar, as is the incoming flow field
The drag coefficient is
For air (Table A.10) at
20oC
CD 
Fm
1
2
ρ V A
2 m m m
kg
ρm  1.21
3
m
μp  1.81  10
 5 N s

2
m
3
2
2
m
CD  2  350  N 
1.21 kg

 s   1  N s
 75 m 
kg m
2


0.1 m
CD  1.028
This is the drag coefficient for model and prototype
For the rig
1
2
Fp   ρp  Vp  Ap  CD
2
2
 Lp 

  100
Am
 Lm 
Ap
with
2
2
Ap  10 m
2
1
kg 
km 1000 m
1  hr 
N s
2
Fp 
 1.21


  90
 10 m  1.028 

2
3 
hr
1  km
3600 s 
kg m
m
For dynamic similarity
ρm Vm Lm
μm

ρp  Vp  Lp
ρp Lp μm
Lp
Vm  Vp 


 Vp 
ρm Lm μp
Lm
μp
km 1000 m
1  hr
10
Vm  90



hr
1  km
3600 s
1
For air at standard conditions, the speed of sound is
c 
Hence we have
M
1.40  286.9 
Vm
c

250
343
c
N m
kg K
 0.729
Fp  3.89 kN
m
Vm  250
s
k R T
 ( 20  273 )  K 
kg m
2
s N
c  343
m
s
which indicates compressibility is significant - this model
speed is impractical (and unnecessary)
Problem 7.83
[Difficulty: 3]
Given:
Recommended procedures for wind tunnel tests of trucks and buses suggest:
-Model frontal area less than 5% of test section area
-Reynolds number based on model width greater than 2,000,000
-Model height less than 30% of test section height
-Model projected width at maximum yaw (20 deg) less than 30% of test section width
-Air speed less than 300 ft/s to avoid compressibility effects
Model of a tractor-trailer to be tested in a tunnel 1.5 ft high x 2 ft wide. Full scale rig is 13'6" high, 8' wide, and 65'
long.
Find:
(a) Max scale for tractor-trailer model in this tunnel
(b) If adequate Reynolds number can be achieved in this facility.
Solution:
Let s be the scale ratio. Then:
Area criterion:
h m  s h p wm  s wp lm  s lp
Am  0.05  1.5 ft  2.0 ft Am  0.15 ft
Height criterion: h m  0.30  1.5 ft h m  0.45 ft
2
0.15
Therefore: s 
Therefore: s 
13.5  8
0.45
13.5
s  0.0373
s  0.0333
Width criterion: we need to account for the yaw in the model. We make a relationship for the maximum width as a
function of the model dimensions and the yaw angle and relate that to the full-scale dimensions.

wm20deg  wm cos( 20 deg)  lm sin( 20 deg)  s wp  cos( 20 deg)  lp  sin( 20 deg)
wm20deg  0.30  2.0 ft
Therefore: s 
wm20deg  0.60 ft
0.60
8  cos( 20 deg)  65  sin( 20 deg)
To determine the acceptable scale for the model, we take the smallest of these scale factors:
1
s
 49.58
We choose a round number to make the model scale easier to calculate:
For the current model conditions: Re 
Re  300 
ft
s

Vm wm
νm
s
1

  8 ft 
 50
 1.57  10 4 ft2

For standard air: νm  1.57  10
 4 ft

s  0.0202
s  0.0202
Model 
1
50
Prototype
2
Substituting known values:
s
5
Re  3.06  10 This is less than the minimum stipulated in the problem, thus:
An adequate Reynolds number can not be achieved.
Problem 7.84
[Difficulty: 4]
Power to drive a fan is a function of ρ, Q, D, and ω.
Given:
3
Find:
Solution:
2
Select primary dimensions M, L, t:
3
P
4
ρ
5
D
Q
M
L
3
t
2
L
D
Q
ρ
3
ρ
D
3
L
n = 5 parameters
ω
ω
1
t
r = 3 dimensions
m = r = 3 repeating parameters
ω
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
a
b
c
Π1  P ρ  D  ω
Thus:
M L
t
Summing exponents:
2  3 a  b  0
t:
3  c  0
b
3

M
 3
L 
a  1
L:
a
2
a
 L  
b
1
c
0 0 0
 M L t

t
The solution to this system is:
M: 1  a  0
c
Π2  Q ρ  D  ω
Thus:
L
3
t
Summing exponents:

M
 3
L 
b  5
a
 L  
b
1
t
Π1 
c  3
a0
L:
3  3 a  b  0
t:
1  c  0
4
Check using F, L, t dimensions:
b  3
Π2 
c  1
1
1
D2  8  in 
5
Q
3
ω D
3
 88 2500 
  1800 
 15

P
Thus the relationship is:
For dynamic similarity we must have geometric and kinematic similarity, and:
 Q2 ω1 
D2  D1  


 Q1 ω2 
3
ρ ω  D
0 0 0
  M L t

F L L
L
1 3
1

 t  1
 t
1
t
2 5
t
3
F t L
L
3
P
c
The solution to this system is:
M: a  0
6
ω2  1800 rpm
We will use the Buckingham pi-theorem.
P
t
ft
Condition 2: Q2  88
s
Fan diameter for condition 2 to insure dynamic similarity
1
M L
3
ft
ω1  2500 rpm Q1  15
s
Condition 1: D1  8  in
3
5
ρ ω  D
Q1
ω1  D1
3

Q2
ω2  D2
3
Q 

3
 ω D 
 f
Solving for D2
3
D2  16.10  in
Problem 7.85 (In Excel)
[Difficulty: 3]
Given: Data on model of aircraft
Find: Plot of lift vs speed of model; also of prototype
Solution:
V m (m/s)
F m (N)
10
2.2
15
4.8
20
8.7
25
13.3
30
19.6
35
26.5
40
34.5
45
43.8
This data can be fit to
Fm 
1
2
 AmCDVm
2
2
Fm  kmVm
or
From the trendline, we see that
N/(m/s)2
k m = 0.0219
(And note that the power is 1.9954 or 2.00 to three signifcant
figures, confirming the relation is quadratic)
Also, k p = 1110 k m
Hence,
kp =
2
24.3 N/(m/s)
F p = k p V m2
V p (m/s)
75
100
125
150
175
200
225
250
F p (kN)
(Trendline)
137
243
380
547
744
972
1231
1519
50
54.0
Lift vs Speed for an Airplane Model
60
y = 0.0219x1.9954
R2 = 0.9999
F m (N)
50
40
30
20
Model
10
Power Curve Fit
0
0
10
20
30
40
50
60
200
250
300
V m (m/s)
Lift vs Speed for an
Airplane Prototype
1600
F p (kN)
1400
1200
1000
800
600
400
200
0
0
50
100
150
V p (m/s)
Lift vs Speed for an Airplane Model
(Log-Log Plot)
F m (N)
100
y = 0.0219x1.9954
R2 = 0.9999
10
Model
Power Curve Fit
1
10
100
V m (m/s)
Lift vs Speed for an Airplane Prototype (Log-Log Plot)
F p (kN)
10000
1000
100
10
1
10
100
V p (m/s)
1000
Problem 7.86
Given:
Find:
Solution:
Information relating to geometrically similar model test for a centrifugal pump.
The missing values in the table
We will use the Buckingham pi-theorem.
1
∆p
2
Select primary dimensions M, L, t:
3
∆p
Q
M
L
L t
4
5
ρ
ρ
Q
3
t
2
ω
ω
ρ
ω
M
1
3
t
L
n = 5 parameters
D
D
L
r = 3 dimensions
m = r = 3 repeating parameters
D
We have n - m = 2 dimensionless groups. Setting up dimensional equations:
a
b
c
Π1  ∆p ρ  ω  D
1  3  a  c  0
t:
2  b  0
a
b
c
Π2  Q ρ  ω  D
t:
1  b  0
 
a  1
L
1
b
Check using F, L, t dimensions:
3

M
b  2
a
 
1
c  2
Π1 
∆p
2
2
ρ ω  D
b
c
0 0 0
L  M L t



t
3
t
L   
Thus:
The solution to this system is:
M: a  0
3  3 a  c  0
a
The solution to this system is:
Summing exponents:
L:
M
c
0 0 0
L  M L t



2
3
t
L t  L   
M: 1  a  0
L:

M
Thus:
Summing exponents:
6
[Difficulty: 2]
a0
F
L
2

L
4
F t
2
b  1
2 1
t 
L
2
1
c  3
L
3
t
 t
1
L
3
1
Π2 
Q
3
ω D
Thus the relationship is:
∆p
2
2
ρ ω  D
Q 

3
 ω D 
 f
The flows are geometrically similar, and we assume kinematic similarity. Thus, for dynamic similarity:
If
Qm
3
ωm Dm
Qp

ωp  Dp
 Dp 
From the first relation: Qp  Qm


ωm Dm
 
ωp
From the second relation:
then
3
2
2
3
∆pp

ρm ωm  Dm
3
m
183
Qp  0.0928


min 367
 ωm Dm 

∆pm  ∆pp 


ρp ωp Dp


ρm
∆pm
2
ρp  ωp  Dp
 150 
 50 


3
3
m
Qp  1.249 
min
2
∆pm  52.5 kPa 
2
999
800

 367  50 


 183 150 
2
∆pm  29.3 kPa
Problem 7.87
CD 
For drag we can use
L=
Model:
 =
For water
 =
D
1
V
2
As a suitable scaling area for A we use L 2
2
3
[Difficulty: 3]
CD 
A
D
1
 V 2 L2
2
ft
1.94
slug/ft3
2.10E-05
lbf·s/ft
Wave Drag
2
The data is:
V (ft/s)
D Wave (lbf)
D Friction (lbf)
Fr
Re
C D(Wave)
C D(Friction)
3.5E-05
10
0
0.022
20
0.028
0.079
30
0.112
0.169
40
0.337
0.281
50
0.674
0.45
60
0.899
0.618
70
1.237
0.731
1.017
2.77E+06
0.00E+00
2.52E-05
2.035
5.54E+06
8.02E-06
2.26E-05
3.052
8.31E+06
1.43E-05
2.15E-05
4.070
1.11E+07
2.41E-05
2.01E-05
5.087
1.39E+07
3.09E-05
2.06E-05
6.105
1.66E+07
2.86E-05
1.97E-05
7.122
1.94E+07
2.89E-05
1.71E-05
3.0E-05
2.5E-05
2.0E-05
CD
1.5E-05
1.0E-05
5.0E-06
0.0E+00
The friction drag coefficient becomes a constant, as expected, at high Re .
The wave drag coefficient appears to be linear with Fr , over most values
0
1
2
3
4
5
6
7
8
Fr
L=
Ship:
V (knot)
V (ft/s)
Fr
Re
15
25.32
0.364
3.51E+08
150
ft
20
33.76
0.486
4.68E+08
D
1
 V 2 L2 C D
2
Friction Drag
3.5E-05
3.0E-05
Hence for the ship we have very high Re , and low Fr .
From the graph we see the friction C D levels out at about 1.9 x 10
From the graph we see the wave C D is negligibly small
2.5E-05
-5
2.0E-05
CD
C D(Wave)
C D(Friction)
0
1.90E-05
0
1.90E-05
D Wave (lbf)
D Friction (lbf)
0
266
0
473
D Total (lbf)
266
473
1.5E-05
1.0E-05
5.0E-06
0.0E+00
0.0.E+00
5.0.E+06
1.0.E+07
1.5.E+07
Re
2.0.E+07
2.5.E+07
Problem 7.88 (In Excel)
[Difficulty: 4]
Given: Data on centrifugal water pump
Find:  groups; plot pressure head vs flow rate for range of speeds
Solution:
We will use the workbook of Example 7.1, modified for the current problem
n
r
m =r
n -m
The number of parameters is:
The number of primary dimensions is:
The number of repeat parameters is:
The number of  groups is:
=5
=3
=3
=2
Enter the dimensions (M, L, t) of
the repeating parameters, and of up to
four other parameters (for up to four  groups).
The spreadsheet will compute the exponents a , b , and c for each.
REPEATING PARAMETERS: Choose , g , d
M
1


D
L
-3
t
-1
1
 GROUPS:
p
 1:
M
1
L
-1
a =
b =
c =
-1
-2
-2
t
-2
Q
 2:
M
0
L
3
a =
b =
c =
0
-1
-3
M
0
L
0
a =
b =
c =
0
0
0
t
-1
The following  groups from Example 7.1 are not used:
 3:
Hence
1 
p
 2 D 2
and
2 
Q
M
0
L
0
a =
b =
c =
0
0
0
t
0
 4:
with 1 = f(2).
D 3
Based on the plotted data, it looks like the relation between 1 and 2 may be parabolic
Hence
p
 Q 
 Q 
 a  b
  c

 D
 D 3 
 D 3 
2
2
2
The data is
Q (ft3/min)
p (psf)
0
50
75
100
120
140
150
165
7.54
7.29
6.85
6.12
4.80
3.03
2.38
1.23
t
0


D =
Q /(D 3)
p /( D )
2
2
1.94
800
1
(D is not given; use D = 1 ft as a scale)
0.00000
0.00995
0.01492
0.01989
0.02387
0.02785
0.02984
0.03283
0.000554
0.000535
0.000503
0.000449
0.000353
0.000223
0.000175
0.000090
Centifugal Pump Data and Trendline
0.0006
2 2
p /( D )
slug/ft3
rpm
ft
0.0005
0.0004
Pump Data
0.0003
Parabolic Fit
0.0002
0.0001
0.0000
0.000
0.005
0.010
0.015
0.020
0.025
0.030
0.035
3
Q /(D )
p /( 2D 2) = -0.6302 (Q /( D 3))2 + 0.006476 (Q /( D 3)) + 0.0005490
The curve fit result is:
From the Trendline analysis
a = 0.000549
b = 0.006476
c = -0.6302
and
2

 Q 
 Q  

p   2 D 2 a  b 
c




 D 3 
 D 3  

Finally, data at 500 and 1000 rpm can be calculated and plotted

Q (ft3/min)
p (kPa)

3
Q (ft /min)
p (kPa)
600
rpm
0
20
40
60
80
100
120
132
4.20
4.33
4.19
3.77
3.08
2.12
0.89
0.00
1200
rpm
0
50
75
100
120
140
150
165
16.82
17.29
16.88
16.05
15.09
13.85
13.12
11.91
p (psf)
Centifugal Pump Curves
Pump Data at 800 rpm
20
18
16
14
12
10
8
6
4
2
0
Pump Curve at 600 rpm
Pump Curve at 1200 rpm
0
20
40
60
80
100
3
Q (ft /min)
120
140
160
180
Problem 7.89
Given:
Model of water pump
Find:
Model head, flow rate and diameter
Solution:
From Buckingham Π
h
2
2
 Q ρ ω D2 


 ω D3
μ 


 f
ω D
Neglecting viscous effects
Qm
3
ωm Dm
Hence if
P
and
3
hm
then
ωp  Dp
3
2
3
ωm
3
2
2
then
and
Pm
We can find Pp from
ωm
Dm
3
5
2
Pm
and
ωp  Dp
2
3
From Eq 2
3
ωp  Dp
5
(2)
5
Dm
kg
(3)
3
J
Pp  ρ Q h  1000
 0.75
 15
 11.25 kW
3
s
kg
m
Pm
Pp
From Eq 1
Pp
(1)
m
1
From Eq 3
5
ωm  Dm

3
5
3
Dm
1000  Dm





 8

3
5
5
Pp
 500  D 5
ωp Dp
Dp
p
ωm
hp

2
2
2
Dm
1000  Dm





 4

2
2
2
hp
 500  D 2
ωp Dp
Dp
p
hm
2
ωm  Dm
 Dm 
 Dm 
1000  Dm 




 2 



Qp
ωp Dp
500
 
 Dp 
 Dp 
Qm
5
 Q ρ ω D2 


 ω D3
μ 


 f
ω D
Qp

[Difficulty: 3]
5
 8
Dm
Dp
 1 Pm 
Dm  Dp   

 8 Pp 
so
 Dm 
Qm  Qp  2  

 Dp 
so
 Dm 
hm  hp 4 

 Dp 
5
 Dm 
 2 

Qp
 Dp 
Qm
 Dm 
 4 

hp
 Dp 
hm
so
3
2
1
5
Dm  0.25 m 
3
5
 1  2.25 
 8 11.25


3
m
Qm  0.75
 2
s
2
h m  15
J
kg
 4
 0.12
 0.25


 0.12


 0.25
3
Dm  0.120 m
3
m
Qm  0.166
s
2
h m  13.8
J
kg
Problem 7.90
[Difficulty: 3]
Given:
Data on model propeller
Find:
Speed, thrust and torque on prototype
Solution:
We will use the Buckingham Pi-theorem to find the functional relationships between these variables. Neglecting the
effects of viscosity:
1
F
2
Select primary dimensions M, L, t:
3
F
T
M L
M L
t
4
5
ρ
T
2
ρ
t
D
2
2
V
ρ
V
M
L
3
t
L
D
ω
D
ω
n = 6 parameters
1
L
r = 3 dimensions
t
m = r = 3 repeating parameters
ω
We have n - m = 3 dimensionless groups. Setting up dimensional equations:
a
b
c
Π1  F ρ  D  ω
Thus:
M L
t
Summing exponents:
2
1  3 a  b  0
t:
2  c  0
a
b
c
Π2  T ρ  D  ω
Thus:
M L
Summing exponents:
t:
2  c  0
a
b
c
Π3  V ρ  D  ω
 L  
 3
L 
b
1
c
0 0 0
  M L t

t
2
2

b  4
M
a
 3
L 
 L  
b
c  2
1
t
a  1
Thus:
L
t

M
 3
L 
b  5
a
 L  
b
1
t
c
Π1 
F
4
2
ρ D  ω
c
0 0 0
  M L t

The solution to this system is:
M: 1  a  0
2  3 a  b  0
a
a  1
t
L:
M
The solution to this system is:
M: 1  a  0
L:

c  2
0 0 0
  M L t

Π2 
T
5
2
ρ D  ω
Summing exponents:
The solution to this system is:
a0
M: 0  a  0
L:
1  3 a  b  0
t:
1  c  0
L
F
6 Check using F, L, t dimensions:
4
F t
2
1

L
4
b  1
2
t  1
F L
V
Π3 
D ω
c  1
L
4
F t
2

1
L
L 1
 t  1
t L
2
5
t  1
For dynamically similar conditions:
Vm
Dm ωm

Vp
Dp  ωp
Fm
4
2

ρm Dm  ωm
Vp Dm
ωp  ωm

Vm Dp
Thus:
Fp
4
ρp  Dp  ωp
Thus:
2
130 1
ωp  1800 rpm 

50
8
 Dp 
Fp  Fm


ρm Dm
 
ρp
4
 ωp 


 ωm 
ωp  585  rpm
2
1
Fp  100  N  
1
4
 8    585 
1 

   1800 
2
Fp  43.3 kN
Tm
5
2
ρm Dm  ωm

Tp
5
ρp  Dp  ωp
Thus:
2
5
 Dp   ωp 
Tp  Tm

  
ρm Dm
   ωm 
ρp
2
Tp  10 N m 
1
1
5

 8    585 
1 

   1800 
2
Tp  34.6 kN m
Problem 7.91
[Difficulty: 3]
Given:
For a marine propeller (Problem 7.40) the thrust force is: FT  FT( ρ D V g ω p μ)
For ship size propellers viscous and pressure effects can be neglected. Assume that power and torque depend on
the same parameters as thrust.
Find:
Solution:
Scaling laws for propellers that relate thrust, power and torque to other variables
We will use the Buckingham pi-theorem. Based on the simplifications given above:
1
FT
2
Select primary dimensions F, L, t:
3
FT
P
P
4
5
ρ
F L
t
ω
D
ρ
T
F L
F
ρ
T
F t
L
D
2
4
L
V
g
ω
V
g
ω
L
L
1
t
2
t
t
n = 8 parameters
r = 3 dimensions
m = r = 3 repeating parameters
D
We have n - m = 5 dimensionless groups (3 dependent, 2 independent). Setting up dimensional equations:
a
a
b
c
Π1  FT ρ  ω  D
Thus:
Summing exponents:
F:
1a0
L:
4  a  c  0
t:
2 a  b  0
 F t2   1 b c 0 0 0
   L  F L t
F 
 L4   t 


The solution to this system is:
a  1
b  2
c  2
Π1 
FT
2
4
ρ ω  D
a
a
b
c
Π2  P ρ  ω  D
Summing exponents:
F:
1a0
L:
1  4 a  c  0
t:
1  2  a  b  0
Thus:
 F t2   1 b c 0 0 0
     L  F  L  t

t  4  t
L 
F L
The solution to this system is:
a  1
b  3
c  5
Π2 
P
3
5
ρ ω  D
a
a
b
c
Π3  T ρ  ω  D
 F t2   1 b c 0 0 0
   L  F L t
F L 
 L4   t 


Thus:
Summing exponents:
F:
1a0
L:
1  4 a  c  0
t:
2 a  b  0
The solution to this system is:
a  1
b  2
Π3 
c  5
T
2
5
ρ ω  D
a
a
b
c
Π4  V ρ  ω  D
 F t2   1 b c 0 0 0
     L  F  L  t

t  4  t
L 
L
Thus:
Summing exponents:
F:
a0
L:
1  4 a  c  0
t:
1  2  a  b  0
a
The solution to this system is:
a0
b
c
Π5  g  ρ  ω  D
 F t2 


t  4 
L 
L
Thus:
Summing exponents:
F:
a0
L:
1  4 a  c  0
t:
1  2  a  b  0
Check using M, L, t dimensions:
a
 
1
t
c  1
b
c
0 0 0
 L  F L t

The solution to this system is:
a0
3
6
b  1
V
Π4 
ω D
b  1
M L L 2 1
 t 
1
4
2 M
L
t
L
t
 t
1
L
1
Π5 
c  1
M L
t
2
3

L
3
M
3 1
t 
L
5
1
M L
t
2
2

g
2
ω D
L
3
M
2 1
t 
L
5
1
L 2 1
t 
1
2
2
L
t
Based on the dependent and independent variables, the "scaling laws" are:
FT
V
g 

 f1 

2 4
ω D 2 
ρ ω  D
ω D 

V
g 

 f2 

ω D 2 
3 5
ρ ω  D
ω D 

P
g 
V
 f3 


2 5
ω D 2 
ρ ω  D
ω D 

T
Problem 7.92
Given:
Water drop mechanism
Find:
Difference between small and large scale drops
[Difficulty: 2]
Solution:

Given relation
3
5
d  D ( We)

 ρ V2 D 

 D 
 σ 

For dynamic similarity
 ρ V 2 D 
m m
Dm 

dm
σ



dp

 ρ V 2 D 
p
p
Dp  

σ


2
Hence
dm
dp
5

 1   5
 20 
1
 
 

3
5
3
5
3
2
 Dm 


 Dp 
5

 Vm 


 Vp 
6
5
where d p stands for dprototype not the original
d p!
5
6
5
The small scale droplets are 4.4% of the size of the large scale
dm
dp
 0.044
Problem 7.93
Given:
Find:
Solution:
[Difficulty: 2]
Kinetic energy ratio for a wind tunnel is the ratio of the kinetic energy flux in the test section to the drive power
Kinetic energy ratio for the 40 ft x 80 ft tunnel at NASA-Ames
nmi 6080 ft
hr
ft
From the text: P  125000 hp Vmax  300 


Vmax  507 
hr
nmi
3600 s
s
2
m
Therefore, the kinetic energy ratio is: KEratio 
V
2
P
2

( ρ V A)  V
2 P
3
3

ρ A V
2 P
Assuming standard conditions
and substituting values:
2
1
slug
ft
hp s
lbf  s
1
KEratio 
 0.00238 
 ( 40 ft  80 ft)   507   


2
3
s
slug ft
125000 hp 550  ft lbf

ft
KEratio  7.22
Problem 7.94
[Difficulty: 3]
Given:
A scale model of a truck is tested in a wind tunnel. The axial pressure gradient and frontal area of the prototype
are known. Drag coefficient is 0.85.
Find:
(a) Horizontal buoyancy correction
(b) Express this correction as a fraction of the measured drag force.
Solution:
The horizontal buoyancy force is the difference in the pressure force between the front and back of the model due
to the pressure gradient in the tunnel:


∆p
FB  p f  p b  A 
L A
∆L m m
where:
lbf
60 ft 110  ft
Thus: FB  0.07


2
16
2
ft  ft
16
Lm 
Lp
16
Am 
Ap
2
16
2
FB  0.113  lbf
The horizontal buoyancy correction should be added to the measured drag force on the model. The measured drag
force on the model is given by:
A
1
1
2
2 p
FDm   ρ V  Am CD   ρ V 
 CD
2
2
2
16
When we substitute in known values we get:
2
2
2
ft 
lbf  s
110  ft

FDm 
 0.00238 
 0.85 
  250   
2
3
s
slug ft
2

ft
16
1
slug
Therefore the ratio of the forces is:
DragRatio 
0.113
27.16
FDm  27.16  lbf
DragRatio  0.42 %
Problem 7.95
Given:
Flapping flag on a flagpole
Find:
Explanation of the flapping
[Difficulty: 4]
Solution:
Discussion: The natural wind contains significant fluctuations in air speed and direction. These fluctuations tend to disturb the flag
from an initially plane position.
When the flag is bent or curved from the plane position, the flow nearby must follow its contour. Flow over a convex surface tends to
be faster, and have lower pressure, than flow over a concave curved surface. The resulting pressure forces tend to exaggerate the
curvature of the flag. The result is a seemingly random "flapping" motion of the flag.
The rope or chain used to raise the flag may also flap in the wind. It is much more likely to exhibit a periodic motion than the flag
itself. The rope is quite close to the flag pole, where it is influenced by any vortices shed from the pole. If the Reynolds number is
such that periodic vortices are shed from the pole, they will tend to make the rope move with the same frequency. This accounts for
the periodic thump of a rope or clank of a chain against the pole.
The vortex shedding phenomenon is characterized by the Strouhal number, St = fD/V∞, where f is the vortex shedding frequency, D is
the pole diameter, and D is the wind speed. The Strouhal number is constant at approximately 0.2 over a broad range of Reynolds
numbers.
Problem 7.96
[Difficulty: 3]
Given:
A 1:16 scale model of a bus (152 mm x 200 mm x 762 mm) is tested in a wind tunnel at 26.5 m/s. Drag force is 6.09
N. The axial pressure gradient is -11.8 N/m2/m.
Find:
(a) Horizontal buoyancy correction
(b) Drag coefficient for the model
(c) Aerodynamic drag on the prototype at 100 kph on a calm day.
Solution:
The horizontal buoyancy force is the difference in the pressure force between the front and back of the model due
to the pressure gradient in the tunnel:


dp
FB  p f  p b  A 
L A
dx m m
2
Am  152  mm  200  mm Am  30400  mm
where:
N
2
Thus: FB  11.8
 762  mm  30400  mm 
2
m m
So the corrected drag force is:
 m 
 1000 mm 


FB  0.273 N
FDc  6.09 N  0.273  N FDc  5.817 N
The corrected model drag coefficient would then be:
CDm 
FDc
1
2
3
CDm  2  5.82 N 
3
Substituting in values:
2
 ρ V  Am
2
m
1.23 kg

2
1000 mm 
1
kg m
 s  

 
 26.5 m 

m
2 
2



30400  mm
N s
CDm  0.443
If we assume that the test was conducted at high enough Reynolds number, then the drag coefficient should be the
same at both scales, i.e.: C
Dp  CDm
1
2
FDp   ρ V  Ap  CDp
2
where
m
2 2

Ap  30400  mm  16  

1000

mm


2
2
2
Ap  7.782  m
2
N s
1
kg 
km 1000 m
hr
2

  100 
FDp 
 1.23


  7.782  m  0.443 
kg m
2
3 
hr
km 3600  s 
m
FDp  1.636  kN
(The rolling resistance must also be included to obtain the total tractive effort needed to propel the vehicle.)
Problem 7.97
[Difficulty: 5]
Discussion: The equation given in Problem 7.2 contains three terms. The first term contains surface tension and gives a
speed inversely proportional to wavelength. These terms will be important when small wavelengths are
considered.
The second term contains gravity and gives a speed proportional to wavelength. This term will be important
when long wavelengths are considered.
The argument of the hyperbolic tangent is proportional to water depth and inversely proportional to
wavelength. For small wavelengths this term should approach unity since the hyperbolic tangent of a large
number approaches one.
The governing equation is:
2
c 
 σ  2 π  g λ   tanh 2 π h 
ρ λ

 λ 
2 π 



The relevant physical parameters are:
g  9.81
m
2
ρ  999 
s
kg
3
σ  0.0728
m
A plot of the wave speed versus wavelength at different depths is shown here:
Wave Speed versus Wavelength
h = 1 mm
h = 5 mm
h = 10 mm
h > 50 mm
Wave Speed (m/s)
0.4
0.3
0.2
0.1
0
0
0.05
Wavelength (m)
0.1
N
m
Problem 8.1
Given:
Air entering duct
Find:
Flow rate for turbulence; Entrance length
Solution:
The basic equations are
The given data is
Re 
V D
π
2
Recrit  2300
Q
D  125 mm
From Table A.10
ν  2.29  10
Llaminar  0.06 Recrit D
or, for turbulent, Lturb = 25D to 40D
ν
Q
π
Hence
[Difficulty: 1]
Recrit 
4
4
D V
2
5 m

s
D
2
D
or
ν
Q 
Recrit π ν  D
4
For laminar flow
Llaminar  0.06 Recrit D
Llaminar  17.3 m
For turbulent flow
Lmin  25 D
Lmin  3.13 m
3
3m
Q  5.171  10
Lmax  40 D
s
Lmax  5.00 m
Problem 8.2
[Difficulty: 2]
Problem 8.3
[Difficulty: 3]
Given:
Air entering pipe system
Find:
Flow rate for turbulence in each section; Which become fully developed
Solution:
From Table A.10
The given data is
ν = 1.69 × 10
2
−5 m
⋅
L = 2⋅ m
s
D1 = 25⋅ mm
D2 = 15⋅ mm
D3 = 10⋅ mm
or
Q=
Recrit = 2300
The critical Reynolds number is
Writing the Reynolds number as a function of flow rate
Re =
V⋅ D
ν
=
Q
π
4
⋅
D
2 ν
⋅D
Re⋅ π⋅ ν⋅ D
4
Then the flow rates for turbulence to begin in each section of pipe are
Q1 =
Q2 =
Q3 =
Recrit⋅ π⋅ ν⋅ D1
Q1 = 7.63 × 10
4
Recrit⋅ π⋅ ν⋅ D2
Q2 = 4.58 × 10
4
Recrit⋅ π⋅ ν⋅ D3
Q3 = 3.05 × 10
4
3
−4m
s
3
−4m
s
3
−4m
s
Hence, smallest pipe becomes turbulent first, then second, then the largest.
For the smallest pipe transitioning to turbulence (Q3)
For pipe 3
Re3 = 2300
Llaminar = 0.06⋅ Re3 ⋅ D3
or, for turbulent,
Lmin = 25⋅ D3
Lmin = 0.25 m
For pipes 1 and 2
Llaminar = 0.06⋅ ⎜
⎛ 4⋅ Q3 ⎞
⎝
π⋅ ν⋅ D1
⎠
⋅ D1
Llaminar = 1.38 m
Lmax = 40⋅ D3
Lmax = 0.4 m
Llaminar = 1.38 m
Llaminar < L: Fully developed
Lmax/min < L: Fully developed
Llaminar < L: Fully developed
⎛ 4⋅ Q3 ⎞
Llaminar = 0.06⋅ ⎜
⎝
π⋅ ν⋅ D2
⎠
⋅ D2
Llaminar = 1.38 m
Llaminar < L: Fully developed
For the middle pipe transitioning to turbulence (Q2)
For pipe 2
Re2 = 2300
Llaminar = 0.06⋅ Re2 ⋅ D2
Llaminar = 2.07 m
or, for turbulent,
Lmin = 25⋅ D2
Lmin = 1.23⋅ ft
Lmax = 40⋅ D2
Llaminar > L: Not fully developed
Lmax = 0.6 m
Lmax/min < L: Fully developed
For pipes 1 and 3
⎛ 4 ⋅ Q2 ⎞
L1 = 0.06⋅ ⎜
⎝
π⋅ ν⋅ D1
L3min = 25⋅ D3
⎠
⋅ D1
L3min = 0.25⋅ m
L1 = 2.07⋅ m
L3max = 40⋅ D3
Llaminar > L: Not fully developed
L3max = 0.4 m
Lmax/min < L: Fully developed
For the large pipe transitioning to turbulence (Q1)
For pipe 1
Re1 = 2300
Llaminar = 0.06⋅ Re1 ⋅ D1
Llaminar = 3.45 m
or, for turbulent,
Lmin = 25⋅ D1
Lmin = 2.05⋅ ft
Lmax = 40⋅ D1
Llaminar > L: Not fully developed
Lmax = 1.00 m
Lmax/min < L: Fully developed
For pipes 2 and 3
L2min = 25⋅ D2
L2min = 1.23⋅ ft
L2max = 40⋅ D2
L2max = 0.6 m
Lmax/min < L: Fully developed
L3min = 25⋅ D3
L3min = 0.82⋅ ft
L3max = 40⋅ D3
L3max = 0.4 m
Lmax/min < L: Fully developed
Problem 8.4
[Difficulty: 2]
Given:
That transition to turbulence occurs at about Re = 2300
Find:
Plots of average velocity and volume and mass flow rates for turbulence for air and water
Solution:
The basic equations are
From Tables A.8 and A.10
For the average velocity
V⋅ D
Re =
Recrit = 2300
ν
kg
ρair = 1.23⋅
3
m
V=
νair = 1.45 × 10
⋅
s
Vair =
2
−5 m
⋅
2
s
Vair =
D
Hence for air
Vw =
⋅
π
4
2
⋅D ⋅V =
π
π
4
2
Vw =
Recrit⋅ ν
D
2
−5 m
Qair =
× 2300 × 1.45⋅ 10
4
⋅
s
=
π⋅ Recrit⋅ ν
4
m
s
D
⋅D
2
⋅D
2
For water
D
2
s
⋅D ⋅
s
0.00262 ⋅
D
Q = A⋅ V =
m
0.0334⋅
2
−6 m
For the volume flow rates
νw = 1.14 × 10
D
2300 × 1.14 × 10
For water
2
−6 m
kg
ρw = 999 ⋅
3
m
Recrit⋅ ν
2300 × 1.45 × 10
Hence for air
2
−5 m
π
−6 m
Qw =
× 2300 × 1.14⋅ 10 ⋅
⋅D
4
s
m
Qair = 0.0262⋅
×D
s
2
m
Qw = 0.00206 ⋅
×D
s
Finally, the mass flow rates are obtained from volume flow rates
mair = ρair⋅ Qair
kg
mair = 0.0322⋅
×D
m⋅ s
mw = ρw⋅ Qw
kg
mw = 2.06⋅
×D
m⋅ s
These results can be plotted in Excel as shown below in the next two pages
⋅
s
From Tables A.8 and A.10 the data required is
◊ air =
1.23
kg/m 3
2
◊ air = 1.45E-05 m /s
◊w =
999
kg/m
3
◊ w = 1.14E-06 m /s
2
0.0001
0.001
0.01
0.05
V air (m/s) 333.500
33.350
3.335
0.667
2.62
0.262
D (m)
V w (m/s)
26.2
1.0
2.5
5.0
7.5
10.0
3.34E-02 1.33E-02 6.67E-03 4.45E-03 3.34E-03
5.24E-02 2.62E-03 1.05E-03 5.24E-04 3.50E-04 2.62E-04
3
Q air (m /s) 2.62E-06 2.62E-05 2.62E-04 1.31E-03 2.62E-02 6.55E-02 1.31E-01 1.96E-01 2.62E-01
Q w (m 3/s) 2.06E-07 2.06E-06 2.06E-05 1.03E-04 2.06E-03 5.15E-03 1.03E-02 1.54E-02 2.06E-02
m air (kg/s) 3.22E-06 3.22E-05 3.22E-04 1.61E-03 3.22E-02 8.05E-02 1.61E-01 2.42E-01 3.22E-01
m w (kg/s) 2.06E-04 2.06E-03 2.06E-02 1.03E-01 2.06E+00 5.14E+00 1.03E+01 1.54E+01 2.06E+01
Average Velocity for Turbulence in a Pipe
1.E+04
V (m/s)
1.E+02
Velocity (Air)
Velocity (Water)
1.E+00
1.E-02
1.E-04
1.E-04
1.E-03
1.E-02
1.E-01
D (m )
1.E+00
1.E+01
Flow Rate for Turbulence in a Pipe
Q (m3/s)
1.E+01
1.E-01
Flow Rate (Air)
Flow Rate (Water)
1.E-03
1.E-05
1.E-07
1 .E-04
1.E-03
1.E-02
1.E-01
1.E+00
1.E+01
D (m)
Mass Flow Rate for Turbulence in a Pipe
m flow (kg/s)
1.E+02
1.E+00
Mas s Flow Rate (Air)
Mas s Flow Rate (Water)
1.E-02
1.E-04
1.E-06
1.E-04
1.E-03
1.E-02
1.E-01
D (m)
1.E+00
1.E+01
Problem 8.5
[Difficulty: 4] Part 1/2
Problem 8.5
[Difficulty: 4] Part 2/2
Problem 8.6
[Difficulty: 2]
Problem 8.7
[Difficulty: 2]
Problem 8.8
[Difficulty: 3]
Problem 8.9
[Difficulty: 3]
D
p1
F
a
L
Given:
Piston cylinder assembly
Find:
Rate of oil leak
Solution:
Q
Basic equation
l
3
3
=
a ⋅ ∆p
Q=
12⋅ μ⋅ L
π⋅ D⋅ a ⋅ ∆p
(from Eq. 8.6c; we assume laminar flow and verify
this is correct after solving)
12⋅ μ⋅ L
F
4⋅ F
∆p = p 1 − p atm =
=
A
2
π⋅ D
For the system
4
∆p =
× 4500⋅ lbf ×
π
⎛ 1 × 12⋅ in ⎞
⎜ 4⋅ in 1⋅ ft
⎝
⎠
2
μ = 0.06 × 0.0209⋅
At 120 oF (about 50oC), from Fig. A.2
∆p = 358 ⋅ psi
lbf ⋅ s
ft
Q =
π
12
× 4 ⋅ in × ⎜⎛ 0.001 ⋅ in ×
Check Re:
⎝
V=
Re =
Q
A
μ = 1.25 × 10
2
− 3 lbf ⋅ s
⋅
ft
2
2
2
3
3
ft
1
− 5 ft
⎞ × 358 ⋅ lbf × 144 ⋅ in ×
× Q = 1.25 × 10 ⋅
12⋅ in ⎠
2
−3
2 ⋅ in
s
2
1 ⋅ ft
1.25 × 10
lbf ⋅ s
in
1 ⋅ ft
=
Q
a ⋅ π⋅ D
V⋅ a
1
π
× 1.25 × 10
ν = 6 × 10
ν
Re = 0.143 ⋅
V =
ft
s
× 0.001 ⋅ in ×
1 ⋅ ft
12⋅ in
−5
×
− 5 ft
3
s
× 10.8
ft
×
1
.001⋅ in
×
1
4 ⋅ in
2
×
⎛ 12⋅ in ⎞
⎜ 1⋅ ft
⎝
⎠
− 4 ft
ν = 6.48 × 10
s
s
−4
6.48 × 10
ft
2
⋅
Re = 0.0184
3
Q = 0.0216⋅
2
V = 0.143 ⋅
Q
⎛ π⋅ D2 ⎞
⎜
⎝ 4 ⎠
3
4
− 5 ft
×
Vp =
× 1.25 × 10
π
s
12⋅ in ⎞
⎛ 1
⎜ 4 ⋅ in × 1 ⋅ ft
⎝
⎠
s
ft
s
2
(at 120 oF, from Fig. A.3)
s
so flow is very much laminar
The speed of the piston is approximately
Vp =
in
2
The piston motion is negligible so our assumption of flow between parallel plates is reasonable
− 4 ft
Vp = 1.432 × 10
⋅
s
Problem 8.10
[Difficulty: 2]
Problem 8.11
[Difficulty: 2]
y
2h
Given:
Laminar flow between flat plates
Find:
Shear stress on upper plate; Volume flow rate per width
Solution:
du
τyx = μ⋅
dy
Then
τyx =
At the upper surface
y=h
The volume flow rate is
⌠
h
2
⌠
⌠
h ⋅ b dp ⎮
Q = ⎮ u dA = ⎮ u ⋅ b dy = −
⋅ ⋅⎮
⌡
2⋅ μ dx ⎮
⌡
−h
⌡
−h
2
2
⋅
u(y) = −
dp
dx
h
2
Basic equation
⋅
⎡
dp
2⋅ μ dx
⋅ ⎢1 −
⎣
b
=−
2
⎛ y ⎞ ⎥⎤
⎜h
⎝ ⎠⎦
(from Eq. 8.7)
2⋅ y ⎞
dp
⎜ 2 = −y⋅ dx
⎝ h ⎠
⋅⎛−
1⋅ m
3 N
τyx = −1.5⋅ mm ×
× 1.25 × 10 ⋅
1000⋅ mm
2
m ⋅m
h
−h
Q
x
2
3
× ⎛⎜ 1.5⋅ mm ×
⎝
1⋅ m
3
⎡
⎢1 −
⎣
2
⎛ y ⎞ ⎥⎤ dy
⎜h
⎝ ⎠⎦
τyx = −1.88Pa
3
Q= −
2
⎞ × 1.25 × 103⋅ N × m
1000⋅ mm ⎠
0.5⋅ N⋅ s
2
m ⋅m
Q
b
2⋅ h ⋅ b dp
⋅
3⋅ μ dx
= −5.63 × 10
2
−6 m
s
Problem 8.12
Given:
Piston-cylinder assembly
Find:
Mass supported by piston
[Difficulty: 3]
Solution:
Basic equation
Q
l
Available data
3

a  ∆p
This is the equation for pressure-driven flow between parallel plates; for a small gap a,
the flow between the piston and cylinder can be modeled this way, with l = πD
12 μ L
L  4  inD  4  in
a  0.001  in
From Fig. A.2, SAE10 oil at 20oF is
Q  0.1 gpm
μ  0.1
N s
68 °F  20 °C
 3 lbf  s
μ  2.089  10
or
2

m
Hence, solving for ∆p
∆p 
12 μ L Q
π D a
2
Note the following
M 
π D
4

2
4
∆p  2.133  10  psi
3
F  ∆p A  ∆p
A force balance for the piston involves the net pressure force
Hence
ft
∆p
π
4
Q
Vave 
a  π D
ft
Vave  2.55
s
Hence an estimate of the Reynolds number in the gap is
W  M g
and the weight
5
M  8331 slug
g
2
D
M  2.68  10  lb
ν  10
Re 
2
4 m

s
a Vave
ν
ν  1.076  10
 3 ft

2
s
Re  0.198
This is a highly viscous flow; it can be shown that the force on the piston due to this motion is much less than that due to ∆p!
Note also that the piston speed is
Vpiston 
Vpiston
Vave
4 Q
2
π D
 0.1 %
ft
Vpiston  0.00255 
s
so the approximation of stationary walls is valid
Problem 8.13
[Difficulty: 3]
Problem 8.14
[Difficulty: 3]
Problem 8.15
[Difficulty: 3]
Given:
Hydrostatic bearing
Find:
Required pad width; Pressure gradient; Gap height
Solution:
Basic equation
Q
l
Available data

h
3
12 μ
 
dp 
 dx 
F  1000 lbf
l  1  ft
p i  35 psi
212 °F  100 °C
At 100 oC from Fig. A.2, for SAE 10-30
(F is the load on width l)
μ  0.01
Q  2.5
N s
gal
per ft
hr
 4 lbf  s
μ  2.089  10
2

m
ft
2
For a laminar flow (we will verify this assumption later), the pressure gradient is constant
2 x 
p ( x )  p i  1 
W

where p i = 35 psi is the inlet pressure (gage), and x = 0 to W/2

F  l   p dx

Hence the total force in the y direction due to pressure is
where b is the pad width into the paper
W
2
2 x 

dx
F  2 l 
p i  1 
W


F
1
2
 p i l W
0
W 
This must be equal to the applied load F. Hence
dp
The pressure gradient is then

dx
∆p
W

2  ∆p
W
 2 
2 F

pi l
35 lbf
2
W  0.397  ft

in
1
0.397  ft
 176 
psia
ft
2
Q
From the basic equation
l
Check Re:
From Fig. A.3
Re 
V D
ν
ν  1.2 10


h
3
12 μ
1
 
dp 
 dx 
Q 

 12 μ l
h  

dp

dx


we can solve for
3
h  2.51  10
3
 in
D Q
h Q
  
ν A
ν l h
2
5 m

s
 4 ft
ν  1.29  10

2
s
Re 
Q
ν l
Re  0.72
so flow is very laminar
Problem 8.16
[Difficulty: 3]
Problem 8.17
Given:
Navier-Stokes Equations
Find:
Derivation of Eq. 8.5
[Difficulty: 2]
Solution:
The Navier-Stokes equations are
4
3
∂u ∂v ∂w
+
+
=0
∂x ∂y ∂z
1
4
5
3
(5.1c)
6
4
3
⎛∂ u ∂ u ∂ u⎞
⎛ ∂u
∂u
∂u
∂u ⎞
∂p
+ µ ⎜⎜ 2 + 2 + 2 ⎟⎟
+u
+v
+ w ⎟⎟ = ρg x −
∂z ⎠
∂y
∂x
∂x
∂y
∂z ⎠
⎝ ∂t
⎝ ∂x
2
ρ ⎜⎜
1
4
5
3
2
4
2
5
(5.27a)
3
⎛ ∂ 2v ∂ 2v ∂ 2v ⎞
⎛ ∂v
∂p
∂v ⎞
∂v
∂v
ρ ⎜⎜ + u + v + w ⎟⎟ = ρg y − + µ ⎜⎜ 2 + 2 + 2 ⎟⎟
∂z ⎠
∂y
∂y
∂z ⎠
∂y
∂x
⎝ ∂t
⎝ ∂x
1
3
3
3
3
3
3
(5.27b)
3
3
⎛∂ w ∂ w ∂ w⎞
⎛ ∂w
∂p
∂w ⎞
∂w
∂w
+ µ ⎜⎜ 2 + 2 + 2 ⎟⎟
⎟⎟ = ρg z −
+w
+v
+u
∂z ⎠
∂y
∂z
∂z ⎠
∂y
∂x
⎝ ∂t
⎝ ∂x
2
ρ ⎜⎜
2
2
(5.27c)
The following assumptions have been applied:
(1) Steady flow (given).
(2) Incompressible flow; ρ = constant.
(3) No flow or variation of properties in the z direction; w= 0 and ∂/∂z = 0.
(4) Fully developed flow, so no properties except pressure p vary in the x direction; ∂/∂x = 0.
(5) See analysis below.
(6) No body force in the x direction; gx = 0
Assumption (1) eliminates time variations in any fluid property. Assumption (2) eliminates space variations in density. Assumption
(3) states that there is no z component of velocity and no property variations in the z direction. All terms in the z component of the
Navier–Stokes equation cancel. After assumption (4) is applied, the continuity equation reduces to ∂v/∂y = 0. Assumptions (3) and (4)
also indicate that ∂v/∂z = 0 and ∂v/∂x = 0. Therefore v must be constant. Since v is zero at the solid surface, then v must be zero
everywhere. The fact that v = 0 reduces the Navier–Stokes equations further, as indicated by (5). Hence for the y direction
∂p
= ρg
∂y
which indicates a hydrostatic variation of pressure. In the x direction, after assumption (6) we obtain
∂ 2u ∂p
µ 2 − =0
∂y
∂x
Integrating twice
u=
1 ∂p 2 c1
y + y + c2
µ
2 µ ∂x
To evaluate the constants, c1 and c2, we must apply the boundary conditions. At y = 0, u = 0. Consequently, c2 = 0. At y = a, u = 0.
Hence
0=
1 ∂p 2 c1
a + a
µ
2 µ ∂x
which gives
c1 = −
1 ∂p
a
2 µ ∂x
and finally
u=
2
a 2 ∂p ⎡⎛ y ⎞ ⎛ y ⎞⎤
⎢⎜ ⎟ − ⎜ ⎟ ⎥
2 µ ∂x ⎢⎣⎝ a ⎠ ⎝ a ⎠⎥⎦
Problem 8.18
[Difficulty: 4]
Problem 8.19
[Difficulty: 5]
Problem 8.20
[Difficulty: 2]
Problem 8.21
[Difficulty: 3]
Given:
Laminar velocity profile of power-law fluid flow between parallel plates
Find:
Expression for flow rate; from data determine the type of fluid
Solution:
⎡⎢
n
h ∆p ⎞
n⋅ h ⎢
u = ⎛⎜ ⋅
⋅
⋅ 1−
⎝ k L ⎠ n + 1 ⎢⎣
n+ 1⎤
1
The velocity profile is
⌠
Q = w⋅ ⎮
⌡
The flow rate is then
⎛y⎞
⎜h
⎝ ⎠
n
⎥
⎥
⎥⎦
h
h
or, because the flow is symmetric
u dy
−h
⌠
⎮
⎮
⎮ 1−
⎮
⌡
The integral is computed as
0
n+ 1
⎛y⎞
⎜h
⎝ ⎠
⌠
Q = 2 ⋅ w⋅ ⎮ u dy
⌡
n
2⋅ n+ 1⎤
⎡⎢
⎥
n
⎢
⎥
n
y
dy = y ⋅ ⎢1 −
⋅ ⎛⎜ ⎞
⎥⎦
⎣ 2⋅ n + 1 ⎝ h ⎠
1
2⋅ n+ 1⎤
⎡
n
⎢
⎥
h ∆p ⎞
n⋅ h
n
n
Q = 2 ⋅ w⋅ ⎛⎜ ⋅
⋅
⋅ h ⋅ ⎢1 −
⋅ ( 1)
⎥
⎝ k L ⎠ n + 1 ⎣ 2⋅ n + 1
⎦
Using this with the limits
An Excel spreadsheet can be used for computation of n.
The data is
dp (kPa)
Q (L/min)
10
0.451
20
0.759
30
1.01
40
1.15
50
1.41
60
1.57
70
1.66
80
1.85
90
2.05
1
n
This must be fitted to
Q=
1
2
⎛ h ⋅ ∆p ⎞ ⋅ 2 ⋅ n⋅ w⋅ h or
⎜ L
⎝k
⎠ 2⋅ n + 1
Q = k ⋅ ∆p
n
100
2.25
1
n
Q=
2
⎛ h ⋅ ∆p ⎞ ⋅ 2 ⋅ n⋅ w⋅ h
⎜k L
⎝
⎠ 2⋅ n + 1
We can fit a power curve to the data
Flow Rate vs Applied Pressure for a
Non-Newtonian Fluid
10.0
Q (L/min)
Data
Power Curve Fit
1.0
Q = 0.0974dp0.677
2
R = 0.997
0.1
10
Hence
dp (kPa)
1/n =
It's a dilatant fluid
0.677
n =
1.48
100
Problem 8.22
[Difficulty: 2]
Given:
Laminar flow between moving plates
Find:
Expression for velocity; Volume flow rate per depth
Solution:
Given data
ft
U1 = 5⋅
s
d = 0.2⋅ in
ft
U2 = 2⋅
s
Using the analysis of Section 8.2, the sum of forces in the x direction is
⎡ ∂ dy ⎛
⎛
dy ⎞⎤
dx
dx ⎞
⎢τ + τ ⋅ − ⎜ τ − ∂ τ ⋅ ⎥ ⋅ b ⋅ dx + ⎜ p − ∂ p ⋅ − p + ∂ p ⋅ ⋅ b ⋅ dy = 0
2 ⎠⎦
∂y
∂x 2
∂x 2 ⎠
⎣ ∂y 2 ⎝
⎝
Simplifying
dτ
dy
=
dp
dx
2
=0
μ⋅
or
d u
dy
Integrating twice
u = c1 ⋅ y + c2
Boundary conditions:
u ( 0 ) = −U1
c2 = −U1
(
)
2
=0
ft
c2 = −5
s
u ( y = d ) = U2
u in ft/s, y in ft
Hence
y
u ( y ) = U1 + U2 ⋅ − U1
d
u ( y ) = 420 ⋅ y − 5
The volume flow rate is
⌠
⌠
Q = ⎮ u dA = b ⋅ ⎮ u dy
⌡
⌡
⌠
Q = b⋅ ⎮
⎮
⌡
d
Q = b⋅ d⋅
(U2 − U1)
Q
2
b
= d⋅
(U2 − U1)
2
3
ft
Q
b
= 0.2⋅ in ×
1 ⋅ ft
12⋅ in
×
1
2
× ( 2 − 5) ×
ft
Q
s
b
= −.025⋅
s
ft
3
ft
Q
b
= −.025⋅
s
ft
×
7.48⋅ gal
1 ⋅ ft
3
×
60⋅ s
Q
1 ⋅ min
b
U1 + U2
d
= −11.2⋅
gpm
ft
−1
c1 = 420 s
U + U2 d 2
⎞
⎡ U + U ⋅ y − U ⎤ dx = b⋅ ⎛⎜ 1
⋅
−
U
⋅
d
⎢( 1
)
⎥
2 d
1
1
2
⎣
⎦
⎝ d
⎠
0
Hence
c1 =
Problem 8.23
[Difficulty: 2]
Problem 8.24
[Difficulty: 3]
Given:
Properties of two fluids flowing between parallel plates; applied pressure gradient
Find:
Velocity at the interface; maximum velocity; plot velocity distribution
Solution:
dp
Given data
=k
dx
k = −50⋅
kPa
m
h = 5 ⋅ mm
μ1 = 0.1⋅
N⋅ s
μ2 = 4 ⋅ μ1
2
μ2 = 0.4⋅
m
N⋅ s
2
m
(Lower fluid is fluid 1; upper is fluid 2)
Following the analysis of Section 8.2, analyse the forces on a differential CV of either fluid
The net force is zero for steady flow, so
dτ dy ⎞⎤
dp dx ⎞⎤
⎡ dτ dy ⎛
⎡ dp dx ⎛
⎢τ + dy ⋅ 2 − ⎜ τ − dy ⋅ 2 ⎥ ⋅ dx⋅ dz + ⎢p − dx ⋅ 2 − ⎜ p + dx ⋅ 2 ⎥ ⋅ dy⋅ dz = 0
⎣
⎝
⎠⎦
⎣
⎝
⎠⎦
dτ
Simplifying
dy
=
dp
dx
=k
μ⋅
so for each fluid
d
2
dy
2
u =k
Applying this to fluid 1 (lower fluid) and fluid 2 (upper fluid), integrating twice yields
u1 =
k
2
⋅ y + c1 ⋅ y + c2
2 ⋅ μ1
u2 =
k
2
2 ⋅ μ2
⋅ y + c3 ⋅ y + c4
For convenience the origin of coordinates is placed at the centerline
y = −h
We need four BCs. Three are obvious
u1 = 0
y=0
(1)
u 1 = u 2 (2)
y=h
u2 = 0
The fourth BC comes from the fact that the stress at the interface generated by each fluid is the same
du1
du2
μ1 ⋅
= μ2 ⋅
dy
dy
y=0
Using these four BCs
0=
k
2 ⋅ μ1
2
⋅ h − c1 ⋅ h + c2
c2 = c4
0=
k
2 ⋅ μ2
(4)
2
⋅ h + c3 ⋅ h + c4
μ1 ⋅ c1 = μ2 ⋅ c3
Hence, after some algebra
c1 =
k⋅ h
2 ⋅ μ1
⋅
(μ2 − μ1)
(μ2 + μ1)
c4 = −
k⋅ h
2
μ2 + μ1
c2 = c4
c3 =
k⋅ h
2 ⋅ μ2
⋅
(μ2 − μ1)
(μ2 + μ1)
(3)
c1 = −750
1
c2 = 2.5
s
m
c3 = −187.5
s
1
c4 = 2.5
s
m
s
The velocity distributions are then
u1( y) =
k
2 ⋅ μ1
⎡
2
⋅ ⎢y + y ⋅ h ⋅
⎣
(μ2 − μ1)⎤
⎥−
(μ2 + μ1)⎦
k⋅ h
2
u2( y) =
μ2 + μ1
k
2 ⋅ μ2
⎡
2
⋅ ⎢y + y ⋅ h ⋅
⎣
(μ2 − μ1)⎤
⎥−
(μ2 + μ1)⎦
k⋅ h
μ2 + μ1
Evaluating either velocity at y = 0, gives the velocity at the interface
u interface = −
k⋅ h
2
u interface = 2.5
μ2 + μ1
m
s
The plots of these velocity distributions can be done in Excel. Typical curves are shown below
5
y (mm)
2.5
0
0.5
1
1.5
2
2.5
3
3.5
− 2.5
−5
u (m/s)
Clearly, u 1 has the maximum velocity, when
(
(
h μ2 − μ1
y max = − ⋅
2 μ2 + μ1
)
)
du1
dy
=0
y max = −1.5 mm
(We could also have used Excel's Solver for this.)
2 ⋅ y max + h ⋅
or
(
)
u max = u 1 y max
u max = 3.06
(μ2 − μ1)
(μ2 + μ1)
m
s
=0
2
Problem 8.25
Given:
Laminar flow of two fluids between plates
Find:
Velocity at the interface
[Difficulty: 3]
Solution:
Using the analysis of Section 8.2, the sum of forces in the x direction is
⎛ ∂ dx
dx ⎞
⎡ ∂ dy ⎛ ∂ dy ⎞⎤
∂
⋅ b ⋅ dy = 0
⎢τ + τ ⋅ − ⎜ τ − τ ⋅ ⎥ ⋅ b ⋅ dx + ⎜ p − p ⋅ − p + p ⋅
∂x 2 ⎠
⎣ ∂y 2 ⎝ ∂y 2 ⎠⎦
⎝ ∂x 2
Simplifying
dτ
dy
=
dp
dx
2
=0
μ⋅
or
dy
y=0
u1 = 0
2
=0
u 1 = c1 ⋅ y + c2
Applying this to fluid 1 (lower fluid) and fluid 2 (upper fluid), integrating twice yields
We need four BCs. Three are obvious
d u
y = h u1 = u2
y = 2⋅ h
u 2 = c3 ⋅ y + c4
u2 = U
The fourth BC comes from the fact that the stress at the interface generated by each fluid is the same
y=h
du1
du2
μ1⋅
= μ2⋅
dy
dy
Using these four BCs
0 = c2
c1⋅ h + c2 = c3⋅ h + c4
Hence
c2 = 0
From the 2nd and 3rd equations
c1⋅ h − U = −c3⋅ h
Hence
μ1
c1⋅ h − U = −c3⋅ h = − ⋅ h ⋅ c1
μ2
and
Hence for fluid 1 (we do not need to complete the analysis for fluid 2)
20⋅
Evaluating this at y = h, where u 1 = u interface
u interface =
ft
s
1⎞
⎛1 +
⎜
3⎠
⎝
U = c3⋅ 2⋅ h + c4
μ1 ⋅ c1 = μ2 ⋅ c3
c1 =
U
⎛
μ1 ⎞
⎝
μ2
⎠
h⋅ ⎜ 1 +
u1 =
U
⎛
h ⋅⎜1 +
⎝
μ1 ⎞
μ2
⎠
u interface = 15⋅
ft
s
⋅y
μ1⋅ c1 = μ2⋅ c3
Problem 8.26
[Difficulty: 2]
Given:
Computer disk drive
Find:
Flow Reynolds number; Shear stress; Power required
Solution:
For a distance R from the center of a disk spinning at speed ω
V  R ω
The gap Reynolds number is
Re 
V  25 mm 
ρ V a
Re  22.3
m
s
1000 mm
 8500 rpm 
2 π rad
rev
1 min

ν  1.45  10
ν
6
 0.25  10

V  22.3
60 s
2
5 m
V a

μ
1 m
5
1.45  10
s
from Table A.10 at 15oC
s
s
m 
m
Re  0.384
2
m
The flow is definitely laminar
The shear stress is then
τ  μ
du
dy
 μ
τ  1.79  10
 5 N s
V
μ  1.79  10
a
 5 N s

2
m
The power required is
P  T ω
T  τ A R
P  τ A R ω
P  1600
N
2
m

from Table A.10 at 15oC
2
m
 22.3
m
s
1

0.25  10
6
τ  1.60 kPa
m
where torque T is given by
A  ( 5  mm)
with
 2.5  10
5
2
 m  25 mm 
2
A  2.5  10
1 m
1000 mm
5
 8500 rpm 
2
m
2  π rad
rev

1  min
60 s
P  0.890 W
Problem 8.27
[Difficulty: 2]
Given:
Velocity profile between parallel plates
Find:
Pressure gradients for zero stress at upper/lower plates; plot
Solution:
From Eq. 8.8, the velocity distribution is
The shear stress is
u=
U⋅ y
a
+
⎛ ∂ ⎞ ⎡⎢⎛ y ⎞ 2
−
p ⋅ ⎜
2 ⋅ μ ⎝ ∂x ⎠ ⎣⎝ a ⎠
a
2
⋅⎜
y⎤
⎥
a⎦
2
1
⎛∂ ⎞
y
τyx = μ⋅
− ⎞
= μ⋅ +
⋅ ⎜ p ⋅ ⎛ 2⋅
⎜
dy
a
2 ⎝ ∂x ⎠
a
2
a
du
U
a
⎝
(a) For τyx = 0 at y = a
The velocity distribution is then
(b) For τyx = 0 at y = 0
The velocity distribution is then
0 = μ⋅
u=
The velocity distributions can be plotted in Excel.
a
U⋅ y
a
0 = μ⋅
u=
U
U
a
U⋅ y
a
+
−
−
+
a ∂
⋅ p
2 ∂x
a
2
2⋅ μ
⋅
∂
∂x
2 ⋅ U⋅ μ
a
2
⎡ y 2
⎞ −
a
⎣⎝ ⎠
⋅ ⎢⎛⎜
y⎤
⎥
a⎦
a ∂
⋅ p
2 ∂x
a
2
2⋅ μ
⋅
⎠
u
U
∂
∂x
2 ⋅ U⋅ μ
a
2
⎡ y ⎞2
−
⎣⎝ a ⎠
⋅ ⎢⎛⎜
y⎤
⎥
a⎦
u
U
p =−
2 ⋅ U⋅ μ
a
= 2⋅
p =
y
−
a
⎛y⎞
⎜a
⎝ ⎠
2 ⋅ U⋅ μ
a
=
2
⎛y⎞
⎜a
⎝ ⎠
2
2
2
y /a
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
(a) u /U
0.000
0.190
0.360
0.510
0.640
0.750
0.840
0.910
0.960
0.990
1.00
(b) u /U
0.000
0.010
0.040
0.090
0.160
0.250
0.360
0.490
0.640
0.810
1.000
Zero-Stress Velocity Distributions
1.00
Zero Stress Upper Plate
Zero Stress Lower Plate
y /a
0.75
0.50
0.25
0.00
0.00
0.25
0.50
u /U
0.75
1.00
Problem 8.28
[Difficulty: 2]
Problem 8.29
[Difficulty: 2]
Problem 8.30
[Difficulty: 3]
Given:
Data on flow of liquids down an incline
Find:
Velocity at interface; velocity at free surface; plot
Solution:
2
Given data
h = 10⋅ mm
θ = 60⋅ deg
ν1
ν2 =
5
m
ν1 = 0.01⋅
s
ν2 = 2 × 10
2
−3m
s
(The lower fluid is designated fluid 1, the upper fluid 2)
From Example 5.9 (or Exanple 8.3 with g replaced with gsinθ), a free body analysis leads to (for either fluid)
d
2
dy
2
u =−
ρ⋅ g ⋅ sin( θ)
μ
Applying this to fluid 1 (lower fluid) and fluid 2 (upper fluid), integrating twice yields
u1 = −
ρ⋅ g ⋅ sin( θ)
2 ⋅ μ1
2
⋅ y + c1 ⋅ y + c2
We need four BCs. Two are
obvious
u2 = −
ρ⋅ g ⋅ sin( θ)
2 ⋅ μ2
2
⋅ y + c3 ⋅ y + c4
y=0
u1 = 0
(1)
y=h
u1 = u2
(2)
The third BC comes from the fact that there is no shear stress at the free surface
y = 2⋅ h
du2
μ2 ⋅
=0
dy
(3)
The fourth BC comes from the fact that the stress at the interface generated by each fluid is the same
du1
du2
μ1 ⋅
= μ2 ⋅
dy
dy
y=h
Using these four BCs
c2 = 0
−
ρ⋅ g ⋅ sin( θ)
2 ⋅ μ1
2
⋅ h + c1 ⋅ h + c2 = −
−ρ⋅ g ⋅ sin( θ) ⋅ 2 ⋅ h + μ2 ⋅ c3 = 0
Hence, after some algebra
c1 =
2 ⋅ ρ⋅ g ⋅ sin( θ) ⋅ h
μ1
c2 = 0
ρ⋅ g ⋅ sin( θ)
2 ⋅ μ2
(4)
2
⋅ h + c3 ⋅ h + c4
−ρ⋅ g ⋅ sin( θ) ⋅ h + μ1 ⋅ c1 = −ρ⋅ g ⋅ sin( θ) ⋅ h + μ2 ⋅ c3
c3 =
2 ⋅ ρ⋅ g ⋅ sin( θ) ⋅ h
μ2
2
c4 = 3 ⋅ ρ⋅ g ⋅ sin( θ) ⋅ h ⋅
(μ2 − μ1)
2 ⋅ μ1 ⋅ μ2
The velocity distributions are then
u1 =
ρ⋅ g ⋅ sin( θ)
2 ⋅ μ1
(
⋅ 4⋅ y⋅ h − y
2
)
u2 =
⎡
ρ⋅ g ⋅ sin( θ)
2
⋅ ⎢3 ⋅ h ⋅
2 ⋅ μ2
(μ2 − μ1)
μ1
⎣
+ 4⋅ y⋅ h − y
2⎤
⎥
⎦
Rewriting in terms of ν1 and ν2 (ρ is constant and equal for both fluids)
u1 =
g ⋅ sin( θ)
2 ⋅ ν1
(
⋅ 4⋅ y⋅ h − y
)
2
u2 =
g ⋅ sin( θ)
2 ⋅ ν2
⎡
2
⋅ ⎢3 ⋅ h ⋅
(ν2 − ν1)
⎣
ν1
+ 4⋅ y⋅ h − y
2⎤
⎥
⎦
(Note that these result in the same expression if ν1 = ν2, i.e., if we have one fluid)
2
u interface =
Evaluating either velocity at y = h, gives the velocity at the interface
Evaluating u 2 at y = 2h gives the velocity at the free surface
2
3 ⋅ g ⋅ h ⋅ sin( θ)
u freesurface = g ⋅ h ⋅ sin( θ) ⋅
u freesurface⋅ h
Note that a Reynolds number based on the free surface velocity is
ν2
2 ⋅ ν1
(3⋅ ν2 + ν1)
= 1.70
2 ⋅ ν1 ⋅ ν2
u interface = 0.127
0.000
0.0166
0.0323
0.0472
0.061
0.074
0.087
0.098
0.109
0.119
0.127
u 2 (m/s)
indicating laminar flow
Velocity Distributions down an Incline
24
Lower Velocity
20
0.127
0.168
0.204
0.236
0.263
0.287
0.306
0.321
0.331
0.338
0.340
y (mm)
0.000
1.000
2.000
3.000
4.000
5.000
6.000
7.000
8.000
9.000
10.000
11.000
12.000
13.000
14.000
15.000
16.000
17.000
18.000
19.000
20.000
Upper Velocity
16
12
8
4
0
0.0
0.1
0.2
u (m/s)
0.3
s
u freesurface = 0.340
The velocity distributions can be plotted in Excel.
y (mm) u 1 (m/s)
m
0.4
m
s
Problem 8.31
[Difficulty: 2]
Given:
Velocity distribution on incline
Find:
Expression for shear stress; Maximum shear; volume flow rate/mm width; Reynolds number
Solution:
u(y) =
From Example 5.9
τ = μ⋅
For the shear stress
τ is a maximum at y = 0
ρ⋅ g ⋅ sin( θ)
μ
du
dy
⎛
y
⎝
2
⋅ ⎜ h⋅ y −
2⎞
⎠
= ρ⋅ g ⋅ sin( θ) ⋅ ( h − y )
τmax = ρ⋅ g ⋅ sin( θ) ⋅ h = SG ⋅ ρH2O⋅ g ⋅ sin( θ) ⋅ h
τmax = 1.2 × 1000
kg
3
× 9.81⋅
m
m
2
2
× sin( 15⋅ deg) × 0.007 ⋅ m ×
s
N⋅ s
τmax = 21.3 Pa
kg⋅ m
This stress is in the x direction on the wall
⌠
h
⎮
⌠
⌠
Q = ⎮ u dA = w⋅ ⎮ u ( y ) dy = w⋅ ⎮
⌡
⌡
⎮
0
⌡
The flow rate is
h
ρ⋅ g ⋅ sin( θ)
μ
2
⎛
y ⎞
dy
⋅ ⎜ h⋅ y −
2 ⎠
⎝
Q=
ρ⋅ g ⋅ sin( θ) ⋅ w⋅ h
3⋅ μ
0
3
m
Q
w
=
1
3
× 1.2 × 1000
kg
3
× 9.81⋅
m
m
2
2
m
3
× sin( 15⋅ deg) × ( 0.007 ⋅ m) ×
s
The gap Reynolds number is
V=
Re =
Q
A
=
Q
V = 217 ⋅
w⋅ h
−4
= 2.18 × 10
s
Q
m
w
= 217
mm
3
s
mm
×
1
V = 31.0⋅
7 ⋅ mm
mm
μ
kg
3
m
× 31⋅
mm
s
2
× 7 ⋅ mm ×
m
1.60⋅ N⋅ s
×
⎛ 1⋅ m ⎞
⎜
⎝ 1000⋅ mm ⎠
3
s
ρ⋅ V⋅ h
Re = 1.2 × 1000
The flow is definitely laminar
N⋅ s
1.60⋅ N⋅ s kg⋅ m
mm
The average velocity is
2
⋅
mm
2
Re = 0.163
s
3
Problem 8.32
[Difficulty: 3]
Given:
Flow between parallel plates
Find:
Shear stress on lower plate; Plot shear stress; Flow rate for pressure gradient; Pressure gradient for zero shear; Plot
Solution:
u(y) =
From Section 8-2
U⋅ y
a
+
⎡ y 2
⎞ −
2 ⋅ μ dx ⎣⎝ a ⎠
a
2
⋅
dp
⋅ ⎢⎛⎜
y⎤
⎥
a⎦
3
ft
a
u = U⋅
For dp/dx = 0
a
⌠
⌠
U⋅ a
y
= ⎮ u ( y ) dy = w⋅ ⎮ U⋅ dy =
⎮
⌡
2
l
a
0
⌡
y
Q
a
1
Q =
2
× 5⋅
ft
×
s
0.1
12
⋅ ft
Q = 0.0208⋅
s
ft
0
τ = μ⋅
For the shear stress
du
dy
=
μ⋅ U
− 7 lbf ⋅ s
μ = 3.79 × 10
when dp/dx =
0
a
⋅
ft
(Table A.9)
2
The shear stress is constant - no need to plot!
τ = 3.79 × 10
− 7 lbf ⋅ s
⋅
ft
2
× 5⋅
ft
s
×
12
0.1⋅ ft
×
⎛ 1⋅ ft ⎞
⎜ 12⋅ in
⎝
⎠
2
−6
τ = 1.58 × 10
Q will decrease if dp/dx > 0; it will increase if dp/dx < 0.
τ = μ⋅
For non- zero dp/dx:
du
dy
=
μ⋅ U
a
+ a⋅
τ( y = 0.25⋅ a) = μ⋅
At y = 0.25a, we get
U
a
dp
dx
⋅ ⎛⎜
+ a⋅
y
−
⎝a
dp
dx
⋅ ⎛⎜
1⎞
2⎠
1
⎝4
−
1⎞
2⎠
= μ⋅
U
a
−
a dp
⋅
4 dx
lbf
Hence this stress is zero when
dp
dx
=
4 ⋅ μ⋅ U
a
2
− 7 lbf ⋅ s
= 4 × 3.79 × 10
⋅
ft
2
× 5⋅
ft
s
×
2
2
⎛ 12 ⎞ = 0.109 ⋅ ft = 7.58 × 10− 4 psi
⎜ 0.1⋅ ft
ft
ft
⎝
⎠
0.1
y (in)
0.075
0.05
0.025
− 1× 10
−4
0
1× 10
−4
Shear Stress (lbf/ft3)
2× 10
−4
−4
3× 10
⋅ psi
Problem 8.33
[Difficulty: 3]
Given:
Flow between parallel plates
Find:
Location and magnitude of maximum velocity; Volume flow in 10 s; Plot velocity and shear stress
Solution:
From Section 8.2
For u max set du/dx = 0
Hence
From Fig. A.2 at
U⋅ y
u(y) =
du
b
=0=
dy
⎡ y 2
⎞ −
2 ⋅ μ dx ⎣⎝ b ⎠
b
+
U
u = u max
dp
⋅
y⎤
⋅ ⎢⎛⎜
⎥
b⎦
2
dp 2 ⋅ y
1
1 dp
U
⋅ ⋅⎛
− ⎞=
+
⋅ ⋅ (2⋅ y − b)
2 ⋅ μ dx ⎜ 2
a
2 ⋅ μ dx
b
⎝b
⎠
b
+
b
2
b
y=
at
59 °F = 15⋅ °C
2
μ = 4⋅
μ⋅ U
−
b⋅
dp
dx
N⋅ s
μ = 0.0835⋅
2
m
y =
Hence
0.1⋅ in
2
+ 0.0835⋅
ft
U⋅ y
u max =
lbf ⋅ s
b
u max = 2 ⋅
ft
s
2
× 2⋅
lbf ⋅ s
ft
ft
s
2
1
×
2
0.1⋅ in
×
in ⋅ ft
y = 0.0834⋅ in
50⋅ lbf
+
⎡ y 2
⎞ −
2 ⋅ μ dx ⎣⎝ b ⎠
×
2
2
2
50⋅ psi ⎛ 12⋅ in ⎞
ft
⎛ .0834 ⎞ + 1 × ⎛ 0.1 ⋅ ft⎞ ×
×
−
×
⎜ 0.1
⎜
⎜ 1 ⋅ ft ×
ft
.0835 ⋅ lbf ⋅ s
⎝
⎠ 2 ⎝ 12 ⎠
⎝
⎠
b
2
⋅
dp
⋅ ⎢⎛⎜
y⎤
⎥
y = 0.0834⋅ in
with
b⎦
2
⎡
⎤
⎢⎛⎜ .0834 ⎞ − ⎛⎜ .0834 ⎞⎥
⎣⎝ 0.1 ⎠ ⎝ 0.1 ⎠⎦
u max = 2.083 ⋅
b
⌠
=⎮
w ⌡0
Q
Q
w
=
1
2
⌠
⎮
u ( y ) dy = w⋅ ⎮
⎮
⌡
b
⎡ U⋅ y b 2 dp ⎡⎛ y ⎞ 2
⎢
+
⋅ ⋅ ⎢⎜
−
2 ⋅ μ dx ⎣⎝ b ⎠
⎣ b
× 2⋅
ft
s
×
0.1
12
⋅ ft −
1
12
3
×
= 0.0125
s
Q
ft
w
2
50⋅ psi ⎞ ⎛ 12⋅ in ⎞
ft
⎛ 0.1 ⎞
× ⎛⎜ −
×⎜
⎜ 12 ⋅ ft ×
ft ⎠ ⎝ 1 ⋅ ft ⎠
.0835 ⋅ lbf ⋅ s ⎝
⎝
⎠
ft
w
3
⎥⎥ dy = U⋅ b − b ⋅ dp
b⎦⎦
12⋅ μ dx
2
0
3
Q
y⎤⎤
= 5.61⋅
gpm
ft
2
ft
s
Flow =
Q
w
⋅ ∆t = 5.61⋅
The velocity profile is
gpm
ft
1 ⋅ min
×
60⋅ s
u
U
=
y
b
× 10⋅ s
+
Flow = 0.935 ⋅
⎡ y 2
⎞ −
2 ⋅ μ⋅ U dx ⎣⎝ b ⎠
b
2
⋅
dp
⋅ ⎢⎛⎜
y⎤
⎥
b⎦
gal
ft
τ = μ⋅
For the shear stress
du
dy
= μ⋅
U
b
+
b dp ⎡ ⎛ y ⎞
⋅ ⋅ ⎢2 ⋅ ⎜
− 1⎤⎥
2 dx ⎣ ⎝ b ⎠
⎦
The graphs below can be plotted in Excel
1
0.8
y/b
0.6
0.4
0.2
0
0.2
0.4
0.6
0.8
1
1.2
u/U
1
0.8
y/b
0.6
0.4
0.2
− 500
0
500
1× 10
3
Shear Stress (Pa)
3
1.5× 10
3
2× 10
2.5× 10
3
Problem 8.34
[Difficulty: 3]
Given:
Flow between parallel plates
Find:
Pressure gradient for no flow; plot velocity and stress distributions; also plot for u = U at y = a/2
Solution:
Basic equations
Available data
From Eq 2 for Q = 0
u(y) =
U⋅ y
a
U = 1.5⋅
dp
dx
=
+
m
⋅
dp
⋅ ⎢⎛⎜
y⎤
Q
⎥ (1)
a⎦
l
=
U⋅ a
2
−
a
3
dp
⋅
12⋅ μ dx
τ = μ⋅
(2)
U
a
μ = 1⋅
From Fig. A.2 for castor oil at 20oC
+ a⋅
dp
dx
N⋅ s
2
m
6 ⋅ μ⋅ U
a
2
a = 5 ⋅ mm
s
2
⎡ y 2
⎞ −
2 ⋅ μ dx ⎣⎝ a ⎠
a
= 6 × 1⋅
N⋅ s
2
× 1.5⋅
m
m
s
×
1
dp
( 0.005 ⋅ m)
dx
2
The graphs below, using Eqs. 1 and 3, can be plotted in Excel
1
y/a
0.75
0.5
0.25
− 0.5
0
0.5
1
1.5
u (m/s)
1
y/a
0.75
0.5
0.25
−1
− 0.5
0
0.5
1
Shear Stress (kPa)
The pressure gradient is adverse, to counteract the flow generated by the upper plate motion
1.5
= 360 ⋅
kPa
m
⋅ ⎜⎛
y
⎝a
−
1⎞
2⎠
(3)
u(y) =
For u = U at y = a/2 we need to adjust the pressure gradient. From Eq. 1
2
a⎤
⎢⎡⎛ a ⎞
⎥
2
2
a dp ⎢⎜ 2
2⎥
U=
⋅ ⋅ ⎜
−
+
a
2 ⋅ μ dx ⎢⎣⎝ a ⎠
a ⎥⎦
U⋅
Hence
U⋅ y
a
⎡ y 2
⎞ −
2 ⋅ μ dx ⎣⎝ a ⎠
a
+
2
⋅
dp
⋅ ⎢⎛⎜
y⎤
⎥
a⎦
a
dp
or
dx
dp
dx
=−
4 ⋅ U⋅ μ
a
2
= −240 ⋅
= −4 × 1 ⋅
N⋅ s
2
× 1.5⋅
m
m
s
kPa
m
1
y/a
0.75
0.5
0.25
0
0.5
1
1.5
2
u (m/s)
1
y/a
0.75
0.5
0.25
−1
− 0.5
0
0.5
1
Shear Stress (kPa)
The pressure gradient is positive to provide the "bulge" needed to satisfy the velocity requirement
1.5
×
1
( 0.005 ⋅ m)
2
Problem 8.35
[Difficulty: 3]
Given:
Flow between parallel plates
Find:
Shear stress on lower plate; pressure gradient for zero shear stress at y/a = 0.25; plot velocity and shear stress
Solution:
u(y) =
Basic equations
U⋅ y
a
q = 1.5⋅
Available data
+
⎡ y 2
⎞ −
2 ⋅ μ dx ⎣⎝ a ⎠
a
2
⋅
dp
gpm
⋅ ⎢⎛⎜
y⎤
⎥ (1)
a⎦
a = 0.05⋅ in
ft
From Fig. A.2, Carbon tetrachloride at 20oC
μ = 0.001 ⋅
Q
=
l
U⋅ a
2
a
−
3
⋅
dp
12⋅ μ dx
τ = μ⋅
(2)
U
a
+ a⋅
U=
From Eq. 3, when y = 0, with
U = 1.60
ft
N⋅ s
s
y
⎝a
−
1⎞
2⎠
(3)
− 5 lbf ⋅ s
μ = 2.089 × 10
2
2⋅ Q
or
a⋅ l
τyx =
dx
⋅ ⎛⎜
68°F = 20°C
⋅
m
From Eq. 2, for zero pressure gradient
dp
ft
U =
2⋅ q
U = 1.60⋅
a
μ⋅ U
ft
s
−5
τyx = 5.58 × 10
a
2
⋅ psi
A mild adverse pressure gradient would reduce the flow rate.
For zero shear stress at y/a = 0.25, from Eq. 3
0 = μ⋅
U
a
+ a⋅
dp
dx
⋅ ⎛⎜
1
⎝4
−
1⎞
2⎠
dp
or
dx
dp
2
dx
a
= 0.0536⋅
0.75
y/a
0.75
y/a
4 ⋅ μ⋅ U
1
1
0.5
0.5
0.25
0.25
− 0.5
=
0
0.5
1
1.5
2
u (ft/s)
Note that the location of zero shear is also where u is maximum!
− 1× 10
−4
0
−4
1× 10
Shear Stress (psi)
−4
2× 10
psi
ft
Problem 8.36
Using the result for average velocity from Example 8.3
[Difficulty: 3]
Problem 8.37
[Difficulty: 3]
Problem 8.38
[Difficulty: 5]
Problem 8.39
[Difficulty: 5]
Problem 8.40
[Difficulty: 2]
Given: Expression for efficiency
Find: Plot; find flow rate for maximum efficiency; explain curve
Solution:
η
0.0%
7.30%
14.1%
20.3%
25.7%
30.0%
32.7%
33.2%
30.0%
20.8%
0.0%
Efficiency of a Viscous Pump
35%
30%
25%
η
q
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
20%
15%
10%
5%
0%
0.00
0.10
0.20
0.30
0.40
q
For the maximum efficiency point we can use Solver (or alternatively differentiate)
q
0.333
η
33.3%
The efficiency is zero at zero flow rate because there is no output at all
The efficiency is zero at maximum flow rate ∆p = 0 so there is no output
The efficiency must therefore peak somewhere between these extremes
0.50
Problem 8.41
Problem 2.66
[Difficulty: 5]
Problem 8.42
[Difficulty: 5] Part 1/2
Problem 8.42
[Difficulty: 5] Part 2/2
Problem 8.43
[Difficulty: 3]
Given:
Data on a journal bearing
Find:
Time for the bearing to slow to 100 rpm; visocity of new fluid
Solution:
The given data is
D = 35⋅ mm
L = 50⋅ mm
δ = 1 ⋅ mm
ωi = 500 ⋅ rpm
ωf = 100 ⋅ rpm
μ = 0.1⋅
2
I = 0.125 ⋅ kg⋅ m
N⋅ s
2
m
I⋅ α = Torque = −τ⋅ A⋅
The equation of motion for the slowing bearing is
D
2
where α is the angular acceleration and τ is the viscous stress, and A = π⋅ D⋅ L is the surface area of the bearing
τ = μ⋅
As in Example 8.2 the stress is given by
U
δ
=
μ⋅ D⋅ ω
2⋅ δ
where U and ω are the instantaneous linear and angular velocities.
Hence
Separating variables
μ⋅ D⋅ ω
I⋅ α = I⋅
dω
dω
μ⋅ π⋅ D ⋅ L
ω
=−
dt
2⋅ δ
⋅ π⋅ D⋅ L⋅
D
2
3
=−
μ⋅ π⋅ D ⋅ L
4⋅ δ
⋅ω
3
=−
4 ⋅ δ⋅ I
⋅ dt
3
−
Integrating and using IC ω = ω0
μ⋅ π⋅ D ⋅ L
ω( t) = ωi⋅ e
4⋅ δ ⋅ I
⋅t
3
−
The time to slow down to ω f
4 ⋅ δ⋅ I
t = −
3
μ⋅ π⋅ D ⋅ L
For the new fluid, the time to slow down is
t = 10⋅ min
Rearranging the equation
μ = −
4 ⋅ δ⋅ I
3
π ⋅ D ⋅ L⋅ t
4⋅ δ ⋅ I
ωf = ωi⋅ e
= 10 rpm is obtained from solving
so
μ⋅ π⋅ D ⋅ L
⎛ ωf ⎞
⋅ ln⎜
⎝ ωi ⎠
Hence
3
t = 1.19 × 10 s
⎛ ωf ⎞
⋅ ln⎜
⎝ ωi ⎠
μ = 0.199
kg
m⋅ s
⋅t
t = 19.9⋅ min
It is more viscous as it slows
down the rotation in a
shorter time
Problem 8.44
Given:
Navier-Stokes Equations
Find:
Derivation of Example 8.3 result
[Difficulty: 2]
Solution:
The Navier-Stokes equations are (using the coordinates of Example 8.3, so that x is vertical, y is horizontal)
4
3
∂u ∂v ∂w
+
+
=0
∂x ∂y ∂z
1
4
5
(5.1c)
3
4
3
⎛ ∂ 2u ∂ 2u ∂ 2u ⎞
⎛ ∂u
∂u
∂u
∂u ⎞
∂p
ρ ⎜⎜ + u + v + w ⎟⎟ = ρg x − + µ ⎜⎜ 2 + 2 + 2 ⎟⎟
∂x
∂y
∂z ⎠
∂x
∂y
∂z ⎠
⎝ ∂t
⎝ ∂x
1
4
5
3
4
6
5
(5.27a)
3
⎛∂ v ∂ v ∂ v⎞
⎛ ∂v
∂v
∂v
∂v ⎞
∂p
+ u + v + w ⎟⎟ = ρg y −
+ µ ⎜⎜ 2 + 2 + 2 ⎟⎟
∂x
∂y
∂z ⎠
∂y
∂z ⎠
∂y
⎝ ∂t
⎝ ∂x
2
ρ ⎜⎜
1
3
3
3
3
2
2
3
3
(5.27b)
3
3
⎛∂ w ∂ w ∂ w⎞
⎛ ∂w
∂p
∂w ⎞
∂w
∂w
+ µ ⎜⎜ 2 + 2 + 2 ⎟⎟
⎟⎟ = ρg z −
+w
+v
+u
∂z ⎠
∂y
∂z
∂z ⎠
∂y
∂x
⎝ ∂t
⎝ ∂x
2
ρ ⎜⎜
2
2
(5.27c)
The following assumptions have been applied:
(1) Steady flow (given).
(2) Incompressible flow; ρ = constant.
(3) No flow or variation of properties in the z direction; w= 0 and ∂/∂z = 0.
(4) Fully developed flow, so no properties except possibly pressure p vary in the x direction; ∂/∂x = 0.
(5) See analysis below.
(6) No body force in the y direction; gy = 0
Assumption (1) eliminates time variations in any fluid property. Assumption (2) eliminates space variations in density. Assumption
(3) states that there is no z component of velocity and no property variations in the z direction. All terms in the z component of the
Navier–Stokes equation cancel. After assumption (4) is applied, the continuity equation reduces to ∂v/∂y = 0. Assumptions (3) and (4)
also indicate that ∂v/∂z = 0 and ∂v/∂x = 0. Therefore v must be constant. Since v is zero at the solid surface, then v must be zero
everywhere. The fact that v = 0 reduces the Navier–Stokes equations further, as indicated by (5). Hence for the y direction
∂p
=0
∂y
which indicates the pressure is a constant across the layer. However, at the free surface p = patm = constant. Hence we conclude that p
= constant throughout the fluid, and so
∂p
=0
∂x
In the x direction, we obtain
µ
∂ 2u
+ ρg = 0
∂y 2
Integrating twice
u=−
c
1
ρgy 2 + 1 y + c2
2µ
µ
To evaluate the constants, c1 and c2, we must apply the boundary conditions. At y = 0, u = 0. Consequently, c2 = 0. At y = a, du/dy =
0 (we assume air friction is negligible). Hence
τ (y = δ ) = µ
which gives
du
dy
=−
y =δ
1
µ
ρgδ +
c1
µ
=0
c1 = ρgδ
and finally
2
1
ρg
ρg 2 ⎡ ⎛ y ⎞ 1 ⎛ y ⎞ ⎤
2
δ ⎢⎜ ⎟ − ⎜ ⎟ ⎥
ρgy +
u=−
y=
µ
µ ⎢⎣⎝ δ ⎠ 2 ⎝ δ ⎠ ⎥⎦
2µ
Problem 8.45
[Difficulty: 3]
Problem 8.46
Given:
Paint flow (Bingham fluid)
Find:
Maximum thickness of paint film before flow occur
[Difficulty: 3]
Solution:
Basic equations:
du
τyx = τy + μp ⋅
dy
Bingham fluid:
Use the analysis of Example 8.3, where we obtain a force balance between gravity and shear stresses:
dτyx
dy
The given data is
τy = 40⋅ Pa
ρ = 1000⋅
= −ρ⋅ g
kg
3
m
From the force balance equation, itegrating
Hence
Motion occurs when
τyx = −ρ⋅ g ⋅ ( δ − y )
τmax ≥ τy
Hence the maximum thickness is
or
τyx = −ρ⋅ g ⋅ y + c
and we have boundary condition
τyx( y = δ) = 0
τmax = ρ⋅ g ⋅ δ
and this is a maximum at the wall
ρ⋅ g ⋅ δ ≥ τy
δ =
τy
ρ⋅ g
δ = 4.08 × 10
−3
m
δ = 4.08 mm
Problem 8.47
[Difficulty: 4]
Given:
Equation for fluid motion in the x-direction.
Find:
Expression for peak pressure
Solution:
Begin with the steady-state Navier-Stokes equation – x-direction
Governing equation:
The Navier-Stokes equations are
4
3
∂u ∂v ∂w
+
+
=0
∂x ∂y ∂z
1
4
5
(5.1c)
3
4
3
⎛ ∂ 2u ∂ 2u ∂ 2u ⎞
⎛ ∂u
∂u
∂p
∂u ⎞
∂u
⎟
⎜
+
u
=
−
v
+
+
g
w
+
ρ⎜
ρ x
µ ⎜⎜ 2 + 2 + 2 ⎟⎟
∂y
∂x
∂x
∂z ⎟⎠
∂y
∂z ⎠
⎝ ∂t
⎝ ∂x
1
4
5
3
4
6
5
(5.27a)
3
⎛ ∂ 2v ∂ 2v ∂ 2v ⎞
⎛ ∂v
∂v ⎞
∂v
∂v
∂p
ρ ⎜⎜ + u + v + w ⎟⎟ = ρg y − + µ ⎜⎜ 2 + 2 + 2 ⎟⎟
∂z ⎠
∂z ⎠
∂y
∂y
∂x
∂y
⎝ ∂t
⎝ ∂x
1
3
3
3
3
3
3
(5.27b)
3
3
⎛∂ w ∂ w ∂ w⎞
⎛ ∂w
∂p
∂w ⎞
∂w
∂w
+w
+v
+u
+ µ ⎜⎜ 2 + 2 + 2 ⎟⎟
⎟⎟ = ρg z −
∂z ⎠
∂y
∂z
∂z ⎠
∂y
∂x
⎝ ∂t
⎝ ∂x
ρ ⎜⎜
2
2
2
(5.27c)
The following assumptions have been applied:
(1) Steady flow (given).
(2) Incompressible flow; ρ = constant.
(3) No flow or variation of properties in the z direction; w= 0 and ∂/∂z = 0.
(4) Fully developed flow, so no properties except possibly pressure p vary in the x direction; ∂/∂x = 0.
(5) See analysis below.
(6) No body force in the y direction; gy = 0
Assumption (1) eliminates time variations in any fluid property. Assumption (2) eliminates space variations in density. Assumption
(3) states that there is no z component of velocity and no property variations in the z direction. All terms in the z component of the
Navier–Stokes equation cancel. After assumption (4) is applied, the continuity equation reduces to ∂v/∂y = 0. Assumptions (3) and (4)
also indicate that ∂v/∂z = 0 and ∂v/∂x = 0. Therefore v must be constant (except of course in a more realistic model v ≠ 0 near the
transition. Since v is zero at the solid surface, then v must be zero everywhere. The fact that v = 0 reduces the Navier–Stokes
equations further, as indicated by (5). Hence for the y direction
∂p
=0
∂y
which indicates the pressure is a constant across the flow. Hence we conclude that p is a function at most of x.
In the x direction, we obtain
0=−
∂p
∂ 2u
+µ 2
∂x
∂y
(1)
Integrating this twice for the first region
u1 =
where
1 dp 2 c1
y + y + c2
µ
2 µ dx 1
dp
denotes the pressure gradient in region 1. Note that we change to regular derivative as p is a function of x only. Note that
dx 1
⎛ ∂p ⎞
⎟ and a function of y only
⎝ ∂x ⎠
Eq 1 implies that we have a function of x only ⎜
⎛ ∂ 2u ⎞
⎜⎜ 2 ⎟⎟ that must add up to be a constant (0); hence
⎝ ∂y ⎠
EACH is a constant! This means that
p
dp
= const = s
L1
dx 1
using the notation of the figure.
To evaluate the constants, c1 and c2, we must apply the boundary conditions. We do this separately for each region.
In the first region, at y = 0, u = U. Consequently, c2 = U. At y = h1, u = 0. Hence
0=
1 dp 2 c1
h1 + h1 + U
2 µ dx 1
µ
so
c1 = −
1 dp
µU
h1 −
2 dx 1
h1
Hence, combining results
u1 =
⎛ y ⎞
1 dp
y 2 − h1 y + U ⎜⎜ − 1⎟⎟
2 µ dx 1
⎝ h1 ⎠
(
)
Exactly the same reasoning applies to the second region, so
u2 =
where
1 dp
2 µ dx
(y
2
2
⎞
⎛ y
− h2 y + U ⎜⎜ − 1⎟⎟
⎝ h2 ⎠
)
p
dp
= const = − s
L2
dx 2
What connects these flow is the flow rate Q.
h1
h2
0
0
q = ∫ u1dy = ∫ u 2 dy = −
1 dp 3 Uh1
1 dp 3 Uh2
h2 −
h1 −
=−
12 µ dx 1
2
12µ dx 2
2
Hence
1 p s 3 Uh1
1 p s 3 Uh2
h1 +
h2 +
=−
12µ L1
2
12µ L2
2
Solving for ps,
p s ⎛ h13 h23 ⎞ Uh2 Uh1
⎜ + ⎟=
−
12 µ ⎜⎝ L1 L2 ⎟⎠
2
2
or
ps =
6 µU (h2 − h1 )
⎛ h13 h23 ⎞
⎜⎜ + ⎟⎟
⎝ L1 L2 ⎠
Problem 8.48
[Difficulty: 2]
Problem 8.49
[Difficulty: 2]
Problem 8.50
Given:
Data on water temperature and tube
Find:
Maximum laminar flow; plot
[Difficulty: 3]
B
Solution:
 5 Ns
A  2.414 10
From Appendix A

B  247.8  K
2
C  140  K
μ ( T)  A  10
in
T C
m
D  7.5 mm
ρ  1000
kg
Recrit  2300
3
m
T1  20 °C
 3 Ns
 
T1  253 K
μ T1  3.74  10
T2  120 °C
2
T2  393 K
 4 Ns
 
μ T2  2.3  10
m
2
m
The plot of viscosity is
0.01
μ
N s
2
m
1 10
3
1 10
4
 20
0
20
40
60
80
100
120
T (C)
For the flow rate
 
ρ Vcrit D
Recrit 
Qmax T1  5.07  10
μ
3.00
5m
s

ρ Qmax D
μ
π
4
 

4  ρ Qmax
2
D
μ π D
Qmax( T) 
π μ( T)  D Recrit
4 ρ
 
L
Qmax T1  182
hr
Qmax T2  3.12  10
3
6m
s
 
L
Qmax T2  11.2
hr
200
Q (L/hr)
150
100
50
 20
0
20
40
60
T (C)
80
100
120
Problem 8.51
[Difficulty: 2]
d
p1 D
F
L
Given:
Hyperdermic needle
Find:
Volume flow rate of saline
Solution:
π⋅ ∆p⋅ d
4
Basic equation
Q=
For the system
F
4⋅ F
∆p = p 1 − p atm =
=
A
2
π⋅ D
4
∆p =
At 68oF, from Table A.7
(Eq. 8.13c; we assume laminar flow and verify this is correct after solving)
128⋅ μ ⋅ L
π
× 7.5⋅ lbf ×
12⋅ in ⎞
⎛ 1
⎜ 0.375⋅ in × 1⋅ ft
⎝
⎠
∆p = 67.9⋅ psi
− 5 lbf ⋅ s
μH2O = 2.1 × 10
⋅
ft
Q =
π
128
× 67.9⋅
lbf
2
Q = 8.27 × 10
V=
Q
A
⋅
Q
=
144 ⋅ in
π⋅ d
2
⋅
1 ⋅ ft
2
4
2
1
12⋅ in
ft
⎞ ×
×
×
12⋅ in ⎠
1 ⋅ in
1 ⋅ ft
−4
1.05 × 10
lbf ⋅ s
1 ⋅ ft
× ⎛⎜ 0.005 ⋅ in ×
⎝
3
3
− 3 in
Q = 1.43 × 10
s
V =
μ = 1.05 × 10
ft
2
×
− 4 lbf ⋅ s
μ = 5⋅ μH2O
2
in
− 7 ft
Check Re:
2
4
π
× 8.27 × 10
− 7 ft
3
s
⋅
Q = 0.0857⋅
s
2
×
3
⎛ 1 ⎞ × ⎛ 12⋅ in ⎞
⎜ .005⋅ in
⎜
⎝
⎠ ⎝ 1⋅ ft ⎠
2
V = 6.07⋅
in
min
ft
s
4
Re =
ρ⋅ V⋅ d
ρ = 1.94⋅
μ
Re = 1.94⋅
slug
ft
slug
ft
3
× 6.07⋅
ft
s
(assuming saline is close to water)
3
× 0.005 ⋅ in ×
1 ⋅ ft
12⋅ in
×
ft
2
1.05 × 10
−4
×
⋅ lbf ⋅ s
slug⋅ ft
2
s ⋅ lbf
Re = 46.7
Flow is laminar
2
Problem 8.52
[Difficulty: 3]
Given:
Data on a tube
Find:
"Resistance" of tube; maximum flow rate and pressure difference for which electrical analogy holds for
(a) kerosine and (b) castor oil
Solution:
L = 250 ⋅ mm
The given data is
D = 7.5⋅ mm
From Fig. A.2 and Table A.2
Kerosene:
μ = 1.1 × 10
− 3 N⋅ s
⋅
ρ = 0.82 × 990 ⋅
2
m
Castor oil:
μ = 0.25⋅
3
= 812 ⋅
m
N⋅ s
ρ = 2.11 × 990 ⋅
2
m
For an electrical resistor
kg
kg
3
3
m
= 2090⋅
m
V = R⋅ I
kg
kg
3
m
(1)
The governing equation for the flow rate for laminar flow in a tube is Eq. 8.13c
4
Q=
or
π⋅ ∆p⋅ D
128 ⋅ μ⋅ L
128 ⋅ μ⋅ L
∆p =
4
⋅Q
(2)
π⋅ D
By analogy, current I is represented by flow rate Q, and voltage V by pressure drop Δp.
Comparing Eqs. (1) and (2), the "resistance" of the tube is
R=
128 ⋅ μ⋅ L
4
π⋅ D
The "resistance" of a tube is directly proportional to fluid viscosity and pipe length, and strongly dependent on the inverse
of diameter
The analogy is only valid for
Re < 2300
ρ⋅
Writing this constraint in terms of flow rate
Q
π
4
or
ρ⋅ V⋅ D
μ
< 2300
⋅D
2
⋅D
μ
< 2300
or
Qmax =
2300⋅ μ⋅ π⋅ D
4⋅ ρ
The corresponding maximum pressure gradient is then obtained from Eq. (2)
∆pmax =
128 ⋅ μ⋅ L
4
π⋅ D
2
⋅ Qmax =
32⋅ 2300⋅ μ ⋅ L
3
ρ⋅ D
Substituting values
(a) For kerosine
(b) For castor oil
3
−5m
Qmax = 1.84 × 10
s
3
−3m
Qmax = 1.62 × 10
s
l
Qmax = 1.10⋅
min
∆pmax = 65.0⋅ Pa
l
Qmax = 97.3⋅
min
∆pmax = 1.30⋅ MPa
The analogy fails when Re > 2300 because the flow becomes turbulent, and "resistance" to flow is then no longer linear with flow
rate
Problem 8.53
[Difficulty: 3]
Problem 8.54
[Difficulty: 3]
Problem 8.56
[Difficulty: 4] Part 1/2
Problem 8.56
[Difficulty: 4] Part 2/2
Problem 8.57
[Difficulty: 4]
Problem 8.52
8.56
Problem 8.58
Given:
Tube dimensions and volumetric flow rate
Find:
Pressure difference and hydraulic resistance
[Difficulty: 2]
Solution:
The flow rate of a fully developed pressure-driven flow in a pipe is Q =
flow rate
π∆pR 4
8µL
. Rearranging it, one obtains ∆p =
8µLQ
. For a
πR 4
Q = 10µl / min , L=1 cm, µ = 1.0 × 10 −3 Pa.s , and R = 1 mm,
∆p =
m
8 10 ×10 −9 m 3
0.01
×
×
×
× 4 ×1.0 ×10 −3 Pa.s = 0.00424 Pa
−12
s 1×10
m
60
π
Similarly, the required pressure drop for other values of R can be obtained.
The hydraulic resistance
Rhyd =
∆p 8µL
=
. Substituting the values of the viscosity, length and radius of the tube, one obtains the
Q πR 4
value of the hydraulic resistance.
R (mm)
1
10-1
10-2
10-3
10-4
∆p
0.00424 Pa
42.4 Pa
424 kPa
4.24 GPa
4.24 x 104 GPa
Rhyd (Pa.s/m3)
2.55 x 107
2.55 x 1011
2.55 x 1015
2.55 x 1019
2.55 x 1023
(3) To achieve a reasonable flow rate in microscale or nanoscale channel, a very high pressure difference is required since ∆p is
proportional to R−4. Therefore, the widely used pressure-driven flow in large scale systems is not appropriate in microscale or
nanoscale channel applications. Other means to manipulate fluids in microscale or nanoscale channel applications are required.
Problem 8.59
.
Given:
Definition of hydraulic resistance
Find:
Hydraulic resistance in a diffuser
Solution:
Basic equation:
Rhyd =
∆p 8µ z2 1
8µ z2
1
dz
dz
=
=
4
∫
∫
z
z
Q
π 1 r
π 1 (ri + αz ) 4
Rhyd =
=
8µ
π
=−
∆p
Q
∫
z
0
1
d (ri + αz )
(ri + αz ) 4
8µ 1
8µ
−3
(ri + αz ) −3 0z = −
[(ri + αz ) −3 − ri ]
πα 3
3πα
Rhyd = −
8µ ⎡
1
1⎤
−
⎢
⎥
3πα ⎣ (ri + αz ) 3 ri 3 ⎦
[Difficulty: 2]
Problem 8.60
[Difficulty: 4]
Given:
Relationship between shear stress and deformation rate; fully developed flow in a cylindrical blood vessel
Find:
Velocity profile; flow rate
Solution:
Similar to the Example Problem described in Section 8.3, based on the force balance, one obtains
τ rx =
r dp
2 dx
(1)
This result is valid for all types of fluids, since it is based on a simple force balance without any assumptions about fluid rheology.
Since the axial pressure gradient in a steady fully developed flow is a constant, Equation (1) shows that τ = 0 < τc at r = 0. Therefore,
there must be a small region near the center line of the blood vessel for which τ < τc. If we call Rc the radial location at which τ = τc,
the flow can then be divided into two regions:
r > Rc: The shear stress vs. shear rate is governed by
τ = τc + µ
du
dr
(2)
r < Rc: τ = 0 < τc.
We first determine the velocity profile in the region r > Rc. Substituting (1) into (2), one obtains:
r dp
du
= τc + µ
2 dx
dr
(3)
Using equation (3) and the fact that du/dr at r = Rc is zero, the critical shear stress can be written as
Rc dp
= τc .
2 dx
(4)
Rearranging eq. (4), Rc is
Rc = 2τ c /
dp
.
dx
Inserting (4) into (3), rearranging, and squaring both sides, one obtains
(5)
µ
du 1 dp
=
[r − 2 rRc + Rc ]
dr 2 dx
(6)
Integrating the above first-order differential equation using the non-slip boundary condition, u = 0 at r = R:
u=−
1 dp ⎡ 2
8
⎤
3/ 2
(R − r 2 ) −
Rc ( Rc − r 3 / 2 ) + 2 Rc ( R − r )⎥ for Rc ≤ r ≤ R
⎢
4µ dx ⎣
3
⎦
(7)
In the region r < Rc, since the shear stress is zero, fluid travels as a plug with a plug velocity. Since the plug velocity must match the
velocity at r = Rc, we set r = Rc in equation (7) to obtain the plug velocity:
u=−
[
]
1 dp 2
2
( R − Rc ) + 2 Rc ( R − Rc ) for r ≤ Rc
4µ dx
(8)
The flow rate is obtained by integrating u(r) across the vessel cross section:
R
Rc
R
Q = ∫ u (r )2πrdr = ∫ u (r )2πrdr + ∫ u (r )2πrdr
0
Rc
0
4
πR 4 dp ⎡ 16 Rc 4 Rc 1 ⎛ Rc ⎞ ⎤
=−
+
− ⎜ ⎟ ⎥
⎢1 −
8µ dx ⎣⎢ 7 R 3 R 21 ⎝ R ⎠ ⎦⎥
Given R = 1mm = 10-3 m, µ = 3.5 cP = 3.5×10-3 Pa⋅s, and τc = 0.05 dynes/cm2 = 0.05×10-1 Pa, and
From eq. (5), Rc = 2τ c /
(9)
dp
= −100 Pa / m .
dx
dp
dx
2 × 0.05 10 ×10 −6 N / m 2
Rc =
= 0.1mm
Pa / m
100
Substituting the values of R, µ, Rc, and
Q=−
π 1× 10−12 m 4
8 3.5 × 10
−3
Pa.s
dp
into eq. (9),
dx
× (−100) Pa / m × [1 −
16 0.1 mm 4 0.1 mm 1 0.1 mm 4
+
− (
) ] = 3.226 × 10 −9 m3 / s
7 1 mm 3 1 mm 21 1 mm
Problem 8.61
[Difficulty: 4]
Given:
Fully developed flow, Navier-Stokes equations; Non-Newtonian fluid
Find:
Velocity profile, flow rate and average velocity
Solution:
According to equation (8.10), we can write the governing equation for Non-Newtonian fluid velocity in a circular tube
n
r ∂p c1
⎛ du ⎞
τ rx = k ⎜ ⎟ =
+
2 ∂x r
⎝ dr ⎠
(1)
However, as for the Newtonian fluid case, we must set c1 = 0 as otherwise we’d have infinite stress at r = 0. Hence, equation (1)
becomes
n
r ∂p
⎛ du ⎞
k⎜ ⎟ =
2 ∂x
⎝ dr ⎠
(2)
The general solution for equation (2), obtained by integrating, is given by
1
1+
1
⎛ 1 ∂p ⎞ n
r n + c2
u =⎜
⎟
⎝ 2k ∂x ⎠ ⎛1 + 1 ⎞
⎜
⎟
⎝ n⎠
1
(3)
Apply the no slip boundary condition at r = R into equation (3), we get
1
1+
1
⎛ 1 ∂p ⎞ n
R n
c 2 = −⎜
⎟
⎝ 2k ∂x ⎠ ⎛1 + 1 ⎞
⎜
⎟
⎝ n⎠
1
(4)
The fluid velocity then is given as
1
n +1
n +1
⎞
n ⎛ 1 ∂p ⎞ n ⎛⎜ n
n ⎟
u (r ) =
r
R
−
⎜
⎟ ⎜
⎟
(n + 1) ⎝ 2k ∂x ⎠ ⎝
⎠
(5)
The volume flow rate is
1
Q = ∫ udA == ∫
A
Hence
R
0
n +1
n +1
⎞
n ⎛ 1 ∂p ⎞ n ⎛ n
2πr
⎟ ⎜⎜ r − R n ⎟⎟dr
⎜
(n + 1) ⎝ 2k ∂x ⎠ ⎝
⎠
(6)
1
1
R
3 n +1
n +1
3 n +1
r2 n ⎤
2nπ ⎛ 1 ∂p ⎞ n ⎡ n
2nπ ⎛ 1 ∂p ⎞ n n ⎛ n
1⎞
n
Q=
=
−
r
R
− ⎟
R
⎜
⎟
⎜
⎟ ⎢
⎜
⎥
(n + 1) ⎝ 2k ∂x ⎠ ⎣ 3n + 1
2
⎝ 3n + 1 2 ⎠
⎦ 0 (n + 1) ⎝ 2k ∂x ⎠
(7)
Simplifying
1
nπ ⎛ 1 ∂p ⎞ n
Q=−
⎜
⎟ R
(3n + 1) ⎝ 2k ∂x ⎠
When n = 1, then k = µ, and
3 n +1
n
(8)
πa 4 ∂p
Q=−
8µ ∂x , just like equation (8.13b) in the textbook.
The average velocity is given by
1
Q
Q
n ⎛ 1 ∂p ⎞ n
V = = 2 =−
⎟ R
⎜
(3n + 1) ⎝ 2k ∂x ⎠
A πR
n +1
n
(9)
Based on Eq. (7), the pressure gradient is
⎤
⎡
∂p
Q(3n + 1) ⎥
= −2k ⎢
3 n +1
⎢
⎥
∂x
⎣ nπR n ⎦
n
(10)
Substituting Q = 1µL/min= 1 × 10-9/60 m3/s, R = 1mm = 10-3 m, and n = 0.5 into eq.(10):
⎡ 10 −9
⎤
(
3(0.5) + 1) ⎥
⎢
∂p
= −2k ⎢ 60 3(0.5 )+1 ⎥
∂x
⎢ 0.5π 0.5
⎥
⎢⎣
⎥⎦
0.5
= −325k Pa/m for n = 0.5
(11)
Similarly, the required pressure gradients for n = 1 and n = 1.5 can be obtained:
∂p
= −42.4k Pa/m, for n = 1
∂x
(12)
∂p
≈ −5.42k Pa/m, for n = 1.5
∂x
(12)
Obviously, the magnitude of the required pressure gradient increases as n decreases. Among the three types of fluids (pseudoplastic
for n = 0.5, Newtonian for n = 1, and dilatant for n = 1.5), the dilatant fluid requires the lowest pressure pump for the same pipe
length.
Problem 8.62
[Difficulty: 2]
Given:
Fully developed flow in a pipe; slip boundary condition on the wall
Find:
Velocity profile and flow rate
Solution:
Similar to the example described in Section 8.3, one obtained
u=
r 2 ∂p
+ c2
4 µ ∂x
(1)
The constant c2 will be determined by the slip velocity boundary condition at r = R:
u =l
∂u
∂r
(2)
and one obtains
c2 =
R 2 ∂p ⎛ l
⎞
⎜ 2 − 1⎟
4 µ ∂x ⎝ R ⎠
(3)
Substituting c2 into Eq.(1), one obtains
u=−
1 ∂p 2
R − r 2 + 2lR
4 µ ∂x
(
)
(4)
The volume flow rate is
R
Q = ∫ u 2πrdr = −
0
l⎤
πR 4 ∂p ⎡
1+ 4 ⎥
⎢
R⎦
8µ ∂x ⎣
Substituting R = 10 µm, µ = 1.84 x 10-5 N⋅s/m2, mean free path l = 68 nm, and −
Q=−
π (10 ×10 −6 ) 4 m 4
8 1.84 × 10 −5 Pa.s
× (−1.0 × 10 6 )Pa/m × [1 + 4
(5)
∂p
= 1.0×106 Pa/m into eq. (5),
∂x
68 ×10 −9 m
] = 2.19 × 10 −10 m 3 /s .
10 × 10 −6 m
Problem 8.63
[Difficulty: 3]
Given:
Fully developed flow, velocity profile, and expression to calculate the flow rate
Find:
Velocity and flow rate
Solution:
For the fully developed flow, the N-S equations can be simplified to
Substituting the trial solution in equation (1), one obtains
⎛ ∂ 2 u ∂ 2 u ⎞ ∂p
= constant
µ ⎜⎜ 2 + 2 ⎟⎟ =
∂z ⎠ ∂x
⎝ ∂y
1 ⎞ ∂p
⎛ 1
− 2µu 0 ⎜ 2 + 2 ⎟ =
b ⎠ ∂x
⎝a
u0 = −
Rearrange it and one obtains
a 2b 2
∂p
2
2
2 µ (a + b ) ∂x
Q = ∫ u ( y, z )dydz = ab ∫
The flow rate is
2π
0
Substituting u ( ρ ,φ ) = u 0 (1 − ρ ) into Eq. (4) and integrating twice:
2
Substituting u0 into (5), one obtains
Q = ab ∫
2π
0
0
1
0
πa 3b 3
1
∂p
Q = πabu0 = −
2
2
2
4 µ (a + b ) ∂x
For a pipe radius R, a = b = R, from equation (6),
1 ⎛ πR 4 ∂p ⎞
⎟
Q pipe = ⎜⎜ −
8 ⎝ µ ∂x ⎟⎠
which is the same as equation (8.13b) in the book.
For a channel with an elliptic cross-section with a = R and b = 1.5R, from equation (6), one has
Q pipe =
29 ⎛ πR 4 ∂p ⎞
⎜−
⎟.
104 ⎜⎝ µ ∂x ⎟⎠
(2)
(3)
1
∫ ρu ( ρ , φ )dρdφ
∫ ρu (1 − ρ
0
(1)
2
(4)
1
)dρdφ = πabu0
2
(6)
(5)
Problem 8.64
Given:
The expression of hydraulic resistance of straight channels with different cross sectional shapes
Find:
Hydraulic resistance
[Difficulty: 2]
Solution:
Based on the expressions of hydraulic resistance listed in the table, one obtains
Using the circle as the example,
Rhyd
1 8 1×10 −3 × 10 × 10 −3 Pa ⋅ s × m
= µL 4 =
= 0.254 ×1012 Pa ⋅ s/m 3
−4 4
4
π
π
(1×10 )
m
a
8
The results are
Shape
Rhyd (1012 Pa·s/m3)
Circle
0.25
Ellipse
3.93
Triangle
18.48
Two plates
0.40
Rectangle
0.51
Square
3.24
Comparing the values of the hydraulic resistances, a straight channel with a circular cross section is the most energy efficient to pump
fluid with a fixed volumetric flow rate; the triangle is the worst.
Problem 8.65
Given:
Two-fluid flow in tube
Find:
Velocity distribution; Plot
[Difficulty: 3]
Solution:
D = 5 ⋅ mm
Given data
L = 5⋅ m
∆p = −5 ⋅ MPa
μ1 = 0.5⋅
N⋅ s
μ2 = 5 ⋅
2
m
N⋅ s
2
m
From Section 8-3 for flow in a pipe, Eq. 8.11 can be applied to either fluid
2
u=
⎛ ∂ ⎞ c1
⋅ ln( r) + c2
p +
4 ⋅ μ ⎝ ∂x ⎠
μ
r
⋅⎜
Applying this to fluid 1 (inner fluid) and fluid 2 (outer fluid)
2
u1 =
r
4 ⋅ μ1
⋅
∆p
L
+
2
c1
⋅ ln( r) + c2
μ1
r=
We need four BCs. Two are obvious
u2 =
D
r
4 ⋅ μ2
u2 = 0
2
⋅
∆p
L
+
(1)
c3
μ2
⋅ ln( r) + c4
r=
D
4
u1 = u2
(2)
The third BC comes from the fact that the axis is a line of symmetry
du1
r= 0
dr
=0
(3)
The fourth BC comes from the fact that the stress at the interface generated by each fluid is the same
r=
du1
du2
μ1 ⋅
= μ2 ⋅
dr
dr
D
4
2
Using these four BCs
⎛ D⎞
⎜2
⎝ ⎠ ⋅ ∆p + c3 ⋅ ln⎛ D ⎞ + c = 0
⎜
4
μ2 ⎝ 2 ⎠
4 ⋅ μ2 L
lim
c1
r → 0 μ1 ⋅ r
(4)
2
⎛ D⎞
⎜4
⎝ ⎠ ⋅ ∆p + c1 ⋅ ln⎛ D ⎞ + c =
⎜
2
μ1 ⎝ 4 ⎠
4 ⋅ μ1 L
2
⎛ D⎞
⎜4
⎝ ⎠ ⋅ ∆p + c3 ⋅ ln⎛ D ⎞ + c
⎜
4
μ2 ⎝ 4 ⎠
4 ⋅ μ2 L
4 ⋅ c1
4 ⋅ c3
D ∆p
D ∆p
⋅
+
= ⋅
+
8 L
D
D
8 L
=0
Hence, after some algebra
c1 = 0
(To avoid singularity)
c2 = −
(
2
D ⋅ ∆p μ2 + 3 ⋅ μ1
64⋅ L
μ1 ⋅ μ2
)
2
c3 = 0
c4 = −
D ⋅ ∆p
16⋅ L⋅ μ2
u 1 ( r) =
The velocity distributions are then
⎡⎢ 2
4 ⋅ μ1 ⋅ L ⎢
⎣
∆p
⋅ r −
2
⎛ D ⎞ ⋅ ( μ2 + 3⋅ μ1 )⎤⎥
⎜2
⎥
4 ⋅ μ2
⎝ ⎠
⎦
u 2 ( r) =
∆p
4 ⋅ μ2 ⋅ L
⎡2
⋅ ⎢r −
⎣
2
⎛ D ⎞ ⎤⎥
⎜2
⎝ ⎠⎦
(Note that these result in the same expression if µ 1 = µ 2, i.e., if we have one fluid)
Evaluating either velocity at r = D/4 gives the velocity at the interface
2
u interface = −
3 ⋅ D ⋅ ∆p
u interface = −
64⋅ μ2 ⋅ L
3
64
× ( 0.005 ⋅ m) × ⎛ −5 × 10 ⋅
2
2
⎞× m × 1
5 ⋅ N⋅ s 5 ⋅ m
2
m ⎠
6 N
⎜
⎝
u interface = 0.234
Evaluating u 1 at r = 0 gives the maximum velocity
2
u max = −
(
D ⋅ ∆p⋅ μ2 + 3 ⋅ μ1
)
64⋅ μ1 ⋅ μ2 ⋅ L
u max = −
1
64
× ( 0.005 ⋅ m) × ⎛ −5 × 10 ⋅
2
2
m
⎞ × 5 + 3 × 0.5 ⋅ m × 1 u
max = 1.02 s
2
N⋅ s 5 ⋅ m
5 × .5
m ⎠
6 N
⎜
⎝
2.5
Inner fluid
Outer fluid
r (mm)
2
1.5
1
0.5
0
0.2
0.4
0.6
Velocity (m/s)
The velocity distributions can be plotted in Excel
0.8
1
1.2
m
s
Problem 8.66
Given:
Turbulent pipe flow
Find:
Wall shear stress
[Difficulty: 2]
Solution:
Basic equation
(Eq. 4.18a)
Assumptions 1) Horizontal pipe 2) Steady flow 3) Fully developed flow
With these assumptions the x momentum equation becomes
2
p1
π D
4
2
π D
 τw π D L  p 2 
0
4
or
τw 
 p2  p1 D
4 L
3
12
1
τw    750  psi 
4
15
Since τw is negative it acts to the left on the fluid, to the right on the pipe wall
τw  3.13 psi

∆p D
4 L
Problem 8.67
Given:
Pipe glued to tank
Find:
Force glue must hold when cap is on and off
[Difficulty: 2]
Solution:
Basic equation
(Eq. 4.18a)
First solve when the cap is on. In this static case
2
π D
Fglue 
4
 p1
where p 1 is the tank pressure
Second, solve for when flow is occuring:
Assumptions 1) Horizontal pipe 2) Steady flow 3) Fully developed flow
With these assumptions the x momentum equation becomes
2
p1
π D
4
2
π D
 τw π D L  p 2 
0
4
Here p1 is again the tank pressure and p 2 is the pressure at the pipe exit; the pipe exit pressure is p atm = 0 kPa gage. Hence
2
Fpipe  Fglue  τw π D L 
π D
4
 p1
We conclude that in each case the force on the glue is the same! When the cap is on the glue has to withstand the tank pressure;
when the cap is off, the glue has to hold the pipe in place against the friction of the fluid on the pipe, which is equal in magnitude
to the pressure drop.
π
lbf
2
Fglue 
 ( 3  in)  30
4
2
in
Fglue  212  lbf
Problem 8.68
[Difficulty: 2]
Given:
Data on pressure drops in flow in a tube
Find:
Which pressure drop is laminar flow, which turbulent
Solution:
Given data
∂
∂x
p 1 = −4.5⋅
kPa
∂
m
∂x
p 2 = −11⋅
kPa
m
D = 30⋅ mm
From Section 8-4, a force balance on a section of fluid leads to
R ∂
D ∂
τw = − ⋅ p = − ⋅ p
2 ∂x
4 ∂x
Hence for the two cases
D ∂
τw1 = − ⋅ p 1
4 ∂x
τw1 = 33.8 Pa
D ∂
τw2 = − ⋅ p 2
4 ∂x
τw2 = 82.5 Pa
Because both flows are at the same nominal flow rate, the higher pressure drop must correspond to the turbulent flow, because, as
indicated in Section 8-4, turbulent flows experience additional stresses. Also indicated in Section 8-4 is that for both flows the
shear stress varies from zero at the centerline to the maximums computed above at the walls.
The stress distributions are linear in both cases: Maximum at the walls and zero at the centerline.
Problem 8.69
Given:
Flow through channel
Find:
Average wall stress
[Difficulty: 2]
Solution:
Basic equation
(Eq. 4.18a)
Assumptions 1) Horizontal pipe 2) Steady flow 3) Fully developed flow
With these assumptions the x momentum equation becomes
H
p 1  W H  τw 2  L ( W  H)  p 2  W H  0
or
W H
τw  p 2  p 1 
2  ( W  H)  L


τw  ∆p
L
2   1 

1
lbf
τw    1 

2
2
in
1  in 
2
144  in
ft
2

H
W
1  ft
12 in
30 ft

1



1  ft
9.5 in 


12 in
1 
30 ft


Since τw < 0, it acts to the left on the fluid, to the right on the channel wall
lbf
τw  0.195 
2
ft
3
τw  1.35  10
 psi
Problem 8.70
[Difficulty: 3]
Problem 8.71
Given:
Data from a funnel viscometer filled with pitch.
Find:
Viscosity of pitch.
[Difficulty: 1]
Solution:
V πD 4 ρg ⎛ h ⎞
(Volume flow rate)
=
⎜1 + ⎟
t
128µ ⎝ L ⎠
where Q is the volumetric flow rate, V flow volume, t is the time of flow, D is the diameter of the funnel stem, ρ is the density of the
Basic equation: Q =
pitch, µ is the viscosity of the pitch, h is the depth of the pitch in the funnel body, and L is the length of the funnel stem.
Assumption:
Viscous effects above the stem are negligible and the stem has a uniform diameter.
The given or available data is:
Calculate the flow rate:
V = 4.7 ×10 −5 m 3
t = 17,708days
D = 9.4mm
h = 75mm
L = 29mm
ρ = 1.1×103
Q=
V
=
t
kg
m3
4.7 × 10 −5 m 3
m3
= 3.702 × 10 −14
24hour 3600s
s
17708day ×
×
day
hour
Solve the governing equation for viscosity:
µ=
µ=
πD 4 ρg ⎛
h⎞
⎜1 + ⎟
128Q ⎝ L ⎠
µ = 2.41×108
N ⋅s
m2
4
m
m
⎛
⎞
3 kg
⎟ ×1.1× 10 3 × 9.81 2
m
s ⎛ 75mm ⎞ N × s 2
⎝ 1000mm ⎠
⎟×
⎜1 +
3
⎝ 29mm ⎠ kg × m
−14 m
128 × 3.702 ×10
s
π × (9.4mm )4 × ⎜
Compare this to the viscosity of water, which is 10-3 N·s/m2!
Relate this equation to 8.13c for flow driven by a pressure gradient:
Q=
π∆pD 4 πD 4 ∆p
.
=
×
128µL 128µ L
For this problem, the pressure (Δp) is replaced by the hydrostatic force of the pitch.
Consider the pressure variation in a static fluid.
∆p
∆p
dp
= − ρg = − ρg =
=
.
∆z L + h
dz
Replacing the term in 8.13c
Q=
Hence
which is the same as the given equation.
πD 4 ∆p πD 4 ρg × (L + h ) πD 4
⎛ h⎞
×
=
×
=
× ρg × ⎜1 + ⎟
L
128µ L 128µ
128µ
⎝ L⎠
V πD 4 ρg ⎛ h ⎞
Q= =
⎜1 + ⎟
t
128µ ⎝ L ⎠
Problem 8.72
[Difficulty: 3]
Problem 8.73
[Difficulty: 3]
Given: Data on mean velocity in fully developed turbulent flow
Find: Trendlines for each set; values of n for each set; plot
Solution:
y/R
0.898
0.794
0.691
0.588
0.486
0.383
0.280
0.216
0.154
0.093
0.062
0.041
0.024
u/U
0.996
0.981
0.963
0.937
0.907
0.866
0.831
0.792
0.742
0.700
0.650
0.619
0.551
y/R
0.898
0.794
0.691
0.588
0.486
0.383
0.280
0.216
0.154
0.093
0.062
0.037
u/U
0.997
0.998
0.975
0.959
0.934
0.908
0.874
0.847
0.818
0.771
0.736
0.690
Equation 8.22 is
Mean Velocity Distributions in a Pipe
u/U
1.0
0.1
0.01
0.10
1.00
y/R
Re = 50,000
Re = 500,000
Power (Re = 500,000)
Power (Re = 50,000)
Applying the Trendline analysis to each set of data:
At Re = 50,000
At Re = 500,000
u/U = 1.017(y/R )0.161
u/U = 1.017(y/R )0.117
2
with R = 0.998 (high confidence)
Hence
1/n = 0.161
n = 6.21
with R 2 = 0.999 (high confidence)
Hence
Both sets of data tend to confirm the validity of Eq. 8.22
1/n = 0.117
n = 8.55
Problem 8.74
[Difficulty: 3]
Problem 8.75
[Difficulty: 3] Part 1/2
Problem 8.75
[Difficulty: 3] Part 2/2
Problem 8.76
Given:
Laminar flow between parallel plates
Find:
Kinetic energy coefficient, α
[Difficulty: 3]
Solution:
Basic Equation: The kinetic energy coefficient, α is given by
∫
α=
A
ρ V 3dA
(8.26b)
m V 2
From Section 8-2, for flow between parallel plates
2
2
⎡ ⎛
⎞ ⎤ 3 ⎡ ⎛ y ⎞ ⎤
y
⎟ ⎥ = V ⎢1 − ⎜
⎟ ⎥
u = umax ⎢1 − ⎜
⎢ ⎜a ⎟ ⎥ 2 ⎢ ⎜a ⎟ ⎥
⎢⎣ ⎝ 2 ⎠ ⎥⎦
⎢⎣ ⎝ 2 ⎠ ⎥⎦
since umax =
3
V .
2
Substituting
α=
∫
A
ρV 3dA
m V 2
=
∫
A
ρu 3dA
ρV A V 2
3
=
1 ⎛u⎞
1
dA =
⎟
⎜
∫
A A⎝V ⎠
wa
a
2
a
2
3
3
2 ⎛u⎞
⎛u⎞
∫a ⎜⎝ V ⎟⎠ wdy = a ∫0 ⎜⎝ V ⎟⎠ dy
−
2
Then
3
3
31
1
3
2 a ⎛ u ⎞ ⎛ umax ⎞ ⎛⎜ y ⎞⎟ ⎛ 3 ⎞
⎜⎜
⎟⎟ ⎜
α=
= ⎜ ⎟ ∫ (1 − η 2 ) dη
⎟ d⎜
∫
a ⎟ ⎝2⎠ 0
a 2 0 ⎝ umax ⎠ ⎝ V ⎠
⎝ 2⎠
where η =
y
a
2
Evaluating,
(1 − η )
2 3
= 1 − 3η 2 + 3η 4 − η 6
The integral is then
⎛ 3⎞
α =⎜ ⎟
⎝2⎠
31
⎛ 3⎞
∫0 (1 − 3η + 3η − η )dη = ⎜⎝ 2 ⎟⎠
2
4
6
3
1
3 5 1 7 ⎤ 27 16
⎡
3
⎢⎣η − η + 5 η − 7 η ⎥⎦ = 8 35 = 1.54
0
Problem 8.77
[Difficulty: 3]
Problem 8.78
[Difficulty: 3]
Given:
Definition of kinetic energy correction coefficient α
Find:
α for the power-law velocity profile; plot
Solution:
Equation 8.26b is
α=
⌠
⎮
3
⎮ ρ⋅ V dA
⌡
2
mrate⋅ Vav
where V is the velocity, mrate is the mass flow rate and Vav is the average velocity
1
n
For the power-law profile (Eq. 8.22)
V = U⋅ ⎛⎜ 1 −
For the mass flow rate
mrate = ρ⋅ π⋅ R ⋅ Vav
Hence the denominator of Eq. 8.26b is
mrate⋅ Vav = ρ⋅ π⋅ R ⋅ Vav
We next must evaluate the numerator of Eq. 8.26b
⎝
⎞
R⎠
r
2.
2
3
⌠
3
⎮
⎮
n
r
3
3
ρ⋅ V dA = ⎮ ρ⋅ 2 ⋅ π⋅ r⋅ U ⋅ ⎛⎜ 1 − ⎞ dr
⎮
R⎠
⎝
⌡
⌠
⎮
⎮
⌡
⌠
⎮
⎮
⎮
⎮
⌡
2
R
3
2 2 3
n
2 ⋅ π⋅ ρ⋅ R ⋅ n ⋅ U
r⎞
⎛
dr =
ρ⋅ 2 ⋅ π⋅ r⋅ U ⋅ ⎜ 1 −
( 3 + n) ⋅ ( 3 + 2⋅ n)
R⎠
⎝
3
0
To integrate substitute
Then
m=1−
r
R
r = R⋅ ( 1 − m)
⌠
⎮
⎮
⎮
⎮
⌡
dm = −
R
dr = −R⋅ dm
R
0
dr
0
⌠
3
⎮
n
⎮
r
n
3
ρ⋅ 2 ⋅ π⋅ r⋅ U ⋅ ⎛⎜ 1 − ⎞ dr = −⎮ ρ⋅ 2 ⋅ π⋅ R⋅ ( 1 − m) ⋅ m ⋅ R dm
⌡
R⎠
⎝
1
3
1
⌠
3 ⎞
⎮
⎛ 3
+1
⎜
⎮
n
3
n
ρ⋅ V dA = ⎮ ρ⋅ 2 ⋅ π⋅ R⋅ ⎝ m − m
⎠ ⋅ R dm
⌡
⌠
⎮
⎮
⌡
Hence
0
2 2
3
⌠
2 ⋅ R ⋅ n ⋅ ρ⋅ π⋅ U
⎮
3
d
ρ
⋅
V
A
=
⎮
( 3 + n) ⋅ ( 3 + 2⋅ n)
⌡
α=
Putting all these results together
⌠
⎮
3
⎮ ρ⋅ V dA
⌡
2
2
=
3
( 3+ n) ⋅ ( 3+ 2⋅ n)
2
3
ρ⋅ π⋅ R ⋅ Vav
mrate⋅ Vav
α=
2
2⋅ R ⋅ n ⋅ ρ⋅ π⋅ U
3
2
2⋅ n
⎛ U ⎞ ⋅
⎜V
⎝ av ⎠ ( 3 + n) ⋅ ( 3 + 2⋅ n )
To plot α versus ReVav we use the following parametric relations
( )
n = −1.7 + 1.8⋅ log Reu
Vav
U
=
2⋅ n
(Eq. 8.23)
2
(Eq. 8.24)
( n + 1) ⋅ ( 2⋅ n + 1)
Vav
ReVav =
⋅ ReU
U
α=
3
2
2⋅ n
⎛ U ⎞ ⋅
⎜V
⎝ av ⎠ ( 3 + n) ⋅ ( 3 + 2⋅ n )
(Eq. 8.27)
A value of ReU leads to a value for n; this leads to a value for Vav/U; these lead to a value for ReVav and α
The plots of α, and the error in assuming α = 1, versus ReVav can be done in Excel.
Re U
1.00E+04
2.50E+04
5.00E+04
7.50E+04
1.00E+05
2.50E+05
5.00E+05
7.50E+05
1.00E+06
2.50E+06
5.00E+06
7.50E+06
1.00E+07
n
5.50
6.22
6.76
7.08
7.30
8.02
8.56
8.88
9.10
9.82
10.4
10.7
10.9
V av/U
0.776
0.797
0.811
0.818
0.823
0.837
0.846
0.851
0.854
0.864
0.870
0.873
0.876
Re Vav
Alpha
7.76E+03 1.09
1.99E+04 1.07
4.06E+04 1.06
6.14E+04 1.06
8.23E+04 1.05
2.09E+05 1.05
4.23E+05 1.04
6.38E+05 1.04
8.54E+05 1.04
2.16E+06 1.03
4.35E+06 1.03
6.55E+06 1.03
8.76E+06 1.03
Error
8.2%
6.7%
5.9%
5.4%
5.1%
4.4%
3.9%
3.7%
3.5%
3.1%
2.8%
2.6%
2.5%
Kinetic Energy Coefficient
vs Reynolds Number
Alpha
1.10
1.08
1.05
1.03
1.00
1E+03
1E+04
1E+05
1E+06
1E+07
1E+06
1E+07
Re Vav
Error in assuming Alpha = 1
vs Reynolds Number
10.0%
Error
7.5%
5.0%
2.5%
0.0%
1E+03
1E+04
1E+05
Re Vav
Problem 8.79
Given:
Data on flow through elbow
Find:
Head loss
Solution:
Basic equation
[Difficulty: 2]
2
2
⎛⎜ p
⎞ ⎜⎛ p
⎞ h
V1
V2
1
lT
2
+
α
⋅
+
z
+
α
⋅
+
z
−
⎜ ρ⋅ g
⎜ ρ⋅ g
1
2 = g = HlT
2
⋅
g
2
⋅
g
⎝
⎠ ⎝
⎠
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1
Then
HlT =
p1 − p2
ρ⋅ g
2
+
V1 − V2
3
2⋅ g
2
+ z1 − z2
2
(
)
2
2
m
kg⋅ m
s
1
m
s
3 N
2
2
HlT = ( 70 − 45) × 10 ⋅
×
×
×
+ × 1.75 − 3.5 ⋅ ⎛⎜ ⎞ ×
+ ( 2.25 − 3 ) ⋅ mHlT = 1.33 m
2
1000⋅ kg
2
9.81⋅ m
2
s
9.81
⋅m
⎝
⎠
m
s ⋅N
In terms of energy/mass
h lT = g ⋅ HlT
h lT = 9.81⋅
m
2
s
2
× 1.33⋅ m ×
N⋅ s
kg⋅ m
h lT = 13.0⋅
N⋅ m
kg
Problem 8.80
[Difficulty: 2]
Given:
Data on flow in a pipe
Find:
Head loss for horizontal pipe; inlet pressure for different alignments; slope for gravity feed
Solution:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
+
α
⋅
+
g
⋅
z
−
⎜ρ
⎜
1 2
1
2 2
2 = h lT
⎝
⎠ ⎝ρ
⎠
The basic equation between inlet (1) and exit (2) is
Given or available data
D = 75⋅ mm
Horizontal pipe data
p 1 = 275 ⋅ kPa
Equation 8.29 becomes
h lT =
V = 5⋅
m
s
p 2 = 0 ⋅ kPa
p1 − p2
ρ = 999 ⋅
kg
m
(Gage pressures)
h lT = 275 ⋅
ρ
3
μ = 0.001 ⋅
(8.29)
N⋅ s
2
m
z1 = z2
V1 = V2
J
kg
For an inclined pipe with the same flow rate, the head loss will be the same as above; in addition we have the following new data
z1 = 0 ⋅ m
Equation 8.29 becomes
z2 = 15⋅ m
(
)
p 1 = p 2 + ρ⋅ g ⋅ z2 − z1 + ρ⋅ h lT
p 1 = 422 ⋅ kPa
For a declining pipe with the same flow rate, the head loss will be the same as above; in addition we have the following new data
z1 = 0 ⋅ m
Equation 8.29 becomes
z2 = −15⋅ m
(
)
p 1 = p 2 + ρ⋅ g ⋅ z2 − z1 + ρ⋅ h lT
p 1 = 128 ⋅ kPa
For a gravity feed with the same flow rate, the head loss will be the same as above; in addition we have the following new data
p 1 = 0 ⋅ kPa
Equation 8.29 becomes
h lT
z2 = z1 −
g
(Gage)
z2 = −28.1 m
Problem 8.81
Given:
Data on flow through elbow
Find:
Inlet velocity
[Difficulty: 2]
Solution:
Basic equation
2
2
⎛⎜ p
⎞ ⎜⎛ p
⎞ h
V1
V2
1
lT
2
⎜ ρ⋅ g + α⋅ 2⋅ g + z1 − ⎜ ρ⋅ g + α⋅ 2 ⋅ g + z2 = g = HlT
⎝
⎠ ⎝
⎠
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1
Then
2
2
(
V2 − V1 = 2 ⋅ V1
)2 − V12 = 3⋅ V12 =
(
2⋅ p1 − p2
)
(
)
+ 2 ⋅ g ⋅ z1 − z2 − 2 ⋅ g ⋅ HlT
ρ
⎡ (p1 − p2)
⎤
+ g ⋅ ( z1 − z2 ) − g ⋅ HlT⎥
3 ⎣
ρ
⎦
V1 =
2
V1 =
2
⋅⎢
3
⎡
× ⎢50 × 10 ⋅
⎢
⎣
3 N
2
m
3
×
m
1000⋅ kg
×
kg⋅ m
2
s ⋅N
+
9.81⋅ m
2
s
× ( −2 ) ⋅ m − 9.81⋅
m
2
s
⎤
× 1 ⋅ m⎥
⎥
⎦
m
V1 = 3.70
s
Problem 8.82
Given:
A given piping system and volume flow rate with two liquid choices.
Find:
Which liquid has greater pressure loss
[Difficulty: 2]
Solution:
Governing equation:
⎞
⎛ P1
⎞ ⎛P
V2
V2
⎜⎜ + α1 1 + gz1 ⎟⎟ − ⎜⎜ 2 + α 2 2 + gz 2 ⎟⎟ = hlT
2
2
⎠
⎠ ⎝ρ
⎝ρ
2
2
LV
V
hlT = hl + hlm = f
+K
D 2
2
Assumption: 1) Steady flow 2) Incompressible 3) Neglect elevation effects 4)Neglect velocity effects
LV2
V2
∆P = ρf
+ ρK
2
D 2
From Table A.8 it is seen that hot water has a lower density and lower kinematic viscosity than cold water.
The lower density means that for a constant minor loss coefficient (K) and velocity the pressure loss due to minor losses will be less
for hot water.
The lower kinematic viscosity means that for a constant diameter and velocity the Reynolds number will increase. From Figure 8.13 it
is seen that increasing the Reynolds number will either result in a decreased friction factor (f) or no change in the friction factor. This
potential decrease in friction factor combined with a lower density for hot water means that the pressure loss due to major losses will
be less for hot water as well.
Cold water has a greater pressure drop
Problem 8.83
Given:
Increased friction factor for water tower flow
Find:
How much flow is decreased
[Difficulty: 2]
Solution:
Basic equation from Example 8.7
V2 =
(
2 ⋅ g ⋅ z1 − z2
f ⋅ ⎛⎜
L
⎝D
)
+ 8⎞ + 1
⎠
where
L = 680 ⋅ ft
D = 4 ⋅ in
With f = 0.0308, we obtain
ft
V2 = 8.97⋅
s
and Q = 351 gpm
We need to recompute with f = 0.035
V2 =
2 × 32.2⋅
ft
2
× 80⋅ ft ×
s
z1 − z2 = 80⋅ ft
1
0.035 ⋅ ⎜⎛
⎜
⎝
680
4
12
+ 8⎞ + 1
ft
V2 = 8.42⋅
s
⎠
2
Hence
π⋅ D
Q = V2 ⋅ A = V2 ⋅
4
Q = 8.42⋅
ft
s
×
π
4
2
×
⎛ 4 ⋅ ft⎞ × 7.48⋅ gal × 60⋅ s
⎜ 12
1 ⋅ min
3
⎝
⎠
1 ⋅ ft
Q = 330 ⋅ gpm
(From Table G.2 1 ft3 = 7.48 gal)
Hence the flow is decreased by
( 330 − 309 ) ⋅ gpm = 21⋅ gpm
Problem 8.84
[Difficulty: 2]
Given:
Increased friction factor for water tower flow, and reduced length
Find:
How much flow is decreased
Solution:
Basic equation from Example 8.7
V2 =
(
2 ⋅ g ⋅ z1 − z2
f ⋅ ⎛⎜
L
⎝D
where now we have
L = 530 ⋅ ft
We need to recompute with f = 0.04
V2 =
)
+ 8⎞ + 1
⎠
D = 4 ⋅ in
2 × 32.2⋅
ft
2
× 80⋅ ft ×
s
z1 − z2 = 80⋅ ft
1
0.035 ⋅ ⎜⎛
⎜
⎝
530
4
12
+ 8⎞ + 1
ft
V2 = 9.51⋅
s
⎠
2
Hence
π⋅ D
Q = V2 ⋅ A = V2 ⋅
4
Q = 9.51⋅
ft
s
×
π
4
2
×
⎛ 4 ⋅ ft⎞ × 7.48⋅ gal × 60⋅ s
⎜ 12
1 ⋅ min
3
⎝
⎠
1 ⋅ ft
Q = 372 ⋅ gpm
(From Table G.2 1 ft3 = 7.48 gal)
Problem 8.85
[Difficulty: 2]
Problem 8.86
Given:
Data on flow through Alaskan pipeline
Find:
Head loss
Solution:
Basic equation
[Difficulty: 2]
2
2
⎛ p
⎞ ⎛ p
⎞ h
V1
V2
lT
⎜ 1
⎜ 2
+ α⋅
+ z1 −
+ α⋅
+ z2 =
= HlT
⎜ ρ ⋅g
⎜ ρ ⋅g
2
⋅
g
2
⋅
g
g
oil
oil
⎝
⎠ ⎝
⎠
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) SG = 0.9 (Table A.2)
Then
p1 − p2
HlT =
+ z1 − z2
SGoil⋅ ρH2O⋅ g
3
2
1
m
kg⋅ m
s
3 N
HlT = ( 8250 − 350 ) × 10 ⋅
×
×
×
×
+ ( 45 − 115 ) ⋅ m
2
0.9 1000⋅ kg
2
9.81⋅ m
m
s ⋅N
In terms of energy/mass h lT = g ⋅ HlT
h lT = 9.81⋅
m
2
s
HlT = 825 m
2
× 825 ⋅ m ×
N⋅ s
kg⋅ m
h lT = 8.09⋅
kN⋅ m
kg
Problem 8.87
[Difficulty: 2]
Problem 8.88
Given:
Data on flow through a tube
Find:
Head loss
Solution:
Basic equation
[Difficulty: 2]
2
2
⎞ ⎜⎛ p
⎞ h
⎛⎜ p
V1
V2
1
lT
2
+
α
⋅
+
z
+
α
⋅
+
z
−
⎜ ρ⋅ g
⎜ ρ⋅ g
1
2 = g = HlT
2
⋅
g
2
⋅
g
⎝
⎠ ⎝
⎠
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1
Given or available data
The basic equation reduces to
Q = 10⋅
h lT =
L
min
∆p
ρ
D = 15⋅ mm
∆p = 85⋅ kPa
ρ = 999 ⋅
kg
3
m
2
h lT = 85.1
m
2
s
HlT =
h lT
g
HlT = 8.68 m
Problem 8.89
[Difficulty: 2]
Problem 8.90
Given:
Data on flow from reservoir
Find:
Head from pump; head loss
Solution:
Basic equations
[Difficulty: 3]
2
2
⎛⎜ p
⎞ ⎜⎛ p
⎞ h
V3
V4
3
lT
4
+
α
⋅
+
z
+
α
⋅
+
z
−
⎜ ρ⋅ g
⎜ ρ⋅ g
3
4 = g = HlT
2
⋅
g
2
⋅
g
⎝
⎠ ⎝
⎠
for flow from 3 to 4
2
2
⎛⎜ p
⎞ ⎜⎛ p
⎞ ∆h
V3
V2
3
pump
2
= Hpump for flow from 2 to 3
⎜ ρ⋅ g + α⋅ 2⋅ g + z3 − ⎜ ρ⋅ g + α⋅ 2 ⋅ g + z2 =
g
⎝
⎠ ⎝
⎠
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) V2 = V3 = V4 (constant area pipe)
Then for the pump
Hpump =
p3 − p2
ρ⋅ g
3
2
m
kg⋅ m
s
3 N
Hpump = ( 450 − 150 ) × 10 ⋅
×
×
×
2
1000⋅ kg
2
9.81⋅ m
m
s ⋅N
In terms of energy/mass
h pump = g ⋅ Hpump
h pump = 9.81⋅
m
2
2
× 30.6⋅ m ×
s
For the head loss from 3 to 4 HlT =
p3 − p4
ρ⋅ g
Hpump = 30.6 m
N⋅ s
kg⋅ m
3
h lT = g ⋅ HlT
N⋅ m
kg
+ z3 − z4
2
m
kg⋅ m
s
3 N
HlT = ( 450 − 0 ) × 10 ⋅
×
×
×
+ ( 0 − 35) ⋅ m
2
1000⋅ kg
2
9.81⋅ m
m
s ⋅N
In terms of energy/mass
h pump = 300 ⋅
h lT = 9.81⋅
m
2
s
HlT = 10.9 m
2
× 10.9⋅ m ×
N⋅ s
kg⋅ m
h lT = 107 ⋅
N⋅ m
kg
Problem 8.91
[Difficulty: 2]
Given:
Data on flow in a pipe
Find:
Friction factor; Reynolds number; if flow is laminar or turbulent
Solution:
Given data
From Appendix A
∆p
D = 75⋅ mm
ρ = 1000⋅
L
kg
= 0.075 ⋅
μ = 4 ⋅ 10
3
Pa
m
kg
mrate = 0.075 ⋅
s
− 4 N⋅ s
⋅
m
2
m
The governing equations between inlet (1) and exit (2) are
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α1⋅ 2 + g ⋅ z1 − ⎜ ρ + α2 ⋅ 2 + g⋅ z2 = h l (8.29)
⎝
⎠ ⎝
⎠
2
hl = f ⋅
L V
⋅
D 2
(8.34)
For a constant area pipe
V1 = V2 = V
Hence Eqs. 8.29 and 8.34 become
f =
2⋅ D
2
⋅
(p1 − p2)
ρ
L⋅ V
For the velocity
mrate
V =
ρ⋅
π
4
f =
The Reynolds number is
Re =
2 ⋅ D ∆p
⋅
2 L
ρ⋅ V
V = 0.017
2
⋅D
2 ⋅ D ∆p
⋅
2 L
ρ⋅ V
Hence
=
ρ⋅ V⋅ D
μ
m
s
f = 0.0390
Re = 3183
This Reynolds number indicates the flow is turbulent.
(From Eq. 8.37, at this Reynolds number the friction factor for a smooth pipe is f = 0.043; the friction factor computed above thus
indicates that, within experimental error, the flow corresponds to turbulent flow in a smooth pipe)
Problem 8.92
[Difficulty: 2]
Problem 8.93
[Difficulty: 3]
Using the above formula for f 0, and Eq. 8.37 for f 1
e/D =
0
0.0001
0.0002
0.0005
0.001
0.002
0.005
0.01
0.02
0.05
Re
1.00E+04
2.50E+04
5.00E+04
7.50E+04
1.00E+05
2.50E+05
5.00E+05
7.50E+05
1.00E+06
5.00E+06
1.00E+07
5.00E+07
1.00E+08
0.0310
0.0244
0.0208
0.0190
0.0179
0.0149
0.0131
0.0122
0.0116
0.0090
0.0081
0.0066
0.0060
0.0311
0.0247
0.0212
0.0195
0.0185
0.0158
0.0145
0.0139
0.0135
0.0124
0.0122
0.0120
0.0120
0.0313
0.0250
0.0216
0.0200
0.0190
0.0167
0.0155
0.0150
0.0148
0.0140
0.0139
0.0138
0.0137
0.0318
0.0258
0.0226
0.0212
0.0204
0.0186
0.0178
0.0175
0.0173
0.0168
0.0168
0.0167
0.0167
0.0327
0.0270
0.0242
0.0230
0.0223
0.0209
0.0204
0.0201
0.0200
0.0197
0.0197
0.0196
0.0196
0.0342
0.0291
0.0268
0.0258
0.0253
0.0243
0.0239
0.0238
0.0237
0.0235
0.0235
0.0234
0.0234
0.0383
0.0342
0.0325
0.0319
0.0316
0.0309
0.0307
0.0306
0.0305
0.0304
0.0304
0.0304
0.0304
0.0440
0.0407
0.0395
0.0390
0.0388
0.0383
0.0381
0.0380
0.0380
0.0379
0.0379
0.0379
0.0379
0.0534
0.0508
0.0498
0.0494
0.0493
0.0489
0.0488
0.0487
0.0487
0.0487
0.0486
0.0486
0.0486
0.0750
0.0731
0.0724
0.0721
0.0720
0.0717
0.0717
0.0716
0.0716
0.0716
0.0716
0.0716
0.0716
0.001
0.002
0.005
0.01
0.02
0.05
0.0338
0.0288
0.0265
0.0256
0.0251
0.0241
0.0238
0.0237
0.0236
0.0235
0.0234
0.0234
0.0234
0.0376
0.0337
0.0322
0.0316
0.0313
0.0308
0.0306
0.0305
0.0305
0.0304
0.0304
0.0304
0.0304
0.0431
0.0402
0.0391
0.0387
0.0385
0.0381
0.0380
0.0380
0.0380
0.0379
0.0379
0.0379
0.0379
0.0523
0.0502
0.0494
0.0492
0.0490
0.0488
0.0487
0.0487
0.0487
0.0486
0.0486
0.0486
0.0486
0.0738
0.0725
0.0720
0.0719
0.0718
0.0716
0.0716
0.0716
0.0716
0.0716
0.0716
0.0716
0.0716
f0
Using the add-in function Friction factor from the Web
e/D =
Re
1.00E+04
2.50E+04
5.00E+04
7.50E+04
1.00E+05
2.50E+05
5.00E+05
7.50E+05
1.00E+06
5.00E+06
1.00E+07
5.00E+07
1.00E+08
0
0.0001
0.0002
0.0005
f
0.0309
0.0245
0.0209
0.0191
0.0180
0.0150
0.0132
0.0122
0.0116
0.0090
0.0081
0.0065
0.0059
0.0310
0.0248
0.0212
0.0196
0.0185
0.0158
0.0144
0.0138
0.0134
0.0123
0.0122
0.0120
0.0120
The error can now be computed
0.0312
0.0250
0.0216
0.0200
0.0190
0.0166
0.0154
0.0150
0.0147
0.0139
0.0138
0.0138
0.0137
0.0316
0.0257
0.0226
0.0212
0.0203
0.0185
0.0177
0.0174
0.0172
0.0168
0.0168
0.0167
0.0167
0.0324
0.0268
0.0240
0.0228
0.0222
0.0208
0.0202
0.0200
0.0199
0.0197
0.0197
0.0196
0.0196
e/D =
0
0.0001
0.0002
0.0005
0.001
0.002
0.005
0.01
0.02
0.05
Re
1.00E+04
2.50E+04
5.00E+04
7.50E+04
1.00E+05
2.50E+05
5.00E+05
7.50E+05
1.00E+06
5.00E+06
1.00E+07
5.00E+07
1.00E+08
0.29%
0.39%
0.63%
0.69%
0.71%
0.65%
0.52%
0.41%
0.33%
0.22%
0.49%
1.15%
1.44%
0.36%
0.24%
0.39%
0.38%
0.33%
0.04%
0.26%
0.41%
0.49%
0.51%
0.39%
0.15%
0.09%
0.43%
0.11%
0.19%
0.13%
0.06%
0.28%
0.51%
0.58%
0.60%
0.39%
0.27%
0.09%
0.06%
0.61%
0.21%
0.25%
0.35%
0.43%
0.64%
0.64%
0.59%
0.54%
0.24%
0.15%
0.05%
0.03%
Error (%)
0.88%
1.27%
0.60%
1.04%
0.67%
1.00%
0.73%
0.95%
0.76%
0.90%
0.72%
0.66%
0.59%
0.47%
0.50%
0.37%
0.43%
0.31%
0.16%
0.10%
0.10%
0.06%
0.03%
0.02%
0.02%
0.01%
1.86%
1.42%
1.11%
0.93%
0.81%
0.48%
0.31%
0.23%
0.19%
0.06%
0.03%
0.01%
0.00%
2.12%
1.41%
0.98%
0.77%
0.64%
0.35%
0.21%
0.15%
0.12%
0.03%
0.02%
0.01%
0.00%
2.08%
1.21%
0.77%
0.58%
0.47%
0.24%
0.14%
0.10%
0.08%
0.02%
0.01%
0.00%
0.00%
1.68%
0.87%
0.52%
0.38%
0.30%
0.14%
0.08%
0.06%
0.05%
0.01%
0.01%
0.00%
0.00%
The maximum discrepancy is 2.12% at Re = 10,000 and e/D = 0.01
0.100
f0
0.010
0.001
1E+04
e/D = 0
e/D = 0.0001
e/D = 0.0002
e/D = 0.0005
e/D = 0.001
e/D = 0.002
e/D = 0.005
e/D = 0.01
e/D = 0.02
e/D = 0.05
1E+05
1E+06
Re
1E+07
1E+08
Problem 8.94
[Difficulty: 3]
Solution:
Using the add-in function Friction factor from the web site
e/D =
Re
500
1.00E+03
1.50E+03
2.30E+03
1.00E+04
1.50E+04
1.00E+05
1.50E+05
1.00E+06
1.50E+06
1.00E+07
1.50E+07
1.00E+08
0
0.0001
0.0002
0.0005
0.001
0.002
0.005
0.01
0.02
0.04
0.1280
0.0640
0.0427
0.0489
0.0338
0.0313
0.0251
0.0246
0.0236
0.0235
0.0234
0.0234
0.0234
0.1280
0.0640
0.0427
0.0512
0.0376
0.0356
0.0313
0.0310
0.0305
0.0304
0.0304
0.0304
0.0304
0.1280
0.0640
0.0427
0.0549
0.0431
0.0415
0.0385
0.0383
0.0380
0.0379
0.0379
0.0379
0.0379
0.1280
0.0640
0.0427
0.0619
0.0523
0.0511
0.0490
0.0489
0.0487
0.0487
0.0486
0.0486
0.0486
0.1280
0.0640
0.0427
0.0747
0.0672
0.0664
0.0649
0.0648
0.0647
0.0647
0.0647
0.0647
0.0647
f
0.1280
0.0640
0.0427
0.0473
0.0309
0.0278
0.0180
0.0166
0.0116
0.0109
0.0081
0.0076
0.0059
0.1280
0.0640
0.0427
0.0474
0.0310
0.0280
0.0185
0.0172
0.0134
0.0130
0.0122
0.0121
0.0120
0.1280
0.0640
0.0427
0.0474
0.0312
0.0282
0.0190
0.0178
0.0147
0.0144
0.0138
0.0138
0.0137
0.1280
0.0640
0.0427
0.0477
0.0316
0.0287
0.0203
0.0194
0.0172
0.0170
0.0168
0.0167
0.0167
0.1280
0.0640
0.0427
0.0481
0.0324
0.0296
0.0222
0.0214
0.0199
0.0198
0.0197
0.0197
0.0196
Friction Factor vs Reynolds Number
1.000
0.100
f
e/D =
0.010
0.001
1.0E+02
0
0.0001
0.0002
0.0005
0.001
0.002
0.005
0.01
0.02
0.04
1.0E+03
1.0E+04
Re
1.0E+05
1.0E+06
1.0E+07
1.0E+08
Problem 8.95
[Difficulty: 2]
Problem 8.96
[Difficulty: 3]
Using the above formula for f 0, and Eq. 8.37 for f 1
e/D =
0
0.0001
0.0002
0.0005
0.001
0.002
0.005
0.01
0.02
0.05
Re
1.00E+04
2.50E+04
5.00E+04
7.50E+04
1.00E+05
2.50E+05
5.00E+05
7.50E+05
1.00E+06
5.00E+06
1.00E+07
5.00E+07
1.00E+08
0.0309
0.0244
0.0207
0.0189
0.0178
0.0148
0.0131
0.0122
0.0116
0.0090
0.0081
0.0066
0.0060
0.0310
0.0245
0.0210
0.0193
0.0183
0.0156
0.0143
0.0137
0.0133
0.0123
0.0122
0.0120
0.0120
0.0311
0.0248
0.0213
0.0197
0.0187
0.0164
0.0153
0.0148
0.0146
0.0139
0.0139
0.0138
0.0138
0.0315
0.0254
0.0223
0.0209
0.0201
0.0183
0.0176
0.0173
0.0172
0.0168
0.0168
0.0167
0.0167
0.0322
0.0265
0.0237
0.0226
0.0220
0.0207
0.0202
0.0200
0.0199
0.0197
0.0197
0.0197
0.0197
0.0335
0.0285
0.0263
0.0254
0.0250
0.0241
0.0238
0.0237
0.0236
0.0235
0.0235
0.0235
0.0235
0.0374
0.0336
0.0321
0.0316
0.0313
0.0308
0.0306
0.0305
0.0305
0.0304
0.0304
0.0304
0.0304
0.0430
0.0401
0.0391
0.0387
0.0385
0.0382
0.0381
0.0381
0.0380
0.0380
0.0380
0.0380
0.0380
0.0524
0.0502
0.0495
0.0492
0.0491
0.0489
0.0488
0.0488
0.0488
0.0487
0.0487
0.0487
0.0487
0.0741
0.0727
0.0722
0.0720
0.0719
0.0718
0.0717
0.0717
0.0717
0.0717
0.0717
0.0717
0.0717
0.001
0.002
0.005
0.01
0.02
0.05
0.0338
0.0288
0.0265
0.0256
0.0251
0.0241
0.0238
0.0237
0.0236
0.0235
0.0234
0.0234
0.0234
0.0376
0.0337
0.0322
0.0316
0.0313
0.0308
0.0306
0.0305
0.0305
0.0304
0.0304
0.0304
0.0304
0.0431
0.0402
0.0391
0.0387
0.0385
0.0381
0.0380
0.0380
0.0380
0.0379
0.0379
0.0379
0.0379
0.0523
0.0502
0.0494
0.0492
0.0490
0.0488
0.0487
0.0487
0.0487
0.0486
0.0486
0.0486
0.0486
0.0738
0.0725
0.0720
0.0719
0.0718
0.0716
0.0716
0.0716
0.0716
0.0716
0.0716
0.0716
0.0716
f0
Using the add-in function Friction factor from the Web
e/D =
Re
1.00E+04
2.50E+04
5.00E+04
7.50E+04
1.00E+05
2.50E+05
5.00E+05
7.50E+05
1.00E+06
5.00E+06
1.00E+07
5.00E+07
1.00E+08
0
0.0001
0.0002
0.0005
f
0.0309
0.0245
0.0209
0.0191
0.0180
0.0150
0.0132
0.0122
0.0116
0.0090
0.0081
0.0065
0.0059
0.0310
0.0248
0.0212
0.0196
0.0185
0.0158
0.0144
0.0138
0.0134
0.0123
0.0122
0.0120
0.0120
0.0312
0.0250
0.0216
0.0200
0.0190
0.0166
0.0154
0.0150
0.0147
0.0139
0.0138
0.0138
0.0137
0.0316
0.0257
0.0226
0.0212
0.0203
0.0185
0.0177
0.0174
0.0172
0.0168
0.0168
0.0167
0.0167
0.0324
0.0268
0.0240
0.0228
0.0222
0.0208
0.0202
0.0200
0.0199
0.0197
0.0197
0.0196
0.0196
The error can now be computed
e/D =
Re
1.00E+04
2.50E+04
5.00E+04
7.50E+04
1.00E+05
2.50E+05
5.00E+05
7.50E+05
1.00E+06
5.00E+06
1.00E+07
5.00E+07
1.00E+08
0
0.01%
0.63%
0.85%
0.90%
0.92%
0.84%
0.70%
0.59%
0.50%
0.07%
0.35%
1.02%
1.31%
0.0001
0.15%
0.88%
1.19%
1.30%
1.34%
1.33%
1.16%
0.99%
0.86%
0.17%
0.00%
0.16%
0.18%
0.0002
0.26%
1.02%
1.32%
1.40%
1.42%
1.25%
0.93%
0.72%
0.57%
0.01%
0.09%
0.18%
0.19%
0.0005
0.001
0.002
0.005
0.01
0.02
0.05
0.46%
1.20%
1.38%
1.35%
1.28%
0.85%
0.48%
0.30%
0.20%
0.11%
0.15%
0.19%
0.20%
Error (%)
0.64%
0.73%
1.22%
1.03%
1.21%
0.84%
1.07%
0.65%
0.94%
0.52%
0.47%
0.16%
0.19%
0.00%
0.07%
0.07%
0.01%
0.10%
0.15%
0.18%
0.18%
0.19%
0.20%
0.20%
0.20%
0.20%
0.55%
0.51%
0.28%
0.16%
0.09%
0.07%
0.13%
0.16%
0.17%
0.19%
0.20%
0.20%
0.20%
0.19%
0.11%
0.00%
0.06%
0.09%
0.15%
0.18%
0.18%
0.19%
0.20%
0.20%
0.20%
0.20%
0.17%
0.14%
0.16%
0.17%
0.18%
0.19%
0.20%
0.20%
0.20%
0.20%
0.20%
0.20%
0.20%
0.43%
0.29%
0.24%
0.23%
0.22%
0.21%
0.20%
0.20%
0.20%
0.20%
0.20%
0.20%
0.20%
The maximum discrepancy is 1.42% at Re = 100,000 and e/D = 0.0002
0.100
f
0.010
0.001
1E+04
e/D = 0
e/D = 0.0001
e/D = 0.0002
e/D = 0.0005
e/D = 0.001
e/D = 0.002
e/D = 0.005
e/D = 0.01
e/D = 0.02
e/D = 0.05
1E+05
1E+06
Re
1E+07
1E+08
Problem 8.97
Given:
Flow through gradual contraction
Find:
Pressure after contraction; compare to sudden contraction
Solution:
[Difficulty: 3]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
+
α
⋅
+
g
⋅
z
−
⎜ρ
⎜
1
2 = h lm
2
2
⎝
⎠ ⎝ρ
⎠
Basic equations
h lm = K⋅
V2
2
2
Q = V⋅ A
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Horizontal
Available data
Q = 25⋅
3
L
Q = 0.025
s
m
D1 = 75⋅ mm D2 = 37.5⋅ mm
s
p 1 = 500⋅ kPa
3
2
⎛ D2 ⎞ ⎛ 37.5 ⎞ 2
For an included angle of 150 and an area ratio
=⎜
= 0.25 we find from Table 8.3
=⎜
A1
⎝ D1 ⎠ ⎝ 75 ⎠
Hence the energy equation becomes
kg
m
A2
o
ρ = 999⋅
2
2
2
⎛⎜ p
V1 ⎞ ⎛⎜ p 2
V2 ⎞
V2
1
⎜ ρ + 2 − ⎜ ρ + 2 = K⋅ 2 with
⎝
⎠ ⎝
⎠
V1 =
K = 0.35
4⋅ Q
π⋅ D1
2
V2 =
4⋅ Q
π⋅ D2
2
2
ρ
8 ⋅ ρ⋅ Q ⎡ ( 1 + K)
1 ⎤
2
2
p 2 = p 1 − ⋅ ⎡( 1 + K) ⋅ V2 − V1 ⎤ = p 2 −
⋅⎢
−
⎥
⎦
2 ⎣
2
4
4
⎢
π
D1 ⎥
⎣ D2
⎦
2
3
2
⎛
m ⎞
1
1
⎤ × N⋅ s p = 170 ⋅ kPa
p 2 = 500 × 10 ⋅
−
× 999 ⋅
−
× ⎜ 0.025 ⋅
× ⎡( 1 + 0.35) ×
⎢
2
2
3 ⎝
s ⎠
4⎥
4
kg⋅ m 2
m
π
m
( 0.075 ⋅ m) ⎦
( 0.0375⋅ m)
⎣
3 N
8
kg
Repeating the above analysis for an included angle of 180 o (sudden contraction)
2
K = 0.41
2
3
⎛
1
m ⎞
1
⎤ × N⋅ s p = 155 ⋅ kPa
× ⎜ 0.025 ⋅
× ⎡( 1 + 0.41) ×
p 2 = 500 × 10 ⋅
−
× 999 ⋅
−
⎢
4
kg⋅ m 2
2
2
3 ⎝
s ⎠
4⎥
( 0.0375⋅ m)
m
π
m
( 0.075 ⋅ m) ⎦
⎣
3 N
8
kg
Problem 8.98
[Difficulty: 3]
Problem 8.99
Given:
Flow through sudden contraction
Find:
Volume flow rate
[Difficulty: 3]
Solution:
Basic equations
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
+
α
⋅
+
g
⋅
z
−
⎜ρ
⎜
1
2 = h lm
2
2
⎝
⎠ ⎝ρ
⎠
h lm = K⋅
V2
2
2
Q = V⋅ A
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Horizontal
Hence the energy equation becomes
2
2
2
⎛⎜ p
V1 ⎞ ⎛⎜ p 2
V2 ⎞
V2
1
⎜ ρ + 2 − ⎜ ρ + 2 = K⋅ 2
⎝
⎠ ⎝
⎠
From continuity
A2
V1 = V2 ⋅
= V2 ⋅ AR
A1
Hence
2
2
2
2
⎛⎜ p
V2 ⋅ AR ⎞ ⎛⎜ p 2
V2 ⎞
V2
1
+
= K⋅
−⎜
⎜ρ +
2
2 ⎠
2
⎝
⎠ ⎝ρ
Solving for V 2
Hence
(
2⋅ p1 − p2
V2 =
(
2
2
)
ρ⋅ 1 − AR + K
V2 =
2 × 0.5⋅
lbf
2
Q = V2 ⋅ A2 =
π
4
π⋅ D2
4
2
×
⎛ D2 ⎞ ⎛ 1 ⎞ 2
AR = ⎜
=⎜
= 0.25
⎝ D1 ⎠ ⎝ 2 ⎠
2
×
in
Q =
)
so from Fig. 8.14
3
1
slug⋅ ft
⎛ 12⋅ in ⎞ × ft
×
×
⎜ 1 ⋅ ft
2
2
1.94⋅ slug
⎝
⎠
1 − 0.25 + 0.4
lbf ⋅ s
(
)
ft
V2 = 7.45⋅
s
2
⋅ V2
3
⎛ 1 ⋅ ft⎞ × 7.45⋅ ft Q = 0.0406⋅ ft
⎜ 12
s
s
⎝
⎠
Q = 2.44⋅
ft
3
min
Q = 18.2⋅ gpm
K = 0.4
Problem 8.100
Given:
Flow through sudden expansion
Find:
Inlet speed; Volume flow rate
[Difficulty: 3]
Solution:
Basic equations
2
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
V1
1
2
⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 = h lm h lm = K⋅ 2
⎝
⎠ ⎝
⎠
Q = V⋅ A
∆p = ρH2O⋅ g ⋅ ∆h
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Horizontal
Hence the energy equation becomes
2
2
2
⎛⎜ p
V1 ⎞ ⎛⎜ p 2
V2 ⎞
V1
1
⎜ ρ + 2 − ⎜ ρ + 2 = K⋅ 2
⎝
⎠ ⎝
⎠
From continuity
A1
V2 = V1 ⋅
= V1 ⋅ AR
A2
Hence
2
2
2
2
⎛⎜ p
V1 ⎞ ⎛⎜ p 2
V1 ⋅ AR ⎞
V1
1
= K⋅
⎜ρ + 2 −⎜ρ +
2
2
⎝
⎠ ⎝
⎠
Solving for V 1
Also
Hence
V1 =
(
2⋅ p2 − p1
(
)
2
2
⎛ D1 ⎞ ⎛ 75 ⎞ 2
AR = ⎜
=⎜
= 0.111
⎝ D2 ⎠ ⎝ 225 ⎠
)
ρ⋅ 1 − AR − K
kg
m
×
1
so from Fig. 8.14
K = 0.8
2
N⋅ s
5
p 2 − p 1 = ρH2O⋅ g ⋅ ∆h = 1000⋅
× 9.81⋅ ×
⋅m ×
= 49.1⋅ Pa
3
2
1000
kg⋅ m
m
s
V1 =
2 × 49.1⋅
3
N
2
m
Q = V1 ⋅ A1 =
π⋅ D1
4
×
m
1.23⋅ kg
2
⋅ V1
(1 − 0.1112 − 0.8)
×
kg⋅ m
2
N⋅ s
3
2
75
m
Q =
⋅ m⎞ × 20.6⋅
× ⎛⎜
4 ⎝ 1000 ⎠
s
π
m
V1 = 20.6
s
Q = 0.0910⋅
m
s
3
Q = 5.46⋅
m
min
Problem 8.101
Given:
Data on a pipe sudden contraction
Find:
Theoretical calibration constant; plot
[Difficulty: 4]
Solution:
Given data
D1 = 45⋅ mm
D2 = 22.5⋅ mm
The governing equations between inlet (1) and exit (2) are
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α1⋅ 2 + g ⋅ z1 − ⎜ ρ + α2 ⋅ 2 + g⋅ z2 = h l
⎝
⎠ ⎝
⎠
where
h l = K⋅
V2
(8.29)
2
(8.40a)
2
Hence the pressure drop is (assuming α = 1)
2
⎛⎜ V 2 V 2
V2 ⎞
2
1
∆p = p 1 − p 2 = ρ⋅ ⎜
−
+ K⋅
2
2 ⎠
⎝ 2
For the sudden contraction
so
π
π
2
2
V1 ⋅ ⋅ D1 = V2 ⋅ ⋅ D2 = Q
4
4
∆p =
4
⎡
⎢⎛ D1 ⎞
⋅ ⎜
( 1 + K) −
2 ⎢ D2
⎣⎝ ⎠
ρ⋅ V1
2
⎛ D1 ⎞
V2 = V1 ⋅ ⎜
⎝ D2 ⎠
or
⎤
⎥
1
⎥
⎦
For the pressure drop we can use the manometer equation
∆p = ρ⋅ g ⋅ ∆h
Hence
In terms of flow rate Q
ρ⋅ g ⋅ ∆h =
4
⎡
⎢⎛ D1 ⎞
⋅ ⎜
( 1 + K) −
2 ⎢ D2
⎣⎝ ⎠
ρ⋅ V1
2
⎤
⎥
1
⎥
⎦
⎡⎛ D ⎞ 4
⎢ 1
ρ⋅ g ⋅ ∆h = ⋅
⋅ ⎜
( 1 + K) −
2 ⎢ D2
2
⎣
⎝
⎠
π
2
⎛ ⋅D ⎞
⎜4 1
⎝
⎠
ρ
2
Q
⎤
⎥
1
⎥
⎦
2
or
⎡ D ⎞4
⎢⎛ 1
g ⋅ ∆h =
⋅ ⎜
( 1 + K) −
2
4 ⎢ D2
⎠
π ⋅ D1 ⎣⎝
Hence for flow rate Q we find
Q = k ⋅ ∆h
2
8⋅ Q
2
k=
where
g ⋅ π ⋅ D1
4
4
⎡
⎢⎛ D1 ⎞
8⋅ ⎜
⎢ D ( 1 + K) −
⎣⎝ 2 ⎠
For K, we need the aspect ratio AR
⎛ D2 ⎞
AR = ⎜
⎝ D1 ⎠
From Fig. 8.15
K = 0.4
⎤
⎥
⎥
⎦
1
⎤
⎥
⎥
⎦
1
2
AR = 0.25
5
2
Using this in the expression for k, with the other given values
k =
g ⋅ π ⋅ D1
4
⎡⎛ D ⎞ 4
⎢ 1
8⋅ ⎜
⎢ D ( 1 + K) −
⎣⎝ 2 ⎠
−3 m
k = 1.52 × 10
⎤
⎥
1
⎥
⎦
L
For Δh in mm and Q in L/min
k = 2.89⋅
min
1
mm
2
The plot of theoretical Q versus flow rate Δh can be done in Excel.
Calibration Curve for a
Sudden Contraction Flow Meter
60
Q (L/mm)
50
40
30
20
10
0
0
50
100
150
200
Dh (mm)
It is a practical device, but is a) Nonlinear and b) has a large energy loss
250
300
350
⋅
2
s
Problem 8.102
Given:
Flow through a reentrant device
Find:
Head loss
[Difficulty: 3]
Solution:
Basic equations
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
+
α
⋅
+
g
⋅
z
−
⎜ρ
⎜
1
2 = h lT
2
2
⎝
⎠ ⎝ρ
⎠
2
V2
L V2
h lT = h l + h lm = f ⋅ ⋅
+ K⋅
2
D 2
2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) L << so ignore hl 5) Reentrant
3
Available data
D1 = 100 ⋅ mm
D2 = 50⋅ mm
Q = 0.01⋅
π
2
A1 = ⋅ D1
4
A1 = 7.85 × 10 mm
3
m
π
2
A2 = ⋅ D2
4
2
K = 0.78
and from Table 8.2
s
3
2
A2 = 1.96 × 10 mm
Hence between the free surface (Point 1) and the exit (2) the energy equation becomes
p1
ρ
From continuity
Hence
Solving for h
+
V1
2
2
−
V2
2
2
−
Q = V1 ⋅ A1 = V2 ⋅ A2
2
p2
ρ
= K⋅
V2
2
2
ρ
2
1 Q ⎞
1 Q ⎞
Q ⎞
g ⋅ h + ⋅ ⎛⎜
− ⋅ ⎛⎜
= K⋅ ⋅ ⎛⎜
2 A1
2 A2
2 A2
⎝ ⎠
⎝ ⎠
⎝ ⎠
1
⎛Q⎞
⎜A
⎝ 2⎠
h =
p1 − p2
and also
=
2
2
2⎤
⎡
⎛ A2 ⎞ ⎥
⎢
⋅ 1+ K− ⎜
⎥
2⋅ g ⎢
⎣
⎝ A1 ⎠ ⎦
h = 2.27 m
ρ⋅ g ⋅ h
ρ
= g⋅ h
where h is the head loss
Q = V⋅ A
Problem 8.103
[Difficulty: 4]
Given:
Contraction coefficient for sudden contraction
Find:
Expression for minor head loss; compare with Fig. 8.15; plot
Solution:
We analyse the loss at the "sudden expansion" at the vena contracta
The governing CV equations (mass, momentum, and energy) are
Assume:
1) Steady flow 2) Incompressible flow 3) Uniform flow at each section 4) Horizontal: no body force 5) No
shaft work 6) Neglect viscous friction 7) Neglect gravity
The mass equation becomes
Vc⋅ Ac = V2 ⋅ A2
The momentum equation becomes
p c⋅ A2 − p 2 ⋅ A2 = Vc⋅ −ρ⋅ Vc⋅ Ac + V2 ⋅ ρ⋅ V2 ⋅ A2
or (using Eq. 1)
Ac
p c − p 2 = ρ⋅ Vc⋅
⋅ V2 − Vc
A2
The energy equation becomes
pc
p2
⎛
⎛
2⎞
2⎞
Qrate = ⎜ u c +
+ Vc ⋅ −ρ⋅ Vc⋅ Ac + ⎜ u 2 +
+ V2 ⋅ ρ⋅ V2 ⋅ A2
ρ
ρ
⎝
⎠
⎝
⎠
or (using Eq. 1)
(1)
(
)
(
h lm = u 2 − u c −
=
mrate
)
)
(2)
(
Qrate
(
)
2
Vc − V2
2
(
2
+
pc − p2
ρ
(3)
)
2
h lm =
Combining Eqs. 2 and 3
Vc − V2
2
⎡
⎢
h lm =
⋅ 1−
2 ⎢
⎣
Vc
Ac
Cc =
From Eq. 1
h lm =
h lm =
Vc
Ac
+ Vc⋅
⋅ V2 − Vc
A2
(
2⎤
A
V
⎤
⎛ V2 ⎞ ⎥
2 c ⎡⎛ 2 ⎞
⎜ V ⎥ + Vc ⋅ A ⋅ ⎢⎜ V − 1⎥
2 ⎣⎝ c ⎠
⎝ c⎠ ⎦
⎦
Vc
2
⋅ ⎛ 1 − Cc
2 ⎝
Vc
2
Vc
2⎞
2
⎠ + Vc ⋅ Cc⋅ ( Cc − 1 )
⋅ ⎛ 1 − C c + 2 ⋅ C c − 2 ⋅ C c⎞
⎝
⎠
2
2
)
V2
=
A2
h lm =
Hence
2
2
2
2
(
⋅ 1 − Cc
2
2
)2
(4)
2
2
2
Vc
⎛ V2 ⎞
2
h lm = K⋅
= K⋅
⋅⎜
= K⋅
⋅ Cc
2
2
2
Vc
⎝ ⎠
V2
But we have
K=
Hence, comparing Eqs. 4 and 5
Vc
(5)
( 1 − Cc) 2
Cc
2
⎛ 1 − 1⎞
⎜C
⎝ c
⎠
2
So, finally
K=
where
⎛ A2 ⎞
Cc = 0.62 + 0.38⋅ ⎜
⎝ A1 ⎠
3
This result,can be plotted in Excel. The agreement with Fig. 8.15 is reasonable.
0.5
0.4
K
0.3
0.2
0.1
0
0.2
0.4
0.6
AR
0.8
1
Problem 8.104
Given:
Flow through short pipe
Find:
Volume flow rate; How to improve flow rate
[Difficulty: 3]
Solution:
Basic equations
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 = h lT
⎝
⎠ ⎝
⎠
2
V2
L V2
h lT = h l + h lm = f ⋅ ⋅
+ K⋅
2
D 2
2
Q = V⋅ A
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) L << so ignore hl 5) Reentrant
Hence between the free surface (Point 1) and the exit (2) the energy equation becomes
V1
2
+ g ⋅ z1 −
2
From continuity
Hence
V2
2
= K⋅
2
V2
2
2
A2
V1 = V2 ⋅
A1
2
2
2
V2
V2
⎛ A2 ⎞
⋅⎜
+ g⋅ h −
= K⋅
2
2
2
⎝ A1 ⎠
V2
2
Solving for V 2
V2 =
Hence
V2 =
2⋅ g⋅ h
2 × 9.81⋅
m
2
× 1⋅ m ×
s
Q = V2 ⋅ A2
Q = 3.33⋅
K = 0.78
and from Table 8.2
2
⎡
⎛ A2 ⎞ ⎤⎥
⎢
⎢1 + K − ⎜ A ⎥
⎣
⎝ 1⎠ ⎦
m
s
1
m
V2 = 3.33
s
2
⎡
350 ⎞ ⎤
⎢1 + 0.78 − ⎛⎜
⎥
⎣
⎝ 3500 ⎠ ⎦
2
× 350 ⋅ mm ×
⎛ 1⋅ m ⎞
⎜
⎝ 1000⋅ mm ⎠
2
3
−3m
Q = 1.17 × 10
s
The flow rate could be increased by (1) rounding the entrance and/or (2) adding a diffuser (both somewhat expensive)
3
Q = 0.070 ⋅
m
min
Problem 8.105
[Difficulty: 3]
Problem 8.106
[Difficulty: 3]
Problem 8.107
[Difficulty: 3]
Given:
Flow out of water tank
Find:
Volume flow rate using hole; Using short pipe section; Using rounded edge
Solution:
Basic equations
2
2
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
V2
1
L V2
2
⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 = h lT h lT = hl + h lm = f ⋅ D ⋅ 2 + K⋅ 2 Q = V⋅ A
⎝
⎠ ⎝
⎠
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Vl << 5) L << so hl = 0
Available data
D = 25⋅ mm
r = 5 ⋅ mm
h = 5⋅ m
Hence for all three cases, between the free surface (Point 1) and the exit (2) the energy equation becomes
2
g ⋅ z1 −
V2
2
2
= K⋅
V2
and solving for V 2
2
From Table 8.2 Khole = 0.5 for a hole (assumed to be square-edged)
Also, for a rounded edge
r
D
Hence for the hole
=
V2 =
5⋅ mm
25⋅ mm
= 0.2 > 0.15
2 × 9.81⋅
m
s
Q = V2 ⋅ A2
Hence for the pipe
V2 =
2 × 9.81⋅
Q = 8.09⋅
m
s
Q = V2 ⋅ A2
V2 =
1
s
×
π
4
× ( 0.025⋅ m)
2 × 9.81⋅
m
2
× 5⋅ m ×
2
1
s
×
π
4
× ( 0.025⋅ m)
L
( 1 + 0.04)
L
s
3
−3m
Q = 3.97 × 10
s
Q = 3.97⋅
L
Q = 3.64⋅
L
s
2
3
−3m
Q = 3.64 × 10
s
s
The pipe leads to a LOWER flow rate
s
1
4.77 − 3.97 = 0.8⋅
Kround = 0.04
m
V2 = 7.42
s
( 1 + 0.78)
m
(1 + K)
m
V2 = 8.09
s
( 1 + 0.5)
m
2⋅ g ⋅ h
Kpipe = 0.78 for a short pipe (rentrant)
so from Table 8.2
3.64 − 3.97 = −0.33⋅
s
Hence the change in flow rate is
2
× 5⋅ m ×
Q = 7.42⋅
Hence the change in flow rate is
Hence for the rounded
2
× 5⋅ m ×
V2 =
m
V2 = 9.71
s
Q = V2⋅ A2
Q = 4.77⋅
L
s
The rounded edge leads to a HIGHER flow rate
Problem 8.108
Given:
Data on inlet and exit diameters of diffuser
Find:
Minimum lengths to satisfy requirements
[Difficulty: 2]
Solution:
Given data
D1 = 100 ⋅ mm
D2 = 150 ⋅ mm
The governing equations for the diffuser are
h lm = K⋅
V1
(
)
V1
= Cpi − Cp ⋅
2
2
1
Cpi = 1 −
and
2
2
(8.44)
(8.42)
2
AR
Combining these we obtain an expression for the loss coefficient K
1
K= 1−
− Cp
2
(1)
AR
The area ratio AR is
⎛ D2 ⎞
AR = ⎜
⎝ D1 ⎠
2
AR = 2.25
The pressure recovery coefficient Cp is obtained from Eq. 1 above once we select K; then, with Cp and AR specified, the minimum
value of N/R1 (where N is the length and R1 is the inlet radius) can be read from Fig. 8.15
(a)
K = 0.2
1
Cp = 1 −
2
−K
Cp = 0.602
AR
From Fig. 8.15
N
R1
= 5.5
R1 =
N = 5.5⋅ R1
(b)
K = 0.35
Cp = 1 −
D1
2
R1 = 50⋅ mm
N = 275 ⋅ mm
1
2
−K
Cp = 0.452
AR
From Fig. 8.15
N
R1
=3
N = 3 ⋅ R1
N = 150 ⋅ mm
Problem 8.109
Given:
Data on geometry of conical diffuser; flow rate
Find:
Static pressure rise; loss coefficient
Solution:
Basic equations
Cp =
p2 − p1
1
2
⋅ ρ⋅ V1
V1
2
[Difficulty: 3]
2
(
(8.41) h lm = K⋅
2
Given data
D1 = 2 ⋅ in
From Eq. 8.41
1
2
∆p = p 2 − p 1 = ⋅ ρ⋅ V1 ⋅ Cp (1)
2
)
V1
2
= Cpi − Cp ⋅
2
D2 = 3.5⋅ in
1
Cpi = 1 −
(8.44)
(8.42)
2
AR
N = 6 ⋅ in
Q = 750 ⋅ gpm
(N = length)
K= 1−
Combining Eqs. 8.44 and 8.42 we obtain an expression for the loss coefficient K
1
2
− Cp
(2)
AR
The pressure recovery coefficient Cp for use in Eqs. 1 and 2 above is obtained from Fig. 8.15 once compute AR and the
dimensionless length N/R1 (where R1 is the inlet radius)
The aspect ratio AR is
⎛ D2 ⎞
AR = ⎜
⎝ D1 ⎠
R1 =
2
D1
R1 = 1 ⋅ in
2
2
⎛ 3.5 ⎞
⎜ 2
⎝ ⎠
AR =
AR = 3.06
N
Hence
R1
=6
From Fig. 8.15, with AR = 3.06 and the dimensionless length N/R1 = 6, we find Cp = 0.6
V1 =
π
To complete the calculations we need V1
4
3
4
gal
1 ⋅ ft
1 ⋅ min
V1 =
× 750 ⋅
×
×
×
π
min 7.48⋅ gal
60⋅ s
Q
⋅ D1
2
∆p =
We can now compute the pressure rise and loss coefficient from Eqs. 1 and 2
∆p =
1
2
K= 1−
× 1.94⋅
slug
ft
1
2
AR
3
− Cp
× ⎛⎜ 76.6⋅
⎝
ft ⎞
s⎠
K = 1−
2
2
× 0.6 ×
1
3.06
2
lbf ⋅ s
slug⋅ ft
− 0.6
×
1
2
⎛ 1 ⎞ V = 76.6⋅ ft
1
⎜ 2
s
⎜ ⋅ ft
⎝ 12 ⎠
2
⋅ ρ⋅ V1 ⋅ Cp
2
⎛ 1 ⋅ ft ⎞
⎜ 12⋅ in
⎝
⎠
2
∆p = 23.7⋅ psi
K = 0.293
Problem 8.110
[Difficulty: 4]
Problem 8.111
[Difficulty: 4]
Given:
Sudden expansion
Find:
Expression for minor head loss; compare with Fig. 8.15; plot
Solution:
The governing CV equations (mass, momentum, and energy) are
Assume:
1) Steady flow 2) Incompressible flow 3) Uniform flow at each section 4) Horizontal: no body force 5) No
shaft work 6) Neglect viscous friction 7) Neglect gravity
The mass equation becomes
V1 ⋅ A1 = V2 ⋅ A2
The momentum equation becomes
p 1 ⋅ A2 − p 2 ⋅ A2 = V1 ⋅ −ρ⋅ V1 ⋅ A1 + V2 ⋅ ρ⋅ V2 ⋅ A2
or (using Eq. 1)
A1
p 1 − p 2 = ρ⋅ V1 ⋅
⋅ V2 − V1
A2
The energy equation becomes
p1
p2
⎛
⎛
2⎞
2⎞
Qrate = ⎜ u 1 +
+ V1 ⋅ −ρ⋅ V1 ⋅ A1 + ⎜ u 2 +
+ V2 ⋅ ρ⋅ V2 ⋅ A2
ρ
ρ
⎝
⎠
⎝
⎠
or (using Eq. 1)
(
)
(
h lm = u 2 − u 1 −
=
mrate
h lm =
V1 − V2
2
⎡
⎢
h lm =
⋅ 1−
2 ⎢
⎣
V1
2
2
)
V1 − V2
2
(
2⎤
(
2
A1
+ V1 ⋅
⋅ V2 − V1
A2
⎛ V2 ⎞
⎜
⎝ V1 ⎠
)
(2)
2
Qrate
(
)
(
2
Combining Eqs. 2 and 3
(1)
+
p1 − p2
ρ
)
A
V
⎤
⎥
2 1 ⎡⎛ 2 ⎞
+ V1 ⋅
⋅ ⎢⎜
− 1⎥
⎥
A2
⎦
⎣⎝ V1 ⎠ ⎦
(3)
)
AR =
From Eq. 1
h lm =
Hence
h lm =
A1
A2
V1
2
V1
2
V1
(
2
(
2
⋅ 1 − AR
2
h lm = K⋅
V2
=
2
)
2
⋅ 1 − AR + 2 ⋅ AR − 2 ⋅ AR
V1
2
2
K = ( 1 − AR)
Finally
) + V12⋅AR⋅(AR − 1)
2
= ( 1 − AR) ⋅
V1
2
2
2
This result, and the curve of Fig. 8.15, are shown below as computed in Excel. The agreement is excellent.
AR
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
K CV
K Fig. 8.15
1.00
0.81
0.64
0.49
0.36
0.25
0.16
0.09
0.04
0.01
0.00
1.00
0.60
0.38
0.25
0.10
0.01
0.00
(Data from Fig. 8.15
is "eyeballed")
Loss Coefficient for a
Sudden Expansion
1.0
Theoretical Curve
0.8
Fig. 8.15
K
0.5
0.3
0.0
0.00
0.25
0.50
Area Ratio AR
0.75
1.00
Problem 8.112
[Difficulty: 3]
Problem 8.113
Given:
Sudden expansion
Find:
Expression for upstream average velocity
[Difficulty: 2]
Solution:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α1⋅ 2 + g⋅ z1 − ⎜ ρ + α2⋅ 2 + g⋅ z2 = h lT
⎝
⎠ ⎝
⎠
The basic equation is
(8.29)
2
V
h lT = h l + K ⋅
2
Assume:
1) Steady flow 2) Incompressible flow 3) h l = 0 4) α1 = α2 = 1 5) Neglect gravity
The mass equation is
V1⋅ A1 = V2⋅ A2
so
V2 = AR⋅ V1
Equation 8.29 becomes
p1
ρ
or (using Eq. 1)
∆p
ρ
Solving for V1
If the flow were frictionless, K = 0, so
+
=
V1 =
V1
2
=
2
(1)
p1
+
ρ
p2 − p1
V1
2
+ K⋅
2
V1
=
ρ
2
2
2 ⋅ ∆p
(
compared to
(
(
2 ⋅ ∆p
2
< V1
)
2
ρ⋅ 1 − AR
∆pinvscid =
∆p =
2
)
2
V1
2
2
(
V1
2
2
(
)
2
⋅ 1 − AR
2
)
⋅ 1 − AR − K
Hence a given flow rate would generate a larger Δp for inviscid flow
2
)
⋅ 1 − AR − K
Hence the flow rate indicated by a given Δp would be lower
If the flow were frictionless, K = 0, so
V1
ρ⋅ 1 − AR − K
Vinviscid =
A1
V2 = V1⋅
A2
Problem 8.114
[Difficulty: 4]
e
d
Flow
Nozzle
Short pipe
Given:
Flow out of water tank through a nozzle
Find:
Change in flow rate when short pipe section is added; Minimum pressure; Effect of frictionless flow
Solution:
Basic equations
2
2
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
V2
1
L V2
2
⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 = h lT h lT = hl + h lm = f ⋅ D ⋅ 2 + K⋅ 2 Q = V⋅ A
⎝
⎠ ⎝
⎠
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Vl << 5) L << so hl = 0
Available data
D2 = 25⋅ mm
r = 0.02⋅ D2
D3 = 50⋅ mm
r = 0.5⋅ mm
z1 = 2.5⋅ m
ρ = 999 ⋅
kg
3
m
Knozzle = 0.28
For a rounded edge, we choose the first value from Table 8.2
Hence for the nozzle case, between the free surface (Point 1) and the exit (2) the energy equation becomes
g ⋅ z1 −
V2
2
V2
2
= Knozzle⋅
2
2
2 ⋅ g ⋅ z1
Solving for V 2
V2 =
(1 + Knozzle)
Hence
V2 =
2 × 9.81⋅
m
2
× 2.5⋅ m ×
s
Q = V2 ⋅ A2
Q = 6.19⋅
m
s
1
m
V2 = 6.19
s
( 1 + 0.28)
×
π
4
× ( 0.025 ⋅ m)
2
3
−3m
Q = 3.04 × 10
s
Q = 3.04
L
When a small piece of pipe is added the energy equation between the free surface (Point 1) and the exit (3) becomes
g ⋅ z1 −
From continuity
V3
2
2
V2
= Knozzle⋅
2
A2
V3 = V2 ⋅
= V2 ⋅ AR
A3
2
V2
+ Ke⋅
2
2
s
V2 =
Solving for V 2
2 ⋅ g ⋅ z1
⎛ AR2 + K
⎞
nozzle + Ke⎠
⎝
2
⎛ D2 ⎞ ⎛ 25 ⎞ 2
AR =
=⎜
=⎜
= 0.25
A3
⎝ D3 ⎠ ⎝ 50 ⎠
A2
We need the AR for the sudden expansion
AR = 0.25
Ke = 0.6
From Fig. 8.15 for AR = 0.25
V2 =
V2 =
Hence
2 ⋅ g ⋅ z1
⎛ AR2 + K
⎞
nozzle + Ke⎠
⎝
2 × 9.81⋅
m
2
× 2.5⋅ m ×
s
Q = V2 ⋅ A2
Q = 7.21⋅
m
s
×
π
4
× ( 0.025 ⋅ m)
1
m
V2 = 7.21
s
(0.252 + 0.28 + 0.6)
3
−3m
2
Q = 3.54 × 10
∆Q
Comparing results we see the flow increases from 3.04 L/s to 3.54 L/s
Q
=
s
3.54 − 3.04
3.04
Q = 3.54
= 16.4⋅ %
The flow increases because the effect of the pipe is to allow an exit pressure at the nozzle LESS than atmospheric!
The minimum pressure point will now be at Point 2 (it was atmospheric before adding the small pipe). The energy equation
between 1 and 2 is
2
2
⎛⎜ p
V2 ⎞
V2
2
g ⋅ z1 − ⎜
+
= Knozzle⋅
2 ⎠
2
⎝ρ
Solving for p 2
Hence
2
⎡⎢
V2
p 2 = ρ⋅ ⎢g ⋅ z1 −
⋅ ( Knozzle +
2
⎣
p 2 = 999 ⋅
kg
3
m
⎡
m
⎢
⎣
s
× ⎢9.81⋅
2
⎥⎤
)⎦
1⎥
× 2.5⋅ m −
1
2
× ⎛⎜ 7.21⋅
⎝
m⎞
s
⎠
2
⎤
lbf ⋅ s
⎥
⎦
slug⋅ ft
× ( 0.28 + 1 )⎥ ×
2
p 2 = −8.736 ⋅ kPa
If the flow were frictionless the the two loss coeffcients would be zero. Instead of
Instead of
V2 =
2 ⋅ g ⋅ z1
⎛ AR2 + K
⎞
nozzle + Ke⎠
⎝
we'd have
If V2 is larger, then p2, through Bernoulli, would be lower (more negative)
V2 =
2 ⋅ g ⋅ z1
2
AR
which is larger
L
s
Problem 8.115
[Difficulty: 2]
Problem 8.116
[Difficulty: 4]
Problem 8.117
Given:
Data on water flow from a tank/tubing system
Find:
Minimum tank level for turbulent flow
[Difficulty: 3]
Solution:
Basic equations:
2
2
⎞ ⎛⎜ p
⎞
⎛⎜ p
V1
V2
1
2
+
α
⋅
+
g
⋅
z
+
α
⋅
+
g
⋅
z
−
⎜ρ
⎜
1 2
1
2 2
2 = h lT =
⎝
⎠ ⎝ρ
⎠
Re =
f =
ρ⋅ V⋅ D
μ
64
2
hl = f ⋅
(8.36)
L V
⋅
D 2
hl +
major
∑
h lm (8.29)
minor
2
h lm = K⋅
(8.34)
(Laminar)
2
g ⋅ d − α⋅
This can be solved expicitly for height d, or solved using Solver
V
(8.40a)
2
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
f
⎝ 3.7 Re⋅ f ⎠
Re
The energy equation (Eq. 8.29) becomes
∑
V
2
2
= f⋅
2
L V
V
⋅
+ K⋅
D 2
2
h lm = f ⋅
(8.37)
Le V2
(8.40b)
⋅
D 2
(Turbulent)
Problem 8.118
[Difficulty: 2]
Problem 8.119
Given:
Data on water flow from a tank/tubing system
Find:
Minimum tank level for turbulent flow
[Difficulty: 3]
Solution:
Basic equations:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α1⋅ 2 + g⋅ z1 − ⎜ ρ + α2⋅ 2 + g⋅ z2 = h lT =
⎝
⎠ ⎝
⎠
Re =
2
ρ⋅ V⋅ D
hl = f ⋅
μ
∑
hl +
major
∑
h lm (8.29)
minor
2
L V
⋅
D 2
h lm = K ⋅
(8.34)
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
(8.37)
V
h lm = f ⋅
(8.40a)
2
Le V2
(8.40b)
⋅
D 2
(Turbulent)
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Velocity at free surface is <<
The available data is
D = 7.5⋅ mm
L = 500 ⋅ mm
From Table A.8 at 10oC
ρ = 1000
− 3 N⋅ s
kg
μ = 1.3⋅ 10
3
m
Re = 10000
From
Re =
Kent = 0.5
ρ⋅ V⋅ D
Re =
μ
π
4
Hence
V =
π⋅ μ⋅ D⋅ Re
2
4⋅ ρ
V = 1.73
⎛ π⋅ D2 ⎞
⎜
⎝ 4 ⎠
f
Q =
⋅D
Q
1
Assuming a smooth tube
or
= −2 ⋅ log⎛⎜
⎞
⎝ Re⋅ f ⎠
2.51
so
m
s
2
2
V
⋅ ⎛⎜ f ⋅
L
+ Kent + Kexit⎞
Solving for d
d =
FOR r > 0.15D)
Kent = 0.04 (Table 8.2)
2⋅ g
⎝ D
⎠
3
−5m
Q = 7.66 × 10
f = 0.0309
g⋅ d = f ⋅
The energy equation (Eq. 8.29) becomes
2
m
Kexit = 1
(Table 8.2)
ρ⋅ Q⋅ D
⋅
2
2
L V
V
V
⋅
+ Kent⋅
+ Kexit ⋅
D 2
2
2
d = 545 ⋅ mm
d = 475 ⋅ mm
s
Q = 0.0766⋅
l
s
Problem 8.120
Given:
Data on a tube
Find:
"Resistance" of tube for flow of kerosine; plot
[Difficulty: 4]
Solution:
The basic equations for turbulent flow are
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
+
α
⋅
+
g
⋅
z
−
⎜ρ
⎜
1 2
1
2 2
2 = hl
⎝
⎠ ⎝ρ
⎠
⎛ e
⎞
2
⎜ D
L V
1
2.51
(8.34)
hl = f ⋅ ⋅
+
= −2.0⋅ log ⎜
D 2
f
⎝ 3.7 Re⋅ f ⎠
The given data is
L = 250 ⋅ mm
From Fig. A.2 and Table A.2
μ = 1.1 × 10
(8.37)
D = 7.5⋅ mm
− 3 N⋅ s
⋅
ρ = 0.82 × 990 ⋅
2
m
For an electrical resistor
(8.29)
kg
3
= 812 ⋅
m
kg
(Kerosene)
3
m
V = R⋅ I
(1)
Simplifying Eqs. 8.29 and 8.34 for a horizontal, constant-area pipe
⎛
⎜
⎜
2
p1 − p2
L V
L ⎝
= f⋅ ⋅
= f⋅ ⋅
ρ
D 2
D
⎞
Q
π
4
2
2
⋅D
⎠
∆p =
or
2
8 ⋅ ρ⋅ f ⋅ L
2
5
2
⋅Q
(2)
π ⋅D
By analogy, current I is represented by flow rate Q, and voltage V by pressure drop Δp . Comparing Eqs. (1) and (2), the
"resistance" of the tube is
R=
∆p
Q
=
8 ⋅ ρ⋅ f ⋅ L⋅ Q
2
5
π ⋅D
The "resistance" of a tube is not constant, but is proportional to the "current" Q! Actually, the dependence is not quite linear,
because f decreases slightly (and nonlinearly) with Q. The analogy fails!
The analogy is hence invalid for
Re > 2300
or
ρ⋅ V⋅ D
μ
> 2300
ρ⋅
Writing this constraint in terms of flow rate
Q
π
4
⋅D
2
⋅D
μ
> 2300
or
Q>
2300⋅ μ⋅ π⋅ D
4⋅ ρ
3
−5m
Q = 1.84 × 10
Flow rate above which analogy fails
s
The plot of "resistance" versus flow rate cab be done in Excel.
"Resistance" of a Tube versus Flow Rate
9
"R"
3
(10 Pa/m /s)
1.E+01
1.0E-05
1.0E-04
1.0E-03
1.E-01
1.E-03
3
Q (m /s)
1.0E-02
Problem 8.121
Given:
Data on tube geometry
Find:
Plot of reservoir depth as a function of flow rate
[Difficulty: 3]
Solution:
Basic equations:
2
2
⎞ ⎛⎜ p
⎞
⎛⎜ p
V1
V2
1
2
⎜ ρ + α1⋅ 2 + g ⋅ z1 − ⎜ ρ + α2 ⋅ 2 + g⋅ z2 = h lT =
⎝
⎠ ⎝
⎠
Re =
f =
ρ⋅ V⋅ D
μ
64
2
hl = f ⋅
(8.36)
L V
⋅
D 2
(Laminar)
g ⋅ d − α⋅
This can be solved expicitly for height d, or solved using Solver
2
In Excel:
V
2⋅ g
⋅ ⎜⎛ α + f ⋅
⎝
L
D
+ K⎞
⎠
major
∑
h lm (8.29)
minor
V
(8.40a)
2
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
2
d=
hl +
2
h lm = K⋅
(8.34)
Re
The energy equation (Eq. 8.29) becomes
∑
V
2
2
= f⋅
2
L V
V
⋅
+ K⋅
D 2
2
h lm = f ⋅
(8.37)
Le V2
(8.40b)
⋅
D 2
(Turbulent)
Required Reservoir Head versus Flow Rate
75
50
d (m)
25
0
0
2
4
6
Q (L/min)
8
10
12
Problem 8.122
Given:
Flow of oil in a pipe
Find:
Percentage change in loss if diameter is reduced
Solution:
Basic equations
Available data
2
hl = f ⋅
ν = 7.5⋅ 10
V=
Here
Re =
Then
L V
⋅
D 2
Q
A
f =
− 4 ft
⋅
s
2
V =
π⋅ D
V⋅ D
4
π
D = 1 ⋅ in
× 0.1⋅
Re = 18.3⋅
ν
hl = f ⋅
Laminar
Re
L = 100 ⋅ ft
2
The flow is LAMINAR
⎛ e
⎞
⎜
1
2.51
D
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
Turbulent
2
4⋅ Q
=
64
[Difficulty: 3]
ft
s
ft
3
×
s
1
×
12
⎛ 12 ⋅ 1 ⎞
⎜ 1 ft
⎝
⎠
hl =
2
V = 18.3⋅
7.5 × 10
−4
⋅ ft
ft
3
s
s
Re = 2033
2
2
⎛ 18.3 ft ⎞
⎜
s⎠
64
100
⎝
hl =
×
×
2033
1
2
64 L V
⋅ ⋅
Re D 2
Q = 0.100
ft
s
⋅ ft ×
2
L V
⋅
D 2
Q = 45⋅ gpm
h l = 6326⋅
ft
2
2
s
12
D = 0.75⋅ in
When the diameter is reduced to
V=
Re =
Q
A
4⋅ Q
=
2
π⋅ D
V⋅ D
V =
4
π
× 0.1⋅
Re = 32.6⋅
ν
The flow is TURBULENT For drawn tubing, from Table 8.1
Given
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
s
3
s
×
×
⎛ 12 ⋅ 1 ⎞
⎜ 0.75 ft
⎝
⎠
0.75
12
2
V = 32.6⋅
ft
s
s
⋅ ft ×
7.5 × 10
−4
⋅ ft
2
L V
⋅
D 2
Re = 2717
e = 0.000005⋅ ft
f = 0.0449
2
⎛ 32.6 ft ⎞
⎜
s⎠
100
⎝
h l = .0449 ×
×
2
0.75
2
hl = f ⋅
ft
ft
12
4
The increase in loss is
3.82 × 10 − 6326
6326
= 504 ⋅ %
4 ft
h l = 3.82 × 10 ⋅
This is a HUGH increase! The main increase is because the
diameter reduction causes the velocity to increase; the loss
goes as V2, and 1/D, so it increases very rapidly
2
2
s
Problem 8.123
Given:
Data on water system
Find:
Minimum tank height; equivalent pressure
[Difficulty: 4]
Solution:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
+
α
⋅
+
g
⋅
z
−
⎜ρ
⎜
1 2
1
2 2
2 = h lT =
⎝
⎠ ⎝ρ
⎠
Basic equations:
2
ρ⋅ V⋅ D
Re =
hl = f ⋅
μ
∑
hl +
major
∑
(8.29)
h lm
minor
2
L V
⋅
D 2
h lm = K⋅
(8.34)
⎛ e
⎞
⎜
D
1
2.51
(8.37)
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
V
(8.40a)
2
h lm = f ⋅
Le V2
(8.40b)
⋅
D 2
(Turbulent)
Available data
D = 7.5⋅ mm
L = 1⋅ m
Re = 100000
From Section 8.7
Kent = 0.5
Lelbow45 = 16⋅ D
Lelbow90 = 30⋅ D
LGV = 8 ⋅ D
Lelbow45 = 0.12 m
Lelbow90 = 0.225 m
LGV = 0.06 m
From Table A.8 at 10oC
ρ = 1000
Q =
π⋅ μ⋅ D⋅ Re
4⋅ ρ
μ = 1.3⋅ 10
3
Q = 7.66 × 10
d =
⎛
2⋅ g ⎝
V
V
2⋅ g
⋅⎜1 + f ⋅
IF INSTEAD the reservoir was pressurized
Q = 0.766
s
2
d−
2
Hence
⋅
2
m
3
−4m
The energy equation becomes
f = 0.0180
− 3 N⋅ s
kg
m
Then
and so
L
D
V =
s
Q
V = 17.3
⎛ π⋅ D2 ⎞
⎜
⎝ 4 ⎠
m
s
Lelbow90
Lelbow45
LGV ⎞
⎛ L
+ 2⋅ f ⋅
+ 2⋅ f ⋅
+ f⋅
2⋅ g ⎝ D
D
D
D ⎠
2
=
l
V
+ 2⋅ f ⋅
⋅⎜f ⋅
Lelbow90
D
+ 2⋅ f ⋅
∆p = ρ⋅ g ⋅ d
Lelbow45
D
+ f⋅
LGV ⎞
D
∆p = 781 ⋅ kPa
⎠
d = 79.6⋅ m
Unrealistic!
which is feasible
at this Re
Problem 8.124
Given:
Flow from pump to reservoir
Find:
Pressure at pump discharge
Solution:
[Difficulty: 2]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
+
α
⋅
+
g
⋅
z
−
⎜ρ
⎜
1
2 = h lT
2
2
⎝
⎠ ⎝ρ
⎠
Basic equations
2
V1
L V1
h lT = h l + h lm = f ⋅ ⋅
+ Kexit ⋅
D 2
2
2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) V2 <<
Hence the energy equation between Point 1 and the free surface (Point 2) becomes
2
2
⎛ p1 V2 ⎞
L V
V
⎜ +
− ( g ⋅ z2 ) = f ⋅ ⋅
+ Kexit ⋅
D 2
2
2 ⎠
⎝ρ
⎛
V
⎝
2
Solving for p 1
p 1 = ρ⋅ ⎜ g ⋅ z2 −
From Table A.7 (68oF)
ρ = 1.94⋅
slug
ft
Re =
2
3
V⋅ D
ν
For commercial steel pipe e = 0.00015 ⋅ ft
2⎞
2
+ f⋅
L V
V
⋅
+ Kexit ⋅
D 2
2
ν = 1.08 × 10
Re = 10⋅
ft
Kexit = 1.0
so we find
ft
4 ⋅ mile
ft
3
⎡
⎢
⎣
2
s
× 50⋅ ft + .0150 ×
12
s
s
⋅ ft ×
−5
1.08 × 10
⋅ ft
Re = 6.94 × 10
2
e
so
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
f
⎝ 3.7 Re⋅ f ⎠
For the exit
× ⎢32.2⋅
9
2
D
Given
slug
⋅
(Table 8.1)
Flow is turbulent:
p 1 = 1.94⋅
s
×
− 5 ft
⎠
0.75⋅ ft
p 1 = ρ⋅ ⎜ g ⋅ z2 + f ⋅
⎝
5280⋅ ft
1mile
×
1
2
× ⎛⎜ 10⋅
⎝
= 0.000200
2⎤
2⎞
L V
⋅
D 2
⎠
2
⎥ × lbf ⋅ s
s ⎠ ⎥ slug⋅ ft
⎦
ft ⎞
Turbulent
f = 0.0150
⎛
×
5
4 lbf
p 1 = 4.41 × 10 ⋅
ft
2
p 1 = 306 ⋅ psi
Problem 8.125
[Difficulty: 3]
Given: Data on reservoir/pipe system
Find: Plot elevation as a function of flow rate; fraction due to minor losses
Solution:
L =
D =
e/D =
K ent =
K exit =
250
50
0.003
0.5
1.0
Required Head versus Flow Rate
m
mm
200
150
z (m)
 = 1.01E-06 m2/s
3
Q (m /s) V (m/s)
0.0000
0.0005
0.0010
0.0015
0.0020
0.0025
0.0030
0.0035
0.0040
0.0045
0.0050
0.0055
0.0060
0.0065
0.0070
0.0075
0.0080
0.0085
0.0090
0.0095
0.0100
0.000
0.255
0.509
0.764
1.02
1.27
1.53
1.78
2.04
2.29
2.55
2.80
3.06
3.31
3.57
3.82
4.07
4.33
4.58
4.84
5.09
Re
0.00E+00
1.26E+04
2.52E+04
3.78E+04
5.04E+04
6.30E+04
7.56E+04
8.82E+04
1.01E+05
1.13E+05
1.26E+05
1.39E+05
1.51E+05
1.64E+05
1.76E+05
1.89E+05
2.02E+05
2.14E+05
2.27E+05
2.40E+05
2.52E+05
100
f
z (m) h lm /h lT
0.000
0.0337 0.562
0.0306 2.04
0.0293 4.40
0.0286 7.64
0.0282 11.8
0.0279 16.7
0.0276 22.6
0.0275 29.4
0.0273 37.0
0.0272 45.5
0.0271 54.8
0.0270 65.1
0.0270 76.2
0.0269 88.2
0.0269 101
0.0268 115
0.0268 129
0.0268 145
0.0267 161
0.0267 179
50
0.882%
0.972%
1.01%
1.04%
1.05%
1.07%
1.07%
1.08%
1.09%
1.09%
1.09%
1.10%
1.10%
1.10%
1.10%
1.11%
1.11%
1.11%
1.11%
1.11%
0
0.0000
0.0025
0.0050
3
Q (m /s)
0.0075
0.0100
Minor Loss Percentage versus Flow Rate
1.2%
1.1%
h lm /h lT
1.0%
0.9%
0.8%
0.0000
0.0025
0.0050
3
Q (m /s)
0.0075
0.0100
Problem 8.126
Given:
Flow through three different layouts
Find:
Which has minimum loss
Solution:
Basic equations
[Difficulty: 3]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
2
V1
V2
1
L V
2
+
α
⋅
+
g
⋅
z
+
α
⋅
+
g
⋅
z
h
=
h
+
h
f
⋅
⋅
−
=
h
=
+
⎜ρ
⎜
1
2
l
lm
lT lT
2
2
D 2
⎝
⎠ ⎝ρ
⎠
∑
Minor
⎛ Le V2 ⎞
⎜f ⋅ ⋅
⎝ D 2⎠
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) Ignore additional length of elbows
For a flow rate of
For water at 20oC
Q = 350 ⋅
L
V=
min
ν = 1.01 × 10
For Case (a)
=
A
⋅
4⋅ Q
2
Re =
s
V⋅ D
ν
e
e = 0.15⋅ mm
2
D
2
p1
s
0.001 ⋅ m
1⋅ L
× 0.05⋅ m ×
2
1 ⋅ min
×
×
60⋅ s
⎛ 1 ⎞ V = 2.97 m
⎜ 0.05⋅ m
s
⎝
⎠
s
−6
1.01 × 10
2
Re = 1.47 × 10
⋅m
−4
= 6.56 × 10
L = 5.81 m
2
p2
m
min
3
×
f = 0.0201
5.25 + 2.5 ⋅ m
Hence the energy equation is
π
L
× 350 ⋅
Re = 2.97⋅
⎛ e
⎞
⎜
1
2.51
D
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
L =
4
V =
π⋅ D
2
−6 m
Flow is turbulent. From Table 8.1
Given
Q
Two 45o miter bends (Fig. 8.16), for each
Le
D
= 13
2
Le V
L V
−
= f⋅ ⋅
+ 2⋅ f ⋅ ⋅
ρ
ρ
D 2
D 2
Le ⎞
⎛L
∆p = p 1 − p 2 = ρ⋅ f ⋅
⋅ ⎜ + 2⋅
2 ⎝D
D⎠
2
Solving for Δp
V
∆p = 1000⋅
kg
3
× .0201 × ⎛⎜ 2.97⋅
⎝
m
For Case (b)
L = ( 5.25 + 2.5) ⋅ m
Hence the energy equation is
p1
ρ
−
p2
ρ
= f⋅
m⎞
s
⎠
2
×
2
⎛ 5.81 + 2⋅ 13⎞ × N⋅ s
⎜
⎝ 0.05
⎠ kg⋅ m
L = 7.75 m
2
Le V2
L V
⋅
+ f⋅ ⋅
D 2
D 2
One standard 90o elbow (Table 8.4)
∆p = 25.2⋅ kPa
Le
D
= 30
5
Solving for Δp
2
Le ⎞
V ⎛L
∆p = p 1 − p 2 = ρ⋅ f ⋅
⋅⎜ +
2 ⎝D
D⎠
∆p = 1000⋅
kg
3
× .0201 × ⎛⎜ 2.97⋅
⎝
m
For Case (c)
Hence the energy equation is
L = ( 5.25 + 2.5) ⋅ m
p1
ρ
Solving for Δp
−
p2
ρ
m⎞
s
2
×
⎠
L = 7.75 m
2
2
⎛ 7.75 + 30⎞ × N⋅ s
⎜
⎝ 0.05
⎠ kg⋅ m
Three standard 90o elbows, for each
∆p = 32.8⋅ kPa
Le
D
= 30
2
Le V
L V
⋅
+ 3⋅ f ⋅ ⋅
D 2
D 2
= f⋅
2
Le ⎞
V ⎛L
∆p = p 1 − p 2 = ρ⋅ f ⋅
⋅ ⎜ + 3⋅
2 ⎝D
D⎠
∆p = 1000⋅
kg
3
m
× .0201 × ⎛⎜ 2.97⋅
⎝
Hence we conclude Case (a) is the best and Case (c) is the worst
m⎞
s
⎠
2
×
2
⎛ 7.75 + 3 × 30⎞ × N⋅ s
⎜
⎝ 0.05
⎠ kg⋅ m
∆p = 43.4⋅ kPa
Problem 8.127
Given:
Flow through rectangular duct
Find:
Pressure drop
Solution:
Basic equations
[Difficulty: 2]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
2
V1
V2
1
L V
2
⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 = h lT h lT = hl + h lm = f ⋅ D ⋅ 2 +
⎝
⎠ ⎝
⎠
∑
Minor
⎛ Le V2 ⎞
⎜f ⋅ ⋅
⎝ D 2⎠
4 ⋅ a⋅ b
Dh =
2⋅ (a + b)
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1
Available data
Q = 1750⋅ cfm
At 50oF, from Table A.9 ρ = 0.00242 ⋅
L = 1000⋅ ft
slug
ft
Hence
1
f
Hence
or, in in water
V = 15.6⋅
ρ⋅ V⋅ Dh
2.51
⎞
⎝ Re⋅ f ⎠
L
Dh
∆p
ρw⋅ g
⋅
ft
5
so
V
2
∆p = 0.031 ⋅ psi
h = 0.848 ⋅ in
a = 0.75⋅ ft
ρw = 1.94⋅
slug
ft
and
2
⋅ ρ⋅
2
s
Re = 1.18 × 10
μ
= −2 ⋅ log⎛⎜
− 7 lbf ⋅ s
ft
a⋅ b
∆p = f ⋅
h =
3
Q
V =
Re =
For a smooth duct
μ = 3.69⋅ 10
b = 2.5⋅ ft
f = 0.017
3
4 ⋅ a⋅ b
Dh =
2⋅ ( a + b)
Dh = 1.15⋅ ft
Problem 8.128
Given:
Data on circuit
Find:
Plot pressure difference for a range of flow rates
[Difficulty: 3]
Solution:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α1⋅ 2 + g ⋅ z1 − ⎜ ρ + α2 ⋅ 2 + g⋅ z2 = h lT =
⎝
⎠ ⎝
⎠
Basic equations:
Re =
f =
ρ⋅ V⋅ D
64
μ
2
hl = f ⋅
(8.36)
L V
⋅
D 2
∑
hl +
major
∑
h lm (8.29)
minor
2
h lm = K⋅
(8.34)
(Laminar)
Re
V
h lm = f ⋅
(8.40a)
2
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
Le V2
(8.40b)
⋅
D 2
(8.37)
(Turbulent)
The energy equation (Eq. 8.29) becomes for the circuit ( 1 = pump inlet, 2 = pump outlet)
p1 − p2
ρ
In Excel:
2
= f⋅
2
2
L V
V
V
⋅
+ 4 ⋅ f ⋅ Lelbow⋅
+ f ⋅ Lvalve⋅
D 2
2
2
2
or
∆p = ρ⋅ f ⋅
V
2
⎛L
⋅⎜
⎝D
+ 4⋅
Lelbow
D
+
Lvalve ⎞
D
⎠
Required Pressure Head for a Circuit
1200
Dp (kPa)
1000
800
600
400
200
0
0.00
0.01
0.02
0.03
Q (m3/s)
0.04
0.05
0.06
0.07
Problem 8.129
[Difficulty: 3]
c
h
LA
d
e
LB
Given:
Pipe friction experiment
Find:
Required average speed; Estimate feasibility of constant head tank; Pressure drop over 5 m
Solution:
Basic equations
2
2
 p
  p

V1
V2
1
2
 ρ  α 2  g z1   ρ  α 2  g z2  h lT

 

2
LA VA
LB VB
h lT  h A  h B  fA

 fB

DA 2
DB 2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) Ignore minor losses
We wish to have
ReB  10
Hence, from
ReB 
5
VB DB
VB 
ν
2
6 m
5
VB  10  1.01  10
We will also need
 DB 
VA  VB 
 DA 
ReA 
VA DA
ν

s
ReB ν
and for water at 20oC
DB

1
m
VA  4.04 
s
 2.5 
 5
 
2
m
ReA  1.01  0.05 m 
s
m
VA  1.01
s
s
1.01  10
4
6
2
m
ReA  5  10
Both tubes have turbulent flow
For PVC pipe (from Googling!) e  0.0015 mm
For tube A
For tube B
Given
Given
2
6 m
m
VB  4.04
s
0.025  m
2
ν  1.01  10

 e

D
1
2.51 
A
 2.0 log

 3.7
fA
ReA fA


 e


D
1
2.51 
B
 2.0 log

 3.7
fB
ReB fB


fA  0.0210
fB  0.0183

s
2
Applying the energy equation between Points 1 and 3

VB

g  LA  h 
LA 
2
2


g 

1
2
LA 
2
2
LA VA
LB VB
 fA

 fB

DA 2
DB 2
LB 

  1  fB
 g h
2
DB


VB
Solving for LA
2
2
fA VA

DA 2
  4.04

m
s
2

9.81

  1  0.0183 
  9.81 m  0.5 m
2
0.025 
s
20

m
2

0.0210
2
s

1
0.05 m
  1.01

m
s
2
LA  12.8 m

Most ceilings are about 3.5 m or 4 m, so this height is IMPRACTICAL
Applying the energy equation between Points 2 and 3
2
2
2
 p
VB   p 3
VB 
2
L VB
 ρ  2   ρ  2  fB D  2

 

B
∆p  1000
kg
3
m

0.0183
2

5 m
0.025  m
  4.04

L
m
s

2
2

VB
∆p  ρ fB

DB 2
or
N s
kg m
∆p  29.9 kPa
2
Problem 8.130
[Difficulty: 3] Part 1/2
Problem 8.130
[Difficulty: 3] Part 2/2
Problem 8.131
Given:
Same flow rate in various ducts
Find:
Pressure drops of each compared to round duct
Solution:
Basic equations
[Difficulty; 3]
2
2
⎞ ⎛⎜ p
⎞
⎛⎜ p
V1
V2
1
2
+
α
⋅
+
g
⋅
z
+
α
⋅
+
g
⋅
z
−
⎜ρ
⎜
1
2 = hl
2
2
⎝
⎠ ⎝ρ
⎠
4⋅ A
Dh =
Pw
e = 0
(Smooth)
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) Ignore minor losses
The energy equation simplifies to
2
L V
∆p = p 1 − p 2 = ρ⋅ f ⋅
⋅
Dh 2
But we have
From Table A.9
Hence
For a round duct
For a rectangular duct
But
Q
V=
V = 1250⋅
A
ν = 1.62 × 10
Re =
V⋅ Dh
− 4 ft
⋅
3
min
×
1 ⋅ min
60⋅ s
b
ft
s
×
ar
2
h =
b⋅ h
ar
2
slug
= ρ⋅
f
2
⋅
V
Dh 2
V = 20.8⋅
ft
s
at 68oF
3
s
5
× Dh = 1.284 × 10 ⋅ Dh
−4 2
1.62 × 10 ⋅ ft
4
Dh =
π
so
1 ⋅ ft
ft
π
4⋅ A
4⋅ b⋅ h
2 ⋅ h ⋅ ar
Dh =
=
=
Pw
2⋅ ( b + h)
1 + ar
h=
1
×
ρ = 0.00234 ⋅
s
4⋅ A
Dh = D =
L
2
Re = 20.8⋅
ν
ft
∆p
or
=
A
× 1 ⋅ ft
2
where
or
ar
(Dh in ft)
Dh = 1.13⋅ ft
ar =
h=
b
h
A
and
ar
2 ⋅ ar
Dh =
⋅ A
1 + ar
The results are:
Round
Given
ar = 1
Dh = 1.13⋅ ft
5 1
Re = 1.284 × 10 ⋅ ⋅ Dh
ft
⎛ e
⎞
⎜
D
h
1
2.51
+
= −2.0⋅ log ⎜
f
⎝ 3.7 Re⋅ f ⎠
2 ⋅ ar
Dh =
⋅ A
1 + ar
f = 0.0167
Dh = 1 ⋅ ft
Re = 1.45 × 10
∆p
L
= ρ⋅
f
5
2
⋅
V
Dh 2
∆p
L
= 7.51 × 10
5 1
5
Re = 1.284 × 10 ⋅ ⋅ Dh Re = 1.28 × 10
ft
− 3 lbf
⋅
ft
3
Given
⎛ e
⎞
⎜
D
h
1
2.51
+
= −2.0⋅ log ⎜
f
⎝ 3.7 Re⋅ f ⎠
f = 0.0171
∆p
L
= ρ⋅
f
2
⋅
∆p
V
Dh 2
L
8.68 × 10
Hence the square duct experiences a percentage increase in pressure drop of
= 8.68 × 10
−3
− 3 lbf
⋅
ft
3
−3
− 7.51 × 10
= 15.6⋅ %
−3
7.51 × 10
ar = 2
Given
2 ⋅ ar
Dh =
⋅ A
1 + ar
Dh = 0.943 ⋅ ft
⎛ e
⎞
⎜
Dh
1
2.51
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
5 1
Re = 1.284 × 10 ⋅ ⋅ Dh
ft
f = 0.0173
∆p
L
Re = 1.21 × 10
= ρ⋅
f
2
⋅
∆p
V
Dh 2
9.32 × 10
Hence the 2 x 1 duct experiences a percentage increase in pressure drop of
5
= 9.32 × 10
L
−3
− 3 lbf
ft
Given
2 ⋅ ar
Dh =
⋅ A
1 + ar
Dh = 0.866 ⋅ ft
= 24.1⋅ %
−3
⎛ e
⎞
⎜
Dh
1
2.51
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
5 1
Re = 1.284 × 10 ⋅ ⋅ Dh
ft
f = 0.0176
∆p
L
Re = 1.11 × 10
= ρ⋅
f
2
⋅
5
∆p
V
Dh 2
L
= 0.01⋅
lbf
ft
3
−3
Hence the 3 x 1 duct experiences a percentage increase in pressure drop of
0.01 − 7.51 × 10
−3
7.51 × 10
Note that f varies only about 7%; the large change in Δp/L is primarily due to the 1/Dh factor
3
−3
− 7.51 × 10
7.51 × 10
ar = 3
⋅
= 33.2⋅ %
Problem 8.132
Given:
Flow down corroded iron pipe
Find:
Pipe roughness; Power savings with new pipe
[Difficulty: 4]
Solution:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
+
α
⋅
+
g
⋅
z
−
⎜ρ
⎜
1
2 = hl
2
2
⎝
⎠ ⎝ρ
⎠
Basic equations
2
hl = f ⋅
L V
⋅
D 2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) No minor losses
3
Available data
D = 50⋅ mm
∆z = 40⋅ m
L = ∆z
p 1 = 750⋅ kPa
p 2 = 250⋅ kPa
Q = 0.015⋅
m
s
ρ = 999⋅
kg
3
m
Hence the energy equation becomes
2
⎛ p1
⎞ ⎛ p2
⎞
L V
⎜ + g ⋅ z1 − ⎜ + g ⋅ z2 = f ⋅ ⋅
D 2
⎝ρ
⎠ ⎝ρ
⎠
Here
V=
Q
=
A
4⋅ Q
V=
2
π⋅ D
4
π
3
× 0.015⋅
m
s
×
1
( 0.05⋅ m)
V = 7.64
2
m
s
In this problem we can compute directly f and Re, and hence obtain e/D
Solving for f
f =
⎛ p1 − p2
2⋅ D
2
L⋅ V
f = 2×
⋅⎜
⎝
0.05
40
From Table A.8 (20oF) ν = 1.01 × 10
Flow is turbulent:
ρ
×
(
+ g z1 − z2
⎞
)⎠
3
2
⎤
kg⋅ m
m
⎛ s ⎞ × ⎡⎢( 750 − 250 ) × 103⋅ N × m
×
+ 9.81⋅ × 40⋅ m⎥ f = 0.0382
⎜ 7.64⋅ m
1000⋅ kg
2
2
⎥
2
⎝
⎠ ⎢⎣
s ⋅N
s
m
⎦
2
−6 m
⋅
s
Re =
⎛ e
⎞
⎜
1
D
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
V⋅ D
ν
Re = 7.64⋅
m
s
× 0.05⋅ m ×
s
−6
1.01 × 10
2
⋅m
Re = 3.78 × 10
5
Solving for e
New pipe (Table 8.1)
⎛ −
⎜
e = 3.7⋅ D⋅ ⎜ 10
⎝
⎞
1
2⋅ f
−
e
e = 0.15⋅ mm
2.51
e
e = 0.507 mm
Re⋅ f ⎠
D
= 0.0101
= 0.003
D
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
f
⎝ 3.7 Re⋅ f ⎠
Given
f = 0.0326
In this problem
Hence
⎡
L V
∆p = p 1 − p 2 = ρ⋅ ⎢g ⋅ z2 − z1 + f ⋅ ⋅
D 2
⎣
(
∆pnew = 1000⋅
kg
3
m
∆pold = p 1 − p 2
Compared to ∆pold = 500 ⋅ kPa we find
⎡
)
× ⎢9.81⋅
⎢
⎣
2⎤
m
2
⎥
⎦
× ( −40⋅ m) +
s
0.0326
2
×
40
0.05
× ⎛⎜ 7.64⋅
⎝
∆pold = 500 kPa
∆pold − ∆pnew
∆pold
= 26.3⋅ %
As power is ΔpQ and Q is constant, the power reduction is the same as the above percentage!
2⎤
2
⎥ × N⋅ s
s ⎠ ⎥ kg⋅ m
⎦
m⎞
∆pnew = 369 ⋅ kPa
Problem 8.133
Given:
Flow through fire hose and nozzle
Find:
Supply pressure
Solution:
Basic equations
[Difficulty: 3]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
+
α
⋅
+
g
⋅
z
−
⎜ρ
⎜
1
2 = h lT
2
2
⎝
⎠ ⎝ρ
⎠
2
L V
h lT = h l + h lm = f ⋅ ⋅
+
D 2
∑
Minor
⎛ V2 ⎞
⎜ K⋅
⎝ 2⎠
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) p 2 = p atm so p 2 = 0 gage
Hence the energy equation between Point 1 at the supply and the nozzle exit (Point n); let the velocity in the hose be V
p1
ρ
2
Vn
−
2
2
(
)
and
V=
2
From continuity
Vn =
⎛ D ⎞ ⋅V
⎜D
⎝ 2⎠
Solving for p 1
⎡⎢ L
p1 =
⋅ f ⋅ + Ke + 4⋅ Kc +
2 ⎢ D
⎣
From Table A.7 (68oF)
ρ = 1.94⋅
2
ρ⋅ V
slug
ft
Re =
For the hose
e
D
Flow is turbulent:
2
2
Vn
L V
V
= f⋅ ⋅
+ Ke + 4⋅ Kc ⋅
+ Kn ⋅
D 2
2
2
V⋅ D
Re = 15.3⋅
ν
A
=
4⋅ Q
V=
2
π⋅ D
4
π
× 0.75⋅
4
⎛ D ⎞ ⋅ 1 + K ⎥⎤
( n)⎥
⎜D
⎝ 2⎠
⎦
ν = 1.08 × 10
3
Q
ft
s
− 5 ft
×
⋅
3
12
ft
3
s
1
×
⎛ 1 ⋅ ft ⎞
⎜4
⎝ ⎠
2
V = 15.3⋅
ft
s
2
s
s
⋅ ft ×
−5
1.08 × 10
⋅ ft
2
Re = 3.54 × 10
5
Turbulent
= 0.004
⎞
⎛ e
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
Given
p1 =
1
2
× 1.94⋅
slug
ft
3
4 lbf
p 1 = 2.58 × 10 ⋅
ft
2
× ⎛⎜ 15.3⋅
⎝
ft ⎞
s⎠
2
⎡⎢
⎢
⎢⎣
× 0.0287 ×
p 1 = 179 ⋅ psi
250
1
4
f = 0.0287
+ 0.5 + 4 × 0.5 +
4
2
⎛ 3 ⎞ × ( 1 + 0.02)⎤⎥ × lbf ⋅ s
⎜
⎥ slug⋅ ft
⎝1⎠
⎥⎦
Problem 8.134
Given:
Proposal for bench top experiment
Find:
Design it; Plot tank depth versus Re
[Difficulty: 4]
Solution:
Basic equations:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
+
α
⋅
+
g
⋅
z
−
⎜ρ
⎜
1 2
1
2 2
2 = h lT =
⎝
⎠ ⎝ρ
⎠
Re =
f =
ρ⋅ V⋅ D
μ
64
2
hl = f ⋅
(8.34)
(8.36)
(Laminar)
The energy equation (Eq. 8.29) becomes
2
2
2
L V
V
g ⋅ H − α⋅
= f⋅ ⋅
+ K⋅
2
D 2
2
This can be solved explicity for reservoir height H
2
H=
In Excel:
V
2⋅ g
⋅ ⎜⎛ α + f ⋅
⎝
L
D
+ K⎞
⎠
major
hl +
∑
h lm
(8.29)
minor
2
L V
⋅
D 2
Re
V
∑
h lm = K ⋅
V
(8.40a)
2
⎛ e
⎞
⎜ D
1
2.51
+
= −2.0⋅ log ⎜
f
⎝ 3.7 Re⋅ f ⎠
h lm = f ⋅
(8.37)
Le V2
(8.40b)
⋅
D 2
(Turbulent)
Computed results:
Q (L/min) V (m/s)
0.200
0.225
0.250
0.275
0.300
0.325
0.350
0.375
0.400
0.425
0.450
0.472
0.531
0.589
0.648
0.707
0.766
0.825
0.884
0.943
1.002
1.061
Re
Regime
f
H (m)
1413
1590
1767
1943
2120
2297
2473
2650
2827
3003
3180
Laminar
Laminar
Laminar
Laminar
Laminar
Laminar
Turbulent
Turbulent
Turbulent
Turbulent
Turbulent
0.0453
0.0403
0.0362
0.0329
0.0302
0.0279
0.0462
0.0452
0.0443
0.0435
0.0428
0.199
0.228
0.258
0.289
0.320
0.353
0.587
0.660
0.738
0.819
0.904
The flow rates are realistic, and could easily be measured using a tank/timer system
The head required is also realistic for a small-scale laboratory experiment
Around Re = 2300 the flow may oscillate between laminar and turbulent:
Once turbulence is triggered (when H > 0.353 m), the resistanc e to flow increases
requiring H >0.587 m to maintain; hence the flow reverts to la minar, only to trip over
again to turbulent! This behavior will be visible: the exit flow will switch back and
forth between smooth (laminar) and chaotic (turbulent)
Required Reservoir Head
versus Reynolds Number
1.00
0.75
H (m)
0.50
Laminar
0.25
Turbulent
0.00
1000
1500
2000
Re
2500
3000
3500
Problem 8.135
[Difficulty: 3]
Problem 8.136
[Difficulty: 3]
Given:
Drinking of a beverage
Find:
Fraction of effort of drinking of friction and gravity
Solution:
Basic equations
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α ⋅ 2 + g ⋅ z1 − ⎜ ρ + α ⋅ 2 + g ⋅ z2 = h l
⎝
⎠ ⎝
⎠
2
hl = f ⋅
L V
⋅
D 2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) No minor losses
Hence the energy equation becomes, between the bottom of the straw (Point 1) and top (Point 2)
g ⋅ z1 −
2
⎛ p2
⎞
L V
⎜ + g ⋅ z2 = f ⋅ ⋅
D 2
⎝ρ
⎠
where p 2 is the gage pressure in the mouth
The negative gage pressure the mouth must create is therefore due to two parts
(
2
)
p grav = −ρ⋅ g ⋅ z2 − z1
p fric = −ρ⋅ f ⋅
12
Q =
Assuming a person can drink 12 fluid ounces in 5 s
Assuming a straw is 6 in long diameter 0.2 in, with roughness
V=
128
4⋅ Q
Re =
Given
Then
− 5 ft
⋅
−5
4
π
Hence the fraction due to friction is
⋅
3
s
in (from Googling!)
− 3 ft
3
s
2
×
⎛ 1 × 12⋅ in ⎞ V = 11.5⋅ ft
⎜ 0.2⋅ in 1 ⋅ ft
s
⎝
⎠
(for water, but close enough)
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
slug
3
slug
ft
− 3 ft
Q = 2.51 × 10
2
ν
p fric = −1.94⋅
3
7.48⋅ gal
× 2.51 × 10
ft
Re = 11.5⋅
p grav = −1.94⋅
1⋅ ft
s
V⋅ D
ft
and
×
e = 5 × 10
π⋅ D
From Table A.7 (68oF) ν = 1.08 × 10
⋅ gal
5⋅ s
V =
2
L V
⋅
D 2
3
ft
× 32.2⋅
2
p fric
p fric + p grav
0.2
12
⋅ ft ×
s
1.08 × 10
4
−5
ft
Re = 1.775 × 10
2
f = 0.0272
×
s
× 0.0272 ×
s
×
1
2
6
0.2
2
⋅ ft ×
×
= 77⋅ %
1
2
lbf ⋅ s
p grav = −31.2⋅
slug⋅ ft
× ⎛⎜ 11.5⋅
⎝
lbf
ft
ft ⎞
s⎠
2
2
×
lbf ⋅ s
slug⋅ ft
and gravity is
These results will vary depending on assumptions, but it seems friction is significant!
p fric = −105 ⋅
lbf
ft
p grav
p fric + p grav
2
2
p grav = −0.217 ⋅ psi
p fric = −0.727 ⋅ psi
= 23⋅ %
Problem 8.137
Given:
Draining of swimming pool with larger hose
Find:
Flow rate and average velocity
Solution:
Basic equations
[Difficulty: 4]
2
2
⎞ ⎛⎜ p
⎞
⎛⎜ p
V1
V2
1
2
⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 = h l
⎝
⎠ ⎝
⎠
2
2
L V
⋅
D 2
hl = f ⋅
V
h lm = Kent⋅
2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) No minor losses
Available data
D = 25⋅ mm
L = 30⋅ m
e = 0.2⋅ mm
h = 3⋅ m
2
g ⋅ ( ∆z + h ) = f ⋅
Hence the energy equation becomes
Solving for V
V=
2 ⋅ g ⋅ ( ∆z + h )
f⋅
We also have
Re =
L
D
∆z = 1.5⋅ m
2
2
−6 m
Kent = 0.5
ν = 1 ⋅ 10
⋅
s
2
L V
V
V
⋅
+ Kent⋅
+
D 2
2
2
(1)
+ Kent + 1
V⋅ D
(2)
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
In addition
ν
(3)
Equations 1, 2 and 3 form a set of simultaneous equations for V, Re and f, which we can solve iteratively
Make a guess for f
f = 0.1
then
V =
2 ⋅ g ⋅ ( ∆z + h )
f⋅
using Eq 3, at this Re
V =
Re =
V⋅ D
ν
Re = 2.13 × 10
4
L
D
V = 1.37
m
V = 1.38
m
+ Kent + 1
s
Re =
V⋅ D
ν
Re = 3.41 × 10
4
Re = 3.46 × 10
4
f = 0.0371
V =
Then, repeating
2 ⋅ g ⋅ ( ∆z + h )
f⋅
f = 0.0371
Q = V⋅
π⋅ D
4
L
D
+ Kent + 1
s
which is the same as before, so we have convergence
2
The flow rate is then
s
+ Kent + 1
2 ⋅ g ⋅ ( ∆z + h )
f⋅
Using Eq 3, at this Re
D
m
f = 0.0382
Then, repeating
using Eq 3, at this Re
L
V = 0.852
3
−4m
Q = 6.79 × 10
Note that we could use Excel's Solver for this problem
s
Q = 0.679
l
s
Re =
V⋅ D
ν
Problem 8.138
Given:
Flow in horizontal pipe
Find:
Flow rate
Solution:
Basic equations
[Difficulty: 4]
2
2
⎞ ⎛⎜ p
⎞
⎛⎜ p
V1
V2
1
2
⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 = h l
⎝
⎠ ⎝
⎠
2
hl = f ⋅
L V
⋅
D 2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) No minor losses
Available data
L = 200 ⋅ m
D = 75⋅ mm
e = 2.5⋅ mm
∆p = 425 ⋅ kPa
ρ = 1000⋅
kg
3
μ = 1.76⋅ 10
− 3 N⋅ s
⋅
m
2
m
Hence the energy equation becomes
p1
ρ
Solving for V
We also have
p2
−
ρ
∆p
ρ
2
= f⋅
L V
⋅
D 2
2 ⋅ D⋅ ∆p
V=
Re =
=
V=
L⋅ ρ⋅ f
ρ⋅ V⋅ D
k
2 ⋅ D⋅ ∆p
f
Re = c⋅ V
or
μ
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
In addition
k =
(1)
(2)
k = 0.565
L⋅ ρ
c =
where
ρ⋅ D
μ
m
s
c = 4.26 × 10
4 s
m
(3)
Equations 1, 2 and 3 form a set of simultaneous equations for V, Re and f
Make a guess for f
Given
Given
f = 0.1
V =
then
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
⎞
⎛ e
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
Q = V⋅
π⋅ D
4
V = 5.86⋅
ft
V = 7.74⋅
ft
f
f = 0.0573
V =
k
f
f = 0.0573
V =
k
V= ⋅
f
2
The flow rate is then
k
3
Q = 0.0104
Note that we could use Excel's Solver for this problem
m
s
Q = 10.42
l
s
s
s
ft
s
Q = 165 ⋅ gpm
Re = c⋅ V
Re = 7.61 × 10
4
Re = c⋅ V
Re = 1.01 × 10
5
Re = c⋅ V
Re = 1.01 × 10
5
Problem 8.139
[Difficulty: 2]
Problem 8.140
[Difficulty: 4]
.
Given:
Two potential solutions to improve flowrate.
Find:
Which solution provides higher flowrate
Solution:
Basic equations:
⎛ p1
⎞ ⎛p
⎞
V2
V2
⎜⎜ + α 1 1 + gz1 ⎟⎟ − ⎜⎜ 2 + α 2 2 + gz 2 ⎟⎟ = hlT
2
2
⎝ρ
⎠ ⎝ ρ
⎠
⎛e/ D
1
2.51
LV2
V2
= −2.0 log⎜
+
hlT = hl + hlm = f
+K
⎜ 3.7 Re f
D 2
2
f
⎝
Assumptions: 1) Steady flow 2) Incompressible 3) Neglect minor losses 4)
Option 1: let z1 = 0
Given data
Q = VA
V1 2
V2
= α2 2
2
2
p 2 = p atm = 0 kPa gage
p1 = 200 kPa gage
The energy equation becomes:
Solving for V:
α1
⎞
⎟
⎟
⎠
V =
D = 0.019 m
p1
ρ
− gz 2 = f
⎛p
⎞
2 ⋅ D ⋅ ⎜⎜ 1 − g ⋅ z 2 ⎟⎟
⎝ρ
⎠
f ⋅L
e
=0
D
z 2 = 15 m
LV2
D 2
V =
k
f
(1)
L = 23 m
k=
⎞
⎛p
2 ⋅ D ⋅ ⎜⎜ 1 − g ⋅ z 2 ⎟⎟
3
⎠ = 2 × 0.019 m × ⎛⎜ 200,000 N × m × kg ⋅ m − 9.81 m × 15 m ⎞⎟ × 1
⎝ρ
⎜
⎟ 23 m
L
m 2 999 kg N ⋅ s 2
s2
⎝
⎠
k = 0.296
m
s
Re =
We also have
ρ ⋅V ⋅ D
µ
Re = c ⋅ V (2)
or
where
c = 999
Assuming water at 20oC (ρ = 999 kg/m3, µ = 1 x 10-3 kg/(m·s)):
⎛ 2.51
= −2.0 log⎜
⎜ Re f
f
⎝
1
In addition:
Given
1
f
Given
1
f
Given
1
f
f
⎛ 2.51
= −2.0 log⎜
⎜ Re f
⎝
⎛ 2.51
= −2.0 log⎜
⎜ Re f
⎝
⎛ 2.51
= −2.0 log⎜
⎜ Re f
⎝
The flowrate is then:
Option 2: let
Given data
f = 0.015
z1 = 0
⎞
⎟
⎟
⎠
⎞
⎟
⎟
⎠
⎞
⎟
⎟
⎠
π
s
m⋅s
kg
= 18981
× 0.019 m ×
3
-3
m
1 × 10 kg
m
V , Re and f
V =
k
f = 0.0213
V =
k
f = 0.0222
V =
k
f = 0.0223
V =
k
then
m
Q1 = × (0.019) m × 1.98 =
s
4
2
ρ⋅D
µ
⎞
⎟ (3)
⎟
⎠
Equations 1, 2 and 3 form a set of simultaneous equations for
Make a guess for
c=
2
f
f
f
f
= 2.42
m
s
Re = c ⋅ V = 4.59 × 10 4
= 2.03
m
s
Re = c ⋅ V = 3.85 × 10 4
= 1.99
m
s
Re = c ⋅ V = 3.77 × 10 4
= 1.98
m
s
Re = c ⋅ V = 3.76 × 10 4
5.61 × 10
−4
m3
s
p 2 = p atm = 0 kPa gage
p1 = 300 kPa gage
D = 0.0127 m
The analysis for Option 2 is identical to Option 1:
The energy equation becomes:
p1
ρ
− gz 2 = f
LV2
D 2
z 2 = 15 m
e
= 0.05
D
L = 16 m
V =
Solving for V:
k=
⎞
⎛p
2 ⋅ D ⋅ ⎜⎜ 1 − g ⋅ z 2 ⎟⎟
⎝ρ
⎠
f ⋅L
V =
k
(4)
f
⎛p
⎞
2 ⋅ D ⋅ ⎜⎜ 1 − g ⋅ z 2 ⎟⎟
3
⎝ρ
⎠ = 2 × 0.0127 m × ⎛⎜ 300,000 N × m × kg ⋅ m − 9.81 m × 15 m ⎞⎟ × 1
⎟ 16 m
⎜
L
m 2 999 kg N ⋅ s 2
s2
⎝
⎠
k = 0.493
We also have
m
s
Re =
ρ ⋅V ⋅ D
µ
c = 999
Re = c ⋅ V (5)
or
ρ⋅D
µ
kg
m⋅s
s
= 12687.3
× 0.0127 m ×
3
-3
m
m
1 × 10 kg
⎛e/ D
2.51
= −2.0 log⎜
+
⎜
f
⎝ 3.7 Re f
1
In addition:
c=
where
⎞
⎛
⎟ = −2.0 log⎜ 0.05 + 2.51
⎟
⎜ 3.7 Re f
⎠
⎝
⎞
⎟
⎟
⎠
(6)
Equations 4, 5 and 6 form a set of simultaneous equations for V , Re and f
Make a guess for
Given
Given
f
f = 0.07
⎛ 0.05
2.51
= −2.0 log⎜
+
⎜ 3.7 Re f
f
⎝
⎛ 0.05
1
2.51
= −2.0 log⎜
+
⎜
f
⎝ 3.7 Re f
1
The flowrate is then:
V =
k
f = 0.0725
V =
k
f = 0.0725
V =
k
then
⎞
⎟
⎟
⎠
⎞
⎟
⎟
⎠
π
f
f
m
s
Re = c ⋅ V = 2.36 × 10 4
= 1.83
m
s
Re = c ⋅ V = 2.32 × 10 4
= 1.83
m
s
Re = c ⋅ V = 2.32 × 10 4
3
m
−4 m
Q2 = × (0.0127 ) m × 1.83 = 2.32 × 10
s
s
4
2
2
This problem can also be solved explicitly:
LV2
− gz 2 = f
ρ
D 2
p1
The energy equation becomes:
or:
f
= 1.86
f =
Plugging this into the Colebrook equation:
1
V
⎛p
⎞
2 D⎜⎜ 1 − gz 2 ⎟⎟
⎝ρ
⎠
L
Option 1 is 2.42 times more effective!
1
f
Noting that the
=V
L
⎛p
⎞
2 D⎜⎜ 1 − gz 2 ⎟⎟
⎝ρ
⎠
⎛
⎛
⎜
⎜
⎜ e / D 2.51µ ⎜
= −2.0 log⎜
+
V
ρV D ⎜⎜
⎜ 3.7
⎜
⎜
⎝
⎝
V s on the right hand side cancel provides:
⎛
⎜
⎜ e / D 2.51µ
V = −2.0 log⎜
+
3
.
7
ρD
⎜
⎜
⎝
⎛
2 D⎜⎜
⎝
⎛
2 D⎜⎜
⎝
⎞⎞
⎟⎟
⎟⎟
L
⎟⎟
p1
⎞
− gz 2 ⎟⎟ ⎟ ⎟
ρ
⎠ ⎟⎠ ⎟⎠
⎞
⎞⎛
⎟⎜ 2 D⎛⎜ p1 − gz 2 ⎞⎟ ⎟
⎜ρ
⎟⎟
⎟⎜
L
⎝
⎠
⎟
⎟⎜
L
p1
⎞ ⎟⎜
⎟
− gz 2 ⎟⎟
⎟
⎜
⎟
ρ
⎠ ⎠⎝
⎠
Assuming water at 20oC (ρ = 999 kg/m3, µ = 1 x 10-3 kg/(m·s)) gives the remaining information needed to perform the calculation.
For Option 1:
⎛
⎞
kg
m3
1
⎜ 2.51 × 1 × 10 −3
⎟
×
×
m ⋅ s 999 kg 0.019 m
⎜
⎟
⎜
⎟
V = −2.0 log⎜
23 m
1
⎟
×
×
⎜
2 × 0.019 m ⎛
⎞⎟
N
m3
kg ⋅ m
m
⎜⎜ 200,000 2 ×
− 9.81 2 × 15 m ⎟⎟ ⎟
×
⎜⎜
2
999
kg
⋅
m
N
s
s
⎝
⎠ ⎟⎠
⎝
⎛ 2 × 0.019 m ⎛
⎞ ⎞⎟
N
m3
kg ⋅ m
m
⎟⎟
9
.
81
15
m
×⎜
× ⎜⎜ 200,000 2 ×
×
−
×
⎜
999 kg N ⋅ s 2
23 m
m
s2
⎝
⎠ ⎟⎠
⎝
m
V = 1.98
s
and:
Q1 =
Option 2: let
Given data
z1 = 0
π
4
× (0.019) m 2 × 1.98
2
m3
m
= 5.61 × 10 − 4
s
s
p 2 = p atm = 0 kPa gage
p1 = 300 kPa gage
z 2 = 15 m
D = 0.0127 m
e
= 0.05
D
The analysis for Option 2 results in the same equations as used in Option 1 once again giving:
⎛
⎜
⎜ e / D 2.51µ
V = −2.0 log⎜
+
ρD
⎜ 3.7
⎜
⎝
⎛
2 D⎜⎜
⎝
⎞
⎞⎛
⎟⎜ 2 D⎛⎜ p1 − gz 2 ⎞⎟ ⎟
⎟⎟
⎜ρ
⎟⎜
L
⎠
⎝
⎟
⎟⎜
L
p1
⎞ ⎟⎜
⎟
− gz 2 ⎟⎟
⎟
⎜
⎟
ρ
⎠ ⎠⎝
⎠
L = 16 m
⎞
⎛ 0.05
kg
m3
1
⎟
⎜
+ 2.51 × 1 × 10 −3
×
×
m ⋅ s 999 kg 0.0127 m
⎟
⎜ 3.7
⎟
⎜
V = −2.0 log⎜
16 m
1
⎟
×
×
3
⎜
2 × 0.0127 m ⎛
⎞⎟
kg ⋅ m
N
m
m
⎜⎜ 300,000 2 ×
×
− 9.81 2 × 15 m ⎟⎟ ⎟
⎜⎜
2
999 kg N ⋅ s
s
m
⎠ ⎟⎠
⎝
⎝
⎛ 2 × 0.0127 m ⎛
⎞ ⎞⎟
N
m3
kg ⋅ m
m
⎟⎟
−
×
×⎜
× ⎜⎜ 300,000 2 ×
×
9
.
81
15
m
⎜
999 kg N ⋅ s 2
16 m
m
s2
⎠ ⎟⎠
⎝
⎝
V = 1.83
The flowrate is then:
Q2 =
π
4
× (0.0127 ) m 2 × 1.83
2
m
=
s
m
s
2.32 × 10 − 4
m3
s
Problem 8.141
Given:
Kiddy pool on a porch.
Find:
Time to fill pool with a hose.
[Difficulty: 3]
Solution:
⎛ p1
⎞ ⎛ p2
⎞
V1 2
V 22
⎜
⎟
⎜
Basic equations:
⎜ ρ + α 1 2 + gz1 ⎟ − ⎜ ρ + α 2 2 + gz 2 ⎟⎟ = hlT
⎝
⎠ ⎝
⎠
2
2
⎛e/ D
1
2.51
LV
V
= −2.0 log⎜
+
hlT = hl + hlm = f
+K
⎜
D 2
2
f
⎝ 3.7 Re f
Assumptions: 1) Steady flow 2) Incompressible 3) Neglect minor losses 4)
Given data
p1 = 60
z1 = 0 ft
D = 0.625 in
z 2 = 20.5 ft
e
=0
D
L = 50 ft
k=
V =
Q = VA
V1 2
V2
= α2 2
2
2
lbf
gage
in 2
The energy equation becomes:
Solving for V:
α1
⎞
⎟
⎟
⎠
p2 = 0
lbf
gage
in 2
LV2
− gz 2 = f
D 2
ρ
p1
⎛p
⎞
2 ⋅ D ⋅ ⎜⎜ 1 − g ⋅ z 2 ⎟⎟
⎝ρ
⎠
f ⋅L
V =
k
(1)
f
⎛p
⎞
2 ⋅ D ⋅ ⎜⎜ 1 − g ⋅ z 2 ⎟⎟
2
3
⎞
slug ⋅ ft
ft
1
⎝ρ
⎠ = 2 × 0.625 in × ft × ⎛⎜ 60 lbf × 144 in × ft
×
− 32.2 2 × 20.5 ft ⎟⎟ ×
2
2
2
⎜
L
12 in ⎝ in
1.94 slug lbf ⋅ s
ft
s
⎠ 50 ft
k = 2.81
ft
s
We also have
Re =
ρ ⋅V ⋅ D
µ
Re = c ⋅ V (2)
or
where
c=
ρ⋅D
µ
Assuming water at 68oF (ρ = 1.94 slug/ft3, µ = 2.1 x 10-5 lbf·s/ft2):
c = 1.94
s
slug
ft
ft 2
lbf ⋅ s 2
= 4811.5
×
×
×
×
0
.
625
in
3
-5
ft
12 in 2.1 × 10 lbf ⋅ s slug ⋅ ft
ft
⎛ 2.51
= −2.0 log⎜
⎜ Re f
f
⎝
1
In addition:
⎞
⎟ (3)
⎟
⎠
Equations 1, 2 and 3 form a set of simultaneous equations for V , Re and f
Given
Given
V =
k
f = 0.0177
V =
k
f = 0.0179
V =
k
f = 0.015
Make a guess for f
⎛ 2.51
= −2.0 log⎜
⎜ Re f
f
⎝
⎛ 2.51
1
= −2.0 log⎜
⎜ Re f
f
⎝
1
then
⎞
⎟
⎟
⎠
⎞
⎟
⎟
⎠
f
f
f
= 22.94
ft
s
Re = c ⋅ V = 1.1 × 10 5
= 21.12
ft
s
Re = c ⋅ V = 1.02 × 10 5
= 21.0
ft
s
Re = c ⋅ V = 1.01 × 10 5
The flowrate is then:
Q =
π
× (0.625) in 2 ×
2
4
ft 2
ft
ft 3
×
=
0
.
0447
21
.
0
s
s
144 in 2
Volume pool = 2.5 ft ×
time =
Volume pool
Q
=
π
4
× (5) ft 2 = 49.1 ft 3
2
49.1 ft 3
= 1097 s =
ft 3
0.0447
s
This problem can also be solved explicitly in the following manner:
p1
The energy equation becomes:
ρ
f =
or:
1
V
− gz 2 = f
LV2
D 2
⎛p
⎞
2 D⎜⎜ 1 − gz 2 ⎟⎟
⎝ρ
⎠
L
Plugging this into the Colebrook equation:
1
=V
f
L
⎞
⎛p
2 D⎜⎜ 1 − gz 2 ⎟⎟
⎝ρ
⎠
⎛
⎛
⎜
⎜
⎜ 2.51µ ⎜
= −2.0 log⎜
⎜V
⎜ ρV D ⎜
⎜
⎜
⎝
⎝
⎛
2 D⎜⎜
⎝
⎞⎞
⎟⎟
⎟⎟
L
⎟⎟
p1
⎞
− gz 2 ⎟⎟ ⎟ ⎟
ρ
⎠ ⎟⎠ ⎟⎠
18.3 min.
Noting that the
V s on the right hand side cancel provides:
⎞
⎞⎛
⎛
⎟⎜ 2 D⎛⎜ p1 − gz 2 ⎞⎟ ⎟
⎜
⎜
⎟
⎟⎜
⎜ 2.51µ
L
⎝ρ
⎠⎟
V = −2.0 log⎜
⎜
⎟
⎟
ρ
D
L
p
⎛
⎞
1
⎟
⎜
⎜
2 D⎜⎜ − gz 2 ⎟⎟ ⎟
⎜
⎟
⎟
⎜
⎝ρ
⎠ ⎠⎝
⎝
⎠
Assuming water at 68oF (ρ = 1.94 slug/ft3, µ = 2 x 10-5 lbf·s/ft2) and g = 32.2 ft/s2 gives the remaining information needed to perform
the calculation finding:
⎞
⎛
slug ⋅ ft
1
12 in
lbf ⋅ s
ft 3
⎟
⎜ 2.51 × 2.1 × 10 −5
×
×
×
×
2
2
0.625 in
ft
1.94 slug lbf ⋅ s
ft
⎟
⎜
⎟
⎜
V = −2.0 log⎜
50 ft
12 in
1
⎟
×
×
×
3
⎟
⎜
2 × 0.625 in
ft
⎞
⎛ lbf 144 in 2
slug ⋅ ft
ft
ft
⎜⎜ 60 2 ×
×
×
− 32.2 2 × 20.5 ft ⎟⎟ ⎟
⎜⎜
2
2
1.94 slug lbf ⋅ s
ft
s
⎠ ⎟⎠
⎝ in
⎝
⎛ 2 × 0.625 in
⎞ ⎞⎟
⎛
ft
lbf 144 in 2
ft 3
slug ⋅ ft
ft
⎟⎟
×⎜
×
× ⎜⎜ 60 2 ×
×
×
−
×
32
.
2
20
.
5
ft
⎜
50 ft
12 in ⎝
1.94 slug lbf ⋅ s 2
in
ft 2
s2
⎠ ⎟⎠
⎝
V = 21.0
ft
s
and:
Q =
π
4
× (0.052) ft 2 × 21.0
2
Volume pool = 2.5 ft ×
time =
Volume pool
Q
=
π
4
ft
ft 3
= 0.0447
s
s
× (5) ft 2 = 49.1 ft 3
2
49.1 ft 3
= 1097 s =
ft 3
0.0447
s
18.3 min.
Problem 8.142
[Difficulty: 3]
Problem 8.143
[Difficulty: 3]
Problem 8.144
[Difficulty: 3]
Problem 8.145
[Difficulty: 3]
Problem 8.146
Given:
Flow from large reservoir
Find:
Flow rates in two pipes
Solution:
Basic equations
[Difficulty: 4]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α ⋅ 2 + g ⋅ z1 − ⎜ ρ + α ⋅ 2 + g ⋅ z2 = h l
⎝
⎠ ⎝
⎠
2
hl = f ⋅
2
L V
⋅
D 2
V
h lm = Kent ⋅
2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1
Available data
D = 50⋅ mm
H = 10⋅ m
L = 10⋅ m
e = 0.15⋅ mm
(Table 8.1)
ν = 1⋅ 10
2
−6 m
⋅
Kent = 0.5
(Table A.8)
s
The energy equation becomes
2
V2
L V2
g ⋅ z1 − z2 − ⋅ V2 = f ⋅ ⋅
+ Kent⋅
2
D 2
2
(
Solving for V
)
2
2
and
V2 = V
z1 − z2 = H
2⋅ g⋅ H
V=
f⋅
We also have
1
L
+ Kent + 1
D
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
f
⎝ 3.7 Re⋅ f ⎠
(1)
Re =
(2)
V⋅ D
(3)
ν
We must solve Eqs. 1, 2 and 3 iteratively.
Make a guess for V
and
Then
V = 1⋅
m
Then
s
⎞
⎛ e
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
2⋅ g⋅ H
V =
f⋅
L
+ Kent + 1
D
Re =
V⋅ D
ν
f = 0.0286
V = 5.21
m
s
Re = 5.00 × 10
4
Repeating
and
Then
Re =
V⋅ D
⎞
⎛ e
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
2⋅ g⋅ H
V =
f⋅
Repeating
and
Then
Re =
L
⎞
⎛ e
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
2⋅ g⋅ H
V =
Q =
π
4
V = 5.36
m
s
Re = 2.68 × 10
ν
L
5
f = 0.0267
+ Kent + 1
D
V⋅ D
f⋅
Hence
Re = 2.61 × 10
ν
5
f = 0.0267
V = 5.36
+ Kent + 1
D
m
s
3
2
⋅D ⋅V
Q = 0.0105
m
Q = 10.5⋅
s
l
s
We repeat the analysis for the second pipe, using 2L instead of L:
Make a guess for V
and
Then
V = 1⋅
m
⎞
⎛ e
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
2⋅ g⋅ H
V =
f⋅
Repeating
and
Then
Then
s
Re =
2⋅ L
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
f
⎝ 3.7 Re⋅ f ⎠
2⋅ g⋅ H
2⋅ L
+ Kent + 1
D
Re = 5.00 × 10
ν
f = 0.0286
V = 3.89
m
s
Re = 1.95 × 10
ν
f⋅
V⋅ D
+ Kent + 1
D
V⋅ D
V =
Re =
f = 0.0269
V = 4.00
m
s
5
4
Repeating
and
Then
Re =
V⋅ D
⎞
⎛ e
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
2⋅ g⋅ H
V =
f⋅
Hence
Re = 2.00 × 10
ν
Q =
π
4
2⋅ L
+ Kent + 1
D
2
⋅D ⋅V
As expected, the flow is considerably less in the longer pipe.
5
f = 0.0268
V = 4.00
m
s
Q = 7.861 × 10
3
−3m
s
Q = 7.86⋅
l
s
Problem 8.147
Given:
Galvanized drainpipe
Find:
Maximum downpour it can handle
Solution:
[Difficulty: 3]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
+
α
⋅
+
g
⋅
z
−
⎜ρ
⎜
1
2 = hl
2
2
⎝
⎠ ⎝ρ
⎠
Basic equations
2
hl = f ⋅
L V
⋅
D 2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) No minor losses
Available data
D = 50⋅ mm
e = 0.15⋅ mm
h=L
(Table 8.1)
(
From Table A.7 (20oC)
2
)
The energy equation becomes
L V
g ⋅ z1 − g ⋅ z2 = g ⋅ z1 − z2 = g ⋅ h = f ⋅ ⋅
D 2
Solving for V
V=
k=
2 ⋅ D⋅ g ⋅ h
L⋅ f
2 ⋅ D⋅ g
=
2 ⋅ D⋅ g
V=
f
k =
k
(1)
f
2 × 0.05⋅ m × 9.81⋅
m
k = 0.99
2
s
Re =
We also have
V⋅ D
Re = c⋅ V
or
ν
s
c = 0.05⋅ m ×
(2)
m
s
c=
where
D
ν
4 s
−6
1.01 × 10
c = 4.95 × 10 ⋅
2
⋅m
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
In addition
ν = 1.01 × 10
m
(3)
Equations 1, 2 and 3 form a set of simultaneous equations for V, Re and f
Make a guess for f
f = 0.01
V =
then
k
V = 9.90
m
f
Given
s
Re = c⋅ V
5
Re = 4.9 × 10
⎛ e
⎞
⎜
1
D
2.51
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
f = 0.0264
V =
k
f
V = 6.09
m
s
Re = c⋅ V
Re = 3.01 × 10
5
2
−6 m
⋅
s
Given
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
f
⎝ 3.7 Re⋅ f ⎠
f = 0.0266
k
V =
V = 6.07
m
V = 6.07
m
s
f
Given
Re = c⋅ V
Re = 3.00 × 10
5
Re = c⋅ V
Re = 3.00 × 10
5
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
f
⎝ 3.7 Re⋅ f ⎠
f = 0.0266
k
V =
s
f
2
The flow rate is then
Q = V⋅
3
π⋅ D
Q = 0.0119⋅
4
m
s
3
The downpour rate is then
Q
Aroof
0.0119⋅
=
m
s
2
×
500 ⋅ m
Note that we could use Excel's Solver for this problem
100 ⋅ cm
1⋅ m
×
60⋅ s
1 ⋅ min
= 0.143 ⋅
cm
min
The drain can handle 0.143 cm/min
Problem 8.148
[Difficulty: 3] Part 1/2
Problem 8.148
[Difficulty: 3] Part 2/2
Problem 8.149
Given:
Flow in a tube
Find:
Effect of tube roughness on flow rate; Plot
[Difficulty: 3]
Solution:
Governing equations:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α1⋅ 2 + g ⋅ z1 − ⎜ ρ + α2 ⋅ 2 + g⋅ z2 = h lT =
⎝
⎠ ⎝
⎠
Re =
f =
ρ⋅ V⋅ D
64
μ
2
hl = f ⋅
L V
⋅
D 2
(8.36)
hl +
major
∑
h lm (8.29)
minor
2
(8.34)
(Laminar)
Re
∑
h lm = K⋅
V
(8.40a)
2
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
h lm = f ⋅
(8.37)
Le V2
(8.40b)
⋅
D 2
(Turbulent)
The energy equation (Eq. 8.29) becomes for flow in a tube
2
L V
p 1 − p 2 = ∆p = ρ⋅ f ⋅ ⋅
D 2
This cannot be solved explicitly for velocity V, (and hence flow rate Q) because f depends on V; solution for a given relative
roughness e/D requires iteration (or use of Solver)
Flow Rate versus Tube Relative Roughness
for fixed Dp
8
6
3
Q (m /s)
x 10
4
4
2
0
0.00
0.01
0.02
e/D
0.03
0.04
0.05
It is not possible to roughen the tube sufficiently to slow the flow down to a laminar flow for this Δp. Even a relative roughness of
0.5 (a physical impossibility!) would not work.
Problem 8.150
Given:
Flow in a tube
Find:
Effect of tube length on flow rate; Plot
[Difficulty: 3]
Solution:
Governing equations:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
+
α
⋅
+
g
⋅
z
−
⎜ρ
⎜
1 2
1
2 2
2 = h lT =
⎝
⎠ ⎝ρ
⎠
Re =
f =
ρ⋅ V⋅ D
64
μ
2
hl = f ⋅
L V
⋅
D 2
(8.36)
hl +
major
∑
h lm (8.29)
minor
2
(8.34)
(Laminar)
Re
∑
h lm = K⋅
V
(8.40a)
2
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
f
⎝ 3.7 Re⋅ f ⎠
h lm = f ⋅
(8.37)
Le V2
(8.40b)
⋅
D 2
(Turbulent)
The energy equation (Eq. 8.29) becomes for flow in a tube
2
L V
p 1 − p 2 = ∆p = ρ⋅ f ⋅ ⋅
D 2
This cannot be solved explicitly for velocity V, (and hence flow rate Q) because f depends on V; solution for a given L requires
iteration (or use of Solver)
Flow Rate vs Tube Length for Fixed Dp
10.0
Laminar
Q (m3/s)
Turbulent
4
x 10 1.0
0.1
0
5
10
15
L (km)
20
25
30
35
The "critical" length of tube is between 15 and 20 km. For this range, the fluid is making a transition between laminar and
turbulent flow, and is quite unstable. In this range the flow oscillates between laminar and turbulent; no consistent solution is
found (i.e., an Re corresponding to turbulent flow needs an f assuming laminar to produce the Δp required, and vice versa!) More
realistic numbers (e.g., tube length) are obtained for a fluid such as SAE 10W oil (The graph will remain the same except for scale)
Problem 8.151
[Difficulty: 5]
Given:
Flow from a reservoir
Find:
Effect of pipe roughness and pipe length on flow rate; Plot
Solution:
Governing equations:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α1⋅ 2 + g ⋅ z1 − ⎜ ρ + α2 ⋅ 2 + g⋅ z2 = h lT =
⎝
⎠ ⎝
⎠
Re =
f =
ρ⋅ V⋅ D
64
μ
2
hl = f ⋅
hl +
major
∑
h lm (8.29)
minor
2
L V
⋅
D 2
(8.36)
∑
h lm = K⋅
(8.34)
V
(8.40a)
2
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
(Laminar)
Re
h lm = f ⋅
(8.37)
Le V2
(8.40b)
⋅
D 2
(Turbulent)
The energy equation (Eq. 8.29) becomes for this flow (see Example 8.5)
⎛
p pump = ∆p = ρ⋅ ⎜ g ⋅ d + f ⋅
⎝
2⎞
L V
⋅
D 2
⎠
We need to solve this for velocity V, (and hence flow rate Q) as a function of roughness e, then length L. This cannot be solved
explicitly for velocity V, (and hence flow rate Q) because f depends on V; solution for a given relative roughness e/D or length L
requires iteration (or use of Solver)
It is not possible to roughen the tube sufficiently to slow
the flow down to a laminar flow for this Δp.
Flow Rate versus Tube Relative Roughness for fixed Dp
0.020
0.015
3
Q (m /s)
0.010
0.005
0.000
0.00
0.01
0.01
0.02
0.02
0.03
0.03
0.04
0.04
0.05
0.05
e/D
Flow Rate versus Tube Length for fixed Dp
0.010
Q (m3/s)
0.005
0.000
0
200
400
600
L (m)
800
1000
1200
Problem 8.152
[Difficulty: 4]
Given:
System for fire protection
Find:
Height of water tower; Maximum flow rate; Pressure gage reading
Solution:
Governing equations:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α1⋅ 2 + g ⋅ z1 − ⎜ ρ + α2 ⋅ 2 + g⋅ z2 = h lT =
⎝
⎠ ⎝
⎠
Re =
f =
∑
hl +
major
∑
h lm(8.29)
minor
2
ρ⋅ V⋅ D
hl = f ⋅
μ
64
(8.36)
L V
⋅
D 2
(8.34)
h lm = 0.1⋅ h l
h lm = f ⋅
⎛ e
⎞
⎜
D
1
2.51
(8.37)
+
= −2.0⋅ log ⎜
3.7
f
Re
⋅
f
⎝
⎠
(Laminar)
Re
Le V2
(8.40b)
⋅
D 2
(Turbulent)
For no flow the energy equation (Eq. 8.29) applied between the water tower free surface (state 1; height H) and pressure gage is
g⋅ H =
p2
or
ρ
H=
p2
ρ⋅ g
(1)
The energy equation (Eq. 8.29) becomes, for maximum flow (and α = 1)
2
g⋅ H −
2
V
= h lT = ( 1 + 0.1) ⋅ h l
2
or
g⋅ H =
V
2
⋅ ⎛⎜ 1 + 1.1⋅ f ⋅
⎝
L⎞
(2)
D⎠
This can be solved for V (and hence Q) by iterating, or by using Solver
The energy equation (Eq. 8.29) becomes, for restricted flow
g⋅ H −
p2
ρ
2
+
The results in Excel are shown below:
V
= h lT = ( 1 + 0.1) ⋅ h l
2
2
p 2 = ρ⋅ g ⋅ H − ρ⋅
V
2
⋅ ⎜⎛ 1 + 1.1⋅ ρ⋅ f ⋅
⎝
L⎞
D⎠
(3)
Problem 8.153
Given:
Syphon system
Find:
Flow rate; Minimum pressure
Solution:
[Difficulty: 4]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
+
α
⋅
+
g
⋅
z
−
⎜ρ
⎜
1
2 = h lT
2
2
⎝
⎠ ⎝ρ
⎠
Basic equations
2
h lT = f ⋅
L V
⋅
+ h lm
D 2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1
Hence the energy equation applied between the tank free surface (Point 1) and the tube exit (Point 2, z = 0) becomes
g ⋅ z1 −
V2
2
2
2
2
2
Le V2
V
L V
V
= g ⋅ z1 −
= f⋅ ⋅
+ Kent⋅
+ f⋅ ⋅
2
D 2
2
D 2
Kent = 0.78
From Table 8.2 for reentrant entrance
For the bend
R
D
=9
The two lengths are
Le = 56⋅ D
We also have
Re =
D
(1)
⎡
⎛ L Le ⎞⎤
⎢1 + Kent + f ⋅ ⎜ +
⎥
⎣
⎝ D D ⎠⎦
Le = 2.8 m
V⋅ D
or
ν
ν = 1.14 × 10
2
−6 m
⋅
s
then
h = 2.5⋅ m
and
L = 4.51 m
Re = c⋅ V (2)
where
c = 0.05⋅ m ×
⎞
⎛ e
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
f = 0.01
= 56
L = ( 0.6 + π⋅ 0.45 + 2.5) ⋅ m
Equations 1, 2 and 3 form a set of simultaneous equations for V, Re and f
Make a guess for f
Le
for a 90o bend so for a 180 o bend
2⋅ g⋅ h
V=
In addition
= 28
D
Solving for V
From Table A.7 (15oC)
Le
so from Fig. 8.16
s
D
ν
4 s
−6
1.14 × 10
(3)
c=
2
c = 4.39 × 10 ⋅
⋅m
e = 0.0015⋅ mm (Table 8.1)
m
V =
2⋅ g⋅ h
⎡
⎛ L Le ⎞⎤
⎢1 + Kent + f ⋅ ⎜ +
⎥
⎣
⎝ D D ⎠⎦
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
Given
2⋅ g⋅ h
V =
⎛ L Le ⎞⎤
⎡
⎢1 + Kent + f ⋅ ⎜ +
⎥
⎣
⎝ D D ⎠⎦
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
f
⎝ 3.7 Re⋅ f ⎠
Given
2⋅ g⋅ h
V =
⎡
⎛L
⎢1 + Kent + f ⋅ ⎜ +
⎥
⎣
⎝ D D ⎠⎦
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
Given
V =
Le ⎞⎤
2⋅ g⋅ h
Le ⎞⎤
⎡
⎛L
⎢1 + Kent + f ⋅ ⎜ +
⎥
⎣
⎝ D D ⎠⎦
V = 3.89
m
Re = c⋅ V
s
Re = 1.71 × 10
5
f = 0.0164
V = 3.43
m
s
Re = c⋅ V
Re = 1.50 × 10
5
Re = c⋅ V
Re = 1.49 × 10
5
Re = c⋅ V
Re = 1.49 × 10
5
f = 0.0168
V = 3.40
m
s
f = 0.0168
V = 3.40
m
s
Note that we could use Excel's Solver for this problem.
2
Q =
The flow rate is then
π⋅ D
4
3
−3m
⋅V
Q = 6.68 × 10
s
The minimum pressure occurs at the top of the curve (Point 3). Applying the energy equation between Points 1 and 3
2
⎛⎜ p
⎞
2
2
V3
⎛ p3 V2
Le V2
⎞
3
V
L V
g ⋅ z1 − ⎜
+
+ g ⋅ z3 = g ⋅ z1 − ⎜
+
+ g ⋅ z3 = f ⋅ ⋅
+ Kent⋅
+ f⋅ ⋅
2
2
D 2
2
D 2
⎝ρ
⎠
⎝ρ
⎠
where we have
Le
D
= 28
for the first 90o of the bend, and
L = ⎜⎛ 0.6 +
⎝
π × 0.45 ⎞
2
⎠
⋅m
L = 1.31 m
2
⎡
⎛ L Le ⎞⎤⎤
V ⎡
⋅ ⎢1 + Kent + f ⋅ ⎜ +
p 3 = ρ⋅ ⎢g ⋅ z1 − z3 −
⎥⎥
2 ⎣
⎣
⎝ D D ⎠⎦⎦
(
)
⎡
⎢
kg ⎢
m
p 3 = 1000⋅
× 9.81⋅ × ( −0.45⋅ m) −
3 ⎢
2
m
s
⎣
2
⎤
⎛ 3.4⋅ m ⎞
⎥
⎜
2
s⎠ ⎡
1.31
⎝
⎥ N⋅ s p = −20.0⋅ kPa
⋅ ⎢1 + 0.78 + 0.0168⋅ ⎛⎜
+ 28⎞⎥⎤ ×
2
⎣
⎝ 0.05
⎠⎦⎥⎦ kg⋅ m 3
Problem 8.154
[Difficulty: 4]
Given:
Tank with drainpipe
Find:
Flow rate for rentrant, square-edged, and rounded entrances
Solution:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
+
α
⋅
+
g
⋅
z
−
⎜ρ
⎜
1
2 = h lT
2
2
⎝
⎠ ⎝ρ
⎠
Basic equations
2
h lT = f ⋅
2
L V
V
⋅
+ Kent⋅
D 2
2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1
Available data
D = 1 ⋅ in
L = 2 ⋅ ft
e = 0.00085 ⋅ ft
h = 3 ⋅ ft
(Table 8.1)
r = 0.2⋅ in
Hence the energy equation applied between the tank free surface (Point 1) and the pipe exit (Point 2, z = 0) becomes
V2
g ⋅ z1 −
Solving for V
V=
We also have
Re =
From Table A.7 (20oC)
In addition
2
2
2
2
2
V
L V
V
= g ⋅ z1 −
= f⋅ ⋅
+ Kent⋅
2
D 2
2
2⋅ g⋅ h
(1)
⎛1 + K + f ⋅ L ⎞
⎜
ent
D⎠
⎝
V⋅ D
Re = c⋅ V (2)
or
ν
ν = 1.01 × 10
2
−6 m
⋅
− 5 ft
ν = 1.09 × 10
s
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
c=
where
2
c =
s
D
ν
Make a guess for f
f = 0.01
(3)
Re = c⋅ V
Given
Kent = 0.78
V =
then
Re = 7.49 × 10
2⋅ g⋅ h
⎛
⎜ 1 + Kent + f ⋅
D⎠
⎝
4
⎛ e
⎞
⎜
1
D
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
L⎞
f = 0.0389
ν
c = 7665⋅
Equations 1, 2 and 3 form a set of simultaneous equations for V, Re and f
For a reentrant entrance, from Table 8.2
D
V = 2.98
m
s
s
ft
2⋅ g⋅ h
V =
Given
L⎞
⎛
⎜ 1 + Kent + f ⋅ D
⎝
⎠
e
⎛
⎞
⎜ D
1
2.51
+
= −2.0⋅ log ⎜
f
⎝ 3.7 Re⋅ f ⎠
V = 2.57
L⎞
⎛
⎜ 1 + Kent + f ⋅ D
⎝
⎠
e
⎛
⎞
⎜ D
1
2.51
+
= −2.0⋅ log ⎜
f
⎝ 3.7 Re⋅ f ⎠
V =
m
s
Re = c⋅ V
Re = 6.46 × 10
4
Re = c⋅ V
Re = 6.46 × 10
4
Re = c⋅ V
Re = 6.46 × 10
4
f = 0.0391
2⋅ g⋅ h
V =
Given
V = 2.57
m
s
f = 0.0391
2⋅ g⋅ h
V = 2.57
L⎞
⎛
⎜ 1 + Kent + f ⋅ D
⎝
⎠
m
s
Note that we could use Excel's Solver for this problem
2
The flow rate is then
Q = V⋅
π⋅ D
Q = 0.0460
4
Given
then
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
f
⎝ 3.7 Re⋅ f ⎠
V =
Given
f = 0.01
V =
2⋅ g⋅ h
⎛
⎜ 1 + Kent + f ⋅
D⎠
⎝
L⎞
2⋅ g⋅ h
V = 2.71
L⎞
m
s
2⋅ g⋅ h
V = 2.71
L⎞
⎛
⎜ 1 + Kent + f ⋅
D⎠
⎝
Q = V⋅
π⋅ D
m
Q = 0.0485
4
For a rounded entrance, from Table 8.2
r
D
m
Re = 8.07 × 10
4
Re = c⋅ V
Re = 6.83 × 10
4
Re = c⋅ V
Re = 6.82 × 10
4
V = 3.21
s
Re = c⋅ V
f = 0.0390
2
The flow rate is then
Q = 20.6⋅ gpm
s
f = 0.0389
⎛
⎜ 1 + Kent + f ⋅ D
⎝
⎠
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
V =
3
Kent = 0.5
For a square-edged entrance, from Table 8.2
Make a guess for f
ft
= 0.2
s
ft
3
s
Q = 21.8⋅ gpm
Kent = 0.04
Make a guess for f f = 0.01
V =
then
V = 3.74
m
V =
V =
2⋅ g⋅ h
V =
m
V = 3.02
L⎞
4
s
Re = c⋅ V
Re = 7.59 × 10
4
Re = c⋅ V
Re = 7.58 × 10
4
Re = c⋅ V
Re = 7.58 × 10
4
f = 0.0389
2⋅ g⋅ h
m
V = 3.01
⎛
⎜ 1 + Kent + f ⋅ D
⎝
⎠
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
Given
Re = 9.41 × 10
f = 0.0388
⎛
⎜ 1 + Kent + f ⋅ D
⎝
⎠
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
Given
⎛1 + K + f ⋅ L ⎞
⎜
ent
D⎠
⎝
Re = c⋅ V
s
⎞
⎛ e
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
Given
2⋅ g⋅ h
L⎞
s
f = 0.0389
2⋅ g⋅ h
m
V = 3.01
⎛
⎜ 1 + Kent + f ⋅ D
⎝
⎠
L⎞
s
Note that we could use Excel's Solver for this problem
2
The flow rate is then
In summary:
Q = V⋅
π⋅ D
4
Renentrant: Q = 20.6⋅ gpm
Q = 0.0539
ft
3
s
Square-edged:
Q = 24.2⋅ gpm
Q = 21.8⋅ gpm
Rounded:
Q = 24.2⋅ gpm
Problem 8.155
[Difficulty: 4]
Given:
Tank with drainpipe
Find:
Flow rate for rentrant, square-edged, and rounded entrances
Solution:
2
2
⎞ ⎛⎜ p
⎞
⎛⎜ p
V1
V2
1
2
⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 = h lT
⎝
⎠ ⎝
⎠
Basic equations
2
h lT = f ⋅
2
L V
V
⋅
+ Kent⋅
D 2
2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1
Available data
D = 1 ⋅ in
L = 2 ⋅ ft
e = 0.00085 ⋅ ft
h = 3 ⋅ ft
(Table 8.1)
r = 0.2⋅ in
Hence the energy equation applied between the tank free surface (Point 1) and the pipe exit (Point 2, z = 0) becomes
V2
g ⋅ z1 −
Solving for V
V=
2
2
2
2
2⋅ g⋅ H
(1)
⎛1 + K + f ⋅ L ⎞
⎜
ent
D⎠
⎝
H = h+L
where now we have
We also have
From Table A.7 (20oC)
In addition
Re =
2
V
L V
V
= g ⋅ z1 −
= f⋅ ⋅
+ Kent⋅
2
D 2
2
V⋅ D
Re = c⋅ V (2)
or
ν
ν = 1.01 × 10
H = 5 ft
2
−6 m
⋅
− 5 ft
ν = 1.09 × 10
s
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
f
⎝ 3.7 Re⋅ f ⎠
c=
where
2
s
c =
D
ν
Make a guess for f
f = 0.01
(3)
Re = c⋅ V
Given
Kent = 0.78
V =
then
Re = 9.67 × 10
2⋅ g⋅ H
L⎞
⎛
⎜ 1 + Kent + f ⋅
D⎠
⎝
4
⎛ e
⎞
⎜
1
D
2.51
= −2.0⋅ log ⎜
+
3.7
f
Re⋅ f ⎠
⎝
f = 0.0388
ν
c = 7665⋅
Equations 1, 2 and 3 form a set of simultaneous equations for V, Re and f
For a reentrant entrance, from Table 8.2
D
V = 3.85
m
s
s
ft
2⋅ g⋅ H
V =
Given
V = 3.32
L⎞
⎛
⎜ 1 + Kent + f ⋅ D
⎝
⎠
e
⎛
⎞
⎜ D
1
2.51
+
= −2.0⋅ log ⎜
f
⎝ 3.7 Re⋅ f ⎠
V =
m
s
Re = c⋅ V
Re = 8.35 × 10
4
Re = c⋅ V
Re = 8.35 × 10
4
f = 0.0389
2⋅ g⋅ H
V = 3.32
L⎞
⎛
⎜ 1 + Kent + f ⋅ D
⎝
⎠
m
s
Note that we could use Excel's Solver for this problem
2
The flow rate is then
Q = V⋅
π⋅ D
Q = 0.0594
4
f = 0.01
Given
Given
then
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
f
⎝ 3.7 Re⋅ f ⎠
V =
V =
2⋅ g⋅ H
⎛
⎜ 1 + Kent + f ⋅ D
⎝
⎠
L⎞
V = 3.51
L⎞
V = 3.51
⎛
⎜ 1 + Kent + f ⋅ D
⎝
⎠
L⎞
π⋅ D
m
s
m
Q = 0.0627
4
For a rounded entrance, from Table 8.2
r
D
Make a guess for f
f = 0.01
V = 4.83
s
s
ft
5
Re = c⋅ V
Re = 8.82 × 10
4
Re = c⋅ V
Re = 8.82 × 10
4
Re = c⋅ V
s
3
Q = 28.2⋅ gpm
s
= 0.2
then
m
m
Re = 1.04 × 10
V = 4.14
f = 0.0388
2⋅ g⋅ H
Q = V⋅
Q = 26.7⋅ gpm
2⋅ g⋅ H
2
The flow rate is then
s
f = 0.0387
⎛
⎜ 1 + Kent + f ⋅ D
⎝
⎠
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
V =
3
Kent = 0.5
For a square-edged entrance, from Table 8.2
Make a guess for f
ft
Kent = 0.04
V =
2⋅ g⋅ H
⎛1 + K + f ⋅ L ⎞
⎜
ent
D⎠
⎝
Re = c⋅ V
Re = 1.22 × 10
5
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
Given
V =
f = 0.0386
2⋅ g⋅ H
L⎞
⎛
⎜ 1 + Kent + f ⋅ D
⎝
⎠
e
⎞
⎛
⎜ D
1
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
Given
V =
V =
s
Re = c⋅ V
Re = 9.80 × 10
4
Re = c⋅ V
Re = 9.80 × 10
4
Re = c⋅ V
Re = 9.80 × 10
4
f = 0.0388
2⋅ g⋅ H
m
V = 3.89
L⎞
⎛
⎜ 1 + Kent + f ⋅ D
⎝
⎠
e
⎛
⎞
⎜ D
1
2.51
+
= −2.0⋅ log ⎜
f
⎝ 3.7 Re⋅ f ⎠
Given
m
V = 3.90
s
f = 0.0388
2⋅ g⋅ H
m
V = 3.89
L⎞
⎛
⎜ 1 + Kent + f ⋅ D
⎝
⎠
s
Note that we could use Excel's Solver for this problem
2
The flow rate is then
In summary:
Q = V⋅
π⋅ D
4
Renentrant: Q = 26.7⋅ gpm
Q = 0.0697
ft
3
s
Square-edged:
Q = 31.3⋅ gpm
Q = 28.2⋅ gpm
Rounded:
Q = 31.3⋅ gpm
Problem 8.156
[Difficulty: 5]
Given:
Tank with drain hose
Find:
Flow rate at different instants; Estimate of drain time
Solution:
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
⎜ ρ + α⋅ 2 + g⋅ z1 − ⎜ ρ + α⋅ 2 + g⋅ z2 = h l
⎝
⎠ ⎝
⎠
Basic equations
2
hl = f ⋅
L V
⋅
D 2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1 4) Ignore minor loss at entrance (L >>; verify later)
Available data
L = 1⋅ m
D = 15⋅ mm
e = 0.2⋅ mm
3
Vol = 30⋅ m
Hence the energy equation applied between the tank free surface (Point 1) and the hose exit (Point 2, z = 0) becomes
g ⋅ z1 −
Solving for V
V=
We also have
Re =
From Fig. A.2 (20oC)
In addition
V2
2
2
2
2
V
L V
= g ⋅ z1 −
= f⋅ ⋅
2
D 2
2⋅ g⋅ h
(1)
⎛1 + f ⋅ L ⎞
⎜
D⎠
⎝
V⋅ D
Re = c⋅ V (2)
or
ν
2
−6 m
ν = 1.8 × 10
⋅
c =
s
D
c = 8333
ν
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
h = 10⋅ m
initially
where
c=
and
D
ν
s
m
(3)
Equations 1, 2 and 3 form a set of simultaneous equations for V, Re and f
Make a guess for f
f = 0.01
V =
then
2⋅ g⋅ h
⎛1 + f ⋅ L ⎞
⎜
D⎠
⎝
V = 10.8
m
s
Re = c⋅ V
Re = 9.04 × 10
4
Given
Given
⎛ e
⎞
⎜
D
1
2.51
+
f = 0.0427
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
V =
⎛ e
⎞
⎜
D
1
2.51
+
f = 0.0427
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
V =
2⋅ g⋅ h
⎛1 + f ⋅
⎜
D⎠
⎝
L⎞
2⋅ g⋅ h
⎛1 + f ⋅
⎜
D⎠
⎝
L⎞
Note that we could use Excel's Solver for this problem
π⋅ D
4
Given
3
−3m
Q = 1.26 × 10
⎛ e
⎞
⎜
D
1
2.51
+
f = 0.0430
= −2.0⋅ log ⎜
f
⎝ 3.7 Re⋅ f ⎠
V =
2
Q = V⋅
π⋅ D
4
Given
Q = 8.9 × 10
⎛1 + f ⋅ L ⎞
⎜
D⎠
⎝
2⋅ g⋅ h
⎛1 + f ⋅
⎜
D⎠
⎝
L⎞
L
D
Re = 5.95 × 10
4
Re = c⋅ V
Re = 5.95 × 10
4
= 2.8
Ke = 0.5
h lm < h l
l
s
V = 5.04
m
V = 5.04
m
Q = 0.890 ⋅
s
V =
⎛ e
⎞
⎜
D
1
2.51
+
f = 0.0444
= −2.0⋅ log ⎜
f
⎝ 3.7 Re⋅ f ⎠
2
Q = V⋅
2⋅ g⋅ h
3
−4m
⎛ e
⎞
⎜
D
1
2.51
+
f = 0.0444
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
The flow rate is then
f⋅
s
Re = c⋅ V
s
s
Re = c⋅ V
Re = 4.20 × 10
4
Re = c⋅ V
Re = 4.20 × 10
4
Re = c⋅ V
Re = 1.85 × 10
4
Re = c⋅ V
Re = 1.85 × 10
4
l
s
h = 1⋅ m
Next we recompute everything for
Given
m
Q = 1.26⋅
s
V =
⎛ e
⎞
⎜
D
1
2.51
+
f = 0.0430
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
The flow rate is then
V = 7.14
s
h = 5⋅ m
Next we recompute everything for
Given
m
Note:
2
Q = V⋅
The flow rate is then
V = 7.14
π⋅ D
4
V =
3
−4m
Q = 3.93 × 10
s
2⋅ g⋅ h
⎛1 + f ⋅
⎜
D⎠
⎝
L⎞
2⋅ g⋅ h
⎛1 + f ⋅ L ⎞
⎜
D⎠
⎝
V = 2.23
m
V = 2.23
m
Q = 0.393 ⋅
s
s
l
s
Initially we have dQ/dt = -1.26L/s, then -.890 L/s, then -0.393 L/s. These occur at h = 10 m, 5 m and 1 m. The corresponding
volumes in the tank are then Q = 30,000 L, 15,000 L, and 3,000 L3. Using Excel we can fit a power trendline to the dQ/dt versus Q
data to find, approximately
1
dQ
dt
= −0.00683 ⋅ Q
t = 293 ⋅ ( 30 −
2
Q)
where dQ/dt is in L/s and t is s. Solving this with initial condition Q = -1.26 L/s when t = 0 gives
Hence, when Q = 3000 L (h = 1 m)
t = 293 ⋅ ( 30000 −
3000) ⋅ s
4
t = 3.47 × 10 s
t = 9.64⋅ hr
Problem 8.157
[Difficulty: 4]
Problem 8.158
[Difficulty: 4] Part 1/2
Problem 8.158
[Difficulty: 4] Part 2/2
Problem 8.159
[Difficulty: 4] Part 1/2
Problem 8.159
[Difficulty: 4] Part 2/2
Problem 8.160
[Difficulty: 5]
Applying the energy equation between inlet and exit:
∆p
ρ
= f
L V 2
D 2
"Old school":
or
∆p ρf V 2
=
D 2
L
∆p ⎛ ∆p ⎞ ⎛ Q0 ⎞
=⎜ ⎟ ⎜ ⎟
L ⎝ L ⎠0 ⎜⎝ Q ⎟⎠
Q (gpm)
20
18
16
Flow (gpm)
14
12
10
8
6
4
2
0
0.00
0.01
Your boss was wrong!
1.25
1.50
1.75
2.00
2.25
2.50
2.75
3.00
3.25
3.50
3.75
4.00
4.25
4.50
4.75
5.00
5.25
5.50
5.75
6.00
6.25
6.50
6.75
7.00
7.25
7.50
7.75
0.02
8.00
8.25
8.50
8.75
9.00
Q (ft3/s)
D=
e=
2
ν =
ρ =
V (ft/s)
Re
f
1 in
0.00015 ft
2
1.08E-05 ft /s
3
1.94 slug/ft
∆p (old
∆p (psi/ft)
school) (psi)
0.00279
0.511
3940
0.0401
0.00085
0.00085
0.00334
0.613
4728
0.0380
0.00122
0.00115
0.00390
0.715
5516
0.0364
0.00166
0.00150
0.00446
0.817
6304
0.0350
0.00216
0.00189
0.00501
0.919
7092
0.0339
0.00274
0.00232
1.021Pressure
7881 Drop 0.0329
0.00338
0.00278
Flow0.00557
Rate versus
0.00613
1.123
8669
0.0321
0.00409
0.00328
0.00668
1.226
9457
0.0314
0.00487
0.00381
0.00724
1.328
10245
0.0307
0.00571
0.00438
0.00780
1.430
11033
0.0301
0.00663
0.00498
0.00836
1.532
11821
0.0296
0.00761
0.00561
0.00891
1.634
12609
0.0291
0.00865
0.00628
0.00947
1.736
13397
0.0286
0.00977
0.00698
0.01003
1.838
14185
0.0282
0.01095
0.00771
0.01058
1.940
14973
0.0278
0.01220
0.00847
0.01114
2.043
15761
0.0275
0.01352
0.00927
0.01170
2.145
16549
0.0272
0.01491
0.01010
0.01225
2.247
17337
0.0268
0.01636
0.01095
0.01281
2.349
18125
0.0265
0.01788
0.01184
0.01337
2.451
18913
0.0263
0.01947
0.01276
0.01393
2.553
19701
0.0260
0.02113
0.01370
0.01448
2.655
20489
0.0258
0.02285
0.01468
School"
0.01504
2.758
21277
0.0255 "Old 0.02465
0.01569
0.01560
2.860
22065
0.0253 Exact0.02651
0.01672
0.01615
2.962
22854
0.0251
0.02843
0.01779
0.01671
3.064
23642
0.0249
0.03043
0.01888
0.01727
3.166
24430
0.0247
0.03249
0.02000
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.10
0.01783
3.268
25218
0.0245
0.03462
0.02115
Pressure Drop (psi/ft)
0.01838
3.370
26006
0.0243
0.03682
0.02233
0.01894
3.472
26794
0.0242
0.03908
0.02354
0.01950
3.575
27582
0.0240
0.04142
0.02477
0.02005
3.677
28370
0.0238
0.04382
0.02604
Problem 8.161
Given:
Flow from large reservoir
Find:
Diameter for flow rates in two pipes to be same
Solution:
Basic equations
[Difficulty: 5]
2
2
⎞ ⎛⎜ p
⎞
⎛⎜ p
V1
V2
1
2
+
α
⋅
+
g
⋅
z
+
α
⋅
+
g
⋅
z
−
⎜ρ
⎜
1
2 = hl
2
2
⎝
⎠ ⎝ρ
⎠
2
hl = f ⋅
2
L V
⋅
D 2
V
h lm = Kent⋅
2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α is approximately 1
Available data
D = 50⋅ mm
H = 10⋅ m
L = 10⋅ m
e = 0.15⋅ mm
(Table 8.1)
ν = 1 ⋅ 10
2
−6 m
⋅
Kent = 0.5
(Table A.8)
s
For the pipe of length L the energy equation becomes
2
V2
L V2
g ⋅ z1 − z2 − ⋅ V2 = f ⋅ ⋅
+ Kent⋅
2
D 2
2
(
Solving for V
)
2
2
and
V2 = V
z1 − z2 = H
2⋅ g⋅ H
V=
f⋅
We also have
1
L
+ Kent + 1
D
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
(1)
Re =
(2)
V⋅ D
(3)
ν
We must solve Eqs. 1, 2 and 3 iteratively.
Make a guess for V
and
Then
V = 1⋅
m
s
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
2⋅ g⋅ H
V =
f⋅
Repeating
Then
Re =
L
+ Kent + 1
D
V⋅ D
ν
Re =
V⋅ D
Re = 5.00 × 10
ν
f = 0.0286
V = 5.21
m
s
Re = 2.61 × 10
5
4
and
Then
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
f
⎝ 3.7 Re⋅ f ⎠
2⋅ g⋅ H
V =
f⋅
Repeating
and
Then
Re =
V = 5.36
L
+ Kent + 1
D
V⋅ D
⎛ e
⎞
⎜
1
2.51
D
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
π
Q =
4
s
5
f = 0.0267
2⋅ g⋅ H
V =
m
Re = 2.68 × 10
ν
f⋅
Hence
f = 0.0267
V = 5.36
L
+ Kent + 1
D
m
s
3
2
⋅D ⋅V
Q = 0.0105
m
Q = 10.5⋅
s
l
s
This is the flow rate we require in the second pipe (of length 2L)
For the pipe of length 2L the energy equation becomes
2
V2
2 ⋅ L V2
g ⋅ z1 − z2 − ⋅ V2 = f ⋅
⋅
+ Kent⋅
2
2
D
2
(
)
2
Hence
H=
V
2⋅ g
⋅ ⎜⎛ f ⋅
⎝
1
2⋅ L
D
+ Kent + 1⎞
and
Re =
V⋅ D
Re = 2.23 × 10
ν
3
and
V2 = V
z1 − z2 = H
Q = 0.0105
m
Q
V = 3.72
m
D = 0.06⋅ m
Then we have
π
4
5
⎛ e
⎞
⎜
D
1
2.51
+
= −2.0⋅ log ⎜
3.7
f
Re⋅ f ⎠
⎝
V =
2
⋅D
f = 0.0256
2
Using Eq 4 to find H
V ⎛ 2⋅ L
⋅⎜f ⋅
Hiterate =
+ Kent + 1⎞
2⋅ g ⎝ D
⎠
Hence the diameter is too large: Only a head of
s
(4)
⎠
We must make a guess for D (larger than the other pipe)
Then
2
2
Hiterate = 7.07 m
Hiterate = 7.07 m
But
H = 10 m
would be needed to generate the flow. We make D smaller
s
Try
Then
and
D = 0.055 ⋅ m
Re =
V⋅ D
Re = 2.43 × 10
ν
Q
V =
Then we have
π
4
5
⎞
⎛ e
⎜ D
2.51
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
1
V = 4.43
2
⋅D
m
s
f = 0.0261
2
Using Eq 4 to find H
2⋅ L
Hiterate =
+ Kent + 1⎞
⋅ ⎛⎜ f ⋅
2⋅ g ⎝ D
⎠
V
Hence the diameter is too small: A head of
Try
Then
and
D = 0.056 ⋅ m
Re =
Q
V =
Re = 2.39 × 10
ν
But
H = 10 m
Hiterate = 10.97 m would be needed. We make D slightly larger
Then we have
V⋅ D
Hiterate = 10.97 m
π
4
5
⎞
⎛ e
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
V = 4.27
2
⋅D
m
s
f = 0.0260
2
Using Eq 4 to find H
V ⎛ 2⋅ L
⋅⎜f ⋅
Hiterate =
+ Kent + 1⎞
2⋅ g ⎝ D
⎠
Hence the diameter is too large A head of
Try
Then
and
Hiterate = 10.02 m
D = 0.05602 ⋅ m Then we have
Re =
V⋅ D
Q
π
4
5
⎞
⎛ e
⎜
1
2.51
D
= −2.0⋅ log ⎜
+
f
⎝ 3.7 Re⋅ f ⎠
But
H = 10 m
would be needed. We can make D smaller
V =
Re = 2.39 × 10
ν
Hiterate = 10.02 m
V = 4.27
2
⋅D
m
s
f = 0.0260
2
Using Eq 4 to find H
V ⎛ 2⋅ L
Hiterate =
+ Kent + 1⎞
⋅⎜f ⋅
2⋅ g ⎝ D
⎠
Hiterate = 10 m
Hence we have
D = 0.05602 m
V = 4.27
3
Check
Q = 0.0105
D = 56.02 ⋅ mm
m
π
s
4
2
⋅ D ⋅ V = 0.0105
3
m
s
m
s
But
H = 10 m
Problem 8.162
Given:
Hydraulic press system
Find:
Minimum required diameter of tubing
Solution:
Basic equations
[Difficulty: 3]
2
2
⎛⎜ p
⎞ ⎛⎜ p
⎞
V1
V2
1
2
+
α
⋅
+
g
⋅
z
+
α
⋅
+
g
⋅
z
−
⎜ρ
⎜
1
2 = hl
2
2
⎝
⎠ ⎝ρ
⎠
L V2
hl = f ⋅ ⋅
D 2
2
Assumptions: 1) Steady flow 2) Incompressible flow 3) α at 1 and 2 is approximately 1 4) Ignore minor losses
The flow rate is low and it's oil, so try assum