Problem 1.1
Problem
1.1
[Difficulty: 3]
1.1
Given:
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
Problem
1.2
[Difficulty: 2]
1.2
1.2
Given:
Five basic conservation laws stated in Section 1-4.
1.2
Write:
A word statement of each, as they apply to a system.
Solution:
Assume that laws are to be written for a system.
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
Problem
1.3
[Difficulty: 3]
1.3
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: 1)
1.4
Given:
Data on oxygen tank.
Find:
Mass of oxygen.
Solution:
Compute tank volume, and then us e oxygen density to find the mass.
The given or available
data is:
D = 16⋅ft
For oxygen the critical temperature and pressure are:
p = 1000⋅psi
T = (77 + 460)⋅R
T = 537⋅R
Tc = 279⋅R
p c = 725.2⋅psi
(data from NIST WebBook)
so the reduced temperature and pressure are:
Using a compressiblity factor chart:
Z = 0.948
Since this number is close to 1, we can assume ideal gas behavior.
Therefore, the governing equation is the ideal gas equation
p = ρ⋅R O2⋅T
3
where V is the tank volume
V = π⋅D
6
V =
π
6
× (16⋅ft)
and
3
ρ=
3
V = 2144.7⋅ft
M
V
Hence:
Problem 1.5
Problem
1.10
[Difficulty: 4]
1.5
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
M⋅
Newton's 2nd law for the sphere (mass M) is (ignoring buoyancy effects)
dV
so
g−
Integrating twice and using limits
3 ⋅ π⋅ μ⋅ d
V( t) =
M
ρgas = 1.40⋅
dV
dt
slug
ft
3
= M ⋅ g − 3 ⋅ π⋅ μ⋅ V⋅ d
= dt
⋅V
M⋅ g
3⋅ π⋅ μ ⋅ d
Replacing M with an expression involving diameter d
− 3⋅ π⋅ μ ⋅ d ⎞
⎛
⋅t
⎜
M
⋅⎝1 − e
⎠
π⋅ d
M = ρgas ⋅
6
⎡
⎛ − 3⋅ π⋅ μ ⋅ d ⋅ t
⎢
⎜
M
M
x( t) =
⋅ ⎢t +
⋅⎝e
−
3⋅ π⋅ μ ⋅ d ⎣
3⋅ π⋅ μ ⋅ d
⎞⎤
⎥
1⎠⎥
⎦
⎡⎢
⎛ − 18⋅ μ ⋅ t
2 ⎜
ρgas ⋅ d ⋅ g ⎢
ρgas ⋅ d ⎜ ρgas⋅ d2
x( t) =
⋅ ⎢t +
⋅⎝e
−
18⋅ μ
18⋅ μ
⎣
⎞⎥⎤
⎥
1⎠⎥
⎦
M⋅ g
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 1.12
1.6
Problem
[Difficulty: 3]
1.6
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
×
= M⋅ g − k⋅ V
k = 1.61 × 10
slug⋅ ft
⋅
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
lbf ⋅ s
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 1.7
(Difficulty: 2)
1.7 A rocket payload with a weight on earth of 2000 𝑙𝑙𝑙 is landed on the moon where the acceleration
due to the moon’s gravity 𝑔𝑚 ≈
payload’s moon weight.
𝑔𝑒
.
6
Find the mass of the payload on the earth and the moon and the
Given: Rocket payload weight on earth 𝑊𝑒 = 2000 𝑙𝑙𝑙. The acceleration due to the moon’s gravity
𝑔
𝑔𝑚 ≈ 𝑒 .
6
Find: The mass of payload on earth 𝑀𝑒 and on moon 𝑀𝑚 in SI and EE units. The payload’s moon weight
𝑊𝑚 .
Solution:
Basic equation: Newton’s law applied to mass and weight
Gravity on the moon relative to that on Earth:
𝑀=
𝑊
𝑔
The value of gravity is:
𝑔𝑚 ≈
𝑔𝑒
6
𝑔𝑒 = 32.2
The mass on earth is:
𝑓𝑓
𝑠2
𝑊𝑒 2000 𝑙𝑙𝑙
=
= 62.1 𝑠𝑠𝑠𝑠
𝑓𝑓
𝑔𝑒
32.2 2 .
𝑠
The mass on moon is the same as it on earth:
𝑀𝑒 =
The weight on the moon is then
𝑀𝑚 = 62.1 𝑠𝑠𝑠𝑠
𝑊𝑚 = 𝑀𝑚 𝑔𝑚 = 𝑀𝑚 �
𝑔𝑒
𝑔𝑒
𝑊𝑒
� = 𝑀𝑒 � � =
= 333 𝑙𝑙𝑙
6
6
6
Problem 1.8
(Difficulty: 1)
1.8 A cubic meter of air at 101 𝑘𝑘𝑘 and 15 ℃ weighs 12.0 𝑁. What is its specific volume? What is the
specific volume if it is cooled to −10 ℃ at constant pressure?
Given: Specific weight 𝛾 = 12.0
𝑁
𝑚3
at 101 𝑘𝑘𝑘 and 15 ℃.
Find: The specific volume 𝑣 at 101 𝑘𝑘𝑘 and 15 ℃. Also the specific volume 𝑣 at 101 𝑘𝑘𝑘 and −10 ℃.
Assume: Air can be treated as an ideal gas
Solution:
Basic equation: ideal gas law:
𝑝𝑝 = 𝑅𝑅
The specific volume is equal to the reciprocal of the specific weight divided by gravity
𝑣1 =
𝑔
𝛾
Using the value of gravity in the SI units, the specific volume is
The temperature conditions are
𝑚
𝑔 9.81 𝑠2
𝑚3
= 0.818
𝑣1 = =
𝛾
12.0 𝑁
𝑘𝑘
𝑇1 = 15 ℃ = 288 𝐾,
𝑇2 = −10 ℃ = 263𝐾
For 𝑣2 at the same pressure of 101 𝑘𝑘𝑘 and cooled to −10 ℃ we have, because the gas constant is the
same at both pressures:
𝑅𝑇1
𝑣1
𝑇1
𝑝
=
=
𝑣2 𝑅𝑇2 𝑇2
𝑝
So the specific volume is
𝑣2 = 𝑣1
𝑚3 263 𝐾
𝑚3
𝑇2
= 0.818
×
= 0.747
𝑇1
𝑘𝑘 288 𝐾
𝑘𝑘
Problem 1.9
(Difficulty: 2)
1.9 Calculate the specific weight, specific volume and density of air at 40℉ and 50 𝑝𝑝𝑝𝑝. What are the
values if the air is then compressed isentropically to 100 psia?
Given: Air temperature: 40℉, Air pressure 50 psia.
Find: The specific weight, specific volume and density at 40℉ and 50 psia and the values at 100 psia
after isentropic compression.
Assume: Air can be treated as an ideal gas
Solution:
Basic equation: 𝑝𝑝 = 𝑅𝑅
The absolute temperature is
The gas constant is
𝑇1 = 40℉ = 500°𝑅
The specific volume is:
𝑅 = 1715
𝑓𝑓 ∙ 𝑙𝑙𝑙
𝑠𝑠𝑠𝑠 ∙ °𝑅
𝑓𝑓 ∙ 𝑙𝑙𝑙
1715
𝑅𝑇1
𝑓𝑓 3
𝑠𝑠𝑠𝑠 ∙ °𝑅
𝑣1 =
=
×
500°𝑅
=
119.1
144𝑖𝑖2
𝑝
𝑠𝑠𝑠𝑠
50𝑝𝑝𝑝𝑝 ×
𝑓𝑓 2
The density is the reciprocal of the specific volume
𝜌1 =
1
𝑠𝑠𝑠𝑠
= 0.0084
𝑣1
𝑓𝑓 3
Using Newton’s second law, the specific weight is the density times gravity:
𝛾1 = 𝜌𝜌 = 0.271
𝑙𝑙𝑙
𝑓𝑓 3
For the isentropic compression of air to 100 psia, we have the relation for entropy change of an ideal gas:
𝑠2 − 𝑠1 = 𝑐𝑝 ln
𝑇2
𝑇1
− 𝑅 ln
𝑝2
𝑝1
The definition of an isentropic process is
𝑠2 = 𝑠1
Solving for the temperature ratio
The values of R and specific heat are
𝑅 = 1715
𝑝2 𝑅/𝑐𝑝
𝑇2
=� �
𝑇1
𝑝1
𝑓𝑓 ∙ 𝑙𝑙𝑙
𝑓𝑓 ∙ 𝑙𝑙𝑙
𝐵𝐵𝐵
= 53.3
= 0.0686
𝑠𝑠𝑠𝑠 ∙ °𝑅
𝑙𝑙 ∙ °𝑅
𝑙𝑙 ∙ °𝑅
𝑐𝑝 = 0.24
The temperature after compression to 100 psia is
𝐵𝐵𝐵
𝑙𝑙𝑙 𝑅
𝑝2 𝑅/𝑐𝑝
100 𝑝𝑝𝑝𝑝 0.0686/0.24
= 500 𝑅 �
�
= 610 °𝑅
𝑇2 = 𝑇1 � �
50 𝑝𝑝𝑝𝑝
𝑝1
𝑝2 = 100 𝑝𝑝𝑝𝑝 = 14400
The specific volume is computed using the ideal gas law:
𝑙𝑙𝑙
𝑓𝑓 2
𝑓𝑓 ∙ 𝑙𝑙𝑙
1715
𝑅𝑇2
𝑓𝑓 3
𝑠𝑠𝑠𝑠 ∙ °𝑅
𝑣2 =
=
×
610.00°𝑅
=
72.6
144𝑖𝑖2
𝑝2
𝑠𝑠𝑠𝑠
100𝑝𝑝𝑝𝑝 ×
𝑓𝑓 2
The density is the reciprocal of the specific volume
The specific weight is:
𝜌2 =
1
𝑠𝑠𝑠𝑠
= 0.0138
𝑣2
𝑓𝑓 3
𝛾2 = 𝜌2 𝑔 = 0.444
𝑙𝑙𝑙
𝑓𝑓 3
Problem 1.10
Problem
1.13
1.10
[Difficulty: 5]
1.6
Given:
Data on sphere and terminal speed from Problem 1.12.
1.6
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
×
slug⋅ ft
V⋅ dV
g−
k
M
(
)
− 13
y = 1 ⋅ 10
2
⋅ slug ⋅
32.2⋅ ft
2
s
+ 1 ⋅ 10
− 13
M⋅
M ⋅g
−3
⎝
k
M⋅ g
dV
⋅ V⎞ −
⎠
M
k
k =
dt
= M⋅
M⋅ g
Vt
dy dV
dV
⋅
= M ⋅ V⋅
= M⋅ g − k⋅ V
dt dy
dy
⋅V
V = 0.198 ⋅
ft
s
2
2
2
1
s
0.198 ⋅ ft slug⋅ ft ⎞
⎞ ⋅ ⎛⎜ lbf ⋅ s ⎞ ⋅ ln⎛⎜ 1 − 1.61⋅ 10− 11⋅ lbf ⋅ s ⋅
⋅⎛
⋅
⋅
⋅
...
⎜
− 11
⎜
− 13
32.2⋅ ft
2
ft
s
slug⋅ ft ⎠
⎝
1 ⋅ 10
lbf ⋅ s ⎠
⋅ slug
⎝ 1.61⋅ 10 ⋅ lbf ⋅ s ⎠
⎝
2
ft
− 11
1.61⋅ 10
y = 4.49 × 10
2
⋅ ln⎛⎜ 1 −
V = 0.99⋅ Vt
ft
⋅ slug ×
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
⋅
ft
lbf ⋅ s
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=
⋅ ⎢t +
⋅⎝e
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 1.11
Problem
1.14
[Difficulty: 4]
1.11
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⋅
so
Mg
dV dy
dV
2
⋅
= M ⋅ V⋅
= M⋅ g − k⋅ V
dy dt
dt
y
M⋅ g
0
2
⎛
k⋅ V ⎞
2⋅ k
ln⎜ 1 −
=−
y
M
⋅
g
M
⎝
⎠
m
Vt = 52.4
s
⌠
dV = ⎮ g dy
⌡
2
0
k⋅ V
V
1−
a = dV/dt
2
or
⎛
−
M⋅ g ⎜
2
V =
⋅⎝1 − e
k
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
⎛⎜
M ⎜
t= ⋅
⋅ ln
2 k⋅ g ⎜
⎜⎝
1
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 1.12
Problem
1.16
[Difficulty: 3]
1.12
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
dt
g
From Eq. 1:
h
V0
(2)
(3)
gR
gR
=
2v 0 2 2gh
∴ u 20 =
gR 2
8h
1
Then
⎛
gR 2
gR 2 ⎞ 2
⎟
+ 2 gh and V0 = ⎜⎜ 2 gh +
V =u +v =
8h
8h ⎟⎠
⎝
2
0
2
0
2
0
(4)
1
⎛
m
9.81 m
1 ⎞2
m
⎟⎟ = 37.7
V0 = ⎜⎜ 2 × 9.81 2 × 10 m +
× 100 2 m 2 ×
2
s
8
s
10
m
s
⎝
⎠
From Eq. 3:
v 0 = 2gh = V0 sin θ , θ = sin −1
2gh
V0
(5)
1
⎡
⎤
2
m
s ⎥
⎛
⎞
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
Problem 1.13
[Difficulty: 2]
1.13
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
M⋅ L
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
=
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 1.14
(Difficulty: 1)
1.14 The density of a sample of sea water is 1.99 𝑠𝑠𝑠𝑠𝑠 ⁄𝑓𝑓 3 . What are the values in SI and EE units?
Given: The density of sea water is 1.99 𝑠𝑠𝑠𝑠𝑠⁄𝑓𝑓 3
Find: The density of sea water in SI and EE units
Solution:
For SI unit:
The relations between the units are 1 𝑚 = 3.28 𝑓𝑓 , 1 𝑘𝑘 = 0.0685 𝑠𝑠𝑠𝑠
For EE units:
1
𝑘𝑘
𝑠𝑠𝑠𝑠 1.99 × 0.0685 𝑘𝑘
= 1026 3
𝜌 = 1.99 3 =
1
𝑚
𝑓𝑓
𝑚3
3.283
The relation between a lbm and a slug is 1 𝑙𝑙𝑙 = 0.0311 𝑠𝑠𝑠𝑠
1
𝑠𝑠𝑠𝑔 1.99 × 0.0311 𝑙𝑙𝑙
𝑙𝑙𝑙
𝜌 = 1.99 3 =
= 64.0
3
𝑓𝑓
𝑓𝑓 3
𝑓𝑓
Problem 1.15
(Difficulty: 1)
1.15 A pump is rated at 50 ℎ𝑝; What is the rating in 𝑘𝑘 and 𝐵𝐵𝐵⁄ℎ𝑟 ?
Given: The pump is rated at 50 ℎ𝑝.
Find: The rating in 𝑘𝑘 and 𝐵𝐵𝐵⁄ℎ𝑟.
Solution:
The relation between the units is
The power is then
1
1 𝑘𝑘 = 1.341 ℎ𝑝
𝐵𝐵𝐵
= 0.000393 ℎ𝑝
ℎ𝑟
𝑃 = 50 ℎ𝑝 = 50 ℎ𝑝 ×
1 𝑘𝑘
= 37.3 𝑘𝑘
1.341 ℎ𝑝
𝐵𝐵𝐵
ℎ𝑟 ℎ𝑝 = 127,200 𝐵𝐵𝐵
𝑃 = 50 ℎ𝑝 = 50 ℎ𝑝 ×
ℎ𝑟
0.000393
1
Problem 1.16
(Difficulty: 1)
1.16 A fluid occupying 3.2 𝑚3 has a mass of 4𝑀𝑀. Calculate its density and specific volume in SI, EE and
BG units.
Given: The fluid volume 𝑉 = 3.2 𝑚3 and mass 𝑚 = 4𝑀𝑀.
Find: Density and specific volume in SI, EE and BG units.
Solution:
For SI units:
The density is the mass divided by the volume
𝜌=
𝑚 4000 𝑘𝑘
𝑘𝑘
=
= 1250 3
3
𝑉
3.2 𝑚
𝑚
The specific volume is the reciprocal of the density:
For EE units:
𝑣=
The density is:
1
𝜌=
And the specific volume is:
𝑣=
𝑚3
1
= 8 × 10−4
𝜌
𝑘𝑘
𝑙𝑙𝑙
𝑘𝑘
=
16.0
𝑓𝑓 3
𝑚3
1250 𝑙𝑙𝑙
𝑙𝑙𝑙
=
78.0
16.0 𝑓𝑓 3
𝑓𝑓 3
1 𝑓𝑓 3
𝑓𝑓 3
1
=
= 0.0128
𝜌 78.0 𝑙𝑙𝑙
𝑙𝑙𝑙
For BG unit, the relation between slug and lbm is:
The density is:
1
𝜌=
𝑠𝑠𝑠𝑠
𝑙𝑙𝑙
= 32.2
3
𝑓𝑓
𝑓𝑓 3
78.0 𝑠𝑠𝑠𝑠
𝑠𝑠𝑠𝑠
= 2.43
3
32.2 𝑓𝑓
𝑓𝑓 3
And the specific volume is
𝑣=
1 𝑓𝑓 3
𝑓𝑓 3
1
=
= 0.412
𝜌 2.43 𝑠𝑠𝑠𝑠
𝑠𝑠𝑠𝑠
Problem 1.17
(Difficulty: 1)
1.17 If a power plant is rated at 2000 𝑀𝑀 output and operates (on average) at 75% of rated power,
how much energy (in 𝐽 and 𝑓𝑓 ∙ 𝑙𝑙𝑙) does it put out a year.
Given: The power plant is rated at = 2000 𝑀𝑀 . Efficiency 𝜂 = 75%.
Find: Energy output per year 𝐸 in SI and EE units.
Solution:
For SI units:
The energy produced is a year is:
𝐸 = 𝑃𝑃 ∙ 𝜂 = 2000 × 106 𝑊 × �365
For EE units:
𝑑𝑑𝑑
ℎ𝑟
𝑠
× 24
× 3600 � 𝑠 × 0.75 = 4.73 × 1016 𝐽
𝑦𝑦
𝑑𝑑𝑑
ℎ𝑟
The relation between ft-lbf and Joules is
1 𝑓𝑓 ∙ 𝑙𝑙𝑙 = 1.356 𝐽
The energy is:
𝐸=
4.73 × 1016
𝑓𝑓 ∙ 𝑙𝑙𝑙 = 3.49 × 1016 𝑓𝑓 ∙ 𝑙𝑙𝑙
1.356
Problem 1.18
Problem
1.18
[Difficulty: 2]
1.18
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
L
2
2
ft
F⋅ t
L
2
2
m
ft
s ⋅K
2
s ⋅R
1
1
1
T
K
R
N⋅ m⋅ s
lbf ⋅ ft⋅ s
⎛ F⋅ t2 ⎞
⎜
⋅T
⎝ L ⎠
=
2
t ⋅T
2
LengthChange
(i) Thermal expansion coefficient ThermalExpansionCoefficient =
(j) Angular momentum
Length
Temperature
=
AngularMomentum = Momentum × Distance = F⋅ t⋅ L
2
Problem 1.19
Problem
1.20
[Difficulty: 1]
1.19
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⋅
= 1⋅
×
×⎜
= 47.9⋅
⋅ ⎠
2
2
2
1⋅ lbf
⎝ 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 1.20
Problem
1.22
[Difficulty: 1]
1.20
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⋅
4047⋅ m
1 ⋅ acre
1 ⋅ ft
3
231 ⋅ in
×
3
= 4.56 × 10 ⋅ m
3
⎛ 0.0254⋅ m ⎞ = 0.00246 ⋅ m
⎜
s
⎝ 1 ⋅ in ⎠
3
hr⋅ s
0.3048⋅ m
3
×
s
mile
×
1 ⋅ gal
×
1609⋅ m
1 ⋅ mile
3
3
⎛ 0.0254⋅ m ⎞ ⋅ 1⋅ min = 0.000189⋅ m
⎜
s
⎝ 1 ⋅ in ⎠ 60⋅ s
×
1 ⋅ hr
3600⋅ s
= 1.34⋅
m
2
s
Problem 1.21
Problem
1.23
[Difficulty: 1]
1.21
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
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)
5.4⋅ acres = 5.4⋅ acre ×
3
⎛ 0.0254⋅ m × 12⋅ in ⎞ × 1 ⋅ min = 0.0472⋅ m
⎜
s
60⋅ s
1 ⋅ ft ⎠
⎝ 1 ⋅ in
×
4047⋅ m
1 ⋅ acre
1 ⋅ hr
3600⋅ s
m
= 29.1⋅
4
s
2
= 2.19 × 10 ⋅ m
Problem 1.22
Problem
1.24
1.22
Given:
Quantities in SI (or other) units.
Find:
Quantities in BG units.
Solution:
Use Table
G.2. Table
appropriate
(a)
50⋅ m = 50⋅ m ×
(b)
250⋅ cc = 250⋅ cm ×
(c)
100⋅ kW = 100⋅ kW ×
(d)
5⋅
2
2
⎛ 1⋅ in × 1⋅ ft ⎞ = 538⋅ ft 2
⎜
⋅
12⋅ in ⎠
⎝ 0.0254m
2
3
kg
2
m
= 5⋅
kg
2
m
×
3
⎛ 1⋅ m × 1⋅ in × 1⋅ ft ⎞ = 8.83 × 10− 3⋅ ft 3
⎜
12⋅ in ⎠
⋅
⎝ 100⋅ cm 0.0254m
1000⋅ W
1⋅ kW
×
1⋅ hp
746⋅ W
= 134⋅ hp
2
⋅
1⋅ slug
slug
12⋅ in ⎞
⎛ 0.0254m
×
×
= 0.0318⋅
⎜
2
14.95⋅ kg
1⋅ ft ⎠
⎝ 1⋅ in
ft
[Difficulty: 1]
Problem
1.26
Problem 1.23
[Difficulty: 2]
1.23
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
0.454 ⋅ kg
Vprop = 17⋅ lbm ×
×
×
×
998 ⋅ kg
0.495
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 1.24
Problem
1.28
[Difficulty: 1]
1.24
Given:
Data in given units
Find:
Convert to different units
Solution:
3
(a)
1⋅
3
in
= 1⋅
min
3
(b)
(c)
(d)
1⋅
1⋅
m
s
3
3
⋅
1⋅ min
mm
1000⋅ mm⎞
⎛ 0.0254m
×
×
= 273⋅
⎜
min ⎝ 1⋅ in
s
60⋅ s
1⋅ m ⎠
in
×
3
= 1⋅
liter
min
m
s
= 1⋅
3
×
4 × 0.000946⋅ m
liter
min
1 ⋅ SCFM = 1 ⋅
1⋅ gal
×
ft
×
1 ⋅ gal
4 × 0.946 ⋅ liter
3
min
3
×
×
60⋅ s
1⋅ min
= 15850⋅ gpm
60⋅ s
1 ⋅ min
= 0.264 ⋅ gpm
3
⎛ 0.0254⋅ m ⎞ × 60⋅ min = 1.70⋅ m
⎜ 1
hr
1 ⋅ hr
⋅ ft
⎜
⎝ 12
⎠
Problem 1.25
Problem
1.30
1.25
Given:
Definition of kgf.
Find:
Conversion from psig to kgf/cm2.
Solution:
Use appropriate
Table G.2.Table
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
⎜
2
100⋅ cm ⎠
0.3048m
⋅
⎝ 1⋅ ft
cm
[Difficulty: 1]
[Difficulty: 2]
Problem 1.26
Problem
1.32
1.26
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 1.27
Problem
1.33
[Difficulty: 2]
1.27
(psia)
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
2
c = 2.38⋅
Hence the constant is actually
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
2
c = 2.38⋅
R ⋅ in ⋅ s
ft
3
1
2
2
= 2.38⋅
R ⋅ in ⋅ s
ft
3
2
×
2
⎛ K ⎞ × ⎛ 1⋅ ft ⎞ × 1⋅ ft
⎜
⎜
0.3048m
⋅
⎝ 1.8⋅ R ⎠
⎝ 12⋅ in ⎠
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 1.28
Problem
1.34
[Difficulty: 1]
1.28
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 1.29
Problem
1.36
[Difficulty: 1]
1.29
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
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 ×
⋅
×
⋅
×
2
32.2 ft
1.93 slug
1⋅ lbf ⋅ s
Vw = 1.24ft
3
Problem 1.30
Problem
1.38
[Difficulty: 1]
1.30
Given:
Specific speed in customary units
Find:
Units; Specific speed in SI units
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 ⎞
3
2
⎜ 12
rpm⋅ gpm
2 ⋅ π⋅ rad 1 ⋅ min ⎛ 4 × 0.000946⋅ m 1 ⋅ min ⎞
NScu = 2000 ×
×
×
×⎜
⋅
×⎜
3
1 ⋅ rev
60⋅ s
1 ⋅ gal
60⋅ s ⎠
⎝
⎝ 0.0254⋅ m ⎠
2
ft
4
1
⎛ m3 ⎞
⋅⎜
s ⎝ s ⎠
NScu = 4.06⋅
3
rad
m
4
2
4
Problem 1.31
Problem
1.40
[Difficulty: 2]
1.31
Given:
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=
p
lbf
lb ⋅o R
1
in 2
= 14.7 2 ×
×
×
144
RT
in 53.3 ft ⋅ lbf 519o R
ft 2
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
1
RT
= 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
uT =
= ± 0.0963%
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 1.32
Problem
1.42
[Difficulty: 2]
1.32
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
V 3 πR
4π (D 2) π D 3
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 4 ⎟ = −3
= −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 1.33
Problem
1.43
[Difficulty: 2]
1.33
Given:
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=
4 m
m
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
=
=
= 1;
um =
= ±0.252%
ρ ∂m ρ π D 2 H ρ πD 2 H
397
D ∂ρ D
±1
4m
1 4m
= ( −2)
= ( −2 )
= −2; u D =
= ±137%
.
3
2
ρ ∂D ρ
ρ
73
πD H
πD H
H ∂ρ H
±1
4m
1 4m
= ( −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
m
kg
×
= 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 1.34
Problem
1.44
[Difficulty: 2]
1.34
Given:
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 m& = ± ⎢⎜
u ∆m ⎟ + ⎜
u ∆t ⎟ ⎥
⎠ ⎝ 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 1.35
Problem
1.45
[Difficulty: 3]
1.35
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 m& = ± ⎢⎜
u ∆m ⎟ + ⎜
u ∆t ⎟ ⎥
⎠ ⎝ m& ∂∆t
⎠ ⎥⎦
⎣⎢⎝ m& ∂∆m
The uncertainties are ± half the least counts of the measuring instruments: δ∆m = ±0.5 g
∆m ∂m&
1
= ∆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 1.36
Problem
1.46
[Difficulty: 2]
1.36
Given:
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
=
3
m
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 = ±
D ∂D D ⎛
6 m⎞
⎛ 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)
SG = 126
. ± 0.0289 (20 to 1)
]
1
2 2
Problem
1.48
Problem 1.37
[Difficulty: 3]
1.37
Given:
Data on water
Find:
Viscosity; Uncertainty in viscosity
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 1.38
Problem
1.50
[Difficulty: 3]
1.38
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
hr 41.1 ft
mi 3600 s
= ±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 = ±
[
]
Problem 1.39
1.39
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 1.40
Problem
1.52
1.40
[Difficulty: 4]
1.32
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 2.1
Problem
2.1
[Difficulty: 1]
2.1
Given:
Velocity fields
Find:
Whether flows are 1, 2 or 3D, steady or unsteady.
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
Problem
2.2
[Difficulty: 1]
2.2
Given:
Velocity fields
Find:
Whether flows are 1, 2 or 3D, steady or unsteady.
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
Problem
2.3
2.3
Given:
Viscous liquid sheared between parallel disks.
Upper disk rotates, lower fixed.
Velocity field is:
r
rω z
V = eˆθ
h
Find:
a.
Dimensions of velocity field.
b.
Satisfy physical boundary conditions.
r
r
To find dimensions, compare to V = V ( x, y , z ) form.
Solution:
r
r
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.
r
At lower disk, V = 0 since stationary.
r
z = 0, so V = eˆθ
2.
rω 0
= 0 , so satisfied.
h
r
At upper disk, V = eˆθ rω since it rotates as a solid body.
r
z = h, so V = eˆθ
rω h
= eˆθ rω , so satisfied.
h
[Difficulty: 2]
Problem 2.4
Problem
2.4
[Difficulty: 1]
2.4
Given:
Velocity field
Find:
Equation for streamlines
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)
2.5 A fluid flow has the following velocity components: 𝑢 = 1 𝑚⁄𝑠 and 𝑣 = 2𝑥 𝑚⁄𝑠. Find an equation
for and sketch the streamlines of this flow.
Given: The velocity components: 𝑢 = 1 𝑚⁄𝑠 and 𝑣 = 2𝑥 𝑚⁄𝑠.
Find: The equation for streamlines and sketch the streamlines.
Assumption: The flow is steady and incompressible
Solution: Use the definition of streamlines in terms of velocity to determine the equation for the
streamlines. By definition, we have:
𝑑𝑑 𝑑𝑑
=
𝑣
𝑢
Or
𝑑𝑑 =
𝑢
𝑑𝑑
𝑣
𝑑𝑑 =
1
𝑑𝑑
2𝑥
Substituting in the velocity components in we obtain:
Integrating both sides, we get:
2𝑥𝑥𝑥 = 𝑑𝑑
𝑥2 = 𝑦 + 𝑐
Where 𝑐 is a constant that can be found for each specific problem.
To plot the streamlines, we write: 𝑦 = 𝑥 2 − 𝑐. The plot is shown in the figure.
Problem 2.6
(Difficulty: 2)
2.6 When an incompressible, non-viscous fluid flows against a plate (two-dimensional) flow, an exact
solution for the equations of motion for this flow is 𝑢 = 𝐴𝐴, 𝑣 = −𝐴𝐴, with 𝐴 > 0. The coordinate
origin is located at the stagnation point 0, where the flow divides and the local velocity is zero. Find the
streamlines.
Given: The velocity components: 𝑢 = 𝐴𝐴, 𝑣 = −𝐴𝐴
Find: The equation for streamlines and sketch the streamlines.
Assumption: The flow is steady and incompressible
Solution: Use the definition of streamlines in terms of velocity to determine the equation for the
streamlines. By definition, we have:
𝑑𝑑 𝑑𝑑
=
𝑣
𝑢
Substituting in for the velocity components we obtain:
𝑑𝑑
𝑑𝑑
=
𝐴𝐴 −𝐴𝐴
𝑑𝑑 𝑑𝑑
=
−𝑦
𝑥
Integrating both sides, we get:
�
𝑑𝑑
𝑑𝑑
= −�
𝑥
𝑦
ln 𝑥 = − ln 𝑦 + 𝑐
where 𝑐 is a constant that can be found for each problem
Using the relation for logarithms, the streamline equation is:
Or we can rewrite as:
ln 𝑥 𝑦 = +𝑐
𝑥𝑥 = 𝑐1
Where 𝑐1 is a constant.
The plot of the streamlines is shown in the figure as an example:
Problem 2.7
(Difficulty: 2)
5
𝑟
2.7 For the free vortex flow the velocities are 𝑉𝑡 = and 𝑉𝑟 = 0. Assume that lengths are in feet or
meters and times are in seconds. Plot the streamlines of this flow. How does the velocity vary with
distance from the origin? What is the velocity at the origin (0,0)?
5
𝑟
Given: The velocity components: 𝑉𝑡 = , 𝑉𝑟 = 0
Find: The streamline, how the velocity varies with distance from the origin, and the velocity at the origin
(0,0).
Assumption: The flow is steady and incompressible
Solution: Use the definition of streamlines in terms of velocity to determine the equation for the
streamlines. By definition, in radial coordinates we have:
1 𝑑𝑑 𝑉𝑟
=
𝑟 𝑑𝑑 𝑉𝑡
Substituting the velocity components we obtain:
Integrating both sides, we get:
𝑑𝑑
=0
𝑑𝑑
𝑟=𝑐
So the streamline can be plotted as:
𝑐 is a constant
The velocity will decrease as the distance to the origin 𝑟 increases as shown in the figure.
vt
The velocity at the origin (0,0) is
r
𝑉𝑟 = 0
𝑉𝑡 = ∞
Problem 2.8
(Difficulty: 2)
2.8 For the forced vortex flow the velocities are 𝑉𝑡 = 𝜔𝜔 and 𝑉𝑟 = 0. Plot the streamlines of this flow.
How does the velocity vary with distance from the origin? What is the velocity at the origin(0,0)?
Given: The velocity components: 𝑉𝑡 = 𝜔𝜔, 𝑉𝑟 = 0
Assumption: The flow is steady and incompressible
Find: The equation for the streamlines, how the velocity varies with distance from origin, the velocity at
origin (0,0).
Solution: Use the definition of streamlines in terms of velocity to determine the equation for the
streamlines. By definition, in radial coordinates we have:
1 𝑑𝑑 𝑉𝑟
=
𝑟 𝑑𝑑 𝑉𝑡
Substituting the velocity components in we obtain:
Integrating both sides, we get:
𝑑𝑑
=0
𝑑𝑑
𝑟=𝑐
So the streamline can be plotted as:
𝑐 is a constant
The velocity will increase as the distance to the origin 𝑟 increases. For example ω = 1.
𝑉𝑡
The velocity at the origin (0,0) is
r
𝑉𝑟 = 0
𝑉𝑡 = 0
Problem 2.9
Problem
2.6
[Difficulty: 1]
2.9
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
m⋅ s
1
2
v = b ⋅ y = −6 ⋅
v
For streamlines
=
u
dy
So, separating variables
y
=
dy
=
dx
× 2⋅ m ×
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
Problem 2.10
2.10
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
Problem 2.11
Problem
2.10
[Difficulty: 2]
2.11
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
−
( 1 ⋅ m) − c
2⋅ A
2
=
2
4⋅ m − 1⋅ m
2
2 × 2⋅
m
s
∆t = 0.75⋅ s
[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
4 ⋅ π⋅ x
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=
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
2⋅ π⋅ x + y
=
)
K⋅ y
−
(2
2 ⋅ π⋅ x + y
So, separating variables
Integrating
)
x
y
2
y ⋅ dy = −x ⋅ dx
y
2
2
The solution is
=−
2
x
=−
2
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: 2)
�⃗ = 𝐴𝐴𝚤⃗ − 𝐴𝐴𝚥⃗, where 𝐴 = 2 𝑠 −1 , which can be interpreted to represent flow
2.13 For the velocity field 𝑉
in a corner, show that the parametric equations for particle motion are given by 𝑥𝑝 = 𝑐1 𝑒 𝐴𝐴 and
𝑦𝑝 = 𝑐2 𝑒 −𝐴𝐴 . Obtain the equation for the pathline of the particle located at the point (𝑥, 𝑦) = (2,2) at
the instant 𝑡 = 0. Compare this pathline with streamline through the same point.
Find: The pathlines and streamlines .
Assumption: The flow is steady and incompressible
Solution: Use the definitions of pathlines and streamlines in terms of velocity. We relate the velocities
to the change in position with time. For the particle motion we have:
𝑑𝑥
= 𝑢 = 𝐴𝐴
𝑑𝑑
Or
𝑑𝑦
= 𝑣 = −𝐴𝐴
𝑑𝑑
𝑑𝑑
= 𝐴 𝑑𝑑
𝑥
Integrating both sides of the equation, we get:
𝑑𝑑
= −𝐴 𝑑𝑑
𝑦
ln 𝑥 = 𝐴𝐴 + 𝑐
ln 𝑦 = −𝐴𝐴 + 𝑐
So the parametric equations for particle motion are given by:
𝑥𝑝 = 𝑒 (𝐴𝐴+𝑐) = 𝑐1 𝑒 𝐴𝐴
With 𝐴 = 2 𝑠 −1 :
𝑦𝑝 = 𝑒 (−𝐴𝐴+𝑐) = 𝑐2 𝑒 −𝐴𝐴
𝑥𝑝 = 𝑒 (𝐴𝐴+𝑐) = 𝑐1 𝑒 2𝑡
𝑦𝑝 = 𝑒 (−𝐴𝐴+𝑐) = 𝑐2 𝑒 −2𝑡
For the pathline:
At 𝑡 = 0, 𝑥𝑝 = 𝑥0 = 2, 𝑦𝑝 = 𝑦0 = 2. So the equation for the pathline is
For the streamline:
𝑥𝑝 𝑦𝑝 = 𝑥0 𝑦0 = 4
𝑑𝑑 𝑣 −𝐴𝐴 −𝑦
= =
=
𝐴𝐴
𝑥
𝑑𝑑 𝑢
Integrating both sides of the equation we get:
𝑑𝑑 −𝑑𝑑
=
𝑥
𝑦
ln 𝑦 = − ln 𝑥 + 𝑐
𝑥𝑥 = 𝑐
For points (𝑥, 𝑦) = (2,2), the constant c = 4 and the equation for the streamline is:
𝑥𝑥 = 4
Comparing the pathline and streamline, it is seen that for steady flow the pathline and streamline coincide
as expected.
Problem 2.14
(Difficulty: 2)
2.14 A velocity field in polar coordinates is given with the radial velocity as 𝑉𝑟 = −
𝐴
𝑟
𝐴
𝑟
and the tangential
velocity as 𝑉𝜃 = , where 𝑟 is in 𝑚 and 𝐴 = 10 𝑚2 . Plot the streamlines passing through the 𝜃 = 0 and
𝑟 = 1 𝑚, 2𝑚, 𝑎𝑎𝑎 3𝑚. What does the flow field model?
Assumption: The flow is steady and incompressible
Solution: Use the definition of streamlines in terms of velocity. The definition of a streamline in radial
coordinates is:
1 𝑑𝑑 𝑑𝑑
=
𝑉𝜃
𝑟 𝑉𝑟
With the velocity components
Or
𝑑𝑑
1 𝑑𝑑
�
�=
𝐴
𝑟 −𝐴
𝑟
𝑟
Integrating both sides:
For the case of 𝜃 = 0 and 𝑟 = 1 𝑚, we have:
So the streamline is:
−
𝑑𝑑
= 𝑑𝑑
𝑟
− ln 𝑟 = 𝜃 + 𝑐
𝑐=0
ln 𝑟 = −𝜃
𝑟 = 𝑒𝑒𝑒(−𝜃)
The plot of the streamline is
90
1
120
60
0.8
0.6
150
30
0.4
0.2
180
0
330
210
300
240
270
For the case 𝜃 = 0 and 𝑟 = 2 𝑚, we have:
𝑐 = − ln 2
So the streamline is:
ln 𝑟 = −𝜃 − 𝑐 = −𝜃 + ln 2
ln 𝑟 − ln 2 = −𝜃
𝑟
ln = −𝜃
2
𝑟 = 2𝑒𝑒𝑒(−𝜃)
90
2
120
60
1.5
1
150
30
0.5
180
0
210
330
240
300
270
For the case 𝜃 = 0 and 𝑟 = 3 𝑚, we have:
So the streamline is:
𝑐 = − ln 3
ln 𝑟 = −𝜃 − 𝑐 = −𝜃 + ln 3
ln 𝑟 − ln 3 = −𝜃
𝑟
ln = −𝜃
3
𝑟 = 3𝑒𝑒𝑒(−𝜃)
The flow field models the circular flow from the center at the origin.
90
3
120
60
2
150
30
1
180
0
330
210
300
240
270
Problem 2.15
(Difficulty: 2)
2.15 The flow of air near the earth’s surface is affected both by the wind and thermal currents. In certain
�⃗ = 𝑎𝚤⃗ + 𝑏 �1 − 𝑦� 𝚥⃗ for 𝑦 < ℎ and by 𝑉
�⃗ = 𝑎𝚤⃗
circumstances the velocity field can be represented by 𝑉
for 𝑦 > ℎ. Plot the streamlines for the flow for
𝑏
𝑎
= 0.01, 0.1 𝑎𝑎𝑎 1.
ℎ
Assumption: The flow is steady and incompressible
Solution: Use the definition of streamlines in terms of velocity to determine the equation for the
streamlines. By definition, we have:
𝑑𝑦 𝑣
=
𝑑𝑑 𝑢
Or, substituting the equations for the velocities:
𝑑𝑑 𝑑𝑑
=
𝑣
𝑢
𝑑𝑑
𝑑𝑑
=
𝑦
𝑎
𝑏 �1 − �
ℎ
And when y < h
𝑥=
And when 𝑦 > ℎ
𝑑𝑑
=0
𝑑𝑑
Integrating both sides of the equation:
when 𝑦 < ℎ.
when 𝑦 > ℎ.
𝑑𝑑
𝑦
𝑏
�1 − �
𝑎
ℎ
𝑥=−
For the critical point 𝑦 = ℎ, we have 𝑐2 = ℎ
𝑎ℎ
ln(𝑏ℎ − 𝑏𝑏) + 𝑐1
𝑏
𝑦 = 𝑐2
For example, the streamline passing through (0,0)：
𝑥=−
when 𝑦 < ℎ.
𝑐1 =
𝑎ℎ
ln(𝑏ℎ)
𝑏
𝑎ℎ
𝑎ℎ
ln(𝑏ℎ − 𝑏𝑏) +
ln(𝑏ℎ)
𝑏
𝑏
Assume ℎ = 𝑏 = 1,. For the first case 𝑎 = 100, the streamline is shown as:
The streamline is shown:
1
0.9
0.8
0.7
y
0.6
0.5
0.4
0.3
0.2
0.1
0
0
200
400
600
x
800
1000
1200
80
100
120
For the second case 𝑎 = 10, the streamline is shown as:
1
0.9
0.8
0.7
y
0.6
0.5
0.4
0.3
0.2
0.1
0
0
20
40
60
x
For the third case 𝑎 = 10, the streamline is shown as:
1
0.9
0.8
0.7
y
0.6
0.5
0.4
0.3
0.2
0.1
0
0
2
4
6
x
8
10
12
Problem 2.16
2.16
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
Problem
2.18
[Difficulty: 2]
2.17
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 2.18
Problem
2.20
[Difficulty: 3]
2.18
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
x
dx
dt
⎛
B⋅ ⎜t+ A⋅
x=e ⎝
2⎞
t
2⎠
+ C1
2
⎠
B⋅ ⎜t+ A⋅
t
⎛
C1
= e ⋅e ⎝
x = c1 ⋅ e ⎝
⎛
x=e ⎝
v
u
So, separating variables
Integrating
=
2⎞
⎝
B⋅ ⎜t+ A⋅
For streamlines
vp =
s
dy
dx
=
( 1 + A⋅ t) ⋅
y
t
⎛
Using given data
1
dy
⎛
B⋅ ⎜t+ A ⋅
The pathlines are
s
B = 1⋅
= B⋅ ( 1 + A⋅ t) ⋅ dt
ln( x ) = B⋅ ⎜ t + A⋅
Integrating
1
+ C1
2⎠
dt
= C⋅ y
C = 1⋅
1
s
= C⋅ dt
ln( y ) = C⋅ t + C2
⎛
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 2.19
Problem
2.22
[Difficulty: 3]
2.19
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
=
=
dy
dx
x0 = 1 m
a⋅ t
2⋅ t
dy
= b ⋅ y ⋅ ( 1 + c⋅ t )
dt
=
⎞ = b ⋅ ⎛ t + 1 ⋅ c⋅ t2⎞
⎜
⎝ 2
⎠
⎝ y0 ⎠
ln⎛⎜
b = 2
1
2
c = 0.4
s
y
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⎠
ln⎛⎜
vp =
y ( t) = e
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 2.20
Problem
2.23
[Difficulty: 3]
2.20
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
y = y 0 ⋅ ⎛⎜
x
⎞
⎝ 0⎠
x
t = 5
y = y 0 ⋅ ⎛⎜
x
⎞
⎝ 0⎠
t = 10
b
= x
b
x
⋅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
Problem 2.21
2.21
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 2.22
Problem
2.26
[Difficulty: 4]
2.22
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
u0
⎣ ⎝
dy =
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
⎣ ⎣ ⎝
ω
= 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
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 2.23
Problem
2.28
2.23
[Difficulty: 4]
2.18
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 2.24
Problem
2.29
[Difficulty: 4]
2.24
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 2.25
Problem
2.30
[Difficulty: 4]
2.25
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
2
⋅ ⎛t − t0
⎝
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⎞
⎠
)
(
)
y = y0 + b⋅ t − t0
⋅ ⎛t − t0
⎝
2⎞
2
(
y − y0 = b⋅ t − t0
⎠
(
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 2.26
Problem
2.31
[Difficulty: 4]
2.26
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
=
⌠
⌠
⎮
2
⎮ a ⋅ y dy = ⎮ b dx
⌡
⌡
or
2
a⋅ y
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⎞⎤
0
3
(
)
3
4
Hence, with x 0 = -3, y 0 = 0 at t0 = 1
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
)
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)
y = 2⋅ ( t − 1)
y = 4⋅ m
⎠⎥⎦
⎠
Problem 2.27
2.32
Problem
[Difficulty: 3]
2.27
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
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
Pathline starting at t = 1 s
Pathline starting at t = 2 s
Streakline at t = 4 s
-1
-2
-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 2.28
Problem
2.34
[Difficulty: 3]
2.28
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
⎝
⎠
4
2⋅ a
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 2.29
Problem
2.35
[Difficulty: 4]
2.29
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
⎤⎥
0
t
2
x − x 0 = a⋅ y 0 ⋅ ( t − t 0 ) + a⋅ ⋅ ⎢ −
− t0 ⋅ ( t − t0 )⎥
3
2 ⎣3
⎦
The pathlines are
⎡⎢ 3 t 3
⎤⎥
0
t
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
⎤⎥
0
t
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
⎤⎥
0
t
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
⎤⎥
0
t
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
dx
y ⋅ dy =
2
Integrating
=
b
a
y − y0
y =
a⋅ y
⋅ t⋅ dx
2
=
2
The streamlines are then
b⋅ t
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
Pathline Plots
Streamline 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 2.30
Problem
2.36
[Difficulty: 4]
2.30
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
(
b
=
dy
So, separating variables
dy =
b
Integrating
b
y − y0 =
⋅ x − x0
a⋅ t
The streamlines are then
b
5⋅ ( x − 2)
y = y0 +
⋅ x − x0 =
+1
t
a⋅ t
For streamlines
u
dx
a⋅ t
=
)
y = y 0 + b ⋅ t − t0 = 3 m
a⋅ t
⋅ 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 2.31
Problem
2.37
[Difficulty: 2]
2.31
A
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
⋅
1
m⋅ s⋅ K
The kinematic viscosity is
where
ν=
b' =
μ
ρ
=
μ⋅ R⋅ T
=
p
S = 110.4 ⋅ K
R = 286.9 ⋅
1
3
3
2
2
2
R⋅ T b ⋅ T
R⋅ b T
b'⋅ T
⋅
=
⋅
=
S
S
S
p
p
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
2
−9
⋅
m
3
101.3 × 10 ⋅ N
2
s⋅ K
3
Hence
p = 101.3 ⋅ kPa
kg⋅ K
2
R⋅ b
b' = 286.9 ⋅
J
2
−9
⋅
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
−5 m
2
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
s⋅ K
ν =
1+
3
× ( 373.1 ⋅ K)
2
2
−5 m
2
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
−5
2× 10
−5
1.5× 10
0
20
40
60
Temperature (C)
80
100
Problem 2.32
Problem
2.38
[Difficulty: 2]
2.32
Given:
Sutherland equation with SI units
Find:
Corresponding equation in BG units
1
Solution:
Governing equation:
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
2
− 8 lbf ⋅ s
lbf ⋅ s
b = 2.27 × 10
slug⋅ ft
⋅
1
2
2
ft ⋅ R
9⋅ R
S = 198.7 ⋅ R
5⋅ K
with T in Rankine, µ in
T
0.3048⋅ m
2
1
and
×
2
slug
ft⋅ s⋅ R
×
lbf ⋅ s
ft
2
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 lbf ⋅ s
2.27 × 10
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 lbf ⋅ s
2.27 × 10
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!
Problem 2.33
2.33
Data:
Using procedure of Appendix A.3:
T (oC)
0
100
200
300
400
µ(x105)
1.86E-05
2.31E-05
2.72E-05
3.11E-05
3.46E-05
T (K)
273
373
473
573
673
T (K)
273
373
473
573
673
T3/2/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
T
3 2
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 2.34
Problem
2.40
[Difficulty: 2]
2.34
Given:
Velocity distribution between flat plates
Find:
Shear stress on upper plate; Sketch stress distribution
Solution:
Basic equation
du
τyx = μ⋅
dy
τyx = −
At the upper surface
Hence
y=
du
=
dy
d
dy
⎡
u max⋅ ⎢1 −
⎣
2
⎛ 2 ⋅ y ⎞ ⎤⎥ = u ⋅ ⎛ − 4 ⎞ ⋅ 2⋅ y = − 8 ⋅ umax⋅ y
⎜
max ⎜ 2
2
⎝ h ⎠⎦
h
⎝ h ⎠
8 ⋅ μ⋅ u max⋅ y
h
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
⋅
2
(Table A.8)
m
2
⎛ 1 × 1000⋅ mm ⎞
⎜
1⋅ m ⎠
⎝ 0.1⋅ mm
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
− 0.01
1
− 0.02
− 0.03
− 0.04
− 0.05
Shear Stress (Pa)
2
3
4
5
Problem 2.35
(Difficulty: 1)
2.35 What is the ratio between the viscosities of air and water at 10℃ ? What is the ratio between their
kinematic viscosities at this temperature and standard barometric pressure ?
Given: The temperature 10℃ .
Find: Ratio between the viscosities of air and water at 10℃ . Ratio of kinematic viscosities at this
temperature and pressure.
Assumption: The standard barometric pressure is sea level pressure. Air can be treated as an ideal gas
Solution:
At 10℃, for the viscosities:
𝜇𝑎𝑎𝑎 = 0.018 × 10−3 𝑃𝑃
For the densities at STP:
𝑎𝑎𝑎
𝜇𝐻2 𝑜 = 1.4 × 10−3 𝑃𝑃 ∙ 𝑠
𝜇𝑎𝑎𝑎 0.018 × 10−3 𝑃𝑃
=
= 0.013
1.4 × 10−3 𝑃𝑃
𝜇𝐻2 𝑜
𝑘𝑘
𝑚3
𝑘𝑘
𝜌𝑎𝑎𝑎 = 1.225 3
𝑚
1
at 15℃, using the ideal gas relation where ρ ∝ at constant pressure
𝜌𝐻2 𝑜 = 1000
𝜌𝑎𝑎𝑎 = 1.225 ×
The ration of kinematic viscosities at 10℃.
𝑣𝑎𝑎𝑎
𝑣𝐻2 𝑜
𝑇
(15 + 273) 𝑘𝑘
𝑘𝑘
=
1.247
(10 + 273) 𝑚3
𝑚3
0.018 × 10−3 𝑃𝑃
𝑘𝑘
1.247 3
𝑚
=
= 10.3
1.4 × 10−3 𝑃𝑃
𝑘𝑘
1000 3
𝑚
The dynamic viscosity of air is much less than that of water but the kinematic viscosity is greater.
Problem 2.36
(Difficulty: 2)
2.36 Calculate the velocity gradients and shear stress for 𝑦 = 0, 0.2, 0.4 and 0.6 𝑚, if the velocity profile
is a quarter-circle center having its center 0.6 𝑚 from the boundary. The fluid viscosity is 7.5 × 10−4
Given: The fluid viscosity 𝜇 = 7.5 × 10−4
Find: The velocity gradient
Solution:
𝑑𝑑
𝑑𝑑
𝑁𝑁
.
𝑚2
The equation for a quarter-circle with y measured up from the surface of the plate is:
Or, expanding the expression:
The velocity gradient is:
�
2
𝑢 2
𝑦
� +�
− 1� = 1
10
0.6
𝑢2 = 278(1.2𝑦 − 𝑦 2 )
2u
And the shear stress is
𝜏=𝜇
𝑑𝑑
= 278(1.2 − 2𝑦)
𝑑𝑑
1.2 − 2𝑦
𝑑𝑑
= 139 �
�
𝑢
𝑑𝑑
𝑑𝑑
𝑁𝑁
1.2 − 2𝑦
1.2 − 2𝑦
= 7.5 × 10−4 2 ∙ 139 �
� = 0.104 �
�
𝑑𝑑
𝑚
𝑢
𝑢
When 𝑦 = 0, from the equation for the velocity we have
𝑢=0
𝑚
𝑠
𝑁𝑁
.
𝑚2
And for the gradient we have
And
When 𝑦 = 0.2
1
𝑑𝑑
=∞
𝑠
𝑑𝑑
𝜏=∞
𝑁
𝑚2
𝑚
𝑠
𝑢 = 7.46
1
𝑑𝑑
= 14.9
𝑠
𝑑𝑑
When 𝑦 = 0.4
𝜏 = 0.0111
𝑁
𝑚2
𝑚
𝑠
𝑢 = 9.43
1
𝑑𝑑
= 5.90
𝑠
𝑑𝑑
When 𝑦 = 0.6
𝜏 = 0.0044
𝑢 = 10
𝑁
𝑚2
𝑚
𝑠
𝑑𝑑
1
=0
𝑑𝑑
𝑠
𝜏=0
𝑁
𝑚2
Problem 2.37
(Difficulty: 2)
2.37 A very large thin plate is centered in a gap of width 0.06 𝑚 with different oils of unknown
viscosities above and below; one viscosity is twice the other. When the plate is pulled at a velocity of
0.3
𝑚
,
𝑠
the resulting force on one square meter of plate due to the viscous shear on both sides is 29 𝑁..
Assuming viscous flow and neglecting all end effects, calculate the viscosities of the oils.
Given: Viscosity: 𝜇2 = 2𝜇1 . Width of gap: ℎ = 0.06 𝑚. Velocity:𝑉 = 0.3
𝐹 = 29
𝑁
𝑚2
𝑚
.
𝑠
Force per square meter:
Find: 𝜇1 and 𝜇2
Assumption: Viscous flow with linear velocity profiles, negligible end effects.
Solution: Use Newton’s law relating shear stress to viscosity and velocity gradient. The relation
between the two viscosities is
𝜇2 = 2𝜇1
Because the gaps are equal and the plate velocity is the same for both fluids, the velocity gradient is the
same for both sides of the plate:
𝑚
0.3 .
𝑉
1
𝑑𝑑
𝑠
=
=
= 10
𝑠
𝑑𝑑 0.5ℎ 0.5 × 0.06 𝑚
For a Newtonian fluid with a linear velocity profile, we have
𝜏=𝜇
𝑑𝑑
∆𝑉
=𝜇
𝑑𝑑
∆𝑦
The force on the plate due to the top layer of fluid is
𝑚
0.3
∆𝑉
𝑠 = 𝜇 101
𝜏1 = 𝜇1
= 𝜇1
1
∆𝑦
0.03 𝑚
Similarly, the force on bottom of the plate is
𝑚
0.3
∆𝑉
𝑠 = 𝜇 10 1
= 𝜇2
𝜏2 = 𝜇2
2
𝑠
∆𝑦
0.03 𝑚
The total force per unit area equals the sum of the two shear stresses, where for the 1 m2 plate the shear
stress is equal to 29 N/m2.
𝑁
𝐹
= 𝜏1 + 𝜏2 = 29 2
𝑚
𝐴
Or, since µ2 = times µ1
𝜇1 =
1
1
𝑁
𝜇1 101 + 𝜇2 10 = 3𝜇1 10 = 29 2
𝑠
𝑠
𝑚
1
𝑁
𝑠
𝑁∙𝑠
× 29 2 ×
= 0.967 2 = 0.967 𝑃𝑃 ∙ 𝑠
3
𝑚
10
𝑚
𝜇2 = 2𝜇1 = 1.934 𝑃𝑃 ∙ 𝑠
Problem 2.38
Problem
2.44
[Difficulty: 2]
2.38
Given:
Ice skater and skate geometry
Find:
Deceleration of skater
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⋅ ×
×
2
dy
h
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 2.39
Problem
2.46
[Difficulty: 2]
2.39
Given:
Block moving on incline on oil layer
Find:
Speed of block when free, pulled, and pushed
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 2.40
Problem
2.48
[Difficulty: 2]
2.40
Given:
Flow data on apparatus
Find:
The terminal velocity of mass m
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
⎜
z (
piston⋅ L)
⎝ 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 2.41
(Difficulty: 2)
2.41 A vertical gap 25 mm wide of infinite extent contains oil of specific gravity 0.95 and viscosity
2.4 Pa ∙ s. A metal plate 1.5 m × 1.5 m × 1.6 mm weighting 45 N is to be lifted through the gap at a
constant speed of 0.06 𝑚⁄𝑠.Estimate the force required.
Given: Plate size:1.5 m × 1.5 m × 1.6 mm .Width of gap: 25 𝑚𝑚. Specific gravity:0.95.
Viscosity: 2.4 Pa ∙ s. Weight: 45 N. Speed: 0.06 𝑚⁄𝑠.
Find: The force required F 𝑇 .
Assumption: Viscous flow. Neglecting all end effects. Linear velocity profile in the gap.
Solution: Make a force balance on the plate. Use Newton’s law of viscosity to relate the viscous force
on the plate to the viscosity and velocity.
We need to calculate all the individual forces. There are the force due to gravity (weight), the buoyancy
force, and the drag force.
Buoyancy force:
𝐹𝐵 = 𝜌𝜌𝜌 = 𝑆𝑆 ∙ 𝜌𝐻2 𝑜 𝑔𝑔 = 0.95 × 998.2 × 9.81 × 1.5 × 1.5 × 0.0016 = 33.5 𝑁
Drag force: The viscous shear stress is given by
𝜏=𝜇
𝑑𝑑
∆𝑢
= 𝜇
𝑑𝑑
∆𝑥
The force on both sides of the plate is
𝑚
0.06
∆𝑢
𝑠 × (1.5 𝑚)2 = 55.4 𝑁
𝐴 = 2 × 2.4 𝑃𝑃 𝑠 ×
𝐹𝜏 = 2 𝜇
∆𝑥
0.0117 𝑚
The force required to maintain motion is 𝐹𝑇 , by the force balance equation we have:
The total force is then
𝐹𝑇 + 𝐹𝐵 = 𝑊 + 𝐹𝜏
𝐹𝑇 = 𝑊 + 𝐹𝜏 − 𝐹𝐵 = 45 𝑁 + 55.4 𝑁 − 33.5 𝑁 = 66.9 𝑁
Problem 2.42
(Difficulty: 2)
2.42 A cylinder 8 in in diameter and 3 ft long is concentric with a pipe of 8.25 in. Between cylinder and
pipe there is an oil film. What force is required to move the cylinder along the pipe at a constant velocity
of 3 fps? The kinematic viscosity of the oil is 0.006
𝑓𝑓 2
.
𝑠
The specific gravity is 0.92.
Given: Cylinder diameter: 𝐷𝑐 = 8 𝑖𝑖. Cylinder length: L = 3 𝑓𝑓. Pipe diameter: 𝐷𝑝 = 8.25 𝑖𝑖. Cylinder
velocity:V = 3
𝑓𝑓
.
𝑠
Oil viscosity: 𝑣 = 0.006
Find:The force required F 𝑇 .
𝑓𝑓 2
.
𝑠
Specific gravity:𝑆𝑆 = 0.92.
Assumption: Viscous flow with linear velocity profile in oil, negligible end effects.
Solution: Use Newton’s law of viscosity to evaluate the viscous force.
The gap ℎ between the cylinder and pipe is:
ℎ=
8.25 − 8
𝑖𝑖 = 0.125 𝑖𝑖 = 0.0104 𝑓𝑓
2
The contact area 𝐴 between cylinder and oil is:
1 𝑓𝑓 = 12 𝑖𝑖
The dynamic viscosity:
𝐴 = 𝜋𝐷𝑐 𝐿 = 𝜋 ×
𝜇 = 𝑣𝑣 = 𝑣 ∙ 𝑆𝑆 ∙ 𝜌𝐻2 𝑜 = 0.006
8
× 3 𝑓𝑓 2 = 6.28 𝑓𝑓 2
12
𝑓𝑓 2
𝑙𝑙𝑙
𝑙𝑙𝑙
× 0.92 × 62.4 3 = 0.344
𝑓𝑓
𝑓𝑓 ∙ 𝑠
𝑠
The drag force is, assuming a linear velocity profile in the fluid
𝑓𝑓
3
𝑑𝑑
𝑉
𝑙𝑙𝑙
𝑙𝑙𝑙 ∙ 𝑓𝑓
𝑠
2
F𝐷 = 𝜇𝜇
= 𝜇𝜇 = 0.344
× 6.28 𝑓𝑓 ×
= 605
= 19 𝑙𝑙𝑙
𝑑𝑑
ℎ
𝑓𝑓 ∙ 𝑠
𝑠2
0.0104 𝑓𝑓
Problem 2.43
(Difficulty: 2)
2.43 Crude oil at 20℃ fills the space between two concentric cylinders 250 𝑚𝑚 high and with diameters
of 150 𝑚𝑚 and 156 𝑚𝑚 . What torque is required to rotate the inner cylinder at 12 𝑟𝑟𝑟, the outer
cylinder remaining stationary?
Given: Temperature: T = 20 ℃. Cylinder height: H = 250 𝑚𝑚. Outer cylinder diameter: 𝐷𝑜 = 156 𝑚𝑚.
Inner cylinder diameter: 𝐷𝐼 = 150 𝑚𝑚. Rotating speed: 12 𝑟𝑟𝑟.
Find: The required torque Γ.
Assumption: Linear velocity profile in fluid, viscous flow, neglect all end effects.
Solution: Use Newton’s law of viscosity to find the force on the surfaces
The torque equals force times radius:
𝑇 = 𝐹𝐷 𝑅𝐼
The velocity of inner cylinder is:
𝑉 = 𝜔𝑅𝐼 = 12
𝑟
1 𝑚𝑚𝑚
1
150
𝑚
×
× �2𝜋 � ×
𝑚 = 0.0942
𝑚𝑚𝑚
60 𝑠
𝑟
2 × 1000
𝑠
The dynamic viscosity of crude oil at Temperature = 20 ℃.:
𝜇 = 0.00718 𝑃𝑃 ∙ 𝑠
Newton’s law of viscosity with a linear velocity profile is
The drag force is:
𝜏=𝜇
F𝐷 = 𝜇
𝑑𝑑
∆𝑢
= 𝜇
𝑑𝑑
∆𝑟
𝑉
𝐴
(𝐷
0.5 × 𝑜 − 𝐷𝐼 )
𝑚
𝑠
F𝐷 = 0.00718 𝑃𝑃 ∙ 𝑠 ×
× (𝜋 × 0.15 𝑚 × 0.25 𝑚) = 0.0266 𝑁
0.5 × (0.006 𝑚)
0.0942
The torque is then
𝑇 = 𝐹𝐷 𝑅𝐼 = 0.0266 𝑁 × 0.075 𝑚 = 0.002 𝑁 ∙ 𝑚
Problem 2.44
Problem
2.49
2.44
[Difficulty: 3]
2.40
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 2.40
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⋅ D
⎥ ( piston⋅ L)
⋅
D
−
D
⎢ ( tube
piston)⎥
⎣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⎜
⎟
g
a
⎜ 1.01⋅ Vt −
a⎠
⎝
1
or
t = 1.93 s
3
Problem 2.45
Problem 2.50
[Difficulty: 3]
2.45
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 2.46
Problem
2.51
[Difficulty: 4]
2.46
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) ×
2
m
0.4⋅ N⋅ s⋅ ( 0.1⋅ m)
2
×
N⋅ s
kg⋅ m
⎛ 0.4⋅ 0.01 ⋅ 0.1⎞⎤
⎡
−⎜
⎢
5⋅ 0.0002
⎠⎥
× ⎣1 − e ⎝
⎦
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
V( t = 0.1⋅ s) =
M ⋅ g ⋅ d ⋅ sin( θ)
μ⋅ A
− μ⋅ A
⎡
⋅ ( t= 0.1⋅ s )⎤
⎢
⎥
M⋅ d
⋅ ⎣1 − e
⎦
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 2.47
(Difficulty: 1)
2.47 A torque of 4 𝑁 ∙ 𝑚 is required to rotate the intermediate cylinder at 30
of the oil. All the cylinders are 450 𝑚𝑚 long. Neglect end effects.
Given: Cylinder height: H = 450 𝑚𝑚. Rotation speed: 30
𝑟
.
𝑚𝑚𝑚
𝑟
.
𝑚𝑚𝑚
Gap between cylinder: ℎ = 0.003 𝑚.
Radius of intermediate cylinder: 𝑅 = 0.15 𝑚. Torque: T = 4 𝑁 ∙ 𝑚.
Find:The oil viscosity 𝜇.
Assumption: Linear velocity profile in fluid, viscous flow, negligible end effects.
Solution: Use Newtons’s viscosity law to evaluate the force on the cylinder
Newton’s law of viscosity for a linear velocity profile is
The velocity of intermediate cylinder:
𝑉 = 𝜔𝜔 = 30
𝜏=𝜇
𝑑𝑑
∆𝑢
= 𝜇
𝑑𝑑
∆𝑟
𝑟
1 𝑚𝑚𝑚
1
𝑚
×
× 2𝜋 × 𝑅 𝑚 = 0.471
𝑚𝑚𝑚
60 𝑠
𝑟
𝑠
The drag force on both sides of the cylinder is:
The torque is given by:
The area is:
𝐹𝐷 = 2𝜇𝜇
𝑉
ℎ
𝑇 = 𝐹𝐷 ∙ 𝑅 = 2𝜇𝜇
A = 2πRH
Calculate the viscosity
𝑉
∙𝑅
ℎ
The viscosity is then:
𝜇=
𝑇ℎ
4 𝑁 ∙ 𝑚 × 0.003 𝑚
𝑁∙𝑆
=
= 0.2
= 0.2 𝑃𝑃 ∙ 𝑠
𝑚
2𝐴𝐴𝐴 2 × 2𝜋 × 0.15 𝑚 × 0.45 𝑚 × 0.471 × 0.15 𝑚
𝑚2
𝑠
Problem 2.48
(Difficulty: 2)
2.48 A circular disk of diameter d is slowly rotated in a liquid of large viscosity µ at a small distance h
from a fixed surface. Derive an expression for the torque T necessary to maintain an angular velocity ω.
Neglect centrifugal effects.
Given: Disk diameter: d. Distance to fixed surface: ℎ. Viscosity: 𝜇. Angular velocity: ω. Radius of
Find: Torque: T.
Assumption: Linear velocity profile in gap between the two disks, viscous flow, negligible end effects,
negligible centrifugal force effects.
Solution: Use Newton’s law of viscosity with a linear velocity profile to find the forces
𝜏=𝜇
∆𝑢
𝑑𝑑
= 𝜇
∆𝑦
𝑑𝑑
The velocity at the interface of the fluid and disk varies with radius
𝑉=𝜔𝑟
The expression for shear stress is then
𝜏= 𝜇
𝜔𝑟
ℎ
The incremental torque the product of the radius and the force per unit area and the area from the center to
the outer radius. The total torque is the integral from the centerline to the outer radius.
𝑑/2
𝑇=�
0
𝑟 �𝜇
𝜔𝑟
� 2𝜋𝜋 𝑑𝑑
ℎ
Or
𝑇=
𝑑/2
𝜇𝜔
2𝜋 � 𝑟 3 𝑑𝑑
ℎ
0
𝑇=
𝜋 𝜇 𝜔 𝑑4
32
ℎ
Problem 2.49
(Difficulty: 2)
2.49 The fluid drive shown transmits a torque T for steady-state conditions (𝜔1 and 𝜔2 constant). Derive
an expression for the slip (𝜔1 − 𝜔2 ) in terms of 𝑇, 𝜇, 𝑑 𝑎𝑎𝑎 ℎ. For values 𝑑 = 6 𝑖𝑖, ℎ = 0.2 𝑖𝑖., SAE 30
oil at 75 𝐹, a shaft rotation of 120 𝑟𝑟𝑟, and a torque of 0.003 ft-lbf, determine the slip.
Given: 𝑑 = 6 𝑖𝑖, ℎ = 0.2 𝑖𝑖 , rotation: 120 𝑟𝑟𝑟. SAE 30 oil at 75 𝐹
Find: The slip (𝜔1 − 𝜔2 ).
Assume: Linear velocity profile in the viscous fluid
Solution: Use Newton’s law of viscosity to relate the viscous forces to the torque and slip
From the force balance equation we have:
𝑑
2
𝑑
2
𝑇 = � 𝜏 ∙ 𝑑𝑑 ∙ 𝑟 = � 𝜏 ∙ 2𝜋𝜋𝜋𝜋 ∙ 𝑟
0
0
Assuming a linear velocity profile in the space between the two disks, Newton’s law of viscosity is
𝑑𝑑
∆𝑢
= 𝜇
𝑑𝑑
∆𝑥
The velocity difference varies with radius
𝜏=𝜇
So the shear stress is
∆𝑢 = (𝜔1 − 𝜔2 )𝑟
The torque is then:
𝜏=𝜇
𝑑
2
𝑇 = � 2𝜋𝜋
0
(𝜔1 − 𝜔2 )𝑟
ℎ
(𝜔1 − 𝜔2 ) 3
(𝜔1 − 𝜔2 ) 4
𝑟 𝑑𝑑 = 𝜋𝜋
𝑑
ℎ
32ℎ
Solving for the slip:
The viscosity for SAE 30 oil at 75 𝐹 is:
The slip is then:
𝜇 = 0.008
(𝜔1 − 𝜔2 ) =
(𝜔1 − 𝜔2 ) =
32 𝑇ℎ
𝜋 𝜇 𝑑4
𝑙𝑙𝑙 ∙ 𝑠2 1
𝑙𝑙𝑙 ∙ 𝑠
𝑠𝑠𝑠𝑠
= 0.008
= 0.008
𝑓𝑓 2
𝑓𝑓 ∙ 𝑠
𝑓𝑓 𝑓𝑓 ∙ 𝑠
32 × 0.003 𝑓𝑓 − 𝑙𝑙𝑙 × 0.0167 𝑓𝑓
𝑟𝑟𝑟
= 1.019
= 9.73 𝑟𝑟𝑟
𝑙𝑙𝑙 ∙ 𝑠
𝑠
4
(0.5
𝜋 × 0.008
×
𝑓𝑓)
𝑓𝑓 2
Problem 2.50
Problem
2.52
[Difficulty: 3]
2.50
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
U0
M⋅ h
⎝ ⎠
ln⎛⎜
dU
Fv = M ⋅
dt
U
U
To solve separate variables
μ⋅ a
so
2
−
Hence taking antilogarithms
U = U0 ⋅ e
t=−
Solving for t
M⋅ h
μ⋅ a
Hence for
U
U0
= 0.05
t = 3.0⋅
2
μ⋅ a
M⋅ h
⋅t
t
⋅ ln⎛⎜
⎞
⎝ U0 ⎠
M⋅ h
μ⋅ a
U
2
Problem
2.54
Problem 2.51
[Difficulty: 3]
2.51
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
⎛
⎞
r
2
2 Ro − Ri
⎜
u z( r) =
⋅
⋅ Ri − r −
⋅ ln⎛⎜ ⎞
4⋅ μ L ⎜
⎝ Ri ⎠ ⎟
⎛ Ri ⎞
ln⎜
⎜
⎝
⎝ Ro ⎠
⎠
∆p
1
2
2
⎛
Ro − Ri
⎛ Ro ⎞ ⎞
2
2
⎜
u z( R o ) =
⋅
⋅ Ri − Ro −
⋅ ln⎜
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 ) =
⋅
⋅ Ri − Ri −
⋅ ln⎜
=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
⎡
⎛
⎞⎤
1 ∆p ⎜ 2
r
2 Ro − Ri
⎢
= μ⋅
⋅
⋅ Ri − r −
⋅ ln⎛⎜ ⎞ ⎥
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⎞
⎛
2 Ro − Ri
⎜
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
⎡
⎢
⎡⎣( 25⋅ mm) 2 − ( 5⋅ mm) 2⎤⎦ × ⎛⎜ 1 ⋅ m ⎞
2
N
1
⋅
m
⎢
3
⎝ 1000⋅ mm ⎠
⎞ −
Fo = 125 × 10 ⋅
× π × −⎛⎜ 25⋅ mm ×
⎢⎝
5
2
1000⋅ mm ⎠
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⎞
⎛
2 Ro − Ri
⎜
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 ×
−
⎢⎝
5
2
1000⋅ mm ⎠
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
Problem 2.52
[Difficulty: 3]
2.52
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
2 ⋅ r ⎞ ⎞⎤⎥
2 D − d
⋅
⋅ d − 4⋅ r −
⋅ ln⎛⎜
⎢ 16⋅ μ L ⎜
⎟⎥
d
⎝ Di ⎠ ⎥
ln⎛⎜ ⎞
⎢⎣
⎜⎝
⎝ D⎠
⎠⎦
1 ∆p ⎛⎜
D −d ⎞
τrx( r) =
⋅
⋅ −8 ⋅ r −
⎟
d
16 L ⎜
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 2.53
(Difficulty: 2)
2.53 The lubricant has a kinematic viscosity of 2.8 × 10−5
piston is 6
𝑚
,
𝑠
𝑉=6
and 𝑆𝑆 of 0.92. If the mean velocity of the
approximately what is the power dissipated in the friction?
Given: The kinematic viscosity: 𝑣 = 2.8 × 10−5
𝑚
.
𝑠
𝑚2
𝑠
The configuration is shown in the figure.
𝑚2
.Specific
𝑠
gravity: 𝑆𝑆 = 0.92 .
Mean velocity:
Assumption: Linear velocity profile in the lubricant, negligible end effects.
Find: Power 𝑃𝑓 dissipated in the friction.
Solution: Use Newton’s law of viscosity to relate the viscous shear stress to the velocities
The shear stress is given by Newton’s law of viscosity
𝜏=𝜇
The density of the lubricant is:
𝑑𝑑
∆𝑢
= 𝜇
𝑑𝑑
∆𝑟
𝜌 = 𝑆𝑆 ∙ 𝜌𝐻2 0 = 0.92 × 998
The dynamic viscosity of the lubricant is:
𝜇 = 𝑣𝑣 = 2.8 × 10−5
The drag force:
𝑚2
𝑘𝑘
𝑘𝑘
× 918 3 = 2.57 × 10−2
= 2.57 × 10−2 𝑃𝑃 ∙ 𝑠
𝑚
𝑚∙𝑠
𝑠
𝐹𝐷 = 𝜇𝜇
𝐹𝐷 = 2.57
× 10−2
𝑘𝑘
𝑘𝑘
= 918 3
3
𝑚
𝑚
∆𝑢
∆𝑟
𝑚
𝑠
𝑃𝑃 ∙ 𝑠 × (𝜋 × 0.15 𝑚 × 0.3 𝑚) ×
= 218 𝑁
0.0001 𝑚
6
The power 𝑃𝑓 dissipated in the friction is the product of the force and velocity:
𝑃𝑓 = 𝐹𝐷 𝑉 = 218 𝑁 × 6
𝑚
= 1308 𝑊
𝑠
Problem 2.54
(Difficulty: 1)
2.54 Calculate the approximate viscosity of the oil.
Given: Velocity: 𝑉 = 0.6
Slope: 𝑠𝑠𝑠𝑠 =
5
.
13
𝑓𝑓
.
𝑠
Gravity: 𝑊 = 25 𝑙𝑙𝑙. Area: 2 𝑓𝑓 × 2𝑓𝑓. Gap: ℎ = 0.05 𝑖𝑖.
Assumption: Linear velocity profile in oil, negligible end effects.
Find: Viscosity of the oil.
Solution: Use Newton’s law of viscosity to find the relation between shear stress and velocity.
The force balance equation is that the drag force equals the component of the weight along the surface:
𝐹𝐷 = 𝑊 ∙ 𝑠𝑠𝑠𝑠 =
The drag force is found using Newton’s law of viscosity
𝜏=𝜇
5
𝑊
13
𝑑𝑑
∆𝑢
= 𝜇𝜇
𝑑𝑑
∆𝑦
The drag force is then, where the velocity profile is assumed linear:
From the force balance
The viscosity of the oil is:
𝐹𝐷 = 𝜇𝜇
𝜇𝜇
∆𝑢
𝑉
= 𝜇𝜇
∆𝑦
ℎ
𝑉
5
=
𝑊
ℎ 13
5
ℎ
5
𝜇=
𝑊∙
=
× 25 𝑙𝑙𝑙 ×
13
𝐴𝐴 13
0.05
𝑓𝑓
12
4 𝑓𝑓 2 × 0.6
𝑓𝑓
𝑠
= 0.0167
𝑙𝑙𝑙 ∙ 𝑠
𝑓𝑓 2
Problem 2.55
(Difficulty: 2)
2.55 Calculate the approximate power lost in friction in this ship propeller shaft bearing.
Given: Rotation speed: 200
Viscosity: 𝜇 = 0.72 𝑃𝑃 ∙ 𝑠.
𝑟
𝑚𝑚𝑚
. Gap: ℎ = 0.23 𝑚𝑚. Length: 𝐿 = 1 𝑚. Shaft diameter: 𝐷 = 0.36 𝑚.
Assumption: Linear velocity profile in fluid, negligible end effects.
Find: Power 𝑃𝑑 lost in friction.
Solution: Use Newton’s law of viscosity to relate the viscous force to the velocity
𝜏=𝜇
𝑑𝑑
∆𝑢
= 𝜇
𝑑𝑑
∆𝑟
The drag force is given by the product of the shear stress and area. For the linear velocity in the fluid:
𝐹𝐷 = 𝐴 𝜇
The velocity is given by:
So we have:
𝑉 = 𝜔𝜔 = 200
∆𝑢
𝑉
=𝐴𝜇
∆𝑟
ℎ
𝑟
1 𝑚𝑚𝑚
𝑟𝑟𝑟
0.36
𝑚
×
× �2𝜋
�×
𝑚 = 3.77
𝑚𝑚𝑚
60 𝑠
𝑟
2
𝑠
𝑚
𝑠 = 13340 𝑁
𝐹𝐷 = 0.72 𝑃𝑃 ∙ 𝑠 × (𝜋 × 0.36 𝑚 × 1 𝑚) ×
0.00023 𝑚
3.77
The power 𝑃𝑑 lost in friction is the product of force and velocity:
𝑃𝑑 = 𝐹. 𝑉 = 13340 𝑁 × 3.77
𝑚
= 50.3 𝑘𝑘
𝑠
Problem 2.56
Problem
2.56
[Difficulty: 2]
2.56
Given:
Flow between two plates
Find:
Force to move upper plate; Interface velocity
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
V − Vi
μ1 ⋅
= μ2 ⋅
h1
h2
or
V
Vi =
1+
μ1 h 2
⋅
μ2 h 1
1⋅
=
1+
)
where V i is the interface velocity
m
s
0.1 0.3
⋅
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
Problem
2.58
2.57
[Difficulty: 2]
Problem 2.58
Problem
2.60
2.58
[Difficulty: 2]
Problem 2.59
Problem
2.62
2.59
Difficulty: [2]
Problem 2.60
Problem
2.64
[Difficulty: 3]
2.60
Given: Shock-free coupling assembly
Find:
Required viscosity
Solution:
du
τrθ = μ⋅
dr
Basic equation
Shear force
F = τ⋅ A
Assumptions: Newtonian fluid, linear velocity
profile
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⋅
δ
dr
∆r
V2 = ω2(R + δ)
δ
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 2.61
Problem
2.66
2.61
[Difficulty: 4]
Problem 2.62
The data is
2
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
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/m2x103)
191
195
Viscosity vs Shear Rate
2
3
η (N.s/m x10 )
1000
Data
Power Trendline
100
η = 44.94(ω/θ)0.2068
R2 = 0.9925
10
100
1000
Shear Rate ω/θ (1/s)
Problem 2.70
Problem
2.63
[Difficulty: 3]
2.63
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
R2 = 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 =
du/dy (s-1) η (N·s/m2)
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 2.64
Problem
2.72
2.64
[Difficulty: 5]
Problem 2.65
Problem
2.74
2.65
[Difficulty: 5]
Problem
2.76
Problem 2.66
[Difficulty: 5]
2.66
Geometry of rotating bearing
Given:
Expression for shear stress; Maximum shear stress; Expression for total torque; Total torque
Find:
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
du
To find the maximum τ set
d ⎡ μ⋅ ω⋅ R⋅ sin( θ) ⎤
⎢
⎥=0
dθ ⎣ a + R⋅ ( 1 − cos( θ) ) ⎦
R⋅ μ⋅ ω⋅ ( R⋅ cos( θ) − R + a⋅ cos( θ) )
so
( R + a − R⋅ cos( θ) )
τ = 79.2⋅
2
⎞ = acos⎛ 75 ⎞
⎜
⎝ R + a⎠
⎝ 75 + 0.5 ⎠
θ = acos⎛⎜
kg
poise
h
a + R⋅ ( 1 − cos( θ) )
R⋅ cos( θ) − R + a⋅ cos( θ) = 0
m⋅ s
h
u
=
μ⋅ ω⋅ R⋅ sin( θ)
τ = μ⋅
dy
=
dy
=
u−0
u = ω⋅ r = ω⋅ R⋅ sin( θ)
Then
τ = 12.5⋅ poise × 0.1⋅
du
r = R⋅ sin( θ)
R
=0
θ = 6.6⋅ deg
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) )
⌡
0
wher
e
This integral is best evaluated numerically using Excel, Mathcad, or a good calculator
⎛ R0 ⎞
θmax = asin⎜
⎝R⎠
T = 1.02 × 10
−3
⋅ N⋅ m
θmax = 15.5⋅ deg
Problem 2.67
Problem
2.77
2.67
[Difficulty: 2]
Problem 2.68
Problem
2.78
[Difficulty: 2]
2.68
Given:
Data on size of various needles
Find:
Which needles, if any, will float
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 2.69
Problem
2.79
[Difficulty: 3]
2.69
Given:
Caplillary rise data
Find:
Values of A and b
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
The fit is a good one (R2 = 0.9919)
Capillary Rise vs. Tube Diameter
∆h (in.)
0.3
∆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 2.70
(Difficulty: 3)
2.70 Calculate and plot the maximum capillary rise of water at 20 C to be expected in a vertical glass
tube as a function of tube diameters from 0.5 to 2.5 mm.
Given: Temperature: 𝑇 = 20 ℃. Diameter: 𝐷 𝑓𝑓𝑓𝑓 0.5 𝑡𝑡 2.5 𝑚𝑚.
Find: Maximum capillary rise ∆ℎ.
Solution: Use the relation for capillary force to find the rise.
For the force balance on the water we have the capillary force and the weight of the volume of water:
� 𝐹𝑧 = 𝜎𝐻2 𝑜 𝜋𝜋 cos 𝜃 − 𝜌𝐻2 𝑜 𝑔∆𝑉 = 0
1
∆𝑉 = 𝜋𝐷 2 ∆ℎ
4
So we have:
∆ℎ =
4𝜎𝐻2 𝑜 𝜋𝜋 cos 𝜃 4𝜎𝐻2 𝑜 cos 𝜃
=
𝜌𝐻2𝑜 𝑔𝑔𝐷 2
𝜌𝐻2 𝑜 𝑔𝑔
When 𝜃 = 0, we have the maximum ∆ℎ for specific tube diameter . The surface tension for water is
given by
The rise is then
For the range
𝜎𝐻2 𝑜 = 0.0728
𝑁
𝑚
𝑁
4 × 0.0728 × 1
4𝜎𝐻2 𝑜 cos 𝜃
2.97 × 10−5
𝑚
=
=
𝑚
∆ℎ =
𝑘𝑘
𝑚
𝐷
𝜌𝐻2 𝑜 𝑔𝐷
998 3 × 9.81 2 × 𝐷
𝑚
𝑠
0.0005 𝑚 ≤ 𝐷 ≤ 0.0025 𝑚
The plot of rise versus diameter is
Problem 2.71
(Difficulty: 2)
2.71 Calculate the maximum capillary rise of water at 20 ℃ to be expected between two vertical, clean
glass plates spaced 1𝑚𝑚 apart.
Given: Temperature: 𝑇 = 20 ℃ . Distance between two plate: 𝐷 = 1 𝑚𝑚.
Find: Maximum capillary rise ∆ℎ.
Solution: Use the relation for capillary force to find the rise
The force balance equation equates the capillary force to the weight of the water:
� 𝐹𝑧 = 𝜎𝐻2 𝑜 ∙ 2𝐿 cos 𝜃 − 𝜌𝐻2 𝑜 𝑔𝑔𝑔∆ℎ = 0
where L is the width of the plate. Solving for ∆h:
For the maximum capillary rise:
∆ℎ =
The surface tension for water is given by
𝜎𝐻2 𝑜 ∙ 2𝐿 cos 𝜃 2𝜎𝐻2 𝑜 cos 𝜃
=
𝜌𝐻2 𝑜 𝑔𝑔𝑔
𝜌𝐻2 𝑜 𝑔𝑔
𝜃=0
𝜎𝐻2 𝑜 = 0.0728
𝑁
𝑚
𝑁
2 × 0.0728
2𝜎𝐻2 𝑜 cos 𝜃
𝑚
∆ℎ =
=
= 0.0149 𝑚 = 14.9 𝑚𝑚
𝑘𝑘
𝑚
𝜌𝐻2 𝑜 𝑔𝑔
998 3 × 9.8 2 × 0.001 𝑚
𝑚
𝑠
Problem 2.72
(Difficulty: 2)
2.72 Calculate the maximum capillary depression of mercury to be expected in the vertical glass tube
1 𝑚𝑚 in diameter at 15.5 ℃.
Given: Temperature: 𝑇 = 15.5 ℃ 𝑜𝑜 60℉ . Distance between two plate: 𝐷 = 1𝑚𝑚 𝑜𝑜 0.04 𝑖𝑖.
Find: Maximum capillary depression ∆ℎ.
Solution: Use the relation for capillary force to find the rise
The force balance equation per width of the plate equates the capillary force to the weight of the water:
� 𝐹𝑧 = 𝜎𝜎𝜎 cos 𝜃 − 𝜌𝜌∆𝑉 = 0
Where the volume is
1
∆𝑉 = 𝜋𝐷 2 ∆ℎ
4
Solving for the depression:
For mercury, the surface tension is
∆ℎ =
4𝜎𝜎𝜎 cos 𝜃 4𝜎 cos 𝜃
=
𝜌𝜌𝜌𝐷 2
𝜌𝜌𝜌
𝜎 = 0.51
And the density is
For the maximum capillary depression:
𝑁
𝑚
𝛾 = 𝜌𝜌 = 133
𝑘𝑘
𝑚3
𝜃 = 130 ° for mercury.
The depression is
∆ℎ =
4𝜎 cos 𝜃 4 × 0.51 × cos(130 °)
=
𝑚 = −9.86 𝑚𝑚
𝛾𝛾
133 × 1000 × 0.001
Problem 2.73
Problem
2.82
2.73
[Difficulty: 2]
Problem 2.74
Problem
2.84
[Difficulty: 2]
2.74
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 ⋅ ⎛⎜
y⎞
+ c⋅ ⎛⎜
⎝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=
dy
d
dy
From 1 and 2
c=−
Hence
u=
Dimensionless Height
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
=
δ
b
δ
+ 3⋅
c
δ
⋅U
y⎞
⎝δ⎠
3
u
U
=
3
2
⋅ ⎛⎜
y⎞
⎝δ⎠
−
1
2
⋅ ⎛⎜
y⎞
3
⎝δ⎠
1
0.75
0.5
0.25
0
0.25
0.5
Dimensionless Velocity
0.75
1
Problem 2.75
Problem
2.86
[Difficulty: 3]
2.75
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
⋅
=
2
μ
π⋅ μ⋅ D
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 2.76
Problem
2.87
[Difficulty: 2]
2.76
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 2.77
Problem
2.88
[Difficulty: 3]
2.77
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
223.5 K ⎠
kg⋅ m
⎝
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
⋅
×
×
⋅
×
3 m
0.0297 kg
2700
1⋅ hr
10
1
x trans = 0.327m
Problem 2.78
Problem
2.89
[Difficulty: 2]
2.78
Given:
Type of oil, flow rate, and tube geometry
Find:
Whether flow is laminar or turbulent
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
×
1 × 10
m
Hence
The specific weight is
SG =
ρ
−5
⋅
s
2
×
⋅
kg⋅ m
ρ = 900
2
γ = ρ⋅ g
γ = 900 ⋅
kg
3
2
Q=
π⋅ D
4
⋅V
V=
so
Then
Hence
V =
Re =
4
π
10
3
⋅m
1 ⋅ mL
× 1.11 × 10
m
2
2
N⋅ s
×
3 N
γ = 8.829 × 10 ⋅
kg⋅ m
s
4⋅ Q
2
×
1 1
⋅
9 s
Q = 1.111 × 10
3
−5 m
2
1 1
1000⋅ mm ⎞
⎛
⋅
×⎜ ⋅
×
s
1⋅ m ⎠
⎝ 12 mm
V = 0.0981
ρ⋅ V⋅ D
μ
Re = 900 ⋅
kg
3
m
Flow is laminar
3
m
π⋅ D
−6
Q = 100 ⋅ mL ×
3
SG = 0.9
× 9.81⋅
m
For pipe flow (Section 2-6)
kg
m
kg
ρwater = 1000⋅
3
m
ρwater
ρ
s
s ⋅N
m
ρ=
so
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 2.79
Problem
2.90
[Difficulty: 2]
2.79
Given:
Data on seaplane
Find:
Transition point of boundary layer
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
2.80
Problem 2.91
[Difficulty: 3]
2.80
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 Table
appropriate
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 3.1
Problem
3.2
[Difficulty: 2]
3.1
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 3.2
Problem
3.3
[Difficulty: 2]
3.2
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
Δp = −ρair ⋅ g ⋅ Δz
Combining
ΔhHg =
ρair
ρHg
ΔhHg =
⋅ Δz =
0.909
13.55 × 999
Δp = −ρHg ⋅ g ⋅ ΔhHg
and also
ρair
SGHg ⋅ ρH2O
SGHg = 13.55 from Table A.2
⋅ Δz
× 100 ⋅ m
ΔhHg = 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 ⋅ Δz8000 = ρair3000 ⋅ g ⋅ Δz3000
where the numerical subscripts refer to conditions at 3000m and 8000m.
Hence
Δz8000 =
ρair3000 ⋅ g
ρair8000 ⋅ g
⋅ Δz3000 =
ρair3000
ρair8000
⋅ Δz3000
Δz8000 =
0.909
× 100 ⋅ m
0.526
Δz8000 = 173 m
Problem
Problem3.3
3.4
[Difficulty: 3]
3.3
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
o
Tsat ( F)
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 3.4
Problem
3.9
[Difficulty: 2]
3.4
Given:
Data on tire at 3500 m and at sea level
Find:
Absolute pressure at 3500 m; pressure at sea level
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 3.5
Problem
3.5
[Difficulty: 2]
3.5
Given:
Data on system
Find:
Force on bottom of cube; tension in tether
Solution:
dp
= − ρ⋅ g
dy
Basic equation
Δp = ρ⋅ g⋅ h
or, for constant ρ
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
FL = 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
Using FU
(
)
ΔF = FL − FU = ρ⋅ g⋅ SGoil⋅ hL − hU ⋅ A
T = ΔF − W
⎡
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
⎣
⎦
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
Problem
3.6
[Difficulty: 2]
3.6
Given:
Data on system before and after applied force
Find:
Applied force
Solution:
Basic equation
dp
= −ρ⋅ g or, for constant ρ
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
(
)
p = patm − ρ⋅ g⋅ y − y0
F1 = p1⋅ A = ρ⋅ g⋅ h⋅ A
and
F1 − W = 0
( )
p y0 = patm
with
(Gage; F1 is hydrostatic upwards force)
W = F1 = ρ⋅ g⋅ h⋅ A
F2 = p2⋅ A = ρ⋅ g⋅ H⋅ A
and
F2 − W − F = 0
(Gage; F2 is hydrostatic upwards force)
F = F2 − W = ρ⋅ g⋅ H⋅ A − ρ⋅ g⋅ h ⋅ A = ρ⋅ g⋅ A⋅ ( H − h )
2
π⋅ D
F = ρH2O⋅ SG⋅ g⋅
⋅ ( H − h)
4
From Fig. A.1
SG = 13.54
F = 1000⋅
kg
3
m
F = 45.6 N
× 13.54 × 9.81⋅
m
2
s
2
×
π
N⋅ s
2
× ( 0.05⋅ m) × ( 0.2 − 0.025) ⋅ m ×
4
kg⋅ m
Problem 3.7
(Difficulty: 1)
3.7 Calculate the absolute pressure and gage pressure in an open tank of crude oil 2.4 𝑚 below the
liquid surface. If the tank is closed and pressurized to 130 𝑘𝑘𝑘, what are the absolute pressure and gage
pressure at this location.
Given: Location: ℎ = 2.4 𝑚 below the liquid surface. Liquid: Crude oil.
Find: The absolute pressure 𝑝𝑎 and gage pressure 𝑝𝑔 for both open and closed tank .
Assumption: The gage pressure for the liquid surface is zero for open tank and closed tank. The oil is
incompressible.
Governing equation: Hydrostatic pressure in a liquid, with z measured upward:
𝑑𝑑
= −𝜌 𝑔 = −𝛾
𝑑𝑑
The density for the crude oil is:
𝜌 = 856
The atmosphere pressure is:
The pressure for the closed tank is:
𝑘𝑘
𝑚3
𝑝𝑎𝑎𝑎𝑎𝑎 = 101000 𝑃𝑃
𝑝𝑡𝑡𝑡𝑡 = 130 𝑘𝑘𝑘 = 130000 𝑃𝑃
Using the hydrostatic relation, the gage pressure of open tank 2.4 m below the liquid surface is:
𝑝𝑔 = 𝜌𝜌ℎ = 856
𝑘𝑘
𝑚
× 9.81 2 × 2.4 𝑚 = 20100 𝑃𝑃
3
𝑚
𝑠
The absolute pressure of open tank at this location is:
𝑝𝑎 = 𝑝𝑔 + 𝑝𝑎𝑎𝑎𝑎𝑎 = 20100 𝑃𝑃 + 101000 𝑃𝑃 = 121100 𝑃𝑃 = 121.1 𝑘𝑘𝑘
The gage pressure of closed tank at the same location below the liquid surface is the same as open tank:
𝑝𝑔 = 𝜌𝜌ℎ = 856
𝑘𝑘
𝑚
× 9.81 2 × 2.4 𝑚 = 20100 𝑃𝑃
3
𝑚
𝑠
The absolute pressure of closed tank at this location is:
𝑝𝑎 = 𝑝𝑔 + 𝑝𝑡𝑡𝑡𝑡 = 20100 𝑃𝑃 + 130000 𝑃𝑃 = 150100 𝑃𝑃 = 150.1 𝑘𝑘𝑘
Problem 3.8
(Difficulty: 1)
3.8 An open vessel contains carbon tetrachloride to a depth of 6 𝑓𝑓 and water on the carbon
tetrachloride to a depth of 5 𝑓𝑓 . What is the pressure at the bottom of the vessel?
Given: Depth of carbon tetrachloride: ℎ𝑐 = 6 𝑓𝑓. Depth of water: ℎ𝑤 = 5 𝑓𝑓.
Find: The gage pressure 𝑝 at the bottom of the vessel.
Assumption: The gage pressure for the liquid surface is zero. The fluid is incompressible.
Solution: Use the hydrostatic pressure relation to detmine pressures in a fluid.
Governing equation: Hydrostatic pressure in a liquid, with z measured upward:
The density for the carbon tetrachloride is:
The density for the water is:
𝑑𝑑
= −𝜌 𝑔 = −𝛾
𝑑𝑑
𝜌𝑐 = 1.59 × 103
𝜌𝑤 = 1.0 × 103
𝑘𝑘
𝑠𝑠𝑠𝑠
= 3.09
3
𝑚
𝑓𝑓 3
𝑘𝑘
𝑠𝑠𝑠𝑠
= 1.940
3
𝑚
𝑓𝑓 3
Using the hydrostatic relation, the gage pressure 𝑝 at the bottom of the vessel is:
𝑝 = 3.09
𝑝 = 𝜌𝑐 𝑔ℎ𝑐 + 𝜌𝑤 𝑔ℎ𝑤
𝑠𝑠𝑠𝑠
𝑓𝑓
𝑠𝑠𝑠𝑠
𝑓𝑓
𝑙𝑙𝑙
× 32.2 2 × 6 𝑓𝑓 + 1.940
× 32.2 2 × 5 𝑓𝑓 = 909 2 = 6.25 𝑝𝑝𝑝
3
3
𝑓𝑓
𝑠
𝑓𝑓
𝑠
𝑓𝑓
Problem
3.8
Problem 3.9
[Difficulty: 2]
3.9
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.
3.7equations:
these
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
(
Δp = pL − pU = ρH2O ⋅ g⋅ 0.9⋅ d + ρSAE10 ⋅ g ⋅ 0.1⋅ d
From Table A.2
SGSAE10 = 0.92
kg
Δp = 999⋅
3
× 9.81⋅
m
m
2
2
× 0.1⋅ m × ( 0.9 + 0.92 × 0.1) ×
s
N⋅s
kg ⋅ m
Δp = 972 Pa
For the cube density, set up a free body force balance for the cube
ΣF = 0 = Δp ⋅ A − W
Hence
W = Δp⋅ A = Δp⋅ d
ρcube =
m
3
d
ρcube = 972⋅
2
W
=
3
=
d ⋅g
N
2
m
2
Δp ⋅ d
3
=
d ⋅g
Δp
d⋅ g
2
×
1
s
kg ⋅ m
×
×
0.1⋅ m 9.81⋅ m N s2
⋅
)
Δp = ρH2O ⋅ g⋅ d ⋅ 0.9 + SGSAE10 ⋅ 0.1
ρcube = 991
kg
3
m
Problem
3.10
Problem
3.1
[Difficulty: 2]
3.10
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
= M⋅ R⋅ T⋅ ⎛⎜
M⋅ R⋅ T
p = 140⋅ lbm × 55.16⋅
p = 3520⋅
6 ⎞
3⎟
⎝ π⋅ D ⎠
V
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 ⋅
4
2
− σc ⋅ π ⋅ D ⋅ t
pπD2/4
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
×
=
σcπDt
p⋅ D
4 ⋅ σc
2.5 ⋅ ft
×
4
2
in
3
30 × 10 ⋅ lbf
t = 0.880⋅ in
2
Problem 3.11
(Difficulty: 2)
3.11 If at the surface of a liquid the specific weight is 𝛾0 , with 𝑧 and 𝑝 both zero, show that, if
𝐸 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐, the specific weight and pressure are given 𝛾 =
𝐸
𝐸
�𝑧+𝛾 �
0
and 𝑝 = −𝐸 ln �1 +
Calculate specific weight and pressure at a depth of 2 𝑘𝑘 assuming 𝛾0 = 10.0
Given: Depth: ℎ = 2 𝑘𝑘. The specific weight at surface of a liquid: 𝛾0 = 10.0
Find: The specific weight and pressure at a depth of 2 𝑘𝑘.
𝑘𝑘
𝑚3
𝛾0 𝑍
�.
𝐸
and 𝐸 = 2070 𝑀𝑀𝑀.
𝑘𝑘
.
𝑚3
Assumption:. Bulk modulus is constant
Solution: Use the hydrostatic pressure relation and definition of bulk modulus to detmine pressures in
a fluid.
Governing equation: Hydrostatic pressure in a liquid, with z measured upward:
Definition of bulk modulus
𝑑𝑑
= −𝜌 𝑔 = −𝛾
𝑑𝑑
𝐸𝑣 =
𝑑𝑑
𝑑𝑑
=
𝑑𝑑�
𝑑𝑑�
𝛾
𝜌
Eliminating dp from the hydrostatic pressure relation and the bulk modulus definition:
Or
Integrating for both sides we get:
At 𝑧 = 0, 𝛾 = 𝛾0 so:
𝑑𝑑 = −𝛾 𝑑𝑑 = 𝐸𝑣
𝑑𝑑 = −𝐸𝑣
𝑧 = 𝐸𝑣
𝑑𝑑
𝛾2
1
+𝑐
𝛾
𝑐 = −𝐸𝑣
1
𝛾0
𝑑𝑑
𝛾
𝑧 = 𝐸𝑣
Solving for 𝛾, we have:
𝛾=
1
1
− 𝐸𝑣
𝛾
𝛾0
𝐸𝑣
𝐸
�𝑧 + 𝑣 �
𝛾0
Solving for the pressure using the hydrostatic relation:
𝑑𝑑 = −𝛾𝛾𝛾 = −
Integrating both sides we to get:
At 𝑧 = 0, 𝑝 = 0 so:
For the specific case
𝐸𝑣
𝑑𝑑
𝐸
�𝑧 + 𝑣 �
𝛾0
𝑝 = −𝐸𝑣 ln �𝑧 +
𝑝 = −𝐸𝑣 ln �𝑧 +
𝐸𝑣
�+𝑐
𝛾0
𝐸𝑣
𝑐 = 𝐸𝑣 ln � �
𝛾0
𝐸𝑣
𝐸𝑣
𝛾0 𝑧
� + 𝐸𝑣 ln � � = −𝐸𝑣 ln �1 +
�
𝐸𝑣
𝛾0
𝛾0
ℎ = 2 𝑘𝑘
𝛾0 = 10.0
The specific weight:
𝛾=
Pressure:
𝑝 = −𝐸𝑣 ln �1 +
𝑘𝑘
𝑚3
𝐸𝑣 = 2070 𝑀𝑀𝑀
𝐸𝑣
2070 × 106 𝑝𝑝
𝑁
𝑘𝑘
=
= 10100 3 = 10.1 3
𝐸
𝑚
𝑚
�𝑧 + 𝑣 �
2070 × 106 𝑃𝑃
𝛾0
�−2000 𝑃𝑃 +
𝑁 �
10 × 103 3
𝑚
𝛾0 𝑧
𝑘𝑘
−2000 𝑚
� = −2070 × 106 𝑃𝑃 × ln �1 + 10000.0 3 × �
�� = 20100 𝑘𝑘𝑘
𝑚
2070 × 106 𝑃𝑃
𝐸𝑣
Problem 3.12
(Difficulty: 2)
3.12 In the deep ocean the compressibility of seawater is significant in its effect on 𝜌 and 𝑝. If
𝐸 = 2.07 × 109 𝑃𝑃, find the percentage change in the density and pressure at a depth of 10000 meters
as compared to the values obtained at the same depth under the incompressible assumption. Let
𝜌0 = 1020
𝑘𝑘
𝑚3
and the absolute pressure 𝑝0 = 101.3 𝑘𝑘𝑘.
Given: Depth: ℎ = 10000 𝑚𝑚𝑚𝑚𝑚𝑚. The density: 𝜌0 = 1020
Find: The percent change in density 𝜌% and pressure 𝑝%.
𝑘𝑘
.
𝑚3
The absolute pressure: 𝑝0 = 101.3 𝑘𝑘𝑘.
Assumption: The bulk modulus is constant
Solution:
Use the relations developed in problem 3.11 for specific weight and pressure for a
compressible liquid:
𝛾=
𝐸
�𝑧 +
𝐸
�
𝛾0
𝑝 = −𝐸 ln �1 +
The specific weight at sea level is:
𝛾0 = 𝜌0 𝑔 = 1020
𝛾0 𝑧
�
𝐸
𝑘𝑘
𝑚
𝑁
× 9.81 2 = 10010 3
3
𝑚
𝑠
𝑚
The specific weight and density at 10000 m depth are
𝛾=
𝐸
�𝑧 +
𝐸
�
𝛾0
The percentage change in density is
=
2.07 × 109
𝑁
𝑁
= 10520 3
9
3
2.07 × 10
𝑚
𝑚
�−10000 +
�
10010
𝜌=
𝜌% =
𝛾 10520 𝑘𝑘
𝑘𝑘
=
= 1072 3
3
𝑔
9.81 𝑚
𝑚
𝜌 − 𝜌0 1072 − 1020
=
= 5.1 %
1020
𝜌0
The gage pressure at a depth of 10000m is:
𝑝 = −𝐸 ln �1 +
𝛾0 𝑧
10010 × (−10000)
� = 101.3 𝑘𝑘𝑘 − 2.07 × 109 × ln �1 +
� 𝑃𝑃 = 102600 𝑘𝑘𝑘
𝐸
2.07 × 109
The pressure assuming that the water is incompressible is:
𝑝𝑖𝑖 = 𝜌𝜌ℎ = 1020
The percent difference in pressure is:
𝑝% =
𝑘𝑘
𝑚
×
9.81
× 10000 𝑚 = 100062 𝑘𝑘𝑘
𝑚3
𝑠2
𝑝 − 𝑝0 102600 𝑘𝑘𝑘 − 100062 𝑘𝑘𝑘
=
= 2.54 %
100062 𝑘𝑘𝑘
𝑝0
Problem 3.13
Problem
3.12
[Difficulty: 4]
3.13
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:
From Table A.2, App. A:
Ev =
Then
dp = ρ⋅ g⋅ dh = Ev⋅
ρ
Ev = 2.42⋅ GPa = 3.51 × 10 ⋅ psi
dp
= ρ⋅ g
dh
Governing Equations:
dρ
5
SGo = 1.025
or
dρ
g
=
dh
2
Ev
ρ
(Hydrostatic Pressure - h is positive downwards)
dp
(Definition of Bulk Modulus)
dρ
ρ
h
ρ
⌠
⌠ g
1
⎮
⎮
d
ρ
=
dh
⎮
2
⎮ Ev
⎮ ρ
⌡0
⌡ρ
Now if we integrate:
o
After integrating:
Now for
ρo⋅ g⋅ h
Ev
ρ − ρo
ρ⋅ ρo
=
g⋅ h
Ev
Therefore: ρ =
Ev⋅ ρo
Ev − g⋅ h⋅ ρo
ρ
=
ρo
and
1−
<<1, the binomial expansion may be used to approximate the density:
1
ρo⋅ g⋅ h
Ev
ρo⋅ g⋅ h
ρ
= 1+
Ev
ρo
2
In other words, ρ = ρo + b⋅ h where b =
Since
ρo ⋅ g
(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 +
(
)
2
2
g⋅ h⋅ 2⋅ ρo + b⋅ h
⎛
b⋅ g⋅ h
b⋅ h ⎞
⎟⋅g
= patm + ρo⋅ g⋅ h +
= patm + ⎜ ρo⋅ h +
2
2
2 ⎠
⎝
Now if we subsitiute in the expression for b and simplify, we get:
2
⎛
ρo ⋅ g h2 ⎞⎟
⎛ ρo⋅ g⋅ h ⎞
⎜
papprox = patm + ρo⋅ h +
⋅
⋅ g = patm + ρo⋅ g⋅ h⋅ ⎜ 1 +
⎟
⎜
Ev 2 ⎟
2⋅ Ev
⎝
⎠
⎝
⎠
⎛ ρo⋅ 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
ρ
Subsitiuting for
ρo
o
If we let x =
ρo⋅ g⋅ h
Ev
For the error to be 0.01%:
we get:
Δpexact − Δpapprox
Δpexact
= 1−
This equation requires an iterative solution, e.g. Excel's Goal Seek. The result is:
h=
x⋅ Ev
ρo⋅ g
⎛ ρo⋅ 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
12⋅ in ⎞
slug⋅ ft
5 lbf
×
×
× ⎛⎜
⎟ ×
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
3.14
Problem 3.14
[Difficulty: 3]
3.14
Given:
Find:
Cylindrical cup lowered slowly beneath pool surface
Air
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
= ρ⋅ g
dh
(Hydrostatic Pressure - h is positive downwards)
p⋅ V = M⋅ R⋅ T
(Ideal Gas Equation)
(1) Constant temperature compression of air inside cup
(2) Static liquid
(3) Incompressible liquid
Assumptions:
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
Problem
3.16
[Difficulty: 2]
3.15
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
= − ρ⋅ g
dy
or, for constant ρ
Δp = 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
s
2
F = 6.94⋅ lbf
2
2
ft ⎞
lbf ⋅ s
⎟ ×
⎝ 12⋅ in ⎠ slug⋅ ft
× 16⋅ ft × 1⋅ in × ⎛⎜
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
3
h = 9⋅ lbf ×
h = 20.7⋅ ft
2
2
1 ft
1 s
1
12⋅ in ⎞
slug⋅ ft
⋅
×
⋅ ×
× ⎛⎜
⎟ ×
2
1.94 slug 32.2 ft in
⎝ ft ⎠ lbf ⋅ s2
Problem 3.16
Problem
3.18
[Difficulty: 2]
3.16
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
pgage = 999⋅
kg
3
× 9.81⋅
m
m
2
2
× ( 13.55 × 0.1 ⋅ m − 1.0 ⋅ m) ×
s
N⋅s
kg ⋅ m
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
pgage = 999⋅
kg
3
m
× 9.81⋅
m
2
s
2
× ( 13.55 × 1 ⋅ m − 1.0 ⋅ m) ×
N⋅s
kg ⋅ m
pgage = 123⋅ kPa
Problem 3.17
Problem
3.20
[Difficulty: 2]
3.17
Given:
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:
Assumptions:
dp
= ρ⋅ g
dh
(Hydrostatic Pressure - h is positive downwards)
ρ = SG⋅ ρwater
(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:
(
)
Δp = p2 − p1 = ρct⋅ g⋅ l − ρwater⋅ g⋅ l = SGct − 1 ⋅ ρwater⋅ g⋅ l
Substituting the known data:
Δp = ( 1.591 − 1) × 1000⋅
kg
3
m
× 9.81⋅
m
2
s
× 10.2⋅ mm ×
m
3
10 ⋅ mm
Δp = 59.1 Pa
Problem 3.18
Problem
3.22
[Difficulty: 2]
3.18
Two fluid manometer contains water and kerosene. With both tubes
open to atmosphere, the difference in free surface elevations is known
Given:
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
= ρ⋅ g
dh
(Hydrostatic Pressure - h is positive downwards)
ρ = SG⋅ ρwater
(Definition of Specific Gravity)
(1) Static liquid
(2) Incompressible liquid
Assumptions:
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 = Ho + 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 = ρk⋅ g⋅ Ho + H1
)
A
pB = ρwater⋅ g⋅ H1
B
Setting these pressures equal:
(
)
ρk⋅ g⋅ Ho + H1 = ρwater⋅ g⋅ H1
Solving for H1
H1 =
ρk⋅ Ho
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 = Δp + ρwater⋅ g⋅ H1 − l
H1 = 91.11⋅ mm
A
B
Setting these pressures equal:
(
)
(
)
Δp
SGk⋅ Ho + H1 + l =
+ H1 − l
ρwater⋅ g
l=
1 ⎡ Δp
+ H1 − SGk⋅ Ho + H1 ⎥⎤
⎢
2 ρwater⋅ g
⎣
(
)
⎦
Substituting in known values we get:
1 ⎡⎢
N
1 m
1 s
kg⋅ m
m ⎥⎤
× 98.0⋅
×
×
⋅ ×
+ [ 91.11⋅ mm − 0.82 × ( 20⋅ mm + 91.11⋅ mm) ] ×
2 999 kg 9.81 m
2
3
2 ⎢
⎥
m
N⋅ s
10 ⋅ mm⎦
⎣
3
l =
2
l = 5.000⋅ mm
Now we solve for H:
H = 20⋅ mm + 2 × 5.000⋅ mm
H = 30.0⋅ mm
Problem 3.19
Problem
3.24
[Difficulty: 2]
3.19
Given:
Data on manometer
Find:
Gage pressure at point a
Assumption:
e
Water, liquids A and B are static and incompressible
c
d
Solution:
Basic equation
dp
= − ρ⋅ g
dy
or, for constant ρ
Δp = ρ⋅ g⋅ Δh
where Δh 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 3.20
(Difficulty: 1)
3.20 With the manometer reading as shown, calculate 𝑝𝑥 .
Given: Oil specific gravity: 𝑆𝑆𝑜𝑜𝑜 = 0.85 Depth: ℎ1 = 60 𝑖𝑖𝑖ℎ. ℎ2 = 30 𝑖𝑖𝑖ℎ.
Find: The pressure 𝑝𝑥 .
Assumption: Fluids are incompressible
Solution: Use the hydrostatic relation to find the pressures in the fluid
Governing equation: Hydrostatic pressure in a liquid, with z measured upward:
𝑑𝑑
= −𝜌 𝑔 = −𝛾
𝑑𝑑
Integrating with respect to z for an incompressible fluid, we have the relation for the pressure difference
over a difference in elevation (h):
Repeated application of this relation yields
The specific weight for mercury is:
The pressure at the desired location is
𝑝𝑥 = 0.85 × 62.4
∆𝑝 = 𝜌𝜌ℎ
𝑝𝑥 = 𝑆𝑆𝑜𝑜𝑜 𝛾𝑤𝑤𝑤𝑤𝑤 ℎ1 + 𝛾𝑀 ℎ2
𝛾𝑀 = 845
𝑙𝑙𝑙
𝑓𝑓 3
𝑙𝑙𝑙
60
𝑙𝑙𝑙
30
𝑙𝑙𝑙
× � � 𝑓𝑓 + 845 3 × � � 𝑓𝑓 = 2380 2 = 16.5 𝑝𝑝𝑝
3
𝑓𝑓
12
𝑓𝑓
12
𝑓𝑓
Problem 3.21
(Difficulty: 2)
3.21 Calculate 𝑝𝑥 − 𝑝𝑦 for this inverted U-tube manometer.
Given: Oil specific gravity: 𝑆𝑆𝑜𝑜𝑜 = 0.90 Depth: ℎ1 = 65 𝑖𝑖𝑖ℎ. ℎ2 = 20 𝑖𝑖𝑖ℎ. ℎ3 = 10 𝑖𝑖𝑖ℎ.
Find: The pressure difference 𝑝𝑥 − 𝑝𝑦 .
Assume: The fluids are incompressible
Solution: Use the hydrostatic relation to find the pressures in the fluid
Governing equation: Hydrostatic pressure in a liquid, with z measured upward:
𝑑𝑑
= −𝜌 𝑔 = −𝛾
𝑑𝑑
Integrating with respect to z for an incompressible fluid, we have the relation for the pressure difference
over a difference in elevation (h):
∆𝑝 = 𝜌𝜌ℎ
Starting at the location of the unknown pressure px, we have the following relations for the hydrostatic
pressure:
𝑝𝑥 − 𝑝1 = 𝛾𝑤𝑤𝑤𝑤𝑤 ℎ1
𝑝1 − 𝑝2 = −𝑆𝑆𝑜𝑜𝑜 𝛾𝑤𝑤𝑤𝑤𝑤 ℎ3
𝑝2 − 𝑝𝑦 = −𝛾𝑤𝑤𝑤𝑤𝑤 (ℎ1 − ℎ2 − ℎ3 )
Adding these three equations together
𝑝𝑥 − 𝑝𝑦 = 𝛾𝑤𝑤𝑤𝑤𝑤 (ℎ2 + ℎ3 ) − 𝑆𝑆𝑜𝑜𝑜 𝛾𝑤𝑤𝑤𝑤𝑤 ℎ3
The pressure difference is then
𝑝𝑥 − 𝑝𝑦 = 62.4
𝑙𝑙𝑙 (10 + 20)
𝑙𝑙𝑙 10
𝑙𝑙𝑙
×
𝑓𝑓 − 0.9 × 62.4 3 ×
𝑓𝑓 = 109.2 2 = 0.758 𝑝𝑝𝑝
𝑓𝑓 3
𝑓𝑓
12
𝑓𝑓
12
Problem 3.22
(Difficulty: 2)
3.22 An inclined gage having a tube of 3 mm bore, laid on a slope of 1:20, and a reservoir of 25 mm
diameter contains silicon oil (SG 0.84). What distance will the oil move along the tube when a pressure
of 25 mm of water is connected to the gage?
Given: Silicon oil specific gravity: 𝑆𝑆𝑜𝑜𝑜 = 0.84. Diameter: 𝐷1 = 3 𝑚𝑚. 𝐷2 = 25 𝑚𝑚.
Depth: ℎ𝑤𝑤𝑤𝑤𝑤 = 25 𝑚𝑚. Slope angle: 1: 20.
Find: The distance 𝑥 of the oil move along the tube.
Assumption: Fluids are incompressible
Solution: Use the hydrostatic relation to find the pressures in the fluid
Governing equation: Hydrostatic pressure in a liquid, with z measured upward:
𝑑𝑑
= −𝜌 𝑔 = −𝛾
𝑑𝑑
Integrating with respect to z for an incompressible fluid, we have the relation for the pressure difference
over a difference in elevation (h):
We have the volume of the oil as constant, so:
or
∆𝑝 = 𝜌𝜌ℎ
𝐴𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 ∆ℎ = 𝐴𝑡𝑡𝑡𝑡 𝑥
𝐴𝑡𝑡𝑡𝑡
𝐷12
9
∆ℎ
=
= 2=
𝑥
𝐴𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝐷2 625
When a pressure of 25 𝑚𝑚 of water is connected with the gage we have:
𝛾𝑤𝑤𝑤𝑤𝑤 ℎ𝑤𝑤𝑤𝑤𝑤 = 𝑆𝑆𝑜𝑜𝑜 𝛾𝑤𝑤𝑤𝑤𝑤 ℎ
ℎ𝑤𝑤𝑤𝑤𝑤
= 29.8 𝑚𝑚
𝑆𝑆𝑜𝑜𝑜
ℎ=
Using these relations, we obtain, accounting for the slope of the manometer:
ℎ = ∆ℎ +
𝑥
√202
ℎ = ∆ℎ +
𝑥=
+ 12
9
1
=�
+
�𝑥
2
625 √20 + 12
9
1
=�
+
�𝑥
625 √401
√401
𝑥
ℎ
9
1
�
+
�
625 √401
= 463 𝑚𝑚
Problem 3.23
Problem
3.26
[Difficulty: 2]
3.23
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:
Assumptions:
dp
= ρ⋅ g
dh
(Hydrostatic Pressure - h is positive downwards)
ρ = SG⋅ ρwater
(Definition of Specific Gravity)
(1) Static liquid
(2) Incompressible liquid
(3) Gravity is constant
Integrating the hydrostatic pressure equation we get:
Δp = ρ⋅ g⋅ Δh
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:
(
)
Δp = pA − pB = ρwater⋅ g⋅ ⎡h ⋅ SGHg − 1 − L⋅ sin( 30⋅ deg)⎤
⎣
⎦
Substituting in values:
2
ft
⎤ lbf⋅ s × ⎛ ft ⎞
Δp = 1.94⋅
× 32.2 × ⎡⎢6⋅ in ×
× ( 13.55 − 1) − 5⋅ ft × sin( 30⋅ deg)⎥ ×
⎜
⎟
3
2 ⎣
12⋅ in
⋅
⎦ slugft
⎝ 12⋅ in ⎠
ft
s
slug
ft
2
Δp = 1.638⋅ psi
Problem 3.24
Problem
3.28
[Difficulty: 2]
3.24
Given:
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:
Assumptions:
dp
= ρ⋅ g
dh
(Hydrostatic Pressure - h is positive downwards)
ρ = SG⋅ ρwater
(Definition of Specific Gravity)
(1) Static liquid
(2) Incompressible liquid
Integrating the hydrostatic pressure equation we get:
Δp = ρ⋅ g⋅ Δh
Beginning at the free surface of the reservoir, and accounting for the changes in pressure with elevation:
patm + Δp + ρoil⋅ g⋅ ( x + L) = patm
Δp
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:
Δh
2
⎡
d⎞ ⎤
SGoil⋅ ⎢1 + ⎛⎜ ⎟ ⎥
D
⎣
(Note:
s =
L
Δh
which yields
=
Δh
SGoil
π 2
π 2
⋅D ⋅x = ⋅d ⋅L
4
4
2
d⎞
⎟ L
⎝D⎠
x = ⎛⎜
Substituting values into the expression:
L =
25⋅ mm
⎡
2
6⋅ mm ⎞ ⎤
⎟⎥
⎝ 18⋅ mm ⎠ ⎦
0.827⋅ ⎢1 + ⎛⎜
⎝ ⎠⎦
s = 1.088
Δp = ρwater⋅ g⋅ Δh where Δh = 25⋅ mm
⎣
for this manometer.)
L = 27.2⋅ mm
Problem 3.25
Problem
3.29
[Difficulty: 2]
3.25
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:
Assumptions:
dp
= ρ⋅ g
dh
(Hydrostatic Pressure - h is positive downwards)
ρ = SG⋅ ρwater
(Definition of Specific Gravity)
(1) Static liquid
(2) Incompressible liquid
Integrating the hydrostatic pressure equation we get:
Δp = ρ⋅ g⋅ Δh
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 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
1.75 −
2
l =
D1 − D2 + d
1
SGoil −
2
l = 1.600 m
c
Problem 3.26
(Difficulty: 2)
3.26 The sketch shows a sectional view through a submarine. Calculate the depth of submarine, y.
Assume the specific weight of the seawater is 10.0
𝑘𝑘
.
𝑚3
Given: Atmos. Pressure: 𝑝𝑎𝑎𝑎𝑎𝑎 = 740 𝑚𝑚 𝐻𝐻. Seawater specific weight:𝛾 = 10.0
dimensional relationship is shown in the figure.
𝑘𝑘
.
𝑚3
All the
Find: The depth 𝑦.
Assumption: Fluids are incompressible
Solution: Use the hydrostatic relation to find the pressures in the fluid
Governing equation: Hydrostatic pressure in a liquid, with z measured upward:
𝑑𝑑
= −𝜌 𝑔 = −𝛾
𝑑𝑑
Integrating with respect to z for an incompressible fluid, we have the relation for the pressure difference
over a difference in elevation (h):
∆𝑝 = 𝜌𝜌ℎ
Using the barometer reading with 760 mm as atmospheric pressure, the pressure inside the submarine is:
𝑝=
840 𝑚𝑚
× 101.3 × 103 𝑃𝑃 = 111.6 × 103 𝑃𝑃
760 𝑚𝑚
However, the actual atmosphere pressure is:
𝑝𝑎𝑎𝑎𝑎𝑎 =
740 𝑚𝑚
× 101.3 × 103 𝑃𝑃 = 98.3 × 103 𝑃𝑃
760 𝑚𝑚
For the manometer, using the hydrostatic relation, we have for the pressure, where y is the depth of the
submarine:
𝑝 = 𝑝𝑎𝑎𝑎𝑎𝑎 + 𝛾𝛾 + 𝛾 × 200 𝑚𝑚 − 𝛾𝐻𝐻 × 400 𝑚𝑚
𝑦=
𝑝 + 𝛾𝐻𝐻 × 400 𝑚𝑚 − 𝛾 × 200 𝑚𝑚 − 𝑝𝑎𝑎𝑎𝑎𝑎
𝛾
The specific weight for mercury is:
So we have for the depth y:
𝑦=
𝛾𝐻𝐻 = 133.1
111.6 × 103 𝑃𝑃 + 133.1 × 1000
𝑘𝑘
𝑚3
𝑁
𝑁
× 0.4 𝑚 − 1000 3 × 0.2 𝑚 − 98.3 × 103 𝑃𝑃
3
𝑚
𝑚
𝑁
1000 3
𝑚
𝑦 = 6.45 𝑚
Problem 3.27
(Difficulty: 1)
3.27 The manometer reading is 6 in. when the tank is empty (water surface at A). Calculate the
manometer reading when the cone is filled with water.
Find: The manometer reading when the tank is filled with water.
Assumption: Fluids are static and incompressible
Solution: Use the hydrostatic relations for pressure
When the tank is empty, we have the equation as:
ℎ𝑀𝑀 ∙ 𝑆𝑆𝑚𝑚𝑚𝑚𝑚𝑚𝑚 ∙ 𝛾𝑤𝑤𝑤𝑤𝑤 = 𝛾𝑤𝑤𝑤𝑤𝑤 ℎ
𝑆𝑆𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 13.57
ℎ = ℎ𝑀𝑀 ∙ 𝑆𝑆𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 150 𝑚𝑚 × 13.57 = 2.04 𝑚
When the tank is filled with water, we assume the mercury interface moves by 𝑥:
𝛾𝑤𝑤𝑤𝑤𝑤 (ℎ𝑡𝑡𝑡𝑡 + ℎ + 𝑥) = 𝛾𝑤𝑤𝑤𝑤𝑤 ∙ 𝑆𝑆𝑚𝑚𝑚𝑚𝑚𝑚𝑚 (ℎ𝑀𝑀 + 2𝑥)
Thus
(3 𝑚 + 2.04 𝑚 + 𝑥) = 13.57(0.15𝑚 + 2𝑥)
The new manometer reading is:
𝑥 = 0.115 𝑚
′
= ℎ𝑀𝑀 + 2𝑥 = 0.15 𝑚 + 2 × 0.115 𝑚 = 0.38 𝑚
ℎ𝑀𝑀
Problem 3.28
Problem
3.30
[Difficulty: 2]
3.28
Given:
Reservoir manometer with dimensions shown. The manometer fluid
specific gravity is given.
D =
5
⋅ in
8
d =
3
⋅ in SGoil = 0.827
16
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.
dp
= − ρ⋅ g
dz
(Hydrostatic Pressure - z is positive upwards)
ρ = SG⋅ ρwater
(Definition of Specific Gravity)
Governing Equations:
Assumptions:
(1) Static liquid
(2) Incompressible liquid
Integrating the hydrostatic pressure equation we get:
Δp = −ρ⋅ g⋅ Δz
h
Beginning at the free surface of the tank, and accounting for the changes in pressure with
elevation:
patm + Δp − ρoil⋅ g⋅ ( x + h) = patm
Upon simplification:
Δp = 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 =
Δp = ρwater⋅ g⋅ l
where
l
SGoil
π 2
π 2
⋅D ⋅x = ⋅d ⋅h
4
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 3.29
Problem
3.32
[Difficulty: 3]
3.29
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.
dp
= − ρ⋅ g
dz
Governing Equation:
Assumptions:
(Hydrostatic Pressure - z is positive upwards)
(1) Static liquid
(2) Incompressible liquid
Integrating the hydrostatic pressure equation we get:
Δp = −ρ⋅ g⋅ Δz
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 2
π 2
⋅D ⋅x = ⋅d ⋅L
4
4
2
d⎞
x = ⎛⎜ ⎟ ⋅ L
⎝D⎠
⎡ 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
⎡⎛ d ⎞ 2
⎤
⎟ + sin ( θ)⎥
⎣⎝ D ⎠
⎦
ρ⋅ g⋅ L⋅ ⎢⎜
=
ρ⋅ g⋅ h
2
Since L/h = 5:
sin ( θ) =
h ⎛d⎞
1
8⋅ mm ⎞
− ⎜ ⎟ = − ⎛⎜
⎟
L ⎝D⎠
5 ⎝ 96⋅ mm ⎠
2
θ = 11.13⋅ deg
The sensitivity of the manometer:
s=
L
L
=
Δhe
SG⋅ h
s=
5
SG
Problem
3.33
Problem 3.30
[Difficulty: 3]
3.30
Given:
Data on inclined manometer
Find:
Angle θ for given data; find sensitivity
Solution:
Basic equation
dp
= − ρ⋅ g
dy
or, for constant ρ
Δp = ρ⋅ g⋅ Δh
where Δh is height difference
Under applied pressure
Δp = SGMer⋅ ρ⋅ g⋅ ( L⋅ sin( θ) + x)
From Table A.1
SGMer = 0.827
and Δp = 1 in. of water, or
Δp = ρ⋅ g⋅ h
Δp = 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
kg⋅ m
= L⋅ ⎛⎜
Δp = 245 Pa
d⎞
2
⎟
⎝ D⎠
(2)
2
2
2
⎡
d⎞ ⎤
Δp = SGMer⋅ ρ⋅ g⋅ ⎢L⋅ sin ( θ) + L⋅ ⎛⎜ ⎟ ⎥
D
⎣
sin ( θ) =
⎝ ⎠⎦
Δp
SGMer⋅ ρ⋅ g⋅ L
d⎞
⎟
D
⎝ ⎠
− ⎛⎜
2
3
m
1 s
1 1 kg⋅ m ⎛ 8 ⎞
sin ( θ) = 245⋅
×
×
⋅
×
⋅ ×
⋅ ×
− ⎜ ⎟ = 0.186
2 0.827 1000 kg 9.81 m 0.15 m
2
m
s ⋅ N ⎝ 76 ⎠
N
1
1
θ = 11⋅ deg
The sensitivity is the ratio of manometer deflection to a vertical water manometer
s=
L
0.15⋅ m
=
h
0.025⋅ m
s=6
Problem 3.31
Problem
3.34
[Difficulty: 4]
3.31
Given:
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:
Assumptions:
dp
= − ρ⋅ g
dh
(Hydrostatic Pressure - h is positive downwards)
ρ = SG⋅ ρwater
(Definition of Specific Gravity)
(1) Static liquid
(2) Incompressible liquid
Water vapor
Water
Integrating the hydrostatic pressure equation we get:
h2
Δp = ρ⋅ g⋅ Δh
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 3.32
Problem
3.36
[Difficulty: 3]
3.32
Given:
Water column standin in glass tube
Δh = 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
= ρ⋅ g
dh
(Hydrostatic Pressure - h is positive downwards)
ΣFz = 0
(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
Assumptions:
Δhc
Δh
Δhp
Δh = Δhc + Δhp
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) −
π
4
4⋅ σ
Therefore: Δhc =
⋅ cos ( θ)
ρ⋅ g⋅ D
2
⋅ D ⋅ ρ⋅ g⋅ Δhc = 0
θ
Substituting values:
⎛ 103⋅ mm ⎞
⎟
Δhc = 4 × 72.8 × 10 ⋅ ×
⋅
×
⋅ ×
⋅
×
×⎜
m 999 kg 9.81 m 2.5 mm N s2 ⎝ m ⎠
⋅
−3 N
3
1
m
1
2
s
1
1
kg⋅ m
2
Δhc
Δhc = 11.89⋅ mm
Therefore: Δhp = Δh − Δhc
Δhp = 50⋅ mm − 11.89⋅ mm
π Dδ
Δhp = 38.1⋅ mm
Mg = ρgV
(result for σ = 0)
For the 1 mm diameter tube:
Δhc = 4 × 72.8 × 10
1 m
1 s
1 1
kg⋅ m ⎛ 10 ⋅ mm ⎞
⎟
⋅ ×
⋅
×
⋅ × ⋅
×
×⎜
m 999 kg 9.81 m 1 mm N s2 ⎝ m ⎠
⋅
−3 N
Δh = 29.7⋅ mm + 38.1⋅ mm
3
2
3
2
Δhc = 29.71⋅ mm
Δh = 67.8⋅ mm
Problem 3.33
Problem
3.38
[Difficulty :2]
3.33
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 = Δp⋅
2
π⋅ D
π⋅ D
− ρ1⋅ g⋅ Δh⋅
+ π⋅ D⋅ σ⋅ cos ( θ)
4
4
where Δp is the pressure difference generated by fluid 2 over height Δh,
2
π⋅ D
Hence
Δp⋅
Solving for Δh
Δh = −
4
σπDcosθ
2
− ρ1⋅ g⋅ Δh⋅
π⋅ D
4
Δp = ρ2⋅ g⋅ Δh
2
= ρ2⋅ g⋅ Δh⋅
π⋅ D
4
2
− ρ1⋅ g⋅ Δh⋅
π⋅ D
4
= −π⋅ D⋅ σ⋅ cos ( θ)
4⋅ σ⋅ cos ( θ)
g⋅ D⋅ ρ2 − ρ1
(
)
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
2
Δh = −4 × 0.375⋅
Δh = 0.360⋅ in
3
3
N
lbf
0.0254m
s
1
ft
12⋅ in ⎞
1
slugft
⋅
×
×
× cos ( 140⋅ deg) ×
×
×
× ⎛⎜
×
⎟ ×
m 4.448⋅ N
in
32.2⋅ ft 0.040⋅ in 1.94⋅ slug ⎝ ft ⎠
( 13.6 − 1) lbf s2
⋅
Problem 3.34
Problem
3.40
[Difficulty: 2]
3.34
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⋅ Δh⋅
π⋅ D
4
Assumption: Neglect meniscus curvature for column height and volume calculations
Solving for Δh
Δh =
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⋅ Δh⋅ w⋅ a
Solving for Δh
Δh =
2⋅ σ⋅ cos ( θ)
ρ⋅ g⋅ a
For water σ = 72.8 mN/m and θ = 0o (Table A.4), so
N
4 × 0.0728⋅
a) Tube
Δh =
999⋅
kg
3
× 9.81⋅
m
m
m
Δh =
999⋅
kg
3
m
2
Δh = 5.94 × 10
−3
m
Δh = 5.94⋅ mm
m
Δh = 2.97⋅ mm
N⋅ s
× 0.005⋅ m
2
kg⋅ m
s
N
2 × 0.0728⋅
b) Parallel Plates
×
× 9.81⋅
m
2
s
m
× 0.005⋅ m
×
kg⋅ m
2
N⋅ s
Δh = 2.97 × 10
−3
Problem 3.35
3.35
p SL =
R =
ρ=
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)
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
T (oC)
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
T (K)
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 3.36
Problem
3.44
[Difficulty: 3]
3.36
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
= − ρ⋅ g
dz
Governing Equations:
Assumptions:
p = ρ⋅ R ⋅ T
(1) Static fluid
(2) Ideal gas behavior
(a) For an incompressible atmosphere:
dp
= − ρ⋅ g
dz
At
z
⌠
p − p0 = −⎮ ρ⋅ g dz
⌡0
becomes
p
(b) For an adiabatic atmosphere:
ρ
dp
= − ρ⋅ g
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 = ρ0⋅ ⎛⎜ ⎟⎞
p0
⎝ ⎠
= const
p
dp = −ρ0⋅ ⎛⎜ ⎟⎞ ⋅ g⋅ dz
⎝ p0 ⎠
287⋅ N ⋅ m
or
1
p
p
But
k−1
⌠
1
k
k
⎮
⋅ p − p0
dp =
⎮
1
k−1
⎮
k
⎮ p
⌡p
(
)
1
×
298⋅ K
N ⋅ s ⎟⎞
kg⋅ m ⎟
2
×
p = 57.5⋅ kPa
⎠
1
k
1
k
becomes
kg⋅ K
(1)
1
k
dp = −
ρ0⋅ g
p0
1
k
⋅ dz
k−1 ⎞
⎛ k−1
ρ0⋅ 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
⎠
⎛
p = 83.2⋅ kPa × ⎜ 1 −
⎜
⎝
k
k−1
or
p
k − 1 g⋅ z ⎞
= ⎛⎜ 1 −
⋅
p0
k R ⋅ T0 ⎟
⎝
k
k−1
(2)
⎠
1.4 − 1
m
kg⋅ K
1
N⋅ s ⎞⎟
× 9.81⋅ × 2690⋅ m ×
×
×
2
1.4
287⋅ N⋅ m 298⋅ K kg⋅ m ⎟
s
2
⎠
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 3.37
(Difficulty: 2)
3.37 If atmospheric pressure at the ground is 101.3 𝑘𝑘𝑘 and temperature is 15 ℃, calculate the
pressure 7.62 𝑘𝑘 above the ground, assuming (a) no density variation, (b) isothermal variation of
density with pressure, and (c) adiabatic variation of density with pressure.
Assumption: Atmospheric air is stationary and behaves as an ideal gas.
Solution: Use the hydrostatic relation to find the pressures in the fluid
Governing equation: Hydrostatic pressure in a liquid, with z measured upward:
𝑑𝑑
= −𝜌 𝑔 = −𝛾
𝑑𝑑
(a) For this case with no density variation, we integrate with respect to z from the ground level pressure
p0 to the pressure at any height h. The pressure is
𝑝 = 𝑝0 − 𝛾ℎ
From Table A.10, the density of air at sea level is
𝜌 = 1.23
Or the specific weight is
𝛾 = 𝜌𝜌 = 1.23
Thus the pressure at 7.62 km is
𝑘𝑘
𝑚3
𝑘𝑘
𝑚
𝑁
×
9.81
=
12.07
𝑚3
𝑠2
𝑚3
𝑝 = 101.3 𝑘𝑘𝑘 − 12.07
𝑁
× 7.62 × 1000 𝑚 = 9.63 𝑘𝑘𝑘
𝑚3
(b) For isothermal condition we have for an ideal gas:
𝑝 𝑝0
=
= 𝑅𝑅 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝜌 𝜌0
Therefore, since ρ = γ g and g is a constant
𝑝 𝑝0 101.3 𝑘𝑘𝑘
=
=
= 8420 𝑚 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝑁
𝛾 𝛾0
12.07 3
𝑚
From the hydrostatic relation we have:
𝑑𝑑 = −𝛾𝛾𝛾
𝛾
𝑑𝑑
= − 𝑑𝑑
𝑝
𝑝
𝑝
𝑧
𝑑𝑑
1
=−
� 𝑑𝑑
8420𝑚 0
𝑝0 𝑝
1
𝑝
𝑧
ln � � = −
8420𝑚
𝑝0
�
Thus the pressure at 7.62 km is
7620 𝑚
𝑝
= 𝑒 − − 8420𝑚 = 𝑒 − 0.905 = 0.4045
𝑝0
𝑝 = 101.3𝑘𝑘𝑘 × 0.4045 = 41.0 𝑘𝑘𝑘
(c) For a reversible and adiabatic variation of density we have:
𝑝
𝑝𝑣 𝑘 = 𝑘 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝜌
Where k is the specific heat ratio
𝑘 = 1.4
Or, since gravity g is constant, we can write in terms of the specific weight
𝑝0
𝑝
= 𝑘 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝑘
𝛾
𝛾0
Or the specific weight is
1
𝑝 �𝑘
𝛾 = 𝛾0 � �
𝑝0
The hydrostatic expression becomes
1
𝑝 �𝑘
𝑑𝑑 = −𝛾0 � � 𝑑𝑑
𝑝0
Separating variables
1/𝑘 𝑝
𝑧
𝑑𝑑
𝑝0
�
=
−
�
𝑑𝑑
𝛾0 𝑝0 (𝑝)1/𝑘
0
Integrating between the limits p=p0 at z=0 and p = p at z = z
1/𝑘
Or
𝑘−1
𝑘−1
𝑝
𝑘
� 0 �𝑝 𝑘 − 𝑝0 𝑘 � = − 𝑧
�
𝑘 − 1 𝛾0
𝑘−1
𝑘
The pressure is then
𝑘 − 1 𝛾0 𝑧
�
�
𝑝 = 𝑝0 �1 − �
𝑘
𝑝0
𝑘�
𝑘−1
𝑝
� �
𝑝0
=1−�
𝑘 − 1 𝛾0 𝑧
�
𝑘
𝑝0
1.4�
1.4−1
𝑁
1.4 − 1
3 × 7620𝑚
𝑚
= 101.3𝑘𝑘𝑘 �1 − �
�×
�
1.4
101.3 × 1000 𝑃𝑃
𝑝 = 35.4 𝑘𝑘𝑘
12.07
The calculation of pressure depends heavily on the assumption we make about how density
changes.
Problem 3.38
(Difficulty: 2)
3.38 If the temperature in the atmosphere is assumed to vary linearly with altitude so T = T0 - αz where
T0 is the sea level temperature and α = - dT / dz is the temperature lapse rate, find p(z) when air is taken
to be a perfect gas. Give the answer in terms of p0, a, g, R, and z only.
Assumption: Atmospheric air is stationary and behaves as an ideal gas.
Solution: Use the hydrostatic relation to find the pressures in the fluid
Governing equation: Hydrostatic pressure in a liquid, with z measured upward:
𝑑𝑑 = −𝛾𝛾𝛾
The ideal gas relation is
Or in terms of the specific weight, the pressure is
𝑝
= 𝑅𝑅
𝜌
𝑝 = 𝜌𝜌𝜌 =
Relating the temperature to the adiabatic lapse rate
𝑝=
𝛾
𝑅𝑅
𝑔
𝛾
𝑅(𝑇0 − 𝛼𝛼)
𝑔
Inserting the expression for specific weight into the hydrostatic equation
𝑑𝑑 = −
Separating variables
𝑔𝑔
𝑑𝑑
𝑅(𝑇0 − 𝛼𝛼)
𝑔
𝑑𝑑
𝑑𝑑
= −
𝑅 (𝑇0 − 𝛼𝛼)
𝑝
Integrating between the surface and any height z
𝑝
Or
𝑑𝑑
𝑔 𝑧
𝑑𝑑
= − �
𝑅 0 (𝑇0 − 𝛼𝛼)
𝑝0 𝑝
�
𝑔
𝑇0 − 𝛼𝛼
𝑝
�
𝑙𝑙 � � = − 𝑙𝑙 �
𝑅
𝑝0
𝑇0
In terms of p
𝑔�
𝛼𝛼
𝛼𝛼
𝑝
= �1 − �
𝑇0
𝑝0
Problem 3.39
Problem
3.46
[Difficulty: 3]
3.39
Given:
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:
b
dp
= ρ⋅ g
dy
(Hydrostatic Pressure - y is positive downwards)
⌠
⎮
FR = ⎮ p dA
⌡
⌠
⎮
y'⋅ FR = ⎮ y⋅ p dA
⌡
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:
(
)
c+ a
1 ⌠
1 ⌠
⎮
y' =
y⋅ p dA =
⋅⎮
y⋅ psg + ρ⋅ g⋅ y ⋅ b dy
FR ⎮
FR ⌡
⌡
(
c
y' =
)
b ⎡ psg ⎡
ρ⋅ g ⎡
2
2
3
3⎤
⎢ ⎣( c + a) − c ⎤⎦ +
⋅ ⎣( c + a) − c ⎤⎦⎥
FR ⎣ 2
3
⎦
)
p = psg + ρ⋅ g⋅ y
( 1)
Simplifying: y' =
(
)
⎤
b ⎡ psg 2
ρ⋅ g ⎡ 3
⋅⎢
a + 2⋅ a⋅ c +
⋅ ⎣a + 3⋅ a⋅ c⋅ ( a + c)⎤⎦⎥
FR ⎣ 2
3
⎦
For part (a) we know
psg = 0
so substituting into (1) we get:
( 2)
Fo =
(
)
ρ⋅ g⋅ b 2
⋅ a + 2⋅ a⋅ c
2
2
Fo =
1
kg
m
N⋅ s
2
× 999⋅
× 9.81⋅ × 1⋅ m × ⎡⎣( 1.5⋅ m) + 2 × 1.5⋅ m × 1⋅ m⎤⎦ ×
3
2
kg⋅ m
2
m
s
y' =
Substituting into (2) for the line of action we get:
Fo = 25.7⋅ kN
ρ⋅ g⋅ b ⎡ 3
⋅ ⎣a + 3⋅ a⋅ c⋅ ( a + c)⎤⎦
3⋅ Fo
2
y' =
1
kg
m
1
1
N⋅ s
3
× 999⋅
× 9.81⋅ × 1⋅ m⋅
⋅ × ⎡⎣( 1.5⋅ m) + 3 × 1.5⋅ m × 1⋅ m × ( 1.5⋅ m + 1⋅ m)⎤⎦ ×
3
2
3 N
3
kg⋅ m
m
s
25.7 × 10
y' = 1.9 m
For part (b) we know
psg = 0.3⋅ atm . Substituting into (1) we get:
⎡
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
N⋅ s ⎤⎥
kg⋅ m⎥
2
⎦
FR = 71.3⋅ kN
Substituting ⎡into (2) for the line of action we get:
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 +
× ⎡⎣( 1.5) + 3⋅ 1.5⋅ 1⋅ ( 1.5 + 1)⎤⎦ ⋅ m ×
1⋅ m ×
⎣
⎦
⎢ 2
2
3
kg⋅ m⎥
m ⋅ atm
⎣
⎦
y' =
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
For the gate
⎡
⎣
b⋅ ⎢psg⋅ a +
yc = c +
(
)
ρ⋅ g 2
⎤
⋅ a + 2⋅ a⋅ c ⎥
2
⎦
(
)
ρ⋅ g⋅ b 2
⋅ a + 2⋅ a⋅ c
2
a
2
= 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
(
2
⋅
c
+
a
)
ρ
⋅
g
2
⎤⎤
⎦
⎡b⋅ ⎡p ⋅ a +
⋅ (a + 2⋅ a⋅ c)⎥⎥
⎢ ⎢ sg
2
⎣ ⎣
⎦⎦
Simplifying this expression we get:
y'
2
=
⋅
yc
( 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
⋅ a + 2⋅ a⋅ c
2
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 3.40
Problem
3.48
[Difficulty: 5]
3.40
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 3.41
Problem
3.50
[Difficulty: 3]
3.41
Given:
Geometry of gate
Find:
Force FA for equilibrium
h
H = 25 ft
FA
A
R = 10 ft
y
y
B
x
z
Solution:
⌠
⎮
FR = ⎮ 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 dA
⎮
⌡
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 ⎝
⎛ 2⋅ H⋅ R π⋅ R ⎞
ρ⋅ g ⎛ 2⋅ H⋅ R
π⋅ R ⎞
⎟ = ρ⋅ g⋅ ⎜
⎟
⋅⎜
−
−
R ⎝ 3
8 ⎠
8 ⎠
⎝ 3
3
FR =
Using given data
FR = 1.94⋅
slug
ft
3
× 32.2⋅
4
2
3
2
2
2 π
3⎤ lbf ⋅ s
× ⎡⎢ × 25⋅ ft × ( 10⋅ ft) − × ( 10⋅ ft) ⎥ ×
2 ⎣3
8
⎦ slug⋅ ft
s
ft
4
FR = 7.96 × 10 ⋅ lbf
Problem 3.42
(Difficulty: 2)
3.42 A circular gate 3 𝑚 in diameter has its center 2.5 𝑚 below a water surface and lies in a plane
sloping at 60°. Calculate magnitude, direction and location of total force on the gate.
Find: The direction, magnitude of the total force 𝐹.
Assumptions: Fluid is static and incompressible
Solution: Apply the hydrostatic relations for pressure, force, and moments, with y measured from the
surface of the liquid:
𝑑𝑑
= 𝜌𝑔=𝛾
𝑑𝑑
𝐹𝑅 = � 𝑝 𝑑𝑑
For the magnitude of the force we have:
A free body diagram of the gate is
𝑦 ′ 𝐹𝑅 = � 𝑦 𝑝 𝑑𝑑
𝐹 = � 𝑝𝑝𝑝
𝐴
The pressure on the gate is the pressure at the centroid, which is yc = 2.5 m. So the force can be
calculated as:
𝐹 = 𝜌𝜌ℎ𝑐 𝐴 = 999
𝑘𝑘
𝑚
𝜋
× 9.81 2 × 2.5 𝑚 × × (3 𝑚)2 = 173200 𝑁 = 173.2 𝑘𝑘
3
𝑚
𝑠
4
The direction is perpendicular to the gate.
For the location of the force we have:
𝑦 ′ = 𝑦𝑐 +
𝐼𝑥�𝑥�
𝐴𝑦𝑐
The y axis is along the plate so the distance to the centroid is:
𝑦𝑐 =
The area moment of inertia is
The area is
𝐼𝑥�𝑥� =
𝐴=
So
𝑦 ′ = 2.89 𝑚 +
2.5 𝑚
= 2.89 𝑚
sin 60°
𝜋𝐷 4
𝜋
=
× (3 𝑚)4 = 3.976 𝑚4
64
64
𝜋 2 𝜋
𝐷 = × (3 𝑚)2 = 7.07 𝑚2
4
4
3.976 𝑚4
= 2.89 𝑚 + 0.1946 𝑚 = 3.08 𝑚
7.07 𝑚2 × 2.89 𝑚
The vertical location on the plate is
ℎ′ = 𝑦 ′ sin 60° = 3.08 𝑚 ×
The force acts on the point which has the depth of 2.67 𝑚.
√3
= 2.67 𝑚
2
Problem 3.43
(Difficulty: 2)
3.43 For the situation shown, find the air pressure in the tank in psi. Calculate the force exerted on the
gate at the support B if the gate is 10 𝑓𝑓 wide. Show a free body diagram of the gate with all the forces
drawn in and their points of application located.
Assumptions: Fluid is static and incompressible
Solution: Apply the hydrostatic relations for pressure and force, and the static relation for moments:
The specfic weight for water is:
𝑑𝑑
= 𝜌𝑔=𝛾
𝑑𝑑
𝛾 = 62.4
𝑙𝑙𝑙
𝑓𝑓 3
The pressure of the air equals that at the surface of the water in the tank. As shown by the manometer,
the pressure at the surface is less than atmospheric due to the three foot head of water. The gage
pressure of the air is then:
𝑝𝑎𝑎𝑎 = −𝛾ℎ = −62.4
A free body diagram for the gate is
𝑙𝑙𝑙
𝑙𝑙𝑙
× 3𝑓𝑓 = −187.2 2
3
𝑓𝑓
𝑓𝑓
For the force in the horizontal direction, we have:
𝐹1 = 𝛾ℎ𝑐 𝐴 = 62.4
𝑙𝑙𝑙
× 3 𝑓𝑓 × (6 𝑓𝑓 × 10 𝑓𝑓) = 11230 𝑙𝑙𝑙
𝑓𝑓 3
𝐹2 = 𝑝𝑎𝑎𝑎 𝐴 = −187.2
𝑙𝑙𝑙
× (8 𝑓𝑓 × 10 𝑓𝑓) = 14980 𝑙𝑙𝑙
𝑓𝑓 2
With the momentume balance about hinge we have:
� 𝑀 = 𝐹1 ℎ𝑐 − 𝑃ℎ − 𝐹2
So the force exerted on B is:
ℎ
= 11230 𝑙𝑙𝑙 × 6𝑓𝑓 − 𝑃 × 8𝑓𝑓 − 14980 𝑙𝑙𝑙 × 4𝑓𝑓 = 0
2
𝑃 = 933 𝑙𝑙𝑙
Problem 3.44
(Difficulty: 3)
3.44 What is the pressure at A? Draw a free body diagram of the 10 ft wide gate showing all forces and
locations of their lines of action. Calculate the minimum force 𝑃 necessary to keep the gate closed.
Given: All the parameters are shown in the figure.
Find: The pressure 𝑝𝐴 . The minimum force 𝑃 necessary to keep the gate closed.
Assumptions: Fluid is static and incompressible
Solution: Apply the hydrostatic relations for pressure, force, and moments, with y measured from the
surface of the liquid:
𝑑𝑑
= 𝜌𝑔=𝛾
𝑑𝑑
𝐹𝑅 = � 𝑝 𝑑𝑑
The specfic weight of the water is:
𝑦 ′ 𝐹𝑅 = � 𝑦 𝑝 𝑑𝑑
𝛾𝑤𝑤𝑤𝑤𝑤 = 62.4
𝑙𝑙𝑙
𝑓𝑓 3
The gage pressure at A is given by integrating the hydrostatic relation:
𝑝𝐴 = 𝛾𝑜𝑜𝑜 ℎ𝐴 = 𝑆𝑆𝛾𝑜𝑖𝑖 ℎ𝐴 = 0.9 × 62.4
𝑙𝑙𝑙
𝑙𝑙𝑙
× 6 𝑓𝑓 = 337 2
3
𝑓𝑓
𝑓𝑓
A free body diagram of the gate is
The horizontal force F1 as shown in the figure is given by the pressure at the centroid of the submerged
area (3 ft):
𝐹1 = 𝛾𝑜𝑜𝑜 ℎ𝑐 𝐴 = 0.9 × 62.4
𝑙𝑙𝑙
× 3 𝑓𝑓 × (6 𝑓𝑓 × 10 𝑓𝑓) = 10110 𝑙𝑙𝑙
𝑓𝑓 3
The vertical force F2 is given by the pressure at the depth of the surface (4 ft)
𝐹2 = 𝑝𝐴 𝐴 = 337
𝑙𝑙𝑙
× (4𝑓𝑓 × 10𝑓𝑓) = 13480 𝑙𝑙𝑙
𝑓𝑓 2
The force F1 acts two-thirds of the distance down from the water surface and the force F2 acts at the
centroid..
Taking the moments about the hinge:
−𝐹1 × 6 𝑓𝑓−𝐹2 × 2 𝑓𝑓 + 𝑃 × 4 𝑓𝑓 = 0
So we have for the force at the support:
𝑃=
10110 𝑙𝑙𝑙 × 6𝑓𝑓 + 13480 𝑙𝑙𝑙 × 2𝑓𝑓
= 21900 𝑙𝑙𝑙
4 𝑓𝑓
Problem 3.45
Problem
3.52
[Difficulty: 3]
3.45
Given:
Geometry of plane gate
Find:
Minimum weight to keep it closed
L=3m
h
y
L/2
dF
W
w=2m
Solution:
⌠
⎮
FR = ⎮ p dA
⌡
Basic equation
or, use computing equations
dp
= ρ⋅ g
dh
ΣMO = 0
FR = pc ⋅ A
Ixx
y' = yc +
A ⋅ yc
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
2
⎮
⎮
⋅
y⋅ p dA =
⋅
y⋅ ρ⋅ g⋅ y⋅ sin ( θ) ⋅ w dy
⎮
L⋅ cos ( θ) ⌡
L⋅ cos ( θ) ⎮
⌡
L
⌠
2
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
kg
m
N⋅ s
2
⋅ 1000⋅
× 9.81⋅ × 2⋅ m × ( 3⋅ m) × tan( 30⋅ deg) ×
3
2
3
kg⋅ m
m
s
W = 68⋅ kN
Problem 3.46
Problem
3.54
[Difficulty: 3]
3.46
Given:
Gate geometry
Find:
Depth H at which gate tips
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
and
Ixx =
w⋅ L
12
with
yc = H −
L
2
where L = 1 m is the plate height and w is the plate width
Hence
L⎞
y' = ⎛⎜ H − ⎟ +
2⎠
⎝
3
2
L⎞
L
= ⎛⎜ H − ⎟ +
L⎞
L⎞
2⎠
⎝
12⋅ w⋅ L⋅ ⎛⎜ H − ⎟
12⋅ ⎛⎜ H − ⎟
2
2⎠
⎝
⎠
⎝
w⋅ L
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
2
L
L
Combining the two equations ⎛⎜ H − ⎞⎟ +
2⎠
⎝
12⋅ ⎛⎜ H −
⎝
Solving for H
H ≤
L
+
2
L
L⎞
⎟
2⎠
≥ H − 0.45⋅ m
2
L
⎞
12⋅ ⎛⎜ − 0.45⋅ m⎟
⎝2
⎠
H ≤
1⋅ m
+
2
2
( 1⋅ m)
1
⋅m
⎞
12 × ⎛⎜
− 0.45⋅ m⎟
⎝ 2
⎠
H ≤ 2.17⋅ m
Problem 3.47
Problem
3.56
[Difficulty: 3]
3.47
Given:
Geometry of lock system
Find:
Force on gate; reactions at hinge
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
and
w =
W
2⋅ cos ( 15⋅ deg)
⌠
⌠
⎮
⎮
FR = ⎮ p dA = ⎮ ρ⋅ g⋅ y dA
⌡
⌡
but
Fn
dA = w⋅ dy
h
Hence
2
⌠
ρ⋅ g⋅ w⋅ h
FR = ⎮ ρ⋅ g⋅ y⋅ w dy =
⌡0
2
Alternatively
FR = pc⋅ A
Using given data
FR =
2
and
h
ρ⋅ g⋅ w⋅ h
FR = pc⋅ A = ρ⋅ g⋅ yc⋅ A = ρ⋅ g⋅ ⋅ h⋅ w =
2
2
2
1
kg
m
34⋅ m
2 N⋅ s
⋅ 1000⋅
× 9.81⋅ ×
× ( 10⋅ m) ×
3
2
2
2⋅ cos ( 15⋅ deg)
kg⋅ m
m
s
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 3.48
(Difficulty: 2)
3.48 Calculate the minimum force 𝑃 necessary to hold a uniform 12 𝑓𝑓 𝑠𝑠𝑠𝑠𝑠𝑠 gate weighing
500 𝑙𝑙𝑙closed on a tank of water under a pressure of 10 𝑝𝑝𝑝. Draw a free body of the gate as part of
your solution.
Given: All the parameters are shown in the figure.
Find: The minimum force 𝑃 to hold the system.
Assumptions: Fluid is static and incompressible
Solution: Apply the hydrostatic relations for pressure, force, and moments, with y measured from the
surface of the liquid:
𝑑𝑑
= 𝜌𝑔=𝛾
𝑑𝑑
𝐹𝑅 = � 𝑝 𝑑𝑑
A free body diagram of the gate is
𝑦 ′ 𝐹𝑅 = � 𝑦 𝑝 𝑑𝑑
The gage pressure of the air in the tank is:
𝑝𝑎𝑎𝑎 = 10 𝑝𝑝𝑝 = 1440
This produces a uniform force on the gate of
𝐹1 = 𝑝𝑎𝑎𝑎 𝐴 = 1440
𝑙𝑙𝑙
𝑓𝑓 2
𝑙𝑙𝑙
× (12 𝑓𝑓 × 12 𝑓𝑓) = 207360 𝑙𝑙𝑙
𝑓𝑓 2
This pressure acts at the centroid of the area, which is the center of the gate. In addition, there is a force
on the gate applied by water. This force is due to the pressure at the centroid of the area. The depth of
the centroid is:
𝑦𝑐 =
The force is them
𝐹2 = 𝛾ℎ𝑐 𝐴 = 62.4
12 𝑓𝑓
× sin 45°
2
𝑙𝑙𝑙 12 𝑓𝑓
×
× sin 45° × 12 𝑓𝑓 × 12 𝑓𝑓 = 38123 𝑙𝑙𝑙
𝑓𝑓 3
2
The force F2 acts two-thirds of the way down from the hinge, or 𝑦 ′ = 8 𝑓𝑓.
Take the moments about the hinge:
𝐿
𝐿
−𝐹𝐵 sin 45° + 𝐹1 + 𝐹2 × 8 𝑓𝑓 − 𝑃 × 12 𝑓𝑓 = 0
2
2
Thus
𝑃=
−500 𝑙𝑙𝑙 × 6 𝑓𝑓 × sin 45° + 207360 𝑙𝑙𝑙 × 6 𝑓𝑓 + 38123 𝑙𝑙𝑙 × 8 𝑓𝑓
= 128900 𝑙𝑙𝑙
12 𝑓𝑓
Problem 3.49
(Difficulty: 2)
3.49 Calculate magnitude and location of the resultant force of water on this annular gate.
Given: All the parameters are shown in the figure.
Find: Resultant force of water on this annular gate.
Assumptions: Fluid is static and incompressible
Solution: Apply the hydrostatic relations for pressure, force, and moments, with y measured from the
surface of the liquid:
𝑑𝑑
= 𝜌𝑔=𝛾
𝑑𝑑
𝐹𝑅 = � 𝑝 𝑑𝑑
For the magnitude of the force we have:
𝑦 ′ 𝐹𝑅 = � 𝑦 𝑝 𝑑𝑑
𝐹 = � 𝑝𝑝𝑝 = 𝜌𝜌ℎ𝑐 𝐴
𝐴
The pressure is determined at the location of the centroid of the area
𝐴=
𝐹 = 999
ℎ𝑐 = 1 𝑚 + 1.5 𝑚 = 2.5 𝑚
𝜋
𝜋 2
(𝐷2 − 𝐷12 ) = ((3 𝑚)2 − (1.5 𝑚)2 ) = 5.3014 𝑚2
4
4
𝑚
𝑘𝑘
× 9.81 2 × 2.5 𝑚 × 5.3014 𝑚2 = 129900 𝑁 = 129.9 𝑘𝑘
3
𝑠
𝑚
The y axis is in the vertical direction. For the location of the force, we have:
𝑦 ′ = 𝑦𝑐 +
Where:
𝐼𝑥�𝑥� =
𝐼𝑥�𝑥�
𝐴𝑦𝑐
𝑦𝑐 = 2.5 𝑚
𝜋(𝐷24 − 𝐷14 )
𝜋
=
× ((3 𝑚)4 − (1.5 𝑚)4 ) = 3.7276 𝑚4
64
64
𝑦 ′ = 𝑦𝑐 +
𝐼𝑥�𝑥�
3.7276 𝑚4
= 2.5 𝑚 +
= 2.78 𝑚
𝐴𝑦𝑐
2.5 𝑚 × 5.3014 𝑚2
So the force acts on the depth of 𝑦 ′ = 2.78 𝑚.
Problem 3.50
(Difficulty: 2)
3.50 A vertical rectangular gate 2.4 𝑚 wide and 2.7 𝑚 high is subjected to water pressure on one side,
the water surface being at the top of the gate. The gate is hinged at the bottom and is held by a
horizontal chain at the top. What is the tension in the chain?
Given: The gate wide: 𝑤 = 2.4 𝑚. Height of the gate: ℎ = 2.7 𝑚.
Find: The tension 𝐹𝑐 in the chain.
Assumptions: Fluid is static and incompressible
Solution: Apply the hydrostatic relations for pressure, force, and moments, with y measured from the
surface of the liquid:
𝑑𝑑
= 𝜌𝑔=𝛾
𝑑𝑑
𝐹𝑅 = � 𝑝 𝑑𝑑
For the magnitude of the force we have:
Where hc is the depth at the centroid
𝑦 ′ 𝐹𝑅 = � 𝑦 𝑝 𝑑𝑑
𝐹 = � 𝑝𝑝𝑝 = 𝜌𝜌ℎ𝑐 𝐴
𝐴
ℎ𝑐 =
2.7 𝑚
= 1.35 𝑚
2
𝐴 = 𝑤ℎ = 2.4 𝑚 × 2.7 𝑚 = 6.48 𝑚2
𝐹 = 999
𝑘𝑘
𝑚
× 9.81 2 × 1.35 𝑚 × 6.48 𝑚2 = 85.7 𝑘𝑘
3
𝑚
𝑠
The y axis is in the vertical direction. For the location of the force, we have:
Taking the momentum about the hinge:
𝐹𝑐 = 𝐹
ℎ𝑝 =
2
× 2.7 𝑚 = 1.8 𝑚
3
𝐹�ℎ − ℎ𝑝 � − 𝐹𝑐 ℎ = 0
�ℎ − ℎ𝑝 �
0.9 𝑚
= 85.7 𝑘𝑘 ×
= 28.6 𝑘𝑘
2.7 𝑚
ℎ
Problem 3.51
Problem
3.58
[Difficulty: 4]
3.51
Given:
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:
Assumptions:
dp
= ρ⋅ g
dh
(Hydrostatic Pressure - h is positive downwards)
⌠
⎮
FR = ⎮ p dA
⌡
(Hydrostatic Force on door)
⌠
⎮
y'⋅ FR = ⎮ y⋅ p dA
⌡
(First moment of force)
ΣM = 0
(Rotational equilibrium)
(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
FD =
− FD ⋅ a +
h
d
a
w
for h > d
for h < d
d = a−c
where
Summing moments around the hinge:
b
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
a−h
=
b
a
h
b
Therefore: w = ( a − h)
a
dF = pdA
a
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
3 kg
3
ρ = 2.5 × 10
m
m
( 1)
d
0.15
=
= 0.375
a
0.4
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
c
= 1−
a
a
d = a−c
FD = 32.9 N
In addition, we can maximize the force by the maximum force
(when c = a or d = 0):
2
1 1⎞
ρ⋅ g⋅ b⋅ a
2
Fmax = ρ⋅ g⋅ b⋅ a ⋅ ⎛⎜ − + ⎟ =
12
⎝ 4 3⎠
and so
4
3
2
⎡ 1⎡
d⎞ ⎤ 1
d⎞ ⎡
d⎞ ⎤ 1 d ⎡
d ⎞ ⎤⎤
= 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 3.52
Problem
3.60
[Difficulty: 2]
3.52
Given:
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:
Assumptions:
dp
=γ
dh
(Hydrostatic Pressure - y is positive downwards)
FR = 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:
p = γ⋅ h
π 2
FR = pc⋅ A = γ⋅ hc⋅ ⋅ D
4
FR = 62.4⋅
lbf
ft
For a circular area:
3
× 12⋅ ft ×
π 4
⋅D
2
π 4
64
D
Ixx =
⋅ D Therefore: y' = yc +
= yc +
64
16⋅ yc
π 2
⋅ D ⋅ yc
4
π
4
× ( 6⋅ ft)
2
4
FR = 2.12 × 10 ⋅ lbf
2
y' = 12⋅ ft +
( 6⋅ ft)
16 × 12⋅ ft
y' = 12.19⋅ ft
The force of water is to the right and
perpendicular to the plug.
Problem 3.53
Problem
3.62
[Difficulty: 2]
3.53
Given:
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 ×
s
π
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:
⎛ FR ⎞
⎟
⎝ 2⋅ π⋅ σ ⎠
A = 8×
π 2
2
⋅ db = 2⋅ π⋅ db
4
1
2
db = ⎜
1
2
2
3
⎛ 1
1
m ⎟⎞
10 ⋅ mm
3
⎜
db =
× 33.3 × 10 ⋅ N ×
⋅
×
6
m
⎜2× π
100 × 10 N ⎟⎠
⎝
db = 7.28⋅ mm
Problem 3.54
Problem
3.64
[Difficulty: 3]
3.54
Given:
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:
Assumptions:
dp
= ρ⋅ g
dh
(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
3
2
Ixx
b⋅ L1
L1
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
2
h'1 = 6⋅ ft +
s
(
h'1 = 8⋅ ft
2
× 12⋅ ft × 6⋅ ft × 6⋅ ft ×
lbf ⋅ s
slug⋅ ft
Since the pressure is uniform over OC, the force acts at the centroid of OC, i.e.,
Summing moments about the hinge gives:
( 12⋅ ft)
12 × 6⋅ ft
)
(
F2 = 27.0⋅ kip
FAB
x'2 = 3⋅ ft
)
FAB⋅ L1 + L3 − F1⋅ L1 − h'1 + F2⋅ x'2 = 0
L1
h1 ’
L1
Solving for the force in the bar:
Substituting in values:
FAB =
(
)
F1⋅ L1 − h'1 − F2⋅ x'2
FAB =
L1 + L3
1
12⋅ ft + 3⋅ ft
FAB = 1800⋅ lbf
3
3
⋅ ⎡⎣27.0 × 10 ⋅ lbf × ( 12⋅ ft − 8⋅ ft) − 27.0 × 10 ⋅ lbf × 3⋅ ft⎤⎦
Thus bar AB is in compression
F1
L2
x2’
F2
Problem 3.55
Problem
3.66
[Difficulty: 3]
3.55
Given:
Geometry of gate
Find:
Force at A to hold gate closed
y
h
Solution:
Basic equation
Computing equations
D
y’
dp
= ρ⋅ g
dh
ΣMz = 0
FR = pc⋅ A
FR
Ixx
y' = yc +
A ⋅ yc
Ixx =
w⋅ L
12
FA
3
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
⎝
⎠
2
3
N⋅ s
⎞
FR = 1000⋅
× 9.81⋅ × ⎛⎜ 1.5 + sin ( 30⋅ deg)⎟ ⋅ m × 3⋅ m × 3⋅ m ×
3
2 ⎝
kg⋅ m
2
⎠
m
s
kg
m
FR = 199⋅ kN
Ixx
The location of this force is given by y' = yc +
where y' and y are measured along the plane of the gate to the free surface
A ⋅ yc
c
yc =
D
L
+
sin ( 30⋅ deg) 2
yc =
1.5⋅ m
3⋅ m
+
sin ( 30⋅ deg)
2
yc = 4.5 m
3
2
2
Ixx
w⋅ L
L
( 3⋅ m)
1 1
y' = yc +
= yc +
⋅
⋅
= yc +
= 4.5⋅ m +
A ⋅ yc
12⋅ yc
12⋅ 4.5⋅ m
12 w⋅ L yc
Taking moments about the hinge
y' = 4.67 m
D
⎞
ΣMH = 0 = FR⋅ ⎛⎜ y' −
⎟ − FA ⋅ L
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 3.56
Problem
3.68
[Difficulty: 4]
3.56
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
D
2
3
12⋅ w⋅ D⋅
2
y
y = D − y' =
Also
m = ρcement⋅ g ⋅ b ⋅ D⋅ w = SG ⋅ ρ⋅ g ⋅ b ⋅ D⋅ w
mg
O
D
so
FH
b
3
b
M 0. = 0 = −FH⋅ y + ⋅ m⋅ g
2
Taking moments about O
∑
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=
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
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.
D
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
D
2
w⋅ D
y' = y c +
=
+
= ⋅D
A⋅ y c
D
2
3
12⋅ w⋅ D⋅
2
D
so
y = D − y' =
Also
FV = ρ⋅ V⋅ g = ρ⋅ g ⋅
3
α⋅ b ⋅ D
2
⋅w =
1
2
⋅ ρ⋅ g ⋅ α⋅ b ⋅ D⋅ w
x = ( b − α⋅ b ) +
2
3
α⎞
⋅ α⋅ b = b ⋅ ⎛⎜ 1 −
⎝
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
D
1
α
1
2
−⎛⎜ ⋅ ρ⋅ g ⋅ D ⋅ w⎞ ⋅ + ⎛⎜ ⋅ ρ⋅ g ⋅ α⋅ b ⋅ D⋅ w⎞ ⋅ b ⋅ ⎛⎜ 1 − ⎞ ...
3⎠
⎝2
⎠ 3 ⎝2
⎠ ⎝
1
2
2⋅ α ⎞ ⎡ 1
⎛
⎛
⎞
+ ⎜ ⋅ SG ⋅ ρ⋅ g ⋅ α⋅ b ⋅ D⋅ w ⋅ b ⋅ ⎜ 1 −
+ ⎢ ⋅ SG ⋅ ρ⋅ g ⋅ ( 1 − α) ⋅ b ⋅ D⋅ w⎤⎥ ⋅ ⋅ b ( 1 − α)
2
3
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=
The cross-section area is
A=
For a general triangle
A=
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
2
A = 0.226 ⋅ D
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 3.57
Problem
3.70
[Difficulty: 2]
3.57
Given:
Geometry of dam
Find:
Vertical force on dam
Assumptions:
(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:
Basic equation:
dp
= ρ⋅ g
dh
For incompressible fluid
p = ρ⋅ g⋅ h
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)
FT = p⋅ A = Σ ρ⋅ g⋅ h⋅ d⋅ w = ρ⋅ g⋅ d⋅ Σ h⋅ w
Starting with the top and working downwards
2
lbf ⋅ s
FT = 1.94⋅
× 32.2⋅ × 3⋅ ft × [ ( 3⋅ ft × 12⋅ ft) + ( 3⋅ ft × 6⋅ ft) − ( 9⋅ ft × 6⋅ ft) − ( 12⋅ ft × 12⋅ ft) ] ×
3
2
slug⋅ ft
ft
s
slug
ft
4
FT = −2.70 × 10 ⋅ lbf
The negative sign indicates a net upwards force (it's actually a buoyancy effect on the three middle sections)
Problem 3.58
Problem
3.72
[Difficulty: 3]
3.58
Given:
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:
Assumptions:
dp
= ρ⋅ g
dh
(Hydrostatic Pressure - h is positive downwards)
ΣMz = 0
(Rotational equilibrium)
⌠
⎮
Fv = ⎮ p dAy
⌡
(Vertical Hydrostatic Force)
⌠
⎮
x'⋅ Fv = ⎮ x dFv
⌡
(Location of line of action)
FH = 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
2
kg
m
⎛ 1 ⋅ m⎞ × N⋅ s
× 999⋅
× 9.81⋅ × 2⋅ m × ( 2⋅ m) × ⎜
⎟
3
2
3
kg⋅ m
⎝ 0.25 ⎠
m
s
Fv =
Substituting values:
1
2
3
2
⌠
1 ⌠
1 ⌠
⎮
⎮
⎮
x'⋅ Fv = ⎮ x dFv Therefore, x' =
⋅ ⎮ x dFv =
⋅ ⎮ x⋅ p dAy
Fv ⌡
Fv ⌡
⌡
To find the line of action of this force:
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 D c ⎛ D ⎞ ⎤
ρ⋅ g⋅ b D
⋅⎢ ⋅ − ⋅⎜ ⎟ ⎥ =
⋅
Fv ⎣ 2 c 4 ⎝ c ⎠ ⎦
Fv 4⋅ c
Evaluating the integral: x' =
(
s
)
Now substituting values into this equation:
2
1 1
1
N⋅ s
2
× × ( 2⋅ m) ×
⋅m ×
3 N 4
0.25
kg⋅ m
73.9 × 10
× 2⋅ m ×
Fv = 73.9⋅ kN
1
⋅
x' = 1.061 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
× 2⋅ m ×
s
hc =
since
D
2
Therefore the horizontal force is:
2
( 2⋅ m)
N⋅ s
×
2
kg⋅ m
FH = 39.2⋅ kN
To calculate the line of action of this force:
3
Ixx
D b⋅ D
1 2
D D
2
h' = hc +
=
+
⋅
⋅ =
+
= ⋅D
12 b⋅ D D
6
A ⋅ hc
2
2
3
h' =
2
3
⋅ 2⋅ m
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
Ox
FA =
FA
x’
1 1
⋅ × [ 73.9⋅ kN × 1.061⋅ m + 39.2⋅ kN × ( 2⋅ m − 1.333⋅ m) ]
3 m
x
Oy
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
L =
c
1
3⋅ m ×
⋅ m L = 3.464 m
0.25
1
1
⋅ × [ 73.9⋅ kN × 1.061⋅ m + 39.2⋅ kN × ( 2⋅ m − 1.333⋅ m) ]
3.464 m
L
x’
FA = 30.2⋅ kN
h’
Ox
FH
Oy
FA
FV
x
Problem 3.59
Problem
3.74
[Difficulty: 2]
3.59
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:
Assumptions:
dp
=γ
dh
(Hydrostatic Pressure - h is positive downwards)
⌠
⎮
Fv = −⎮ p dAy
⌡
(Vertical Hydrostatic Force)
⌠
⎮
x'⋅ Fv = ⎮ x dFv
⌡
(Moment of vertical force)
x’
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
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 ⎥
⎢
p dAy = −⎮ γ⋅ ⎣L − R − x ⎦ ⋅ b dx = −γ⋅ b ⋅ ⎮
L−
⌡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
x'⋅ Fv =
=
2
⎝
3
⌠
⎮
x dFv
⎮
⌡
(
dFv = −γ⋅ b ⋅ L −
where
(
⌠
⋅ ⎮ x⋅ γ⋅ b ⋅ L −
π ⎞ ⌡0
γ⋅ b ⋅ R⋅ ⎛⎜ L − R⋅ ⎟
4⎠
⎝
2
Fv = −17.12 × 10 ⋅ lbf
R
1
)
R − x dx = −γ⋅ b ⋅ R⋅ ⎛⎜ L − R⋅
R
2
− x ) dx =
(negative indicates downward)
2
2
)
R − x ⋅ dx
R
1
2
π⎞
⎟
4⎠
⌠
⋅⎮
π ⎞ ⌡0
R⋅ ⎛⎜ L − R⋅ ⎟
4⎠
⎝
( L⋅ x − x⋅
2
2
R −x
) dx
2
Evaluating the integral:
Substituting known values:
4
1
4⋅ R
L R⎞
4⋅ R
L R⎞
2 1 3⎞
x' =
⋅ ⎛⎜ ⋅ L⋅ R − ⋅ R ⎟ =
⋅ ⎛⎜ − ⎟ =
⋅ ⎛⎜ − ⎟
R⋅ ( 4⋅ L − π⋅ R) ⎝ 2
3
⎠ R⋅ ( 4⋅ L − π⋅ R) ⎝ 2 3 ⎠ 4⋅ L − π⋅ R ⎝ 2 3 ⎠
x' =
4⋅ 4⋅ ft
10⋅ ft 4⋅ ft ⎞
⋅ ⎛⎜
−
⎟
4⋅ 10⋅ ft − π⋅ 4⋅ ft ⎝ 2
3 ⎠
x' = 2.14⋅ ft
Problem 3.60
Problem
3.76
[Difficulty: 3]
3.60
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
= ρ⋅ g
dh
(Hydrostatic Pressure - h is positive downwards from
free surface)
⌠
⎮
Fv = ⎮ p dAy
⌡
(Vertical Hydrostatic Force)
FH = pc⋅ A
(Horizontal Hydrostatic Force)
⌠
⎮
x'⋅ Fv = ⎮ x dFv
⌡
Ixx
h' = hc +
hc⋅ A
(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
Into the vertical force equation:
h’
FH
x
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:
B
x⋅ y − A⋅ y = B Therefore: y =
x−A
2
and
10⋅ ft
xA =
+ 1⋅ ft
9⋅ ft
xA = 2.11⋅ ft
xB
⌠
Fv = ρ⋅ g⋅ b⋅ ⎮
⎮
⌡x
Into the force equation:
A
Fv = 1.94⋅
slug
ft
3
× 32.2⋅
ft
2
⎡
⎣
s
x' =
Fv
xB
1 ⌠
=
⋅⎮
Fv ⎮
⌡x
2
⎛ 7.0 − 1 ⎞⎤ ⋅ lbf ⋅ s
⎟⎥
⎝ 2.11 − 1 ⎠⎦ slug⋅ ft
2
⌠
⎮
x'⋅ Fv = ⎮ x dFv where
⌡
B ⎞
⎛
x⋅ ρ⋅ g⋅ b⋅ ⎜ H −
⎟ dx =
x
−
A⎠
⎝
A
⎛
⎝
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:
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
(
⎟
⎝ A − A⎠
)
7−1 ⎞
9⋅ ft
2
2 2
2
2
× 7 − 2.11 ⋅ ft − 10⋅ ft × ( 7 − 2.11) ⋅ ft − 10⋅ ft × 1⋅ ft × ln ⎛⎜
⎟
2
⎝ 2.11 − 1 ⎠
x' =
7−1 ⎞
2
9⋅ ft × ( 7 − 2.11) ⋅ ft − 10⋅ ft × ln ⎛⎜
⎟
2.11
− 1⎠
⎝
x' = 4.96⋅ ft
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
Substituting values:
For the line of action:
2
slug
Ixx
h' = hc +
hc⋅ A
ft
where
2
hc =
H
2
3
Therefore: h' =
2
lbf ⋅ s
FH = × 1.94⋅
× 32.2⋅ × 160⋅ ft × ( 9⋅ ft) ×
3
2
slug⋅ ft
2
ft
s
1
H b⋅ H 2 1
H H
2
+
⋅ ⋅
=
+
= ⋅H
2
2
3
12 H b⋅ H
6
h' =
A = H⋅ b
5
FH = 4.05 × 10 ⋅ lbf
Ixx =
2
⋅ 9⋅ ft
3
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 3.61
(Difficulty: 2)
3.61 The quarter cylinder 𝐴𝐴 is 10 𝑓𝑓 long. Calculate magnitude, direction, and location of the resultant
force of the water on 𝐴𝐴.
Given: All the parameters are shown in the figure.
Assumptions: Fluid is incompressible and static
Find: The resultant force.
Solution: Apply the hydrostatic relations for pressure as a function of depth and for the location of
forces on submerged objects.
A freebody diagram for the cylinder is:
∆𝑝 = 𝜌𝜌ℎ
The force balance in the horizontal direction yields thathorizontal force is due to the water pressure:
𝐹𝐻 = 𝑃𝐻
Where the depth is the distance to the centroid of the horizontal area (8 + 5/2 ft):
𝐹𝐻 = 𝛾ℎ𝑐 𝐴 = 62.4
𝑙𝑙𝑙
5 𝑓𝑓
× �8 𝑓𝑓 +
� × (5 𝑓𝑓 × 10 𝑓𝑡) = 32800 𝑙𝑙𝑙
𝑓𝑓 3
2
𝑃𝐻 = 32800 𝑙𝑙𝑙
The force in the vertical direction can be calculated as the weight of a volume of water that is 8 ft + 5 ft =
13 ft deep less the weight of water that would be in the quarter cylinder. This force is then:
𝑃𝑉 = 𝐹𝑉 − 𝑊 = 62.4
𝑙𝑙𝑙
𝑙𝑙𝑙 𝜋
× 13 𝑓𝑓 × (5 𝑓𝑓 × 10 𝑓𝑓) − 62.4 3 × × (5 𝑓𝑓)2 × (10 𝑓𝑓) = 28308 𝑙𝑙𝑙
3
𝑓𝑓
𝑓𝑓
4
The total resultant force is the vector sum of the two forces:
𝑃 = �𝑃𝐻 2 + 𝑃𝑉 2 = �(32800 𝑙𝑙𝑙)2 + (28308 𝑙𝑙𝑙)2 = 43300 𝑙𝑙𝑙
The angle with respect to the horizontal is:
𝑃𝑉
28308 𝑙𝑙𝑙
� = 40.9°
𝜃 = tan−1 � � = tan−1 �
32800 𝑙𝑙𝑙
𝑃𝐻
So the force acts on the quarter cylinder surface point at an angle of 𝜃 = 40.9 ° with respect to the
horizontal.
Problem 3.62
(Difficulty: 2)
3.62 Calculate the magnitude, direction (horizontal and vertical components are acceptable), and line of
action of the resultant force exerted by the water on the cylindrical gate 30 𝑓𝑓 long.
Assumptions: Fluid is incompressible and static
Find: The resultant forces.
Solution: Apply the hydrostatic relations for pressure as a function of depth and for the location of
forces on submerged objects.
A free body diagram of the gate is
∆𝑝 = 𝜌𝜌ℎ
The horizontal force is calculated as:
𝑃𝐻 = 𝐹𝐻
Where the depth is the distance to the centroid of the horizontal area (5 ft):
𝐹𝐻 = 𝛾ℎ𝑐 𝐴 = 62.4
𝑙𝑙𝑙
× 5𝑓𝑓 × (10 𝑓𝑓 × 30 𝑓𝑓) = 93600 𝑙𝑙𝑙
𝑓𝑓 3
𝑃𝐻 = 93600 𝑙𝑙𝑙
The force in the vertical direction can be calculated as the weight of a volume of water that is 10 ft deep
less the weight of water that would be in the quarter cylinder. This force is then:
𝑃𝑉 = 62.4
𝑃𝑉 = 𝐹𝑉 − 𝑊 = 𝛾ℎ𝑐 𝐴 − 𝛾∀
𝑙𝑙𝑙
𝑙𝑙𝑙
(10
×
10
𝑓𝑓
×
𝑓𝑓
×
30
𝑓𝑓)
−
62.4
𝑓𝑓 3
𝑓𝑓 3
𝜋
× �10 𝑓𝑓 × (10 𝑓𝑓 × 30 𝑓𝑓) − × (10 𝑓𝑡)2 × 30 𝑓𝑓� = 147000 𝑙𝑙𝑙
4
The total resultant force is the vector sum of the two forces:
𝑃 = �𝑃𝐻 2 + 𝑃𝑉 2 = �(93600 𝑙𝑙𝑙)2 + (147000 𝑙𝑙𝑙)2 = 174200 𝑙𝑙𝑙
The direction can be calculated as:
𝑃𝑉
147000 𝑙𝑙𝑙
𝜃 = tan−1 � � = tan−1 �
� = 57.5°
93600 𝑙𝑙𝑙
𝑃𝐻
Problem 3.63
(Difficulty: 2)
3.63 A hemispherical shell 1.2 𝑚 in diameter is connected to the vertical wall of a tank containing water.
If the center of the shell is 1.8 𝑚 below the water surface, what are the vertical and horizontal force
components on the shell? On the top half of the shell?
Assumptions: Fluid is incompressible and static
Find: The resultant forces.
Solution: Apply the hydrostatic relations for pressure as a function of depth and for the location of
forces on submerged objects.
∆𝑝 = 𝜌𝜌ℎ
A free body diagram of the system is
The force in the horizontal direction can be calculated using the distance to the centroid (1.8 m) as:
𝐹𝐻 = 𝛾ℎ𝑐 𝐴 = 9.81
𝑘𝑁
1
× 1.8 𝑚 × � × 𝜋 × (1.2 𝑚)2 � = 19.97 𝑘𝑘
3
𝑚
4
The force in the vertical direction is the buoyancy force due to the volume displaced by the shell:
𝐹𝑉 = 𝛾𝛾 = 9.81
𝑘𝑘 1 1
× × × 𝜋 × (1.2 𝑚)3 = 4.44 𝑘𝑘
𝑚3 2 6
For the top shell, the horizontal force acts at:
𝑦𝑐 = 1.8 𝑚 −
4 × 0.6 𝑚
= 1.545 𝑚
3𝜋
The horizontal force on the top half of the shell is then:
𝐹𝐻 = 𝛾𝑦𝑐 𝐴 = 9.81
𝑘𝑘
𝜋
× 1.545 𝑚 × × (1.2 𝑚)2 = 8.57 𝑘𝑘
3
𝑚
8
The vertical force on the top half of the shell is the buoyancy force:
𝐹𝑉 = 𝑝𝑝 = 9.81
𝑘𝑘
𝜋
𝑘𝑘 1 1
× 1.8 𝑚 × × (1.2 𝑚)2 − 9.81 3 × × × 𝜋 × (1.2 𝑚)3 = 7.77 𝑘𝑘
3
𝑚
8
𝑚
4 6
Problem 3.64
Problem
3.78
[Difficulty: 4]
3.64
Given:
Gate geometry
Find:
Force on stop B
x
y’
Solution:
Basic equations
4R/3π
R/2
D
FV
dp
= ρ⋅ g
dh
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
4
2
= 1000⋅
FV = W1 − W2
with x given by
3
× 9.81⋅
m
2
s
or
397 3⋅ m 208
4
×
−
×
× 3⋅ m
189
2
189 3⋅ π
Computing equations
m
2
× 3⋅ m ×
π
2 N⋅ s
× ( 3⋅ m) ×
4
kg⋅ m
FV = 189⋅ kN
R
4⋅ R
FV⋅ x = W1⋅ − W2⋅
2
3⋅ π
x =
For FH
kg
FH = pc⋅ A
W1 = 397⋅ kN
x=
W1 R W2 4⋅ R
⋅ −
⋅
Fv 2
Fv 3⋅ π
x = 1.75 m
Ixx
y' = yc +
A ⋅ yc
W2 = 208⋅ kN
Hence
R⎞
⎛
FH = pc⋅ A = ρ⋅ g⋅ ⎜ D − ⎟ ⋅ w⋅ R
2⎠
⎝
kg
FH = 1000⋅
3
× 9.81⋅
m
2
⎛
⎝
m
× ⎜ 4.5⋅ m −
2
s
3⋅ m ⎞
N⋅ s
⎟ × 3⋅ m × 3⋅ m ×
2 ⎠
kg⋅ 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
kg⋅ m
F1 = 397⋅ kN
For the concrete gate (SG = 2.4 from Table A.2)
2
WGate = SG⋅ ρ⋅ g⋅ w⋅
FB⋅ R + F1⋅
Hence, taking moments about A
FB =
FB =
4
3⋅ π
⋅ WGate +
4
3⋅ π
x
R
× 499⋅ kN +
FB = 278⋅ kN
2
π⋅ R
kg
m
π
2 N⋅ s
= 2.4⋅ 1000⋅
× 9.81⋅ × 3⋅ m × × ( 3⋅ m) ×
3
2
4
kg⋅ m
4
m
s
R
− WGate⋅
2
⋅ FV +
1.75
3
4⋅ R
3⋅ π
[ y' − ( D − R) ]
R
× 189⋅ kN +
− FV⋅ x − FH⋅ [ y' − ( D − R) ] = 0
⋅ FH −
1
⋅F
2 1
[ 3.25 − ( 4.5 − 3) ]
3
× 265⋅ kN −
1
2
× 397⋅ kN
WGate = 499⋅ kN
Problem
3.80
Problem 3.65
[Difficulty: 3]
3.65
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.
dp
= ρ⋅ g
dh
⎯→
⎯
→
dFR = −p ⋅ dA
Governing Equations:
Assumptions:
(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
⎮
⎮
⌠
⎮
FRx = ⎮
p⋅ L⋅ R⋅ sin ( θ) dθ =
ρ⋅ g⋅ R⋅ ( 1 − cos ( θ) ) ⋅ L⋅ R⋅ sin ( θ) dθ − ⎮
ρ⋅ g⋅ R⋅ cos ( θ) ⋅ L⋅ R⋅ sin ( θ) dθ
⌡0
⌡0
⌡π
3⋅ π
⌠ 2
π
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
⎮
1⎞
3
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
⎦
2
Substituting known values:
FRx =
3
kg
m
N⋅ s
2
× 999⋅
× 9.81⋅ × ( 1.5⋅ m) × 6⋅ m ×
3
2
2
kg⋅ m
m
s
FRx = 198.5⋅ kN
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
⎮
π 3⋅ π π ⎞
3⋅ π
2 ⎢⌠
2 ⎥
2
2
FRy = −ρ⋅ g⋅ R ⋅ L⋅ ⎮ ( 1 − cos ( θ) ) ⋅ cos ( θ) dθ − ⎮
( cos ( θ) ) dθ = ρ⋅ g⋅ R ⋅ L⋅ ⎛⎜ +
− ⎟ =
⋅ ρ⋅ g⋅ R ⋅ L
⎢⌡
⎥
⌡
4
2
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 3.66
Problem
3.82
[Difficulty: 3]
3.66
Given:
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
= ρ⋅ g
dh
(Hydrostatic Pressure - h is positive downwards from
free surface)
⌠
⎮
Fv = ⎮ p dAy
⌡
(Vertical Hydrostatic Force)
FH = pc⋅ A
(Horizontal Hydrostatic Force)
⌠
⎮
x'⋅ Fv = ⎮ x dFv
⌡
Ixx
h' = hc +
hc⋅ A
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 dA y =
ρ⋅ g⋅ h⋅ sin ( θ) dA = ⎮ ρ⋅ g⋅ ( H − R⋅ sin ( θ) ) ⋅ sin ( θ) ⋅ w⋅ R dθ
⎮
⎮
⌡0
⌡
⌡
θ
⌠ 1
⎡
2
Fv = ρ⋅ g⋅ w⋅ R⋅ ⎮ ⎡⎣H⋅ sin ( θ) − R⋅ ( sin ( θ) ) ⎤⎦ dθ = ρ⋅ g⋅ w⋅ R⋅ ⎢H⋅ 1 − cos θ1
⌡0
⎣
(
⎛ θ1
( )) − R⋅ ⎜⎝ 2
−
(
)
sin 2⋅ θ1 ⎞⎤
⎟⎥
4
⎠⎦
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⋅
=
kg
3
ρ ⋅ g⋅ w ⋅ R
Fv
× 9.81⋅
m
2
m
2
( ( ))
2 R
3⎤
⎡H
⋅ ⎢ ⋅ sin θ1 − ⋅ sin θ1 ⎥
2
3
⎣
⎦
( ( ))
( ( ))
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
3
⎝ 0.750 ⎠
⎝ 0.750 ⎠ ⎦ kg⋅ m
2.47 × 10 N ⎣ 2
1
⋅
1
×⎢
x' = 0.645 m
2
For the horizontal force:
FH =
1
2
× 999⋅
kg
3
H
ρ⋅ g⋅ H ⋅ w
FH = pc⋅ A = ρ⋅ g⋅ hc⋅ H⋅ w = ρ⋅ g⋅ ⋅ H⋅ w =
2
2
× 9.81⋅
m
m
2
2
s
For the line of action of the horizontal force:
Ixx
2
× ( 0.650⋅ m) × 3.55⋅ m ×
N⋅ s
kg⋅ m
Ixx
h' = hc +
hc⋅ A
where
3
w⋅ H 2 1
H H
2
h' = hc +
=
+
⋅ ⋅
=
+
= ⋅H
12 H w⋅ H
6
hc⋅ A
2
2
3
H
FH = 7.35⋅ kN
h' =
2
3
Ixx =
w⋅ H
12
× 0.650⋅ m
3
A = w⋅ H
Therefore:
h' = 0.433 m
Problem 3.67
Problem
3.83
[Difficulty: 2]
3.67
Given:
Canoe floating in a pond
Find:
What happens when an anchor with too short of a line is thrown from canoe
Solution:
Governing equation:
FB = ρ 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 = ρ w gVcanoe1
The anchor weight can be expressed as
Wanchor = ρ a gVa
so the initial volume displaced by the canoe can be written as
Vcanoe1 =
Wcanoe ρ a
+
Va
ρw g ρw
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 = ρ w gVcanoe2 + ρ w gVa
Vcanoe 2 =
Wcanoe Wa
+
− Va
ρw g ρw g
Vcanoe 2 =
Wcanoe ρ a
Va − Va
+
ρw g ρw
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 3.68
Problem
3.86
[Difficulty: 4]
3.68
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
= ρ⋅ g
dh
(Hydrostatic Pressure - h is positive downwards from free surface)
⌠
⎮
Fv = ⎮ p dAy
⌡
(Vertical Hydrostatic Force)
ΣM = 0
(Rotational Equilibrium)
H = αR
(1) Static fluid
(2) Incompressible fluid
(3) Atmospheric pressure acts on free surface of the liquid.
Assumptions:
h
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
Therefore
H = h + R⋅ ( 1 − cos ( θ) )
h = H − R⋅ ( 1 − cos ( θ) ) and into the vertical force equation:
2 H
⎤
dFv = ρ⋅ g⋅ [ H − R⋅ ( 1 − cos ( θ) ) ] ⋅ w⋅ R⋅ cos ( θ) ⋅ dθ = ρ⋅ g⋅ w⋅ R ⋅ ⎡⎢ − ( 1 − cos ( θ) )⎥ ⋅ cos ( θ) ⋅ dθ
R
⎣
⎦
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:
θ
θ
⌠ max
⌠ max
Fv = ⎮
1 dF v = ⎮
2 dFv
⌡0
⌡− θ
max
where
(
)
cos θmax =
R−H
= 1−α
R
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 3.69
Problem
*3.89
[Difficulty: 2]
3.69
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:
(Buoyant force is equal to weight of displaced fluid)
M = ρw ⋅ V w
M=
Fbuoy
g
= ρ⋅ V d
When immersed in nitric acid:
ρw⋅ Vw = ρn⋅ Vn
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
ρn = SG⋅ ρw
Solving for the height h:
Vw⋅ ( 1 − SG) = −SG⋅
π 2
⋅d ⋅h
4
1.5 − 1 ⎞
4
10⋅ mm ⎞
× ⎛⎜
⎟×
⎟
2
⎝ 1.5 ⎠ π × ( 6⋅ mm) ⎝ cm ⎠
h = 15⋅ cm × ⎛⎜
3
M = ρn⋅ Vn
3
h = 177⋅ mm
Problem 3.70
(Difficulty: 2)
3.70 A cylindrical can 76 𝑚𝑚 in diameter and 152 𝑚𝑚 high, weighing 1.11 𝑁, contains water to a depth
of 76 𝑚𝑚. When this can is placed in water, how deep will it sink?
Find: The depth it will sink.
Assumptions: Fluid is incompressible and static
Solution: Apply the hydrostatic relations:
Pressure as a function of depth
Buoyancy force:
∆𝑝 = 𝜌𝜌ℎ
A free body diagram on the can is
𝐹𝑏 = 𝜌 𝑔 𝑉
We have the force balance equation in the vertical direction as:
𝐹𝑏 − 𝑊𝑐𝑐𝑐 − 𝑊𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 0
The buoyancy force can be calculated as:
We also have:
𝐹𝑏 = 𝛾𝑤𝑤𝑤𝑤𝑤 𝑉𝑐𝑐𝑐 = 9810
𝑁 𝜋
× × (0.076 𝑚)2 × 𝑥 𝑚 = 44.50𝑋𝑋 𝑁
𝑚3 4
𝑊𝑐𝑐𝑐 = 1.11 𝑁
𝑊𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝛾𝑤𝑤𝑤𝑤𝑤 𝑉𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 9810
𝑁 𝜋
× × (0.076 𝑚)3 = 3.38 𝑁
𝑚3 4
Thus making a force balance for which the net force is zero at equilibrium
44.50𝑥 = 1.11 𝑁 + 3.38 𝑁 = 4.49 𝑁
𝑥 = 0.1009 𝑚 = 100.9 𝑚𝑚
So this can will sink to depth of 100.9 𝑚𝑚.
Problem 3.71
(Difficulty: 1)
3.71 If the 10 𝑓𝑓 long box is floating on the oil water system, calculate how much the box and its
contents must weigh.
Find: The weight of the box and its contents.
Assumptions: Fluid is incompressible and static
Solution: Apply the hydrostatic relations:
Pressure as a function of depth
∆𝑝 = 𝜌𝜌ℎ
Buoyancy force:
𝐹𝑏 = 𝜌 𝑔 𝑉
The force balance equation in the vertical diretion:
𝐹𝐵 − 𝑊𝐵 = 0
𝐹𝐵 = 𝛾𝑜𝑜𝑜 𝑉 + 𝛾𝑤𝑤𝑤𝑤𝑤 𝑉
Thus
𝐹𝐵 = 0.8 × 62.4
𝑙𝑙𝑙
𝑙𝑙𝑙
× 2𝑓𝑓 × 8𝑓𝑓 × 10𝑓𝑓 + 62.4 3 × 1𝑓𝑓 × 8𝑓𝑓 × 10𝑓𝑓 = 12980 𝑙𝑙𝑙
3
𝑓𝑓
𝑓𝑓
So the box and its contents must weigh:
𝑊𝐵 = 12980 𝑙𝑙𝑙
Problem 3.72
(Difficulty: 2)
3.72 The timber weighs 40
𝑙𝑙𝑙
𝑓𝑓 3
and is held in a horizontal position by the concrete �150
Calculate the minimum total weight which the anchor may have.
Find: The minimum total weight the anchor may have.
Assumptions: Fluid is incompressible and static
Solution: Apply the hydrostatic relations:
Pressure as a function of depth
Buoyancy force:
∆𝑝 = 𝜌𝜌ℎ
For the buoyancy force we have:
𝐹𝑏 = 𝜌 𝑔 𝑉
𝐹𝑏𝑏 = 62.4
The weight of the timber is:
𝑊𝑡 = 40
𝐹𝑏𝑏 = 𝛾𝑤𝑤𝑤𝑤𝑤 𝑉𝑡
𝑙𝑙𝑙
6
6
×
�
𝑓𝑓�
×
�
𝑓𝑓� × (20 𝑓𝑓) = 312 𝑙𝑙𝑙
𝑓𝑓 3
12
12
𝑊𝑡 = 𝛾𝑡 𝑉𝑡
𝑙𝑙𝑙
6
6
�
𝑓𝑓� × � 𝑓𝑓� × (20 𝑓𝑓) = 200 𝑙𝑙𝑙
3
𝑓𝑓
12
12
At the horizontal position we take moments about the pivot:
𝐹𝑎 𝐿 + 𝑊𝑡
𝐿
𝐿
− 𝐹𝑏𝑏 = 0
2
2
1
1
1
𝐹𝑎 = 𝐹𝑏𝑏 − 𝑊𝑡 = × (312 𝑙𝑙𝑙 − 200𝑙𝑙𝑙) = 56 𝑙𝑙𝑙
2
2
2
𝑙𝑙𝑙
�
𝑓𝑓 3
anchor.
𝐹𝑎 = 𝐹𝑏𝑏 − 𝑊𝑎
The weight of the anchor is:
The buoyancy force on the anchor is:
𝑊𝑎 = 𝛾𝑎 𝑉𝑎
𝐹𝑏𝑏 = 𝛾𝑤𝑤𝑤𝑤𝑤 𝑉𝑎
𝑉𝑎 =
So the weight is:
𝛾𝑎 𝑉𝑎 − 𝛾𝑤𝑤𝑤𝑤𝑤 𝑉𝑎 = 56 𝑙𝑙𝑙
56 𝑙𝑙𝑙
= 0.64 𝑓𝑓 3
𝑙𝑙𝑙
𝑙𝑙𝑙
�150 3 − 62.4 3 �
𝑓𝑓
𝑓𝑓
𝑊𝑎 = 𝛾𝑎 𝑉𝑎 = 150
𝑙𝑙𝑙
× 0.64 𝑓𝑓 3 = 96 𝑙𝑙𝑙
𝑓𝑓 3
Problem 3.73
Problem
*3.90
[Difficulty: 3]
3.73
Given:
Data on sphere and weight
Find:
SG of sphere; equilibrium position when freely floating
T
Solution:
Basic equation
FB
FB = ρ⋅ g⋅ V
where
Hence
ΣFz = 0
and
T = M⋅ g
M⋅ g + ρ⋅ g⋅
V
2
M = 10⋅ kg
− SG⋅ ρ⋅ g⋅ V = 0
3
SG = 10⋅ kg ×
The specific weight is
γ =
Weight
Volume
ΣFz = 0 = T + FB − W
m
×
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
where h is submerged depth and R is the sphere radius
FB = ρ⋅ g⋅ Vsubmerged
Vsubmerged =
From references (trying Googling "partial sphere volume")
3⋅ V ⎞
R = ⎛⎜
⎟
⎝ 4⋅ π ⎠
1
3
π⋅ h
2
3
2
Hence
π⋅ h
W = SG⋅ ρ⋅ g⋅ V = FB = ρ⋅ g⋅
⋅ ( 3⋅ R − h)
3
3
3⋅ 0.9⋅ .025⋅ m
h ⋅ ( 3⋅ 0.181⋅ m − h) =
π
2
⋅ ( 3⋅ R − h)
3
3⎞
R = ⎛⎜
⋅ 0.025⋅ m ⎟
⎝ 4⋅ π
⎠
2
h ⋅ ( 3⋅ R − h) =
1
3
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 3.74
Problem
*3.91
[Difficulty: 2]
3.74
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.
Governing Equation:
Assumptions:
Fbuoy = ρ⋅ g⋅ Vd
(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
⎟
w4C ⎠
SG
Fair = ⎝
Fair − Fnet) =
⋅ (F − Fnet)
(
SGw air
⎛ ρw ⎞
⎜
⎟
⎝ ρw4C ⎠
Solving this expression for the specific gravity of the person SG, we get:
SG = SGw⋅
F
Fair
air − Fnet
Problem *3.93
3.75
Problem
[Difficulty: 2]
3.75
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
m
2
δ = 1⋅ mm
× 3.22 × 10
−4
3
⋅m ×
s
−4
3
⋅m
2
N⋅ s
kg⋅ m
W = 24.7 N
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
2
N⋅ s
Vdisplaced = 2.52 L
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 ⎠
Hence, the cylinder must be made to sink an additional distance
x2 = H − x1
x2 = 0.679 m
2
We deed to add n weights so that
π⋅ D
1⋅ kg⋅ n⋅ g = ρ⋅ g⋅
⋅ x2
4
2
2
ρ⋅ π⋅ D ⋅ x2
kg π
1
N⋅ s
2
n=
= 999⋅
× × ( 0.1⋅ m) × 0.679⋅ m ×
×
3 4
1⋅ kg kg⋅ m
4 × 1⋅ kg
m
Hence we need n = 6 weights to sink the cylinder
n = 5.33
Problem 3.76
(Difficulty: 2)
3.76 If the timber weights 670 𝑁, calculate its angle of inclination when the water surface is 2.1 𝑚
above the pivot. Above what depth will the timber stand vertically?
Find: Above what depth will the timber stand vertically.
Assumptions: Fluid is incompressible and static
Solution: Apply the hydrostatic relations:
Pressure as a function of depth
Buoyancy force:
∆𝑝 = 𝜌𝜌ℎ
The buoyancy force is:
𝐹𝑏 = 𝜌 𝑔 𝑉
𝐹𝑏 = 𝛾𝑤𝑤𝑤𝑤𝑤 𝑉 = 0.152 𝑚 × 0.152 𝑚 × 𝑥 𝑚 × 9810
Take the moment about pivot we have:
𝑀 = 𝑊 × 0.5 × 3.6 𝑚 cos 𝜃 −
670 𝑁 × 0.5 × 3.6 𝑚 × cos 𝜃 −
Soving this equation we have:
The angle when water surface 𝑦 = 2.1 𝑚 is:
𝑁
= 226.7𝑥 (𝑁)
𝑚3
𝑥
𝑚 × 𝐹𝑏 cos 𝜃 = 0
2
𝑥
𝑚 × 226.7𝑥 × cos 𝜃 = 0
2
𝑥 = 3.26 𝑚
2.1 𝑚
� = 40.1 °
𝜃 = sin−1 �
3.26 𝑚
We have the following relation:
Substitute in to the momentum we have:
𝑥=
670 𝑁 × 0.5 × 3.6 𝑚 −
If the timber is vertically, we have:
𝑦
sin 𝜃
𝑦
𝑦
𝑚 × 226.7
=0
2sin 𝜃
sin 𝜃
𝜃 = 90°
So we have:
Solving this equation we have:
sin 90° = 1
670 𝑁 × 0.5 × 3.6 𝑚 −
𝑦
𝑚 × 226.7𝑦 = 0
2
𝑦 = 3.26 𝑚
When the water surface is 𝑦 = 3.26 𝑚, the timber will stand vertically.
Problem 3.77
(Difficulty: 2)
3.77 The barge shown weights 40 𝑡𝑡𝑡𝑡 and carries a cargo of 40 𝑡𝑡𝑡𝑡. Calculate its draft in freshwater.
Find: The draft, where the draft is the depth to which the barge sinks.
Assumptions: Fluid is incompressible and static
Solution: Apply the hydrostatic relations:
Pressure as a function of depth
∆𝑝 = 𝜌𝜌ℎ
Buoyancy force:
𝐹𝑏 = 𝜌 𝑔 𝑉
For the barge floating in water we have the buoyancy force as:
𝐹𝐵 = 𝛾𝑤𝑤𝑤𝑤𝑤 𝑉 = 𝑊
The weight of the barge is:
𝑊 = (40 + 40)𝑡𝑡𝑡𝑡 = 80 𝑡𝑡𝑡𝑡 ×
The volume of water displaced is then:
𝑉=
The volume in terms of the draft d is:
𝑊
𝛾𝑤𝑤𝑤𝑤𝑤
=
2000 𝑙𝑙𝑙
= 160000 𝑙𝑙𝑙
𝑡𝑡𝑡
160000 𝑙𝑙𝑙
= 2564 𝑓𝑓 3
𝑙𝑙𝑙
62.4 3
𝑓𝑓
5
𝑑
∀= 𝐴𝑐 𝐿 = �40𝑓𝑓 + 40𝑓𝑓 + 2 × 𝑑� × × 20𝑓𝑓 = 800𝑑 + 12.5𝑑 2
8
2
Thus we have the relation:
800𝑑 + 12.5𝑑 2 = 2564
Solving this equation we have for the draft:
𝑑 = 3.06 𝑓𝑓
Problem
Problem *3.94
3.78
[Difficulty: 2]
3.78
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.
Fb = ρ⋅ g⋅ Vd
(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 *3.96
3.79
Problem
[Difficulty: 2]
3.79
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:
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
Assumptions:
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
lbm⋅ R
Tg = Ta
and therefore:
now if we take the ideal gas equation and
Ra⋅ Ta ⎞
Ra⋅ Ta ⎞
⎛
⎛
L
= ρa⋅ g⋅ ⎜ 1 −
⎟ = γa⋅ ⎜ 1 −
⎟
V
⎝ Rg⋅ Tg ⎠
⎝ 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
Tg = Ta + 250⋅ R and therefore:
LVair250
= 1.450
LVair150
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 3.80
Problem
*3.98
[Difficulty: 3]
3.80
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
FB = ρatm⋅ g⋅ V
Basic equation
ΣFy = M⋅ ay
and
Wload
ΣFy = 0 = FB − Whotair − Wload = ρatm⋅ g⋅ V − ρhotair⋅ g⋅ V − M⋅ g
Hence
(
)
M = V⋅ ρatm − ρhotair =
3
M = 320000⋅ ft × 14.7⋅
lbf
2
in
V⋅ patm
R
for neutral buoyancy
⋅ ⎛⎜
1
1 ⎞
−
Tatm Thotair ⎟
⎝
⎠
12⋅ in ⎞
2
lbm⋅ R
1
1
⎤
× ⎡⎢
−
⎟ ×
⎥
(
160
+
460
)
⋅
R
ft
53.33
⋅
ft
⋅
lbf
(
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
2⋅ a ⎤
1 ⎞
−
−
⎢
⎜
⎟
⎥
a+g
⎣ ⎝ Tatm Thotair ⎠ Thotair⎦
a+g
2
2
lbf ⎛ 12⋅ in ⎞
lbm⋅ R
s
1
1
1
⎤ − 2⋅ 2.5⋅
⎤ ft
Mnew = 320000⋅ ft ⋅ 14.7⋅
⋅⎜
⋅
⋅ ⎡⎢32.2⋅ ⎡⎢
−
⎟ ⋅
⎥
⎥⋅ 2
2 ⎝ ft ⎠ 53.33⋅ ft⋅ lbf ( 2.5 + 32.2) ⋅ ft ⎣
( 160 + 460)⎦ s
⎣ ( 48 + 460) ( 160 + 460)⎦
in
⋅R
3
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 3.81
(Difficulty: 2)
3.81 The opening in the bottom of the tank is square and slightly less than 2 𝑓𝑓 on each side. The
opening is to be plugged with a wooden cube 2 𝑓𝑓 on a side.
(a) What weight 𝑊 should be attached to the wooden cube to insure successful plugging of the hole?
The wood weighs 40
𝑙𝑙𝑙
,
𝑓𝑓 3
(b) What upward force must be exerted on the block to lift it and allow water to drain from the tank?
Find: The weight of the block and the force needed to lift it
Assumptions: Fluid is incompressible and static
Solution: Apply the hydrostatic relations:
Pressure as a function of depth
Buoyancy force:
∆𝑝 = 𝜌𝜌ℎ
𝐹𝑏 = 𝜌 𝑔 𝑉
(a) Because the wood bottom surface is in the atmosphere so the pressure on the bottom surface is
zero in this case and there is no buoyancy force. The force acting on the wood cube in the
vertical direction is:
𝐹𝑉 = 𝐹𝑝 + 𝐺
𝐹𝑉 = 𝛾ℎ1 𝐴 + 𝐺 = 62.4
𝑙𝑙𝑙
𝑙𝑙𝑙
×
5
𝑓𝑓
×
2𝑓𝑓
×
2𝑓𝑓
+
40
× (2 𝑓𝑓)3 = 1568 𝑙𝑙𝑙
𝑓𝑓 3
𝑓𝑓 3
The direction of 𝐹𝑉 is downward. So we do not need any weight 𝑊 attached to the wood cube.
(b) To lift the block, we need a force larger than 𝐹𝑉 , so we have:
𝐹𝑢𝑢 ≥ 𝐹𝑉 = 1568 𝑙𝑙𝑙
Problem 3.82
(Difficulty: 2)
3.82 A balloon has a weight (including crew but not gas) of 2.2 𝑘𝑘 and a gas-bag capacity of 566 𝑚3 . At
the ground it is (partially) inflated with 445 𝑁 of helium. How high can this balloon rise in the U.S
standard atmosphere if the helium always assumes the pressure and temperature of the atmosphere?
Find: How high this balloon will rise.
Assumptions: Fluid is incompressible and static
Solution: Apply the hydrostatic relations:
Pressure as a function of depth
Buoyancy force:
∆𝑝 = 𝜌𝜌ℎ
At the sea level, for helium we have:
𝐹𝑏 = 𝜌 𝑔 𝑉
𝑝 = 101.3 𝑘𝑘𝑘
𝑇 = 288 𝐾
𝑅 = 2076.8
According to the ideal gas law:
𝜌ℎ =
The volume of the helium is:
𝐽
𝑘𝑘 ∙ 𝐾
𝑝
101.3 𝑘𝑘𝑘
𝑘𝑘
=
= 0.1694 3
𝐽
𝑅𝑅 2076.8
𝑚
× 288 𝐾
𝑘𝑔 ∙ 𝐾
𝛾ℎ = 𝜌𝜌 = 0.1694
𝑉ℎ =
The buoyancy force is calculated by:
The weight of the whole balloon is:
𝑘𝑘
𝑚
𝑁
× 9.81 2 = 1.662 3
3
𝑚
𝑠
𝑚
𝑊ℎ
445 𝑁
=
= 268 𝑚3
𝑁
𝛾ℎ
1.662 3
𝑚
𝐹𝐵 = 𝛾𝑎𝑎𝑎 𝑉𝑏
𝑊 = 2.2 𝑘𝑘 + 𝑊ℎ
We have the following table as (the helium always has the same temperature and pressure as the
atmosphere):
Altitude
(km)
Pressure
(kPa)
Temperature
(K)
∀ (𝑚3 )
𝛾𝑎𝑎𝑎 �
𝑁
�
𝑚3
𝑊ℎ
(𝑘𝑘)
𝐹𝐵
(𝑘𝑘)
𝑊
(𝑘𝑘)
6
47.22
249.2
497
6.46
0.445
3.21
2.65
8
35.70
236.3
566
5.14
0.402
2.91
2.60
10
26.50
223.4
566
4.04
0.317
2.29
2.52
When the maximum volume of the helium is reached, the volume will become a constant for helium.
Equilibrium is reached as:
At 8 𝑘𝑘 we have:
At 10 𝑘𝑘 we have:
𝐹𝐵 = 𝑊
𝐹𝐵 − 𝑊 = 0.31 𝑘𝑘
𝐹𝐵 − 𝑊 = −0.23 𝑘𝑘
With the interpolation we have the height for equilibrium as:
ℎ = 8𝑘𝑘 + 2𝑘𝑘 ×
0.31
= 9.15 𝑘𝑘
0.31 + 0.23
Problem *3.100
3.83
Problem
[Difficulty: 3]
3.83
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
Dmax =
∆p
Solving this expression for the diameter:
−3
Dmax = 4 × 0.015 × 10
6 N
⋅ m × 62 × 10 ⋅
2
m
2
1
×
πD 2∆p/4
4⋅ t⋅ σ
×
−3
0.45⋅ 10
⋅ 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 ×
kg⋅ K
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
287⋅ N ⋅ m
×
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*3.102
3.84
Problem
[Difficulty: 3]
3.84
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⋅ σ
4
=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
5 × 10
−3
⋅m ×
⋅m
kg⋅ m
2
−3
m
s ⋅N
0.1
−3
2 × 10
∆h = 1.53 × 10
∆SG = 0.0765
⋅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
Problem*3.103
3.85
[Difficulty:4]
3.85
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:
θ
θ
θmax
⌠ max 2
⌠ max 2
2
3⌠
3
Vd = ⎮
π r dh = π⋅ ⎮
R ⋅ ( sin( θ) ) ⋅ R⋅ sin( θ) dθ = π⋅ R ⋅ ⎮
( sin( θ) ) dθ
⌡
⌡
⌡
0
0
Evaluating the integral we get:
0
⎡⎢ ( cos( θ ) ) 3
max
Vd = π⋅ R ⋅ ⎢
− cos( θmax) +
3
⎣
3
Thus the buoyant force is:
h
θmax
⎤
3
d
d⎞
d
2⎤
3 ⎡1 ⎛
Now
since:
we
get:
⎢
cos
θ
=
1
−
V
=
π
⋅
R
⋅
1
−
− ⎛⎜ 1 − ⎞ + ⎥
(
)
⎜
⎥
max
d
R
R⎠
R ⎠ 3⎦
3⎦
⎣3 ⎝
⎝
2⎥
3
d⎞
d
2⎤
⎛
⎢
Fbuoy = ρw⋅ SG⋅ g ⋅ π⋅ R ⋅ ⋅ ⎜ 1 −
− ⎛⎜ 1 − ⎞ + ⎥
R⎠
R ⎠ 3⎦
⎣3 ⎝
⎝
3 ⎡1
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
⎤
d
d
⎢ ⋅ ⎛⎜ 1 − ⎞ − ⎛⎜ 1 − ⎞ + 2⎥
R⎠
R ⎠ 3⎦
4 ⎣3 ⎝
⎝
3 ⎡1
d
d
2⎤
Fd = ⎢ ⋅ ⎛⎜ 1 − ⎞ − ⎛⎜ 1 − ⎞ + ⎥
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 *3.106
3.86
Problem
[Difficulty: 4]
3.86
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
FL = p atm⋅ π⋅ a
where
and
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
⎝
This expression simplifies to
⎠
4 3
2
ΣFy = π⋅ ρ⋅ g ⋅ ⎡⎢( 1 − SG ) ⋅ ⋅ R − H⋅ a ⎤⎥
3
⎣
⎦
3
2
2
⎡4
ft ⎞
ft ⎞ ⎤ lbf ⋅ s
⎛
⎛
⎢
⎥
ΣFy = π × 1.94⋅
× 32.2⋅ ×
× ( 1 − 0.95) × ⎜ 1 ⋅ in ×
− 2.5⋅ ft × ⎜ 0.075 ⋅ in ×
×
3
2 ⎣3
12⋅ in ⎠
12⋅ in ⎠ ⎦ slug⋅ ft
⎝
⎝
ft
s
slug
ΣFy = −0.012 ⋅ lbf
ft
Therefore, the sphere stays at the bottom of the tank.
Problem *3.108
3.87
Problem
[Difficulty: 3]
3.87
Given:
Data on boat
Floating
Find:
Sinking
H = 8 ft
Effective density of water/air bubble mix if boat sinks
Solution:
Basic equations
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
ΣFy = 0 = FB − W
2
2⋅ H ⎞
L⋅ H
⎛
Vsubsink = ⋅ H⋅ ⎜
⋅L =
2 ⎝ tan⋅ θ ⎠
tan( θ)
1
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
Problem *3.112
3.88
[Difficulty: 2]
3.88
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 (∆V < 0 because SGballs > 1)
Problem 3.89
Problem
*3.113
[Difficulty: 4]
3.89
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 5.1
Problem
5.2
[Difficulty: 2]
5.1
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)
x
y
z
t
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 5.2
(Difficulty 1)
5.2 Which of the following sets of equations represent possible two-dimensional incompressible flow
cases?
(a) 𝑢 = 2𝑥𝑥; 𝑣 = −𝑥 2 𝑦
(b) 𝑢 = 𝑦 − 𝑥 + 𝑥 2 ; 𝑣 = 𝑥 + 𝑦 − 2𝑥𝑥
(c) 𝑢 = 𝑥 2 𝑡 + 2𝑦; 𝑣 = 2𝑥 − 𝑦𝑡 2
(d) 𝑢 = −𝑥 2 − 𝑦 2 − 𝑥𝑥𝑥; 𝑣 = 𝑥 2 + 𝑦 2 + 𝑥𝑥𝑥
Find: The sets of equations represent possible two-dimensional incompressible flow.
Solution:
From the continuity equation for two-dimensional incompressible flow, we have:
(a)
𝜕𝜕 𝜕𝜕
+
=0
𝜕𝜕 𝜕𝜕
𝜕𝜕 𝜕𝜕
+
= 2𝑦 − 𝑥 2 ≠ 0
𝜕𝜕 𝜕𝜕
This is not incompressible flow.
(b)
𝜕𝜕 𝜕𝜕
+
= −1 + 2𝑥 + 1 − 2𝑥 = 0
𝜕𝜕 𝜕𝜕
This represents the incompressible flow.
(c)
𝜕𝜕 𝜕𝜕
+
= 2𝑥𝑥 − 𝑡 2 ≠ 0
𝜕𝜕 𝜕𝜕
This is not incompressible flow.
(d)
𝜕𝜕 𝜕𝜕
+
= −2𝑥 − 𝑦𝑦 + 2𝑦 + 𝑥𝑥 ≠ 0
𝜕𝜕 𝜕𝜕
This is not incompressible flow.
Problem 5.3
(Difficulty 1)
5.3 In an incompressible three-dimensional flow field, the velocity components are given by 𝑢 = 𝑎𝑎 +
𝑏𝑏𝑏; 𝑣 = 𝑐𝑐 + 𝑑𝑑𝑑. Determine the form of the 𝑧 component of velocity. If the 𝑧 component were not a
function of 𝑥 or 𝑦, what would the form be?
Find: The 𝑧 component of velocity.
Assumptions: The flow is steady and incompressible
Solution: Use the continuity equation for incompressible flow is:
�⃗ = 0
∇∙𝑉
or
𝜕𝜕 𝜕𝜕 𝜕𝜕
+
+
=0
𝜕𝜕 𝜕𝜕 𝜕𝜕
Thus
𝜕𝜕
=𝑎
𝜕𝜕
𝜕𝜕
=𝑐
𝜕𝜕
As 𝑤 is not a function of 𝑥 and 𝑦 we have:
𝜕𝜕
= −𝑎 − 𝑐
𝜕𝜕
𝑤 = −(𝑎 + 𝑐)𝑧 + 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
Problem 5.4
(Difficulty 1)
5.4 In a two-dimensional incompressible flow field, the 𝑥 component of velocity is given by 𝑢 = 2𝑥.
Determine the equation for 𝑦 component of velocity if 𝑣 = 0 along the x-axis.
Find: The equation for 𝑦 component of velocity.
Assumptions: The flow is steady and incompressible
Solution: Use the continuity equation for incompressible two-dimensional flow:
𝜕𝜕 𝜕𝜕
+
=0
𝜕𝜕 𝜕𝜕
We have:
𝜕𝜕
=2
𝜕𝜕
Thus
𝜕𝜕
= −2
𝜕𝜕
We know:
𝑣 = −2𝑦 + 𝑓(𝑥)
At 𝑦 = 0, 𝑣 = 0, so we have:
The equation for 𝑦 component of velocity is:
𝑓(𝑥) = 0
𝑣 = −2𝑦
Problem 5.5
Problem
5.4
[Difficulty: 1]
5.5
The velocity field provided above
Given:
The conditions under which this fields could represent incompressible flow
Find:
We will check this flow field against the continuity equation
Solution:
Governing

u    v    w    0 (Continuity equation)
Equations:
x
y
z
t
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 5.6
Problem
5.6
[Difficulty: 2]
5.6
The x-component of velocity in a steady, incompressible flow field
Given:
The simplest y-component of velocity for this flow field
Find:
We will check this flow field against the continuity equation
Solution:
Governing

u    v    w    0 (Continuity equation)
Equations:
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:
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
Problem
5.8
[Difficulty: 3]
5.7
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 ) 

 dx 


2
x y
The simplest form is
 2 x x  3 y
2 x y 


 
u   v   

2
3
x
y
y  2

2 
2
2
x

y

y
x




 2 x x2  3 y2

3

2
2
x

y

1
2
1
2
x y
2
2
or



2 y
x



x2  y2
2
2
y
2 y
2
x2  y2

2
2
 f (y) 

2
x  y  2 y
x2  y2
2
2
 f ( y)
2
2
2
2
x y

 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
 2  y2

x


  0 However, this does not verify

the solution is the simplest.


2
2



Problem 5.8
(Difficulty 1)
5.8 The velocity components for an incompressible steady flow field are 𝑢 = 𝑎(𝑥 2 + 𝑧 2 ) and
𝑣 = 𝑏(𝑥𝑥 + 𝑦𝑦). Determine the general expression for the 𝑧 component of velocity. If the flow were
unsteady, what would be the expression for 𝑧 component ?
Find: The expression for 𝑧 component velocity.
Assumptions: The flow is steady and incompressible
Solution: Use the continuity equation:
�⃗ +
∇ ∙ 𝜌𝑉
For incompressible flow:
𝜕𝜕
=0
𝜕𝜕
𝜌 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
�⃗ = 0
∇∙𝑉
Thus
𝜕𝜕 𝜕𝜕 𝜕𝜕
+
+
=0
𝜕𝜕 𝜕𝜕 𝜕𝜕
𝜕𝜕
= 2𝑎𝑎
𝜕𝜕
𝜕𝜕
= 𝑏(𝑥 + 𝑧)
𝜕𝜕
𝜕𝜕
= −2𝑎𝑎 − 𝑏𝑏 − 𝑏𝑏 = (−2𝑎 − 𝑏)𝑥 − 𝑏𝑏
𝜕𝜕
𝑏
𝑤 = (−2𝑎 − 𝑏)𝑥𝑥 − 𝑧 2 + 𝑓(𝑥, 𝑦)
2
𝑓(𝑥, 𝑦) is a general function of 𝑥 and 𝑦.
If the flow were non-steady, the expression for 𝑧 component will be the same because 𝜌 is constant. The
term respect to time in the continuity equation is always zero for incompressible flow.
Problem 5.9
(Difficulty 2)
5.9 The radial component of velocity in an incompressible two-dimensional flow is given by 𝑉𝑟 = 3𝑟 −
2𝑟 2 cos(𝜃). Determine the general expression for the 𝜃 component of velocity. If the flow were nonsteady, what would be the expression for the 𝜃 component?
Find: The expression for 𝜃 component velocity.
Assumptions: The flow is steady and incompressible
Solution: Use the continuity equation:
For incompressible flow:
�⃗ +
∇ ∙ 𝜌𝑉
𝜕𝜕
=0
𝜕𝜕
𝜌 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
Thus
�⃗ = 0
∇∙𝑉
𝜕(𝑟𝑉𝑟 ) 𝜕𝑉𝜃
+
=0
𝜕𝜕
𝜕𝜕
We can get:
𝜕(𝑟𝑉𝑟 ) 𝜕(3𝑟 2 − 2𝑟 3 cos(𝜃))
=
= 6𝑟 − 6𝑟 2 cos(𝜃)
𝜕𝜕
𝜕𝜕
𝜕𝑉𝜃
= 6𝑟 2 cos(𝜃) − 6𝑟
𝜕𝜕
𝑉𝜃 = 6𝑟 2 sin(𝜃) − 6𝑟𝑟 + 𝑓(𝑟)
If the flow were non-steady, the expression for 𝜃 component will be the same because 𝜌 is constant. The
term respect to time in the continuity equation is always zero for incompressible flow.
Problem 5.10
Problem
5.10
[Difficulty: 2]
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
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
Uy
u

  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 
v
δ
The maximum value of v/U is where y = δ: v ratmax 

u
4 x
v ratmax 
dy  f ( x ) 
3
4  c x
U y
3
4  c x
U y


y
1 4 x
2
5  10
2
c x

 f ( x)
2
u y
4 x
(Q.E.D.)
v
u y

4 x
2
3
m
4  0.5 m
v ratmax  0.0025
Problem 5.11
Problem
5.11
[Difficulty: 3]
5.11
Given:
Approximate (parabolic) profile for a laminar boundary layer:
u
U
 2  
y
y
 
δ δ
2
δ  c x
(c is constant)
(a) Show that the simplest form of v for incompressible flow is
Find:
v
U

1 y
1 y
        
3 δ
x 2  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)
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:
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


v
u Uc 2


x  2
y
2
2

c
and thus
δ
 y   y  2 
     
      
 y   y  2
     dy  f ( x )
δ  δ   δ  
U c
 
1
3
2
2
3
 y2
U c  1  y 
1 y 
y 
2

δ
v


 f ( x) 
           f ( x)
2 2
Since
δ  c x c 
2  2 δ
2
3 δ 
δ 2  δ 
δ 
3 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
        
3 δ
x 2  δ 
2
δ
3


v
(Q.E.D.)
U
1 y
1 y
        
3 δ
x 2  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
4
5 10
0.001
0.0015
0.002
Dimensionless Velocity (v/U)
v
δ 1
δ
1
The maximum value of v/U is where y = δ: v ratmax 
     
x 2
U
3  6 x
v ratmax 
5  10
3
m
6  0.5 m
v ratmax  0.00167
Problem 5.12
Problem
5.13
[Difficulty: 3]
5.12
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 )  


3
4
and
du
dx

3
4
 y3 x5 y x3 

   dy
 c3 2 c 2 


 U 
4 
 y2
y


v ( x y )   U

3
5
8 
 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 5.13
Problem
5.14
[Difficulty: 3]
5.13
Steady, incompressible flow in x-y plane:
Given:
2 2
u  A x  y
3 1
A  0.3 m
s
(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)
Find:
We will check this flow field using the continuity equation
Solution:
Governing

u    v    w    0 (Continuity equation)
Equations:
x
y
z
t
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 5.14
Problem
5.16
[Difficulty: 5]
5.14
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 5.15
Problem
5.18
[Difficulty: 2]
5.15
The list of velocity fields provided above
Given:
Find:
Solution:
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
r r
r 
z
t
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 cos  1      U cos  1      0
r

  r  
  r  
This could be an incompressible flow field.
Problem 5.16
Problem
5.20
[Difficulty: 3]
5.16
Given:
r component of velocity
Find:
θ component for incompressible flow; How many θ components
Solution:
Basic equation:
Assumptions:
1 
1 


ρ Vθ 
 ρ r Vr  
ρ Vz  ρ  0
r r
r θ
z
t






Incompressible flow
Flow in r-θ plane
Hence
1 
1 
V 0
 r Vr  
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
5.22
Problem 5.17
[Difficulty: 2]
5.17
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
r r
r 
z
t

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 5.18
Problem
5.24
[Difficulty: 3]
5.18
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 dθ
r dr
Hence
1 d
1 d
 r Vr   Vθ  0
r dθ
r dr
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 5.19
Problem
*5.26
[Difficulty: 3]
5.19
Given:
Velocity field
Find:
Stream function ψ
Solution:
Basic equations:  ( ρ u )   ( ρ v)   ( ρ w)   ρ  0
x
Assumptions:
Hence
y
z
t
u

y

v ψ
x
ψ
Incompressible flow
Flow in x-y plane

x
u 

y
v 0

u  2 y ( 2 x  1) 
and
2

v  x ( x  1)  2 y   ψ
x
Comparing these
f ( x)  
The stream function is
ψ( x y )  y  2  x  y 
Checking
u ( x y ) 
3
3

2
x
y
x
[ 2  y  ( 2x  1 ) ] 
ψ


ψ( x y )  

2
g(y)  y
and
2
2

y
x ( x  1 )  2  y2  0

2
2
ψ( x y )   2  y  ( 2  x  1 ) dy  2  x  y  y  f ( x )

Hence
x

or
x
2
2

x
3
2
x ( x  1 )  2  y2 dx   x  x  2  x y 2  g( y
2
3
3
2
3
 2
x
x 
 y  2 x y2 
  u( x y )  2  y  4  x y

2
3 
y 
2
3
 2
x
x 
2

  v ( x y)  x2  x  2 y 2
v ( x y )    y  2  x  y 

2
3 
x 

3
2
Problem 5.20
Problem
*5.28
[Difficulty: 2]
5.20
Stream function for an incompressible flow field:
Given:
ψ  U r sin( θ) 
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.
Find:
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 0


 2  π U 
0  0
ψstagnation  0
Problem 5.21
(Difficulty 2)
5.21 Determine the stream functions for the following flow fields. For the value of 𝛹 = 2, plot the
streamline in the region between 𝑥 = −1 and 𝑥 = 1.
(a) 𝑢 = 4; 𝑣 = 3
(b) 𝑢 = 4𝑦, 𝑣 = 0
(c) 𝑢 = 4𝑦, 𝑣 = 4𝑥
(d) 𝑢 = 4𝑦, 𝑣 = −4𝑥
Find: Determine the stream functions for the flow fields.
Assumptions: The flow is steady and incompressible
Solution: Use the definitions of stream function:
𝑢=
(a) In this case, we have:
𝜕𝜕
𝜕𝜕
𝑣=−
𝑢=
𝜕𝜕
𝜕𝜕
𝜕𝜕
=4
𝜕𝜕
𝛹 = 4𝑦 + 𝑓(𝑥)
𝑣=−
So the stream function is:
𝜕𝜕
𝜕𝜕
=−
=3
𝜕𝜕
𝜕𝜕
𝜕𝜕
= −3
𝜕𝜕
𝑓(𝑥) = −3𝑥 + 𝑐
𝛹 = 4𝑦 − 3𝑥 + 𝑐
The plot for the streamline is shown by (𝑐 = 0):
𝛹 = 4𝑦 − 3𝑥 = 2
𝑦 = 0.75𝑥 + 0.5
1.4
1.2
1
0.8
y
0.6
0.4
0.2
0
-0.2
-0.4
-1
-0.8
-0.6
-0.4
-0.2
0
x
0.2
0.4
0.6
(b) In this case we have:
𝑢=
𝜕𝜕
= 4𝑦
𝜕𝜕
𝛹 = 2𝑦 2 + 𝑓(𝑥)
𝑣=−
The streamline is showing by (𝑐 = 0):
𝜕𝜕
𝜕𝜕
=−
=0
𝜕𝜕
𝜕𝜕
𝑓(𝑥) = 𝑐
𝛹 = 2𝑦 2 + 𝑐
2𝑦 2 = 2
𝑦2 = 1
𝑦 = ±1
0.8
1
2
1.5
1
y
0.5
0
-0.5
-1
-1.5
-2
-1.5
-1
-0.5
0
x
(c) In this case we have:
𝑢=
0.5
𝜕𝜕
= 4𝑦
𝜕𝜕
𝛹 = 2𝑦 2 + 𝑓(𝑥)
𝑣=−
𝜕𝜕
𝜕𝜕
=−
= 4𝑥
𝜕𝜕
𝜕𝜕
𝑓(𝑥) = −2𝑥 2
The streamline is showing by (𝑐 = 0):
𝛹 = 2𝑦 2 − 2𝑥 2 + 𝑐
2𝑦 2 − 2𝑥 2 = 2
𝑦2 = 𝑥2 + 1
𝑦 = ±�𝑥 2 + 1
1
1.5
1.5
1
y
0.5
0
-0.5
-1
-1.5
-1
-0.8
-0.6
-0.4
-0.2
(d) In this case we have:
𝑢=
0
x
0.2
𝜕𝜕
=4
𝜕𝜕
𝛹 = 4𝑦 + 𝑓(𝑥)
𝑣=−
The streamline is showing by (𝑐 = 0):
0.4
𝜕𝜕
𝜕𝜕
=−
= −4𝑥
𝜕𝜕
𝜕𝜕
𝑓(𝑥) = 2𝑥 2
𝛹 = 4𝑦 + 2𝑥 2 + 𝑐
4𝑦 + 2𝑥 2 = 2
−𝑥 2 + 1
𝑦=
2
0.6
0.8
1
0.5
0.45
0.4
0.35
y
0.3
0.25
0.2
0.15
0.1
0.05
0
-1
-0.8
-0.6
-0.4
-0.2
0
x
0.2
0.4
0.6
0.8
1
Problem 5.22
(Difficulty 1)
5.22 Determine the stream function for the steady incompressible flow between parallel plates. The
velocity profile is parabolic and given by 𝑢 = 𝑢𝑐 + 𝑎𝑦 2, where 𝑢𝑐 is the centerline velocity and 𝑦 is the
distance measured from the centerline. The plate spacing is 2𝑏 and the velocity is zero at each plate.
Explain why the stream function is not a function of x.
Find: Determine the stream function 𝛹 and explain why it is not a function of x.
Assumptions: The flow is steady and incompressible
Solution: Use the definition of stream function
For this flow we have the velocity as:
𝑢 = 𝑢𝑐 + 𝑎𝑦 2
𝑣=0
For the stream function, we have:
𝑢=
𝜕𝜕
𝜕𝜕
𝑣=−
Thus
𝑢=
𝜕𝜕
𝜕𝜕
𝜕𝜕
= 𝑢𝑐 + 𝑎𝑦 2
𝜕𝜕
1
𝛹 = 𝑢𝑐 𝑦 + 𝑎𝑦 3 + 𝑓(𝑥)
3
𝑣=−
c is a constant. So we get:
𝜕𝑓(𝑥)
𝜕𝜕
=−
=0
𝜕𝜕
𝜕𝜕
𝑓(𝑥) = 𝑐
1
𝛹 = 𝑢𝑐 𝑦 + 𝑎𝑦 3 + 𝑐
3
The reason that stream function is independent of 𝑥 is because this is steady flow between two plates.
This is unidirectional flow and the velocity profile is the same at all 𝑥 locations and the 𝑦 component of
velocity 𝑣 is zero.
Problem 5.23
Problem
*5.30
[Difficulty: 3]
5.23
Stream function for an incompressible flow field:
Given:
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:
10
ψ =0 is in red; ψ = 5 is in blue
y (m)
5
4
2
v  5 A

V  4iˆ  10 ˆj
At the point (4, 1) the stream function value is:
The flow rate between these two streamlines is:
ψa  5  2 
ψb  5  2 
m
s
m
s
Q  ψb  ψa
 2 m  2  2
 4 m  2  2
4
 10
Therefore, the velocity vector at (0, 0) is:
At the point (2, 2) the stream function value is:
2
5
Generating the velocity components from the stream function derivatives:
u  2  A
0
x (m)
m
s
m
s
2
2
 2  m ψa  28
m
s
2
 1  mψb  44
2


m
m
   28 
Q   44
s  
s 

m
s
3
Q  16
m
s m
Flow rate is 16 m3/s per meter of depth
Problem *5.32
5.24
Problem
[3]
5.24
Approximate profile for a laminar boundary layer:
Given:
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:
 
 
 
3 δ 
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


 
 
3 δ 
 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
      or    3    1  0
3 δ 
 δ 
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 5.25
(Difficulty 2)
5.25 A flow field is characterized by the stream function 𝛹 = 3𝑥 2 𝑦 − 𝑦 3 . Demonstrate that the flow
field represents a two-dimensional incompressible flow. Show that the magnitude of the velocity
depends only on the distance from the origin of the coordinates. Plot the stream line 𝛹 = 2.
Find: Demonstrate two-dimensional incompressible flow and that the magnitude only depends on
distance from the origin. Plot stream line 𝛹 = 2.
Assumptions: The flow is steady and incompressible
Solution: Use the definition of stream function
For this flow, the stream function is:
The velocity field is given by:
𝛹 = 3𝑥 2 𝑦 − 𝑦 3
𝑢=
𝜕𝜕
= 3𝑥 2 − 3𝑦 2
𝜕𝜕
𝑣=−
𝜕𝜕
= −6𝑥𝑥
𝜕𝜕
For the two-dimensional incompressible flow, we should satisfy the continuity equation as:
So we have:
𝜕𝜕 𝜕𝜕
+
=0
𝜕𝜕 𝜕𝜕
𝜕𝜕
= 6𝑥
𝜕𝜕
Thus
𝜕𝜕
= −6𝑥
𝜕𝜕
𝜕𝜕 𝜕𝜕
+
= 6𝑥 − 6𝑥 = 0
𝜕𝜕 𝜕𝜕
So this is two-dimensional incompressible flow.
The magnitude of the velocity is:
𝑉 = �𝑢2 + 𝑣 2 = �(3𝑥 2 − 3𝑦 2 )2 + (−6𝑥𝑥)2
𝑉 = �9𝑥 4 − 18𝑥 2 𝑦 2 + 9𝑦 4 + 36𝑥 2 𝑦 2 = �9𝑥 4 + 18𝑥 2 𝑦 2 + 9𝑦 4 = �9(𝑥 2 + 𝑦 2 )2 = 3(𝑥 2 + 𝑦 2 )
As we know the distance from the origin is:
𝑟2 = 𝑥 2 + 𝑦 2
Thus
𝑉 = 3𝑟 2
So the magnitude of the velocity depends only on the distance from the origin.
The stream line for 𝛹 = 2 is shown by:
𝑥2 =
2 + 𝑦3
3𝑦
2 + 𝑦3
𝑥 = ±�
3𝑦
Stream line for Ψ=2
5
4.5
4
3.5
y
3
2.5
2
1.5
1
0.5
0
-10
-8
-6
-4
-2
0
x
2
4
6
8
10
Problem 5.26
(Difficulty 1)
5.26 A flow field is characterized by the stream function 𝛹 = 𝑥𝑥. Plot sufficient streamlines to represent
the flow field. Determine the location of any stagnation points. Give at least two possible physical
interpretations of this flow.
Find: Plot sufficient streamlines to represent the flow field. Determine the stagnation points.
Assumptions: The flow is steady and incompressible
Solution: Use the definition of stream function
For this flow, the stream function is:
𝛹 = 𝑥𝑥
The plot of the streamlines is then
250
200
150
100
y
50
0
-50
-100
-150
-200
-250
-1
-0.8
-0.6
-0.4
-0.2
0
x
0.2
0.4
0.6
0.8
1
The velocity field is:
𝑢=
For the stagnation points, we have:
𝑣=−
𝜕𝜕
=𝑥
𝜕𝜕
𝜕𝜕
= −𝑦
𝜕𝜕
𝑢=𝑣=0
Thus
The stagnation point is the origin.
This flow can represent
(1) a jet hitting a wall;
(2) flow in a corner.
𝑢 = 0 𝑎𝑎 𝑥 = 0 𝑎𝑎𝑎 𝑣 = 0 𝑎𝑎 𝑦 = 0
Problem 5.27
Problem
*5.34
[Difficulty: 3]
5.27
5.12
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        
2  δ  2  δ 
y
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        
8 δ
4  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 5.28
(Difficulty 2)
5.28 A flow field is characterized by the stream function
𝛹=
1
1
𝑦−𝑎
𝑦+𝑎
�𝑡𝑡𝑡−1
− 𝑡𝑡𝑡−1
�−
ln �𝑥 2 + 𝑦 2
2𝜋
2𝜋
𝑥
𝑥
Locate the stagnation points and sketch the flow field. Derive an expression for the velocity at (𝑎, 0).
Find: Locate stagnation points and sketch the flow. Determine the velocity at (𝑎, 0).
Assumptions: The flow is steady and incompressible
Solution: Use the definition of stream function
The stream function for this flow is given by:
𝛹=
1
1
𝑦−𝑎
𝑦+𝑎
�𝑡𝑡𝑡−1
− 𝑡𝑡𝑡−1
�−
ln �𝑥 2 + 𝑦 2
2𝜋
2𝜋
𝑥
𝑥
The velocity field is related to the stream function by:
𝑢=
𝜕𝜕
1
1
𝑥
𝑥
𝑦
=−
+
�
−
�
𝜕𝜕
2𝜋 𝑥 2 + 𝑦 2 2𝜋 𝑥 2 + (𝑦 − 𝑎)2 𝑥 2 + (𝑦 + 𝑎)2
𝑣=−
𝜕𝜕
1
1
𝑎−𝑦
𝑎+𝑦
𝑥
=
−
� 2
+ 2
�
2
2
2
𝜕𝜕 2𝜋 𝑥 + 𝑦
2𝜋 𝑥 + (𝑦 − 𝑎)
𝑥 + (𝑦 + 𝑎)2
The velocity expression at (𝑎, 0) is:
𝑣=−
𝑢=
𝜕𝜕
1
𝑎
𝑎
=
� 2
− 2
�=0
2
𝜕𝜕 2𝜋 𝑎 + (0 − 𝑎)
𝑎 + (𝑎)2
𝜕𝜕
1 1
1
𝑎
𝑎
1 1
1 2𝑎
1 1
1 1
=
−
� 2
+ 2
�=
−
=
−
=0
2
2
2
𝜕𝜕 2𝜋 𝑎 2𝜋 𝑎 + (−𝑎)
𝑎 + (𝑎)
2𝜋 𝑎 2𝜋 2𝑎
2𝜋 𝑎 2𝜋 𝑎
So the stagnation point is (𝑎, 0). The fluid flow is sketched as:
Problem 5.29
Problem
*5.36
[Difficulty: 3]
5.29
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
x

u 
y
y
v 0

Hence
u  U
and

v0 ψ
x
Comparing these
f ( x)  0
The stream function is
h
ψ( x y ) 

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
U 
1 U h  U h

u dy   
y dy 
  

2 h
h 0
2  2 
4
2 0
Q
2
h half 
1
2
h
2
h half 
1
2
h 
1.5 m
2
 1.06 m
Problem 5.30
Problem
5.38
[Difficulty: 2]
5.30
The velocity field provided above
Given:
Find:
(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)
We will check this flow field against the continuity equation, and then apply the definition of acceleration
Solution:
Governing

u    v    w    0 (Continuity equation)
Equations:
x
y
z
t






V V (Particle acceleration)
V
DV
V

w
v
ap 
u
t
z
y
Dt
x
(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)
Assumptions:
The flow is two dimensional.
Based on assumption (2), we may state that:
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 5.31
Problem
5.40
[Difficulty: 2]
5.31
The velocity field provided above
Given:
Find:
(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)
We will check this flow field against the continuity equation, and then apply the definition of acceleration
Solution:
Governing

u    v    w    0 (Continuity equation)
Equations:
x
y
z
t






V V (Particle acceleration)
V
DV
V

w
v
ap 
u
t
z
y
Dt
x
(1) Incompressible flow (ρ is constant)
(2) Steady flow (velocity is not a function of t)
Assumptions:
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
v
ap  u
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 5.32
Problem
5.42
[Difficulty: 2]
5.32
Given:
Find:
The velocity field provided above
(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
We will check this flow field against the continuity equation, and then apply the definition of acceleration
Solution:
Governing

u    v    w    0 (Continuity equation)
Equations:
x
y
z
t






V V (Particle acceleration)
V
DV
V

w
v
ap 
u
t
z
y
Dt
x
(1) Incompressible flow (ρ is constant)
Assumptions:
(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:

AUy
3 AUy 2 ˆ
V
  3 2 iˆ 
j
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
x  2x
8x
2x
4x
4x

 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 5.33
(Difficulty 1)
�⃗ = 10𝑡𝚤̂ −
5.33 A velocity field is given by 𝑉
10
𝚥̂.
𝑡3
Show that the flow field is a two-dimensional flow and
determine the acceleration as a function of time.
Find: Show that this is two-dimensional flow. Determine the acceleration.
Assumptions: The flow is steady and incompressible
Solution: Use the expression for acceleration
The velocity is given by:
�⃗
�⃗ 𝜕𝑉
�⃗
�⃗
�⃗
𝑑𝑉
𝐷𝑉
𝜕𝑉
𝜕𝑉
= 𝑎⃗𝑝 =
=
+𝑢
+𝑣
𝑑𝑑
𝐷𝐷
𝜕𝜕
𝜕𝜕
𝜕𝜕
�⃗ = 10𝑡𝚤̂ − 103 𝚥̂.
𝑉
𝑡
Thus
𝑢 = 10𝑡
𝑣=−
10
𝑡3
𝑤=0
This is two-dimensional flow with u and v depending only on the time t. The x- and y-accelerations are
given by
𝑎𝑥 =
So the acceleration is:
𝑎𝑦 =
𝐷𝐷
𝜕𝜕
𝜕𝜕 𝜕𝜕 𝜕𝜕
=𝑢
+𝑣
+
=
= 10
𝐷𝐷
𝜕𝜕
𝜕𝜕 𝜕𝜕 𝜕𝜕
𝐷𝐷
𝜕𝜕
𝜕𝜕 𝜕𝜕 𝜕𝜕 30
=𝑣
+𝑣
+
=
=
𝐷𝐷
𝜕𝜕
𝜕𝜕 𝜕𝜕 𝜕𝜕 𝑡 4
𝑎⃗ = 10𝚤̂ +
30
𝚥̂
𝑡4
Problem 5.34
Problem
5.44
[Difficulty: 2]
5.34
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)
x
y
z
t






V V (Particle acceleration)
V
DV
V

w
v
ap 
u
t
z
y
Dt
x
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
u v
u
v

 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:
1
Lt
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 5.35
(Difficulty 1)
5.35 A 4 𝑚 diameter tank is filled with water and then rotated at a rate of 𝜔 = 2𝜋(1 − 𝑒 −𝑡 )
𝑟𝑟𝑟
.
𝑠
At the
tank walls, viscosity prevents relative motion between the fluid and the wall. Determine the speed and
acceleration of the fluid particles next to the tank walls as a function of time.
Find: The speed and acceleration of the fluid particles next to tank walls.
Assumptions: The flow is steady and incompressible
Solution: Use the expression for acceleration
�⃗
�⃗ 𝜕𝑉
�⃗
�⃗
�⃗
𝑑𝑉
𝐷𝑉
𝜕𝑉
𝜕𝑉
= 𝑎⃗𝑝 =
=
+𝑢
+𝑣
𝑑𝑑
𝐷𝐷
𝜕𝜕
𝜕𝜕
𝜕𝜕
The fluid particles velocity next to the tank wall are the same as the tank because of the viscosity. The
particle velocity is
1
𝑟𝑟𝑟
𝑚
1
𝑉 = 𝜔𝜔 = 𝜔𝜔 = × 2𝜋(1 − 𝑒 −𝑡 )
× 4 𝑚 = 4𝜋(1 − 𝑒 −𝑡 )
2
𝑠
𝑠
2
The tangential acceleration is given by:
The normal acceleration is given by:
𝑎𝑡 =
𝑑𝑑
𝑚
= 4𝜋𝑒 −𝑡 2
𝑑𝑑
𝑠
𝑚2
16𝜋 2 (1 − 𝑒 −𝑡 )2 2
𝑉2
𝑠 = −8𝜋 2 (1 − 𝑒 −𝑡 )2 𝑚
𝑎𝑛 = −
=−
𝑠2
𝑟
2𝑚
Problem 5.36
Problem
5.46
[Difficulty: 2]
5.36
Given:
Duct flow with incompressible, inviscid liquid
U  5
m
s
L  0.3 m
u ( x )  U  1 

x


2 L 
Find:
Expression for acceleration along the centerline of the duct
Solution:
We will apply the definition of acceleration to the velocity.
Governing
Equation:
Assumptions:






V V (Particle acceleration)
V
DV
V

w
v
ap 
u
t
z
y
Dt
x
(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
    U    U   1  x 
 



2 L 
2 L 
2 L   2 L 
x
2
apx  
U
2 L
  1 

x


2 L 
Problem 5.37
(Difficulty 2)
5.37 Sketch the following flow fields and derive general expressions for the acceleration.
a)
b)
c)
d)
𝑢 = 2𝑥𝑥; 𝑣 = −𝑥 2 𝑦.
𝑢 = 𝑦 − 𝑥 + 𝑥 2 ; 𝑣 = 𝑥 + 𝑦 − 2𝑥𝑥.
𝑢 = 𝑥 2 𝑡 + 2𝑦; 𝑣 = 2𝑥 − 𝑦𝑡 2 .
𝑢 = −𝑥 2 − 𝑦 2 − 𝑥𝑥𝑥; 𝑣 = 𝑥 2 + 𝑦 2 + 𝑥𝑥𝑥
Find: Sketch the flow fields and derive the general expressions for acceleration.
Assumptions: The flow is steady and incompressible
Solution: Use the expression for acceleration
�⃗
�⃗ 𝜕𝑉
�⃗
�⃗
�⃗
𝑑𝑉
𝐷𝑉
𝜕𝑉
𝜕𝑉
= 𝑎⃗𝑝 =
=
+𝑢
+𝑣
𝑑𝑑
𝐷𝐷
𝜕𝜕
𝜕𝜕
𝜕𝜕
(a) The flow field is shown in the figure:
The acceleration can be calculated as:
𝐷𝐷 𝜕𝜕
𝜕𝜕
𝜕𝜕
𝑎𝑥 =
=
+𝑢
+𝑣
= 2𝑥𝑥(2𝑦) + (−𝑥 2 𝑦)(2𝑥) = 4𝑥𝑦 2 − 2𝑥 3 𝑦
𝐷𝐷 𝜕𝜕
𝜕𝜕
𝜕𝜕
𝐷𝐷 𝜕𝜕
𝜕𝜕
𝜕𝜕
=
+𝑢
+𝑣
= 2𝑥𝑥(−2𝑥𝑥) + (−𝑥 2 𝑦)(−𝑥 2 ) = −4𝑥 2 𝑦 2 + 𝑥 4 𝑦
𝑎𝑦 =
𝐷𝐷 𝜕𝜕
𝜕𝜕
𝜕𝜕
(b) The flow field is shown in the figure:
The acceleration can be calculated as:
𝐷𝐷 𝜕𝜕
𝜕𝜕
𝜕𝜕
𝑎𝑥 =
=
+𝑢
+𝑣
= (𝑦 − 𝑥 + 𝑥 2 )(−1 + 2𝑥) + (𝑥 + 𝑦 − 2𝑥𝑥)
𝐷𝐷
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝑎𝑥 = 2𝑥 − 3𝑥 2 + 2𝑥 3
𝐷𝐷 𝜕𝜕
𝜕𝜕
𝜕𝜕
𝑎𝑦 =
=
+𝑢
+𝑣
= (𝑦 − 𝑥 + 𝑥 2 )(1 − 2𝑦) + (𝑥 + 𝑦 − 2𝑥𝑥)(1 − 2𝑥)
𝐷𝐷 𝜕𝜕
𝜕𝜕
𝜕𝜕
(c) The flow field at 𝑡 = 1𝑠 is shown in the figure:
The acceleration can be calculated as:
𝐷𝐷 𝜕𝜕
𝜕𝜕
𝜕𝜕
𝑎𝑥 =
=
+𝑢
+𝑣
= 𝑥 2 + (𝑥 2 𝑡 + 2𝑦)(2𝑥𝑥) + (4𝑥 − 2𝑦𝑡 2 )
𝐷𝐷
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝐷𝐷 𝜕𝜕
𝜕𝜕
𝜕𝜕
𝑎𝑦 =
=
+𝑢
+𝑣
= (−2𝑦𝑦) + (2𝑥 2 𝑡 + 4𝑦) + (2𝑥 − 𝑦𝑡 2 )(−𝑡 2 )
𝐷𝐷 𝜕𝜕
𝜕𝜕
𝜕𝜕
(d) The flow field at 𝑡 = 1𝑠 is shown in the figure:
The acceleration can be calculated as:
𝐷𝐷 𝜕𝜕
𝜕𝜕
𝜕𝜕
𝑎𝑥 =
=
+𝑢
+𝑣
𝐷𝐷 𝜕𝜕
𝜕𝜕
𝜕𝜕
= (−𝑥𝑥) + (−𝑥 2 − 𝑦 2 − 𝑥𝑥𝑥)(−2𝑥 − 𝑦𝑦) + (𝑥 2 + 𝑦 2 + 𝑥𝑥𝑥)(−2𝑦 − 𝑥𝑥)
𝑎𝑦 =
𝑎𝑥 = (−𝑥𝑥) + (𝑥 2 + 𝑦 2 + 𝑥𝑥𝑥)(−2𝑦 − 𝑥𝑥 + 2𝑥 + 𝑦𝑦)
𝐷𝐷 𝜕𝜕
𝜕𝜕
𝜕𝜕
=
+𝑢
+𝑣
𝐷𝐷 𝜕𝜕
𝜕𝜕
𝜕𝜕
= (𝑥𝑥) + (−𝑥 2 − 𝑦 2 − 𝑥𝑥𝑥)(2𝑥 + 𝑦𝑦) + (𝑥 2 + 𝑦 2 + 𝑥𝑥𝑥)(2𝑦 + 𝑥𝑥)
𝑎𝑦 = (𝑥𝑥) + (𝑥 2 + 𝑦 2 + 𝑥𝑥𝑥)(2𝑦 + 𝑥𝑥 − 2𝑥 − 𝑦𝑦)
Problem 5.38
Problem
5.48
[Difficulty: 2]
5.38
Incompressible, inviscid flow of air between parallel disks
Given:
Find:
(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


We will apply the conservation of mass and the definition of acceleration to the velocity.
Solution:

Governing
1 
1 V









rV


V


V

 0 (Continuity Equation)
r

z
Equations:
r r 
r 
t
 z

 V

DV
ap 
 V  V 
(Particle acceleration)
Dt
t
(1) Incompressible flow (ρ is constant)
Assumptions:


(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
Thus: Vr 
should be the form of the solution. Now since at r = R:
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
3
V R
 R
   V 2    R   
r
r
 r 
R
Therefore, at r = ri:
Therefore, at r = R:
apr   15
m
apr   15
m



apr  Vr Vr
r
2
1
75
  
 
s
0.075  m  25 
2
1
75
  
 
s
0.075  m  75 
3
4m
apr  8.1  10
3
apr  3  10
2
s
3m
2
s
Problem 5.39
Problem
5.50
[Difficulty: 4]
5.39
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
x
y
z t
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 A  e
 u
Dt
v0
dx
 

c( x )  A  e
w 0
x
2 a
dx
We need to convert this to a function of time. For this motion u = U so
 


e
Dt
a 
Dc
U A
U t
a

1
2

e
2 a
 
U A 
a 
e
e
 
a 

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
5
1 10
6
c (ppm)
8 10
6
6 10
6
4 10
6
2 10
0
10
20
x (ft)
30
4
Dc/Dt (ppm/s)
4 10
4
3 10
4
2 10
4
1 10
4
0
0.1
0.2
0.3
0.4
0.5
 1 10
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
tmax  0.119  s
70 ft

1
2
4
x max  8.32 ft
s
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 5.40
Problem
5.52
[Difficulty: 2]
5.40
Given:
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:
DT
T
T
T T
Governing
(Substantial Derivative)
u
v
w

Equation:
Dt
x
y
z t
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 
 675.56 ft    41.67  ft 

 

s
s


So the rate of change of temperature through the cold front is:
δTx 
0.7 Δ°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
Δ°F
δTx  0.0914
mi
Problem 5.41
Problem
5.54
[Difficulty: 4]
5.41
Given:
Z component of an axisymmetric transient flow.
Find:
Radial component of flow and total acceleration.
Solution:
Governing
Equations:
1  rVr  1 V Vz


 0 (Continuity Equation for an Incompressible Fluid)
r r
r 
z
V V Vr V2
V V
a r , p  Vr r  

 Vz r  r
r
r 
r
z
t (Particle acceleration)
V V
V V Vz
a z , p  Vr z  
 Vz z  z
r
r 
z
t
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:
V
rVr 
1 rVr 
 2t 
 z 
 r   A  sin 

r r
z
r
 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
2
2  T 
r
z
 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 

2
r
z
 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 5.42
Problem
5.56
[Difficulty: 3]
5.42
Given:
Find:
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
Dt
x
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 5.43
Problem
5.58
[Difficulty: 3]
5.43
Given:
Find:
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)
x
y
z
t






V V (Particle acceleration)
V
DV
V

w
v
ap 
u
t
z
y
Dt
x
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
At (x,y) = (3m, 2m)

Therefore:
y   x 

B
ln x 
  constant
A

v
u

C y
A x  B


m
a p  4iˆ  8 ˆj  5kˆ 2
s
Thus:
dx
1 dy
 

or
A x  B
A y
dx
x
B

dy
y
0
A
B
  ln( y)  const
A
Here is a plot of the streamlines
passing through (3, 2):
3
Y (m)
Solving this ODE by integrating:

C  2  s
Substituting in the field:
2
 2  2

2
mˆ  2 
m
3
m
4
a p    
 
i      2 mˆj  5 2 kˆ
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 5.44
Problem
5.60
[Difficulty: 4]
5.44
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) 

1 3
1
   λ   λ 
3
x 2

δ

dδ
dx

1
1
2

 c x

1
2
1
λ  1
λ 1
2
2
 U ( 2  2  λ)   
   c x
   c x
1 2
δ 2

λ
2 x


U λ  λ
2
 c x 2 

x


2
2
  2 U   y   y    2 U λ  λ
 
δ
2
δ δ  δ  
y
y
δ 

2
 2 U λ λ2 
δ 1
1
2  U λ  λ 


  U    λ   λ3     
ax  u  u  v  u  U 2  λ  λ 
x 2
3
x
y




x
y
2
2
2
3
4
U  2 4 3 1 4 U   y 
4 y
1 y 
ax 
  λ   λ   λ  
           
x 
3
3
3 δ
3 δ 
 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
3
 0.634 
2
1
3
 0.634
4


2
 0.116 
1
 
0.8 m
 s
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 5.45
(Difficulty: 3)
5.45 A cubic approximate velocity profile was used in problem 5.12 to model flow in a laminar
incompressible boundary layer on a flat plate. For this profile, obtain an expression for the x and y
components of acceleration of a fluid particle in the boundary layer. Plot 𝑎𝑥 and 𝑎𝑦 at location 𝑥 = 3 𝑓𝑓,
where 𝛿 = 0.04 𝑖𝑖., for a flow with 𝑈 = 20 𝑓𝑓⁄𝑠 . Find the maxima of 𝑎𝑥 at this x location.
Given: Cubic profile for two-dimensional boundary layer
Find: (1) x and y components of acceleration of a fluid particle
(2) plot components as functions of 𝑦⁄𝛿 for 𝑈 = 20 𝑓𝑓⁄𝑠, 𝑥 = 3 𝑓𝑓, 𝛿 = 0.04 𝑖𝑖
(3) maximum values of acceleration at this x location
Assumptions: (1) two dimensional flow (velocity is not a function of z)
(2) incompressible flow
(3) steady flow
Solution: We will apply the acceleration definition.
Based on the above assumptions the particle acceleration reduces to:
𝑎⃗𝑝 = 𝑢
�⃗
�⃗
𝜕𝑉
𝜕𝑉
+𝑣
𝜕𝜕
𝜕𝜕
According to problem 5.12 we have for the velocity profile:
𝑢 3 𝑦
1 𝑦 3
= � �− � �
𝑈 2 𝛿
2 𝛿
1
To make the analysis easier, define 𝜂:
𝛿 = 𝑐𝑥 2
𝜂=
𝑦
= 𝜂(𝑥, 𝑦)
𝛿
𝛿
𝑑𝑑 𝑐 −1
= 𝑥2 =
2𝑥
𝑑𝑑 2
𝑦 𝛿
𝑦
𝜕𝜕 𝜕𝜕 𝑑𝑑
=
∙
=− 2∙
=−
𝛿 2𝑥
2𝑥𝑥
𝜕𝜕 𝜕𝜕 𝑑𝑑
The velocities are given as:
From the continuity equation we have:
3
1
𝑢 = 𝑈 � 𝜂 − 𝜂3�
2
2
𝜕𝜕 𝜕𝜕
+
=0
𝜕𝜕 𝜕𝜕
So we have:
3 3
𝑦
𝜕𝜕 𝜕𝜕 𝜕𝜕
=
∙
= 𝑈 � − 𝜂 2 � ∙ �−
�
2 2
2𝑥𝑥
𝜕𝜕 𝜕𝜕 𝜕𝜕
Integrating
3 3
𝑦
3 3 𝑦 2
𝑦
𝜕𝜕
= 𝑈 � − 𝜂2� ∙ �
� = 𝑈� − � � �∙�
�
2 2
2𝑥𝑥
2 2 𝛿
2𝑥𝑥
𝜕𝜕
𝑣 = 𝑈�
3 𝑦4
3𝑦 2
−
� + 𝑓(𝑥)
8𝑥𝑥 16 𝑥𝛿 2
Apply the boundary condition for 𝑣 = 0 𝑎𝑎 𝑦 = 0 then
𝑓(𝑥) = 0
3
1
𝑢 = 𝑈 � 𝜂 − 𝜂3�
2
2
3𝑦 2
3 𝑦4
3𝑦 2
3 𝑦4
𝑣 = 𝑈�
−
�
=
𝑈
�
−
�
3
16 𝑐𝑥 2
8𝑥𝑥 16 𝑥𝛿 2
8𝑥 2 𝑐
3 3
𝜂
𝜕𝜕
= 𝑈 � − 𝜂 2 � ∙ �− �
2 2
2𝑥
𝜕𝜕
3 3
1
𝜕𝜕 𝜕𝜕 𝜕𝜕
=
∙
= 𝑈 � − 𝜂2� ∙
2 2
𝛿
𝜕𝜕 𝜕𝜕 𝜕𝜕
𝜕𝜕
9 𝑦2 3 𝑦4
= 𝑈 �−
+
�
𝜕𝜕
16 52
8 𝑐𝑥 3
𝑥 𝑐
So the accelerations are:
𝑎𝑝𝑝 = 𝑢
3𝑦
3 𝑦3
𝜕𝜕
= 𝑈� 3 −
�
4 𝑐𝑥 2
𝜕𝜕
2
4𝑥 𝑐
𝜕𝜕
𝜕𝜕
3
1
3 3
𝜂
3𝑦 2
3 𝑦4
3 3
1
+𝑣
= 𝑈 � 𝜂 − 𝜂 3 � 𝑈 � − 𝜂 2 � ∙ �− � + 𝑈 � 3 −
� 𝑈 � − 𝜂2� ∙
2
𝜕𝑥
𝜕𝜕
2
2
2 2
2𝑥
16 𝑐𝑥
2 2
𝛿
8𝑥 2 𝑐
𝑎𝑝𝑝 = 𝑢
3 3
3
1
𝜂
3𝑦 2
3 𝑦4 1
𝑎𝑝𝑝 = 𝑈 2 � − 𝜂 2 � �� 𝜂 − 𝜂 3 � ∙ �− � + � 3 −
�∙ �
2 2
2
2
2𝑥
16 𝑐𝑥 2 𝛿
8𝑥 2 𝑐
𝜕𝜕
𝜕𝜕
3
1
9 𝑦2 3 𝑦4
3𝑦 2
3 𝑦4
3𝑦
3 𝑦3
+𝑣
= 𝑈 � 𝜂 − 𝜂 3 � ∙ 𝑈 �−
+
�
+
𝑈
�
−
�
∙
𝑈
�
−
�
3
3
𝜕𝜕
𝜕𝜕
2
2
16 52
8 𝑐𝑥 3
16 𝑐𝑥 2
4 𝑐𝑥 2
4𝑥 2 𝑐
8𝑥 2 𝑐
𝑥 𝑐
3
1
9 𝑦2 3 𝑦4
3𝑦 2
3 𝑦4
3𝑦
3 𝑦3
+
�
+
�
−
�
∙
�
−
��
𝑎𝑝𝑝 = 𝑈 2 �� 𝜂 − 𝜂 3 � ∙ �−
3
3
2
2
16 52
8 𝑐𝑥 3
16 𝑐𝑥 2
4 𝑐𝑥 2
2
2
4𝑥 𝑐
8𝑥 𝑐
𝑥 𝑐
When 𝑥 = 3 𝑓𝑓, 𝛿 = 0.04 𝑖𝑖 = 0.0033 𝑓𝑓, 𝑈 = 20 𝑓𝑓⁄𝑠 , we have the folllowing plot:
-3
3.5
x 10
3
2.5
y (ft)
2
1.5
1
0.5
0
-35
-30
-25
-20
-15
-10
-5
0
X-accleration (ft/s 2)
-3
3.5
x 10
3
2.5
y (ft)
2
1.5
1
0.5
0
-1.6
-1.4
-1.2
-1
-0.8
-0.6
Y-accleration (ft/s 2)
The maximum 𝑎𝑝𝑝 = −31.5
𝑓𝑓
𝑠2
-0.4
-0.2
0
-4
x 10
at 𝑦 = 0.00222 𝑓𝑓
Problem 5.46
Problem
5.64
[Difficulty: 3]
5.46
Steady inviscid flow over a circular cylinder of radius R
Given:
Find:
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
ap 
 V  V 
Dt
t


(Particle Accleration)
(1) Steady flow
(2) Inviscid flow
(3) No flow in z-direction, velocity is not a function of z
Assumptions:
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 θ
r
r
Vθ
Vθ

apr  Vr Vr 
 Vr 
r
r θ
r


Vr  U 1 

When θ = π:
So the accelerations are:
2
 R   V  0
 
θ
r

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 
 
 
3
R r 
r
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
2 U
2
3
4
2
d
apr 
 3  η  1  η  η  2  η 
5  η  3η  0 Therefore:
R
R
dη
η 
3
2
1 
1  


aprmax 

  1  

R  1.291  
 1.291  
3
5
η
or
R
r
r  1.291  R
2
The maximum acceleration would then be:
2 U
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)
apθ 
4 U
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:
2
aθmin  2 
U
0.3
0.2
0.1
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 5.47
Problem
5.66
[Difficulty: 3]
5.47
Given:
Velocity field and nozzle geometry
Find:
Acceleration along centerline; plot
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 5.48
Problem
5.68
[Difficulty: 3]
5.48
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
ap 
 V  V 
Dt
t

Governing
Equations:
0

(Particle Accleration)
 

dV   V  dA

t CV
CS
(Continuity equation)
(1) Incompressible flow
(2) One-dimensional flow
Assumptions:
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
 R  ΔR x 
 1

L

2

U0  U1 sin( ω t) 
1  ΔR   x 
 

R1  L 


2
 R1 
u  U


A
 r 
A1
2
x
x
r  R1  R1  R2   R1  ΔR Therefore the flow speed is:
L
L


Based on the above assumptions the particle acceleration reduces to:

 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)
1  ΔR   x 
 

R1  L 


2  ΔR
R1 L

2

1  ΔR   x 
 

R1  L 


U0  U1 sin( ω t) 2
1  ΔR   x 
 

R1  L 


5

3
  ω U1  cos( ω t)

 R1  L   ΔR  x  2
1  R   L 
1


 ( 2 )   
ω U1  cos( ω t)
1  ΔR   x 
 

R1  L 


2
ΔR
When we simplify this expression we get:
Now we substitute the given values into this expression we get:
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
5
 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
 0.3
rad
s
 2
m
s
 cos( ω t) 
1
2
 1  0.1 m  1


0.2 m


Acceleration in Converging Channel
4
2 10
Acceleration (m/s^2)
ax  2  0.1 m 
4
1.5 10
4
1 10
3
5 10
  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 5.49
Problem
5.70
[Difficulty: 4]
5.49
150.
5.12a .
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 5.12a.
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 
 
 
 V eˆ  V z kˆ   eˆr
 eˆ
 kˆ  Vr eˆr  V eˆ  V z kˆ
r 
z 
 r




  V 
  Vr

 V z  Vr eˆr  V eˆ  V z kˆ
z 
 r r 
V 


 Vr
Vr eˆr  V eˆ  V z kˆ  
Vr eˆr  V eˆ  V z kˆ  V z
Vr eˆr  V eˆ  V z kˆ
r
z
r 
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 eˆ
V V
V eˆ
V V z ˆ
Vr
V z ˆ V Vr
eˆr  Vr  eˆ  Vr
k
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
V   V  V
  
r
Collecting terms:
V   V  V
V V VrV
V
Vr V Vr V2
V 
 V


 V z r eˆr  Vr   

 Vz 
r
r 
r
r
r 
r
z 
r
z


V 
 V z V V z
  Vr

 V z z kˆ
r
r 
z 


eˆ

The three terms in parentheses are the three components of convective acceleration given in Equations 5.12a.
Problem 5.50
(Difficulty 1)
5.50 Determine the velocity potential for:
(a) a flow field characterized by the stream function 𝛹 = 3𝑥 2 𝑦 − 𝑦 3 .
(b) a flow field characterized by the stream function 𝛹 = 𝑥𝑥.
Find: The velocity potential 𝜙
Assumptions: The flow is steady and incompressible
Solution: Use the definitions of stream function and velocity potential
(a) The stream function is given by:
We have for the x-component of velocity:
𝑢=
𝛹 = 3𝑥 2 𝑦 − 𝑦 3
𝜕𝜕
𝜕𝜕
=−
= 3𝑥 2 − 3𝑦 2
𝜕𝜕
𝜕𝜕
𝜕𝜕
= −3𝑥 2 + 3𝑦 2
𝜕𝜕
𝜙 = −𝑥 3 + 3𝑥𝑦 2 + 𝑓(𝑦)
Also we have for the y-component of velocity:
𝑣=−
𝜕𝜕
𝜕𝜕
=−
𝜕𝜕
𝜕𝜕
𝑑𝑑(𝑦)
𝜕𝜕 𝜕𝜕
=
= 6𝑥𝑥 = 6𝑥𝑥 +
𝜕𝜕 𝜕𝜕
𝑑𝑑
𝑑𝑑(𝑦)
=0
𝑑𝑑
𝑓(𝑦) = 𝑐
So the velocity potential can be given by:
Where 𝑐 is a constant.
𝜙 = −𝑥 3 + 3𝑥𝑦 2 + 𝑐
(b) The stream function is:
𝛹 = 𝑥𝑥
We have for the x-component of velocity:
𝑢=
𝜕𝜕
𝜕𝜕
=−
=𝑥
𝜕𝜕
𝜕𝜕
𝜕𝜕
= −𝑥
𝜕𝜕
1
𝜙 = − 𝑥 2 + 𝑓(𝑦)
2
Also we have for the y-component of velocity:
𝑣=−
𝜕𝜕
𝜕𝜕
=−
𝜕𝜕
𝜕𝜕
𝜕𝜕 𝜕𝜕
=
=𝑦
𝜕𝜕 𝜕𝜕
𝑑𝑑(𝑦)
=𝑦
𝑑𝑑
1
𝑓(𝑦) = 𝑦 2 + 𝑐
2
So the velocity potential can be given by:
Where 𝑐 is a constant.
1
1
𝜙 = − 𝑥2 + 𝑦2 + 𝑐
2
2
Problem 5.51
(Difficulty 1)
5.51 Determine whether the following flow fields are irrotational.
(a) 𝑢 = 2𝑥𝑥; 𝑣 = −𝑥 2 𝑦
(b) 𝑢 = 𝑦 − 𝑥 + 𝑥 2 ; 𝑣 = 𝑥 + 𝑦 − 2𝑥𝑥
(c) 𝑢 = 𝑥 2 𝑡 + 2𝑦; 𝑣 = 2𝑥 − 𝑦𝑡 2
(d) 𝑢 = −𝑥 2 − 𝑦 2 − 𝑥𝑥𝑥; 𝑣 = 𝑥 2 + 𝑦 2 + 𝑥𝑥𝑥
Find: Determine whether the flow fields are irrotational.
Assumptions: The flows are steady and incompressible
Solution: Use the definition for irrotational flow :
1
�⃗ = 0
𝜔
�⃗ = ∇ × 𝑉
2
�⃗ = 0
∇×𝑉
(a) The velocity field is:
�⃗ = �
∇×𝑉
𝑢 = 2𝑥𝑥
𝑣 = −𝑥 2 𝑦
𝜕𝜕 𝜕𝜕
− � 𝑘� = (−2𝑥𝑥 − 2𝑥)𝑘� ≠ 0
𝜕𝜕 𝜕𝜕
This flow is not irrotational.
(b) The velocity field is:
𝑢 = 𝑦 − 𝑥 + 𝑥2
𝑣 = 𝑥 + 𝑦 − 2𝑥𝑥
𝜕𝜕 𝜕𝜕
�⃗ = � − � 𝑘� = (1 − 2𝑦 − 1)𝑘� ≠ 0
∇×𝑉
𝜕𝜕 𝜕𝜕
This flow is not irrotational.
(c) The velocity field is:
𝑢 = 𝑥 2 𝑡 + 2𝑦
𝑣 = 2𝑥 − 𝑦𝑡 2
𝜕𝜕 𝜕𝜕
�⃗ = � − � 𝑘� = (2 − 2)𝑘� = 0
∇×𝑉
𝜕𝜕 𝜕𝜕
This flow is irrotational.
(d) The velocity field is:
This flow is not irrotational.
𝑢 = −𝑥 2 − 𝑦 2 − 𝑥𝑥𝑥
𝑣 = 𝑥 2 + 𝑦 2 + 𝑥𝑥𝑥
𝜕𝜕 𝜕𝜕
�⃗ = � − � 𝑘� = (2𝑥 + 𝑦𝑦 + 2𝑦 + 𝑥𝑥)𝑘� ≠ 0
∇×𝑉
𝜕𝜕 𝜕𝜕
Problem 5.52
(Difficulty 1)
5.52 The velocity profile for steady flow between parallel is parabolic and given by 𝑢 = 𝑢𝑐 + 𝑎𝑦 2 ,
where 𝑢𝑐 is the centerline velocity and 𝑦 is the distance measured from the centerline. The plate spacing
is 2𝑏 and the velocity is zero at each plate. Demonstrate that the flow is rotational. Explain why your
answer is correct even though the fluid does not rotate but moves in straight parallel paths.
Find: Demonstrate the flow is rotational and explain why it is rotational
Assumptions: The flow is steady and incompressible
Solution: Use the definition of the rotation vector to evaluate whether the flow is irrotational:
1
�⃗
𝜔
�⃗ = ∇ × 𝑉
2
The rotation vector for this velocity profile is then
1
𝜕𝜕 𝜕𝜕
�⃗ = � − � 𝑘� = (0 − 2𝑎𝑎)𝑘� = −2𝑎𝑎𝑘� ≠ 0
𝜔
�⃗ = ∇ × 𝑉
2
𝜕𝜕 𝜕𝜕
So the fluid flow is rotational.
The answer is correct because the rotation of fluid particles and circular streamlines are two different
concepts. We should not be confused about them. An irrotational flow can also have circular streamlines.
Similarly, the rotational flow can also only have straight streamlines. In this case the flow is rotational
because the velocity profile represents viscous flow between parallel plates, and the effect of viscosity is
to introduce rotation into the flow.
Problem 5.53
Problem
5.72
[Difficulty: 2]
5.53
Given:
Find:
Solution:
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
 d
 c
 b
 a
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
x
y
x
y
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 5.54
Problem
5.74
[Difficulty: 2]
5.54
Given:
Find:
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:
x
y
z
t

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:
x
y
x
y
 b
 c
 d
 a
2
2
Γ   A x dx   B x  y dy   A x dx   B x  y dy
x
y
x
y
a
Γ
b
c
  x  xa  xd  xc
3  b
A
Γ
3
3
 x   y  yb
2 c c
B
2
3
2

Evaluating the integrals:
d
3
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:
Γ 
1
2


  1  ft  12  0 2  ft2

 ft s 
  
1
Γ  0.500 
ft
2
s
Problem 5.55
Problem
5.76
[Difficulty: 3]
5.55
Given:
Stream function
Find:
If the flow is incompressible and irrotational
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 5.56
(Difficulty 2)
5.56 Fluid passes through the set of thin closely space blades at a velocity of 3
circulation for the flow.
Find: The circulation 𝛤.
Assumptions: The flow is steady and incompressible
Solution: Use the definition of circulation:
The angular velocity is given as:
�⃗ ∙ 𝑑𝑠⃗ = � �∇ × 𝑉
�⃗ � 𝑑𝑑
𝛤 = �𝑉
𝑧
𝑐
𝐴
𝑚
3 × 0.5
1
𝑉 sin 30°
𝑠
=
=5
𝜔=
1
𝑠
0.5
×
0.6
𝑚
𝐷
2
The angular velocity equals that of the blades at the outer radius:
𝑉𝜃 = 𝜔𝜔
And the radial component of velocity is
𝑉𝑟 = 𝑉 cos 30°
Thus the cross-product in the integral for circulation is:
The circulation is then
1 𝜕𝜕𝑉𝜃 1 𝜕𝑉𝑟
�⃗ � = �
�∇ × 𝑉
−
� = 2𝜔
𝑧
𝑟 𝜕𝜕
𝑟 𝜕𝜕
𝛤 = 2𝜔𝜔𝑟 2 = 2 × 5
1
𝑚2
× 𝜋 × (0.3 𝑚)2 = 2.83
𝑠
𝑠
𝑚
.
𝑠
Determine the
Problem 5.57
(Difficulty 2)
5.57 A two-dimensional flow field is characterized as 𝑢 = 𝐴𝑥 2 and 𝑣 = 𝐵𝐵𝐵 where 𝐴 =
𝐵 = −1
1
,
𝑚∙𝑠
1 1
2 𝑚∙𝑠
and
and 𝑥 and 𝑦 are in meters. Demonstrate that the velocity field represents a possible
incompressible flow field. Determine the rotation at the location (1,1). Evaluate the circulation about
the “curve” bounded by 𝑦 = 0, 𝑥 = 1, 𝑦 = 1, and 𝑥 = 0.
Find: Demonstrate the flow is possibly incompressible.
Determine the rotation 𝜔
�⃗ at (1,1).
Evaluate the circulation 𝛤 bounded by the curve.
Assumptions: The flow is steady and incompressible
Solution: Use the continuity equation and the relations for circulation
From the two-dimensional continuity equation, for incompressible flow we have:
𝜕𝜕 𝜕𝜕
+
=0
𝜕𝜕 𝜕𝜕
For this velocity profile we have
𝜕𝜕
= 2𝐴𝐴
𝜕𝜕
𝜕𝜕
= 𝐵𝐵
𝜕𝜕
𝜕𝜕 𝜕𝜕
1 1
1
+
=2×
𝑥−1
𝑥=0
𝜕𝜕 𝜕𝜕
2𝑚 ∙𝑠
𝑚∙𝑠
So this is a possible two-dimensional incompressible flow.
The rotation is calculated for this velocity profile as:
1
1
1 𝜕𝜕 𝜕𝜕
1
�⃗ = � − � 𝑘� = (𝐵𝐵 − 0)𝑘� = 𝐵𝐵𝑘�
𝜔
�⃗ = ∇ × 𝑉
2
2
2 𝜕𝜕 𝜕𝜕
2
So the rotation at the location (1,1) will be:
𝜔
�⃗ =
1
1
1
× �−1
� × 1 𝑚 𝑘� = −0.5 𝑘�
2
𝑚∙𝑠
𝑠
𝜔𝑧 = −0.5
1
𝑠
The flow and the curve y=0, x=1, y=1, and x = 0 is shown in the figure:
The circulation about the curve is:
1
𝑚2
�⃗ ∙ 𝑑𝑠⃗ = � �∇ × 𝑉
�⃗ � 𝑑𝑑 = 2𝜔𝑧 𝐴 = 2 × �−0.5 � × 1 𝑚2 = −1
𝛤=�𝑉
𝑧
𝑠
𝑠
𝑐
𝐴
Problem 5.58
(Difficulty 1)
5.58 A flow field is represented by the stream function 𝛹 = 𝑥 4 − 2𝑥 3 𝑦 + 2𝑥𝑦 3 − 𝑦 4 . Is this a possible
two-dimensional flow? Is this flow irrotational?
Find: Whether the flow field is two-dimensional and whether the flow is irrotational.
Assumptions: The flow is steady and incompressible
Solution: For the flow to be two-dimensional flow, it needs to satisfy the continuity equation:
𝜕𝜕 𝜕𝜕
+
=0
𝜕𝜕 𝜕𝜕
The velocities u and v are
𝑢=
𝜕𝜕
= −2𝑥 3 + 6𝑥𝑦 2 − 4𝑦 3
𝜕𝜕
𝑣=−
Using the continuity equation:
𝜕𝜕
= −6𝑥 2 + 6𝑦 2
𝜕𝑥
𝜕𝜕
= −4𝑥 3 + 6𝑥 2 𝑦 − 2𝑦 3
𝜕𝜕
𝜕𝜕
= 6𝑥 2 − 6𝑦 2
𝜕𝜕
𝜕𝜕 𝜕𝜕
+
= −6𝑥 2 + 6𝑦 2 + 6𝑥 2 − 6𝑦 2 = 0
𝜕𝜕 𝜕𝜕
So this flow is possible two-dimensional flow.
For the vorticity we have:
�⃗
𝜉⃗ = ∇ × 𝑉
𝜕𝜕 𝜕𝜕
�⃗ = � − � 𝑘� = −12𝑥 2 + 12𝑥𝑥 − 12𝑥𝑥 + 12𝑦 2 𝑘� = (−12𝑥 2 + 12𝑦 2 )𝑘� ≠ 0
∇×𝑉
𝜕𝜕 𝜕𝜕
So this flow is not irrotational.
Problem 5.59
Problem
5.78
[Difficulty: 2]
5.59
Given:
Find:
Velocity field for motion in the x-direction with constant shear
(a) Expression for the velocity field
(b) Rate of rotation
(c) Stream function
We will apply the definition of circulation to the given velocity field.
Solution:




Governing

u   v   w 
 0 (Continuity equation)
Equations:
x
y
z
t

1
2

   V
Assumptions:
(Definition of rotation)
(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



2
x
y
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
g ( x )  0 Therefore, the derivatives of both f and g are zero, and thus f and g are constants:
v   ψ   f ( x)  y 
d
dx
x
x
ψ
1
2
2
 A y  c1  y  c2
Problem 5.60
Problem
5.80
[Difficulty: 2]
5.60
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 5.61
Problem
5.82
[Difficulty: 2]
5.61
Given:
Find:
Flow field represented by a velocity function.
(a) Fluid rotation
(b) Circulation about the curve shown
(c) Stream function
(d) Plot several streamlines in first quadrant
We will apply the definition of rotation and circulation to the given velocity field.
Solution:

 1
Governing
    V (Definition of rotation)
Equation:
2
 
(Definition of circulation)
   V  ds
Assumption:
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
 b
 c
 d
 a
2
2
Γ   A x dx   B x  y dy   A x dx   B x  y dy
x
y
x
y
a
Γ
b
c
  x  xa  xd  xc
3  b
A
Γ
3
3
3
 x   y  yb
2 c c
B
2
2

1
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:
ft s
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 5.62
Problem
5.84
[Difficulty: 3]
5.62
Given:
Find:
Viscometric flow of Example 5.7, V = U(y/h)i, where U = 4 mm/s and h = 4 mm
(a) Average rate of rotation of two line segments at +/- 45 degrees
(b) Show that this is the same as in the example
We will apply the definition of rotation to the given velocity field.
Solution:

 1
Governing
    V (Definition of rotation)
Equation:
2
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 
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 5.63
Problem
5.86
[Difficulty: 3]
5.63
Given:
Find:
Velocity field approximation for the core of a tornado
(a) Whether or not this is an irrotational flow
(b) Stream function for the flow
We will apply the definition of rotation to the given velocity field.
Solution:

 1
Governing
(Definition of rotation)
   V
Equation:
2


1 

  eˆr
 eˆ
 kˆ
r 
z
r
(Definition of "del" operator)
eˆ
 eˆr

eˆr
 eˆ

(Hints from text)
(1) Steady flow
(2) Two-dimensional flow (no z velocity, velocity is not a function of θ or z)
Assumptions:
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 

2
r  
 
r
r



1
eˆr  eˆr  Vr  eˆr  eˆ  V  1 Vr  V   eˆ  eˆ  1 V  Vr

2
r
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
ψ   r Vr dθ  f ( r)  
 f ( r)
dθ  f ( r)  


2

π
2
π



K
K
Vθ dr  g ( θ)  
 ln( r)  g ( θ) Comparing these two expressions:
dr  g ( θ)  
 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 5.64
(Difficulty 1)
�⃗ = 2𝚤̂ − 4𝑥𝚥̂
5.64 A velocity field is given by 𝑉
vorticity of the flow.
𝑚
.
𝑠
Determine an equation for the streamline. Calculate the
Find: The streamline Ψ and vorticity 𝜉⃗.
Assumptions: The flow is steady and incompressible
Solution: Apply the relations for streamlines and vorticity
The velocity field is:
𝑢=2
𝑣 = 4𝑥
𝑚
𝑠
𝑚
𝑠
The x-component of the velocity is given in terms of the stream function as:
𝑢=
Integrating both sides we get:
𝜕𝜕
=2
𝜕𝜕
𝛹 = 2𝑦 + 𝑓(𝑥)
The y-component of velocity is given by
𝑣=−
𝜕𝜕
𝑑𝑑(𝑥)
=−
= −4𝑥
𝜕𝜕
𝑑𝑑
𝑑𝑑(𝑥)
= 4𝑥
𝑑𝑑
𝑓(𝑥) = 2𝑥 2 + 𝑐
So the stream function is:
𝛹 = 2𝑦 + 2𝑥 2 + 𝑐
where c is a constant.
The vorticity of the flow is calculated as:
𝜕𝜕 𝜕𝜕
�⃗ = � − � 𝑘� = (−4 − 0) 𝑘� = −4 𝑘�
𝜉⃗ = ∇ × 𝑉
𝜕𝜕 𝜕𝜕
Problem 5.65
Problem
5.88
[Difficulty: 3]
5.65
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
 V 


0
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 5.66
Problem
5.89
[Difficulty: 2]
5.66
Given:
Flow between infinite plates
Find:
Prove that u = 0, dP/dy = constant
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
   u  v  w   g y 
   2  2  2 

t

x

y

z

y
y
z 


 x
(Navier-Stokes Equations)
 2w 2w 2w 
 w
w
w
w 
P
u
v
 w   g z 
   2  2  2 
x
y
z 
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 5.67
Problem
5.90
[Difficulty: 3]
5.67
Given:
temperature profile and temperature-dependent viscosity expression
Find:
Velocity Profile
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
x
y
z 
y
z 
 t
 x
 
  2v  2v  2v 
 v
v
v
v 
P
 u  v  w   g y 
   2  2  2 
x
y
z 
y
y
z 
 t
 x
 
 2w 2w 2w 
 w
w
w
w 
P
u
v
 w   g z 
   2  2  2 
 
x
y
z 
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))
0
dy
(6)
Integrating equation (6) once
u
y
y y2
gh sin 
( y (1  )  a(Tw  T0 ) y (1   2 ))  C2
2h
h 3h
0
(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 ))
2h
0
h 3h
When a=0, eq. (8) can be simplified to
u
(8)
y
gh sin 
y (1  ) , and it is exactly the same velocity profile in Example 5.9.
2h
0
Problem 5.68
(Difficulty 2)
5.68 Consider a steady, laminar, fully developed incompressible flow between two infinite parallel plates
�⃗ ≠ 𝑉
�⃗ (𝑧), 𝑤 = 0
as shown. The flow is due to a pressure gradient applied in the 𝑥-direction. Given that 𝑉
and that gravity points in the negative 𝑦 direction, prove that 𝑣 = 0 and that the pressure gradients in
the 𝑥- and 𝑦-directions are constant.
Find: Prove that 𝑣 = 0 and pressure gradient in the 𝑥- and 𝑦-directions are constant.
Assumptions: The flow is steady and incompressible
Solution:
For 2D incompressible steady flow we have the following governing equations:
Continuity equation:
𝜕𝜕 𝜕𝜕 𝜕𝜕
+
+
=0
𝜕𝜕 𝜕𝜕 𝜕𝜕
Momentum equation for the x, y, and z directions
𝜌 �𝑢
𝜌 �𝑢
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕 2𝑢 𝜕 2𝑢 𝜕 2𝑢
𝜕𝜕
+𝑣
+𝑤 �=−
+ 𝜇 � 2 + 2 + 2�
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕 2𝑣 𝜕 2𝑣 𝜕 2𝑣
+𝑣
+ 𝑤 � = −𝜌𝜌 −
+ 𝜇 � 2 + 2 + 2�
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜌 �𝑢
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕 2𝑤 𝜕 2 𝑤 𝜕 2 𝑤
+𝑣
+𝑤
�=−
+ 𝜇� 2 +
+
�
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝑥
𝜕𝑦 2 𝜕𝑧 2
For this steady fully developed flow, we have:
𝑤=0
�⃗ ≠ 𝑉
�⃗ (𝑧)
𝑉
𝜕𝜕
=0
𝜕𝜕
From the continuity equation:
𝜕𝜕
=0
𝜕𝜕
𝑣=𝑐
We know that 𝑣 = 0 at the plate, so
𝑐=0
So we get:
𝑣=0
From the z component momentum equation, because the velocity w and its derivatives are zero:
𝜕𝜕
=0
𝜕𝜕
Thus from the y component momentum equation:
𝜌(𝑢 × 0 + 0 × 0 + 0 × 0) = −𝜌𝜌 −
0 = −𝜌𝜌 −
For incompressible the density 𝜌 is constant, so
From the x component momentum equation:
𝜌 �𝑢 × 0 + 0 ×
𝜕𝜕
𝜕𝜕
𝜕𝜕
+ 𝜇(0 + 0 + 0)
𝜕𝜕
𝜕𝜕
= −𝜌𝜌
𝜕𝜕
𝜕𝜕
= 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝜕𝜕
𝜕𝜕
𝜕 2𝑢
𝜕𝜕
+ 0 × 0� = −
+ 𝜇 � 2 + 0 + 0�
𝜕𝜕
𝜕𝑥
𝜕𝜕
0=−
𝜕𝜕
𝜕𝜕
𝜕 2𝑢
+ 𝜇 � 2�
𝜕𝜕
𝜕𝑦
𝜕 2𝑢
𝜕𝜕
= 𝜇 � 2�
𝜕𝑦
𝜕𝜕
𝜕𝜕
Because 𝑢 is not a function of 𝑥 or 𝑧, so is also not a function of 𝑥 and 𝑧. However, could be a
𝜕𝜕
𝜕𝜕
function of y. Also we already have:
Differentiating
Therefore
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
= 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝜕𝜕
with respect to y and interchanging the order of differentiation
does not vary with 𝑦. Then
𝜕 𝜕𝜕
𝜕 𝜕𝜕
� �=
� �=0
𝜕𝜕 𝜕𝜕
𝜕𝜕 𝜕𝜕
𝜕𝜕
𝜕𝜕
is not a function of 𝑥, 𝑦, 𝑜𝑜 𝑧. So
𝜕𝜕
= 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐.
𝜕𝜕
Problem 5.69
(Difficulty 3)
5.69 Consider a steady, laminar, fully developed incompressible flow between two infinite parallel plates
separated by a distance 2ℎ as shown below. The top plate moves with a velocity 𝑉0 . Derive an
expression for the velocity profile. Determine the pressure gradient for which the flow rate is zero. Plot
the profile for that condition.
Find: The expression for the velocity profile. Determine the pressure gradient for which the flow rate is
zero and plot the velocity profile.
Assumptions: The flow is steady and incompressible. The effect of gravity is neglected or gravity acts
in the z-direction
Solution:
For 2D incompressible steady flow we have the following governing equations:
Continuity equation:
𝜕𝜕 𝜕𝜕
+
=0
𝜕𝜕 𝜕𝜕
Momentum equations in the x and y directions
𝜌 �𝑢
𝜌 �𝑢
𝜕𝜕
𝜕𝜕
𝜕 2𝑢 𝜕 2 𝑢
𝜕𝜕
+𝑣 �=−
+ 𝜇 � 2 + 2�
𝜕𝜕
𝜕𝜕
𝜕𝑥
𝜕𝑦
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕 2𝑣 𝜕 2 𝑣
+ 𝑣 � = 𝜌𝜌 −
+ 𝜇 � 2 + 2�
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝑥
𝜕𝑦
This flow is steady and fully developed and the vertical velocity v is zero at the walls. Thus we have:
𝑣=0
𝜕𝜕
=0
𝜕𝜕
𝜕𝜕
=0
𝜕𝜕
Neglecting the body force due to gravity in the y-direction, we have from the y-momentum equation:
𝜌(𝑢 × 0 + 0 × 0) = 0 −
𝜕𝜕
=0
𝜕𝜕
𝜕𝜕
+ 𝜇(0 + 0)
𝜕𝜕
The pressure 𝑝 is not a function of 𝑦 and can only be a function of y.
We have from the x-momentum equation:
𝜌 �𝑢 × 0 + 0 ×
𝜕𝜕
𝜕 2𝑢
𝜕𝜕
�=−
+ 𝜇 � 2 + 0�
𝜕𝜕
𝜕𝑥
𝜕𝜕
𝜕 2 𝑢 1 𝑑𝑑
=
𝜕𝑦 2 𝜇 𝑑𝑑
Where we can use the total differential d since the velocity and pressure only vary with x.
Integrating with respect to x
Applying the boundary condition:
𝑑𝑑 1 𝑑𝑑
=
𝑦 + 𝐶1
𝑑𝑑 𝜇 𝑑𝑑
𝑢=
1 𝑑𝑑 2
𝑦 + 𝐶1 𝑦 + 𝐶2
2𝜇 𝑑𝑑
𝑢 = 0 at 𝑦 = −ℎ
1 𝑑𝑑 2
ℎ − 𝐶1 ℎ + 𝐶2 = 0
2𝜇 𝑑𝑑
𝑢 = 𝑉0 at 𝑦 = ℎ
So we have:
1 𝑑𝑑 2
ℎ + 𝐶1 ℎ + 𝐶2 = 𝑉0
2𝜇 𝑑𝑑
2𝐶1 ℎ = 𝑉0
𝐶1 =
𝑉0
2ℎ
2𝐶2 = 𝑉0 −
𝐶2 =
The velocity profile is then
𝑢=
or
ℎ
1
𝑉0
1 𝑑𝑑 2
−
ℎ
2 2𝜇 𝑑𝑑
1 𝑑𝑑 2 𝑉0
𝑉0
1 𝑑𝑑 2
𝑦 +
𝑦+ −
ℎ
2𝜇 𝑑𝑑
2ℎ
2 2𝜇 𝑑𝑑
𝑢=
The flow rate per unit depth is:
𝑦
ℎ
𝑉0 𝑦
ℎ2 𝑑𝑑 𝑦 2
�� � − 1� + � + 1�
2𝜇 𝑑𝑑 ℎ
2 ℎ
1
𝑄 = ∫−ℎ 𝑢𝑢𝑢 = ∫−1 𝑢 ℎ 𝑑 � � = ∫−1 ℎ �
𝑄 = ��
𝑄 = ��
− 1) +
𝑉0
(𝑧
ℎ
+ 1)� 𝑑𝑑
1
where 𝑧 =
1
ℎ3 𝑑𝑑 𝑧 3
𝑉0 𝑧 2
� � − 𝑧� + � + 𝑧��
2𝜇 𝑑𝑑
3
2 2
−1
2ℎ3
𝑑𝑑
�− � + 𝑉0 ℎ = 0
𝑑𝑑
3𝜇
2ℎ3 𝑑𝑑
= 𝑉0 ℎ
3𝜇 𝑑𝑑
or
Substituting this value of
ℎ2 𝑑𝑑
(𝑧 2
2𝜇 𝑑𝑑
𝑉0 𝑧 2
ℎ3 𝑑𝑑 𝑧 3
� � − 𝑧� + � + 𝑧��
2𝜇 𝑑𝑑
3
2 2
−1
𝑄=
Now, the flow will be zero when
1 𝑑𝑑 2
ℎ
𝜇 𝑑𝑑
𝑑𝑑
𝑑𝑑
𝑑𝑑 3𝜇𝑉0
=
𝑑𝑑
2ℎ2
into the velocity equation, we have for the velocity profile:
𝑢=
We can plot the velocity profile as
𝑉0 𝑦
ℎ2 3𝜇𝑉0 𝑦 2
�� � − 1� + � + 1�
2
ℎ
2𝜇 2ℎ
2 ℎ
𝑢=
𝑉0
𝑦 2
𝑦
�3 � � + 2 � � − 1�
ℎ
ℎ
4
𝑦
ℎ
Note the velocity is u = 0 at h = -h and u = V0 at y = h, which are the boundary conditions.
Problem 5.70
Problem
5.92
[Difficulty: 3]
5.70
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
2


   U  1  3   y  
1
1 
8 x 

2
2
c x 
2  c x
1 1 3 U y
1 U
1  3 U y
ωz     
 
   

5
1
5
2
2 8
2 4 2
2
c x
2
2
c x
2
Computing the partial derivatives:


c x
U 
3 y
ωz  
 1    
8 x
2 δ 
U
2
2
c x
2
2


 c x 2 


angdef 
y 
1 U y
Linear deformation:  u    U
 
1
3
2
x
x 
 c x 2 
c x

x
u 
x
U
δ
2
x

3

8
 1 

y
v 
 
U y
Maximum value at y = δ

2δ x
U y

2δ x


2
2
 U y2  2 U y
 
v   
3
3
4
y
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    
3  y 
1
1
c 4 2 5
x
y
x  4


2
Maximum value at y = δ
y
U
1
c x
2

x 

3

8
 1 
 
y
2

x 
2
Maximum value at y = 0
The shear stress is
 3 y 2

  μ U 
v  u 
 1     
δ 
8 x 
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
3  μ U y
3 2 y 
 
 dy  
 dy
2
δ  8 2
x 
4  δ x

μ U
Maximum value at y = δ
Problem 5.71
(Difficulty 3)
5.71 A cylinder of radius 𝑟𝑖 rotates at a speed 𝜔 coaxially inside a fixed cylinder of radius 𝑟0 . A viscous
fluid fills the space between the two cylinders. Determine the velocity profile in the space between the
cylinders and the shear stress on the surface of each cylinder. Explain why the shear stresses are not
equal.
Find: The velocity profile and stress on each cylinder.
Assumptions: The flow is steady and incompressible
Solution:
For this two dimensional steady incompressible flow with circular streamlines we have the following
governing equations:
Continuity:
Momentum equation:
In this particular case, we have:
1 𝜕(𝑟𝑉𝑟 ) 1 𝜕𝑉𝜃
+
=0
𝑟 𝜕𝜕
𝑟 𝜕𝜕
𝜌(𝑢
�⃗ ∙ ∇𝑢
�⃗) = −∇𝑝 + 𝜇∇2 𝑢
�⃗
𝑉𝑟 = 0
Thus
𝜕𝑉𝜃
=0
𝜕𝜕
𝑢
�⃗ ∙ ∇𝑢
�⃗ = 𝑉𝜃 𝑒̂𝜃 ∙ ∇𝑢
�⃗ =
∇2 𝑢
�⃗
As we have:
𝑉𝜃 𝜕(𝑉𝜃 𝑒̂𝜃 ) 𝑉𝜃 𝜕𝑉𝜃
𝑉𝜃2
𝜕𝑒̂𝜃
= �
𝑒̂𝜃 + 𝑉𝜃
� = − 𝑒̂𝑟
𝑟
𝑟 𝜕𝜕
𝜕𝜕
𝑟
𝜕𝜕
1𝜕
𝜕𝑉𝜃 𝑒̂𝜃
1 𝜕 2 (𝑉𝜃 𝑒̂𝜃 )
=
�𝑟
�+ 2
𝑟 𝜕𝜕
𝑟
𝜕𝜕
𝜕𝜃 2
𝜕𝑒̂𝜃
=0
𝜕𝜕
𝜕 2 𝑒̂𝜃 𝜕(−𝑒̂𝑟 )
=
= −𝑒̂𝜃
𝜕𝜃 2
𝜕𝜕
Thus
The momentum equation becomes:
So we have:
∇2 𝑢
�⃗ =
1 𝜕
𝜕𝑉𝜃
𝑉𝜃
�𝑟
� 𝑒̂𝜃 − 2 𝑒̂𝜃
𝑟 𝜕𝜕
𝜕𝜕
𝑟
𝑉𝜃2
𝜕𝜕
1 𝜕𝜕
1𝜕
𝜕𝑉𝜃
𝑉𝜃
−𝜌 𝑒̂𝑟 = − 𝑒̂𝑟 −
𝑒̂𝜃 + 𝜇
�𝑟
� 𝑒̂𝜃 − 𝜇 2 𝑒̂𝜃
𝜕𝜕
𝑟 𝜕𝜕
𝑟 𝜕𝜕
𝑟
𝜕𝜕
𝑟
−
1 𝜕𝜕
1 𝜕
𝜕𝑉𝜃
𝑉𝜃
𝑒̂𝜃 + 𝜇
�𝑟
� 𝑒̂𝜃 − 𝜇 2 𝑒̂𝜃 = 0
𝑟 𝜕𝜕
𝑟 𝜕𝜕
𝜕𝜕
𝑟
As the pressure at 𝜃 = 0 and 𝜃 = 2𝜋 are the same.
𝜕𝜕
=0
𝜕𝜕
Thus
𝜕𝑉𝜃
𝑉𝜃
1 𝜕
�𝑟
�− 2 =0
𝑟 𝜕𝜕
𝜕𝜕
𝑟
𝜕 2 𝑉𝜃 1 𝜕𝑉𝜃 𝑉𝜃
+
−
=0
𝑟 𝜕𝜕 𝑟 2
𝜕𝜕 2
This differential equation is easily solved with:
Apply the boundary conditions:
𝑉𝜃 = 𝐶1 𝑟 + 𝐶2
1
𝑟
𝑉𝜃 = 𝑉 = 𝜔𝜔𝑖 at 𝑟 = 𝑟𝑖
The constants are evaluated as
𝑉𝜃 = 0 at 𝑟 = 𝑟0
𝐶1 𝑟𝑖 + 𝐶2
𝐶1 𝑟0 + 𝐶2
The velocity profile becomes
For the shear stress we have:
𝑉𝜃 =
For the rotating surface:
For the fixed surface:
1
=0
𝑟0
𝑉𝑟0 𝑟𝑖 𝑟 𝑟0
𝜔𝑟0 𝑟𝑖 2 𝑟 𝑟0
�
−
�
=
� − �
𝑟𝑖2 − 𝑟02 𝑟0 𝑟
𝑟𝑖2 − 𝑟02 𝑟0 𝑟
𝜏𝑟𝑟 = 𝜇𝜇
𝜏𝑟𝑟 =
1
=𝑉
𝑟𝑖
𝑑 𝑉𝜃
𝑑 𝜔𝑟0 𝑟𝑖 2 1 𝑟0
� � = 𝜇𝜇 � 2
� − ��
𝑑𝑑 𝑟
𝑑𝑑 𝑟𝑖 − 𝑟02 𝑟0 𝑟 2
𝜇𝜇𝜇𝑟0 2 𝑟𝑖 2 𝑑
1
𝜇𝜇𝜇𝑟0 2 𝑟𝑖 2 2
𝜇𝜇𝑟0 2 𝑟𝑖 2 2
�−
�
=
=
𝑟2
𝑟𝑖2 − 𝑟02 𝑑𝑑
𝑟𝑖2 − 𝑟02 𝑟 3
𝑟𝑖2 − 𝑟02 𝑟 2
𝜏𝑟𝑟 =
2𝜇𝜇𝑟0 2 𝑟𝑖 2 1
𝑟𝑖2 − 𝑟02 𝑟 2
𝜏𝑟𝑟𝑟 =
2𝜇𝜇𝑟0 2 𝑟𝑖 2 1
2𝜇𝜇𝑟0 2
=
𝑟𝑖2 − 𝑟02 𝑟𝑖 2 𝑟𝑖2 − 𝑟02
𝜏𝑟𝑟0 =
2𝜇𝜇𝑟0 2 𝑟𝑖 2 1
2𝜇𝜇𝑟𝑖 2
=
𝑟𝑖2 − 𝑟02 𝑟0 2 𝑟𝑖2 − 𝑟02
We see that the shear stress is a function of radius 𝑟, and that the shear stresses are not equal. Because the
torque is the same on both cylinders, the larger radius 𝑟 cylinder will have more surface area and a
smaller shear stress.
Problem 5.72
(Difficulty: 2)
5.72 The velocity profile for fully developed laminar flow in a circular tube is 𝑢 = 𝑢𝑚𝑚𝑚 [1 − (𝑟⁄𝑅 )2 ].
Obtain an expression for the shear force per unit volume in the x direction for this flow. Evaluate its
maximum value for a pipe radius of 75 mm and a maximum velocity of 3 m/s. The fluid is water.
Given: Velocity profile for fully developed flow in a tube
Find: (a) Express shear force per unit volume in the x direction
(b) Maximum value at these conditions
Assumptions: Steady incompressible flow
Solution:
The differential of shear force would be:
𝑑𝐹𝑠ℎ𝑒𝑒𝑒 = (𝜏 + 𝑑𝑑)2𝜋𝜋𝜋𝜋𝜋𝜋 − 𝜏2𝜋𝜋𝑑𝑧𝑧𝑧 = 2𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋
In cylindrical coordinates:
From the given profile:
1𝑑
1
𝑑𝑑
𝑑
𝑑𝐹𝑠𝑠
(𝑟𝑟)2𝜋𝜋𝜋𝜋𝜋 =
=
�𝑟𝑟 �
𝑟 𝑑𝑑
2𝜋𝜋𝜋𝜋𝜋𝜋 𝑑𝑑
𝑑𝑑
𝑑𝑑
2𝑟
𝑑𝑑
= −𝑢𝑚𝑚𝑚 2
𝑅
𝑑𝑑
Therefore:
𝑑2𝑢
2
= −𝑢𝑚𝑚𝑚 2
2
𝑑𝑑
𝑅
𝑑2𝑢
1 𝑑𝑑
1
4𝜇𝑢𝑚𝑚𝑚
𝑑𝐹𝑠𝑠
2
2𝑟
=𝜇 2 +𝜇
= −𝜇𝑢𝑚𝑚𝑚 2 − 𝜇 𝑢𝑚𝑚𝑚 2 = −
𝑟 𝑑𝑑
𝑟
𝑑𝑑
𝑅
𝑅
𝑑𝑑
𝑅2
𝐹𝑉𝑉𝑉𝑉 = −
For water we have:
4𝜇𝑢𝑚𝑚𝑚
𝑅2
𝜇 = 1.002 × 10−3 𝑃𝑃 ∙ 𝑠
𝐹𝑉𝑉𝑉𝑉 = −
4 × 1.002 × 10−3 𝑃𝑃 ∙ 𝑠 × 3
(0.075 𝑚)2
𝑚
𝑠 = 2.14 𝑁
𝑚3
Problem 5.73
Problem
5.94
[Difficulty: 3]
5.73
Given:
Horizontal, fully developed flow
Find:
Velocity Profile and pressure gradient
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
v
v
v 
P
 u  v  w   g y 
   2  2  2 
x
y
z 
y
y
z 
 t
 x
 
 2w 2w 2w 
 w
w
w
w 
P
u
v
 w   g z 
   2  2  2 
x
y
z 
z
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 ' 
 dx

h
2
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 5.74
Problem
5.96
[Difficulty: 2]
5.74
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
P
u
u
u 
   2  2  2 
   u  v  w   g x 
x
x
y
z 
y
z 
 t
 x
  2v  2v  2v 
 v
v
v
v 
P
 u  v  w   g y 
   2  2  2 
x
y
z 
y
y
z 
 t
 x
 
(Navier-Stokes Equations)
 2w 2w 2w 
 w
w
w
w 
P
u
v
 w   g z 
   2  2  2 
x
y
z 
z
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
5.75
Problem 5.98
[Difficulty: 3]
5.75
ρ =
3
250
1
4
µ =
999
0.001
h =
12
mm
mm
m
m
Draining a Tank
1
1
min
h =
dy
d 4 ρg
=−
y
dt
32 D 2 µL
y (0 ) = y 0
y n +1 = y n + hkyn
k=−
d 4 ρg
32 D 2 µL
y Exact (t ) = y0 e
6
−
d 4 ρg
32 D 2 µL
min
t
t n+1 = t n + h
t n (min)
0
12
24
36
48
60
72
84
96
108
120
Error:
1
0
-1
k = 0.000099 s
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
Depth y (m)
d =
D =
y0 =
L =
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 5.76
Problem
5.100
[Difficulty: 3]
5.76
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
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
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
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 5.77
(Difficulty 2)
5.77 For a small spherical particle of styrofoam 𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = 16
𝑘𝑘
𝑚3
with a diameter of 5 𝑚𝑚 falling in air,
the drag is given by 𝐹𝐷 = 3𝜋𝜋𝜋𝜋, where 𝜇 is the air viscosity and 𝑉 is the sphere velocity. Derive the
differential equation that describes the motion. Using the Euler method, find the maximum speed
starting from rest, and the time it takes to reach 95 percent of this speed. Plot the speed as function of
time.
Find: Derive the differential equation that describes the motion. Maximum velocity: 𝑉𝑚𝑚𝑚 . Time to reach
95 percent of maximum velocity: 𝑡.
Assumptions: The air is quiescent and the only drag is due to viscous friction.
Solution: Use Newton’s second law of motion. For motion in the y-direction
� 𝐹𝑦 = 𝑚 𝑎
The forces are the body force FB acting downward and the drag force FD acting upward. For positive y in
the direction of motion (downward) we have:
𝐹𝐵 − 𝐹𝐷 = 𝑚 𝑎
Or, the acceleration is
𝑎=
The mass can be calculated by:
𝑚 = 𝜌∀
The volume of the particle is:
The body force is :
∀=
𝐹𝐵 − 𝐹𝐷
𝑚
𝜋 3 𝜋
𝑑 = × (0.005 𝑚)3 = 6.55 × 10−8 𝑚3
6
6
This motion is along the vertical direction.
𝐹𝐵 = 𝑚𝑚 = 𝜌𝜌∀
So the differential equation for the motion (𝐸𝐸𝐸𝐸𝐸 𝑚𝑚𝑚ℎ𝑜𝑜) is:
𝑎=
𝑑𝑑 𝜌𝜌∀ − 3𝜋𝜋𝜋𝜋
=
𝑑𝑑
𝜌∀
When the acceleration 𝑎 = 0, the particle reaches the maximum velocity, so we have:
𝜌𝜌∀ − 3𝜋𝜋𝑉𝑚𝑚𝑚 𝑑 = 0
The viscosity of air is:
The density of the particle is:
𝜇 = 1.827 × 10−5
𝜌 = 16
Thus
𝑉𝑚𝑚𝑚
𝑘𝑘
𝑚∙𝑠
𝑘𝑘
𝑚3
𝑘𝑘
𝑚
16 3 × 9.81 2 × 6.55 × 10−8 𝑚3
𝜌𝜌∀
𝑚
𝑠
𝑚
=
=
= 11.93
3𝜋𝜋𝜋 3 × 𝜋 × 1.827 × 10−5 𝑘𝑘 × 0.005 𝑚
𝑠
𝑚∙𝑠
From the differential equation of motion, we have:
𝑑𝑑 𝜌𝜌∀ − 3𝜋𝜋𝜋𝜋
=
𝜌∀
𝑑𝑑
The finite difference equation for the Euler method is
𝜌𝜌∀ − 3𝜋𝜋𝜋𝜋
∆𝑉 = �
� ∆𝑡
𝜌∀
Using the Euler method in an equation solver, the time it takes reach 95 percent of the maximum velocity,
which is 11.4 m/s, is 3.8 s. The velocity versus time is plotted as:
Problem 5.102
5.78
Problem
[Difficulty: 3]
5.78
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
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
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
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 5.79
Problem
5.104
5.79
ui 
x 
[Difficulty: 3]
ug i 1  x ug2i
1  2x ug i
0.333
x
Iteration
0
1
2
3
4
5
6
Exact
0.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
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 5.106
5.80
Problem
5.80
[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 4.1
Problem
4.2
[Difficulty: 2]
4.1
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
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
T2 = (− 5 + 273) K = 268 K
kJ
K
u 2 − u1 = cv (T2 − T1 )
∆U = 999
or
∆U = mcv (T2 − T1 ) = ρVcv ∆T
kg
10 −6 m 3
kcal
J
×
×
×1
× (− 268 − 288)K × 4190
250
mL
3
m
mL
kg ⋅ K
kcal
∆U = −20.9 kJ
Problem 4.2
(Difficulty: 2)
4.2 A hot air balloon with an initial volume of 2600 𝑚3 rises from sea level to 1000 𝑚 elevation. The
temperature of the air inside the balloon is 100 ℃ at the start and drops to 90 ℃ at 1000 𝑚. What are
the net amounts of heat and work transferred between the balloon and the atmosphere?
Find: The heat and work transfers
Assumption: The air in the balloon is an ideal gas
Solution: Apply an energy balance to the air and use the ideal gas relations.
From the first law of thermodynamics for the balloon we have:
𝛿𝛿 − 𝛿𝛿 = 𝑑𝑑
Where
𝛿𝛿 = 𝑚𝑚 �𝑢 +
𝑉2
+ 𝑧�
2
We need to find the mass of the air in the balloon. The density of the air at sea level at 100 ℃ is:
𝜌1 =
𝑝1
101.3 𝑘𝑘𝑘
𝑘𝑘
=
= 0.946 3
𝑚
𝑅𝑇1 286.9 𝑘𝑘 × 373.2 𝐾
𝑘𝑘 𝐾
The mass of the hot air in the balloon is then:
𝑚 = 𝜌𝜌 = 0.946
The change in internal energy is given by
𝑘𝑘
× 2600 𝑚3 = 2460 𝑘𝑘
𝑚3
𝑚𝑚𝑚 = 𝑚𝑐𝑣 𝑑𝑑 = 2460 𝑘𝑘 × 717
𝐽
× (90 𝐶 − 100 𝐶) = −17,650 𝑘𝑘
𝑘𝑘 𝐾
The change in kinetic energy is assumed to be zero since the velocities are low. The change in potential
energy is
𝑚 𝑔 𝑑𝑑 = 2460 𝑘𝑘 × 9.8
The work done by the balloon is given by
𝑚
𝑘𝑘
× (1000 𝑚 − 0 𝑚) ×
= 24,120 𝑘𝑘
2
𝑠
1000 𝑁 𝑚
𝛿𝛿 = � 𝑝 𝑑𝑑
The pressure varies as the balloon rises, but we will assume that the pressure varies linearly and use the
average pressure for the process. The work is then given by
𝛿𝛿 = 𝑝𝑎𝑎𝑎 ∆𝑉
The volume at 1000 m depends on the pressure at that elevation. From Appendix A, the pressure at
1000 m is 0.887x101.3 kPa = 89.8 kPa. Using the ideal gas law, the density at 1000 m is
𝜌2 =
The volume at 1000 m is then
𝑝2
89.8 𝑘𝑘𝑘
𝑘𝑘
=
= 0.823 3
𝑚
𝑅𝑇2 286.9 𝑘𝑘 × 363.2 𝐾
𝑘𝑘 𝐾
𝑉2 =
The work is then
𝑚
2460 𝑘𝑘
=
= 2854 𝑚3
𝜌2 0.823 𝑘𝑘
𝑚3
𝛿𝛿 = 𝑝𝑎𝑎𝑎 ∆𝑉 = 95.55 𝑘𝑘𝑘 × (2854 − 2460)𝑚3 = 24,301 𝑘𝑘
The balloon does work on the atmosphere as it increases its volume
The heat transfer is determined using the energy balance
𝛿𝛿 = 𝛿𝛿 + 𝑑𝑑 = 24,301 𝑘𝑘 − 17,650 𝑘𝑘 + 24,120 𝑘𝑘 = 30,770 𝑘𝑘
Heat needed to be transferred to the balloon as it rose.
Problem 4.3
Problem
4.4
[Difficulty: 2]
4.3
Given:
Data on Boeing 777-200 jet
Find:
Minimum runway length for takeoff
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
1
N⋅ s
⎞ ×
⋅ ×
3 N
3600⋅ s ⎠
kg⋅ m
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 4.4
(Difficulty: 3)
4.4 On the Milford Trek in New Zealand, there is a pass with a cliff known as the “12 second drop” for
the time it takes a rock to hit the ground below from the pass. Estimate the height assuming that you
throw a 5 𝑐𝑐 diameter rock that weighs 200 𝑔 rock over the edge for (a) no air resistance and (b)
assuming that the drag force is 𝐾𝐾 �
𝑁∙𝑠 2
𝑚2
�, where 𝑉 is instantaneous velocity and K = 0.01. Explain why
there is a difference in the calculated height.
Find: The height of the pass
Solution:
From the Newton’s second law we have:
𝑑𝑑
𝑑𝑑
𝐹=𝑚
(a) If there is no air resistance, we only have:
𝑚
𝑑𝑑
= 𝑚𝑚
𝑑𝑑
Integrating
𝑑𝑑
=𝑔
𝑑𝑑
Or
𝑉 = 𝑔𝑔
𝑉=
Integrating again
𝑠
𝑑𝑑
= 𝑔𝑔
𝑑𝑑
12
� 𝑑𝑑 = � 𝑔𝑔𝑔𝑔
0
0
1
𝑠(𝑡) = [ 𝑔𝑡 2 ]12
0 = 706 𝑚
2
(b) From Newton’s second law, we now have the additional force of the aerodynamic drag:
𝑚
𝑑𝑑
= 𝑚𝑚 − 𝐾𝑉
𝑑𝑑
Or
𝑚
𝐾𝐾
𝐾
𝑑𝑑
=𝑔−
= − �𝑉 − 𝑔�
𝐾
𝑚
𝑚
𝑑𝑑
Separating variables
𝑚
𝑚
𝐾
𝑑
�𝑉 − 𝑔� = − �𝑉 − 𝑔�
𝐾
𝐾
𝑚
𝑑𝑑
𝑚
𝑑 �𝑉 − 𝑔�
𝐾
𝐾
=
−
𝑑𝑑
𝑚
𝑚
�𝑉 − 𝑔�
𝐾
Integrating both sides from the start, where V = 0 at t = 0, we have:
�
0
Or
ln �𝑉 −
Or, using the relation for logarithms
Which can be written as
𝑚
𝑡
− 𝑔�
𝐾
𝐾
�
=
− 𝑑𝑑
𝑚
𝑚
0
�𝑉 − 𝑔�
𝐾
𝑉 𝑑 �𝑉
𝑚
𝑚
𝐾
𝑔� − ln �− 𝑔� = − 𝑡
𝐾
𝐾
𝑚
𝑚
𝑉− 𝑔
𝐾𝐾
𝐾
�=− 𝑡
ln � 𝑚𝐾 � = 𝑙𝑙 �1 −
𝑚𝑚
𝑚
− 𝑔
𝐾
1−
Solving for V
𝑉=
The distance is then found from
𝑉=
Integrating both sides:
𝑠
Then
� 𝑑𝑑 =
0
𝐾
𝐾𝐾
= 𝑒 − 𝑚𝑡
𝑚𝑚
𝐾
𝑚𝑚
�1 − 𝑒 − 𝑚𝑡 �
𝐾
𝐾
𝑚𝑚
𝑑𝑑
�1 − 𝑒 − 𝑚𝑡 �
=
𝐾
𝑑𝑑
𝑚𝑚 𝑡 − 𝐾 𝑡
𝑚𝑚 𝑡
� 𝑑𝑑 −
� 𝑒 𝑚 𝑑𝑑
𝐾 0
𝐾 0
𝑠=
𝐾
𝑚𝑚
𝑚2 𝑔
𝑡 − 2 �1 − 𝑒 −𝑚𝑡 �
𝐾
𝐾
With values of 𝑡 = 12 𝑠, 𝑚 = 0.2 𝑘𝑘 and 𝐾 = 0.01, we have
𝑠=
0.05
0.22 × 9.8
0.2 × 9.81
�1 − 𝑒 − 0.2 ×12 � = 483 𝑚
× 12 −
2
0.01
0.01
The distance is less because the drag slows the rock and it takes more time to go the same distance.
Problem 4.5
Problem
4.6
[Difficulty: 2]
4.5
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
W
W
ΣFx = −Ff = −μ⋅ W = M ⋅ ax =
⋅ ax =
⋅
g
g
d
M = 2 ⋅ kg
L = 2⋅ m
2
dt
2
x
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 4.6
(Difficulty: 3)
4.6 For a small particle of Styrofoam �𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = 19.2
𝑘𝑘
𝑚3
, 𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑤𝑤𝑤ℎ 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑑 = 1.0 𝑚𝑚�
falling in standard air at speed 𝑉, the drag is given by 𝐹𝐷 = 3𝜋𝜋𝜋𝜋, where 𝜇 is the air viscosity. Find the
maximum speed of the particle starting from rest and the time it takes to reach 95 percent of this speed.
Plot the speed
𝑚
𝑠
as a function of time.
Find: The maximum velocity: 𝑉𝑚𝑚𝑚 . The time 𝑡 to reach 0.95𝑉𝑚𝑚𝑚 .
Solution:
When the partical reaches the maximum speed, the force is balanced. We have the following force
balance equation:
The body force is the weight of the Styrofoam
𝐹𝐵 = 𝐹𝐷
𝐹𝐵 = 𝑊 = 𝑔𝜌𝑠𝑠 𝑉𝑉𝑉
The volume of the Styrofoam is calculated as:
𝑉𝑉𝑉 =
The weight is then
The air viscosity is:
𝑊 = 9.81
The drag force is given by:
𝑚
𝑘𝑘
× 19.2 3 × 5.24 × 10−10 𝑚3 = 9.86 × 10−8 𝑁
2
𝑠
𝑚
𝜇 = 1.827 × 10−5
𝑘𝑘
𝑚∙𝑠
𝐹𝐷 = 3𝜋𝜋𝜋𝜋
The force balance is then
Or the maximum velocity is
𝑉𝑚𝑚𝑚 =
1 3 𝜋
𝜋𝑑 = × (0.001 𝑚)3 = 5.24 × 10−10 𝑚3
6
6
𝑊 = 3𝜋𝜋𝑉𝑚𝑚𝑚 𝑑
𝑊
9.86 × 10−8 𝑁
𝑚
=
= 0.573
𝑘𝑘
3𝜋𝜋𝜋 3 × 𝜋 × 1.827 × 10−5
𝑠
× 0.0015 𝑚
𝑚∙𝑠
From Newton’s second law, we have:
𝐹𝐵 − 𝐹𝐷 = 𝑚 𝑎 = 𝑚
Or
𝑑𝑑
𝑑𝑑
𝑑𝑑 𝐹𝐵 − 𝐹𝐷 𝑊 − 3𝜋𝜋𝜋𝜋
=
=
𝑚
𝑚
𝑑𝑑
𝑚
𝑑𝑑 = 𝑑𝑑
𝑊 − 3𝜋𝜋𝜋𝜋
Integrating this equation, we have:
�
𝑉
0
−
𝑉
𝑡
𝑚
𝑑(𝑊 − 3𝜋𝜋𝜋𝜋)
𝑚
��
𝑑𝑑 = − �
= � 𝑑𝑑
𝑊 − 3𝜋𝜋𝜋𝜋
3𝜋𝜋𝜋 0 𝑊 − 3𝜋𝜋𝜋𝜋
0
(𝑊 − 3𝜋𝜋𝜋𝜋)
𝑚
𝑚
ln(𝑊 − 3𝜋𝜋𝜋𝜋)𝑉0 = −
ln
=𝑡
3𝜋𝜋𝜋
3𝜋𝜋𝜋
𝑊
The time as a function of velocity is
𝑡=
The velocity at 95 % of the maximum is:
𝑚
𝑊
�
ln �
3𝜋𝜋𝜋
𝑊 − 3𝜋𝜋𝜋𝜋
𝑉 = 0.95𝑉𝑚𝑚𝑚 = 0.95 × 0.573
The mass of the Styrofoam sphere is
𝑚 = 𝜌𝑠𝑠 𝑉𝑉𝑉 = 19.2
𝑚
𝑚
= 0.544
𝑠
𝑠
𝑘𝑘
× 5.24 × 10−10 𝑚3 = 1.006 × 10−8 𝑘𝑘
𝑚3
The time to reach this velocity is calculated as follows.
ln �
1.006 × 10−8 𝑘𝑘
𝑡=
×
𝑘𝑘
× 0.001 𝑚
3 × 𝜋 × 1.827 × 10−5
𝑚∙𝑠
9.83 × 10−8 𝑁
9.83 × 10−8 𝑁 − 3 × 𝜋 × 1.827 × 10−5
𝑡 = 0.17 𝑠
The plot for the velocity as function of time is shown:
𝑘𝑘
𝑚
× 0.544 × 0.001𝑚
𝑚∙𝑠
𝑠
�
Problem 4.7
Problem
4.8
[Difficulty: 2]
4.7
Given:
Data on air compression process
Find:
Work done
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
cv
−k
⋅ dv
1 −k
⎝
cp
1 −k ⎞
− v2
⎠
=
1
( k − 1)
⋅ ⎛ p2⋅ v2 v2
⎝
k
1 −k
R ⋅ T1 ⎛ T2
⎞
⋅ T2 − T1 =
⋅⎜
−1
( k − 1)
( k − 1) T1
R
(
)
⎝
k
1 −k ⎞
− p1⋅ v1 ⋅ v2
⎠
(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 4.8
Problem
4.10
[Difficulty: 2]
4.8
Given:
Data on heating and cooling a copper block
Find:
Final system temperature
Solution:
Basic equation
Q − W = ∆E
Assumptions: 1) Stationary system ∆E = ∆U 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 (copper)
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
m
Solving Eq. 1 for Tf
Tf =
so
J
kg⋅ K
Tw = ( 10 + 273 ) ⋅ K
M w = ρ⋅ V
M copper ⋅ ccopper ⋅ Tcopper + M w⋅ cw⋅ Tw
(Mcopper⋅ ccopper + Mw⋅ cw)
Tf = 291 K
cw = 4186⋅
Tf = 18.1⋅ °C
M w = 4.00 kg
V = 4⋅ L
(1)
Problem
4.11
Problem 4.9
[Difficulty: 2]
4.9
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 ∆E =∆U 2) No work W = 0
W
Then for the air
∆U = Q = 85⋅
For the air and people
∆U = Qsurroundings = 0
person
× 6000⋅ people × 15⋅ min ×
60⋅ s
∆U = 459 ⋅ MJ
min
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 ⋅
This is the temperature change in 15 min. The rate of change is then
kg⋅ K
2
1
6
× 459 × 10 ⋅ J × ( 20 + 273 ) K ×
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 4.10
Problem
4.12
[Difficulty: 3]
4.10
Given:
Data on velocity field and control volume geometry
Find:
Several surface integrals
Solution:
r
dA1 = wdzˆj − wdykˆ
r
dA1 = dzˆj − dykˆ
r
dA2 = − wdykˆ
r
dA2 = − dykˆ
(
r
V = aˆj + bykˆ
(a)
(b)
)
(
r
V = 10 ˆj + 5 ykˆ
(
)(
)
)
r
V ⋅ dA1 = 10 ˆj + 5 ykˆ ⋅ dzˆj − dykˆ = 10dz − 5 ydy
1
1
1
r
5
1
V ⋅ dA1 = 10dz − 5 ydy = 10 z 0 − y 2 = 7.5
A1
2 0
∫
∫
∫
0
0
(
)(
)
(c)
r
V ⋅ dA2 = 10 ˆj + 5 ykˆ ⋅ − dykˆ = −5 ydy
(d)
r r
V V ⋅ dA2 = − 10 ˆj + 5 ykˆ 5 ydy
(e)
z
(
) (
)
) ∫(
)
1
1
r r
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
d
c
Control
volume
y
Problem 4.11
Problem
4.14
[Difficulty: 3]
4.11
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
r
dA = dydziˆ + dxdzˆj
r
r
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 )
r
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 =
r r r
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 4.12
[Difficulty 2]
Problem 4.13
(Difficulty: 2)
4.13 A 0.3 𝑚 by 0.5 𝑚 rectangular air duct carries a flow of 0.45
𝑚3
𝑠
at a density of 2
𝑘𝑘
𝑚3
. Calculate the
mean velocity in the duct. If the duct tapers to 0.15 𝑚 by 0.5 𝑚 size, what is the mean velocity in this
section if the density is 1.5
𝑘𝑘
𝑚3
there?
Given: Duct size: 𝑤1 = 0.3 𝑚; 𝐿1 = 0.5 𝑚; 𝑤2 = 0.3 𝑚; 𝐿2 = 0.5 𝑚. Density: 𝜌1 = 2
Volumetric flow rate: 𝑄 = 0.45
𝑚3
𝑠
.
Find: The mean velocity 𝑉1 and 𝑉2 .
Assumption: The density is constant
Solution: Use the continuity equation
𝑚̇ = 𝜌 𝑉 𝐴
For the duct entrance, we have for the flow area:
𝐴1 = 𝑤1 𝐿1 = 0.3 𝑚 × 0.5 𝑚 = 0.15 𝑚2
The mean velocity can be calculated:
𝑚3
0.45
𝑄
𝑠 =3 𝑚
𝑉1 =
=
𝐴1 0.15 𝑚2
𝑠
For the tapered section, the mass flow rate is the same as:
The flow area is:
𝑚̇ = 𝜌1 𝑉1 𝐴1 = 2
𝑘𝑘
𝑚
𝑘𝑘
× 3 × 0.15 𝑚2 = 0.9
3
𝑚
𝑠
𝑠
𝐴2 = 𝑤2 𝐿2 = 0.15 𝑚 × 0.5 𝑚 = 0.075 𝑚2
So the mean velocity in the tapered section is:
𝑘𝑘
0.9
𝑚̇
𝑚
𝑠
=
=8
𝑉2 =
𝜌2 𝐴2 1.5 𝑘𝑘 × 0.075 𝑚2
𝑠
𝑚3
𝑘𝑘
𝑚3
; 𝜌2 = 1.5
𝑘𝑘
𝑚3
.
Problem 4.14
(Difficulty: 1)
4.14 Across a shock wave in a gas flow there is a great change in gas density 𝜌. If a shock wave occurs in
a duct such that 𝑉 = 660
after the shock?
𝑚
𝑠
and 𝜌 = 1.0
𝑘𝑘
𝑚3
before the shock and 𝑉 = 250
Given: The velocity before the shock: 𝑉1 = 660
density before the shock: 𝜌1 = 1.0
𝑘𝑘
𝑚3
.
𝑚
𝑠
Solution:
From the continuity equation we have:
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
For steady flow there is no change with time and we have:
or
𝑠
after the shock, what is 𝜌
. The velocity after the shock: 𝑉2 = 250
Find: The density after the shock 𝜌2 .
0=
𝑚
0 = � 𝜌𝑉� ∙ 𝑑𝐴̅
𝐶𝐶
𝜌1 𝑉1 𝐴 = 𝜌2 𝑉2 𝐴
𝑘𝑘
𝑚
𝜌1 𝑉1 1.0 𝑚3 × 660 𝑠
𝑘𝑘
=
= 2.64 3
𝜌2 =
𝑚
𝑚
𝑉2
250
𝑠
𝑚
𝑠
. The
Problem 4.15
(Difficulty: 1)
4.15 Water flows in a pipeline composed of 75 𝑚𝑚 and 150 𝑚𝑚 in pipe. Calculate the mean velocity in
the 75 𝑚𝑚 pipe when that in the 150 𝑚𝑚 pipe is 2.5
150 𝑚𝑚 pipe?
Given: The velocity in the 150 𝑚𝑚: 𝑉1 = 2.5
𝑚
𝑠
𝑚
𝑠
. What is its ratio to the mean velocity in the
.
Find: Its ratio 𝛾 to the mean velocity in 150 𝑚𝑚 pipe.
Solution:
From the continuity equation we have:
0=
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
For steady flow there is no change with time and we have:
0 = � 𝜌𝑉� ∙ 𝑑𝐴̅
𝐶𝐶
or
Since the density is constant
𝜌𝑉1 𝐴1 = 𝜌𝑉2 𝐴2
𝜋𝜋12
𝑉1 𝐴1
𝑚 (150 𝑚𝑚)2
𝑚
𝐷12
= 𝑉1 4 2 = 𝑉1 2 = 2.5 ×
= 10
𝑉2 =
2
(75 𝑚𝑚)
𝑠
𝑠
𝐴2
𝜋𝜋2
𝐷2
4
𝛾=
𝑉2
=4
𝑉1
Problem 4.16
[Difficulty 2]
Problem 4.17*
Problem
4.20
[Difficulty: 1]
*Please Note: this solution shows the solution to the problem with one particular set of values. These values may not match those in the problem you have been given to solve.
4.17
Given:
Data on flow through nozzles
Find:
Average velocity in head feeder; flow rate
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
3
12⋅ in ⎠
1 ⋅ min
1 ⋅ ft
Q = 3.82⋅ gpm
Problem 4.18
Problem
4.22
[Difficulty: 1]
4.18
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 4.19
(Difficulty: 2)
4.19 Hydrogen is being pumped through a pipe system whose temperature is held at 273 𝐾. At a
section where the pipe diameter is 10 𝑚𝑚., the pressure and average velocity are 200 𝑘𝑘𝑘 and 30
Find all possible velocities and pressure at a downstream whose diameter is 20 𝑚𝑚.
Given: Temperature: 𝑇 = 273 𝐾. The upstream diameter: 𝐷1 = 10 𝑚𝑚. The upstream pressure:
𝑚
𝑝1 = 200 𝑘𝑘𝑘. The upstream velocity: 𝑉1 = 30 . The downstream diameter: 𝐷2 = 20 𝑚𝑚
𝑠
Find: The possible downstream velocity 𝑉2 and pressure 𝑝2 .
Solution:
From the continuity equation we have:
0=
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
For steady flow there is no change with time and we have:
or
For idea gas, we have:
0 = � 𝜌𝑉� ∙ 𝑑𝐴̅
𝐶𝐶
𝜌1 𝑉1 𝐴1 = 𝜌2 𝑉2 𝐴2
𝑝
= 𝑅𝑅
𝜌
Thus,
𝜌=
𝑝
𝑅𝑅
𝑝2
𝑝1
𝑉1 𝐴1 =
𝑉𝐴
𝑅𝑅
𝑅𝑅 2 2
𝑝1 𝑉1 𝐴1 = 𝑝2 𝑉2 𝐴2
𝑚 𝜋
3
2
𝑝1 𝑉1 𝐴1 200 × 10 𝑃𝑃 × 30 𝑠 × 4 × (10 𝑚𝑚)
𝑃𝑃 ∙ 𝑚
𝑃𝑃 ∙ 𝑚
𝑝2 𝑉2 =
=
= 1500000
= 1.5 × 106
𝜋
𝑠
𝑠
𝐴2
× (20 𝑚𝑚)2
4
Any combination of 𝑝2 and 𝑉2 giving the above result will be acceptable.
𝑚
𝑠
.
Problem 4.20
(Difficulty: 2)
4.20 Calculate the mean velocity for these two-dimensional velocity profiles if 𝑉𝑐 = 3
𝑚
Given: All the velocity profiles are shown in the figure with 𝑉𝑐 = 3 .
𝑠
Find: The mean velocity 𝑉𝑚 .
Solution: Use the definition for the mean velocity 𝑉𝑚 :
𝑉𝑚 =
(a) Parabola
1
(b) Linear
𝑉𝑚 = 𝑉𝑐 � (1 − 𝑥
0
1
(c) Circle
2)
1
� 𝑉𝑉𝑉
𝐴 𝐴
1
𝑥3
2
𝑚
𝑑𝑑 = 𝑉𝑐 �𝑥 − � = 𝑉𝑐 = 2
𝑠
3 0 3
1
𝑥2
1
𝑚
𝑉𝑚 = 𝑉𝑐 � (1 − 𝑥) 𝑑𝑑 = 𝑉𝑐 �𝑥 − � = 𝑉𝑐 = 1.5
𝑠
2 0 2
0
1 𝐴
𝑉
2 𝑐2
𝐴 = 𝜋𝑅 2
𝑅=1
1 𝜋
𝜋
𝑚
𝑉𝑚 = × × 𝑉𝑐 = 𝑉𝑐 = 2.36
2 2
4
𝑠
𝑉𝑚 =
(d) Linear and Flat
𝑉𝑚 =
(e) Parabola and Flat
𝑉𝑚 =
1
3
1
1 1
�� (1 − 𝑥)𝑉𝑐 𝑑𝑑 + 𝑉𝑐 � = � 𝑉𝑐 + 𝑉𝑐 � = 𝑉𝑐
4
2 0
2 2
𝑚
𝑉𝑚 = 2.25
𝑠
1
5
1
1 2
�� (1 − 𝑥 2 ) 𝑉𝑐 𝑑𝑑 + 𝑉𝑐 � = � 𝑉𝑐 + 𝑉𝑐 � = 𝑉𝑐
6
2 0
2 3
𝑚
𝑉𝑚 = 2.5
𝑠
𝑚
𝑠
.
Problem 4.21
(Difficulty: 2)
4.21 If the velocity profile in a passage of width 2𝑅 is given by the equation
expression for
𝑉
𝑣𝑐
𝑣
𝑣𝑐
=
1
𝑦 𝑛
� � ,
𝑅
derive an
in terms of n: (a) for a two-dimensional passage, and (b) for a cylindrical passage.
Find: The expression for
Solution:
𝑉
𝑣𝑐
in terms of n.
We have the equation:
1
(a) For the two dimensional passage
1
𝑦 𝑛
𝑣 = 𝑣𝑐 � �
𝑅
𝑛+1 𝑅
𝑅
1
𝑣𝑐
𝑦 𝑛
𝑛
𝑦 𝑛
� 1
𝑉=
∙ 2 � 𝑣𝑐 � � 𝑑𝑑 = �
2𝑅
𝑅 𝑛+1
𝑅
0
𝑅𝑛
So we have:
(b) For the axisymmetric passage
1
0
=
𝑣𝑐
𝑛
𝑛
�
� 𝑅 = 𝑣𝑐 �
�
𝑅 𝑛+1
𝑛+1
𝑉
𝑛
�
=�
𝑣𝑐
𝑛+1
𝑛+1
2𝑛+1
𝑅
𝑅
1
𝑛+1
2𝑣𝑐 𝑅𝑦 𝑛
1
2𝑣𝑐 𝑅
𝑦 𝑛
𝑦 𝑛
�
�
�
� �𝑅𝑦 𝑛 − 𝑦 𝑛 � 𝑑𝑑 =
𝑣
𝑉=
2𝜋(𝑅
−
𝑦)𝑑𝑑
=
�
−
�
𝑐
1
1
2
𝑛+1
2𝑛 + 1
𝜋𝑅 0
𝑅
2+
2+
0
𝑛
𝑛
𝑅
𝑅
𝑛
𝑛
0
2𝑛
2𝑛
�
−
𝑉 = 𝑣𝑐 �
𝑛 + 1 2𝑛 + 1
So we have:
2𝑛
2𝑛
𝑉
=
−
𝑣𝑐 𝑛 + 1 2𝑛 + 1
Problem 4.22*
Problem
4.24
[Difficulty: 1]
*Please Note: this solution shows the solution to the problem with one particular set of values. These values may not match those in the problem you have been given to solve.
4.22
Given:
Data on flow through box
Find:
Velocity at station 3
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 4.23
Problem
4.26
[Difficulty: 1]
4.23
Given:
Water needs of farmer
Find:
Number of supply pipes needed
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 4.24*
Problem
4.28
[Difficulty: 1]
*Please Note: this solution shows the solution to the problem with one particular set of values. These values may not match those in the problem you have been given to solve.
4.24
Given:
Data on filling of a sink
Find:
Time to half fill; rate at which level drops
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 4.25
(Difficulty: 1)
4.25 Fluid passes through this set of thin closely spaced blades. What flow rate 𝑞 is required for the
velocity 𝑉 to be 10
𝑓𝑓
𝑠
?
Given: The velocity 𝑉 = 10
Find: The flow rate 𝑞.
𝑓𝑓
𝑠
Solution: Use the continuity equation
The velocity vertical to the blade surface is:
𝑉𝑒 = 𝑉 cos 30°
The lateral surface area is
The volumetric flow rate can be calculated by:
𝐴 = 2𝜋𝜋ℎ
𝑄 = 𝑉𝑒 𝐴 = 𝑉 cos 30° 2𝜋𝜋ℎ
The volumetric flow rate per unit blade height is:
𝑞=
𝑞 = 10
𝑄
= 𝑉 cos 30° 2𝜋𝜋
ℎ
𝑟 = 1 𝑓𝑓 = 0.305 𝑚
𝑓𝑓
𝑓𝑓 2
× cos 30° × 2 × 𝜋 × 1 𝑓𝑓 = 54.4
𝑠
𝑠
Problem 4.26
(Difficulty: 2)
4.26 A pipeline 0.3 𝑚 in diameter divides at a 𝑌 into two branches 200 𝑚𝑚 and 150 𝑚𝑚 in diameter. If
the flow rate in the main line is 0.3
flow rate in the 150 𝑚𝑚 pipe?
𝑚3
𝑠
and the mean velocity in the 200 𝑚𝑚 pipe is 2.5
𝑚
𝑠
, what is the
Given: The diameter for the main line: 𝐷1 = 0.3 𝑚. The diameter for the two branches: 𝐷2 = 200 𝑚𝑚
and 𝐷3 = 150 𝑚𝑚. The flow rate in the main line is: 𝑞1 = 0.3
pipe: 𝑉2 = 2.5
𝑚
𝑚3
𝑠
𝑠
. The mean velocity in the 200 𝑚𝑚
Find: The flow rate 𝑞3 in the 150 𝑚𝑚 pipe.
Solution:
From the continuity equation we have:
0=
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
For steady flow there is no change with time and we have:
or
Since the density of water is constant
0 = � 𝜌𝑉� ∙ 𝑑𝐴̅
𝐶𝐶
𝜌𝑉1 𝐴1 = 𝜌𝑉2 𝐴2 + 𝜌𝑉3 𝐴3
𝑉1 𝐴1 = 𝑉2 𝐴2 + 𝑉3 𝐴3
where
𝑄1 = 𝑄2 + 𝑄3
𝑄1 = 𝑉1 𝐴1
𝑄2 = 𝑉2 𝐴2
Thus
𝑄3 = 𝑉3 𝐴3
𝑄2 = 𝑉2 𝐴2 =
𝜋 2
𝑚 𝜋
𝑚3
𝐷2 𝑉2 = 2.5 × × (0.2 𝑚)2 = 0.0785
4
𝑠 4
𝑠
So we can find the flow rate in the 150 mm pipe:
𝑄3 = 𝑄1 − 𝑄2 = 0.3
𝑚3
𝑚3
𝑚3
− 0.0785
= 0.222
𝑠
𝑠
𝑠
Problem 4.27
(Difficulty: 2)
4.27 A manifold pipe of 3 𝑖𝑖 diameter has four openings in its walls spaced equally along the pipe and is
closed at the downstream end. If the discharge from each opening is 0.5
velocities in the pipe between the openings?
𝑓𝑓 3
𝑠
, what are the mean
Given: The pipe diameter 𝐷 = 3 𝑖𝑖. The mass flow rate from each opening is: 𝑞𝑜 = 0.5
Find: The mean velocities in the pipe at A,B,C and D.
Solution:
From the continuity equation we have:
0=
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
For steady flow there is no change with time and we have:
0 = � 𝜌𝑉� ∙ 𝑑𝐴̅
𝐶𝐶
The density is constant and the volume flow rate in each pipe is
The area of the main pipe is:
𝑄 = 𝐴𝐴
2
𝜋 2 𝜋
3
𝑓𝑓� = 0.0491 𝑓𝑓 2
𝐴= 𝐷 = ×�
4
4
12
So we have the mass flow rate and mean velocity at cross section A are:
𝑄𝐴 = 4𝑄𝑜 = 2
𝑓𝑓 3
𝑠
𝑓𝑓 3
𝑄𝐴
𝑓𝑓
𝑠
𝑉𝐴 =
=
= 40.7
2
𝑠
𝐴
0.0491 𝑓𝑓
2
𝑓𝑓 3
𝑠
The mass flow rate and mean velocity at cross section B are:
𝑄𝐵 = 𝑄𝐴 − 𝑄𝑜 = 2
𝑓𝑓 3
𝑓𝑓 3
𝑓𝑓 3
− 0.5
= 1.5
𝑠
𝑠
𝑠
𝑓𝑓 3
1.5
𝑄𝐵
𝑓𝑓
𝑠
𝑉𝐵 =
=
= 30.5
2
𝑠
𝐴
0.0491 𝑓𝑓
The mass flow rate and mean velocity at cross section C are:
𝑓𝑓 3
𝑓𝑓 3
𝑓𝑓 3
− 0.5
= 1.0
𝑄𝐶 = 𝑄𝐵 − 𝑄𝑜 = 1.5
𝑠
𝑠
𝑠
𝑓𝑓 3
1.0
𝑄𝐶
𝑓𝑓
𝑠
𝑉𝐶 =
=
= 20.4
2
𝑠
𝐴
0.0491 𝑓𝑓
The mass flow rate and mean velocity at cross section D are:
𝑄𝐷 = 𝑄𝐶 − 𝑄𝑜 = 1.0
𝑓𝑓 3
𝑓𝑓 3
𝑓𝑓 3
− 0.5
= 0.5
𝑠
𝑠
𝑠
𝑓𝑓 3
0.5
𝑄
𝑓𝑓
𝑠
𝑉𝐷 = =
= 10.18
2
𝐴 0.0491 𝑓𝑓
𝑠
Problem
4.30
Problem 4.28
[Difficulty: 1]
4.28
Given:
Data on filling of a basement during a storm
Find:
Flow rate of storm into basement
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
⎜
3
60⋅ min
⎝ 12 hr ⎠
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 4.29*
Problem
4.32
[Difficulty: 2]
*Please Note: this solution shows the solution to the problem with one particular set of values. These values may not match those in the problem you have been given to solve.
4.29
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 4.30
Problem
4.34
[Difficulty: 2]
4.30
y
2h
c
Given:
Data on flow at inlet and outlet of channel
Find:
Find u max
x
d
CS
Solution:
Basic equation
r
r
∫ 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
⎡ h3
⎛ h3 ⎞⎤⎤
⎢
⎥⎥ = 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 4.31
(Difficulty: 1)
4.31 Find the average efflux velocity 𝑉 if the flow exists from a hole of area 1 𝑚2 in the side of the duct
as shown.
Given: The area of the hole 𝐴 = 1 𝑚2 . All the other parameters are shown in the figure.
Find: The average efflux velocity 𝑉.
Solution:
From the continuity equation we have:
0=
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
For steady flow there is no change with time and we have:
Assuming that the density is constant:
0 = � 𝜌𝑉� ∙ 𝑑𝐴̅
𝐶𝐶
𝑄𝑖𝑖 = 𝑄ℎ𝑜𝑜𝑜 + 𝑄𝑜𝑜𝑜
So we have:
𝑄ℎ𝑜𝑜𝑜 = 𝑄𝑖𝑖 − 𝑄𝑜𝑜𝑜
𝑄ℎ𝑜𝑜𝑜 = 10
For the flux through the hole we have:
𝑚3
𝑚3
𝑚3
−5
=5
𝑠
𝑠
𝑠
𝑄ℎ𝑜𝑜𝑜 = 𝑉 sin 30°𝐴
So the average efflux velocity exit from the hole is:
𝑚3
𝑄ℎ𝑜𝑜𝑜
𝑚
𝑠
𝑉=
=
= 10
2
𝑠
𝐴 sin 30° 1 𝑚 × 0.5
5
Problem 4.32
(Difficulty: 2)
4.32 Find 𝑉 for this mushroom cap on a pipeline.
Given: All the other parameters are shown in the figure.
Find: The velocity 𝑉.
Assume: The density is constant
Solution: From the continuity equation we have:
0=
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
For steady flow there is no change with time and we have:
The density is constant we have:
0 = � 𝜌𝑉� ∙ 𝑑𝐴̅
𝐶𝐶
𝑄𝑖𝑖 = 𝑄𝑜𝑜𝑜
Also we can get:
The outlet area is:
𝑄𝑜𝑜𝑜 = 𝑉 cos 45° 𝐴𝑜𝑜𝑜
𝐴𝑜𝑜𝑜 = 𝜋(𝑟22 − 𝑟12 ) = 𝜋[(2 𝑚)2 − (1.8 𝑚)2 ] = 2.39 𝑚2
The velocity is calculated to be:
𝑚3
3
𝑚
𝑄𝑜𝑜𝑜
𝑠
=
= 1.78
𝑉=
2
𝑠
𝐴𝑜𝑜𝑜 cos 45° 2.39 𝑚 × cos 45°
Problem 4.33
Problem
4.36
[Difficulty: 2]
4.33
Given:
Data on flow at inlet and outlet of channel
Find:
Find u max
Solution:
r
r
∫ 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⋅
Vm⋅ cos⎛⎜
⋅ ⎜ sin⎜
− sin( 0 ) ⎞ =
π ⎝ ⎝2⎠
π
⎝ 2⋅ H ⎠
⎠
Problem 4.34
Problem
4.38
[Difficulty: 2]
4.34
Given:
Data on flow at inlet and outlet of a reducing elbow
Find:
Find the maximum velcoity at section 1
Solution:
r
r
∫ 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
⋅ ⎮ y dy =
⋅
= V2 ⋅ h 2 + V3 ⋅ h 3
h 1 ⌡0
2
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 4.35
Problem
4.39
[Difficulty: 2]
4.35
Given:
Data on flow at inlet and outlet of channel
Find:
Find u max
Solution:
Basic equation
r
r
∫ 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 4.36
(Difficulty: 2)
4.36 Viscous liquid from a circular tank, 𝐷 = 300 𝑚𝑚 in diameter, drains through a long circular tube of
𝑟 2
radius 𝑅 = 50 𝑚𝑚. The velocity profile at the tube discharge is 𝑢 = 𝑢𝑚𝑚𝑚 �1 − � � �. Show that the
𝑅
1
average speed of flow in the drain tube is 𝑉� = 𝑢𝑚𝑚𝑚 . Evaluate the rate of change of liquid level in the
tank at the instant when 𝑢𝑚𝑚𝑚 = 0.155
𝑚
𝑠
2
.
Given: Tank diameter: 𝐷 = 300 𝑚𝑚. Tube radius: 𝑅 = 50 𝑚𝑚.
𝑟 2
Velocity profile at the tube discharge: 𝑢 = 𝑢𝑚𝑚𝑚 �1 − � � �. Maximum velocity: 𝑢𝑚𝑚𝑚 = 0.155
𝑅
1
𝑑ℎ
Find: Average velocity: 𝑉� = 𝑢𝑚𝑚𝑚 . Rate of change of liquid level in tank: .
2
𝑚
𝑠
𝑑𝑑
Assume: The liquid density is constant. The mass flow of air that enters the CV is neglected.
Solution:
a) The average velocity 𝑉� is defined as
𝑉=
Since 𝑄 = ∫ 𝑢𝑢𝑢, 𝑑𝑑 = 2𝜋𝜋𝜋𝜋 and 𝐴 = 𝜋𝑅 2 , then
𝑉� =
𝑄
=
𝐴
𝑅
𝑟 2
∫0 𝑢𝑚𝑚𝑚 �1 − �𝑅 � � 2𝜋𝜋𝜋𝜋
𝜋𝑅 2
=
2𝑢𝑚𝑚𝑚 𝑅
2𝑢𝑚𝑚𝑚 𝑅2 1
𝑟 2
�
�1
�
�
�
−
𝑟𝑟𝑟
=
= 𝑢𝑚𝑚𝑚
2
𝑅2 0
𝑅2 4
𝑅
b) Apply conservation of mass to the CV shown:
Basic equation:
𝑄
𝐴
.
0=
Then
As we have
0 = 𝜌𝑐
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝜕
𝜕 𝜋𝐷2
∀ + 𝜌𝑐 𝑉� 𝐴 = 𝜌𝑐 �
ℎ + 𝐿𝐿𝑅2 � + 𝜌𝑐 𝑉� 𝜋𝑅2
𝜕𝜕
𝜕𝜕 4
The change rate of the liquid level is:
0=
𝑑𝑑
=0
𝑑𝑑
𝜋𝐷 2 𝑑ℎ
+ 𝑉� 𝜋𝑅2
4 𝑑𝑑
4𝑉� 𝑅2
2𝑢𝑚𝑚𝑚 𝑅2
𝑚
𝑚
𝑚𝑚
𝑑ℎ
0.05 𝑚 2
� = −0.00861
�
=− 2 =−
=
−2
×
0.155
×
= 8.61
2
𝑠
𝑠
𝑠
𝐷
𝐷
𝑑𝑑
0.3 𝑚
Problem 4.37
(Difficulty: 2)
4.37 A rectangular tank used to supply water for a Reynolds flow experiment is 230 𝑚𝑚 deep. Its width
and length are 𝑊 = 150 𝑚𝑚 and 𝐿 = 230 𝑚𝑚. Water flows from the outlet tube (inside diameter
𝐷 = 6.35 𝑚𝑚) at Reynolds number 𝑅𝑅 = 2000, when the tank is half full. The supply valve is closed.
Find the rate of change of water level in the tank at this instant.
Given: Tank width: 𝑊 = 150 𝑚𝑚. Tank length: 𝐿 = 230 𝑚𝑚. Tube diameter: 𝐷 = 6.35 𝑚𝑚. Reynolds
number 𝑅𝑅 = 2000.
Find: Rate of change of water level in tank:
𝑑ℎ
𝑑𝑑
.
Assumption: (1) uniform flow at exit of tube.
(2) incompressible flow.
(3) neglect mass flow of air entering the control volume.
Solution:
From the continuity equation we have:
0=
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
For the control volume shown in the figure:
We have:
0=𝜌
𝜋𝐷2 𝐿1
𝜋𝐷2
𝜕
�𝑤𝑤ℎ +
� + 𝜌𝜌
𝜕𝜕
4
4
𝑑𝐿1
=0
𝑑𝑑
0 = 𝑤𝑤
At exit:
For water at 20 ℃
𝑅𝑅 =
𝜕ℎ
𝜋𝐷2
+𝑉
𝜕𝜕
4
𝑉𝑉
= 2000
𝑣
𝑣 = 1 × 10−6
𝑚2
𝑠
2
−6 𝑚
𝑅𝑅𝑅 2000 × 1 × 10
𝑠 = 0.315 𝑚
𝑉=
=
−3
𝐷
𝑠
6.35 × 10 𝑚
The change rate of the water level in the tank can be calculated by:
𝑚
−3
2
𝑉𝑉𝐷2 0.315 𝑠 × 𝜋 × (6.35 × 10 𝑚)
𝑚
𝑚𝑚
𝜕ℎ
=−
=
= 2.89 × 10−4
= −0.289
𝑠
𝑠
4𝑤𝑤
4 × 0.15 𝑚 × 0.23 𝑚
𝜕𝜕
Problem 4.38
(Difficulty: 2)
4.38 A cylindrical tank, 0.3 𝑚 in diameter, drains through a hole in its bottom. At the instant when the
depth is 0.6 𝑚, the flow rate from the tank is observed to be 4
water level at this instant.
𝑘𝑘
𝑠
. Determine the rate of change of
Given: Tank diameter: 𝐷 = 0.3 𝑚. Flow rate from the tank: 𝑚̇1 = 4.0
𝑑ℎ
Find: Rate of change of water level in tank:
𝑑𝑑
.
Assumption: (1) uniform flow at exit of tube.
(2) incompressible flow.
(3) the control volume is fixed
Solution:
From the continuity equation we have:
0=
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
For the control volume shown in the figure:
We have:
0=𝜌
𝜕
𝜋𝐷2
[∀] + 𝜌𝑉2
+ 𝑚̇1
4
𝜕𝜕
𝑑∀
=0
𝑑𝑑
𝑘𝑘
𝑠
.
̇ 1
4𝑚
𝜌𝜌𝐷2
The density of water is:
𝑉2 = −
The rate of change of water level is:
𝜌 = 999
𝑘𝑘
𝑚3
𝑘𝑘
4 × 4.0
̇ 1
4𝑚
𝑚
𝑚𝑚
𝑑ℎ
𝑠
= 𝑉2 = −
=−
= −0.0566
= −56.6
2
𝑘𝑘
𝑠
𝑠
𝜌𝜌𝐷
𝑑𝑑
999 3 × 𝜋 × (0.3 𝑚)2
𝑚
Problem 4.39*
(Difficulty: 2)
*Please Note: this solution shows the solution to the problem with one particular set of values. These values may not match those in the problem you have been given to solve.
4.39 Air enters a tank through an area of 0.2 𝑓𝑓 2 with a velocity of 15
leaves with a velocity of 5
tank is 0.02
𝑠𝑠𝑠𝑠
𝑓𝑓 3
𝑓𝑓
𝑠
𝑓𝑓
𝑠
𝑠𝑠𝑠𝑠
and a density of 0.03
𝑓𝑓 3
. Air
and a density equal to that in the tank. The initial density of the air in the
. The total tank volume is 20 𝑓𝑓 3 and the exit area is 0.4 𝑓𝑓 2 . Find the initial rate of
change of density in the tank.
Given: Tank inlet area: 𝐴1 = 0.2 𝑓𝑓 2 . Inlet air velocity: 𝑉1 = 15
Tank outlet area: 𝐴2 = 0.4 𝑓𝑓 2 . Outlet air velocity: 𝑉2 = 5
Tank volume: 𝑉 = 20 𝑓𝑓 3 .
Find: Initial change rate of density in the tank
Assumption: (1) density is uniform in the tank.
𝜕𝜌0
𝜕𝜕
𝑓𝑓
𝑠
𝑓𝑓
𝑠
. Inlet air density: 𝜌1 = 0.03
. Initial air density: 𝜌0 = 0.02
.
(2 flow is uniform at inlet and outlet sections.
Solution:
For the control volume shown in the figure:
At the initial time we have:
0=
0=
0=∀
𝜕
�⃗ ∙ 𝑑𝐴⃗
� 𝜌𝜌∀ + � 𝜌𝑉
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝜕
(𝜌 ∀) − 𝜌1 𝑉1 𝐴1 + 𝜌2 𝑉2 𝐴2
𝜕𝜕 0
𝜕𝜌0
𝜕∀
+ 𝜌0
− 𝜌1 𝑉1 𝐴1 + 𝜌2 𝑉2 𝐴2
𝜕𝜕
𝜕𝜕
𝑠𝑠𝑠𝑠
𝑓𝑓 3
𝑠𝑠𝑠𝑠
.
𝑓𝑓 3
.
So we have:
𝜕∀
=0
𝜕𝜕
The initial rate of density change in the tank is:
𝜌2 = 𝜌0
𝜕𝜌0 𝜌1 𝑉1 𝐴1 − 𝜌2 𝑉2 𝐴2
=
𝜕𝜕
∀
𝑓𝑓
𝑠𝑠𝑠𝑠
𝑓𝑓
𝑠𝑠𝑠𝑠
2
2
𝑠𝑠𝑠𝑠
𝜕𝜌0 0.03 𝑓𝑓 3 × 15 𝑠 × 0.2 𝑓𝑓 − 0.02 𝑓𝑓 3 × 5 𝑠 × 0.4 𝑓𝑓
=
= 0.0025 3
3
𝑓𝑓 ∙ 𝑠
𝜕𝜕
20 𝑓𝑓
So mass in the tank increases.
Problem 4.40*
Problem
4.48
[Difficulty: 3]
*Please Note: this solution shows the solution to the problem with one particular set of values. These values may not match those in the problem you have been given to solve.
4.40
Given:
Data on draining of a tank
Find:
Depth at various times; Plot of depth versus time
Solution:
Basic equation
r r
∂
ρ
d
V
+
ρ
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
dt
+ ρ⋅
π
4
2
⋅d ⋅V = 0
1
2⋅ g⋅ y
and simplifying
dy
dt
2
=
4
2 dy
⋅D ⋅
⎛ 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 4.41
(Difficulty: 3)
4.41 A conical flask contains water to height 𝐻 = 36.8 𝑚𝑚, where the flask diameter is 𝐷 = 29.4 𝑚𝑚.
Water drains out through a smoothly rounded hole of diameter 𝑑 = 7.35 𝑚𝑚 at the apex of the cone.
The flow speed at the exit is 𝑉 = �2𝑔𝑔, where 𝑦 is the height of the liquid free surface above the hole.
A stream of water flows into the top of the flask at constant volume flow rate, 𝑄 = 3.75 × 10−7
𝑚3
ℎ𝑟
.
Find the volume flow rate from the bottom of the flask. Evaluate the direction and rate of change of
water surface level in the flask at this instant.
Given: Water Height: 𝐻 = 36.8 𝑚𝑚. Flask diameter: 𝐷 = 29.4 𝑚𝑚.
Diameter of round hole: 𝑑 = 7.35 𝑚𝑚. The speed at exit: 𝑉 = �2𝑔𝑔.
Volumetric flow rate into the flask: 𝑄𝑖𝑖 = 3.75 × 10−10
𝑚3
ℎ𝑟
.
Find: The volume flow rate from the bottom of the flask 𝑄𝑜𝑜𝑜 . The direction and rate of change of
𝑑𝑑
water surface level .
𝑑𝑑
Assumption: 1) uniform flow at each section.
2) neglect mass of air.
3) Density is constant
Solution:
For the control volume shown in the figure:
Then we have:
0=
𝜕
�⃗ ∙ 𝑑𝐴⃗
� 𝜌𝜌∀ + � 𝜌𝑉
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑄𝑜𝑜𝑜 = 𝑉0
0=𝜌
𝑑∀
+ 𝜌𝑄𝑜𝑜𝑜 − 𝜌𝑄𝑖𝑖
𝑑𝑑
𝑚
𝜋 × (0.00735 𝑚)2
𝑚3
𝜋𝑑 2
𝜋𝑑 2
= �2𝑔𝑔
= �2 × 9.81 2 × 0.0368 𝑚 ×
= 3.61 × 10−5
𝑠
4
4
4
𝑠
As we know:
𝑄𝑜𝑜𝑜 = 3.61 × 10−5
𝑚3
𝑚3
= 0.130
𝑠
ℎ𝑟
𝑑∀
= 𝑄𝑖𝑖 − 𝑄𝑜𝑜𝑜
𝑑𝑑
∀=
1 2
𝜋𝑅 𝑦
3
𝑅 = 𝑦 tan 𝜃
So we have the following equation:
1
𝑑 𝜋𝑦 3 tan 𝜃 tan 𝜃
3
= 𝑄𝑖𝑖 − 𝑄𝑜𝑜𝑜
𝑑𝑑
𝜋𝜋 2 tan 𝜃 tan 𝜃
The change rate of the water surface level is:
𝑑𝑑 𝑄𝑖𝑖 − 𝑄𝑜𝑜𝑜 𝑄𝑖𝑖 − 𝑄𝑜𝑜𝑜
=
=
=
𝐷2
𝜋𝑅 2
𝑑𝑑
𝜋
4
𝑑𝑑
𝑑𝑑
= 𝜋𝑅 2
= 𝑄𝑖𝑖 − 𝑄𝑜𝑜𝑜
𝑑𝑑
𝑑𝑑
𝑚3
𝑚3
�
− 0.130
𝑚
𝑚
ℎ𝑟
ℎ𝑟
= −191.5
= −0.0532
ℎ𝑟
𝑠
𝜋(0.0294 𝑚)2
4 × �3.75 × 10−7
Problem 4.42
(Difficulty: 2)
4.42 Water flows steadily past a porous flat plate. Constant suction is applied along the porous section.
The velocity profile at section cd is
𝑢
𝑦
= 3� � −
𝑢∞
𝛿
Given: The velocity profile at section cd is:
are shown in the figure.
𝑢
𝑢∞
3
𝑦 2
2� � .
𝛿
𝑦
Evaluate the mass flow rate across the section bc.
= 3� � −
𝛿
3
𝑦 2
2� � .
𝛿
All the other dimensions and parameters
Find: Evaluate the mass flow rate 𝑚̇𝑏𝑏 across the section bc.
Assumption: (1) steady flow.
(2) incompressible flow.
Solution:
�⃗ = −0.2 𝚥⃗
(3) 𝑉
𝑚𝑚
𝑠
along da.
Basic Equations: The continuity equation:
0=
For steady state we have:
𝜕
�⃗ ∙ 𝑑𝐴⃗
� 𝜌𝜌∀ + � 𝜌𝑉
𝜕𝜕 𝐶𝐶
𝐶𝐶
�⃗ ∙ 𝑑𝐴⃗ = � 𝜌𝑉
�⃗ ∙ 𝑑𝐴⃗ + 𝑚̇𝑏𝑏 + � 𝜌𝑉
�⃗ ∙ 𝑑𝐴⃗ + � 𝜌𝑉
�⃗ ∙ 𝑑𝐴⃗
0 = � 𝜌𝑉
𝐶𝐶
So we have:
𝑎𝑎
𝛿
𝑐𝑐
3
𝑑𝑑
𝑦 2
𝑦
0 = −𝜌𝜌∞ 𝛿𝛿 + 𝑚̇𝑏𝑏 + � 𝜌 𝑢∞ �3 � � − 2 � � � 𝑤𝑤𝑤 + 𝜌𝑉0 𝑤𝑤
𝛿
𝛿
0
𝑚̇𝑏𝑏
1
3
𝑦 2
𝑦
𝑦
= 𝜌𝜌∞ 𝛿𝛿 − 𝜌𝑉0 𝑤𝑤 − 𝜌𝜌∞ 𝑤𝑤 � �3 � � − 2 � � � 𝑑 � �
𝛿
𝛿
𝛿
0
2 𝑦 2.5 1
3 𝑦 2
� � � − 𝑉0 𝐿� = 𝜌𝜌(0.3𝑢∞ 𝛿 − 𝑉0 𝐿)
𝑚̇𝑏𝑏 = 𝜌𝜌 �𝑢∞ 𝛿 − 𝑢∞ 𝛿 � � � −
2.5 𝛿
2 𝛿
0
The density for the water is:
So the mass flow rate across section bc is:
𝑚̇𝑏𝑏 = 999
𝜌 = 999
𝑘𝑘
𝑚3
𝑘𝑘
𝑚
𝑚
𝑘𝑘
× 1.5 𝑚 × �0.3 × 3 × 0.0015 𝑚 − 0.0002 × 2 𝑚� = 1.42
3
𝑚
𝑠
𝑠
𝑠
The mass flow rate is out of the control volume.
Problem 4.43
(Difficulty: 3)
4.43 A tank of fixed volume contains brine with initial density, 𝜌𝑖 greater than water. Pure water enters
the tank steadily and mixes thoroughly with the brine in the tank. The liquid level in the tank remains
constant. Derive expressions for (a) the rate of change of density of the liquid mixture in the tank and (b)
the time required for the density to reach the value 𝜌𝑓 , where 𝜌𝑖 > 𝜌𝑓 > 𝜌𝐻2 𝑜 .
Given: The initial density: 𝜌𝑖 .
Find: (a) The rate of density change of liquid mixture. (b) The time required for the density to reach 𝜌𝑓 .
Assumption: (1) 𝑉𝑡𝑡𝑡𝑡 = 𝑐𝑜𝑜𝑜𝑜𝑜𝑜𝑜.
(2) 𝜌 uniform in the tank.
(3) uniform flows at inlet and outlet sections.
Solution:
(a)
For the control volume shown in the figure, the continuity equation is:
0=
𝜕
�⃗ ∙ 𝑑𝐴⃗
� 𝜌𝜌∀ + � 𝜌𝑉
𝜕𝜕 𝐶𝐶
𝐶𝐶
The tank volume remains constant and so the volume flow rate in and out are related as:
𝑉1 𝐴1 = 𝑉2 𝐴2
The continuity equation is:
0=∀
Or the rate of change of density with time is
𝜕𝜕
+ 𝜌𝜌𝜌 − 𝜌𝐻2 𝑜 𝑉𝑉
𝜕𝜕
�𝜌 − 𝜌𝐻2 𝑜 �𝑉𝑉
𝑑𝑑
=−
∀
𝑑𝑑
(b) We have the relation as:
𝑑𝑑
�𝜌 − 𝜌𝐻2 𝑜 �
=−
Integrating for both sides from the initial state we have:
�
𝜌𝑓
𝜌𝑖
𝑑𝑑
�𝜌 − 𝜌𝐻2 𝑜 �
𝑉𝑉
𝑑𝑑
∀
𝑡
=� −
0
𝑉𝑉
𝑑𝑑
∀
ln�𝜌𝑓 − 𝜌𝐻2 𝑜 � − ln�𝜌𝑖 − 𝜌𝐻2 𝑜 � = −
𝜌𝑓 − 𝜌𝐻2 𝑜
𝑉𝑉
�=−
𝑡
∀
𝜌𝑖 − 𝜌𝐻2 𝑜
ln �
𝑉𝑉
(𝑡 − 0)
∀
Finally we have for the time required for the density to become ρf :
𝑡=−
Note that 𝜌𝑓 → 𝜌𝐻2 𝑜 asymptotically as 𝑡 → ∞.
𝜌𝑓 − 𝜌𝐻2 𝑜
∀
ln �
�
𝜌𝑖 − 𝜌𝐻2 𝑜
𝑉𝑉
Problem 4.44
Problem
4.55
[Difficulty: 4]
*Please Note: this solution shows the solution to the problem with one particular set of values. These values may not match those in the problem you have been given to solve.
4.44
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
r r
∂
ρ
d
V
+
ρ
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
For the funnel
Hence
Then
2
Afunnel = π⋅ r = π⋅ ( y ⋅ tan( θ) )
2∂⌠
2
y
π 2
2
ρ⋅ π⋅ ( tan( θ) ) ⋅ ⎮ y dy + ρ⋅ V⋅ ⋅ d = 0
⌡
4
∂t 0
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 = ⋅
2
5
2⋅ g⋅ d
y 1 = 6 ⋅ in
2
5
2
8 tan( θ) ⋅ y 1
− ⋅
2
5
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 4.45*
Problem
4.58
[Difficulty: 3]
*Please Note: this solution shows the solution to the problem with one particular set of values. These values may not match those in the problem you have been given to solve.
4.45
4.22
Given:
Data on flow through a control surface
Find:
Net rate of momentum flux
Solution:
Basic equation: We need to evaluate
∫
CS
r r
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
r r
r r r r r r
r r r
VρV ⋅ dA = V1ρV1 ⋅ A1 + V2 ρV2 ⋅ A2 + V3 ρV3 ⋅ A3
CS
r r
r r
r r
= 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
lbf ⋅ s
) ft2 × lbm
= −2406⋅ lbf
⋅ ft
4
2
× −10 × 0.5 + 5 × 0.6 × sin( 60⋅ deg) ⋅
s
ρ [V22 A2 − V32 A3 cos(60 )] =
65⋅
lbm
ft
3
(
2
2
lbf ⋅ s
) ft2 × lbm
= 2113⋅ lbf
⋅ ft
4
× 20 × 0.1 − 5 × 0.6 × cos( 60⋅ deg) ⋅
s
2
Problem 4.46
Problem
4.60
4.46
Problem
4.61
Problem 4.47
[Difficulty: 3]
4.47
4.34
Given:
Data on flow through a bend
Find:
Find net momentum flux
Solution:
Basic equations
r
r
∫ 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
2
h1
⌡
⎮
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
⎛
⎞
2 h1
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 4.48
(Difficulty: 2)
4.48 Evaluate the net momentum flux through the channel of Problem 4.35. Would you expect the
outlet pressure to be higher, lower, or the same as the inlet pressure? Why?
Find: Would you expect the outlet pressure to be higher, lower or the same as inlet one.
Assumption: (1) incompressible flow
(2) uniform flow at ①
Solution:
The momentum flux is defined as
𝑚1 𝑓 = � 𝑉� (𝜌𝑉� ∙ 𝑑𝐴̅)
The net momentum flux through the CV is
where
𝑚1 𝑓 = � 𝑉� (𝜌𝑉� ∙ 𝑑𝐴̅) + � 𝑉� (𝜌𝑉� ∙ 𝑑𝐴̅)
𝐴1
𝐴2
𝑉�1 = 𝑈 𝚤̂
𝑥
𝑉�2 = �𝑉𝑚𝑚𝑚 − (𝑉𝑚𝑚𝑚 − 𝑉𝑚𝑚𝑚 ) � 𝚥̂
ℎ
𝑥
𝑥
𝑉�2 = �2𝑉𝑚𝑚𝑚 − (𝑉𝑚𝑚𝑚 ) � 𝚥̂ = 𝑉𝑚𝑚𝑚 �2 − � 𝚥̂
ℎ
ℎ
� 𝑉� (𝜌𝑉� ∙ 𝑑𝐴̅) = 𝑉�1 {−|𝜌𝑉1 𝐴1 |} = −𝜌𝑈 2 ℎ2 𝚤̂
𝐴1
ℎ
ℎ
𝑥
𝑥
𝑥 𝑥2
2
� 𝑉� (𝜌𝑉� ∙ 𝑑𝐴̅) = � 𝑉𝑚𝑚𝑚 �2 − � 𝚥̂𝜌𝑉𝑚𝑚𝑚 �2 − � ℎ𝑑𝑑 = 𝚥̂𝜌𝑉𝑚𝑚𝑚
ℎ � �4 − 4 + 2 � 𝑑𝑑
ℎ
ℎ
ℎ ℎ
𝐴2
0
0
7 2 2
� 𝑉� (𝜌𝑉� ∙ 𝑑𝐴̅) = 𝚥̂ 𝜌𝑉𝑚𝑚𝑚
ℎ
3
𝐴2
Evaluating
7 2 2
7 2
ℎ 𝚥̂ = 𝜌ℎ2 �−𝑈 2 𝚤̂ + 𝑉𝑚𝑚𝑚
𝚥̂�
𝑚1 𝑓 = −𝜌𝑈 2 ℎ2 𝚤̂ + 𝜌𝑉𝑚𝑚𝑚
3
3
𝑚1 𝑓 = 999
𝑚 2
𝑚 2
𝑘𝑘
7
𝑁 ∙ 𝑠2
2 �− �7.5
(0.0755
�
�5
�
�
×
𝑚)
𝚤̂
+
𝚥̂
×
𝑠
𝑠
𝑚3
3
𝑘𝑘 ∙ 𝑚
𝑚1 𝑓 = −320 𝚤̂ + 332 𝚥̂ 𝑁
For viscous (real) flow, friction causes a pressure drop in the direction or flow.
For flow in a bend, the streamline curvature results in a pressure gradient normal to the flow.
Problem 4.49
(Difficulty: 3)
4.49 A conical enlargement in a vertical pipeline is 5 𝑓𝑓 long and enlarges the pipe from 12 𝑖𝑖 to 24 𝑖𝑖
diameter. Calculate the magnitude and direction of the vertical force on this enlargement when 10
𝑓𝑓 3
𝑠
of water flow upward through the line and the pressure at the smaller end of the enlargement is 30 𝑝𝑝𝑝.
Given: The flow rate: 𝑄 = 10
𝑓𝑓 3
𝑠
. All the other parameters are shown in the figure.
Find: The magnitude and direction of the vertical force on the enlargement.
Assumptions: The water density is constant.
The flow is steady
Solution:
Basic equations:
Continuity:
Bernoulli equation
Momentum equation
0=
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
2
𝜌
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑣𝑣𝑣∀ + � 𝑣𝑣𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
Thus for steady, incompressible flow
𝑄 = 𝑉1 𝐴1 = 𝑉2 𝐴2
The area for the inlet section and outlet section are:
2
𝜋 2 𝜋
12
𝐴1 = 𝐷1 = × � 𝑓𝑓� = 0.785 𝑓𝑓 2
4
4
12
𝐴2 =
2
𝜋 2 𝜋
24
𝐷2 = × �
𝑓𝑓� = 3.14 𝑓𝑓 2
4
4
12
The velocities for the inlet and outlet section are:
𝑓𝑓 3
10
𝑄
𝑠 = 12.73 𝑓𝑓
𝑉1 =
=
𝐴1 0.785 𝑓𝑓 2
𝑠
𝑓𝑓 3
10
𝑄
𝑠 = 3.18 𝑓𝑓
𝑉2 =
=
𝐴2 3.14 𝑓𝑓 2
𝑠
The pressure at the smaller end is:
𝑝1 = 30 𝑝𝑝𝑝 = 4320
From the Bernoulli equation:
𝑙𝑙𝑙
𝑓𝑓 2
𝑝2 𝑉22
𝑝1 𝑉12
+
+ 𝑔(0) = +
+ 𝑔ℎ
𝜌
2
𝜌
2
𝜌 = 1.94
ℎ = 5 𝑓𝑓
𝑠𝑠𝑠𝑠
𝑙𝑙𝑙 ∙ 𝑠 2
=
1.94
𝑓𝑡 3
𝑓𝑓 4
𝜌
𝑙𝑙𝑙
𝑙𝑙𝑙 1.94 𝑙𝑙𝑙 ∙ 𝑠 2
𝑓𝑓 2
2
2)
(12.73
×
−
3.18
− 62.4 3 × 5𝑓𝑓
𝑝2 = 𝑝1 + (𝑉12 − 𝑉22 ) − 𝛾ℎ = 4320 2 +
4
2
2
2
𝑓𝑓
𝑓𝑓
𝑠
𝑓𝑓
𝑝2 = 4156
The volume for the enlargement is:
𝑉=
𝑙𝑙𝑙
𝑓𝑓 2
1
�𝐴 + 𝐴2 + �𝐴1 𝐴2 �ℎ = 9.16 𝑓𝑓 3
3 1
From the momentum equation, we have:
𝑅𝑦 + 𝐹𝑠𝑠 + 𝐹𝐵𝐵 = −𝑉1 2 𝜌𝐴1 + 𝑉2 2 𝜌𝐴2
𝑅𝑦 + 𝑝1 𝐴1 − 𝑝2 𝐴2 − 𝛾𝛾 = −𝑉1 2 𝜌𝐴1 + 𝑉2 2 𝜌𝐴2
𝑅𝑦 = −𝑉1 2 𝜌𝐴1 + 𝑉2 2 𝜌𝐴2 + 𝑝2 𝐴2 − 𝑝1 𝐴1 + 𝛾𝛾
So we have:
𝑅𝑦 = 10060 𝑙𝑙𝑙
𝐾𝑦 = −𝑅𝑦 = −10060 𝑙𝑙𝑙 (direction is going down)
Problem 4.50
(Difficulty: 2)
4.50 A 100 𝑚𝑚 nozzle is bolted with 6 bolts to the flange of a 300 𝑚𝑚 horizontal pipeline and
discharges water into the atmosphere. Calculate the tension load on each bolt when the pressure in the
pipe is 600 𝑘𝑘𝑘. Neglect vertical forces.
Given: All the parameters are shown in the figure.
Find: The tension load on each bolt.
Assumption:
Density is constant
Flow is steady
Solution:
Basic equations:
Continuity:
Bernoulli equation
Momentum equation
0=
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
2
𝜌
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
Thus for steady incompressible flow
𝑄 = 𝑉1 𝐴1 = 𝑉2 𝐴2
The area for the inlet section and outlet section are:
𝐴1 =
𝐴2 =
𝜋 2 𝜋
𝐷 = × (0.3 𝑚)2 = 0.071 𝑚2
4 1 4
𝜋 2 𝜋
𝐷 = × (0.1 𝑚)2 = 0.0079 𝑚2
4 2 4
𝑉1 =
The pressure at the inlet and outlet is:
𝑉2 𝐴2 1
= 𝑉2
9
𝐴1
𝑝1 = 600 𝑘𝑘𝑘
𝑝2 = 0 𝑃𝑃
From the Bernoulli equation:
0 𝑉22
𝑝1 𝑉12
+
+ 𝑔(0) = +
+ 𝑔(0)
𝜌
𝜌
2
2
𝑝1 𝑉12 𝑉22 81 𝑉12
+
=
=
𝜌
2
2
2
𝜌 = 999
𝑉12 =
𝑉1 = �
The mass flow rate is
𝑘𝑘
𝑚3
𝑝1
40𝜌
𝑝1
𝑚
600 × 103 𝑃𝑃
=�
= 3.87
𝑘𝑘
𝑠
40𝜌
40 × 999 3
𝑚
𝑉2 = 9𝑉1 = 9 × 3.87
𝑚̇ = 𝜌𝑉1 𝐴1 = 999
From the momentum equation, we have:
𝑚
𝑚
= 34.8
𝑠
𝑠
𝑘𝑘
𝑚
𝑘𝑘
× 3.87 × 0.071 𝑚2 = 275
3
𝑚
𝑠
𝑠
𝑅𝑥 + 𝐹𝑠𝑠 + 𝐹𝐵𝐵 = −𝑉1 𝑚̇ + 𝑉2 𝑚̇
𝑅𝑥 + 𝑝1 𝐴1 − 𝑝2 𝐴2 = −𝑉1 𝑚̇ + 𝑉2 𝑚̇
𝑅𝑥 = 𝑚̇(𝑉2 − 𝑉1 ) − 𝑝1 𝐴1
𝑅𝑥 = −34100 𝑁
So we have:
𝐾𝑥 = −𝑅𝑥 = 34100 𝑁
The force on each bolt is (direction to the right):
𝐹𝑥 =
𝐾𝑥 34100 𝑁
=
= 5680 𝑁 = 5.68 𝑘𝑘
6
6
Problem 4.51
(Difficulty: 3)
4.51 The projectile partially fills the end of the 0.3 𝑚 pipe. Calculate the force required to hold the
projectile in position when the mean velocity in the pipe is 6
Given: The mean velocity in the pipe 𝑉1 = 6
𝑚
𝑠
𝑚
𝑠
.
. All the other parameters are shown in the figure.
Find: The force required to hold the projectile in position.
Assumptions: Flow is steady
Density is constant
Solution:
Basic equations:
Continuity
0=
Bernoulli equation
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
2
𝜌
Momentum equation
For steady incompressible flow
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑄 = 𝑉1 𝐴1 = 𝑉2 𝐴2
The area for the inlet section and outlet section are:
𝐴1 =
𝐴2 = 𝐴1 −
𝜋 2 𝜋
𝐷 = × (0.3 𝑚)2 = 0.071 𝑚2
4 1 4
𝜋 2
𝜋
𝐷2 = 0.071 𝑚2 − × (0.25 𝑚)2 = 0.0219 𝑚2
4
4
𝑉1 = 6
𝑚
𝑠
The pressure at the outlet is:
𝑚
2
𝑉1 𝐴1 6 𝑠 × 0.071 𝑚
𝑚
=
=
19.45
𝑉2 =
𝑠
𝐴2
0.0219 𝑚2
From the Bernoulli equation:
𝑝2 = 0 𝑃𝑃
𝑝2 𝑉22
𝑝1 𝑉12
+
+ 𝑔(0) = +
+ 𝑔(0)
𝜌
2
𝜌
2
𝑝1 𝑉12 𝑉22
+
=
𝜌
2
2
𝜌 = 999
𝑘𝑘
𝑚3
𝑘𝑘
999 3
𝜌 2
𝑚 2
𝑚 2
𝑚
𝑝1 = (𝑉2 − 𝑉12 ) =
× ��19.45 � − �6 � � = 180 𝑘𝑘𝑘
2
𝑠
𝑠
2
The mass flow rate is
𝑚̇ = 𝜌𝑉1 𝐴1 = 999
From the momentum equation, we have:
𝑘𝑘
𝑚
𝑘𝑘
× 6 × 0.071 𝑚2 = 426
3
𝑚
𝑠
𝑠
𝑅𝑥 + 𝐹𝑠𝑠 + 𝐹𝐵𝐵 = −𝑉1 𝑚̇ + 𝑉2 𝑚̇
𝑅𝑥 + 𝑝1 𝐴1 − 𝑝2 𝐴2 = −𝑉1 𝑚̇ + 𝑉2 𝑚̇
𝑅𝑥 = 𝑚̇(𝑉2 − 𝑉1 ) − 𝑝1 𝐴1
So we have:
𝑅𝑥 = −7050 𝑁
𝐹𝑥 = −𝑅𝑥 = 7050 𝑁
Problem 4.52
Problem
4.64
[Difficulty: 1]
4.52
Given:
Fully developed flow in pipe
Find:
Why pressure drops if momentum is constant
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 4.53
Problem
4.66
[Difficulty: 2]
4.53
Given:
Nozzle hitting stationary cart
Find:
Value of M to hold stationary; plot M versu θ
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
Hence
Rx = −M ⋅ g = V⋅ ρ⋅ ( −V⋅ A) + V⋅ cos( θ) ⋅ ( V⋅ A) = ρ⋅ V ⋅ A⋅ ( cos( θ) − 1 )
When θ = 40o
M =
2
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 4.54
Problem
4.68
[Difficulty: 2]
4.54
Given:
Water flowing past cylinder
Find:
Horizontal force on cylinder
V
y
c
x
Solution:
CS
Rx
Basic equation: Momentum flux in x direction
d
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 4.55
(Difficulty: 3)
4.55 A 6 𝑖𝑖 horizontal pipeline bends through 90° and while bending changes its diameter to 3 𝑖𝑖. The
pressure in the 6 𝑖𝑖 pipe is 30 𝑝𝑝𝑝. Calculate the magnitude and direction of the force on the bend when
2.0
𝑓𝑓 3
𝑠
of water flow therein. Both pipes are in the same horizontal plane.
Given: The pressure at inlet: 𝑝1 = 30 𝑝𝑝𝑝. The flow rate: 𝑄 = 2.0
shown in the figure.
𝑓𝑓 3
𝑠
. All the other parameters are
Find: The force on the bend.
Assumptions: Flow is steady
Density is constant
Solution:
Basic equation:
Continuity equation
Bernoulli equation
0=
Momentum equation in the x-direction
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
2
𝜌
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
Momentum equation in the y-direction
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
For steady incompressible flow we have
𝜕
� 𝑣𝑣𝑣∀ + � 𝑣𝑣𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑄 = 𝑉1 𝐴1 = 𝑉2 𝐴2
The area for the inlet section and outlet section are:
𝐴1 =
𝐴2 =
2
𝜋 2 𝜋
6
𝐷1 = × � 𝑓𝑓� = 0.1963 𝑓𝑓 2
4
4
12
2
𝜋 2 𝜋
3
𝐷2 = × � 𝑓𝑓� = 0.0491 𝑓𝑓 2
4
4
12
𝑓𝑓 3
2.0
𝑄
𝑓𝑓
𝑠
𝑉1 =
=
= 10.19
2
𝐴1 0.1963 𝑓𝑓
𝑠
𝑓𝑓 3
𝑄
𝑓𝑓
𝑠
𝑉2 =
=
= 40.7
2
𝐴2 0.0491 𝑓𝑓
𝑠
2.0
The pressure at the inlet is:
𝑝1 = 30 𝑝𝑝𝑝 = 4320
From the Bernoulli equation:
𝑙𝑙𝑙
𝑓𝑓 2
𝑝2 𝑉22
𝑝1 𝑉12
+
+ 𝑔(0) = +
+ 𝑔(0)
𝜌
2
𝜌
2
𝑠𝑠𝑠𝑠
1.94
𝜌 2
𝑓𝑓 2
𝑓𝑓 2
𝑙𝑙𝑙
𝑓𝑓 3
× ��10.19 � − �40.7 � �
𝑝2 = 𝑝1 + (𝑉1 − 𝑉22 ) = 4320 2 +
2
𝑠
𝑠
2
𝑓𝑓
𝑝2 = 4320
The mass flow rate is
𝑙𝑙𝑙
+
𝑓𝑓 2
1.94
𝑚̇ = 𝜌𝑉1 𝐴1 = 1.94
𝑙𝑙𝑙 ∙ 𝑠 2
𝑙𝑙𝑙
𝑓𝑓 2
𝑓𝑓 2
𝑓𝑓 4
× ��10.19 � − �40.7 � � = 2810 2
𝑓𝑓
𝑠
𝑠
2
𝑙𝑙𝑙 ∙ 𝑠 2
𝑓𝑓
𝑙𝑙𝑙 ∙ 𝑠
× 10.19
× 0.1963 𝑓𝑓 2 = 3.88
4
𝑠
𝑓𝑓
𝑓𝑓
From the x momentum equation, we have:
𝑅𝑥 + 𝐹𝑠𝑠 + 𝐹𝐵𝐵 = −𝑉1 𝑚̇
𝑅𝑥 + 𝑝1 𝐴1 = −𝑉1 𝑚̇
𝑅𝑥 = −𝑉1 𝑚̇ − 𝑝1 𝐴1 = −10.19
𝑓𝑓
𝑙𝑙𝑙 ∙ 𝑠
𝑙𝑙𝑙
× 3.88
− 4320 2 × 0.1963 𝑓𝑓 2 = −888 lbf
𝑠
𝑓𝑓
𝑓𝑓
𝐹𝑥 = −𝑅𝑥 = 888 𝑙𝑙𝑙
From the y momentum equation, we have:
𝑅𝑦 + 𝐹𝑠𝑠 + 𝐹𝐵𝐵 = 𝑉2 𝑚̇
𝑅𝑦 = 𝑉2 𝑚̇ + 𝑝2 𝐴2 = 40.7
So the force can be computed by:
𝑅𝑦 − 𝑝2 𝐴2 = 𝑉2 𝑚̇
𝑓𝑓
𝑙𝑙𝑙 ∙ 𝑠
𝑙𝑙𝑙
× 3.88
+ 2810 2 × 0.0491 𝑓𝑓 2 = 296 𝑙𝑙𝑙
𝑠
𝑓𝑓
𝑓𝑓
𝐹𝑦 = −𝑅𝑦 = −296 𝑙𝑙𝑙
𝐹 = �𝐹𝑥2 + 𝐹𝑦2 = �(888 𝑙𝑙𝑙)2 + (−296 𝑙𝑙𝑙)2 = 936 𝑙𝑙𝑙
The direction is calculated by (shown in the figure):
tan 𝛼 =
𝐹𝑦
= 0.3333
𝐹𝑥
𝛼 = 18.74°
𝛼 = 18.74°
𝐹𝑦
𝐹
𝐹𝑥
Problem 4.56
(Difficulty: 2)
4.56 The axes of the pipes are in a vertical plane. The flow rate is 2.83
𝑚3
𝑠
of water. Calculate the
magnitude, direction, and location of the resultant force of the water on the pipe bend.
Given: The flow rate: 𝑄 = 2.83
Find: The force on the bend.
𝑚3
𝑠
. All the other parameters are shown in the figure.
Assumptions: Flow is steady
Density is constant
Solution:
Basic equations:
Continuity equation
Bernoulli equation
0=
Momentum equation in the x-direction
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
2
𝜌
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
Momentum equation in the y-direction
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
For steady incompressible flow
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑣𝑣𝑣∀ + � 𝑣𝑣𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑄 = 𝑉1 𝐴1 = 𝑉2 𝐴2
The area for the inlet section and outlet section are:
𝐴1 =
𝐴2 =
𝜋 2 𝜋
𝐷 = × (0.9 𝑚)2 = 0.636 𝑚2
4 1 4
𝜋 2 𝜋
𝐷 = × (0.9 𝑚)2 = 0.636 𝑚2
4 2 4
𝑚3
2.83
𝑄
𝑠 = 4.45 𝑚
=
𝑉1 =
𝐴1 0.636 𝑚2
𝑠
The pressure at the inlet is:
𝑚3
2.83
𝑄
𝑠 = 4.45 𝑚
𝑉2 =
=
𝐴2 0.636 𝑚2
𝑠
𝑝1 = 34.5 𝑘𝑘𝑘
From the Bernoulli equation:
𝑝2 𝑉22
𝑝1 𝑉12
+
+ 𝑔(0) = +
+ 𝑔ℎ
𝜌
2
𝜌
2
ℎ=𝑅+
The mass flow rate is
𝐷2
= 0.6 𝑚 + 0.45 𝑚 = 1.05 𝑚
2
𝑝2 = 𝑝1 − 𝛾ℎ = 34.5 𝑘𝑘𝑘 − 9.81
𝑚̇ = 𝜌𝑉1 𝐴1 = 999
From the x momentum equation, we have:
𝑘𝑘
× 1.05𝑚 = 24.2 𝑘𝑘𝑘
𝑚3
𝑘𝑘
𝑚
𝑘𝑘
× 4.45 × 0.636 𝑚2 = 2827
3
𝑚
𝑠
𝑠
𝑅𝑥 + 𝐹𝑠𝑠 + 𝐹𝐵𝐵 = 𝑉2 𝑚̇
𝑅𝑥 = 𝑉2 𝑚̇ + 𝑝2 𝐴2 = 4.45
𝑅𝑥 − 𝑝2 𝐴2 = 𝑉2 𝑚̇
𝑚
𝑘𝑘
× 2827
+ 24.2 × 103 𝑃𝑃 × 0.636 𝑚2 = 28000 N
𝑠
𝑠
𝐹𝑥 = −𝑅𝑥 = −28000 N
From the y momentum equation, we have:
𝑅𝑦 + 𝐹𝑠𝑠 − 𝐹𝐵𝐵 = −𝑉1 𝑚̇
𝑅𝑦 + 𝑝1 𝐴1 − 𝐹𝐵𝐵 = −𝑉1 𝑚̇
𝐹𝐵𝐵 = 𝜌𝜌𝐴1
𝑅𝑦 = −𝑉1 𝑚̇ − 𝑝1 𝐴1 + 𝐹𝐵𝐵 = −4.45
2𝜋
ℎ = 10290 𝑁
4
𝑚
𝑘𝑘
× 2827
− 34.5 × 103 𝑃𝑃 × 0.636 𝑚2 + 10290 𝑁
𝑠
𝑠
𝑅𝑦 = −24200 𝑁
𝐹𝑦 = 24200 𝑁
So the force can be computed by:
𝐹 = �𝐹𝑥2 + 𝐹𝑦2 = �(−28000 𝑁)2 + (24200)2 = 37000 𝑁
The direction is calculated by (shown in the figure):
tan 𝛼 =
𝐹
𝐹𝑥
𝐹𝑦
= 0.8643
𝐹𝑥
𝛼 = 49.52°
𝐹𝑦
𝛼
Problem 4.57
(Difficulty: 2)
4.57 Water flows through a tee in a horizontal pipe system. The velocity in the stem of the tee is 15
𝑓𝑓
𝑠
,
and the diameter is 12 𝑖𝑖. Each branch is of 6 𝑖𝑖 diameter. If the pressure in the stem is 20 𝑝𝑝𝑝,
calculate magnitude and direction of the force of the water on the tee if the flow rate in the branches
are the same.
Given: The diameter: 𝐷1 = 12 𝑖𝑖. 𝐷2 = 𝐷3 = 6 𝑖𝑖 The pressure in the stem is: 𝑝1 = 20 𝑝𝑝𝑝. All the other
parameters are shown in the figure.
Find: The force on the tee.
Assumptions: Flow is steady
Density is constant
Solution:
Basic equation:
Continuity
0=
Momentum equation for the x-direction
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑣𝑣𝑣∀ + � 𝑣𝑣𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
Momentum equation for the y-direction
The volume flow rates for steady incompressible flow are related as
𝑄 = 𝑉2 𝐴2 + 𝑉3 𝐴3 = 2𝑉2 𝐴2
The area for the inlet section and outlet section are:
2
𝜋 2 𝜋
12
𝐴1 = 𝐷1 = × � 𝑓𝑓� = 0.785 𝑓𝑓 2
4
4
12
𝐴2 =
2
𝜋 2 𝜋
6
𝐷2 = × � 𝑓𝑓� = 0.196 𝑓𝑓 2
4
4
12
𝐴3 = 𝐴2 = 0.196 𝑓𝑓 2
The volumetric flow rate:
𝑓𝑓
𝑓𝑓 3
2
× 0.785 𝑓𝑓 = 11.78
𝑄 = 𝑉1 𝐴1 = 15
𝑠
𝑠
Thus
𝑓𝑓 3
11.78
𝑄
𝑓𝑓
𝑠
𝑉2 =
=
=
30
2𝐴2 2 × 0.196 𝑓𝑓 2
𝑠
𝑉3 = 30
The pressure at the inlet and outlet are:
𝑓𝑓
𝑠
𝑝1 = 20 𝑝𝑝𝑝 = 2880
𝑝2 = 0
𝑙𝑙𝑙
𝑓𝑓 2
𝑝3 = 0
The mass flow rate is
𝑚̇1 = 𝜌𝑉1 𝐴1 = 1.94
𝑙𝑙𝑙 ∙ 𝑠 2
𝑓𝑓
𝑙𝑙𝑙 ∙ 𝑠
× 15
× 0.785 𝑓𝑓 2 = 22.8
4
𝑠
𝑓𝑓
𝑓𝑓
From the x momentum equation, we have:
𝑚̇2 = 𝑚̇3 =
1
𝑚̇
2 1
𝑅𝑥 + 𝐹𝑠𝑠 + 𝐹𝐵𝐵 = −𝑉1 𝑚̇
𝑅𝑥 = −𝑉1 𝑚̇ − 𝑝1 𝐴1 = −15
𝑅𝑥 + 𝑝1 𝐴1 = −𝑉1 𝑚̇
𝑓𝑓
𝑙𝑙𝑙 ∙ 𝑠
𝑙𝑙𝑙
× 22.8
− 2880 2 × 0.785 𝑓𝑓 2 = −2602 lbf
𝑠
𝑓𝑓
𝑓𝑓
𝐹𝑥 = −𝑅𝑥 = 2602 𝑙𝑙𝑙
From the y momentum equation, we have:
𝑅𝑦 + 𝐹𝑠𝑠 + 𝐹𝐵𝐵 = 𝑉2 𝑚̇2 − 𝑉3 𝑚̇3
𝑅𝑦 = 0
So we have:
The direction is to the right.
𝐹𝑦 = 0
𝐹 = 𝐹𝑥 = 2602 𝑙𝑙𝑙
Problem 4.58*
Problem
4.70
[Difficulty: 4]
*Please Note: this solution shows the solution to the problem with one particular set of values. These values may not match those in the problem you have been given to solve.
4.58
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⋅
×
×
1000⋅ kg
π
s
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
⎜
2
s⎠
⎝
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 ∆W 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 4.59
Problem
4.72
[Difficulty: 4]
4.59
Given:
Gate held in place by water jet
Find:
Required jet speed for various water depths
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
h
2
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 ⋅
3
2
3
4
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
⎜
1⋅ m
⎝ 0.05⋅ 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 4.60
(Difficulty: 2)
4.60 Water flows steadily thorough a fire hose and nozzle. The hose is 75 𝑚𝑚 inside diameter, and the
nozzle tip is 25 𝑚𝑚 ID: water gage pressure in the hose is 510 𝑘𝑘𝑘, and the stream leaving the nozzle is
uniform. The exit speed and pressure are 32
𝑚
𝑠
and atmospheric, respectively. Find the force
transmitted by the coupling between the nozzle and hose. Indicate whether the coupling is in tension or
compression.
Given: All the parameters are shown in the figure.
Find: The force transmitted by the coupling between the nozzle and hose.
Assumption: (1) steady flow.
(2) uniform flow at each section.
(3) incompressible flow.
Solution:
(4) 𝐹𝑒𝑒 = 0
Basic equations:
Continuity
Momentum equation in x-direction
The pressure force is
0=
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝑐𝑐
𝑅𝑥 + 𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝑐𝑐
𝐹𝑠𝑠 = 𝑝1𝑔 𝐴1
For incompressible steady flow we have:
𝐹𝐵𝐵 = 0
𝑉1 = 𝑉2
0 = −𝜌1 𝐴1 𝑉1 + 𝜌2 𝐴2 𝑉2
𝑚
𝑚
𝐴2
𝐷2 2
25 𝑚𝑚 2
� = 3.56
= 𝑉2 � � = 32 × �
𝑠
𝑠
𝐴1
𝐷1
75 𝑚𝑚
𝑅𝑥 + 𝑝1𝑔 𝐴1 = 𝑢1 (−𝜌𝑉1 𝐴1 ) + 𝑢2 (𝜌𝑉2 𝐴2 ) = 𝑉1 (−𝜌𝑉1 𝐴1 ) + 𝑉2 (𝜌𝑉2 𝐴2 )
𝑅𝑥 = −𝑝1𝑔 𝐴1 − 𝑉1 (𝜌𝑉1 𝐴1 ) + 𝑉2 (𝜌𝑉2 𝐴2 ) = −𝑝1𝑔 𝐴1 − 𝑉1 (𝜌𝑉2 𝐴2 ) + 𝑉2 (𝜌𝑉2 𝐴2 )
The density of the water is:
So we get:
𝑅𝑥 = −510000
𝑅𝑥 = −𝑝1𝑔 𝐴1 + 𝜌𝑉2 𝐴2 (𝑉2 − 𝑉1 )
𝜌 = 999
𝑘𝑘
𝑚3
𝑁 𝜋
𝑘𝑘
𝑚 𝜋
𝑚
𝑚
2
2
(0.075
(0.025
�
�32
×
×
𝑚)
+
999
×
32
×
×
𝑚)
×
−
3.56
𝑚2 4
𝑚3
𝑠 4
𝑠
𝑠
𝑅𝑥 = −1.81 𝑘𝑘
The force is on CV to the left, so the coupling must be in tenstion.
Problem 4.61
(Difficulty: 3)
4.61 Two types of gasoline are blended by passing them through a horizontal “wye” as shown. Calculate
the magnitude and direction of the force exerted on the “wye” by the gasoline. The pressure 𝑝3 =
145 𝑘𝑘𝑘.
Given: The pressure 𝑝3 = 145 𝑘𝑘𝑘.All the other parameters are shown in the figure.
Find: The force on the bend.
Assumptions: Flow is steady
Density is constant
Solution:
Basic equations: Continuity
Bernoulli equation;
0=
Momentum equation for the x-direction
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
2
𝜌
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑣𝑣𝑣∀ + � 𝑣𝑣𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
Momentum equation for the y-direction
The area for the inlet section and outlet section are:
𝐴1 =
𝜋 2 𝜋
𝐷 = × (0.2 𝑚)2 = 0.0314 𝑚2
4 1 4
𝐴3 =
𝜋 2 𝜋
𝐷 = × (0.2 𝑚)2 = 0.0314 𝑚2
4 3 4
𝐴2 =
𝜋 2 𝜋
𝐷 = × (0.1 𝑚)2 = 0.0079 𝑚2
4 2 4
The velocity at each section can be calculated by:
𝐿
𝑚3
30
30 × 10−3
𝑄1
𝑠
𝑠 = 0.955 𝑚
=
=
𝑉1 =
2
2
𝑠
𝐴1 0.0314 𝑚
0.0314 𝑚
𝐿
𝑚3
3.4
3.4 × 10−3
𝑄2
𝑠 =
𝑠 = 0.430 𝑚
𝑉2 =
=
2
2
𝑠
𝐴2 0.0079 𝑚
0.0079 𝑚
𝑉3 =
𝑄3 𝑄1 + 𝑄2
=
=
𝐴3
𝐴3
30 × 10−3
The pressure at the outlet is:
𝑚3
𝑚3
+ 3.4 × 10−3
𝑠
𝑠 = 1.064 𝑚
2
𝑠
0.0314 𝑚
The density of the gas:
𝑝3 = 145 𝑘𝑘𝑘
From the Bernoulli equation:
𝜌 = 680.3
𝑘𝑘
𝑚3
𝑝1 𝑉12 𝑝2 𝑉22 𝑝3 𝑉32
+
= +
= +
𝜌
2
𝜌
2
𝜌
2
𝑘𝑘
680.3 3
𝜌 2
𝑚 2
𝑚 2
𝑚
𝑝1 = 𝑝3 + (𝑉3 − 𝑉12 ) = 145 𝑘𝑘𝑘 +
× ��1.064 � − �0.955 � � = 145.08 𝑘𝑘𝑘
2
𝑠
𝑠
2
𝑘𝑘
680.3 3
2
2
𝜌 2
𝑚 × ��1.064 𝑚� − �0.43 𝑚� � = 145.33 𝑘𝑘𝑘
𝑝2 = 𝑝3 + (𝑉3 − 𝑉22 ) = 145 𝑘𝑘𝑘 +
2
𝑠
𝑠
2
The mass flow rates are
𝑚̇1 = 𝜌𝑉1 𝐴1 = 680.3
𝑚̇2 = 𝜌𝑉2 𝐴2 = 680.3
𝑘𝑘
𝑚
𝑘𝑘
× 0.955 × 0.0314 𝑚2 = 20.40
3
𝑚
𝑠
𝑠
𝑘𝑘
𝑚
𝑘𝑘
× 0.430 × 0.0079 𝑚2 = 2.311
3
𝑚
𝑠
𝑠
𝑚̇3 = 𝜌𝑉3 𝐴3 = 680.3
From the x momentum equation, we have:
𝑘𝑘
𝑚
𝑘𝑘
× 1.064 × 0.0314 𝑚2 = 22.72
3
𝑚
𝑠
𝑠
𝑅𝑥 + 𝐹𝑠𝑠 + 𝐹𝐵𝐵 = −𝑉1 cos 30° 𝑚̇1 −𝑉2 cos 45° 𝑚̇2 + 𝑉3 𝑚̇3
𝑅𝑥 + 𝑝1 𝐴1 cos 30° + 𝑝2 𝐴2 cos 45° − 𝑝3 𝐴3 = −𝑉1 cos 30° 𝑚̇1 −𝑉2 cos 45° 𝑚̇2 + 𝑉3 𝑚̇3
𝑅𝑥 = −𝑉1 cos 30° 𝑚̇1 −𝑉2 cos 45° 𝑚̇2 + 𝑉3 𝑚̇3 + 𝑝3 𝐴3 − 𝑝1 𝐴1 cos 30° − 𝑝2 𝐴2 cos 45°
𝑅𝑥 = −197.4 𝑁
From the y momentum equation, we have:
𝐹𝑥 = 197.4 𝑁
𝑅𝑦 + 𝐹𝑠𝑠 − 𝐹𝐵𝐵 = 𝑉1 𝑚̇1 sin 30° − 𝑉2 𝑚̇2 sin 45°
𝑅𝑦 − 𝑝1 𝐴1 sin 30° + 𝑝2 𝐴2 sin 45° = 𝑉1 𝑚̇1 sin 30° − 𝑉2 𝑚̇2 sin 45°
𝑅𝑦 = 𝑉1 𝑚̇1 sin 30° − 𝑉2 𝑚̇2 sin 45° + 𝑝1 𝐴1 sin 30° − 𝑝2 𝐴2 sin 45°
𝑅𝑦 = 1475 𝑁
So the force can be computed by:
𝐹𝑦 = −1475 𝑁
𝐹 = �𝐹𝑥2 + 𝐹𝑦2 = �(197.4 𝑁)2 + (−1475 𝑁)2 = 1488 𝑁
The direction is calculated by (shown in the figure):
tan 𝛼 =
𝐹𝑦
= 7.4721
𝐹𝑥
𝛼 = 82.4°
𝐹𝑥
𝛼
𝐹𝑦
𝐹
Problem 4.62
(Difficulty: 2)
4.62 The lower tank weighs 224 𝑁, and the water in it weighs 897 𝑁. If this tank is on a platform scale,
what weight will register on the scale beam?
Given: Tank weight: 𝐹𝑡𝑡𝑡𝑡 = 224 𝑁. Water weight: 𝐹𝑤𝑤𝑤𝑤𝑤 = 897 𝑁. All the other parameters are
shown in the figure.
Find: The weight on the scale beam.
Assumptions: Flow is steady
Density is constant
Solution:
Basic equation:
Continuity
Bernoulli equation;
0=
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
2
𝜌
Momentum equation for the y-direction
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑣𝑣𝑣∀ + � 𝑣𝑣𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
For the upper surface of the lower tank, from Bernoulli equation we have:
𝑉12
− 𝑔ℎ1 = 0
2
ℎ1 = 7.8 𝑚
𝑉1 = �2𝑔ℎ1 = �2 × 9.81
For the bottom of the lower tank, we have:
𝑚
𝑚
× 7.8 𝑚 = 12.36
2
𝑠
𝑠
𝑉22
− 𝑔ℎ2 = 0
2
ℎ2 = 1.8 𝑚
𝑉2 = �2𝑔ℎ2 = �2 × 9.81
The mass flow rate of the lower tank is:
𝑚
𝑚
×
1.8
𝑚
=
5.94
𝑠2
𝑠
𝑘𝑘
𝑚 𝜋
𝑘𝑘
𝜋
𝑚̇ = 𝜌𝑉2 𝐴2 = 𝜌𝑉2 𝐷22 = 999 3 × 5.94 × × (0.075 𝑚)2 = 26.22
𝑚
𝑠 4
𝑠
4
Force on scale:
𝐹𝑦 = −𝑉1 𝑚̇ + 𝑉2 𝑚̇ = 26.22
Direction is going down.
𝑘𝑘
𝑚
𝑚
× �5.94 − 12.36 � = −168.3 𝑁
𝑠
𝑠
𝑠
Weight on the scale beam:
𝐹𝑤 = 𝐹𝑦 + 𝐹𝑡𝑡𝑡𝑡 + 𝐹𝑤𝑤𝑤𝑤𝑤 = 168.3 𝑁 + 224 𝑁 + 897 𝑁 = 1289 𝑁
Problem 4.63
(Difficulty: 2)
4.63 The pressure difference results from head loss caused by eddies downstream from the orifice plate.
Wall friction is negligible. Calculate the force exerted by the water on the orifice plate. The flow rate is
7.86
𝑓𝑓 3
𝑠
.
Given: The flow rate: 𝑄 = 7.86
𝑓𝑓 3
𝑠
. All the other parameters are shown in the figure.
Find: The force exerted by the water.
Assumptions: Flow is steady
Density is constant
Solution:
Basic equation:
Continuity
0=
Bernoulli equation;
Momentum equation for the x-direction
The cross section area is:
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
2
𝜌
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝐴1 =
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
2
𝜋 2 𝜋
12
𝐷1 = × � 𝑓𝑓� = 0.785 𝑓𝑓 2
4
4
12
𝐴2 =
The pressure at inlet and outlet are:
2
𝜋 2 𝜋
12
𝐷2 = × � 𝑓𝑓� = 0.785 𝑓𝑓 2
4
4
12
𝑝1 = 24.0 𝑝𝑝𝑝 = 3456
𝑝2 = 20.1 𝑝𝑝𝑝 = 2894
𝑙𝑙𝑙
𝑓𝑓 2
𝑙𝑙𝑙
𝑓𝑓 2
The velocity can be calculated by:
𝑓𝑓 3
7.86
𝑄
𝑠 = 10.01 𝑓𝑓
=
𝑉1 =
𝐴1 0.785 𝑓𝑓 2
𝑠
From the x momentum equation:
𝑓𝑓 3
7.86
𝑄
𝑠 = 10.01 𝑓𝑓
𝑉2 =
=
𝐴2 0.785 𝑓𝑓 2
𝑠
𝑅𝑥 + 𝑝1 𝐴1 − 𝑝2 𝐴2 = −𝑉1 𝑚̇ + 𝑉2 𝑚̇ = 0
𝑅𝑥 = 𝑝2 𝐴2 − 𝑝1 𝐴1 = 2894
The force on the orifice place is:
The direction is to the right.
𝑙𝑙𝑙
𝑙𝑙𝑙
× 0.785 𝑓𝑓 2 − 3456 2 × 0.785 𝑓𝑓 2 = −441 𝑙𝑙𝑙
2
𝑓𝑓
𝑓𝑓
𝐹𝑥 = −𝑅𝑥 = 441 𝑙𝑙𝑙
Problem 4.64
Problem
4.76
[Difficulty: 3]
4.64
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⋅ V⋅ A) =
+ ρ⋅ ( −V1 ⋅ A1 − V2 ⋅ A2 + V3 ⋅ A3 + V4 ⋅ A4 ) = 0
∑
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
3
12
+ ⋅ V3 ⋅ ρ⋅ V3 ⋅ A3 −
⋅ V ⋅ ρ⋅ V3 ⋅ A3
5
13 3
⋅ −ρ⋅ V1 ⋅ A1 −
(
Fy = −
(
5
⋅ V ⋅ −ρ⋅ V2 ⋅ A2 ...
13 2
5
4
+ ⋅ V3 ⋅ ρ⋅ V3 ⋅ A3 +
⋅ V ⋅ ρ⋅ V3 ⋅ A3
13 3
5
(
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 4.65*
Problem
4.78
[Difficulty: 2]
*Please Note: this solution shows the solution to the problem with one particular set of values. These values may not match those in the problem you have been given to solve.
4.65
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
2
⎤
⎡⎛ ft ⎞ 2
ft
1⋅ ft ⎞
lbf ⋅ s
2
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 4.66
Problem
4.79
[Difficulty: 2]
4.66
Given:
Water flow through nozzle
Find:
Force to hold nozzle
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 ⎞
V2 = V1 ⋅
= 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 4.67
(Difficulty: 2)
4.67 The pump, suction pipe, discharge pipe, and nozzle are all welded together as a single unit.
Calculate the horizontal component of force (magnitude and direction) exerted by the water on the unit
when the pump is developing a head of 22.5 m.
Given: All the parameters are shown in the figure.
Find: The horizontal component force exerted by the water.
Assumptions: Flow is steady
Density is constant
Solution:
Basic equation:
Continuity
Bernoulli equation;
0=
Momentum equation for the x-direction
The cross section area is:
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
2
𝜌
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝐴=
From the Bernoulli equation we have:
𝜋 2 𝜋
𝐷 = × (0.3 𝑚)2 = 0.071 𝑚2
4
4
𝑉2
− 𝑔ℎ = 0
2
ℎ = 22.5 𝑚 − 1.2 𝑚 − 1.8 𝑚 = 19.5 𝑚
𝑉 = �2𝑔ℎ = �2 × 9.81
The mass flow rate is:
𝑚̇ = 𝜌𝜌𝜌 = 999
From the x momentum equation:
𝑅𝑥 = 19.55
𝑚
𝑚
× 19.5 𝑚 = 19.55
2
𝑠
𝑠
𝑚
𝑘𝑘
𝑘𝑘
× 19.55 × 0.071 𝑚2 = 1387
3
𝑠
𝑠
𝑚
𝑅𝑥 = 𝑉 cos 20° 𝑚̇
𝑚
𝑘𝑘
× cos 20° × 1387
= 25500 𝑁 = 25.5 𝑘𝑘
𝑠
𝑠
The horizontal component force exerted by water is :
The direction is to the left.
𝐹𝑥 = −𝑅𝑥 = −25.5 𝑘𝑘
Problem 4.68
(Difficulty: 2)
4.68 The passage is 1.2 𝑚 wide normal to the paper. What will be the horizontal component of force
exerted by the water on the structure?
Given: The width of the passage: 𝑤 = 1.2 𝑚. All the other parameters are shown in the figure.
Find: The horizontal component force exerted by the water.
Assumptions: Flow is steady
Density is constant
Solution: Basic equations are
Continuity
Bernoulli equation;
0=
Momentum equation for the x-direction
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
2
𝜌
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
The heights of the passage inlet and outlet are:
ℎ1 = 1.5 𝑚
The area of inlet and outlet are:
ℎ2 = 0.9 𝑚
𝐴1 = 𝑤ℎ1 = 1.2 𝑚 × 1.5 𝑚 = 1.8 𝑚2
𝐴2 = 𝑤ℎ2 = 1.2 𝑚 × 0.9 𝑚 = 1.08 𝑚2
From continuity equation for steady incompressible flow:
𝐴1 𝑉1 = 𝐴2 𝑉2
From the Bernoulli equation we have:
𝑉1 =
𝐴2
𝑉 = 0.6𝑉2
𝐴1 2
𝑉2 2
𝑉1 2
+ 𝑔ℎ1 =
+ 𝑔ℎ2
2
2
𝑉2 2
0.36𝑉2 2
+ 𝑔ℎ1 =
+ 𝑔ℎ2
2
2
0.32𝑉2 2 = 𝑔(ℎ1 − ℎ2 )
𝑚
𝑔(ℎ1 − ℎ2 ) �9.81 𝑠 2 × (1.5 𝑚 − 0.9 𝑚)
𝑚
=
= 4.29
𝑉2 = �
0.32
0.32
𝑠
𝑉1 = 0.6𝑉2 = 2.57
The mass flow rate is:
𝑚̇ = 𝜌𝑉1 𝐴1 = 999
From the x momentum equation:
𝑚
𝑠
𝑘𝑘
𝑚
𝑘𝑘
× 2.57 × 1.8 𝑚2 = 4621
3
𝑚
𝑠
𝑠
𝑅𝑥 + 𝑝1 𝐴1 − 𝑝2 𝐴2 = −𝑉1 𝑚̇ + 𝑉2 𝑚̇
𝑅𝑥 = 𝑚̇(𝑉2 −𝑉1 ) + 𝑝2 𝐴2 − 𝑝1 𝐴1 = 𝑚̇(𝑉2 −𝑉1 ) +
𝛾 = 9800
𝑁
𝑚3
𝑅𝑥 = −503 𝑁
The horizontal component force exerted by water is :
The direction is to the right.
𝐹𝑥 = −𝑅𝑥 = 503 𝑁
𝛾ℎ22 𝑤 𝛾ℎ12 𝑤
−
2
2
Problem 4.69
(Difficulty: 2)
4.69 If the two-dimensional flow rate through this sluice gate is 50
vertical components of force on gate, neglecting wall friction.
Given: The flow rate:𝑞 = 50
𝑓𝑓 2
𝑠
𝑓𝑓 2
𝑠
, calculate the horizontal and
. All the other parameters are shown in the figure.
Find: The horizontal and vertical component force exerted by the water.
Assumptions: Flow is steady
Density is constant
Solution: Basic equations are:
Continuity
0=
Momentum equation for the x-direction
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
From the continuity equation for steady incompressible flow we have:
𝑄 = 𝑉1 ℎ1 = 𝑉2 ℎ2
𝑓𝑓 2
𝑄 50 𝑠
𝑓𝑓
𝑉1 =
=
= 6.25
ℎ1
𝑠
8 𝑓𝑓
𝑓𝑓 3
50
𝑄
𝑠 = 12.5 𝑓𝑓
𝑉2 =
=
𝐴2
𝑠
4 𝑓𝑓 2
The mass flow rate per width is:
𝑚̇ = 𝜌𝜌 = 1.94
𝑠𝑠𝑠𝑠
𝑓𝑓 2
𝑙𝑙𝑙 ∙ 𝑠 2
𝑓𝑓 2
𝑙𝑙𝑙 ∙ 𝑠
×
50
=
1.94
×
50
= 97
3
4
𝑓𝑓
𝑓𝑓 2
𝑠
𝑓𝑓
𝑠
From the x momentum equation per width:
𝑅𝑥 + 𝑝1 ℎ1 − 𝑝2 ℎ2 = −𝑉1 𝑚̇ + 𝑉2 𝑚̇
𝑅𝑥 = 𝑚̇(𝑉2 −𝑉1 ) + 𝑝2 ℎ2 − 𝑝1 ℎ1 = 𝑚̇(𝑉2 −𝑉1 ) +
𝑅𝑥 = 97
𝛾 = 62.4
𝑙𝑙𝑙
𝑓𝑓 3
𝛾ℎ2 ℎ2 𝛾ℎ1 ℎ1
−
2
2
𝑙𝑙𝑙
𝑙𝑙𝑙 ∙ 𝑠
𝑓𝑓
𝑓𝑓
4 𝑓𝑓 × 4 𝑓𝑓 8 𝑓𝑓 × 8𝑓𝑓
�
× �12.5
− 6.25 � + 62.4 3 × �
−
2
𝑓𝑓
𝑓𝑓
𝑠
𝑠
2
2
𝑅𝑥 = −891
Or the force on the gate is
𝑙𝑙𝑙
𝑓𝑓
𝐹𝑥 = −𝑅𝑥 = 891
𝑙𝑙𝑙
𝑓𝑓
The total force F must be normal to the gate surface. The forces on the gate are then related as:
Fx is the horizontal component force per width, and the y component of the force is then
𝐹𝑦 = 𝐹𝑥 tan 30° = 514
The width for the gate is:
So the total force can be calculated by:
𝑙𝑙𝑙
𝑓𝑓
𝑤 = 6 𝑓𝑓
𝐹𝑡𝑡 = 𝐹𝑥 𝑤 = 891
𝐹𝑡𝑡 = 𝐹𝑦 𝑤 = 514
𝑙𝑙𝑙
× 6 𝑓𝑓 = 5350 𝑙𝑙𝑙
𝑓𝑓
𝑙𝑙𝑙
× 6 𝑓𝑓 = 3080 𝑙𝑙𝑙
𝑓𝑓
Problem 4.70
(Difficulty: 2)
4.70 Assume the bend of Problem 4.35 is a segment of a larger channel and lies in a horizontal plane.
The inlet pressure is 170 𝑘𝑘𝑘 (𝑎𝑎𝑎), and the outlet pressure is 130 𝑘𝑘𝑘 (𝑎𝑎𝑎). Find the force required
to hold the bend in place.
Given: The inlet pressure: 𝑝1 = 170 𝑘𝑘𝑘. The outlet pressure: 𝑝2 = 130 𝑘𝑘𝑘.
Find: The force required to hold the bend in place.
Assumption: (1) steady flow.
(2) 𝐹𝐵𝐵 = 𝐹𝐵𝐵 = 0.
(3) incompressible flow.
(4) atmosphere pressure acts on the outside surfaces.
Solution:
Basic equations: Momentum equation in the x-direction
𝐹�𝑠 + 𝐹�𝐵 =
Momentum equation for the y-direction
𝜕
� 𝑉� 𝜌𝜌∀ + � 𝑉� (𝜌𝑉� ∙ 𝑑𝐴̅)
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
The x-momentum equation becomes
𝜕
� 𝑣𝑣𝑣∀ + � 𝑣𝑣𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑅𝑥 + 𝑝1 𝐴1 = � 𝑢 (𝜌𝑉� ∙ 𝑑𝐴̅) = 𝑈{−|𝜌𝜌𝐴1 |}
𝐶𝐶
𝑅𝑥 = −𝑝1 𝐴1 − 𝜌𝑈 2 𝐴1 = −ℎ2 (𝑝1 + 𝜌𝑈 2 )
𝑅𝑥 = −(0.0755 𝑚)2 × �(170 − 101) × 103
The y-momentum equation becomes
𝑁
𝑘𝑘
𝑚 2
𝑁∙𝑠
� ×
� = −714 𝑁
�7.5
+
999
×
2
3
𝑚
𝑚
𝑠
𝑘𝑘 ∙ 𝑚
𝑅𝑦 − 𝑝2 𝐴2 = � 𝑣 (𝜌𝑉� ∙ 𝑑𝐴̅)
𝑈2 = 𝑉2 = 𝑉𝑚𝑚𝑚 − (𝑉𝑚𝑚𝑚 − 𝑉𝑚𝑚𝑚 )
𝐶𝐶
𝑥
𝑥
𝑥
= 2𝑉𝑚𝑚𝑚 − 𝑉𝑚𝑚𝑚 = 𝑉𝑚𝑚𝑚 �2 − �
ℎ
ℎ
ℎ
ℎ
𝑥
𝑥
𝑅𝑦 − 𝑝2 𝐴2 = � 𝑉𝑚𝑚𝑚 �2 − � 𝜌𝑉𝑚𝑚𝑚 �2 − � ℎ𝑑𝑑
ℎ
ℎ
0
ℎ
𝑥 𝑥2
2
𝑅𝑦 = 𝑝2 𝐴2 + 𝜌𝑉𝑚𝑚𝑚
ℎ � �4 − 4 + 2 � 𝑑𝑑
ℎ ℎ
0
ℎ
7 2 2
2
𝑅𝑦 = 𝑝2 𝐴2 + 𝜌𝑉𝑚𝑚𝑛
ℎ �4ℎ − 2ℎ + � = 𝑝2 𝐴2 + 𝜌𝑉𝑚𝑚𝑚
ℎ
3
3
7 2
𝑁
7
𝑘𝑘
𝑚 2
𝑁∙𝑠
� = (0.0755 𝑚)2 �(130 − 110) × 103 2 + × 999 3 × �5 � ×
�
𝑅𝑦 = ℎ2 �𝑝2 + 𝜌𝑉𝑚𝑚𝑚
3
𝑚
3
𝑚
𝑠
𝑘𝑘 ∙ 𝑚
So we get:
𝑅𝑦 = 498 𝑁
𝑅� = −714𝚤̂ + 498 𝚥̂ 𝑁
Problem 4.71*
Problem
4.82
[Difficulty: 2]
*Please Note: this solution shows the solution to the problem with one particular set of values. These values may not match those in the problem you have been given to solve.
4.71
CS
c
d
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⋅
×
s
A1
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
s
⎝d⎠
2
⎛ 4 ⎞ = 1628⋅ ft
⎜
s
⎝ 1.5 ⎠
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
1 ⋅ ft ⎞
lbf ⋅ s
⎢⎜ 163 ⋅ ⎞ × π⋅ ( 1.5⋅ in) − ⎛⎜ 22.9⋅ ⎞ × π⋅ ( 4 ⋅ in) ⎥ × ⎛⎜
×
s⎠
s⎠
4
4
slug⋅ ft
⎣⎝
⎝
⎦ ⎝ 12⋅ in ⎠
2
2
⎛ 1 ⋅ ft ⎞ × lbf ⋅ s
⎜
slug⋅ ft
⎝ 12⋅ in ⎠
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 4.72
Problem
4.84
[Difficulty: 2]
4.72
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 4.73
(Difficulty: 2)
4.73 Flow from the end of a two-dimensional open channel is deflected vertically downward by the gate
AB. Calculate the force exerted by the water on the gage. At (and downstream from) B the flow may be
considered a free jet.
Given: All the parameters are shown in the figure.
Find: The force exerted by the water on the gage.
Assumptions: Flow is steady
Density is constant
Solution:
Basic equation: Continuity
Bernoulli equation:
0=
Momentum equation for the x-direction
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
2
𝜌
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
From the continuity equation for incompressible flow we have:
𝑄 = 𝑉1 ℎ1 = 𝑉2 ℎ2
ℎ1 = 1.55 𝑚
ℎ2 = 0.52 𝑚
From the Bernoulli equation:
𝑉1 = 𝑉2
ℎ2
= 0.3355 𝑉2
ℎ1
𝑔𝑔1 +
𝑉12 𝑉22
=
2
2
𝑧1 = 1.55 𝑚 + 1.13 𝑚 = 2.68 𝑚
(0.3355 𝑉2 )2 𝑉22
=
2
2
𝑔𝑔1 +
0.4437𝑉22 = 𝑔𝑔1
𝑚
9.81 2 × 2.68 𝑚
𝑚
𝑔𝑔1
𝑠
𝑉2 = �
=�
= 7.69
0.4437
𝑠
0.4437
𝑉1 = 2.58
The mass flow rate per width is:
𝑚̇ = 𝜌𝑉1 ℎ1 = 999
From the x momentum equation per width:
𝑚
𝑠
𝑘𝑘
𝑚
𝑘𝑘
× 2.58 × 1.55 𝑚 = 3995
3
𝑚
𝑠
𝑚∙𝑠
𝑅𝑥 + 𝑝1 ℎ1 = −𝑉1 𝑚̇
𝑅𝑥 +
𝜌𝜌ℎ1
ℎ1 = −𝑉1 𝑚̇
2
𝑅𝑥 = −𝑉1 𝑚̇ −
𝜌𝜌ℎ1
ℎ1
2
𝑘𝑘
𝑚
999 3 × 9.81 2 × 1.55 𝑚
𝑚
𝑘𝑘
𝑁
𝑠
𝑚
−
× 1.55 𝑚 = −22080
𝑅𝑥 = −2.58 × 3995
2
𝑠
𝑚∙𝑠
𝑚
The direction is to the right.
𝐹𝑥 = −𝑅𝑥 = 22080
𝑁
𝑚
Problem 4.74
(Difficulty: 2)
4.74 Calculate the magnitude and direction of the vertical and horizontal components and the total
𝑚
force exerted on this stationary blade by a 50 𝑚𝑚 jet of water moving at 15 .
𝑠
𝑚
Given: The diameter is: 𝐷 = 50 𝑚𝑚. The velocity is: 𝑉 = 15 . All the parameters are shown in the
𝑠
figure.
Find: The vertical and horizontal and the total force.
Assumptions: Flow is steady
Density is constant
Solution:
Basic equations: Continuity
0=
Momentum equation for the x-direction
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑣𝑣𝑣∀ + � 𝑣𝑣𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
Momentum equation for the y-direction
From the continuity equation for incompressible flow we have:
𝑄 = 𝑉𝑉
𝑄 = 𝑉𝑉 = 15
The mass flow rate is:
𝐴=
𝑚 𝜋
𝑚3
× × (0.05 𝑚)2 = 0.0295
𝑠 4
𝑠
𝑚̇ = 𝜌𝜌 = 999
From the x-momentum equation:
𝑅𝑥 = −29.47
𝜋 2
𝐷
4
𝑘𝑘
𝑚3
𝑘𝑘
×
0.0295
= 29.47
3
𝑚
𝑠
𝑠
𝑅𝑥 = −𝑉 cos 30° 𝑚̇ − 𝑉 cos 45° 𝑚̇
𝑘𝑘
𝑚
𝑘𝑘
𝑚
× 15 × cos 30° − 29.47
× 15 × cos 45° = −695 𝑁
𝑠
𝑠
𝑠
𝑠
𝐹𝑥 = 695 𝑁
From the y-momentum equation:
The total force is:
𝐹𝑦 = −𝑉 sin 30° 𝑚̇ + 𝑉 sin 45° 𝑚̇ = 91.6 𝑁
The direction is:
𝐹 = �𝐹𝑥2 + 𝐹𝑦2 = �(695 𝑁)2 + (91.6 𝑁)2 = 701 𝑁
tan 𝛼 =
𝐹𝑦 91.6 𝑁
=
= 0.1318
𝐹𝑥 695 𝑁
𝐹𝑦
𝛼 = 7.5°
𝛼
𝐹𝑥
𝐹
Problem 4.75
(Difficulty: 2)
4.75 This water jet of 50 𝑚𝑚 diameter moving at 30
𝑚
𝑠
is divided in half by a “splitter” on the
stationary flat plate. Calculate the magnitude and direction of the force on the plate. Assume that flow is
in a horizontal plane.
𝑚
Given: The diameter is: 𝐷1 = 50 𝑚𝑚. The velocity is: 𝑉1 = 30 . All the parameters are shown in the
𝑠
figure.
Find: The vertical and horizontal and the total force.
Assumptions: Flow is steady
Density is constant
Solution:
Basic equations: Continuity
0=
Momentum equation for the x-direction
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑣𝑣𝑣∀ + � 𝑣𝑣𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
Momentum equation for the y-direction
From the continuity equation for incompressible flow we have:
𝑄 = 𝑉1 𝐴1
The mass flow rate is:
𝑄 = 𝑉1 𝐴1 = 30
𝐴1 =
𝜋 2
𝐷
4 1
𝑚 𝜋
𝑚3
× × (0.05 𝑚)2 = 0.0589
𝑠 4
𝑠
𝑚̇ = 𝜌𝜌 = 999
𝑘𝑘
𝑚3
𝑘𝑘
×
0.0589
= 58.84
3
𝑚
𝑠
𝑠
From the x-momentum equation, one-half the flow out is in the positive x-direction and one-half in the
negative x-direction. The x-momentum equation becomes:
𝑅𝑥 =
The direction is to the right.
𝑚̇
𝑚̇
𝑉2 𝑐𝑜𝑠(60) − 𝑉2 𝑐𝑐𝑐(60) − 𝑉1 𝑚̇ = −𝑉1 𝑚̇
2
2
𝑅𝑥 = −30
𝑚
𝑘𝑘
× 58.84
= −1765 𝑁
𝑠
𝑠
𝐹𝑥 = 1765 𝑁
For the y-direction, the momentum flows leaving upwards and downwards are equal. So there is no net
force in the y-direction.
Problem 4.76
(Difficulty: 3)
4.76 If the splitter is removed from the plate of the preceding problem and sidewalls are provided on
the plate to keep the flow two-dimensional, how will the jet divide after striking the plate?
𝑚
Given: The diameter is: 𝐷1 = 50 𝑚𝑚. The velocity is: 𝑉1 = 30 . All the parameters are shown in the
𝑠
figure.
Find: The vertical and horizontal and the total force.
Assumptions: Flow is steady
Density is constant
Solution:
Basic equations: Continuity
Bernoulli equation
0=
Momentum equation for the x-direction
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
2
𝜌
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑣𝑣𝑣∀ + � 𝑣𝑣𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
Momentum equation for the y-direction
From the continuity equation for steady incompressible flow we have:
𝑄 = 𝑉1 𝐴1
𝑄 = 𝑉1 𝐴1 = 30
𝐴1 =
𝜋 2
𝐷
4 1
𝑚 𝜋
𝑚3
× × (0.05 𝑚)2 = 0.0589
𝑠 4
𝑠
From the Bernoulli equation, there are no pressure or elevation changes. Therefore the flow velocity
remains constant:
𝑉1 = 𝑉2 = 𝑉3
From the continuity equation for steady incompressible flow
We also have the velocity components:
𝑄 = 𝑄2 + 𝑄3
𝑚
𝑠
𝑉1𝑥 = 30
𝑉2𝑥 = 15
𝑉3𝑥 = −15
𝑉1𝑦 = 0
𝑚
𝑠
𝑚
𝑠
𝑉2𝑦 = 25.98
From x-momentum equation:
𝑚
𝑠
𝑉3𝑦 = −25.98
𝑚
𝑠
𝑚
𝑠
𝑅𝑥 = −𝑉1𝑥 𝜌𝜌 + 𝑉2𝑥 𝜌𝑄2 + 𝑉3𝑥 𝜌𝑄3
From y-momentum equation:
𝐹𝑥 = −𝑅𝑥 = 𝑉1𝑥 𝜌𝜌 − 𝑉2𝑥 𝜌𝑄2 − 𝑉3𝑥 𝜌𝑄3
𝑅𝑦 = 𝑉2𝑦 𝜌𝑄2 + 𝑉3𝑦 𝜌𝑄3
𝐹𝑦 = −𝑉2𝑦 𝜌𝑄2 − 𝑉3𝑦 𝜌𝑄3
Because the net force is normal to the plate, we also have:
𝐹𝑦 = 𝐹𝑥 tan 30°
−𝑉2𝑦 𝜌𝑄2 − 𝑉3𝑦 𝜌𝑄3 = (𝑉1𝑥 𝜌𝜌 − 𝑉2𝑥 𝜌𝑄2 − 𝑉3𝑥 𝜌𝑄3 ) tan 30°
−25.98𝑄2 + 25.98𝑄3 = 17.32𝑄 − 8.66𝑄2 + 8.66𝑄3
−2𝑄2 + 2𝑄3 = 𝑄
−𝑄2 + 𝑄3 = 0.5𝑄
𝑄2 + 𝑄3 = 𝑄
𝑄2 = 0.25𝑄 = 0.015
𝑚3
𝑠
and 𝑄3 = 0.75𝑄 = 0.044
𝑚3
𝑠
Problem 4.77
(Difficulty: 2)
4.77 Consider flow through the sudden expansion shown. If the flow is incompressible and friction is
neglected, show that the pressure rise, ∆𝑝 = 𝑝2 − 𝑝1 , is given by
𝑑 2
𝑑 2
= 2 � � �1 − � � �
1 �2
𝐷
𝐷
𝜌𝑉
2 1
∆𝑝
Plot the nondimensional pressure rise versus diameter ratio to determine the optimum value of
𝑑
𝐷
and
the corresponding value of the nondimensional pressure rise. Hint: Assume the pressure is uniform and
equal to 𝑝1 on the vertical surface of the expansion.
Given: The inlet pressure: 𝑝1 = 170 𝑘𝑘𝑘. The outlet pressure: 𝑝2 = 130 𝑘𝑘𝑘.
Find: The nondimensional pressure rise versus diameter ratio.
Assumption: (1) no friction, so surface force due to pressure only.
(2) 𝐹𝐵𝐵 = 0.
(3) steady flow.
(4) incompressible flow.
(5) uniform flow at section 1 and 2.
Solution:
(6) uniform pressure 𝑝1 on vertical surface of expansion.
Basic equations: Continuity
0=
Momentum equation for the x-direction
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
Applying the x component of momentum equation using fixed CV shown, we have:
𝑝1 𝐴2 − 𝑝2 𝐴2 = 𝑉1 (−𝜌𝑉1 𝐴1 ) + 𝑉2 (𝜌𝑉2 𝐴2 )
From the continuity for the steady incompressible uniform flow,
𝜌𝑉1 𝐴1 = 𝜌𝑉2 𝐴2
𝑉2 =
The pressure difference is then:
∆𝑝 = 𝑝2 − 𝑝1 = 𝜌𝑉1
𝐴1
𝑉
𝐴2 1
𝐴1
𝐴1
𝐴1
𝐴1
𝑉1 − 𝜌𝑉2 𝑉2 = 𝜌𝑉1 𝑉1 − 𝜌𝑉1 𝑉2 = 𝜌𝑉1 (𝑉1 − 𝑉2 )
𝐴2
𝐴2
𝐴2
𝐴2
∆𝑝 = 𝜌𝑉1 2
𝑉2
𝐴1
𝐴1
𝐴1
�1 − � = 𝜌𝑉1 2 �1 − �
𝐴2
𝑉1
𝐴2
𝐴2
Or the nondimensional pressure difference is:
∆𝑝
1
𝜌𝑉 2
2 1
From the plot below we see that 1
2
∆𝑝
=2
𝜌𝑉1 2
Note: As expected:
For 𝑑 = 𝐷, ∆𝑝 = 0 for straight pipe.
𝐴1
𝐴1
𝑑 2
𝑑 2
�1 − � = 2 � � �1 − � � �
𝐴2
𝐴2
𝐷
𝐷
has an optimum value of 0.5 at
𝑑
𝐷
= 0.7.
For
𝑑
𝐷
→ 0, ∆𝑝 = 0 for freejet.
Also note that the location of section 2 would have to be chosen with care to make assumption (5)
reasonable.
Problem 4.78
(Difficulty: 2)
4.78 A conical spray head is shown. The fluid is water and the exit stream is uniform. Evaluate (a) the
thickness of the spray sheet at 400 𝑚𝑚 radius and (b) the axial force exerted by the spray head on the
supply pipe.
Given: All the parameters are shown in the figure.
Find: The thickness of the spray sheet 𝑡 at 400 𝑚𝑚. The axial force 𝑘𝑥 .
Assumption: (1) 𝐹𝐵𝐵 = 0.
(2) steady flow.
(3) incompressible flow.
(4) uniform flow at each section.
Solution:
Basic equations: Continuity
0=
Momentum equation for the x-direction
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
From continuity for incompressible flow:
𝑉1 =
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑄
4𝑄
4
𝑚3
1
𝑚
=
=
×
0.03
×
= 0.424
2
2
(0.3
𝐴1 𝜋𝐷1 𝜋
𝑚)
𝑠
𝑠
Assume velocity in the jet sheet is constant at 𝑉 = 10
𝑚
𝑠
. Then
𝑡=
𝑄 = 2𝜋𝜋𝜋𝜋
𝑄
1
𝑚3
1
1
=
× 0.03
×
×
= 0.00119 𝑚 = 1.19 𝑚𝑚
2𝜋𝜋𝜋 2𝜋
0.4 𝑚 10 𝑚
𝑠
𝑠
From the x-momentum equation:
𝐹𝐵𝐵 = 0
𝑅𝑥 + 𝑝1𝑔 𝐴1 = 𝑢1 [−𝜌𝜌] + 𝑢2 [−𝜌𝜌]
𝑢1 = 𝑉1
The density of water is:
So we have:
𝑢2 = −𝑉 sin 𝜃
𝜌 = 999
𝑘𝑘
𝑚3
𝑅𝑥 = −𝑝1𝑔 𝐴1 − (𝑉1 + 𝑉 sin 𝜃)𝜌𝜌
𝑅𝑥 = −(150 − 101) × 103
𝑁 𝜋
𝑚
𝑚
𝑘𝑘
𝑚3
2
(0.3
�0.424
×
×
𝑚)
−
+
10
×
sin
30°�
×
999
×
0.03
𝑚2 4
𝑠
𝑠
𝑚3
𝑠
𝑅𝑥 = −3.63 𝑘𝑘
𝑅𝑥 is the force on CV, so the force on supply pipe is:
The direction is to the right.
𝑘𝑥 = −𝑅𝑥 = 3.63 𝑘𝑘
Problem
4.90
Problem 4.79
[Difficulty: 2]
4.79
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
(
A1 ⋅ V1 = A2 ⋅ V2
Hence
Rx = 1000⋅
⎝
3
m⎞
s
⎠
2
π
×
4
2
m
V2 = 2 ⋅ ×
s
(
π⋅ D1
2
+ W + ρ⋅ Vol ⋅ g +
)
4
W = 4.5⋅ kg × 9.81⋅
m
2
3 N
2
×
kg
3
m
Ry = 554 ⋅ N
×
π
4
kg⋅ m
4
×
m
V2 = 18
s
Rx = 138 ⋅ N
kg⋅ m
(
)
⋅ ⎛ V ⋅ D1 − V2 ⋅ D2 ⋅ sin( θ) ⎞
⎠
4 ⎝ 1
2
2
2
2
3
W = 44.1 N
π⋅ ( 0.075 ⋅ m)
m
+ 1000⋅
N⋅ s
2
N⋅ s
2
×
s
Ry = 125 × 10 ⋅
ρ⋅ π
⎛ 7.5 ⎞
⎜
⎝ 2.5 ⎠
2
2
× ( 0.025 ⋅ m) × cos( 30⋅ deg) ×
Ry − p 1 ⋅ A1 − W − ρ⋅ Vol ⋅ g = −V1 ⋅ −ρ⋅ V1 ⋅ A1 − V2 ⋅ sin( θ) ⋅ ρ⋅ V2 ⋅ A2
Ry = p 1⋅
Hence
⋅ cos( θ)
⎛ D1 ⎞
V2 = V1 ⋅
= V1 ⋅ ⎜
A2
⎝ D2 ⎠
× ⎛⎜ 18⋅
kg
4
2
A1
m
where
)
Rx = V2 ⋅ cos( θ) ⋅ ρ⋅ V2 ⋅ A2 = ρ⋅ V2 ⋅
From continuity
For y momentum
2 π⋅ D2
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 4.80
(Difficulty: 2)
4.80 The pump maintains a pressure of 10 𝑝𝑝𝑝 at the gage. The velocity leaving the nozzle is 34 ft/s.
Calculate the tension force in the cable.
Given: The pressure at the gage 𝑝1 = 10 𝑝𝑝𝑝. All the parameters are shown in the figure.
Find: The tension force in the cable.
Assumptions: Flow is steady and density is constant
Solution:
Basic equations: Continuity
Bernoulli equation
0=
Momentum equation for the x-direction
From the Bernoulli equation,
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
2
𝜌
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑝1 𝑉22
=
+ 𝑔ℎ2
𝜌
2
𝑉2 = �2 �
For water we have:
𝜌 = 1.94
𝑝1
− 𝑔ℎ2 �
𝜌
𝑙𝑙𝑙 ∙ 𝑠 2
𝑠𝑠𝑠𝑠
=
1.94
𝑓𝑓 4
𝑓𝑓 3
𝑔 = 32.2
𝑓𝑓
𝑠2
𝑝1 = 10 𝑝𝑝𝑝 = 1440
ℎ2 = 5 𝑓𝑓
𝑙𝑙𝑙
𝑓𝑓 2
𝑙𝑙𝑙
1440 2
𝑓𝑓
𝑓𝑓
𝑝1
𝑓𝑓
𝑉2 = �2 � − 𝑔ℎ2 � = �2 × �
− 32.2 2 × 5 𝑓𝑓� = 34.0
2
𝑙𝑙𝑙 ∙ 𝑠
𝑠
𝑠
𝜌
1.94
𝑓𝑓 4
The flow area is:
The mass flow rate is:
𝐴2 =
2
𝜋 2 𝜋
6
𝐷2 = × � 𝑓𝑓� = 0.196 𝑓𝑓 2
4
4
12
𝑚̇ = 𝜌𝑉2 𝐴2 = 1.94
𝑙𝑙𝑙 ∙ 𝑠 2
𝑓𝑓
𝑙𝑙𝑙 ∙ 𝑠
× 34.0
× 0.196 𝑓𝑓 2 = 12.93
4
𝑠
𝑓𝑓
𝑓𝑓
There is no entering x-momentum and the x-momentum equation becomes:
𝑅𝑥 = 𝑉2 𝑚̇ = 34.0
The tension in the cable is 440 lbf.
𝑓𝑓
𝑙𝑙𝑙 ∙ 𝑠
× 12.93
= 440 𝑙𝑙𝑙
𝑠
𝑓𝑓
Problem 4.81
(Difficulty: 2)
4.81 A motorboat moves up a river at a speed of 9
1.5
𝑚
𝑠
𝑚
𝑠
relative to the land. The river flows at a velocity of
. The boat is powered by a jet-propulsion unit which takes in water at the bow and discharges it
beneath the surface at the stern. Measurements in the jet show its velocity relative to the boat to be
18
𝑚
𝑠
. For a flow rate through the unit of 0.15
𝑚3
𝑠
, calculate the propulsive force produced.
Given: All the parameters are shown in the figure.
Find: The propulsive force produced.
Assumptions: Flow is steady
Density is constant
Solution:
Basic equations: Continuity
Bernoulli equation
0=
Momentum equation for the x-direction
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
2
𝜌
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
The boat velocity relative to the river flow is:
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑉𝑏𝑏𝑏𝑏−𝑟𝑟𝑟𝑟𝑟 = 9
𝑚
𝑚
𝑚
+ 1.5
= 10.5
𝑠
𝑠
𝑠
Now we can assume the river flow is constant and so the jet flow velocity relative to the river flow is:
𝑉𝑗𝑗𝑗−𝑟𝑟𝑟𝑟𝑟 = 𝑉𝑗𝑗𝑗−𝑏𝑏𝑏𝑏 − 𝑉𝑏𝑏𝑏𝑏−𝑟𝑟𝑟𝑟𝑟 = 18
𝑚
𝑚
𝑚
− 10.5
= 7.5
𝑠
𝑠
𝑠
From x momentum equation:
𝑅𝑥 = 𝜌𝑉𝑗𝑗𝑗−𝑟𝑟𝑟𝑟𝑟 𝑄 = 999
𝑘𝑘
𝑚
𝑚3
×
7.5
×
0.15
= 1124 𝑁
𝑚3
𝑠
𝑠
Problem 4.82
(Difficulty: 2)
4.82 A 30° reducing flow elbow is shown. The fluid is water. Evaluate the components of force that must
be provided by the adjacent pipes to keep the elbow from moving.
Given: All the parameters are shown in the figure.
Find: The force 𝑅𝑥 and 𝑅𝑦 must be provided by pipes to keep the elbow from moving.
Assumption: (1) steady flow.
(2) uniform flow at each section
(3) use gage pressures
(4) x horizontal
Solution:
Apply the x and y components of the momentum equation using the CS and CV shown.
Basic equations: Continuity
0=
Momentum equation for the x-direction
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑣𝑣𝑣∀ + � 𝑣𝑣𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
Momentum equation for the y-direction
The density for the water is:
x component:
𝜌 = 999
𝑘𝑘
𝑚3
𝑅𝑥 + 𝑝1𝑔 𝐴1 − 𝑝2𝑔 𝐴2 cos 𝜃 = 𝑢1 [−𝜌𝜌] + 𝑢2 [𝜌𝜌]
𝑢1 = 𝑉1
𝑢2 = 𝑉2 cos 𝜃
The velocities are calculated using the continuity relation
𝑚3
0.11
𝑄
𝑠 = 6.04 𝑚
=
𝑉1 =
𝐴1 0.0182 𝑚2
𝑠
𝑚3
0.11
𝑄
𝑠 = 13.6 𝑚
𝑉2 =
=
𝐴2 0.0081 𝑚2
𝑠
𝑅𝑥 = (−𝑉1 + 𝑉2 cos 𝜃)𝜌𝜌 − 𝑝1𝑔 𝐴1 + 𝑝2𝑔 𝐴2 cos 𝜃
𝑚
𝑚
𝑘𝑘
𝑚3
𝑁
𝑅𝑥 = �−6.04 + 13.6 cos 30°� × 999 3 × 0.11
− (200 − 101) × 1000 2 × 0.0182 𝑚2
𝑠
𝑠
𝑚
𝑚
𝑠
𝑁
+ (120 − 101) × 1000 2 × 0.0081𝑚2 × cos 30°
𝑚
y component:
𝑅𝑥 = −1040 𝑁
𝑅𝑦 + 𝑝2𝑔 𝐴2 sin 𝜃 − 𝑀𝑀 − 𝜌∀𝑔 = −𝑣1 [−𝜌𝜌] + 𝑣2 [𝜌𝜌]
𝑣1 = 0
So we have:
𝑅𝑦 = −13.6
𝑣2 = −𝑉2 sin 𝜃
𝑅𝑦 = −𝑉2 sin 𝜃 𝜌𝜌 + 𝑀𝑀 + 𝜌∀𝑔 − 𝑝2𝑔 𝐴2 sin 𝜃
𝑚
𝑘𝑘
𝑚3
𝑚
𝑘𝑘
𝑚
× 0.5 × 999 3 × 0.11
+ 10 𝑘𝑘 × 9.81 2 + 999 3 × 0.006 𝑚3 × 9.81 2
𝑠
𝑚
𝑠
𝑚
𝑠
𝑠
𝑁
− (120 − 101) × 1000 2 × 0.0081 𝑚2 × sin 30°
𝑚
𝑅𝑦 = −667 𝑁
𝑅𝑥 and 𝑅𝑦 are the horizontal and vertical components of force that must be supplied by the adjacent pipes
to keep the elbow from moving.
Problem 4.83
(Difficulty: 2)
*Please Note: this solution shows the solution to the problem with one particular set of values. These values may not match those in the problem you have been given to solve.
4.83 A monotube boiler consists of a 20 𝑓𝑓 length of tubing with 0.375 𝑖𝑖 inside the diameter. Water
enters at the rate of 0.3
𝑙𝑙𝑙
𝑠
at 500 𝑝𝑝𝑝𝑝. Steam leaves at 400 𝑝𝑝𝑝𝑝 with 0.024
magnitude and direction of the force exerted by the flowing fluid on the tube.
Given: All the parameters are shown in the figure.
Find: The magnitude and direction of the force exerted by the fluid on the tube.
Assumption: (1) 𝐹𝐵𝐵 = 0.
(2) steady flow of an incompressible fluid
(3) uniform flow at each section
Solution:
Basic equations: Continuity
0=
Momentum equation for the x-direction
From continuity we have:
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
With constant A, we get:
𝑚̇ = 𝜌1 𝑉1 𝐴1 = 𝜌2 𝑉2 𝐴2
The density of water is:
𝜌1 𝑉1 = 𝜌2 𝑉2
𝑠𝑠𝑠𝑠
𝑓𝑓 3
density. Find the
The density of steam is:
𝜌1 = 1.94
𝑠𝑠𝑠𝑠
𝑓𝑓 3
We obtain for the velocity:
𝜌2 = 0.024
𝑠𝑠𝑠𝑠
𝑓𝑓 3
𝑙𝑙𝑙
0.3 𝑠𝑠𝑠𝑠
0.3
𝑚̇
𝑓𝑓
𝑠
32.2
𝑠
=
=
= 6.26
𝑉1 =
2
𝜌1 𝐴 1.94 𝑠𝑠𝑠𝑠 × 𝜋 × (0.375 𝑖𝑖)2
𝑠
𝑠𝑠𝑠𝑠 𝜋
0.375
1.94 ×
× ×�
𝑓𝑓�
𝑓𝑓 3 4
12
𝑓𝑓 3 4
𝑠𝑠𝑠𝑠
1.94
𝜌1
𝑓𝑓
𝑓𝑓
𝑓𝑓 3
× 6.26
= 506
𝑉2 = 𝑉1 =
𝑠𝑠𝑠𝑠
𝑠
𝑠
𝜌2
0.024
𝑓𝑓 3
Apply the x component of the momentum equation, using the CV and coordinate shown. The pressures
are gage pressures with the atmospheric pressure then zero.
𝑅𝑥 + 𝑝1𝑔 𝐴1 − 𝑝2𝑔 𝐴2 = 𝑢1 [−𝑚̇] + 𝑢2 [𝑚̇] = (𝑉2 − 𝑉1 )𝑚̇
𝑅𝑥 = �𝑝2𝑔 − 𝑝1𝑔 �𝐴 + (𝑉2 − 𝑉1 )𝑚̇
𝑅𝑥 = �400 − (500 − 14.7)�
𝑙𝑙𝑙
𝑙𝑙𝑙 𝜋
𝑓𝑓
𝑓𝑓
× × (0.375 𝑖𝑖)2 + �506
− 6.26 � × 0.3
2
𝑠
𝑖𝑖
4
𝑠
𝑠
𝑅𝑥 = −9.42 𝑙𝑙𝑙 + 150
𝑙𝑙𝑙 ∙ 𝑓𝑓
150
= −9.42 𝑙𝑙𝑙 +
𝑙𝑙𝑙 = −4.76 𝑙𝑙𝑙
2
𝑠
32.17
But 𝑅𝑥 is force on CV; force on pipe is 𝐹𝑥 .
The direction is to right.
𝐹𝑥 = −𝑅𝑥 = 4.76 𝑙𝑙𝑙
Problem 4.84
Problem
4.96
[Difficulty: 3]
4.84
Given:
Data on flow out of pipe device
Find:
Velocities at 1 and 2; force on coupling
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 4.85
(Difficulty: 2)
4.85 A nozzle for a spray system is designed to produce a flat radial sheet of water. The sheet leaves the
nozzle at 𝑉2 = 10
𝑚
𝑠
, covers 180° of arc, and has thickness 𝑡 = 1.5 𝑚𝑚. The nozzle discharge radius is
𝑅 = 50 𝑚𝑚. The water supply pipe is 35 𝑚𝑚 in diameter and the inlet pressure is 𝑝1 = 150 𝑘𝑘𝑘 (abs).
Evaluate the axial force exerted by the spray nozzle on the coupling.
Given: The sheet leaves nozzle at: 𝑉2 = 10
𝑚
𝑠
. Thickness: 𝑡 = 1.5 𝑚𝑚. Radius: 𝑅 = 50 𝑚𝑚.
Pipe diameter: 𝐷1 = 35 𝑚𝑚. Inlet pressure: 𝑝1 = 150 𝑘𝑘𝑘.
Find: The axial force exerted by the spray nozzle on the coupling.
Assumption: (1) 𝐹𝐵𝐵 = 0.
(2) steady flow of an incompressible fluid
(3) uniform flow at each section
Solution:
Basic equations: Continuity
0=
Momentum equation for the x-direction
From continuity we have:
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑄 = 𝑉1 𝐴1 = 𝑉2 𝐴2 = 𝑉2 𝜋𝜋𝜋 = 10
𝑚
𝑚3
× 𝜋 × 0.05 𝑚 × 0.0015 𝑚 = 0.00236
𝑠
𝑠
𝑚3
0.00236
×4
𝑄
𝑄
𝑚
𝑠
𝑉1 =
=𝜋
=
= 2.45
2
𝐴1
𝑠
𝜋 × (0.035 𝑚)
𝐷2
4 1
Apply the x component of the momentum equation, using the CV and coordinate shown. The pressures
are gage pressures.
𝑅𝑥 + 𝑝1𝑔 𝐴1 = 𝑢1 (−𝜌𝜌) + � 𝑢2 𝜌𝑉2 𝑑𝐴2
𝐴2
𝑢1 = 𝑉1
𝑢2 = 𝑉2 cos 𝜃
𝜋
2
𝑑𝐴2 = 𝑅𝑅𝑅𝑅
� 𝑢2 𝜌𝑉2 𝑑𝐴2 = � 𝑉2 cos 𝜃𝜃𝑉2 𝑅𝑅𝑅𝑅 =
𝜋
−
2
𝐴2
The density of the water is:
𝑅𝑥 = −𝑝1𝑔 𝐴1 − 𝑉1 𝜌𝜌 + 2𝜌𝑉22 𝑅𝑅
𝜌 = 999
Thus
𝜋
2
2
2𝜌𝑉2 𝑅𝑅 � cos 𝜃𝜃𝜃
0
= 2𝜌𝑉22 𝑅𝑅
𝑘𝑘
𝑚3
𝑁 𝜋 × (0.035 𝑚)2
𝑚
𝑘𝑘
𝑚3
×
−
2.45
×
999
×
0.00236
+2
𝑚3
𝑠
𝑚3
4
𝑠
2
𝑘𝑘
𝑚
× 999 3 × �10 � × 0.05 𝑚 × 0.0015 𝑚
𝑚
𝑠
𝑅𝑥 = −(150 − 101) × 1000
𝑅𝑥 = −37.9 𝑁
But 𝑅𝑥 is force on CV. Force on coupling is 𝐹𝑥 .
The direction is to the right.
𝐹𝑥 = −𝑅𝑥 = 37.9 𝑁
Problem 4.86
Problem
4.100
[Difficulty: 2]
4.86
Given:
Data on wake behind object
Find:
An expression for the drag
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 ⎞
dr⎥
2 ⋅ ⎮ r⋅ ⎜ 1 − cos⎛⎜
⎮ ⎝
2
⎝ ⎠ ⎠ ⎥
⌡
⎛
⎜
2
F = π ρ⋅ U ⋅ ⎜ 1 −
⎜
⎝
⎞
⌠
2
4
⎮
π
⋅
r
π
⋅
r
⎞ + r⋅ cos⎛ ⎞ dr⎟
2 ⋅ ⎮ r − 2 ⋅ r⋅ cos⎛⎜
⎜
2
⎮
⎝ ⎠
⎝ 2 ⎠
⌡
F = π ρ⋅ U ⋅ ⎡1 −
⎛ 3 + 2 ⎞⎤
⎜8
2 ⎥
π ⎠⎦
⎝
2
⎢
⎣
1
0
⎦
1
0
⎠
F=
⎛ 5 ⋅ π − 2 ⎞ ⋅ ρ⋅ U2
⎜
π⎠
⎝ 8
Problem 4.87
Problem
4.102
[Difficulty: 3]
4.87
Given:
Data on flow in 2D channel
Find:
Maximum velocity; Pressure drop
y
2h
x
Solution:
c
Basic equations: Continuity, and momentum flux in x direction
d
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
⎛
r
⎜
⎝
R
u max⋅ ⎜ 1 −
⎞
2
2
⎠
⎛ R2
⋅ 2 ⋅ π⋅ r dr = 2 ⋅ π⋅ u max⋅ ⎜
⎜ 2
⎝
−
R
⎞
4
4⋅ R
2
⎠
= 2 ⋅ π⋅ u max⋅
R
2
2
R
= π⋅ u max⋅
4
2
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
ρ⋅ u max ⋅ ⎜ 1 −
⋅ 2 ⋅ π⋅ r dr = −ρ⋅ π⋅ R ⋅ U1 + 2 ⋅ π⋅ ρ⋅ u max ⋅ ⎜
− 2⋅
+
⎜
2
⎜ 2
2
4
R ⎠
4⋅ R
6⋅ R ⎠
⎝
⎝
)
2
0
1
1
2 1
2
2 1
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 4.88
(Difficulty: 2)
4.88 Consider the incompressible flow of fluid in a boundary layer as depicted in Example 4.2. Show that
the friction drag force of the fluid on the surface is given by
𝛿
𝐹𝑓 = � 𝜌𝜌(𝑈 − 𝑢) 𝑤𝑤𝑤
0
Evaluate the drag force for the conditions of Example 4.2
Given: All the parameters are shown in the figure.
Find: The drag force 𝐷𝐷𝐷𝐷.
Assumption: (1) steady flow
(2) no net pressure force
(3) 𝐹𝐵𝐵 = 0
(4) uniform flow at section AB
Solution:
Basic equations: Continuity equation
0=
Momentum equation in the x-direction
Then from continuity:
𝜕
�⃗ ∙ 𝑑𝐴⃗
� 𝜌𝜌∀ + � 𝜌𝑉
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝛿
0 = −𝜌𝜌𝜌𝜌 + � 𝜌𝜌𝜌𝜌𝜌 + 𝑚̇𝐵𝐵
0
𝛿
𝛿 = � 𝑑𝑑
0
The mass flow rate is
𝛿
𝑚̇𝐵𝐵 = 𝜌 � (𝑈 − 𝑢)𝑤𝑤𝑤
0
Apply the x component of momentumto the controlvolume. The pressure is the same on all surfaces.
𝛿
𝛿
−𝐹𝑓 = 𝑈{−|𝜌𝜌𝜌𝜌|} + �� 𝜌𝑢2 𝑤𝑤𝑤� + 𝑈𝑚̇𝐵𝐵 = 𝜌 � [−𝑈 2 + 𝑢2 + 𝑈(𝜎 − 𝑢)]𝑤𝑤𝑤
0
𝛿
0
𝐷𝐷𝐷𝐷 = 𝐹𝑓 = � 𝜌𝜌(𝑈 − 𝑢)𝑤𝑤𝑤
0
At location CD:
𝑦 2
𝑢
𝑦
= 2 � � − � � = 2𝜂 − 𝜂2
𝛿
𝑈
𝛿
𝑦
𝑑𝑑 = 𝛿𝛿 � � = 𝛿𝛿𝛿
𝛿
𝛿
1
𝑦
𝑦 2
𝑦
𝑦 2
𝐷𝐷𝐷𝐷 = � 𝜌𝜌 �2 � � − � � � �𝑈 − 𝑈 �2 � � − � � �� 𝑤𝑤𝑤 = 𝜌𝑈 2 𝑤𝑤 � (2𝜂 − 𝜂2 ) (1 − 2𝜂 + 𝜂2 )𝑑𝑑
𝛿
𝛿
𝛿
𝛿
0
0
1
1
5
1
= 𝜌𝑈 2 𝑤𝑤 � (2𝜂 − 5𝜂2 + 4𝜂3 − 𝜂4 )𝑑𝑑 = 𝜌𝑈 2 𝑤𝑤 �𝜂2 − 𝜂3 + 𝜂4 − 𝜂5 �
3
5
0
0
𝐷𝐷𝐷𝐷 =
The drag force is
𝐷𝐷𝐷𝐷 =
2
𝜌𝑈 2 𝑤𝑤
15
𝑘𝑘
𝑚 2
2
× 1.24 3 × �30 � × 0.6 𝑚 × 0.005 𝑚 = 0.446 𝑁
𝑚
𝑠
15
Problem 4.89
Problem
4.106
[Difficulty: 4]
4.89
CS
b
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
⌡
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
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 ⌠
ρ⋅ U0 ⋅ δ⋅
⋅ ⎛⎜ 1 −
dη = ρ⋅ U0 ⋅ δ⋅ ⎮ η⋅ ( 1 − η) dη
⌡
U0
U0
0
⎝
⎠
2
u
0
Ff
1
1
1
2
2
= ρ⋅ U0 ⋅ δ⋅ ⎛⎜ − ⎞ = ⋅ ρ⋅ U0 ⋅ δ
w
2
6
3
⎝
⎠
Hence
2
2
m⎞
2
N⋅ s
⎛
= × 1.225 ⋅
× ⎜ 20⋅
×
⋅m ×
w
3 ⎝
s⎠
1000
6
kg⋅ m
m
Ff
Ff
w
kg
1
= 0.163 ⋅
N
m
(using standard atmosphere density)
Problem 4.90
(Difficulty: 2)
4.90 Gases leaving the propulsion nozzle of a rocket are modeled as following radially outward from a
point upstream from the nozzle throat. Assume the speed of the exit flow, 𝑉𝑒 , has a constant magnitude.
Develop an expression for the axial thrust, 𝑇𝑎 , developed by flow leaving the nozzle exit plane. Compare
your result to the one-dimensional approximation, 𝑇 = 𝑚̇𝑉𝑒 . Evaluate the present error for 𝛼 = 15°.
Plot the percent error versus 𝛼 for 0 ≤ 𝛼 ≤ 22.5°.
Given: All the parameters are shown in the figure.
Assumptions: The flow is steady
Find: The expression for the axial thrust 𝑇𝑎 . Compare it with the one-dimensional approximation.
Solution:
Basic equations: Continuity
0=
Momentum equation for the x-direction
The mass flow rate is given by
And the thrust is given by
For spherically symmetric flow:
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑚̇ = � 𝜌𝜌𝜌𝜌
𝐴
𝑇𝑎 = � 𝑢𝑢𝑢𝑢𝑢
𝐴
The mass flow rate is [𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝜌𝑒 ≠ 𝑓(𝜃)]
𝛼
𝑚̇ = � 𝜌𝜌𝜌𝜌 = � 𝜌𝑒 𝑉𝑒 (2𝜋𝜋 sin 𝜃)𝑅𝑅𝑅 = 2𝜋𝜌𝑒 𝑉𝑒 𝑅2 (1 − cos 𝛼)
𝐴
0
The one-dimensional approximation for thrust is then
𝑇 = 𝑚̇𝑉𝑒 = 2𝜋𝜌𝑒 𝑉𝑒2 𝑅2 (1 − cos 𝛼)
The axial thrust is given by
𝛼
𝛼
𝑇𝑎 = � 𝑢𝑢𝑢𝑢𝑢 = � 𝑉𝑒 cos 𝜃 𝜌𝑒 𝑉𝑒 (2𝜋𝜋 sin 𝜃)𝑅𝑅𝑅 = 2𝜋𝜌𝑒 𝑉𝑒2 𝑅2 � sin 𝜃 cos 𝜃 𝑑𝑑
0
𝑇𝑎 = 𝜋𝑝𝑒 𝑉𝑒2 𝑅2 𝑠𝑠𝑠2 𝛼
0
The error in the one-dimensional approximation is
𝑒=
𝑇1−𝐷 − 𝑇𝑎 𝑇1−𝐷
2𝜋𝜌𝑒 𝑉𝑒2 𝑅2 (1 − cos 𝛼)
2(1 − cos 𝛼)
=
−1=
−1=
−1
2
2
2
𝑠𝑠𝑠2 𝛼
𝑇𝑎
𝑇𝑎
𝜋𝜌𝑒 𝑉𝑒 𝑅 𝑠𝑠𝑠 𝛼
The present error is plotted as a function of 𝛼.
For 𝛼 = 15°
𝑒15 =
2(1 − cos 15°)
− 1 = 1.73%
𝑠𝑠𝑠2 15°
Eq (1)
Problem 4.91
(Difficulty: 2)
4.91 Two large tanks containing water have small smoothly contoured orifices of equal area. A jet of
liquid issues from the left tank. Assume the flow is uniform and unaffected by friction. The jet impinges
on a vertical flat plate covering the opening of the right tank. Determine the minimum value for the
height, ℎ, required to keep the plate in place over the opening of the right tank.
Given: All the parameters are shown in the figure.
Find: The minimum value for the height ℎ to keep the plate.
Assumption: (1) steady flow
(2) incompressible flow
(3) No friction
Solution:
(4) 𝐹𝐵𝐵 = 0
Basic equations: Continuity
Bernoulli equation
0=
Momentum equation for the x-direction
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
2
𝜌
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
Apply Bernoulli equation along a streamsline from the water surface to jet:
𝑝 𝑉2
𝑝 𝑉𝑠2
+
+ 𝑔ℎ = +
+ 𝑔(0)
𝜌
2
2
𝜌
𝑉𝑠 ≪ 𝑉
So that:
𝑉 = �2𝑔ℎ
The pressure is related to depth using the fluid statics relation:
Applying the x-momentum equation:
𝑝3𝑔 = 𝜌𝜌𝜌
−𝑝3𝑔 𝐴 = −𝜌𝜌𝜌𝜌 = 𝑢1 [−𝜌𝜌𝜌] + 𝑢2 [𝜌𝜌𝜌]
𝑢1 = 𝑉
𝑢2 = 0
So we have:
Thus, using the Bernoulli relation:
−𝑝3𝑔 𝐴 = −𝜌𝜌𝜌𝜌 = −𝜌𝑉 2 𝐴
𝜌𝜌𝜌𝜌 = 𝜌𝑉 2 𝐴 = 𝜌2𝑔ℎ𝐴
The necessary height to keep the plate in place is
𝐻 = 2ℎ
ℎ=
𝐻
2
Problem *4.112
4.92
Problem
[Difficulty: 3]
4.92
d
CS
y
x
V, A
Rx
c
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 *4.114
4.93
Problem
[Difficulty: 3]
4.93
CS
c
Given:
Water jet striking disk
Find:
Expression for speed of jet as function of height; Height for stationary disk
d
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⋅ ⎞ −
9.81⋅ m ⎢⎝
2
s⎠
⎢⎣
1
h = 4.28 m
2
s
3
⎡
⎤
s
4
⎢2⋅ kg × 9.81⋅ m × m
⎥
×
×
2
1000⋅ kg 10⋅ m
2⎥
⎢
25
s
π⋅ ⎛⎜
⋅ m⎞ ⎥
⎢
⎣
⎝ 1000 ⎠ ⎦
2⎤
⎥
⎥
⎥
⎥⎦
Problem *4.116
4.94
Problem
[Difficulty: 3]
4.94
Given:
Stream of water striking a vane
Find:
Water speed; horizontal force on vane
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 *4.118
4.95*
Problem
[Difficulty: 3]
*Please Note: this solution shows the solution to the problem with one particular set of values. These values may not match those in the problem you have been given to solve.
4.95
Given:
Nozzle flow striking inclined plate
Find:
Mimimum gage pressure
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
ρ
2
2
⋅ V2 = p atm + ⋅ V3 − ρ⋅ g ⋅ H
2
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 *4.99
4.96
Problem *4.120
4.96
Problem
[Difficulty: 4]
Problem 4.97
(Difficulty: 2)
4.97 Incompressible fluid of negligible viscosity is pumped, at total volume flow rate 𝑄, through a porous
surface into the small gap between closely spaced parallel plates as shown. The fluid has only horizontal
motion in the gap. Assume uniform flow across any vertical section. Obtain an expression for the
pressure variation as a function of 𝑥. Hint: Apply conservation of mass and the momentum equation to a
differential control volume of thickness dx, located at position x.
Given: All the parameters are shown in the figure.
Find: Obtain the pressure variation as function of 𝑥.
Assumption: (1) steady flow
(2) incompressible flow
(3) uniform flow at each section
(4) neglect friction
Solution:
(5) 𝐹𝐵𝐵 = 0
Basic equations: Continuity
0=
Momentum equation for the x-direction
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
Then from the continuity equation
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
0 = � 𝑉� ∙ 𝑑𝐴̅ = −𝑉𝑉ℎ + �−
𝐶𝐶
𝑤ℎ𝑑𝑑 =
With 𝑉(𝑥 = 0) = 0, we have 𝑐 = 0, so that:
Applying the x- momentum equation:
𝑉=
𝑄
𝑤𝑤𝑤� + (𝑉 + 𝑑𝑑)𝑤ℎ
𝑤𝑤
𝑄
𝑑𝑑
𝐿
𝑄 𝑥
+𝑐
𝑤ℎ 𝐿
𝑉=
𝑄 𝑥
𝑤ℎ 𝐿
𝑝𝑝ℎ − (𝑝 + 𝑑𝑑)𝑤ℎ = 𝑢𝑥 (−𝜌𝜌𝜌ℎ) + 𝑢𝑑𝑑 �−𝜌
𝑢𝑥 = 𝑉
𝑄
𝑤𝑤𝑤� + 𝑢𝑥+𝑑𝑑 [𝜌(𝑉 + 𝑑𝑑)𝑤ℎ]
𝑤ℎ
The u-component of velocity at the lower surface is zero: 𝑢𝑑𝑑 = 0
𝑢𝑥+𝑑𝑑 = 𝑉 + 𝑑𝑑
From continuity equation, the velocity at the exit of the CV is:
The momentum equation becomes:
(𝑉 + 𝑑𝑑)𝑤ℎ = 𝑉𝑉ℎ + 𝑄
𝑑𝑑
𝐿
−𝑑𝑑𝑑ℎ = −𝜌𝑉 2 𝑤ℎ + 0 + (𝑉 + 𝑑𝑑) �𝑉𝑉ℎ + 𝑄
−𝑑𝑑𝑑ℎ = 𝜌𝜌𝜌ℎ𝑑𝑑 + 𝑉𝑉𝑉
𝑑𝑑
�𝜌
𝐿
𝑑𝑑
𝑑𝑑
+ 𝜌𝜌𝜌𝜌
𝐿
𝐿
The products of differentials are neglected (𝑖𝑖. 𝑑𝑑𝑑𝑑 ≪ 𝑑𝑑), and with the expression for dV
The momentum equation becomes
−𝑑𝑑 = 𝜌𝜌𝜌𝜌 +
For 𝑝(0) = 𝑝0 , then
𝑑𝑑 =
𝑄 𝑑𝑑
𝑤ℎ 𝐿
𝑉𝑉𝑉 𝑑𝑑
𝑄 𝑑𝑑 𝑉𝑉𝑉 𝑑𝑑
𝑄 𝑥 𝑄 𝑑𝑑
= 𝜌𝜌
+
= 2𝜌
𝑤ℎ 𝐿
𝑤ℎ 𝐿
𝑤ℎ 𝐿
𝑤ℎ 𝐿 𝑤ℎ 𝐿
−𝑑𝑑 = 2𝜌 �
𝑝(𝑥) = −𝜌 �
𝑄 2
� 𝑥𝑥𝑥
𝑤ℎ𝐿
𝑄 2 2
� 𝑥 +𝑐
𝑤ℎ𝐿
𝑝(𝑥) = 𝑝0 − 𝜌 �
𝑄 2 𝑥 2
� � �
𝐿
𝑤ℎ
Problem
*4.124
Problem 4.98
[Difficulty: 5]
4.98
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 4.99
(Difficulty: 2)
4.99 Design a clepsydra(Egyptian water clock)-a vessel from which water drains by gravity through a
hole in the bottom and which indicates time by the level of the remaining water. Specify the dimensions
of the vessel and the size of the drain hole; indicate the amount of water needed to fill the vessel and
the interval at which it must be filled. Plot the vessel radius as a function of elevation.
Discussion: The original Egyptian water clock was an open water-filled vessel with an orifice in the
bottom. The vessel shape was designed so that the water level dropped at a constant rate during use.
Water leaves the orifice at higher speed when the water level within the vessel is high, and at lower
speed when the water level within the vessel is low. The size of the orifice is constant. Thus the
instantaneous volume flow rate depends on the water level in the vessel.
The rate at which the water level falls in the vessel depends on the volume flow rate and the area of the
water surface. The surface area at each water level must be chosen so that the water level within the
vessel decreases at a constant rate. The continuity and Bernoulli equations can be applied to determine
the required vessel shape so that the water surface level drops at a constant rate.
Use the CV and notation shown.
Find: The vessel radius as a function of elevation.
Assumption: (1) quasi-steady flow
(2) incompressible flow
(3) uniform flow at each cross-section
(4) flow along a streamline
(5) No friction
Solution:
(6) 𝜌𝑎𝑎𝑎 ≪ 𝜌𝐻2 𝑜
Basic equations: Continuity
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
0=
Bernoulli equation
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
2
𝜌
writing Bernoulli equation from the liquid surface to the jet exit.
For ∆≪ 𝑉, then 𝑉 = �2𝑔ℎ.
𝑝𝑎𝑎𝑎 𝑉 2
𝑝𝑎𝑎𝑎 ∆2
+
+ 𝑔ℎ =
+
+ 𝑔(0)
𝜌
2
𝜌
2
For the CV,
0=
𝜕
𝜕
� 𝜌𝑎𝑎𝑎 𝑑∀ + � 𝜌𝐻2 𝑜 𝑑∀ + {−|𝜌𝑎𝑎𝑎 𝑉1 𝐴1 |} + �𝜌𝐻2 𝑜 𝑉𝑉�
𝜕𝜕 ∀𝑎𝑎𝑎
𝜕𝜕 ∀𝐻2 𝑜
With 𝜌𝑎𝑎𝑎 ≪ 𝜌𝐻2 𝑜 we have:
But ℎ decreases, so
𝑑ℎ
𝑑𝑑
0=𝜌
= −∆. Thus
𝑑∀
𝑑ℎ
+ 𝜌𝜌𝜌 = 𝜌𝜌𝑟 2
+ 𝜌�2𝑔ℎ𝐴
𝑑𝑑
𝑑𝑑
𝜋Ω2 ∆= �2𝑔ℎ𝐴
For n hours operation, 𝐻 = 𝑛∆, and
𝐴 1
ℎ4
𝜋∆
Ω = 4�2𝑔�
𝐻
∀= � 𝜋Ω2 𝑑ℎ = �
0
𝑛∆
0
�2𝑔ℎ
3
2𝐴�2𝑔𝑛2 1
∀=
∆2
3
𝐴
𝑑ℎ
∆
Evaluating and plotting:
Problem 4.100
Problem
4.128
[Difficulty: 3]
4.100
d
CS (moves
at speed U)
c
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⋅
× ⎡⎢( 30 − 10) ⋅
kg
⎣
3
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
4.130
Problem 4.101
[Difficulty: 3]
4.101
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
⎜
k
⎝ 2⋅ k ⎠
+
π
2
Q = Vj⋅ ⋅ D
4
α=
ρ⋅ Q
(1)
3
Q = 0.0663
m
s
2
2
2
2
For
k =
ρ⋅ Q
⎛ m⎞
⎜s
⎝ ⎠
m
Vj = 25⋅
s
α =
or
(
V
)
2 ⋅ Vj − V
α = 10
m
s
N
k = 3.31
2⋅ α
α + 2 ⋅ α⋅ Vj
2
( V + α) = V + 2 ⋅ α⋅ V + α = α + 2 ⋅ α⋅ Vj
Hence
2
V = −α +
then
2⋅ k
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
V = −
+ ⎜
+
k ⎦
s
⎣ 2⋅ k
⎝ 2⋅ k ⎠
Problem 4.102
(Difficulty: 2)
4.102 The Canadair CL-215T amphibious aircraft is specially designed to fight fires. It is the only
production aircraft that can scoop water-1620 gallons in 12 seconds-from any lake, river, or ocean.
Determine the added thrust required during water scooping, as a function of aircraft speed, for a
reasonable range of speeds.
Find: The added thrust required during water scooping.
Assumption: (1) horizontal motion, so 𝐹𝐵 = 0.
(2) neglect 𝑢𝑥𝑥𝑥 within the CV.
(3) uniform flow at inlet cross-section
(4) neglect hydrostatic pressure
Solution:
Use a CV that moves with the aircraft, as shown.
Basic equation: Momentum equation in the x-direction
Then
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑢 𝜌𝜌∀ + � 𝑢𝑥𝑥𝑥 𝜌𝑉�𝑥𝑥𝑥 ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶 𝑥𝑥𝑥
𝐶𝐶
𝑅𝑥 = 𝑢1 {−|𝜌𝜌|} = −𝑈(−𝜌𝜌) = 𝑈𝑈𝑈
𝑢1 = −𝑈
From data given:
The density for water is:
𝑄=
𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑓𝑓 3
∆∀ 1620 𝑔𝑔𝑔𝑔𝑔𝑔𝑔
=
= 135
= 18.0
12 𝑠𝑠𝑠
𝑠𝑠𝑠
∆𝑡
𝑠
For an aircraft speed of 𝑈 = 75 𝑚𝑚ℎ �110
𝜌 = 1.94
𝑓𝑓
𝑠
�
𝑠𝑠𝑠𝑠
𝑓𝑓 3
𝑙𝑙𝑙 ∙ 𝑠 2
𝑓𝑓
𝑠𝑠𝑠𝑠
𝑓𝑓
𝑓𝑓
𝑓𝑓 3
𝑓𝑓
𝑅𝑥 = 110
× 1.94
×
18.0
=
110
×
1.94
×
18.0
= 3840 𝑙𝑙𝑙
𝑠
𝑓𝑓 3
𝑠
𝑠
𝑓𝑓 3
𝑠
3
For a range of aircraft speeds:
Thus at 60 mph the added thrust is 3070 lbf, while at 125 mph the added thrust is 6400 lbf.
Problem 4.103*
Problem
4.134
[Difficulty: 3]
*Please Note: this solution shows the solution to the problem with one particular set of values. These values may not match those in the problem you have been given to solve.
4.103
CS (moves to
left at speed Vc) d
Vj + Vc
Vj + Vc
c
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
Rx = 1.94⋅
slug
ft
3
× ⎡⎢( 100 + 45) ⋅
⎣
ft⎤
2
⎥ ×
s⎦
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 4.104
(Difficulty: 2)
4.104 Consider a series of turning vanes struck by a continuous jet of water that leaves a 50-mm
diameter nozzle at constant speed, 𝑉 = 86.6
𝑚
𝑠
. The vanes move with constant speed, 𝑈 = 50
𝑚
𝑠
. Note
that all the mass flow leaving the jet crosses the vanes. The curvature of the vanes is described by angles
𝜃1 = 30° and 𝜃2 = 45°, as shown. Evaluate the nozzle angle, 𝛼, required to ensure that the jet enters
tangent to the leading edge of each vane. Calculate the force that must be applied to maintain the vane
speed constant.
Find: The force must be applied to maintain the vane speed constant.
Assumption: (1) no pressure forces
(2) horizontal 𝐹𝐵𝐵 = 0
(3) steady flow
(4) uniform flow at each section
(5) no change in relative velocity on vane
(6) flow enters and leaves tangent to vanes.
Solution: The nozzle angle may be obtained from trigonometry. inlet velocity relationship is shown in
the sketch. Apply momentum equation using CV moving with vanes, as shown.
Basic equation: Momentum equation in the x-direction
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑢 𝜌𝜌∀ + � 𝑢𝑥𝑥𝑥 𝜌𝑉�𝑥𝑥𝑥 ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶 𝑥𝑥𝑥
𝐶𝐶
From the relation between the angles for the velocity vectors as shown on the sketch:
sin 𝛼 sin(90 + 𝜃1 ) 𝑠𝑠𝑠(𝛽)
=
=
𝑈
𝑉
𝑉𝑟𝑟
Where
𝑈
𝛽 = 𝑠𝑠𝑛−1 � sin(90 + 𝜃1 )� = 𝑠𝑠𝑠−1 �
𝑉
50
86.6
sin(120°)� = 30°
From the sketch, 90° = 𝛼 + 𝛽 + 𝜃1 , so 𝛼 = 90° − 𝛽 − 𝜃1 = 90° − 30° − 30° = 30°
Also
𝑉𝑟𝑟 = 𝑉
𝑉𝑟𝑟 cos 𝜃1 = 𝑉 sin 𝛼
sin 𝛼
𝑚 sin 30°
𝑚
= 86.6 ×
= 50.0
cos 𝜃1
𝑠 cos 30°
𝑠
From momentum equation, as all of 𝑚̇ flows across vanes
𝑅𝑥 = 𝑢1 {−𝑚̇} + 𝑢2 {𝑚̇} = 𝑉𝑟𝑟 sin 𝜃1 (−𝑚̇) − 𝑉𝑟𝑟 sin 𝜃2 (𝑚̇) = 𝑉𝑟𝑟 𝑚̇(− sin 𝜃1 − sin 𝜃2 )
The velocities are given by.
𝑢1 = 𝑉𝑟𝑟 sin 𝜃1
𝑢2 = −𝑉𝑟𝑟 sin 𝜃2
Thus, since 𝑚̇ = 𝜌𝜌,
𝑅𝑦 = 𝑚̇𝑉𝑟𝑟 (− cos 𝜃1 + cos 𝜃2 )
𝑅𝑥 = 𝑉𝑟𝑟 𝜌𝜌(− sin 𝜃1 − sin 𝜃2 ) = 50
The net force on the CV in the x-direction is
𝑘𝑘
𝑚3
𝑚
𝑁∙𝑠
(− sin 30° − sin 45°)
× 999 3 × 0.170
𝑚
𝑠
𝑠
𝑘𝑘 ∙ 𝑚
𝑅𝑥 = −10.3 𝑘𝑘 (to left)
And the net force on the CV in the y-direction
𝑅𝑦 = −1.35 𝑘𝑘.
Problem 4.105
(Difficulty: 2)
4.105 A steady jet of water is used to propel as a small cart along a horizontal track as shown. Total
resistance to motion of the cart assembly is given by 𝐹𝐷 = 𝑘𝑈 2 , where 𝑘 = 0.92
acceleration of the cart at the instant when its speed is 𝑈 = 10
𝑚
𝑠
Find: The acceleration of the cart when the instant speed 𝑈 = 10
Assumption: (1) Only resistance is 𝐹𝐷
.
𝑚
𝑠
𝑁∙𝑠 2
𝑚2
. Evaluate the
.
(2) horizontal 𝐹𝐵𝐵 = 0
(3) neglect
𝜕𝜕
𝜕𝜕
of mass of water in CV
(4) no change in speed with respect to vane.
(5) uniform flow at each cross-section
Solution: Apply the momentum equation using control volume and control surface shown.
Basic equation: Momentum equation in x-direction
Then
So
𝐹𝑠𝑠 + 𝐹𝐵𝐵 − � 𝑎𝑟𝑟𝑟 𝜌𝜌∀ =
𝐶𝐶
𝜕
� 𝑢 𝜌𝜌∀ + � 𝑢𝑥𝑥𝑥 𝜌𝑉�𝑥𝑥𝑥 ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶 𝑥𝑥𝑥
𝐶𝐶
−𝑘𝑈 2 − 𝑎𝑟𝑟𝑟 𝑀𝐶𝐶 = −𝜌(𝑉 − 𝑈)2 𝐴 − 𝜌(𝑉 − 𝑈)2 𝐴 sin 𝜃 = −𝜌(𝑉 − 𝑈)2 𝐴(1 + sin 𝜃)
𝑎𝑟𝑟𝑟 =
1
[𝜌(𝑉 − 𝑈)2 𝐴(1 + sin 𝜃) − 𝑘𝑈2 ]
𝑀
𝑎𝑟𝑟𝑟
1
𝑘𝑘
𝑚2 𝜋
𝑁 ∙ 𝑠2
2
2
=
× × (0.025 𝑚) × (1 + sin 30°) − 0.92
�999 3 × (30 − 10)
15 𝑘𝑘
𝑚
𝑠2 4
𝑚2
𝑚 2 𝑘𝑘 ∙ 𝑚
× �10 � ×
�
𝑠
𝑁 ∙ 𝑠2
The direction is to the right.
𝑎𝑟𝑟𝑟 = 13.5
𝑚
𝑠2
Problem 4.106
(Difficulty: 2)
4.106 The hydraulic catapult of Problem 4.105 is accelerated by a jet of water that strikes the curved
vane. The cart moves along a level track with negligible resistance. At any time its speed is 𝑈. Calculate
𝑉
the time required to accelerate the cart from rest to 𝑈 = .
2
𝑉
Find: The time 𝑡 required to accelerate the cart from rest to 𝑈 = .
2
Assumption: (1) 𝐹𝑠𝑠 = 0, since no pressure forces, no resistance.
(2) 𝐹𝐵𝐵 = 0, since horizontal
(3) neglect mass of water inside control volume
(4) uniform flow in jet
(5) no change in relative velocity on vane
Solution:
Apply x component of momentum equation to accelerating CV.
Basic equation: Momentum equation in x-direction
Then
𝐹𝑠𝑠 + 𝐹𝐵𝐵 − � 𝑎𝑟𝑟𝑟 𝜌𝜌∀ =
𝐶𝐶
𝜕
� 𝑢 𝜌𝜌∀ + � 𝑢𝑥𝑥𝑥 𝜌𝑉�𝑥𝑥𝑥 ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶 𝑥𝑥𝑥
𝐶𝐶
−𝑎𝑟𝑟𝑟 𝑀𝐶𝐶 = 𝑢1 {−𝜌(𝑉 − 𝑈)𝐴} + 𝑢2 {𝜌(𝑉 − 𝑈)𝐴}
𝑢1 = 𝑉 − 𝑈
So we have:
𝑢2 = −(𝑉 − 𝑈) sin 𝜃
−𝑎𝑟𝑟𝑟 𝑀𝐶𝐶 = −(1 + sin 𝜃)𝜌(𝑉 − 𝑈)2 𝐴
𝑑𝑑 𝜌𝜌(1 + sin 𝜃)
(𝑉 − 𝑈)2
=
𝑀
𝑑𝑑
To integrate, since 𝑉 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 we can replace 𝑑𝑑 = −𝑑(𝑉 − 𝑈). Separating variables
Or
𝑑(𝑉 − 𝑈) 𝜌𝜌(1 + sin 𝜃)
𝑑𝑑
=−
=
𝑑𝑑
(𝑉 − 𝑈)2
(𝑉 − 𝑈)2
𝑀
−�
𝑉
2 𝑑(𝑉
0
𝑡
− 𝑈)
𝜌𝜌(1 + sin 𝜃)
�
=
𝑡
2
(𝑉 − 𝑈)
𝑀
0
Or, integrating and evaluating the integral at the limits
𝑉
Thus the time is
1
1 1 𝜌𝜌(1 + sin 𝜃)
1 𝑈= 2
1 𝑈=0
�
�
�
−�
=
− = =
𝑡
𝑉 𝑉 𝑉
𝑀
𝑉−𝑈
𝑉−𝑈
𝑉−
2
𝑡=
𝑡 = 15.0 𝑘𝑘 ×
𝑀
𝜌𝜌𝜌(1 + sin 𝜃)
𝑠
4
1
𝑚3
×
×
×
2
(1
+ sin 30°)
999 𝑘𝑘 30 𝑚 𝜋(0.025 𝑚)
𝑡 = 0.680 𝑠
Problem 4.107
(Difficulty: 2)
4.107 A vane/slider assembly moves under the influence of a liquid jet as shown. The coefficient of
kinetic friction for motion of the slider along the surface is 𝜇𝑘 = 0.30. Calculate the terminal speed of
the slider.
Find: The terminal speed of the slider 𝑈𝑡 .
Assumption: (1) horizontal motion, so 𝐹𝐵𝐵 = 0
(2) neglect mass of liquid on vane, 𝑢 ≈ 0 on vane
(3) uniform flow at each section
(4) measure velocities relative to CV
Solution:
Apply x momentum equation to linearly accelerating CV.
Basic equation: Momentum equation in x-direction
Then
𝐹𝑠𝑠 + 𝐹𝐵𝐵 − � 𝑎𝑟𝑟𝑟 𝜌𝜌∀ =
𝐶𝐶
𝜕
� 𝑢 𝜌𝜌∀ + � 𝑢𝑥𝑥𝑥 𝜌𝑉�𝑥𝑥𝑥 ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶 𝑥𝑥𝑥
𝐶𝐶
−𝑀𝑀𝜇𝑘 − 𝑎𝑟𝑟𝑟 𝑀 = 𝑢1 {−|𝜌(𝑉 − 𝑈)𝐴|} + 𝑢2 {𝑚̇ 2 } + 𝑢3 {𝑚̇ 3 }
𝑢1 = 𝑉 − 𝑈
𝑢2 = 0
or
−𝑀𝑀𝜇𝑘 − 𝑀
𝑢3 = 0
𝑑𝑑
= −𝜌(𝑉 − 𝑈)2 𝐴
𝑑𝑑
At terminal speed,
𝑑𝑑 𝜌(𝑉 − 𝑈)2 𝐴
=
− 𝑔𝜇𝑘
𝑑𝑑
𝑀
𝑑𝑑
=0
𝑑𝑑
So we have:
𝑈 = 𝑈𝑡
𝜌(𝑉 − 𝑈𝑡 )2 𝐴
− 𝑔𝜇𝑘 = 0
𝑀
𝑀𝑔𝜇𝑘
𝜌𝜌
(𝑉 − 𝑈𝑡 ) = �
0.5
𝑀𝑔𝜇𝑘
𝑚
𝑚
𝑚3
1
�
𝑈𝑡 = 𝑉 − �
= 20 − �30 𝑘𝑘 × 9.81 2 × 0.3 ×
×
𝜌𝜌
𝑠
999 𝑘𝑘 0.005 𝑚2
𝑠
= 15.8
𝑚
𝑠
Problem 4.108
(Difficulty: 2)
4.108 A cart is propelled by a liquid jet issuing horizontally from a tank as shown. The track is horizontal;
resistance to motion may be neglected. The tank is pressurized so that the jet speed may be considered
constant. Obtain a general expression for the speed of the cart as it accelerates from rest. If 𝑀0 =
100 𝑘𝑘, 𝜌 = 999
1.5
𝑚
𝑠
𝑘𝑘
𝑚3
, and 𝐴 = 0.005 𝑚2 , find the jet speed 𝑉 required for the cart to reach a speed of
after 30 seconds. For this condition, plot the cart speed 𝑈 as a function of time. Plot the cart
speed after 30 seconds as a function of jet speed.
Find: Plot the cart speed after 30 seconds as a function of jet speed.
Assumption: (1) no resistance.
(2) 𝐹𝐵𝐵 = 0 since track is horizontal
(3) neglect change in fluid velocities within CV
(4) uniform flow at jet exit
Solution:
a) Apply x component of momentum equation using linearly accelerating CV shown.
Basic equation: Momentum equation in x-direction
Then
𝐹𝑠𝑠 + 𝐹𝐵𝐵 − � 𝑎𝑟𝑟𝑟 𝜌𝜌∀ =
𝐶𝐶
𝜕
� 𝑢 𝜌𝜌∀ + � 𝑢𝑥𝑥𝑥 𝜌𝑉�𝑥𝑥𝑥 ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶 𝑥𝑥𝑥
𝐶𝐶
−𝑎𝑟𝑟𝑟 𝑀 = 𝑢{|𝜌𝜌𝜌|} = −𝜌𝑉 2 𝐴
From continuity, the flow rate is given by
𝑢 = −𝑉
𝑚̇ = 𝜌𝜌𝜌
So the momentum flow is
Using 𝑎𝑟𝑟𝑟 =
𝑑𝑑
𝑑𝑑
𝑀 = 𝑀0 − 𝜌𝜌𝜌𝜌
,
Separating the variables and integrating,
𝑈
𝜌𝑉 2 𝐴
𝑑𝑑
=
𝑑𝑑 𝑀0 − 𝜌𝜌𝜌𝜌
� 𝑑𝑑 = 𝑈 = �
0
or
Check dimensions:
𝑡
0
𝜌𝑉 2 𝐴
𝑀0
�
= 𝑉 ln �
𝑀0 − 𝜌𝜌𝜌𝜌
𝑀0 − 𝜌𝜌𝜌𝜌
𝑈
𝑀0
�
= ln �
𝑀0 − 𝜌𝜌𝜌𝜌
𝑉
[𝜌𝜌𝜌𝜌] =
𝑀𝐿 2
𝐿 𝑡=𝑀
𝐿3 𝑡
b) Using the given data in Excel (with solver) the jet speed for 𝑈 = 1.5
𝑚
𝑉 = 0.61
𝑠
𝑚
𝑠
at 𝑡 = 30 𝑠 is
Problem 4.109
Problem
4.143
4.109
[Difficulty: 4]
4.107
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
U= V−
so
and finally the identity
g ⋅ μk ⋅ M
ρ⋅ A
⎛ g⋅ μk ⋅ ρ⋅ A
⋅ tanh⎜
⎜
⎜
⎝
M
⋅ t + atanh⎛
⎞ − π ⋅ i⎞
2 ⎟
1
⎜
ρ⋅ A
⋅V
⎜
g ⋅ μk ⋅ M
⎝
⎠
π
1
tanh⎛⎜ x − ⋅ i⎞ =
2
tanh
( x)
⎝
⎠
⎠
g ⋅ μk ⋅ M
to obtain
ρ⋅ A
U( t) = V −
⎛ g⋅ μk ⋅ ρ⋅ A
⎛ g⋅ μk ⋅ M 1 ⎞ ⎞
⋅ 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 4.110
(Difficulty: 2)
4.110 If the cart of Problem 4.105 is released at 𝑡 = 0, when would you expect the acceleration to be
maximum? Sketch what you would expect for the curve of acceleration versus time. What value of 𝜃
would maximize the acceleration at any time? Why? Will the cart speed ever equal the jet speed?
Explain briefly.
Find: The value of the angle 𝜃 that would maximize the acceleration.
Assumption: (1) 𝐹𝑠𝑠 = −𝐹𝐷 = −𝑘𝑈 2 , where 𝑘 = 0.92
(2) 𝐹𝐵𝐵 = 0, since horizontal
𝑁∙𝑠 2
𝑚2
.
(3) neglect mass of water on vane
(4) uniform flow in jet
(5) no change in relative velocity on vane
Solution:
Apply x component of momentum equation to accelerating CV.
Basic equation: Momentum equation in x-direction
Then
𝐹𝑠𝑠 + 𝐹𝐵𝐵 − � 𝑎𝑟𝑟𝑟 𝜌𝜌∀ =
𝐶𝐶
𝜕
� 𝑢 𝜌𝜌∀ + � 𝑢𝑥𝑥𝑥 𝜌𝑉�𝑥𝑥𝑥 ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶 𝑥𝑥𝑥
𝐶𝐶
−𝑘𝑈 2 − 𝑎𝑟𝑟𝑟 𝑀𝐶𝐶 = 𝑢1 {−𝜌(𝑉 − 𝑈)𝐴} + 𝑢2 {𝜌(𝑉 − 𝑈)𝐴}
𝑢1 = 𝑉 − 𝑈
𝑢2 = −(𝑉 − 𝑈) sin 𝜃
−𝑘𝑈 2 − 𝑎𝑟𝑟𝑟 𝑀𝐶𝐶 = −(1 + sin 𝜃)𝜌(𝑉 − 𝑈)2 𝐴
So we have for the acceleration of the cart
𝑘𝑈 2
𝑑𝑑 𝜌𝜌(1 + sin 𝜃)
(𝑉 − 𝑈)2 −
=
𝑀
𝑀
𝑑𝑑
(a) Acceleration is maximum at 𝑡 = 0, when 𝑈 = 0.
(b) Acceleration versus time will be
(c) From the equation for acceleration,
(d)
𝑑𝑑
𝑑𝑑
𝑑𝑑
is maximum when sin 𝜃 = 1, which is 𝜃 =
𝜋
2
= 90°
will go to zero when 𝑈 = 𝑉. This will be the terminal speed for the cart, 𝑈𝑡 . From the equation
for acceleration
or
𝑑𝑑
𝑑𝑑
𝑑𝑑
= 0 when
𝜌𝜌(1 + sin 𝜃)(𝑉 − 𝑈)2 = 𝑘𝑈 2
𝑈=
𝑈 will be asymptotic to 𝑉.
1
𝜌𝜌(1 + sin 𝜃) 2
�
�
𝑘
1+�
1𝑉
2
𝜌𝜌(1 + sin 𝜃)
�
𝑘
= 0.472𝑉
Problem 4.111
Problem
4.146
[Difficulty: 3]
4.111
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 4.112
(Difficulty: 2)
4.112 A rocket sled is to be slowed from an initial speed of 300
𝑚
𝑠
by lowering a scoop into a water
trough. The scoop is 0.3 𝑚 wide; it deflects the water trough 150° . The trough is 800 𝑚 long. The mass
of the sled is 8000 𝑘𝑘. At the initial speed it experiences an aerodynamic drag force of 90 𝑘𝑘. The
aerodynamic force is proportional to the square of the sled speed. It is desired to show the sled to
100
𝑚
𝑠
. Determine the depth 𝐷 to which the scoop must be lowered into the water.
Find: The depth 𝐷 to which the scoop must be lowered into the water.
Assumption: (1) 𝐹𝐵𝐵 = 0
(2) Neglect rate of change of u in CV
(3) uniform flow at each section
(4) no change in relative speed of liquid crossing scoop
Solution:
Apply x component of momentum equation using linearly acceleration CV shown.
Basic equation: Momentum equation for the x-direction
Then
𝐹𝑠𝑠 + 𝐹𝐵𝐵 − � 𝑎𝑟𝑟𝑟 𝜌𝜌∀ =
𝐶𝐶
𝜕
� 𝑢 𝜌𝜌∀ + � 𝑢𝑥𝑥𝑥 𝜌𝑉�𝑥𝑥𝑥 ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶 𝑥𝑥𝑥
𝐶𝐶
−𝐹𝐷 − 𝑀𝑎𝑟𝑟𝑟 = 𝑢1 {−|𝜌𝜌𝜌ℎ|} + 𝑢2 {|𝜌𝜌𝜌ℎ|}
Where h = the depth of the scoop immersion. The product of mass and acceleration is
We also have
𝑀𝑎𝑟𝑟𝑟 = 𝑀
𝑑𝑑
𝑑𝑑
𝑢1 = −𝑈
𝑢2 = 𝑈 cos 𝜃
The drag force is given by
𝐹𝐷 = 𝑘𝑈 2
The constant k is evaluated as
𝑘=
𝑠2
103 𝑁 𝑘𝑘 ∙ 𝑚
𝑘𝑘
𝐹𝐷0
=
90𝑘𝑘
×
×
×
= 1.00
2
2
2
(300 𝑚)
𝑁∙𝑠
𝑚
𝑘𝑘
𝑈0
−𝑘𝑈 2 − 𝑀
The momentum equation becomes:
−𝑀
𝑑𝑑
= 𝜌𝑈 2 𝑤ℎ(1 + cos 𝜃)
𝑑𝑑
𝑑𝑑
= [𝑘 + 𝜌𝜌ℎ(1 + cos 𝜃)]𝑈 2
𝑑𝑑
The rate of change of velocity with time can be rewritten using the chain rule as
𝑑𝑑 𝑑𝑑 𝑑𝑑 𝑑𝑑
=
=
𝑈
𝑑𝑑 𝑑𝑑 𝑑𝑑
𝑑𝑑
The momentum can then be re-written and the variables separated as
𝑑𝑑
= −𝑐𝑐𝑐
𝑈
Where the constant c contains the terms
𝑐=
𝑘 + 𝜌𝜌ℎ(1 + cos 𝜃)
𝑀
Integrating the equation from the initial velocity where x = 0, we get:
𝑈
= −𝑐𝑐
𝑈0
So
ln
The value of c is then
1 𝑈
𝑐 = − ln
𝑥 𝑈0
Solving for h,
𝑐=−
1
100
� = 1.37 × 10−3 𝑚−1
ln �
800 𝑚
300
ℎ=
𝑀𝑀 − 𝑘
𝜌𝜌(1 + cos 𝜃)
ℎ = �8000 𝑘𝑘 ×
1.37 × 10−3
𝑘𝑘 𝑚3
1
1
− 1.00 �
×
×
= 0.0179 𝑚
𝑚 999 𝑘𝑘 0.3 𝑚 (1 + cos 30°)
𝑚
ℎ = 17.9 𝑚𝑚
Problem 4.113
(Difficulty: 2)
4.113 Starting from rest, the cart shown is propelled by a hydraulic catapult (liquid jet). The jet strikes
the curved surface and makes a 180 deg turn, leaving horizontally. Air and rolling resistance may be
neglected. If the mass of the cart is 100 kg and the jet of water leaves the nozzle (of area 0.001 m2) with
a speed of 35 m/s an aerodynamic drag force proportional to the square of cart speed, FD 5 kU2, with k
5 2.0 N_ s2/m2. Derive an expression for the cart acceleration as a function of cart speed and other
given parameters. Evaluate the acceleration of the cart at U = 10 m/s. What fraction is this speed of the
terminal speed of the cart?
Find: The fraction this speed of the terminal speed of the cart.
Assumption: (1) Horizontal, 𝐹𝐵𝐵 = 0
(2) Neglect mass of liquid in CV (components of u cancel)
(3) uniform flow at each section
(4) measure all velocities relative to the CV
(5) No change in stream area or speed on vane
Solution:
Apply x momentum for CV with linear acceleration.
Basic equation:
Then
𝐹𝑠𝑠 + 𝐹𝐵𝐵 − � 𝑎𝑟𝑟𝑟 𝜌𝜌∀ =
𝐶𝐶
𝜕
� 𝑢 𝜌𝜌∀ + � 𝑢𝑥𝑥𝑥 𝜌𝑉�𝑥𝑥𝑥 ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶 𝑥𝑥𝑥
𝐶𝐶
−𝑘𝑈 2 − 𝑎𝑟𝑟𝑟 𝑀 = 𝑢1 {−|𝜌(𝑉 − 𝑈)𝐴|} + 𝑢2 {|𝜌(𝑉 − 𝑈)𝐴|}
𝑢1 = 𝑉 − 𝑈
𝑢2 = −(V − U)
−𝑘𝑈 2 − 𝑎𝑟𝑟𝑟 𝑀 = −2𝜌(𝑉 − 𝑈)2 A
or
At 𝑈 = 10
𝑎𝑟𝑟𝑟 =
𝑎𝑟𝑟𝑟 =
𝑚
𝑠
2 × 999
𝑑𝑑
𝑑𝑑
=
2𝜌(𝑉 − 𝑈)2 𝐴 − 𝑘𝑈2
𝑀
2
2
𝑘𝑘
𝑁 ∙ 𝑠2
𝑘𝑘 ∙ 𝑚
2𝑚
2
2𝑚
(
)
(
)
×
30
−
10
×
0.001
𝑚
−
2.0
×
10
×
𝑚3
𝑁 ∙ 𝑠2 = 5.99 𝑚
𝑠2
𝑚2
𝑠2
𝑠2
100 𝑘𝑘
At terminal speed, 𝑎𝑟𝑟𝑟 = 0. Then 2𝜌(𝑉 − 𝑈𝑡 )2 A = 𝑘𝑈𝑡2 , or
𝑘
2𝜌𝜌
𝑉 − 𝑈𝑡 = 𝑈𝑡 �
Solving,
𝑈𝑡 =
𝑈𝑡 = 30
Finally,
𝑚
×
𝑠
𝑉
𝑘
1+�
2𝜌𝜌
1
1
2
𝑘𝑘 ∙ 𝑚
1
𝑁 ∙ 𝑠2
𝑚3
1
�
1 + � × 2.0
×
×
×
2
2
2
999
𝑘𝑘
0.001 𝑚
𝑁 ∙ 𝑠2
𝑚
𝑚
10
𝑈
𝑠 = 0.667
𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 =
=
𝑈𝑡 15.0 𝑚
𝑠
= 15.0
𝑚
𝑠
Problem 4.114
(Difficulty: 2)
4.114 Solve Problem 4.107 if the vane and slider ride on a film of oil instead of sliding in contact with the
surface. Assume motion resistance is proportional to speed, 𝐹𝑅 = 𝑘𝑘, with 𝑘 = 7.5
𝑁∙𝑆
𝑚
Assumption: (1) Horizontal, 𝐹𝐵𝐵 = 0
(2) Neglect mass of liquid in on vane, 𝑢 ≈ 0 on vane
(3) uniform flow at each section
(4) measure all velocities relative to the CV
Solution:
Apply x momentum equation to linearly accelerating CV.
Basic equation:
Then
𝐹𝑠𝑠 + 𝐹𝐵𝐵 − � 𝑎𝑟𝑟𝑟 𝜌𝜌∀ =
𝐶𝐶
𝜕
� 𝑢 𝜌𝜌∀ + � 𝑢𝑥𝑥𝑥 𝜌𝑉�𝑥𝑥𝑥 ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶 𝑥𝑥𝑥
𝐶𝐶
−𝑘𝑘 − 𝑎𝑟𝑟𝑟 𝑀 = 𝑢1 {−|𝜌(𝑉 − 𝑈)𝐴|} + 𝑢2 {𝑚̇2 } + 𝑢3 {𝑚̇3 }
𝑢1 = 𝑉 − 𝑈
𝑢2 = 0
or
−𝑘𝑘 − 𝑀
𝑢3 = 0
𝑑𝑑
= −𝜌(𝑉 − 𝑈)2 𝐴
𝑑𝑑
.
𝑑𝑑 𝜌(𝑉 − 𝑈)2 𝐴 𝑘𝑘
=
−
𝑀
𝑑𝑑
𝑀
𝑘𝑘
1
𝑁∙𝑠
𝑚
1
𝑘𝑘 ∙ 𝑚
𝑑𝑑
𝑚2
= 999 3 × (20 − 10)2 2 × 0.005 𝑚2 ×
− 7.5
× 10 ×
×
𝑚
30 𝑘𝑘
𝑚
𝑠 30 𝑘𝑘 𝑁 ∙ 𝑠 2
𝑑𝑑
𝑠
at 𝑈 = 10
𝑚
𝑠
𝑚
𝑑𝑑
= 14.2 2
𝑠
𝑑𝑑
.
At terminal speed, 𝑈 = 𝑈𝑡 and
𝑑𝑑
𝑑𝑑
or
= 0, so
0=
𝜌(𝑉 − 𝑈)2 𝐴 𝑘𝑘
−
𝑀
𝑀
𝑉 2 − 2𝑈𝑈 + 𝑈 2 −
𝑈 2 − �2𝑉 +
𝑈=
1+
2𝑉 +
𝑘
𝑘 2
± ��2𝑉 + � − 4𝑉 2
𝜌𝜌
𝜌𝜌
2
𝑘
𝑈=0
𝜌𝜌
𝑘
� 𝑈 + 𝑉2 = 0
𝜌𝜌
𝑘 2
𝑘
�
� ± �1 +
� − 1�
= 𝑉 ��1 +
2𝜌𝜌𝜌
2𝜌𝜌𝜌
𝑘
1
𝑁∙𝑠
𝑚3
𝑠
1
𝑘𝑘 ∙ 𝑚
= 1 + × 7.5
×
×
×
×
= 1.0375
2
2𝜌𝜌𝜌
2
𝑚
𝑁 ∙ 𝑠2
999 𝑘𝑘 20 𝑚 0.005 𝑚
𝑈 = 𝑉 �1.0375 ± �(1.0375)2 − 1� = 0.761𝑉 = 0.761 × 20
The negative root was chosen so 𝑈𝑡 < 𝑉, as required.
𝑚
𝑚
= 15.2
𝑠
𝑠
Problem 4.115
Problem
4.153
4.115
[Difficulty: 4]
4.114
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
−k ⋅ U − M ⋅
or
dU
dt
dU
u1 = V − U
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
so
dx
dt
=U
x ( n + 1 ) = x ( n ) + U( n ) ⋅ ∆t
The final set of equations is
⎡ ρ⋅ ( V − U( n) ) 2⋅ A k⋅ U( n )⎤
⎥ ⋅ ∆t
U( n + 1 ) = U( n ) + ⎢
−
M
M ⎦
⎣
2
a( n ) =
ρ⋅ ( V − U( n ) ) ⋅ A
M
−
x ( n + 1 ) = x ( n ) + U( n ) ⋅ ∆t
k ⋅ U( n )
M
where ∆t 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 4.116
(Difficulty: 2)
4.116 A rectangular block of mass 𝑀, with vertical faces, rolls without resistance along a smooth
horizontal plane as shown. The block travels initially at speed 𝑈0 . At 𝑡 = 0 the block is struck by a liquid
jet and its speed begins to slow. Obtain an algebraic expression for the acceleration of the block for
𝑡 > 0. Solve the equation to determine the time at which 𝑈 = 0.
Find: The time 𝑡 at which 𝑈 = 0.
Assumption: (1) no pressure for friction forces, so 𝐹𝑠𝑠 = 0.
(2) horizontal, so 𝐹𝐵𝐵 = 0.
(3) neglect mass of liquid in CV, 𝑢 = 0 in CV
(4) uniform flow at each section
(5) measure velocities relative to CV
Solution:
Apply x momentum equation to linearly accelerating CV.
Basic equation:
Then
𝐹𝑠𝑠 + 𝐹𝐵𝐵 − � 𝑎𝑟𝑟𝑟 𝜌𝜌∀ =
𝐶𝐶
−𝑀𝑎𝑟𝑟𝑟 = −M
𝜕
� 𝑢 𝜌𝜌∀ + � 𝑢𝑥𝑥𝑥 𝜌𝑉�𝑥𝑥𝑥 ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶 𝑥𝑥𝑥
𝐶𝐶
𝑑𝑑
= 𝑢1 {−|𝜌(𝑉 + 𝑈)𝐴|} + 𝑢2 {𝑚̇2 } + 𝑢3 {𝑚̇3 }
𝑑𝑑
𝑢1 = −(𝑉 + 𝑈)
𝑢2 = 0
𝑢3 = 0
or
𝜌(𝑉 + 𝑈)2 𝐴
𝑑𝑑
=−
𝑑𝑑
𝑀
But, since 𝑉 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐, 𝑑𝑑 = 𝑑(𝑉 + 𝑈), so
Integrating from 𝑈0 at 𝑡 = 0 to 𝑈 = 0 at 𝑡
Solving,
𝜌𝜌
𝑑(𝑉 + 𝑈)
=
−
𝑑𝑑
(𝑉 + 𝑈)2
𝑀
𝑉
1
1
−𝑈0
𝜌𝜌𝜌
𝑑(𝑉 + 𝑈)
=− +
=
=−
2
𝑉 𝑉 + 𝑈0 𝑉(𝑉 + 𝑈0 )
𝑀
𝑉+𝑈0 (𝑉 + 𝑈)
�
𝑡=
𝑀
𝑀𝑈0
=
𝜌𝜌𝜌(𝑉 + 𝑈0 ) 𝜌𝜌𝜌 �1 + 𝑉 �
𝑈0
Problem 4.117
Problem
4.156
[Difficulty: 3]
4.117
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 +
M
ρ⋅ A
(
⎡
ρ⋅ A⋅ V + U0
⎣
M
⋅ ln⎢1 +
)
⎤
⎦
⋅ 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 4.118
Problem
*4.158
[Difficulty: 3]
4.118
d
c
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 ×
2
s⎦
12.9 30⋅ kg
s
−g
arfz = 2.28
using
m
2
s
V1 ⋅ A1 = V0 ⋅ A0
Problem 4.119
(Difficulty: 2)
4.119 A rocket sled traveling on a horizontal track is slowed by a retro-rocket fired in the direction of
travel. The initial speed of the sled is 𝑈0 = 500
𝑚
𝑠
retro-rocket consumes fuel at the rate of 7.75
atmospheric pressure and a speed of 2500
𝑚
𝑠
. The initial mass of the sled is 𝑀0 = 1500 𝑘𝑘. The
𝑘𝑘
𝑠
, and the exhaust gases leave the nozzle at
relative to the rocket. The retro-rocket fires for 20 𝑠.
Neglect aerodynamic drag and rolling resistance. Obtain an plot an algebraic expression for sled speed 𝑈
as a function of firing time. Calculate the sled speed at the end of retro-rocket firing.
Find: Sled speed 𝑈(𝑡). The sled speed at the end of retro-rocket firing 𝑈(𝑡∞) .
Assumption: (1) no pressure, drag, or rolling resistance, so 𝐹𝑠𝑠 = 0.
(2) horizontal motion, so 𝐹𝐵𝐵 = 0.
(3) neglect unsteady effects within CV
(4) uniform flow at nozzle exit plane
Solution:
(5) 𝑝𝑒 = 𝑝𝑎𝑎𝑎
Apply x-component of momentum equation to the linearly accelerating CV shown.
From continuity,
𝑀𝐶𝐶 = 𝑀0 − 𝑚̇𝑡, 𝑡 < 𝑡∞
Basic equation:
Then
or
𝐹𝑠𝑠 + 𝐹𝐵𝐵 − � 𝑎𝑟𝑟𝑟 𝜌𝜌∀ =
𝐶𝐶
𝜕
� 𝑢 𝜌𝜌∀ + � 𝑢𝑥𝑦𝑦 𝜌𝑉�𝑥𝑥𝑥 ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶 𝑥𝑥𝑥
𝐶𝐶
−𝑎𝑟𝑟𝑟 𝑀𝐶𝐶 = 𝑢𝑒 {𝑚̇} = V𝑒 𝑚̇
V𝑒 𝑚̇
V𝑒 𝑚̇
𝑑𝑑
=−
=−
𝑑𝑑
𝑀𝐶𝐶
𝑀0 − 𝑚̇𝑡
Thus
𝑑𝑑 = V𝑒 �
and
−𝑚̇𝑑𝑑
�
𝑀0 − 𝑚̇𝑡
𝑈 − 𝑈0 = V𝑒 ln(𝑀0 − 𝑚̇𝑡)𝑡0 = V𝑒 ln �1 −
𝑈(𝑡) = 𝑈0 + V𝑒 ln �1 −
for 𝑡 < 𝑡∞ .
At 𝑡 = 𝑡∞ ,
𝑈(𝑡∞ ) = 500
𝑚̇𝑡
�
𝑀0
𝑚̇𝑡
�
𝑀0
𝑚
𝑘𝑘
1
𝑚
�
+ 2500 × ln �1 − 7.75
× 20 𝑠 ×
𝑠
𝑠
1500 𝑘𝑘
𝑠
𝑈(𝑡∞ ) = 227
𝑚
𝑠
Problem 4.120
(Difficulty: 2)
4.120 A rocket sled accelerates from rest on a level track with negligible air and rolling resistances. The
initial mass of the sled is 𝑀0 = 600 𝑘𝑘. The rocket initially contains 150 𝑘𝑘 of fuel. The rocket motor
burns fuel at constant rate 𝑚̇ = 15
𝑉𝑒 = 2900
𝑚
𝑠
𝑘𝑘
𝑠
. Exhaust gases leave the rocket nozzle uniformly and axially at
relative to the nozzle, and the pressure is atmospheric. Find the maximum speed reached
by the rocket sled. Calculate the maximum acceleration of the sled during the run.
Find: The maximum speed 𝑈𝑚𝑚𝑚 and the maximum acceleration
Assumption: (1) no net pressure forces (𝑝𝑒 = 𝑝𝑎𝑎𝑎 )
𝑑𝑑
𝑑𝑑 𝑚𝑚𝑚
during the run.
(2) horizontal motion, so 𝐹𝐵𝐵 = 0
(3) neglect
𝜕
𝜕𝜕
in CV
(4) uniform axial jet
Solution:
Apply the momentum equation to linearly accelerating CV shown.
Basic equation:
From continuity,
𝐹𝑠𝑠 + 𝐹𝐵𝐵 − � 𝑎𝑟𝑟𝑟 𝜌𝜌∀ =
𝐶𝐶
−𝑎𝑟𝑟𝑟 M = −
Separating variables,
𝜕
� 𝑢 𝜌𝜌∀ + � 𝑢𝑥𝑥𝑥 𝜌𝑉�𝑥𝑥𝑥 ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶 𝑥𝑥𝑥
𝐶𝐶
𝑀 = 𝑀0 − 𝑚̇t
𝑑𝑑
(𝑀0 − 𝑚̇t) = 𝑢𝑒 {𝑚̇} = −𝑉𝑒 𝑚̇
𝑑𝑑
Eq (1)
𝑑𝑑 = 𝑉𝑒
Integrating from 𝑈 = 0 at 𝑡 = 0 to 𝑈 at 𝑡 gives
𝑚̇𝑑𝑑
𝑀0 − 𝑚̇t
𝑈 = −𝑉𝑒 ln(𝑀0 − 𝑚̇t)𝑡0 = −𝑉𝑒 ln
(𝑀0 − 𝑚̇t)
𝑀0
= 𝑉𝑒 ln
(𝑀0 − 𝑚̇t)
𝑀0
The speed is a maximum at burnout. At burnout 𝑀𝑓 = 0 and 𝑀 = 𝑀0 − 𝑚̇t = 450 kg.
At burnout,
𝑡=
Then from Eq (2)
From Eq (1) the acceleration is:
𝑀𝑓𝑓𝑓𝑓𝑓𝑓𝑓
𝑠
= 150 𝑘𝑘 ×
= 10 𝑠
15 𝑘𝑘
𝑚̇𝑓𝑓𝑓𝑓
𝑈𝑚𝑚𝑚 = 2900
𝑚
600 𝑘𝑘
𝑚
× ln
= 834
𝑠
450 𝑘𝑘
𝑠
𝑚̇𝑉𝑒
𝑑𝑑
=
𝑑𝑑 𝑀0 − 𝑚̇t
The maximum acceleration occurs at the instant prior to burnout
𝑘𝑘
𝑚
1
𝑚
𝑑𝑑
= 15
× 2900 ×
= 96.7 2
𝑠
𝑠 450 𝑘𝑘
𝑠
𝑑𝑑 𝑚𝑚𝑚
The sled speed as a function of time is
for 0 ≤ 𝑡 ≤ 10 𝑠.
for 𝑡 > 10 𝑠 (neglecting resistance).
𝑈 = 𝑉𝑒 ln
𝑀0
(𝑀0 − 𝑚̇t)
𝑈 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 834
The sled accelerating is given by
for 0 ≤ 𝑡 ≤ 10 𝑠.
𝑚̇𝑉𝑒
𝑑𝑑
=
𝑑𝑑 𝑀0 − 𝑚̇t
𝑑𝑑
=0
𝑑𝑑
𝑚
𝑠
Eq (2)
for 𝑡 > 10 𝑠.
Problem
4.164
Problem 4.121
[Difficulty: 3]
4.121
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
⎠
t=
M0
mrate
⎛⎜
−
⎜
⋅⎝1 − e
⎞
⎠
mf = 82.7 kg
U
Ve
⎞
⎠
Problem 4.122
(Difficulty: 2)
4.122 A rocket sled with initial mass of 3 metric tons, including 1 ton of fuel, rests on a level section of
track. At 𝑡 = 0, the solid fuel of the rocket is ignited and the rocket burns fuel at the rate of 75
exit speed of the exhaust gas relative to the rocket is 2500
𝑚
𝑠
𝑘𝑘
𝑑𝑑
𝑑𝑑
and speed of sled 𝑈 at 𝑡 = 10 𝑠.
Assumption: (1) 𝐹𝑆𝑆 = 0, no resistance (given).
(2) 𝐹𝐵𝐵 = 0, horizontal
(3) neglect
𝜕
𝜕𝜕
inside CV
(4) uniform flow at nozzle exit
Solution:
(5) 𝑝𝑒 = 𝑝𝑎𝑎𝑎 (given)
Apply the x component of momentum to linearly accelerating CV. Use continuity to find 𝑀(𝑡).
Basic equation:
0=
From continuity,
𝜕
� 𝜌𝜌∀ + � 𝜌 𝑉�𝑥𝑥𝑥 ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝐹𝑠𝑠 + 𝐹𝐵𝐵 − � 𝑎𝑟𝑟𝑟 𝜌𝜌∀ =
𝐶𝐶
0=
𝜕
� 𝑢 𝜌𝜌∀ + � 𝑢𝑥𝑥𝑥 𝜌𝑉�𝑥𝑥𝑥 ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶 𝑥𝑦𝑦
𝐶𝐶
𝑑𝑑
𝜕𝜕
+ �𝑀̇� =
+ 𝑀̇
𝑑𝑑
𝜕𝜕
𝑑𝑑 = −𝑀̇𝑑𝑑
. The
, and the pressure is atmospheric.
Neglecting friction and air resistance, calculate the acceleration and speed of the sled at 𝑡 = 10 𝑠.
Find: The acceleration
𝑠
Integrating,
𝑀
𝑡
� 𝑑𝑑 = 𝑀 − 𝑀0 = � − 𝑀̇𝑑𝑑 = −𝑀̇𝑡
𝑀0
0
𝑀 = 𝑀0 − 𝑀̇𝑡
From the momentum equation
Thus
−𝑎𝑟𝑟𝑟 𝑀 = −𝑎𝑟𝑟𝑟 �𝑀0 − 𝑀̇𝑡� = 𝑢1 ��𝑀̇�� = −𝑉𝑒 𝑀̇
𝑎𝑟𝑟𝑟 =
At 𝑡 = 10 𝑠
From Eq (1),
𝑑𝑑
𝑉𝑒 𝑀̇
=
𝑑𝑑 �𝑀0 − 𝑀̇𝑡�
𝑚
𝑘𝑘
1
𝑚
𝑑𝑑
= 2500 × 75
×
= 83.3 2
𝑘𝑘
𝑠
𝑠
𝑠
𝑑𝑑
3000 𝑘𝑘 − 75
× 10 𝑠
𝑠
𝑑𝑑 =
Integrating from 𝑈 = 0 at𝑡 = 0 to 𝑈 at 𝑡 gives
𝑉𝑒 𝑀̇𝑑𝑑
�𝑀0 − 𝑀̇𝑡�
𝑡
𝑈 = −𝑉𝑒 ln�𝑀0 − 𝑀̇𝑡�0 = −𝑉𝑒 ln
𝑈 = 𝑉𝑒 ln
At 𝑡 = 10 𝑠
Eq (1)
𝑈 = 2500
𝑀0
�𝑀0 − 𝑀̇𝑡�
�𝑀0 − 𝑀̇𝑡�
𝑀0
3000 𝑘𝑘
𝑚
𝑚
× ln
= 719
𝑘𝑘
𝑠
𝑠
3000 𝑘𝑘 − 75
× 10 𝑠
𝑠
Note that all fuel will be expended at
𝑡𝑏𝑏 =
The sled speed as a function of time is then
𝑀𝑓 1000 𝑘𝑘
=
= 13.3 𝑠
𝑘𝑘
𝑀̇
75
𝑠
Eq (2)
for 𝑡 ≤ 13.3 𝑠.
for 𝑡 ≥ 13.3 𝑠.
𝑈 = 𝑉𝑒 ln
𝑀0
�𝑀0 − 𝑀̇𝑡�
𝑈 = 𝑈𝑚𝑚𝑚 = 1010
The sled acceleration is given by:
for 𝑡 ≤ 13.3 𝑠.
for 𝑡 ≥ 13.3 𝑠.
𝑉𝑒 𝑀̇
𝑑𝑑
=
𝑑𝑑 �𝑀0 − 𝑀̇𝑡�
𝑑𝑑
=0
𝑑𝑑
𝑚
𝑠
Problem 4.123*
(Difficulty: 2)
*Please Note: this solution shows the solution to the problem with one particular set of values. These values may not match those in the problem you have been given to solve.
4.123 A “home-made” solid propellant rocket has an initial mass of 20 𝑙𝑙𝑙; 15 𝑙𝑙𝑙 of this is fuel. The
rocket is directed vertically upward from rest, burns fuel at a constant rate of 0.5
exhaust gas at a speed of 6500
𝑓𝑓
𝑠
𝑙𝑙𝑙
𝑠
, and ejects
relative to the rocket. Assume that the pressure at the exit is
atmospheric and that air resistance may be neglected. Calculate the rocket speed after 20 𝑠 and the
distance traveled by the rocket in 20 𝑠. Plot the rocket speed and the distance traveled as function of
time.
Find: The speed 𝑉 after 20 𝑠. The distance 𝑌 traveled in 20 𝑠.
Assumption: (1) neglect air resistance; 𝑝𝑒 = 𝑝𝑎𝑎𝑎
(2) neglect 𝑉𝑥𝑥𝑥 and
𝜕
𝜕𝜕
within CV
(3) uniform flow at nozzle exit section
Solution:
Apply the y component of momentum equation to accelerating CV using CS shown.
Basic equation:
Then
and
𝐹𝑠𝑠 + 𝐹𝐵𝐵 − � 𝑎𝑟𝑟𝑟 𝜌𝜌∀ =
𝐶𝐶
𝜕
� 𝑉 𝜌𝜌∀ + � 𝑉𝑥𝑥𝑥 𝜌𝑉�𝑥𝑥𝑥 ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶 𝑥𝑥𝑥
𝐶𝐶
𝐹𝐵𝐵 − 𝑎𝑟𝑟𝑟 𝑀 = −𝑀𝑀 − 𝑀𝑎𝑟𝑟𝑟 = −𝑉𝑒 𝑚̇
𝑎𝑟𝑟𝑟 =
𝑑𝑑 𝑉𝑒 𝑚̇
=
−𝑔
𝑑𝑑
𝑀
Introducing 𝑀 = 𝑀0 − 𝑚̇𝑡 and separating variables,
𝑑𝑑 = �
Integrating from rest at 𝑡 = 0
𝑡
𝑉=� �
0
At 𝑡 = 20 𝑠,
𝑉 = 6500
To find height , note 𝑉 =
Let Ω = 1 −
𝑚̇𝑡
𝑀0
𝑑𝑑
𝑑𝑑
𝑓𝑓
𝑓𝑓
20 𝑙𝑙𝑙
� − 32.2 2 × 20 𝑠
× ln �
𝑙𝑙𝑙
𝑠
𝑠
× 20 𝑠
20 𝑙𝑙𝑙 − 0.5
𝑠
𝑚̇𝑡
𝑑𝑑
𝑀0
� − 𝑔𝑔 = −𝑉𝑒 ln �1 −
� − 𝑔𝑔
= 𝑉𝑒 ln �
𝑀0
𝑀0 − 𝑚̇𝑡
𝑑𝑑
𝑚̇
𝑀0
𝑑𝑑, then
𝑑𝑑 = −𝑉𝑒 ln Ω 𝑑𝑑 − 𝑔𝑔𝑔𝑔 =
𝑌=�
𝑡
0
𝑌=
𝑉𝑒 𝑀0
ln Ω 𝑑Ω − 𝑔𝑔𝑔𝑔
𝑚̇
1
𝑉𝑒 𝑀0
1
𝑉𝑒 𝑀0
[Ω ln Ω − Ω]𝑡0 − 𝑔𝑡 2
ln Ω 𝑑Ω − 𝑔𝑡 2 =
2
2
𝑚̇
𝑚̇
𝑌=
𝑌 = 6500
𝑓𝑓
𝑠
. Substitute into Eq (1) to obtain:
Integrating from 𝑌 = 0 at 𝑡 = 0,
So we have:
𝑉𝑒 𝑚̇
𝑀0
� − 𝑔𝑔
− 𝑔� 𝑑𝑑 = 𝑉𝑒 ln �
𝑀0 − 𝑚̇𝑡
𝑀0 − 𝑚̇𝑡
𝑉(20 𝑠) = 3860
, and 𝑑Ω = −
At 𝑡 = 20 𝑠,
𝑉𝑒 𝑚̇
− 𝑔� 𝑑𝑑
𝑀0 − 𝑚̇𝑡
𝑡
𝑚̇𝑡
𝑚̇𝑡
1
𝑉𝑒 𝑀0
� �ln �1 − � − 1�� − 𝑔𝑡 2
��1 −
𝑀0
𝑀0
2
𝑚̇
0
𝑚̇𝑡
1
𝑉𝑒 𝑀0
𝑚̇𝑡
� �ln �1 −
� − 1� + 1� − 𝑔𝑡 2
��1 −
𝑀0
2
𝑀0
𝑚̇
1−
𝑚̇𝑡
𝑙𝑙𝑙
1
1
= 1 − 0.5
× 20 𝑠 ×
=
𝑀0
𝑠
20 𝑙𝑙𝑙 2
1
𝑓𝑓
𝑠
𝑓𝑓
1
1
× 20 𝑙𝑙𝑙 ×
�� � �ln � � − 1� + 1� − × 32.2 2 × (20)2 𝑠 2
2
𝑠
0.5 𝑙𝑙𝑙 2
𝑠
2
Eq (1)
𝑌 = 33500 𝑓𝑓
Problem 4.124
Problem
4.170
[Difficulty: 3]
4.124
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
⎞
− g ⋅ dt
⎜
⎝ M0 − mrate⋅ t
⎠
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
⎝
Vmax = −2500⋅ ⋅ ln⎛⎜ 1 − 10⋅
×
× 5 ⋅ s⎞ − 9.81⋅ × 5 ⋅ s
2
s ⎝
s
350 ⋅ kg
⎠
s
m
kg
mrate⋅ t ⎞
1
m
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 4.125
Problem
4.172
[Difficulty: 4]
4.125
y
x
d
CS (moves
at speed U)
c
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 4.126
(Difficulty: 2)
4.126 The moving tank shown is to be slowed by lowering a scoop to pick up water from a trough. The
initial mass and speed of the tank and its contents are 𝑀0 and 𝑈0 , respectively. Neglect external forces
due to pressure or friction and assume that the track is horizontal. Apply the continuity and momentum
equations to show that at any instant 𝑈 =
𝑈0 𝑀0
𝑀
. Obtain a general expression for
𝑈
𝑈0
as a function of time.
CV
Water
Find: The expression for
Assumption: (1) 𝐹𝑠𝑠 = 0
𝑈
𝑈0
as a function of time.
(2) 𝐹𝐵𝐵 = 0
(3) neglect u within CV
(4) uniform flow across inlet section
Solution:
Apply continuity and momentum equations to linearly accelerating CV shown.
Basic equation:
0=
From continuity,
𝜕
� 𝜌𝜌∀ + � 𝜌 𝑉�𝑥𝑥𝑥 ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝐹𝑠𝑠 + 𝐹𝐵𝐵 − � 𝑎𝑟𝑟𝑟 𝜌𝜌∀ =
𝐶𝐶
𝜕
� 𝑢 𝜌𝜌∀ + � 𝑢𝑥𝑥𝑥 𝜌𝑉�𝑥𝑥𝑥 ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶 𝑥𝑥𝑥
𝐶𝐶
0=
or
𝜕
𝑀 + {−|𝜌𝜌𝜌|}
𝜕𝜕 𝐶𝐶
𝑑𝑑
= 𝜌𝜌𝜌
𝑑𝑑
From momentum
−𝑎𝑟𝑟𝑟 𝑀 = −
−𝑎𝑟𝑟𝑟 𝑀 = −
But from continuity,
𝑑𝑑
𝑀 = 𝑢{−|𝜌𝜌𝜌|}
𝑑𝑑
𝑢 = −𝑈
𝑑𝑑
𝑀 = 𝑢{−|𝜌𝜌𝜌|} = 𝑈𝑈𝑈𝑈
𝑑𝑑
𝜌𝜌𝜌 =
So we have:
𝑀
or
𝑑𝑑
𝑑𝑑
𝑑𝑑
𝑑𝑑
+𝑈
=0
𝑑𝑑
𝑑𝑑
𝑈𝑈 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝑈0 𝑀0
Substituting 𝑀 =
𝑈0 𝑀0
𝑈
Integrating,
into momentum, −
𝑈=
𝑑𝑑 𝑀0 𝑈0
𝑑𝑑
𝑈
𝑈0 𝑀0
𝑀
= 𝜌𝑈 2 𝐴, or
𝜌𝜌
𝑑𝑑
=−
𝑑𝑑
3
𝑈0 𝑀0
𝑈
𝑡
1 1𝑈
1 1
1
𝜌𝜌
𝑑𝑑
𝜌𝜌
� 3 =−
�
=
−
�
−
�
=
−
𝑑𝑑 = −
𝑡
2
2
2
2 𝑈 𝑈0
2 𝑈
𝑈0 𝑀0
𝑈0
𝑈0 𝑈
0 𝑈0 𝑀0
𝑈
Solving for 𝑈,
𝑈=
𝑈0
1
2𝜌𝑈0 𝐴 2
�1 +
𝑡�
𝑀0
Problem 4.127
Problem
4.176
[Difficulty: 4]
4.127
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
− g ⋅ dt
⎜
⎝ M0 − mrate⋅ t
⎠
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
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
mrate⋅ tb ⎞
⎛
Vmax = −Ve⋅ ln⎜ 1 −
− g ⋅ tb
M0
⎝
⎠
m
1
m
− 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
+−
1
2
× 9.81⋅
m
2
mrate⋅ t ⎞
⎠⎝ ⎝
m
Yb = 782 ⋅ × 0.0696⋅ kg ×
s
At t = tb
⎛ ⎛
⋅ ⎜ 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 4.128
(Difficulty: 2)
4.128 The 90° reducing elbow of Example 4.6 discharges to atmosphere. Section (2) is located 0.3 𝑚 to
the right of section (1). Estimate the moment exerted by the flange on the elbow.
�𝑓𝑓𝑓𝑓𝑓𝑓 exerted by the flange on the elbow.
Find: The moment 𝑀
Assumption: (1) neglect body forces
(2) no shafts, so 𝑇�𝑠ℎ𝑎𝑎𝑎 = 0
(3) steady flow (given)
(4) uniform flow across each across section
(5) incompressible flow
Solution:
Apply moment of momentum, using the CV and CS shown.
From example problem:
Steady flow,
𝑉�2 = −16 𝚥̂
𝑚
𝑠
𝐴1 = 0.01 𝑚2
𝐴2 = 0.0025 𝑚2
Basic equation:
Then we have:
𝑟̅ × 𝐹�𝑠 + � 𝑟̅ × 𝑔̅ 𝜌𝜌∀ + 𝑇�𝑠ℎ𝑎𝑎𝑎 =
𝐶𝐶
𝜕
� 𝑟̅ × 𝑉� 𝜌𝜌∀ + � 𝑟̅ × 𝑉� 𝜌 𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
�𝑓𝑓𝑓𝑓𝑓𝑓 = 𝑟̅ × 𝐹�𝑠 )𝑓𝑓𝑓𝑓𝑓𝑓 = 𝑟̅1 × 𝑉�1 {−𝜌𝑉1 𝐴1 } + 𝑟̅2 × 𝑉�2 {𝜌𝑉2 𝐴2 }
𝑀
𝑟̅1 = 0
𝑟̅2 = 𝑎𝚤̂ + 𝑏𝚥̂
𝑉�2 = −𝑉2 𝚥̂
𝑟̅2 × 𝑉�2 = −𝑎𝑉2 𝑘� + 0
Substituting into equation (1)
�𝑓𝑓𝑓𝑓𝑓𝑓 = −𝑎𝑉2 𝑘�{𝜌𝑉2 𝐴2 } = −𝑎𝑎𝑉22 𝐴2 𝑘�
𝑀
�𝑓𝑓𝑓𝑓𝑓𝑓 = 0.3𝑚 × 999
𝑀
𝑘𝑘
𝑁 ∙ 𝑠2
𝑚2
2
2
(16)
×
×
0.0025
𝑚
×
�−𝑘��
𝑚3
𝑠2
𝑘𝑘 ∙ 𝑚
�𝑓𝑓𝑓𝑓𝑓𝑓 = −192 𝑘� 𝑁 ∙ 𝑚
𝑀
This is the torque that must be exerted on the CV by the flange.
�𝑓𝑓𝑓𝑓𝑓𝑓 is in the −𝑘� direction, it must act CW in the x-y plane.
Since 𝑀
Eq (1)
Problem 4.129
(Difficulty: 2)
4.129 Crude oil 𝑆𝑆 = 0.95 from a tanker dock flows through a pipe of 0.25 𝑚 diameter in the
configuration shown. The flow rate is 0.58
𝑚3
𝑠
, and the gage pressures are shown in the diagram.
Determine the force and torque that are exerted by the pipe assembly on its supports.
𝑄 = 0.58
𝑚3
𝑠
Find: The force and torque exerted by the pipe for support.
Assumption: (1) 𝐹𝐵𝐵 = 0;𝑔̅ acts in the 𝛿 direction
(2) steady flow
(3) uniform flow at each section
(4) no 𝛿 component of 𝑟̅ × 𝑔̅
Solution:
(5) 𝑇�𝑠ℎ𝑎𝑎𝑎 = 0
No momentum components exist in the y direction. Apply x component of linear momentum and the
moment of momentum equations using the CV shown. Location of coordinates is arbitrary. For
simplicity, choose as shown.
Basic equation:
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
The area is:
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑟̅ × 𝐹�𝑠 + � 𝑟̅ × 𝑔̅ 𝜌𝜌 ∀ + 𝑇�𝑠ℎ𝑎𝑎𝑎 =
𝐶𝐶
𝜕
� 𝑟̅ × 𝑉� 𝜌𝜌∀ + � 𝑟̅ × 𝑉� 𝜌 𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
From momentum equation,
𝐴=
𝜋𝐷 2 𝜋
= (0.25 𝑚)2 = 0.049 𝑚2
4
4
𝑅𝑥1 + 𝑅𝑥2 + 𝑝1 𝐴 − 𝑝2 𝐴 = 𝑢1 {−𝑚̇} + 𝑢2 {𝑚̇} = 0
From momentum of momentum,
𝑅𝑥1 + 𝑅𝑥2 = (𝑝2 − 𝑝1 )𝐴
𝑟̅1 × (𝑅𝑥1 + 𝑝1 𝐴)𝚤̂ = 𝑟̅1 × 𝑉1 𝚤̂{−𝑚̇}
𝑟̅1 = 𝐿𝚥̂
𝑟̅1 × 𝚤̂ = −𝐿𝑘�
−𝐿(𝑅𝑥1 + 𝑝1 𝐴)𝑘� = −𝐿𝑉1 (−𝑚̇)𝑘� = 𝐿𝑉1 𝑚̇𝑘� = 𝐿
𝑅𝑥1 = −0.95 × 999
𝑅𝑥2
𝑅𝑥1 = −𝜌
𝑄2
− 𝑝1 𝐴
𝐴
𝑘𝑘
1
𝑁 ∙ 𝑠2
𝑚6
𝑁
2
(0.58)
×
×
×
− 3.45 × 105 2 × 0.049 𝑚2
3
2
2
𝑚
0.049 𝑚
𝑠
𝑘𝑘 ∙ 𝑚
𝑚
𝑅𝑥1 = −23.4 𝑘𝑘
= (𝑝2 − 𝑝1 )𝐴 − 𝑅𝑥1 = 𝑝2 𝐴 − 𝑝1 𝐴 + 𝜌
𝑅𝑥2 = 3.32 × 105
𝑄
𝑄2
(𝜌𝜌)𝑘� = 𝐿𝐿
𝑘�
𝐴
𝐴
𝑄2
𝑄2
+ 𝑝1 𝐴 = 𝑝2 𝐴 + 𝜌
𝐴
𝐴
𝑘𝑘
1
𝑁 ∙ 𝑠2
𝑁
𝑚6
2
2
(0.58)
×
0.049
𝑚
+
0.95
×
999
×
×
×
𝑚3
𝑠 2 0.049 𝑚2 𝑘𝑘 ∙ 𝑚
𝑚2
𝑅𝑥2 = 22.8 𝑘𝑘
𝑟̅ × 𝐹�𝑠 = 𝑟̅1 × 𝑅𝑥1 𝚤̂ = 𝐿𝚥̂ × 𝑅𝑥1 𝚤̂ = −𝐿𝑅𝑥1 𝑘� = −20 𝑚 × (−46.0) 𝑘𝑘 𝑘� = 468 𝑘� 𝑘𝑘 ∙ 𝑚
These are forces and torque on CV. The corresponding reactions are:
Force:
𝐾𝑥1 = −𝑅𝑥1 = 23.4 𝑘𝑘
Torque:
𝐾𝑥2 = −𝑅𝑥2 = −22.8 𝑘𝑘
� = −𝑟̅ × 𝐹�𝑠 = −468 𝑘� 𝑘𝑘 ∙ 𝑚
𝑀
Problem 4.130
Problem
4.188
[Difficulty: 3]
4.130
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
1
L
R = 225⋅ mm
min
2 π
4
⋅d
ρ = 999⋅
kg
3
m
Q
⋅
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
r
⋅
α
⋅
ρ
d
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 4.131
(Difficulty: 2)
4.131 For the configuration below calculate the torque about the pipe’s centerline in the plane of the
bolted flange that is caused by the flow through the nozzle. The nozzle center line is 0.3 𝑚 above the
flange centerline. What is the effect of this torque on the force on the bolts? Neglect the effects of the
weights of the pipe and the fluid in the pipe.
Given: All the parameters are shown in the figure.
Find: The effect of the torque.
Solution:
Basic equation:
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
The flow rate is:
𝐴2 =
The mass flow rate:
From x-momentum equation:
𝑀 = 𝑅𝑥 𝑟
𝑄 = 56.5
𝐿
𝑚3
= 0.0565
𝑠
𝑠
𝜋 2 𝜋
𝐷 = × (0.05 𝑚)2 = 0.00196 𝑚2
4 2 4
𝑚3
0.0565
𝑄
𝑠 = 28.83 𝑚
=
𝑉2 =
𝐴2 0.00196 𝑚2
𝑠
𝑚̇ = 𝜌𝜌 = 999
𝑚3
𝑘𝑘
𝑘𝑘
×
0.0565
= 56.4
3
𝑠
𝑚
𝑠
𝑅𝑥 = 𝑉2 𝑚̇ = 28.83
𝑚
𝑘𝑘
× 56.4
= 1626 𝑁
𝑠
𝑠
The torque can be calculated by:
𝑀 = 𝑅𝑥 𝑟 = 1626 𝑁 × 0.3 𝑚 = 488 𝑁 ∙ 𝑚
The direction is counter-clockwise.
The momentum increases the force on the upper bolts, the momentum decreases the force on the lower
bolts, so the total force is unchanged.
Problem 4.132
(Difficulty: 2)
4.132 A fire truck is equipped with a 66 ft long extension ladder which is attached at a pivot and raised
to an angle of 45°. A 4 in. diameter fire hose is laid up the ladder and a 2 in. diameter nozzle is attached
to the top of the ladder so that the nozzle directs the stream horizontally into the window of a burning
building. If the flow rate is 1 ft3/s. Compute the torque exerted about the ladder pivot point. The ladder,
hose and the water in the hose weigh about 10 lbf/ft.
Given: All the parameters are shown in the figure.
Find: Torque exerted about the ladder pivot point.
Solution:
Basic equations: Conservation of mass
0=
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
Conservation of momentum in x-direction
The flow rate is:
The nozzle velocity is
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝐴𝑁 =
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑄=1
𝑓𝑓 3
𝑠
2
𝜋 2 𝜋
2
𝐷𝑁 = × � 𝑓𝑓� = 0.0218 𝑓𝑓 2
4
4
12
𝑓𝑓 3
1
𝑄
𝑓𝑓
𝑠
=
= 45.8
𝑉𝑁 =
2
𝐴𝑁 0.0218 𝑓𝑓
𝑠
𝐴ℎ =
The mass flow rate:
2
𝜋 2 𝜋
4
𝐷ℎ = × �
𝑓𝑓� = 0.00873 𝑓𝑓 2
4
4
12
𝑚̇ = 𝜌𝜌 = 1.94
From x-momentum equation:
𝐹𝐵 = 10
𝑙𝑙𝑙 ∙ 𝑠 2
𝑓𝑓 3
𝑙𝑙𝑙 ∙ 𝑠
×
1
= 1.94
4
𝑓𝑓
𝑓𝑓
𝑠
𝑙𝑙𝑙
× 66 𝑓𝑓 = 660 𝑙𝑙𝑙
𝑓𝑓
𝐿
𝑅𝑅 + 𝐹𝐵 cos 45° = 𝑉𝑁 𝑚̇𝐿 sin 45°
2
𝑅𝑅 = 45.8
𝐿
𝑅𝑅 = 𝑉𝑁 𝑚̇𝐿 sin 45° − 𝐹𝐵 cos 45°
2
𝑙𝑙𝑙 ∙ 𝑠
𝑓𝑓
× 1.94
× 66 𝑓𝑓 × sin 45° − 660 𝑙𝑙𝑙 × 33 𝑓𝑓 × cos 45°
𝑓𝑓
𝑠
So the moment on the base by water is:
The direction is clockwise.
𝑅𝑅 = −11250 𝑙𝑙𝑙 ∙ 𝑓𝑓
𝑀 = −𝑅𝑅 = 11250 𝑙𝑙𝑙 ∙ 𝑓𝑓
Problem 4.133
(Difficulty: 2)
4.133 Calculate the torque exerted on the flange joint by the fluid flow as a function of the pump flow
rate. Neglect the weight of the 100 𝑚𝑚 diameter pipe and the fluid in the pipe.
Given: All the parameters are shown in the figure.
Find: Torque exerted on the flange.
Solution:
Basic equation:
0=
The area is:
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝐴𝑝 =
The mass flow rate:
From x-momentum equation:
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝜋 2 𝜋
𝐷 = × (0.1 𝑚)2 = 0.0079 𝑚2
4 𝑝 4
𝑉𝑝 =
𝑄
𝑄 𝑚
=
𝐴𝑝 0.0079 𝑠
𝑚̇ = 𝜌𝜌 = 999 𝑄
𝑅𝑀 = −𝑉𝑝 𝑚̇𝐿
𝑘𝑘
𝑠
𝑅𝑀 = −
𝑄 𝑚
𝑘𝑘
× 999 𝑄
× 2.5 𝑚 = −316000 𝑄2 𝑁 ∙ 𝑚 = −316𝑄2 𝑘𝑘 ∙ 𝑚
0.0079 𝑠
𝑠
So the torque on the flange is:
The direction is clockwise.
𝐹𝑀 = −𝑅𝑀 = 316𝑄2 𝑘𝑘 ∙ 𝑚
Problem 4.134
Problem
4.189
4.134
[Difficulty: 3]
4.130
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
⋅d
ρ = 999 ⋅
kg
3
m
Q
⋅
d = 5 ⋅ mm
V = 6.37
2
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 =
⌡
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
⎛
2
−⎜ ω⋅ R ⋅ V + α⋅
⎝
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 4.135
[Difficulty 4]
Problem 4.136
(Difficulty: 2)
4.136 The lawn sprinkler shown is supplied with water at a rate of 68
𝐿
𝑚𝑚𝑚
. Neglecting friction in the
pivot, determine the steady-state angular speed for 𝜃 = 30°. Plot the steady-state angular speed of the
sprinkler for 0 ≤ 𝜃 ≤ 90°.
Find: The angular speed 𝜔 of the sprinkler.
Assumption: (1) 𝐹𝑠 = 0
(2) Body torques cancel
(3) 𝑇�𝑠ℎ𝑎𝑎𝑎 = 0
(4) neglect aerodynamic drag
(5) no 𝑘� component of centripetal acceleration
(6) steady flow
Solution:
(7) 𝐿 ≪ 𝑅
Choose rotating CV. Apply angular momentum principle, Eq.4.53.
Basic equation:
� × 𝑉�𝑥𝑥𝑥 + 𝜔
� × (𝜔
� × 𝑟̅ ) + 𝜔
� × 𝑟̅ �𝜌𝜌∀
𝑟̅ × 𝐹�𝑠 + � 𝑟̅ × 𝑔̅ 𝜌𝜌∀ + 𝑇�𝑠ℎ𝑎𝑎𝑎 − � 𝑟̅ × �2𝜔
𝐶𝐶
𝐶𝐶
𝜕
= � 𝑟̅ × 𝑉�𝑥𝑥𝑥 𝜌𝜌∀ + � 𝑟̅ × 𝑉�𝑥𝑥𝑥 𝜌 𝑉�𝑥𝑥𝑥 ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
Analyze one arm of sprinkler. From geometry, 𝑟̅ = 𝑟𝚤̂ in CV. 𝑟̅ = 𝑅𝚤̂ at jet.
Then
− � 𝑟̅ × �2𝜔
� × 𝑉�𝑥𝑥𝑥 �𝜌𝜌∀= 𝑅𝚤̂ × (−𝑉 sin 𝜃𝚥̂) 𝜌
𝐶𝐶
𝑟𝚤̂ × �2𝜔𝑘� × 𝑉𝚤̂� = 2𝜔𝜔𝜔𝑘�
𝑄
𝑄𝑄𝑄
= −𝜌
sin 𝜃𝑘�
3
3
− � = −𝜔𝜔𝑅 2 𝜌𝜌𝑘�
𝐶𝐶
Dropping 𝑘� ,
−𝜔𝜔𝑅2 𝜌𝜌 = −
So with
𝑉𝑉 =
𝑉=
𝜔=
𝜌𝜌𝜌𝜌
sin 𝜃
3
𝑄
3
𝑉
sin 𝜃
𝑅
𝑄
4𝑄
4
1
𝑚𝑚𝑚
𝑚
𝑚3
−3
=
=
×
68
×
10
×
×
= 11.9
2
2
2
3𝐴 3𝜋𝑑
3𝜋
60 𝑠
𝑠
𝑚𝑚𝑚 (0.00635) 𝑚
𝜔 = 11.9
𝑚
1
𝑟𝑟𝑟
×
× sin 𝜃 = 78.3 sin 𝜃
𝑠 0.152 𝑚
𝑠
Problem 4.137
(Difficulty: 2)
4.137 A small lawn sprinkler is shown. The sprinkler operates at a gage pressure of 140 𝑘𝑘𝑘. The total
flow rate of water through the sprinkler is
4𝐿
𝑚𝑚𝑚
. Each jet discharges at 17
𝑚
𝑠
(relative to the sprinkler arm)
in a direction inclined 30° above the horizontal. The sprinkler rotates about a vertical axis. Friction in the
bearing causes a torque of 0.18 𝑁 ∙ 𝑚 opposing rotation. Evaluate the torque required to hold the
sprinkler stationary.
Find: The torque required to hold the sprinkler stationary.
Assumption: (1) neglect torque due to surface forces
(2) torques due to body forces cancel by symmetry
(3) steady flow
(4) uniform flow leaving each jet
Solution:
Apply moment of momentum using fixed CV enclosing sprinkler arms.
Basic equation:
Then
𝑟̅ × 𝐹�𝑠 + � 𝑟̅ × 𝑔̅ 𝜌𝜌∀ + 𝑇�𝑠ℎ𝑎𝑎𝑎 =
𝐶𝐶
𝜕
� 𝑟̅ × 𝑉� 𝜌𝜌∀ + � 𝑟̅ × 𝑉� 𝜌 𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
1
−𝑇𝑓 𝑘� = (𝑟̅ × 𝑉� )𝑖𝑖 {−𝜌𝜌} + 2(𝑟̅ × 𝑉� )𝑗𝑗𝑗 � 𝜌𝜌�
2
(𝑟̅ × 𝑉� )𝑖𝑖 ≈ 0
𝑉� = (𝑅𝑅 − 𝑉𝑟𝑟𝑟 cos 𝛼)𝚤̂𝜃 + 𝑉𝑟𝑟𝑟 sin 𝛼 𝚤̂𝑧
The absolute velocity of the jet leaving sprinkler is
𝑉� = 𝑉𝑟𝑟𝑟 [cos 𝛼 (−𝚤̂𝜃 ) + sin 𝛼𝚤̂𝑧 ]
Then
(𝑟̅ × 𝑉� )𝑧 = {𝑅𝚤̂𝑟 × 𝑉𝑟𝑟𝑟 [cos 𝛼(−𝚤̂𝜃 ) + sin(𝚤̂𝑧 )]}𝑧 = {𝑅𝑉𝑟𝑟𝑟 cos 𝛼(−𝚤̂𝑧 ) + 𝑅𝑉𝑟𝑟𝑟 sin 𝛼 (−𝚤̂𝜃 )}𝑧
(𝑟̅ × 𝑉� )𝑧 = −𝑅𝑉𝑟𝑟𝑟 cos 𝛼
Substituting,
1
𝑇�𝑠ℎ𝑎𝑎𝑎 = 𝑇𝑒𝑒𝑒 − 𝑇𝑓 = 2(−𝑅𝑉𝑟𝑟𝑟 cos 𝛼) � 𝜌𝜌�
2
Thus
𝑇𝑒𝑒𝑒 = 0.18 𝑁 ∙ 𝑚 − 999
to hold sprinkler stationary.
𝑇𝑒𝑒𝑒 = 𝑇𝑓 − 𝜌𝜌𝑅𝑉𝑟𝑟𝑟 cos 𝛼
𝑘𝑘
𝐿
𝑚
𝑚3
𝑚𝑚𝑚 𝑁 ∙ 𝑠 2
×
4
×
0.2
𝑚
×
17
×
0.866
×
×
×
𝑚3
𝑚𝑚𝑚
𝑠
1000 𝐿 60 𝑠 𝑘𝑘 ∙ 𝑚
𝑇𝑒𝑒𝑒 = −0.0161 𝑁 ∙ 𝑚
Since 𝑇𝑒𝑒𝑒 < 0, it must be applied in the minus z direction to oppose motion.
Problem *4.178
4.138
Problem 4.138
Problem
4.198
[Difficulty: 4]
Problem 4.139
(Difficulty: 2)
4.139 A pipe branches symmetrically into two legs of length 𝐿, and the whole system rotates with
angular speed 𝜔 around its axis of symmetry. Each branch is inclined at angle 𝛼 to the axis of rotation.
Liquid enters the pipe steadily, with zero angular momentum, at volume flow rate 𝑄. The pipe diameter,
𝐷, is much smaller than 𝐿. Obtain an expression for the external torque required to turn the pipe. What
additional torque would be required to impart angular acceleration 𝜔̇ ?
Find: The torque required to hold the sprinkler stationary.
Assumption: (1) no surface forces
(2) body forces produce no torque about axis
(3) flow steady in the rotating frame
Solution:
(4) 𝑟̅ and 𝑉�𝑥𝑥𝑥 are: 𝑟̅ × 𝑉�𝑥𝑥𝑥 = 0.
Apply moment of momentum equation using rotating CV.
Basic equation:
� × 𝑉�𝑥𝑥𝑥 + 𝜔
� × (𝜔
� × 𝑟̅ ) + 𝜔
�̇ × 𝑟̅ � 𝜌𝜌∀
𝑟̅ × 𝐹�𝑠 + � 𝑟̅ × 𝑔̅ 𝜌𝜌∀ + 𝑇�𝑠ℎ𝑎𝑎𝑎 − � 𝑟̅ × �2𝜔
𝐶𝐶
Then
𝐶𝐶
𝜕
= � 𝑟̅ × 𝑉�𝑥𝑥𝑥 𝜌𝜌∀ + � 𝑟̅ × 𝑉�𝑥𝑥𝑥 𝜌 𝑉�𝑥𝑥𝑥 ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑇�𝑠ℎ𝑎𝑎𝑎 = � 𝑟̅ × �2𝜔
� × 𝑉�𝑥𝑥𝑥 + 𝜔
� × (𝜔
� × 𝑟̅ ) + 𝜔
�̇ × 𝑟̅ � 𝜌𝜌∀
Using the coordinates above:
𝐶𝐶
𝜔
� = 𝜔𝑘�
𝜔
�̇ = 𝜔̇ 𝑘�
𝑟̅ = 𝑟�cos 𝛼𝑘� + sin 𝛼 𝚤̂� (upper tube)
𝑉�𝑥𝑥𝑥 =
𝑄
2𝐴
�cos 𝛼𝑘� + sin 𝛼 𝚤̂� (upper tube)
𝐴=
And
𝜋𝐷 2
4
𝜔
�̇ × 𝑟̅ = 𝜔̇ 𝑟 sin 𝛼 𝚥̂
𝜔
� × (𝜔
� × 𝑟̅ ) = 𝜔𝑘� × 𝜔𝜔 sin 𝛼 𝚥̂ = −𝜔2 𝑟 sin 𝛼 𝚤̂
Thus for the upper tube,
2𝜔
� × 𝑉�𝑥𝑥𝑥 = 2𝜔
𝑄
𝜔𝜔
sin 𝛼𝚥̂ =
sin 𝛼 𝚥̂
2𝐴
𝐴
𝐿
𝜔𝜔
𝑇�𝑠ℎ𝑎𝑎𝑎 = � �𝑟�cos 𝛼𝑘� + sin 𝛼 𝚤̂� × ��
+ 𝜔̇ 𝑟� sin 𝛼 𝚥̂ − 𝜔2 𝑟 sin 𝛼 𝚤̂�� 𝜌𝜌𝜌𝜌
𝐴
0
𝑇�𝑠ℎ𝑎𝑎𝑎 = ��
𝐿2 𝜔𝜔 𝜔̇ 𝐿3
𝐿2 𝜔𝜔 𝜔̇ 𝐿3
𝜔2 𝐿3
+
� sin 𝛼 cos 𝛼 𝚤̂ + �
+
� sin2 𝛼𝑘� +
sin 𝛼 cos 𝛼 (−𝚥̂)� 𝜌𝜌
3
3
3
2𝐴
2𝐴
For the lower tube:
𝜔
� = 𝜔𝑘�
𝜔
�̇ = 𝜔̇ 𝑘�
𝑟̅ = 𝑟�cos 𝛼𝑘� − sin 𝛼 𝚤̂� (lower tube)
And
𝑉�𝑥𝑥𝑥 =
𝑄
2𝐴
�cos 𝛼𝑘� − sin 𝛼 𝚥̂� (lower tube)
𝜔
�̇ × 𝑟̅ = −𝜔̇ 𝑟 sin 𝛼 𝚥̂
𝜔
� × (𝜔
� × 𝑟̅ ) = 𝜔𝑘� × (−𝜔𝜔 sin 𝛼 𝚥̂) = 𝜔2 𝑟 sin 𝛼 𝚤̂
So for lower tube:
2𝜔
� × 𝑉�𝑥𝑥𝑥 = 2𝜔
𝑄
𝜔𝜔
(−sinα)𝚥̂ = −
sin 𝛼 𝚥̂
2𝐴
𝐴
𝐿
𝑇�𝑠ℎ𝑎𝑎𝑎 = � �𝑟�cos 𝛼𝑘� − sin 𝛼 𝚤̂� × ��
0
𝜔𝜔
+ 𝜔̇ 𝑟� sin 𝛼 (−𝚥̂) + 𝜔2 𝑟 sin 𝛼 𝚤̂�� 𝜌𝜌𝜌𝜌
𝐴
𝐿2 𝜔𝜔 𝜔̇ 𝐿3
𝐿2 𝜔𝜔 𝜔̇ 𝐿3
𝜔2 𝐿3
+
� sin 𝛼 cos 𝛼 𝚤̂ + �
+
� sin2 𝛼𝑘� +
sin 𝛼 cos 𝛼 (𝚥̂)� 𝜌𝜌
3
3
3
2𝐴
2𝐴
𝑇�𝑠ℎ𝑎𝑎𝑎 = ��
Summing these expressions gives:
𝐿2 𝜔𝜔 2𝜔̇ 𝐿3
+
� sin2 𝛼𝛼𝛼𝑘�
3
𝐴
𝑇�𝑠ℎ𝑎𝑎𝑎 = �
The steady state portion of the torque is:
𝑇�𝑠ℎ𝑎𝑎𝑎 =
𝐿2 𝜔𝜔 2
sin 𝛼𝛼𝛼𝑘� = 𝐿2 𝜌𝜌𝜌sin2 𝛼𝑘�
𝐴
The additional torque need to provide angular acceleration 𝜔̇ is:
𝑇�𝑠ℎ𝑎𝑎𝑎 =
2𝜔̇ 𝜌𝐿3 𝐴 2
sin 𝛼𝑘�
3
Torques of individual tubes about the x and y axis are reacted internally. They must be considered in
design of the tube.
(b) Using fixed CV:
Assumption: (1) no surface forces
(2) body forces symmetric
(3) no change in angular momentum within CV
(4) symmetric in two branches
(5) uniform flow at each cross-section
𝑄
𝑄
𝑄
𝑇�𝑠 = 𝑟̅1 × 𝑉�1 {−𝜌𝜌} + 𝑟̅2 × 𝑉�2 �𝜌 � + 𝑟̅3 × 𝑉�3 �𝜌 � = 2𝑟̅2 × 𝑉�2 �𝜌 �
2
2
2
𝑟̅1 = 0
𝑟̅2 = 𝐿 sin 𝛼 𝚥̂
𝑉�2 = 𝜔𝑟2 𝑘�
Or
steady state torque.
𝑟̅2 × 𝑉�2 = 𝜔𝐿2 sin2 𝛼 𝚤̂
𝑇𝑠 = 𝜌𝜌𝜌𝐿2 sin2 𝛼
The torque required for acceleration is:
𝑇𝑎𝑎𝑎 = 𝐼𝜔̇
Where
𝐼 = � 𝑟 2 𝑑𝑑
For one leg of the branch,
𝐿
(b) Neglect mass of pipe
𝐼 = � 𝑟 2 𝑑𝑑 = � (𝑠 sin 𝛼)2 𝜌𝜌𝜌𝜌 =
0
𝜌𝜌𝐿3 2
sin 𝛼
3
For both sides,
𝐼=
Thus
Torque required for angular acceleration.
𝑇𝑎𝑎𝑎 =
2𝜌𝜌𝐿3 2
sin 𝛼
3
2𝜌𝜔̇ 𝐴𝐿3 2
sin 𝛼
3
The total torque that must be applied is:
𝑇 = 𝑇𝑠 + 𝑇𝑎𝑎𝑎 = 𝜌𝜌𝜌𝐿2 sin2 𝛼 +
2𝜌𝜔̇ 𝐴𝐿3 2
sin 𝛼
3
Problem 4.140
(Difficulty: 2)
4.140 For the rotating sprinkler of Example 4.14, what value of 𝛼 will produce the maximum rotational
speed? What angle will provide the maximum area of coverage by the spray? Draw a velocity diagram
(using an 𝑟, 𝜃, 𝑧 coordinate system) to indicate the absolute velocity of the water jet leaving the nozzle.
What governs the steady rotational speed of the sprinkler? Does the rotational speed of the sprinkler
affect the area covered by the spray? How would you estimate the area? For fixed 𝛼, what might be
done to increase or decrease the area covered by the spray?
Solution:
The results of Example Problem 4.14 were computed assuming steady flow of water and constant
frictional retarding torque at the sprinkler pivot.
From these results,
𝑇𝑓 = 𝑅(𝑉𝑟𝑟𝑟 cos 𝛼 − 𝜔𝜔)𝜌𝜌
𝜔=
𝑇𝑓
𝑉𝑟𝑟𝑟 cos 𝛼
−
𝜌𝜌𝑅2
𝑅
Thus rotational speed of the sprinkler increases as cos 𝛼 increases, i.e. as 𝛼 decreases. The maximum
rotational speed occurs when 𝛼 = 0. Then cos 𝛼 = 1 and the rotational speed is
𝜔=
𝑇𝑓
𝑉𝑟𝑟𝑟
−
𝑅
𝜌𝜌𝑅2
For the conditions of Example Problem 4.14 the maximum rotational speed is
𝜔 = 4.97
1
𝑚3
𝑚𝑚𝑚
1
𝐿
𝑠
𝑚
×
− 0.0718 𝑁 ∙ 𝑚 ×
×
×
× 1000 3 × 60
2
𝑚
𝑚𝑚𝑚
999 𝑘𝑘 7.5 𝐿 (0.150 𝑚)
𝑠 0.150 𝑚
𝜔 = 7.58
𝑟𝑟𝑟
𝑠
The steady rotation speed 𝜔 of the sprinkler is governed by torque 𝑇𝑓 and angle 𝛼.
Maximum coverage by the spray occurs when the “carry” of each jet stream is the longest. When
aerodynamic drag on the stream is neglected, maximum carry occurs when the absolute velocity of the
stream leaves the sprinkler at 𝛽 = 45°, as shown in the velocity diagram below.
Note:
�⃗𝑎𝑎𝑎 = 𝑉
�⃗𝑟𝑟𝑟 − 𝜔𝜔𝚤̂𝜃
𝑉
�⃗𝑎𝑎𝑎 vary with 𝜔.
Both the magnitude and direction of 𝑉
For 𝜔 = 0, the relative velocity angle 𝛼 and absolute velocity angle 𝛽 are equal. Therefore maximum
carry occurs when 𝛼 = 45° (see graph on next page).
Any rotation rate 𝜔 reduces the magnitude 𝑉𝑎𝑎𝑎 and increases the angle 𝛽 of the absolute velocity leaving
the sprinkler jet. When 𝜔 > 0, then 𝛽 > 𝛼, so for maximum carry 𝛼 must be less than 45°. Consequently
rotation reduces the carry of the stream and the area of coverage; at specified 𝛼 the area of coverage
decreases with increasing 𝜔.
For the conditions of Example Problem 4.14 (𝜔 = 30 𝑟𝑟𝑟), optimum carry occurs at 𝛼 = 42°, and the
coverage area is reduced from approximately 20 𝑚2 with a fixed sprinkler to 15 𝑚2 with 30 𝑟𝑟𝑟
rotation. If the rotation speed is increased (by decreasing pivot friction or decreasing nozzle angle 𝛼),
coverage area may be reduced still further, to 9 𝑚2 or less.
𝐴 ≈ 𝜋(𝑥𝑚𝑚𝑚 )2
Problem 4.141
Problem
4.204
[Difficulty: 3]
4.141
Given:
Compressed air bottle
Find:
Rate of temperature change
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
kg⋅ K
M CV + mexit = 0
T = 293K
cv = 717.4⋅
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
(
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
Hence
kg
3
m
2
1
L
kg⋅ K
3 N
= −0.01⋅
× 500 × 10 ⋅
×
×
×
×
2
−3 3
s
100 ⋅ L
717.4 ⋅ N⋅ m
dt
m
10 ⋅ m
dT
kg
⎛ m3 ⎞
K
C
⎜
= −1.97⋅ = −1.97⋅
s
s
⎝ 5.95⋅ kg ⎠
Problem 4.142
(Difficulty: 2)
4.142 A turbine is supplied with 0.6
𝑚3
𝑠
of water from a 0.3 − 𝑚 − 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 pipe; the discharge pipe
has a 0.4 𝑚 diameter. Determine the pressure drop across the turbine if it delivers 60 𝑘𝑘.
𝑄1 = 0.6
𝑚3
𝑠
𝐷1 = 0.3 𝑚
Find: The pressure drop 𝑝1 − 𝑝2 .
Assumption: (1) steady flow
(2) uniform flow at each section
(3) incompressible flow
(4) 𝑄̇ = 0
(5) 𝑊̇𝑠ℎ𝑒𝑒𝑒 = 0
(6) neglect ∆𝑢
Solution:
(7) neglect ∆𝛿
Choose rotating CV. Apply angular momentum principle, Eq.4.53.
Basic equation:
0=
Then
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑄̇ − 𝑊̇𝑠 − 𝑊̇𝑠ℎ𝑒𝑒𝑒 − 𝑊̇𝑜𝑜ℎ𝑒𝑒 =
𝑉2
𝜕
� 𝑒𝑒𝑒∀ + � �𝑢 +
+ 𝑔𝑔 + 𝑝𝑝� 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
2
𝐶𝐶
0 = {−|𝜌𝑉1 𝐴1 |} + {|𝜌𝑉2 𝐴2 |}
or
𝑉2 = 𝑉1
and
𝐴1
𝐷1 2
= 𝑉1 � �
𝐴2
𝐷2
𝑉12
𝑉22
−𝑊̇𝑠 = � + 𝑝1 𝑣� {−|𝜌𝑉1 𝐴1 |} + � + 𝑝2 𝑣� {|𝜌𝑉2 𝐴2 |}
2
2
−𝑊̇𝑠 = − �
or
𝑉12
𝐷1 4
𝑉12 − 𝑉22
+ (𝑝1 − 𝑝2 )𝑣� 𝜌𝜌 = − � �1 − � � � + (𝑝1 − 𝑝2 )𝑣� 𝜌𝜌
2
2
𝐷2
𝑝1 − 𝑝2 =
But
𝑊̇𝑠 𝜌𝜌12
1 𝑊̇𝑠 𝑉12
𝐷1 4
𝐷1 4
�1 − � � �� =
�1 − � � �
� −
−
𝑣 𝜌𝜌
2
𝐷2
𝑄
2
𝐷2
𝑉1 =
So we have:
𝑄
𝑚3 4
1
𝑚
= 0.6
× ×
= 8.49
2
𝐴1
𝜋 (0.3 𝑚)
𝑠
𝑠
𝑊̇𝑠 = 𝑊̇𝑜𝑜𝑜 = 60 𝑘𝑘
𝑠
1
𝑘𝑘
𝑚 2
𝑁∙𝑚
0.3 𝑚 4 𝑁 ∙ 𝑠 2
� �
�8.49
�
�1
�
𝑝1 − 𝑝2 = 60 × 10
×
− × 999 3 ×
× −
0.6 𝑚3 2
𝑚
𝑠
𝑘𝑘 ∙ 𝑚
𝑠
0.4 𝑚
3
𝑝1 − 𝑝2 = 75.4 𝑘𝑘𝑘
Problem 4.143
(Difficulty: 2)
4.143 Air is drawn from atmosphere into a turbo-machine. At the exit, conditions are 500 𝑘𝑘𝑘 (gage)
and 130℃. The exit speed is 100
𝑚
𝑠
and the mass flow rate is 0.8
𝑘𝑘
𝑠
. Flow is steady and there is no heat
transfer. Compute the shaft work interaction with the surroundings.
Find: The shaft work 𝑊̇𝑠 interaction with the surroundings.
Assumption: (1) ideal gas, constant specific heat
(2) 𝑤̇𝑠ℎ𝑒𝑒𝑒 = 0 by choice of CV. 𝑤̇𝑜𝑜ℎ𝑒𝑒 = 0
(3) steady flow
(4) uniform flow at each section
(5) neglect ∆𝛿
(6) 𝑉1 = 0
Solution:
(7) 𝑄̇ = 0
Apply energy equation, using CV shown.
Basic equation:
𝑝 = 𝜌𝜌𝜌
𝑄̇ − 𝑊̇𝑠 − 𝑊̇𝑠ℎ𝑒𝑒𝑒 − 𝑊̇𝑜𝑜ℎ𝑒𝑒 =
By definition,
∆ℎ = 𝑐𝑝 ∆𝑇
𝑉2
𝜕
� 𝑒𝑒𝑒∀ + � �𝑢 +
+ 𝑔𝑔 + 𝑝𝑝� 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
2
𝐶𝐶
ℎ = 𝑢 + 𝑝𝑝
so we have:
Or
−𝑊̇𝑠 = �ℎ1 +
−𝑊̇𝑠 = 0.8
𝑉12
𝑉22
𝑉22
� {−|𝑚̇|} + �ℎ2 + � {|𝑚̇|} = 𝑚̇ �ℎ2 − ℎ1 + �
2
2
2
−𝑊̇𝑠 = 𝑚̇ �ℎ2 − ℎ1 +
𝑉22
𝑉22
� = 𝑚̇ �𝑐𝑝 (𝑇2 − 𝑇1 ) + �
2
2
𝑘𝑘
𝑘𝑘
1
𝑚 2
× �1.0
× (405 − 288) 𝐾 + × �100 � � = 96.0 𝑘𝑘
𝑠
𝑘𝑘 ∙ 𝐾
2
𝑠
Power is into CV because 𝑊̇𝑠 < 0.
𝑊̇𝑠 = −96.0 𝑘𝑘
Problem 4.144
(Difficulty: 2)
4.144 At high speeds the compressor and turbine of the jet engine may be eliminated entirely. The
result is called a ramjet (a subsonic configuration is shown). Here the incoming air is slowed and the
pressure increases; the air is heated in the widest part by the burning of injected fuel. The heated air
exhausts at high velocity from the converging nozzle. What nozzle area 𝐴2 is needed to deliver a 90 𝑘𝑘
thrust at an air speed of 270
𝑚
𝑠
if the exhaust velocity is the sonic velocity for the heated air, which is at
1000 𝐾. Assume that the jet operates at an altitude of 12 𝑘𝑘 and neglect the fuel mass and pressure
differentials.
Given: All the parameters are shown in the figure.
Find: Nozzle area 𝐴2 .
Solution:
Basic equation:
0=
The parameters are:
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑢𝑢𝑢∀ + � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑉1 = 270
𝑚
𝑠
𝐹𝑝 = 90 𝑘𝑘
𝑝 = 19.4 𝑘𝑘𝑘
The mass flow rate:
𝑇 = 1000 𝐾
𝑚̇ = 𝜌2 𝑉2 𝐴2
From x-momentum equation:
𝐹𝑝 = 𝑚̇(𝑉2 − 𝑉1 ) = 𝜌2 𝑉2 𝐴2 (𝑉2 − 𝑉1 )
𝐴2 =
𝐹𝑝
𝜌2 𝑉2 (𝑉2 − 𝑉1 )
1
𝑉2 = (𝑘𝑘𝑘)2 = 633.7
𝐴2 =
𝜌2 =
𝑚
𝑠
𝑝
𝑘𝑘
= 0.0676 3
𝑅𝑅
𝑚
90 × 103 𝑁
= 5.78 𝑚2
𝑘𝑘
𝑚
𝑚
𝑚
0.0676 3 × 633.7 × �633.7 − 270 �
𝑠
𝑠
𝑠
𝑚
Problem 4.145
(Difficulty: 2)
4.145 Transverse thrusters are used to make large ships fully maneuverable at low speeds without
tugboat assistance. A transverse thruster consists of a propeller mounted in a duct; the unit is then
mounted below the waterline in the bow or stern of the ship. The duct runs completely across the ship.
Calculate the thrust developed by a 1865 𝑘𝑘 unit (supplied to the propeller) if the duct is 2.8 𝑚 in
diameter and the ship is stationary.
Given: All the parameters are shown in the figure.
Find: The thrust developed.
Solution:
𝑃𝑢𝑢𝑢𝑢 = 1865 𝑘𝑘
𝐴=
𝑃𝑢𝑢𝑢𝑢 =
Since
3
𝑉4 = �
𝜋 2
𝐷
4
𝐴𝐴𝐴 2
(𝑉4 − 𝑉12 )
2
𝑉1 = 0
2𝑃𝑢𝑢𝑢𝑢 3 2 × 1865 × 1000 𝑤
𝑚
=�
= 8.46
𝑘𝑘
𝜋
𝑠
𝐴𝐴
× (2.8 𝑚)2 × 999 3
4
𝑚
𝐹=
𝑃𝑢𝑢𝑢𝑢 1865 𝑘𝑘
=
𝑚 = 220000 𝑁 = 220 𝑘𝑘
𝑉
8.46
𝑠
Problem 4.146*
Problem
4.209
[Difficulty: 3]
*Please Note: this solution shows the solution to the problem with one particular set of values. These values may not match those in the problem you have been given to solve.
4.146
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
s
⎝2
⎠
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 4.147
(Difficulty: 2)
4.147 A pump draws water from a reservoir through a reservoir through a 150 − 𝑚𝑚 − 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
suction pipe and delivers it to a 75 − 𝑚𝑚 − 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 discharge pipe. The end of the suction pipe is
2 𝑚 below the free surface of the reservoir. The pressure gage on the discharge pipe (2 𝑚 above the
reservoir surface) reads 170 𝑘𝑘𝑘. The average speed in the discharge pipe is 3
is 75 percent, determine the power required to drive it.
𝑚
𝑠
. If the pump efficiency
Find: The power 𝑊̇𝑠,𝑎𝑎𝑎𝑎𝑎𝑎 required to drive it.
Assumption: (1) 𝑊̇𝑠ℎ𝑒𝑒𝑒 = 𝑊̇𝑜𝑜ℎ𝑒𝑒 = 0
(2) steady flow
(3) 𝑉1 ≅= 0
(4) 𝛿1 = 0
(5) 𝑝1 = 0 (gage)
(6) uniform flow at each section
Solution:
(7) incompressible flow; 𝑉1 𝐴1 = 𝑉2 𝐴2
Apply the first law of CV shown. Noting that flow enters with negligible velocity at section ①
Basic equation:
or
𝑄̇ − 𝑊̇𝑠 = �𝑢1 +
𝑉12
𝑝1
𝑉22
𝑝2
+ 𝑔𝑧1 + � {−𝑚̇} + �𝑢2 +
+ 𝑔𝑧2 + � {𝑚̇}
2
𝜌
2
𝜌
−𝑊̇𝑠 = 𝑚̇ �
𝛿𝛿
𝑝2 𝑉22
��
+
+ 𝑔𝑧2 + �𝑢2 − 𝑢1 −
𝑑𝑑
𝜌
2
Obtain the ideal or minimum power input by neglecting terminal effects, thus
𝑝2 𝑉22
+ 𝑔𝑧2 �
−𝑊̇𝑠,𝑖𝑖𝑖𝑖𝑖 = 𝑚̇ � +
𝜌
2
For the system,
and
𝑚̇ = 𝜌𝑉2 𝐴2 = 999
−𝑊̇𝑠,𝑖𝑖𝑖𝑖𝑖 = 13.2
𝑘𝑘
𝑚 𝜋
𝑘𝑘
× 3 × × (0.075 𝑚)2 = 13.2
3
𝑚
𝑠 4
𝑠
𝑘𝑘
𝑚3
1
𝑚 2
𝑚
𝑁
× �1.70 × 108 2 ×
+ × �3 � + 9.81 2 × 2𝑚�
𝑠
𝑠
𝑠
999 𝑘𝑘 2
𝑚
𝑊̇𝑠,𝑖𝑖𝑖𝑖𝑖 = −2.56 𝑘𝑘
Finally
𝑊̇𝑠,𝑎𝑎𝑎𝑎𝑎𝑎 =
𝑊̇𝑠,𝑖𝑖𝑖𝑖𝑖 −2.56 𝑘𝑘
=
= −3.41 𝑘𝑘
0.75
𝜂
Problem 4.148
(Difficulty: 2)
4.148 Liquid flowing at high speed in a wide, horizontal open channel under some conditions can
undergo a hydraulic jump, as shown. For a suitable chosen control volume, the flows entering and
leaving the jump may be considered uniform with hydrostatic pressure distributions (see Example 4.7).
Consider a channel of width 𝑤, with water flow at 𝐷1 = 6 𝑚𝑚 and 𝑉1 = 5
𝐷2 =
1
8𝑉12
𝐷1 ��1 +
− 1�
2
𝑔𝐷1
𝑚
𝑠
. Show that in general,
Evaluate the change in mechanical energy through the hydraulic jump. If heat transfer to the
surroundings is negligible, determine the change in water temperature through the jump.
Find: The change in mechanical energy and water temperature through the pump.
Assumption: (1) steady flow
(2) incompressible flow
(3) uniform flow at each section
(4) hydrostatic pressure distribution at section ① and ②, so 𝑝 = 𝜌𝜌(𝐷 − 𝑧)
(5) neglect friction force, 𝐹𝑓 , on CV.
(6) 𝑄̇ = 0
(7) 𝑊̇𝑠 = 𝑊̇𝑠ℎ𝑒𝑒𝑒 = 𝑊̇𝑜𝑜ℎ𝑒𝑒 = 0
Solution:
(8) 𝐹𝐵𝐵 = 0, since channel is horizontal.
Apply continuity, x component of momentum, and energy equations using CV shown.
Basic equation:
0=
𝜕
� 𝜌𝜌∀ + � 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝐹𝑠𝑠 + 𝐹𝐵𝐵 =
𝜕
� 𝑉 𝜌𝜌∀ + � 𝑉𝑥 𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶 𝑥
𝐶𝐶
𝑄̇ − 𝑊̇𝑠 − 𝑊̇𝑠ℎ𝑒𝑒𝑒 − 𝑊̇𝑜𝑜ℎ𝑒𝑒 =
𝜕
� 𝑒𝑒𝑒∀ + � (𝑒 + 𝑝𝑝)𝜌𝑉� ∙ 𝑑𝐴̅
𝜕𝜕 𝐶𝐶
𝐶𝐶
𝑒=𝑢+
From continuity,
𝑉2
+ 𝑔𝑔
2
0 = {−|𝜌𝑉1 𝐴1 |} + {|𝜌𝑉2 𝐴2 |} = −𝜌𝑉1 𝑤𝑤1 + 𝜌𝑉2 𝑤𝑤2
𝑉1 𝐷1 = 𝑉2 𝐷2
From momentum,
𝐹𝑠𝑠 = 𝜌𝜌
𝐷1
𝐷2
𝑤𝐷1 − 𝜌𝜌 𝑤𝐷2 = 𝑉𝑥1 {−|𝜌𝑉1 𝑤𝑤1 |} + 𝑉𝑥2 {−|𝜌𝑉2 𝑤𝑤2 |}
2
2
𝑉𝑥1 = 𝑉1
or
𝑉𝑥2 = 𝑉2
or
𝑔 2
𝑉2
𝐷1
(𝐷1 − 𝐷22 ) = 𝑉1 𝐷1 (𝑉2 − 𝑉1 ) = 𝑉12 𝐷1 � − 1� = 𝑉12 𝐷1 � − 1�
2
𝑉1
𝐷2
Thus
𝑔
𝐷1
(𝐷1 + 𝐷2 ) = 𝑉12
2
𝐷2
or
𝐷2
𝑔𝐷1
𝐷1
�1 + � = 𝑉12
2
𝐷1
𝐷2
or
𝐷2
2𝑉12
𝐷2
�1 + � =
𝐷1
𝐷1
𝑔𝐷1
Using the quadratic equation,
𝐷2 2 𝐷2 2𝑉12
� � +
−
=0
𝐷1
𝐷1 𝑔𝐷1
8𝑉12
𝐷2 1
= �−1 ± �1 +
�
𝐷1 2
𝑔𝐷1
or
𝐷1
8𝑉12
�
𝐷2 =
� 1+
− 1�
2
𝑔𝐷1
Solving for 𝐷2 ,
𝐷2 =
1
𝑚 2
𝑠2
1
× 0.6 𝑚 ��1 + 8 × �5 � ×
×
− 1� = 1.47 𝑚
2
𝑠
9.81 𝑚 0.6 𝑚
From the energy equation with
𝑉2 =
𝐷1
0.6
𝑚
𝑚
𝑉1 =
× 5 = 2.04
1.47
𝑠
𝑠
𝐷2
𝑒𝑚𝑚𝑚ℎ =
The mechanical energy fluxes are
𝑚𝑚𝑚1 = �
𝐷1
0
𝑚𝑚𝑚2 = �
𝐷2
0
and
�
𝑉2
𝑝
+ 𝑔𝑔 +
𝜌
2
𝑑𝑑 = 𝑤𝑤𝑤
1
𝑉12
𝑉12
+ 𝑔𝑔 + 𝜌𝜌(𝐷 − 𝑧)� 𝜌𝑉1 𝑤𝑤𝑤 = � + 𝑔𝐷1 � 𝜌𝑉1 𝑤𝐷1
𝜌
2
2
1
𝑉22
𝑉22
� + 𝑔𝑔 + 𝜌𝜌(𝐷 − 𝑧)� 𝜌𝑉2 𝑤𝑤𝑤 = � + 𝑔𝐷2 � 𝜌𝑉2 𝑤𝐷2
𝜌
2
2
Since
𝑉22 − 𝑉12
+ 𝑔(𝐷2 − 𝐷1 )� 𝜌𝑉1 𝑤𝐷1
∆𝑚𝑚𝑚 = 𝑚𝑚𝑚2 − 𝑚𝑚𝑚1 = �
2
Thus
𝑉1 𝐷1 = 𝑉2 𝐷2
∆𝑚𝑚𝑚 1 2
= [𝑉2 − 𝑉12 + 2𝑔(𝐷2 − 𝐷1 )]
2
𝑚̇
𝑚 2
𝑚 2
𝑚
𝑁 ∙ 𝑠2
𝑁∙𝑚
∆𝑚𝑚𝑚 1
(1.47
��2.04
�
�
�5
=
−
+ 2 × 9.81 2 ×
𝑚 − 0.6 𝑚)�
= −1.88
𝑠
2
𝑠
𝑠
𝑘𝑘
𝑘𝑘 ∙ 𝑚
𝑚̇
From the energy equation
or
0 = �𝑢1 +
𝑉12
1
𝑉22
1
+ 𝑔𝑔 + 𝜌𝜌(𝐷 − 𝑧)� {−|𝜌𝑉1 𝑤𝑤1 |} + �𝑢2 +
+ 𝑔𝑔 + 𝜌𝜌(𝐷 − 𝑧)� {|𝜌𝑉2 𝑤𝑤2 |}
𝜌
𝜌
2
2
0 = (𝑢2 − 𝑢1 )𝑚̇ + ∆𝑚𝑚𝑚
Thus
∆𝑇 = 𝑇2 − 𝑇1 = −
𝑢2 − 𝑢1 = 𝑐𝑣 (𝑇2 − 𝑇1 ) = −
∆𝑚𝑚𝑚
𝑚̇
∆𝑚𝑚𝑚
𝑁 ∙ 𝑚 𝑘𝑘 ∙ 𝐾
𝑘𝑘𝑘𝑘
�
= − �−1.88
×
= 4.49 × 10−4 𝐾
𝑚̇𝑐𝑣
𝑘𝑘 1 𝑘𝑘𝑘𝑘 4187 𝐽
This small temperature change would be almost impossible to measure.
Problem 6.1
Problem
6.2
[Difficulty: 2]
6.1
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
2
m
N⋅s
kg
× 20⋅
×
p = ρ⋅ g x − ρ⋅ ax = −999⋅
3
2
kg⋅ m
∂x
s
m
∂
∂
∂x
2
m
N⋅s
kg
× ( −9.81 − 20) ⋅
×
p = −ρ⋅ g y − ρ⋅ ay = 999⋅
3
2
kg⋅ m
∂y
s
m
∂
∂
∂y
p = −20000⋅
Pa
p = −29800⋅
Pa
m
m
= −20.0⋅
kPa
= −29.8⋅
kPa
m
m
Problem 6.2
Problem
6.4
[Difficulty: 2]
6.2
Given:
Velocity field
Find:
Pressure gradient at (1,1) at 1 s
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 6.3
Problem
6.6
[Difficulty: 2]
6.3
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
v ( x, y ) = −⎮
u dy = −⎮ A dy = −A ⋅ y + c
u +
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
∂
2
∂y
( −A⋅ y ) + ( −A⋅ y ) ⋅
2
( A ⋅ x) = A ⋅ x
∂
∂y
ax = A ⋅ x
2
ay = A ⋅ y
( −A⋅ y )
2
⎛ 2 ⎞ × 1⋅ m
⎜
⎝ s⎠
⎛ ay ⎞
θ = atan⎜
⎝ ax ⎠
ay =
ax = 8
m
ay = 4
2
s
a = 8.94
m
θ = 26.6⋅ deg
2
s
2
m
N⋅ s
kg
× 8⋅ ×
p = ρ⋅ g x − ρ⋅ ax = −1.50⋅
3
2
kg⋅ m
∂x
s
m
∂
∂x
2
m
N⋅ s
kg
× 4⋅ ×
p = ρ⋅ g y − ρ⋅ ay = −1.50⋅
3
2
kg⋅ m
∂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) ⋅
dp =
∂x
p
p ( x ) = 190 ⋅ kPa −
⌠
p − p0 = ⎮
⌡
x
0
1
2
⋅ 1.5⋅
2
s
m
∂
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 6.4
(Difficulty 2)
�⃗ = 3𝚤̂ + 5𝑡𝚥̂ + 8𝑡 2 𝑘�, where the velocity is in 𝑚
6.4 Consider the flow field with the velocity given by 𝑉
and 𝑡 is in seconds. The fluid density is 800
𝑘𝑘
𝑚3
𝑠
and gravity acts in the negative 𝑧 direction. Determine
the velocity, acceleration, and pressure gradient of the fluid at one second time increments from
time= 0,1 to time= 5 seconds.
Find: The velocity, acceleration and pressure gradient at different time.
Assumption: Flow is frictionless and incompressible
Solution: Use Euler’s equation to find the pressure gradient
𝜌
�⃗
𝐷𝑉
= 𝜌𝑔⃗ − ∇𝑝
𝐷𝐷
For the velocity field we have the expression as:
�⃗ = 3𝚤̂ + 5𝑡𝚥̂ + 8𝑡 2 𝑘�
𝑉
The velocity is found at the different times. For 𝑡 = 1 𝑠:
At 𝑡 = 2 𝑠:
At 𝑡 = 3 𝑠:
At 𝑡 = 4 𝑠:
At 𝑡 = 5 𝑠:
�⃗ = 3𝚤̂ + 5𝚥̂ + 8𝑘�
𝑉
𝑚
𝑠
�⃗ = 3𝚤̂ + 10𝚥̂ + 32𝑘�
𝑉
𝑚
𝑠
�⃗ = 3𝚤̂ + 15𝚥̂ + 72𝑘�
𝑉
𝑚
𝑠
�⃗ = 3𝚤̂ + 20𝚥̂ + 128𝑘�
𝑉
�⃗ = 3𝚤̂ + 25𝚥̂ + 200𝑘�
𝑉
For the acceleration, we have the following definition:
Where the acceleration scalar values are then
𝑎𝑥𝑝 =
𝑎𝑦𝑝 =
Thus
𝑎𝑧𝑝 =
𝑎⃗𝑝 =
�⃗
𝐷𝑉
𝐷𝐷
𝐷𝐷
𝜕𝜕
𝜕𝜕
𝜕𝜕 𝜕𝜕
=𝑢
+𝑣
+𝑤
+
𝐷𝐷
𝜕𝜕
𝜕𝜕
𝜕𝜕 𝜕𝜕
𝐷𝐷
𝜕𝜕
𝜕𝜕
𝜕𝜕 𝜕𝜕
=𝑢
+𝑣
+𝑤
+
𝐷𝐷
𝜕𝜕
𝜕𝜕
𝜕𝜕 𝜕𝜕
𝐷𝐷
𝜕𝜕
𝜕𝜕
𝜕𝜕 𝜕𝜕
=𝑢
+𝑣
+𝑤
+
𝐷𝐷
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝑎𝑥𝑝 = 0
𝑎𝑦𝑝 = 5
𝑚
𝑠2
𝑚
𝑠2
𝑎𝑧𝑝 = 16𝑡
𝑚
𝑠2
𝑎⃗𝑝 = 5𝚥̂ + 16𝑡𝑘�
At 𝑡 = 1 𝑠:
At 𝑡 = 2 𝑠:
At 𝑡 = 3 𝑠:
At 𝑡 = 4 𝑠:
At 𝑡 = 5 𝑠:
𝑎⃗𝑝 = 5𝚥̂ + 16 𝑘�
𝑚
𝑠2
𝑎⃗𝑝 = 5𝚥̂ + 32 𝑘�
𝑚
𝑠2
𝑎⃗𝑝 = 5𝚥̂ + 48 𝑘�
𝑚
𝑠2
𝑎⃗𝑝 = 5𝚥̂ + 64 𝑘�
𝑚
𝑠2
𝑎⃗𝑝 = 5𝚥̂ + 80 𝑘�
𝑚
𝑠2
For the frictionless flow we have Euler’s equation:
�⃗
𝐷𝑉
= 𝜌𝑔⃗ − ∇𝑝
𝐷𝐷
Thus pressure gradient is
𝜌
The density is
∇𝑝 = 𝜌𝑔⃗ − 𝜌
𝜌 = 800
As the gravity acts in the negative 𝑧 direction, we have:
�⃗
𝐷𝑉
𝐷𝐷
𝑘𝑘
𝑚3
∇𝑝 = −𝜌𝜌𝑘� − 𝜌�5𝚥̂ + 16𝑡𝑘��
∇𝑝 = −5𝜌𝚥̂ − 𝜌(𝑔 + 16𝑡)𝑘�
At 𝑡 = 1 𝑠:
At 𝑡 = 2 𝑠:
At 𝑡 = 3 𝑠:
At 𝑡 = 4 𝑠:
At 𝑡 = 5 𝑠:
𝑔 = 9.81
𝑚
𝑠2
∇𝑝 = −4000𝚥̂ − 20648𝑘�
𝑃𝑃
𝑚
∇𝑝 = −4000𝚥̂ − 33448𝑘�
𝑃𝑃
𝑚
∇𝑝 = −4000𝚥̂ − 46248𝑘�
𝑃𝑃
𝑚
∇𝑝 = −4000𝚥̂ − 59048𝑘�
𝑃𝑃
𝑚
∇𝑝 = −4000𝚥̂ − 71848𝑘�
𝑃𝑃
𝑚
Problem 6.5
(Difficulty 2)
�⃗ = 4𝑦𝚤̂ + 3𝑥𝚥̂, where the velocity is in
6.5 Consider the flow field with the velocity given by 𝑉
coordinate are in feet. The fluid density is 𝜌 = 1.5
𝑠𝑠𝑠𝑠
𝑓𝑓 3
𝑓𝑓
𝑠
and the
and gravity acts in the negative 𝑦 direction.
Determine the general expressions for the acceleration and pressure gradient. Plot the acceleration and
pressure gradient in the 𝑦 direction for x = 0 and x = 2 ft.
Find: The acceleration and pressure gradient.
Assumption: Flow is frictionless and incompressible
Solution: Use Euler’s equation to find the pressure gradient
𝜌
�⃗
𝐷𝑉
= 𝜌𝑔⃗ − ∇𝑝
𝐷𝐷
For the velocity field we have the expression as:
�⃗ = 4𝑦𝚤̂ + 3𝑥𝚥̂
𝑉
For the acceleration, we have the following definition of acceleration:
𝑎⃗𝑝 =
Or in terms of the scalar accelerations
𝑎𝑥𝑝 =
𝑎𝑦𝑝 =
Thus
𝑎𝑧𝑝 =
𝑎𝑥𝑝 = 𝑢
𝑎𝑦𝑝 = 𝑢
�⃗
𝐷𝑉
𝐷𝐷
𝐷𝐷
𝜕𝜕
𝜕𝜕
𝜕𝜕 𝜕𝜕
=𝑢
+𝑣
+𝑤
+
𝐷𝐷
𝜕𝜕
𝜕𝜕
𝜕𝜕 𝜕𝜕
𝐷𝐷
𝜕𝜕
𝜕𝜕
𝜕𝜕 𝜕𝜕
=𝑢
+𝑣
+𝑤
+
𝐷𝐷
𝜕𝜕
𝜕𝜕
𝜕𝜕 𝜕𝜕
𝐷𝐷
𝜕𝜕
𝜕𝜕
𝜕𝜕 𝜕𝜕
=𝑢
+𝑣
+𝑤
+
𝐷𝐷
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝑓𝑓
+𝑣
= 4𝑦 × 0 + 3𝑥 × 4 = 12𝑥 2
𝜕𝜕
𝜕𝜕
𝑠
𝜕𝜕
𝜕𝜕
𝑓𝑓
+𝑣
= 4𝑦 × 3 + 3𝑥 × 0 = 12𝑦 2
𝜕𝜕
𝜕𝜕
𝑠
𝑎𝑧𝑝 = 0
𝑓𝑓
𝑠2
So the general expression for acceleration is:
𝑎⃗𝑝 = 12𝑥𝚤̂ + 12𝑦𝚥̂
For the pressure gradient we use Euler’s equation:
�⃗
𝐷𝑉
= 𝜌𝑔⃗ − ∇𝑝
𝐷𝐷
Thus the pressure gradient is
𝜌
Where the density is
∇𝑝 = 𝜌𝑔⃗ − 𝜌
𝜌 = 1.5
�⃗
𝐷𝑉
𝐷𝐷
𝑙𝑙𝑙 ∙ 𝑠2
𝑠𝑠𝑠𝑠
=
1.5
𝑓𝑓 3
𝑓𝑓 4
As the gravity acts in the negative 𝑧 direction, we have:
∇𝑝 = −𝜌𝜌𝚥̂ − 𝜌(12𝑥𝚤̂ + 12𝑦𝚥̂)
Thus the general expression for pressure gradient is:
∇𝑝 = −12𝜌𝜌𝚤̂ − 𝜌(𝑔 + 12𝑦)𝚥̂
For the acceleration and pressure gradient in the 𝑦 direction we have:
𝑎𝑦𝑝 = 12𝑦
𝑓𝑓
𝑠2
𝜕𝜕
𝑙𝑙𝑙 ∙ 𝑠 2
𝑓𝑓
𝑓𝑓
= −𝜌(𝑔 + 12𝑦) = −1.5
× �32.2 2 + 12𝑦 2 �
4
𝜕𝜕
𝑠
𝑠
𝑓𝑓
The plots are shown in the figures.
𝜕𝜕
𝑙𝑙𝑙
= −(48.3 + 18𝑦) 3
𝜕𝜕
𝑓𝑓
120
Acceleration in y direction (ft/s2)
100
80
60
40
20
0
0
1
2
3
4
5
y (ft)
6
7
8
9
10
0
1
2
3
4
5
y (ft)
6
7
8
9
10
-40
Pressure gradient in y direction (ft/s2)
-60
-80
-100
-120
-140
-160
-180
-200
-220
-240
Problem 6.6
Problem
6.8
[Difficulty: 3]
6.6
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
u=
(
2
π⋅ x + h
∂x
u + v⋅
∂
∂y
2
u =−
⎡
(2
q ⋅ x⋅ ⎣ x + y
)
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⎤
− h ⋅ h + 4⋅ x ⎦
q ⋅ y⋅ ⎣ x + y
∂
∂
ay = u ⋅ v + v ⋅ v = −
2
2
∂x
∂y
2 2
2
2
2
π ⋅ ⎡⎣x + ( y + h ) ⎤⎦ ⋅ ⎡⎣x + ( y − h ) ⎤⎦
(
)
2
2
)
2
(
)
y = 0⋅ m
For motion along the wall
q⋅ x
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 6.7
Problem
6.10
[Difficulty: 2]
6.7
Given:
Velocity field
Find:
Expression for pressure field; evaluate at (2,2)
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 6.8
(Difficulty 2)
6.8 Consider a two-dimensional incompressible flow flowing downward against a plate. The velocity is
�⃗ = 𝐴𝐴𝚤̂ − 𝐴𝐴𝚥̂, where 𝐴 = 2 𝑠 −1 and 𝑥 and 𝑦 are in meters. Determine general expressions
given by 𝑉
for the acceleration and pressure gradient in the 𝑥 − and 𝑦 −directions. Plot the pressure gradient along
the plate (𝑦 = 0) from 𝑥 = 0 to 𝑥 = 3 m, and the pressure gradient along the centerline (𝑥 = 0) from
𝑦 = 0 to 𝑦 = 3 𝑚.
Find: The acceleration and pressure gradient.
Assumption: Flow is frictionless and incompressible
Solution: Use Euler’s equation to find the pressure gradient
𝜌
�⃗
𝐷𝑉
= 𝜌𝑔⃗ − ∇𝑝
𝐷𝐷
For the velocity field we have the expression as:
�⃗ = 𝐴𝐴𝚤̂ − 𝐴𝐴𝚥̂
𝑉
𝐴 = 2 𝑠 −1
�⃗ = 2𝑥𝚤̂ − 2𝑦𝚥̂
𝑉
For the acceleration, we have the following definition of acceleration:
𝑎⃗𝑝 =
Or in terms of the scalar accelerations
𝑎𝑥𝑝 =
𝑎𝑦𝑝 =
Thus
𝑎𝑧𝑝 =
𝑎𝑥𝑝 = 𝑢
�⃗
𝐷𝑉
𝐷𝐷
𝐷𝐷
𝜕𝜕
𝜕𝜕
𝜕𝜕 𝜕𝜕
=𝑢
+𝑣
+𝑤
+
𝐷𝐷
𝜕𝜕
𝜕𝜕
𝜕𝜕 𝜕𝜕
𝐷𝐷
𝜕𝜕
𝜕𝜕
𝜕𝜕 𝜕𝜕
=𝑢
+𝑣
+𝑤
+
𝐷𝐷
𝜕𝜕
𝜕𝜕
𝜕𝑧 𝜕𝜕
𝐷𝐷
𝜕𝜕
𝜕𝜕
𝜕𝜕 𝜕𝜕
=𝑢
+𝑣
+𝑤
+
𝐷𝐷
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝑚
+𝑣
= 2𝑥 × 2 + (−2𝑦) × 0 = 4𝑥 2
𝜕𝜕
𝜕𝜕
𝑠
𝑎𝑦𝑝 = 𝑢
𝜕𝜕
𝜕𝜕
𝑚
+𝑣
= 2𝑥 × 0 + (−2𝑦) × (−2) = 4𝑦 2
𝜕𝜕
𝜕𝜕
𝑠
𝑎𝑧𝑝 = 0
So the general expression for acceleration is:
𝑚
𝑠2
𝑎⃗𝑝 = 4𝑥𝚤̂ + 4𝑦𝚥̂
For the pressure gradient we use Euler’s equation:
Thus the pressure gradient is
𝜌
�⃗
𝐷𝑉
= 𝜌𝑔⃗ − ∇𝑝
𝐷𝐷
∇𝑝 = 𝜌𝑔⃗ − 𝜌
𝜌 = 998
As the gravity acts in the negative 𝑧 direction, we have:
�⃗
𝐷𝑉
𝐷𝐷
𝑘𝑘
𝑚3
∇𝑝 = −𝜌𝜌𝚥̂ − 𝜌(4𝑥𝚤̂ + 4𝑦𝚥̂)
Thus the general expression for pressure gradient is:
∇𝑝 = −4𝜌𝜌𝚤̂ − 𝜌(𝑔 + 4𝑦)𝚥̂
∇𝑝 = −4𝜌𝜌𝚤̂ − 𝜌(𝑔 + 4𝑦)𝚥̂
∇𝑝 = −3992𝑥𝚤̂ − (9810 + 3992𝑦)𝚥̂
The pressure gradient along the plate (𝑦 = 0) is:
𝑃𝑃
𝜕𝜕
= −3992𝑥
𝑚
𝜕𝜕
𝑃𝑃
𝑚
The plot of pressure gradient along the plate from 𝑥 = 0 to 𝑥 = 3 m is shown as:
Pressure gradient dp/dx along the plate (Pa/m)
0
-2000
-4000
-6000
-8000
-10000
-12000
0
0.5
1
1.5
x (m)
2
2.5
3
The pressure gradient along the centerline (𝑥 = 0) is:
𝑃𝑃
𝜕𝜕
= −(9810 + 3992𝑦)
𝑚
𝜕𝜕
The plot of pressure gradient along the centerline from 𝑦 = 0 to 𝑦 = 3 m is shown as:
4
Pressure gradient dp/dy along the centerline (Pa/m)
-0.8
x 10
-1
-1.2
-1.4
-1.6
-1.8
-2
-2.2
0
0.5
1
1.5
y (m)
2
2.5
3
Problem 6.9
Problem
6.12
[Difficulty: 2]
6.9
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
U ⋅ ρ⋅ ⎝ e
− 1⎠
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 6.10
(Difficulty 2)
6.10 Consider a flow of water in pipe. What is the pressure gradient required to accelerate the water at
20
𝑓𝑓
𝑠2
if the pipe is (a) horizontal, (b) vertical with the water flowing upward, and (c) vertical with the
water flowing downward. Explain why the pressure gradient depends on orientation and why the
pressure gradient differs in sign between case (b) and (c).
Assumption: Frictionless, incompressible, and unidirectional flow in the pipe.
Solution: Use Euler’s equation to find the pressure gradient:
�⃗
𝐷𝑉
= 𝜌𝑔⃗ − ∇𝑝
𝐷𝐷
The pressure gradient is then
𝜌
Or in terms of acceleration
∇𝑝 = 𝜌𝑔⃗ − 𝜌
The density is:
∇𝑝 = 𝜌𝑔⃗ − 𝜌𝑎⃗𝑝
𝜌 = 1.9
�⃗
𝐷𝑉
𝐷𝐷
𝑙𝑙𝑙 ∙ 𝑠2
𝑠𝑠𝑠𝑠
=
1.9
𝑓𝑓 3
𝑓𝑓 4
(a) For the horizontal pipe there is no effect of gravity, and Euler’s equation becomes
∇𝑝 = −𝜌𝑎⃗𝑝
The acceleration is
The pressure gradient we need is:
𝑎⃗𝑝 = 20𝚤̂
𝑓𝑓
𝑠2
𝑙𝑙𝑙 ∙ 𝑠2
𝑓𝑓
𝑙𝑙𝑙
𝜕𝜕
= −𝜌𝑎⃗𝑝 = −1.9
× 20 2 = −38
4
𝑠
𝑓𝑓 3
𝜕𝜕
𝑓𝑓
(b) For the vertical pipe with water flowing upward, the upward direction is positive y coordinate and
gravity acts downward. Euler’s equation is
Where the acceleration is
The pressure gradient is then
∇𝑝 = 𝜌𝑔⃗ − 𝜌𝑎⃗𝑝
𝑎⃗𝑝 = 20𝚥̂
𝑓𝑓
𝑠2
𝑙𝑙𝑙 ∙ 𝑠2
𝑓𝑓
𝑙𝑙𝑙 ∙ 𝑠2
𝑓𝑓
𝑙𝑙𝑙
𝜕𝜕
= 𝜌𝜌 − 𝜌𝑎⃗𝑝 = 1.9
×
�−32.2
�
−
1.9
× 20 2 = −99.2
4
2
4
𝑠
𝑠
𝑓𝑓 3
𝜕𝜕
𝑓𝑓
𝑓𝑓
(c) For the vertical pipe with water flowing downward:
𝑎⃗𝑝 = −20𝚥̂
𝑓𝑓
𝑠2
𝑙𝑙𝑙 ∙ 𝑠2
𝑓𝑓
𝑙𝑙𝑙 ∙ 𝑠2
𝑓𝑓
𝑙𝑙𝑙
𝜕𝜕
= 𝜌𝜌 − 𝜌𝑎⃗𝑝 = 1.9
×
�−32.2
�
−
1.9
× �−20 2 � = −23
4
2
4
𝑠
𝑠
𝑓𝑓 3
𝜕𝜕
𝑓𝑓
𝑓𝑓
The pressure gradient is the driven force in this fluid flow, when we have different acceleration with
different orientation, the pressure gradient will be different.
For the case (b) and case (c), the fluid velocity is opposite, the acceleration is opposite, then the pressure
gradient will be different.
Problem 6.11
Problem
6.14
[Difficulty: 3]
6.11
Given:
Velocity field
Find:
The acceleration at several points; evaluate pressure gradient
Solution:
The given data is
3
3
m
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 ∂θ
r
∂r
aθ = 0
Evaluating at point (1,0)
ar = −0.127
r - component
Evaluating at point (1,π/2)
Evaluating at point (2,0)
From Eq. 6.3, pressure gradient is
∂
ar = −0.127
m
∂
∂r
s
m
aθ = 0
2
s
m
2
s
Evaluating at point (1,π/2)
Evaluating at point (2,0)
∂
∂r
∂
∂r
∂
∂r
p = 127 ⋅
p = 127 ⋅
∂r
Pa
m
Pa
p = 15.8⋅
aθ = 0
∂
p = −ρ⋅ ar
1 ∂
⋅ p = −ρ⋅ aθ
r ∂θ
Evaluating at point (1,0)
m
Pa
m
2 3
4⋅ π ⋅ r
aθ = 0
2
ar = −0.0158
2
q +K
p =
(2
2 3
4⋅ π ⋅ r
1 ∂
⋅ p =0
r ∂θ
1 ∂
⋅ p =0
r ∂θ
1 ∂
⋅ p =0
r ∂θ
1 ∂
⋅ p =0
r ∂θ
)
2
ρ⋅ q + K
Problem 6.12
Problem
6.16
[Difficulty: 3]
6.12
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
⎤ Ai ⋅ L ⋅ u i ⋅ ( Ae − Ai)
⎡
⎥=
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
∂
⎮
⎮
p − pi =
p dx =
−
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 6.13
(Difficulty: 2)
6.13 Consider water flowing in a circular section of a two-dimensional channel. Assume the velocity is
𝑚
.
𝑠
uniform across the channel at 12
The pressure is 120 𝑘𝑘𝑘 at centerline (point 1). Determine the
pressure at point 2 and 3 for the case of (a) flow in the horizontal plane (b) gravity acting in the direction
of 2 to 3
.
Find: The pressures of the fluid
Assumption: The flow is frictionless and steady
Solution: Apply Euler’s equation
𝜌
�⃗
𝐷𝑉
= 𝜌𝑔⃗ − ∇𝑝
𝐷𝐷
In cylindrical coordinates, for this steady two-dimensional flow we have:
𝜌𝑎𝑟 = 𝜌 �𝑉𝑟
For this flow we have:
𝜌𝑎𝜃 = 𝜌 �𝑉𝑟
𝜕𝜕
𝜕𝑉𝑟 𝑉𝜃 𝜕𝑉𝑟 𝑉𝜃2
+
− � = 𝜌𝑔𝑟 −
𝜕𝜕
𝜕𝜕
𝑟 𝜕𝜕
𝑟
1 𝜕𝜕
𝜕𝑉𝜃 𝑉𝜃 𝜕𝑉𝜃 𝑉𝜃 𝑉𝑟
+
−
� = 𝜌𝑔𝜃 −
𝑟 𝜕𝜕
𝜕𝜕
𝑟 𝜕𝜕
𝑟
𝑉𝑟 = 0
𝜕𝑉𝜃
=0
𝜕𝜕
The governing equation can be simplified to:
𝑉𝜃 = 12
𝑚
𝑠
𝜌 �−
𝑉𝜃2
𝜕𝜕
� = 𝜌𝑔𝑟 −
𝜕𝜕
𝑟
𝑉𝜃2
𝜕𝜕
= 𝜌𝑔𝑟 + 𝜌 � �
𝜕𝜕
𝑟
(a) For the case flow in the horizontal plane, we have:
𝑉𝜃2
𝜕𝜕
= 𝜌� �
𝑟
𝜕𝜕
1
𝜕𝜕
𝑉𝜃2
𝑝1 − 𝑝2 = �
𝑑𝑑 = � �𝜌 � �� 𝑑𝑑
𝑟
2 𝜕𝜕
2
1
𝑟1
𝑝1 − 𝑝2 = 𝜌𝑉𝜃2 ln � �
𝑟2
𝑟1
𝑝2 = 𝑝1 − 𝜌𝑉𝜃2 ln � �
𝑟2
𝑘𝑘
𝑚 2
6.75
×
�12
� × ln �
� = 103.1 𝑘𝑘𝑘
3
𝑚
𝑠
6
𝑝2 = 120 𝑘𝑘𝑘 − 998
3
𝑝3 − 𝑝1 = �
1
3
𝜕𝜕
𝑉𝜃2
𝑑𝑑 = � �𝜌 � �� 𝑑𝑑
𝜕𝜕
𝑟
1
𝑟3
𝑝3 − 𝑝1 = 𝜌𝑉𝜃2 ln � �
𝑟1
𝑟3
𝑝3 = 𝑝1 + 𝜌𝑉𝜃2 ln � �
𝑟1
𝑘𝑘
𝑚 2
7.5
×
�12
� × ln �
� = 135.1 𝑘𝑘𝑘
3
𝑚
𝑠
6.75
𝑝3 = 120 𝑘𝑘𝑘 + 998
(b) Gravity normal to the flow, we have:
𝑉𝜃2
𝜕𝜕
= −𝜌𝜌 + 𝜌 � �
𝜕𝜕
𝑟
1
𝑝1 − 𝑝2 = �
2
1
𝜕𝜕
𝑉𝜃2
𝑑𝑑 = � �−𝜌𝜌 + 𝜌 � �� 𝑑𝑑
𝜕𝜕
𝑟
2
𝑟1
𝑝1 − 𝑝2 = −𝜌𝜌(𝑟1 − 𝑟2 ) + 𝜌𝑉𝜃2 ln � �
𝑟2
𝑟1
𝑝2 = 𝑝1 + 𝜌𝜌(𝑟1 − 𝑟2 ) − 𝜌𝑉𝜃2 ln � �
𝑟2
𝑝2 = 120 𝑘𝑘𝑘 + 9.81
𝑘𝑘𝑘
𝑘𝑘
𝑚 2
6.75
× 0.75 𝑚 − 998 3 × �12 � × ln �
�
𝑚
𝑚
𝑠
6
Also we have:
𝑝2 = 110.4 𝑘𝑘𝑘
3
𝜕𝜕
𝑉𝜃2
𝑝3 − 𝑝1 = �
𝑑𝑑 = � �−𝜌𝜌 + 𝜌 � �� 𝑑𝑑
𝑟
1 𝜕𝜕
1
3
𝑟3
𝑝3 − 𝑝1 = −𝜌𝜌(𝑟3 − 𝑟1 ) + 𝜌𝑉𝜃2 ln � �
𝑟1
𝑟3
𝑝3 = 𝑝1 − 𝜌𝜌(𝑟3 − 𝑟1 ) + 𝜌𝑉𝜃2 ln � �
𝑟1
𝑝3 = 120 𝑘𝑘𝑘 − 9.81
𝑘𝑘𝑘
𝑘𝑘
𝑚 2
7.5
× 0.75 𝑚 + 998 3 × �12 � × ln �
�
𝑚
𝑚
𝑠
6.75
𝑝3 = 127.8 𝑘𝑘𝑘
Problem 6.14
(Difficulty 2)
6.14 Consider a tornado as air moving in a circular pattern in the horizontal plane. If the wind speed is
200 𝑚𝑚ℎ and the diameter of the tornado is 200 𝑓𝑓, determine the radial pressure gradient. If it is
desired to model the tornado using water in a 6 in diameter tube, what speed is needed to give the
same radial pressure gradient?
Find: The pressure gradient for prototype and model case.
Assumption: Flow is frictionless and incompressible
Solution: Use Euler’s equation to find the pressure gradient
𝜌
The pressure gradient is
�⃗
𝐷𝑉
= 𝜌𝑔⃗ − ∇𝑝
𝐷𝐷
∇𝑝 = 𝜌𝑔⃗ − 𝜌
�⃗
𝐷𝑉
𝐷𝐷
For this two-dimensional flow we have in cylindrical coordinates:
𝜌𝑎𝑟 = 𝜌 �
𝜕𝑉𝑟
𝜕𝜕
𝜕𝑉𝑟 𝑉𝜃 𝜕𝑉𝑟 𝑉𝜃2
+ 𝑉𝑟
+
− � = 𝜌𝑔𝑟 −
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝑟 𝜕𝜕
𝑟
𝜕𝑉𝜃
1 𝜕𝜕
𝜕𝑉𝜃 𝑉𝜃 𝜕𝑉𝜃 𝑉𝜃 𝑉𝑟
+ 𝑉𝑟
+
−
� = 𝜌𝑔𝜃 −
𝜌𝑎𝜃 = 𝜌 �
𝑟 𝜕𝜕
𝜕𝜕
𝜕𝜕
𝑟 𝜕𝜕
𝑟
For this specific flow we have:
𝑉𝑟 = 0
𝜕𝑉𝜃
=0
𝜕𝜕
The tangential velocity is
𝑉𝜃 = 200 𝑚𝑚ℎ = 293
1
𝑟 = 𝐷 = 100 𝑓𝑓
2
There is no gravity force in the horizontal plane.
𝑓𝑓
𝑠
The governing equation are then simplified as:
𝜌 �−
𝑉𝜃2
𝜕𝜕
�=−
𝜕𝜕
𝑟
𝑉𝜃2
𝜕𝜕
= 𝜌� �
𝜕𝜕
𝑟
The density of the air is:
𝜌 = 0.0024
Thus the radial pressure gradient is
𝑙𝑙𝑙 ∙ 𝑠2
𝑠𝑠𝑠𝑠
=
0.0024
𝑓𝑓 3
𝑓𝑓 4
𝑠2
For the model, we have:
𝑙𝑙𝑙 ∙
𝜕𝜕
= 0.0024
𝜕𝜕
𝑓𝑓 4
×
𝑓𝑓 2
�
𝑙𝑙𝑙
𝑠
= 2.06 3
𝑓𝑓
100 𝑓𝑓
�293
𝑠𝑠𝑠𝑠
𝑙𝑙𝑙 ∙ 𝑠2
= 1.94
𝜌𝑚 = 1.94
𝑓𝑓 3
𝑓𝑓 4
The model radial pressure gradient is
𝑟𝑚 = 3 𝑖𝑖 = 0.25 𝑓𝑓
2
𝜕𝜕
𝑉𝜃𝜃
= 𝜌𝑚 �
�
𝜕𝑟𝑚
𝑟𝑚
Or the velocity is
2
𝑉𝜃𝜃
=
𝑟𝑚 𝜕𝜕
𝜌𝑚 𝜕𝑟𝑚
Where the radial gradient is the same as for the prototype
The tangential velocity is
𝑙𝑙𝑙
𝜕𝜕
= 2.06 3
𝑓𝑓
𝜕𝑟𝑚
0.25 𝑓𝑓
𝑙𝑙𝑙
𝑓𝑓
𝑟𝑚 𝜕𝜕
=
× 2.06 3 = 0.515
𝑉𝜃𝜃 = �
2
�
𝑙𝑙𝑙
∙
𝑠
𝑓𝑓
𝑠
𝜌𝑚 𝜕𝑟𝑚
1.94
𝑓𝑓 4
This velocity is easily attainable in water.
Problem 6.15
Problem
6.18
[Difficulty: 3]
6.15
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
Do − Di
L
π 2
2
⋅ D = Vi⋅ ⋅ Di
4
4
3
m
⋅x
Q = 0.00785
2
⋅ D( x ) = Q
m
s
4⋅ Q
V( x ) =
Do − Di ⎞
⎛
⋅x
π⋅ ⎜ Di +
L
⎝
⎠
Vi
Do − Di ⎞
⎛
⋅x
⎜1 +
L⋅ Di
⎝
⎠
kg
ρ = 1000⋅
3
π
V( x ) =
m
Vi = 1 ⋅
s
L = 0.5⋅ m
2
2
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
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
)
⎡ ( Do − Di) ⎤
Di⋅ L⋅ ⎢1 +
⋅ x⎥
Di⋅ L
⎣
⎦
5
2
ax ( x ) = −
(
2 ⋅ Vi ⋅ Do − Di
)
⎡ ( Do − Di) ⎤
Di⋅ L⋅ ⎢1 +
⋅ x⎥
Di⋅ L
⎣
⎦
5
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
)
⎡ ( Do − Di) ⎤
Di⋅ L⋅ ⎢1 +
⋅ x⎥
Di⋅ L
⎣
⎦
5
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 6.16
Problem
6.19
[Difficulty: 3]
6.16
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
ρ = 1000⋅
4⋅ Q
m
s
Vi
V( x) =
or
Do − Di ⎞
⎛
⋅x
π⋅ ⎜ Di +
L
⎝
⎠
2
Do − Di ⎞
⎛
⋅x
⎜1 +
L⋅ Di
⎝
⎠
The governing equation for this flow is
ax = V⋅
or, for steady 1D flow, in the notation of the problem
2
Hence
ax ( x ) = −
(
d
V=
dx
Vi
d
2 dx
Vi
⋅
Do − Di ⎞
⎛
⋅x
⎜1 +
L⋅ Di
⎝
⎠
Do − Di ⎞
⎛
⋅x
⎜1 +
L⋅ Di
⎝
⎠
)
⎡ ( Do − Di) ⎤
Di⋅ L⋅ ⎢1 +
⋅ x⎥
Di⋅ L
⎣
⎦
5
This can be plotted in Excel (see below)
From Eq. 6.2a, pressure gradient is
∂
∂x
3
m
⋅x
Q = 0.245
V( x) =
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
)
⎡ ( Do − Di) ⎤
Di⋅ L⋅ ⎢1 +
⋅ x⎥
Di⋅ L
⎣
⎦
5
2
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
)
⎡ ( Do − Di) ⎤
Di⋅ L⋅ ⎢1 +
⋅ x⎥
Di⋅ L
⎣
⎦
5
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
6.17
Problem 6.17
Problem
6.22
[Difficulty: 3]
Problem 6.18
(Difficulty: 3)
6.18 Consider problem 6.15 with the nozzle directed upward. Assuming that the flow is uniform at each
section, derive and plot the acceleration of a fluid particle for an inlet speed of 𝑉𝑖 = 2
𝑚
.
𝑠
Plot the
pressure gradient through the nozzle, and its maximum absolute value. If the pressure gradient must be
no greater than 7
𝑀𝑀𝑀
𝑚
in absolute value, how long would the nozzle have to be?
Find: the flow properties of the nozzle
Assumption: The flow is ideal
Solution: Apply the continuity and Euler’s equation
From the continuity equation we have:
𝑄 = 𝑉𝑉 =
𝜋
𝜋
𝑚
𝑚3
𝜋 2
𝐷 𝑣 = 𝐷𝑖2 𝑉𝑖 = × (0.1 𝑚)2 × 2
= 0.0157
4
4
𝑠
4
𝑠
The velocity v in t he y-direction is
𝑄
4𝑄
𝑉𝑖
𝑣=𝜋
=
=
2
𝐷𝑜 − 𝐷𝑖 2
𝐷 2 𝜋 �𝐷 + 𝐷𝑜 − 𝐷𝑖 𝑦�
�1
+
𝑦�
4
𝑖
𝐿
𝐷𝐿
For this 1D flow in the positive y direction, we have the acceleration as:
𝑎𝑦 =
𝑖
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝑑𝑑
+𝑢
+𝑣
+𝑤
=𝑣
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝑑𝑑
For a linear decrease in diameter, the diameter of the nozzle at any location is
𝐷(𝑥) = 𝐷𝑖 +
We have the following parameters for this nozzle
𝐷𝑜 − 𝐷𝑖
𝑦
𝐿
𝐷𝑖 = 0.1 𝑚, 𝐷𝑜 = 0.02 𝑚, 𝐿 = 0.5 𝑚 ,𝑉𝑖 = 2
The acceleration at any location is given by
𝑎𝑦 =
�1 +
𝑉𝑖
𝐷𝑜 − 𝐷𝑖
𝑦�
𝐷𝑖 𝐿
2
𝑑
𝑉𝑖
𝐷𝑜 − 𝐷𝑖 2
�1 +
𝑦�
𝐷𝑖 𝐿
=−
𝑑𝑑
𝑚
,
𝑠
and 𝜌 = 1000
2𝑉𝑖2 (𝐷𝑜 − 𝐷𝑖 )
𝐷𝑖 𝐿 �1 +
𝐷𝑜 − 𝐷𝑖
𝑦�
𝐷𝑖 𝐿
5
=
𝑘𝑘
𝑚3
12.8
𝑚
5
(1 − 1.6𝑦) 𝑠2
The plot is shown as:
4
4.5
x 10
4
3.5
a (m/s 2)
3
2.5
2
1.5
1
0.5
0
0
0.05
0.1
0.15
0.2
0.25
x (m)
0.3
0.35
0.4
0.45
0.5
To find the pressure gradient, we use the momentum equation:
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜌� +𝑢
+𝑣
+ 𝑤 � = 𝜌𝑔𝑦 −
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
For steady flow in the vertical direction, the equation reduces to
The pressure gradient is
𝜌𝜌
𝜕𝜕
𝜕𝜕
= −𝜌𝜌 −
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝑑𝑑
𝜕𝜕
= −𝜌𝜌 − 𝜌𝜌
= −𝜌𝜌 − 𝜌𝜌
= −𝜌𝜌 + 𝜌
𝜕𝜕
𝑑𝑑
𝜕𝜕
2𝑉𝑖2 (𝐷𝑜 − 𝐷𝑖 )
𝐷 − 𝐷𝑖 5
𝑦�
𝐷𝑖 𝐿 �1 + 𝑜
𝐷𝑖 𝐿
12.8
𝑘𝑘𝑘
𝜕𝜕
= −9.8 −
5
(1 − 1.6𝑦) 𝑚
𝜕𝜕
= −𝜌𝜌 − 𝜌𝑎𝑦
The plot for the pressure gradient is:
4
0
x 10
-0.5
-1
dp/dx (kPa/m)
-1.5
-2
-2.5
-3
-3.5
-4
-4.5
0
0.05
0.1
0.2
0.15
0.25
x (m)
0.3
0.4
0.35
0.45
At the inlet:
𝜕𝜕
= −22.6 𝑘𝑘𝑘
𝜕𝜕
At the outlet:
𝜕𝜕
= −40.1 𝑀𝑀𝑀
𝜕𝜕
The maximum absolute value of pressure gradient is 40.1 𝑀𝑀𝑀.
If the pressure gradient must be no great than 7
𝜕𝜕
= �−𝜌𝜌 + 𝜌
𝜕𝜕
𝑀𝑀𝑀
,
𝑚
we have:
2𝑉𝑖2 (𝐷𝑜 − 𝐷𝑖 )
𝐷𝑖 𝐿 �1 +
𝐷𝑜 − 𝐷𝑖
𝑦�
𝐷𝑖 𝐿
�≤7
5�
At the outlet we have the maximum pressure gradient so:
Or
�−9800 −
9800 +
6400
� ≤ 7000000
𝐿(1 − 0.8)5
6400
≤ 7000000
𝐿(1 − 0.8)5
𝑀𝑀𝑀
𝑚
0.5
Or
6400
≤ 6990200
𝐿(1 − 0.8)5
So the length must be
𝐿≥
6400
𝑚 = 2.86 𝑚
6990200 × (1 − 0.8)5
Problem 6.19
(Difficulty: 3)
6.19 Consider problem 6.16 with the diffuser directed upward. Assuming that the flow is uniform at
each section, derive and plot the acceleration of a fluid particle for an inlet speed of 𝑉𝑖 = 12
𝑚
.
𝑠
Plot the
pressure gradient through the diffuser, and its maximum absolute value. If the pressure gradient must
be no greater than 20
𝑘𝑘𝑘
𝑚
, how long would the diffuser have to be?
Find: the flow properties of the diffuser
Assumption: The flow is ideal
Solution: Apply the continuity and Euler’s equation
From the continuity equation we have:
𝑄 = 𝑉𝑉 =
𝜋
𝜋
𝑚
𝑚3
𝜋 2
𝐷 𝑣 = 𝐷𝑖2 𝑉𝑖 = × (0.25 𝑚)2 × 12
= 0.589
4
4
𝑠
4
𝑠
The velocity v in the y-direction is
𝑄
4𝑄
𝑉𝑖
𝑣=𝜋
=
=
2
𝐷𝑜 − 𝐷𝑖 2
𝐷 2 𝜋 �𝐷 + 𝐷𝑜 − 𝐷𝑖 𝑦�
�1
+
𝑦�
4
𝑖
𝐿
𝐷𝐿
For this 1D flow in the positive y direction, we have the acceleration as:
𝑎𝑦 =
𝑖
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝑑𝑑
+𝑢
+𝑣
+𝑤
=𝑣
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝑑𝑑
For the linear decrease in diameter, the diameter at any location is:
𝐷(𝑥) = 𝐷𝑖 +
The acceleration is then
𝑎𝑦 =
�1 +
𝑉𝑖
𝐷𝑜 − 𝐷𝑖
𝑦�
𝐷𝑖 𝐿
2
𝑑
𝐷𝑜 − 𝐷𝑖
𝑦
𝐿
𝑉𝑖
𝐷 − 𝐷𝑖 2
�1 + 𝑜
𝑦�
𝐷𝑖 𝐿
=−
𝑑𝑑
For this problem we have the following parameters:
2𝑉𝑖2 (𝐷𝑜 − 𝐷𝑖 )
𝐷𝑖 𝐿 �1 +
𝐷𝑖 = 0.25 𝑚, 𝐷𝑜 = 0.75 𝑚, 𝐿 = 1 𝑚, 𝑉𝑖 = 5
𝐷𝑜 − 𝐷𝑖
𝑦�
𝐷𝑖 𝐿
5
=−
576
𝑚
5
(1 + 2𝑦) 𝑠2
𝑚
𝑘𝑘
, 𝑎𝑎𝑎 𝜌 = 1000 3
𝑠
𝑚
The plot for acceleration is:
0
-100
a (m/s 2)
-200
-300
-400
-500
-600
0
0.1
0.2
0.3
0.4
0.5
x (m)
0.6
0.7
0.8
0.9
1
To find the pressure gradient, we use the momentum equation:
This reduces to
The pressure gradient is then
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜌� +𝑢
+𝑣
+ 𝑤 � = 𝜌𝑔𝑦 −
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜌𝜌
𝜕𝜕
𝜕𝜕
= −𝜌𝜌 −
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝑑𝑑
𝜕𝜕
= −𝜌𝜌 − 𝜌𝜌
= −𝜌𝜌 − 𝜌𝜌
= −𝜌𝜌 + 𝜌
𝜕𝜕
𝑑𝑑
𝜕𝜕
The plot is shown as:
2𝑉𝑖2 (𝐷𝑜 − 𝐷𝑖 )
𝐷 − 𝐷𝑖 5
𝑦�
𝐷𝑖 𝐿 �1 + 𝑜
𝐷𝑖 𝐿
576
𝑘𝑘𝑘
𝜕𝜕
= −9.8 +
5
(1 + 2𝑦) 𝑚
𝜕𝜕
= −𝜌𝜌 − 𝜌𝑎𝑦
600
500
dp/dx (kPa/m)
400
300
200
100
0
-100
0
0.1
0.2
0.3
0.4
0.5
x (m)
0.6
0.8
0.7
0.9
At the inlet:
𝜕𝜕
= 566.2 𝑘𝑘𝑘
𝜕𝜕
At the outlet:
𝜕𝜕
= −7.43 𝑘𝑘𝑘
𝜕𝜕
The maximum absolute value of pressure gradient is 566.2 𝑘𝑘𝑘.
If the pressure gradient must be no great than 20
𝜕𝜕
= �−𝜌𝜌 + 𝜌
𝜕𝜕
𝑘𝑘𝑘
,
𝑚
we have:
2𝑉𝑖2 (𝐷𝑜 − 𝐷𝑖 )
𝐷𝑖 𝐿 �1 +
𝐷𝑜 − 𝐷𝑖
𝑦�
𝐷𝑖 𝐿
� ≤ 20
5�
At the inlet we have the maximum pressure gradient so:
Or
�−9800 +
576000
� ≤ 20000
𝐿(1)5
Or
−9800 +
576000
≤ 20000
𝐿
576000
≤ 29800
𝐿
𝑘𝑘𝑘
𝑚
1
So the length must be
𝐿≥
576000
𝑚 = 19.33 𝑚
29800
Problem 6.20
Problem
6.23
[Difficulty: 4]
6.20
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
q
x
du
∂
d
= − ⎛⎜ q ⋅ ⎞ = −
v =− u =−
h
dx
dx ⎝ h ⎠
∂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
y⎞
so
v ( y ) = q ⋅ ⎛⎜ 1 −
u
x q
y
ax = q ⋅ ⋅ ⎛⎜ ⎞ + q ⋅ ⎛⎜ 1 − ⎞ ⋅ ( 0 )
h ⎝h⎠
h⎠
⎝
⎝
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 ( x ) − p atm =
p dx = ⎮ −ρ⋅ ⋅ x dx = −ρ⋅
⋅ x + ρ⋅
⋅b
⎮ ∂x
2
2
2
⎮
h
2⋅ h
8⋅ h
⎮
⎮b
⌡b
⌡
2
2 2
p ( x ) = p atm + ρ⋅
q ⋅b
⎡
⋅ ⎢1 − 4 ⋅ ⎛⎜
⎥
⎝b⎠ ⎦
8⋅ h ⎣
2
x⎞
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
pg( x) =
b
ρ⋅ q ⋅ b
8⋅ h
2
⎡
⋅ ⎢1 − 4 ⋅ ⎛⎜
⎣
x⎞
2⎤
⎥
⎝b⎠ ⎦
b
⌠2
⌠2
2
2 2
⎮
⎮ ρ⋅ q 2⋅ b 2 ⎡
ρ⋅ q ⋅ b ⋅ L ⎛ b
x ⎤
1 b
Fnet = 2 ⋅ L⋅ ⎮ p g ( x ) dx = 2 ⋅ L⋅ ⎮
⋅ ⎢1 − 4 ⋅ ⎛⎜ ⎞ ⎥ dx =
⋅⎜ − ⋅ ⎞
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
q =
Solving for q for the given mass/length
12⋅ h ⋅ g ⋅ M length
ρ⋅ b
q = 0.0432⋅
3
s
2
m
b
The maximum speed
b⎞
b⋅ q
Umax =
2⋅ h
Umax = u ⎛⎜ x =
= q⋅
h
2⎠
⎝
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 6.21
Problem
6.24
6.21
[Difficulty: 3]
Problem 6.22
Problem
6.26
6.22
[Difficulty: 4] Part 1/2
Problem 6.26
[Difficulty: 4] Part 2/2
Problem 6.23
Problem
6.28
[Difficulty: 2]
6.23
Given:
Velocity field for doublet
Find:
Expression for pressure gradient
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 ⋅ sin( θ) ⎞
⎜ 2
⎝ r
⎠
r
2⎤
⎥
⎥
⎥
⎥⎦
∂
∂r
2
p =
2⋅ Λ ⋅ ρ
5
r
Vθ
Vr⋅ Vθ ⎞
⎛ ∂
1 ∂
∂
⋅ p = ρ⋅ ⎜ Vr⋅ Vθ +
⋅ Vθ +
r ∂θ
r ∂θ
r ⎠
⎝ ∂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 ⋅ sin( θ) ⎞
⎜ 2
⎝ r
⎠ ⋅ ∂ ⎛ − Λ ⋅ sin( θ)⎞ +
r
∂θ ⎜ r2
⎝
⎠
⎛ − Λ ⋅ sin( θ)⎞ ⋅ ⎛ − Λ ⋅ cos( θ) ⎞ ⎤
⎥
⎜ 2
⎜ 2
⎝ r
⎠⎝ r
⎠⎥
⎥
r
⎦
∂
∂θ
p =0
Problem 6.24
Problem
6.30
[Difficulty: 2]
6.24
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 6.25
Problem
6.32
[Difficulty: 3]
6.25
P6.25
.
Given:
Velocity field for free vortex flow in elbow
Find:
Similar solution to Example 6.1; find k (above)
Solution:
Basic equation
∂
∂r
2
p =
ρ⋅ V
c
V = Vθ =
r
with
r
Assumptions: 1) Frictionless 2) Incompressible 3) free vortex
2
∂
p =
2
ρ⋅ V
ρ⋅ c
d
=
p =
3
r
dr
r
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 = ⎮
⋅⎜
−
=
3
2
2 2
2 ⎜ 2
r
r1
r2
2⋅ r1 ⋅ r2
⎮
⎝
⎠
⌡r
so
∂r
r
(1)
1
Next we obtain c in terms of Q
⌠ →→
⎮
Q = ⎮ V dA =
⌡
r
r
⌠2
⌠ 2 w⋅ c
⎛ r2 ⎞
dr = w⋅ c⋅ ln⎜
⎮ V⋅ w dr = ⎮
r
⎮
⌡r
⎝ r1 ⎠
⌡r
1
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 6.26
Problem
6.34
6.26
[Difficulty: 2]
6.25,
r2/r1
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 6.25 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 ⎞
⎝
⎠
or
⋅ ∆p
Q⋅ ρ
w⋅ r1 ⋅ ∆p
=
(1)
⎛ r2 ⎞ ⎛ r2 ⎞
2
⎜ ⋅ ln⎜ r ⋅
⎝ r1 ⎠ ⎝ 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 6.27
Problem
6.36
[Difficulty: 4]
6.27
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
(x2 + y2)
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
(x2 + y2)
3
2
) dy
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 +
(x2 + y2)
∂
∂y
v =0
)
2
2
∂
∂x
u =
(2
2⋅ Λ⋅ x⋅ x − 3⋅ y
(x2 + y2)
3
2
)
v=−
2⋅ Λ⋅ x⋅ y
(x2 + y2)
∂
2
∂y
v =−
(2
2⋅ Λ⋅ x⋅ x − 3⋅ y
(x2 + y2)
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
⎥ ⋅⎢
3
⎢ 2 2 2⎥ ⎢
2
2
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 ⋅
(
)
2
2
2
2⋅ Λ⋅ x⋅ y ⎤ ⎡ 2⋅ Λ⋅ y⋅ 3⋅ x − y ⎤
2Λ
⎥ax = − ⋅ ⋅ x
⋅⎢
⎢
⎥
3
3
⎥
⎢ x 2 2 2⎥ ⎢
2
2
2
2
x +y
x +y
+y ⎦⎣
⎣
⎦
⎥
⎦
(
)
(
)
m
s
v = 0⋅
)⎤⎥ + ⎡−
y = 1m
)
(
)
2
2
2
2⋅ Λ⋅ x⋅ y ⎤ ⎡ 2⋅ Λ⋅ y⋅ 3⋅ y − x ⎤
2Λ
⎥ay = − ⋅ ⋅ y
⋅⎢
⎢
⎥
3
3
⎥
⎢ x 2 + y2 2⎥ ⎢
2
2
2
2
x +y
x +y
⎣
⎦⎣
⎦
⎥
⎦
m
(
)
ax = 0 ⋅
s
(
m
2
)
s
m
ax = 0 ⋅
s
m
2
ax = 0 ⋅
s
m
2
⎛ 2⋅ m ⎞
⎜ s
⎝
⎠
r =
or
m
2
ay = −0.25⋅
m
2
s
ay = −0.0333⋅
s
u
aradial = −ay = −
r
)
s
s
m
8⋅
(
ay = −8 ⋅
m
2
s
2
The instantaneous radius of curvature is obtained from
For the three points
(
v
⎡ Λ⋅ x2 − y2 ⎤ ⎡ 2⋅ Λ⋅ y⋅ 3⋅ x2 − y2
⎥ ⋅⎢
3
⎢ 2 2 2⎥ ⎢
2
2
x +y
⎣ x +y ⎦⎣
ay = ⎢−
)⎤⎥ + ⎡−
r= −
u
2
ay
2
r = 0.5 m
m
2
s
y = 2m
⎛ 0.5⋅ m ⎞
⎜
s⎠
⎝
r =
0.25⋅
2
r = 1m
m
2
s
y = 3m
⎛ 0.2222⋅ m ⎞
⎜
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
2⋅ Λ⋅ x⋅ y
−
(x2 + y2) = 2⋅x⋅y
=
=
2
2
dx
u
(x2 − y2)
Λ⋅ (x − y )
−
2
(x2 + y2)
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
) ⋅dy = 0
2
The equation becomes an exact integral
x
x −y
−2 ⋅ ⋅ dx +
2
y
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 6.28
Problem
6.38
[Difficulty: 1]
6.28
Given:
Water at speed 25 ft/s
Find:
Dynamic pressure in in. Hg
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 6.29
Problem
6.40
[Difficulty: 2]
6.29
Given:
Air speed
Find:
Plot dynamic pressure in mm Hg
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 6.30
(Difficulty 2)
6.30 Water flows in a pipeline. At a point in the line where the diameter is 7 𝑖𝑖, the velocity is 12
𝑓𝑓
𝑠
and
the pressure is 50 𝑝𝑝𝑝. At a point 40 𝑓𝑓 away the diameter reduces to 3 𝑖𝑖. Calculate the pressure here
when the pipe is (a) horizontal, and (b) vertical with flow downward, and (c) vertical with the flow
upward. Explain why there is a difference in the pressure for the different situations.
Find: Calculate the pressure 𝑝2 .
Assumption: The flow is steady, incompressible, uniform, and frictionless.
Solution: Apply the continuity and Bernoulli equations to find the pressure
The continuity equation is:
𝑄 = 𝑉1 𝐴1 = 𝑉2 𝐴2
The Bernoulli equation along a streamline is
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝜌 2
The velocity is the same for each orientation and is calculated using the continuity equation. The flow
areas are:
𝐴1 =
𝜋𝐷12
4
𝜋𝐷22
𝐴2 =
4
The velocity V2 is
2
𝑓𝑓
7
𝑉1 𝐴1 𝑉1 𝐷12 12 𝑠 × �12 𝑓𝑓�
𝑓𝑓
𝑉2 =
= 2 =
= 65.3
2
𝑠
𝐴2
𝐷2
3
� 𝑓𝑓�
12
(a) For the horizontal pipe, the height cancels out. The Bernoulli equation along the center
streamline is:
𝑝1 𝑉12 𝑝2 𝑉22
+
= +
𝜌
2
𝜌
2
The inlet pressure is
𝑙𝑙𝑙
𝑝1 = 50 𝑝𝑝𝑝 = 7200 2
𝑓𝑓
𝑠𝑠𝑠𝑠
𝑙𝑙𝑙 ∙ 𝑠2
𝜌 = 1.938
=
1.938
𝑓𝑓 3
𝑓𝑓 4
So the pressure is computed as:
𝑝2 = 7200
𝑙𝑙𝑙
+
𝑓𝑓 2
1.938
𝜌
𝑝2 = 𝑝1 + (𝑉12 − 𝑉22 )
2
𝑙𝑙𝑙 ∙ 𝑠2
𝑓𝑓 2
𝑓𝑓 2
𝑙𝑙𝑙
𝑓𝑓 4
× ��12 � − �65.3 � � = 3208 2 = 22.3 𝑝𝑝𝑝
𝑠
𝑠
𝑓𝑓
2
(b) For the vertical pipe with the flow downward, we have for the Bernoulli equation:
𝑝2 𝑉22
𝑝1 𝑉12
+
+ 𝑔𝑧1 = +
+ 𝑔𝑧2
𝜌
2
𝜌
2
𝜌
𝑝2 = 𝑝1 + (𝑉12 − 𝑉22 ) + 𝜌𝑔(𝑧1 − 𝑧2 )
2
Using the same values for velocity and pressure p1 as for the horizontal situation, we have the additional
pressure due to the height difference
𝑝2 = 3208
𝑙𝑙𝑙
𝑙𝑙𝑙
𝑙𝑙𝑙 ∙ 𝑠2
𝑓𝑓
)
=
3208
+
𝜌𝜌(𝑧
−
𝑧
+
1.938
× 32.2 2 × (40 − 0)𝑓𝑓
1
2
2
2
4
𝑓𝑓
𝑓𝑓
𝑠
𝑓𝑓
𝑝2 = 5704
𝑙𝑙𝑙
= 39.6 𝑝𝑝𝑝
𝑓𝑓 2
(c) For the vertical pipe with the flow upward, we have for the Bernoulli equation, where the flow is now
from 2 to 1:
𝑝1 𝑉12
𝑝2 𝑉22
+
+ 𝑔𝑧2 = +
+ 𝑔𝑧1
𝜌
2
𝜌
2
The pressure at the small end (2) is then
𝜌
𝑝2 = 𝑝1 + (𝑉12 − 𝑉22 ) + 𝜌𝜌(𝑧1 − 𝑧2 )
2
Using the same values for velocity and pressure p1 as for the horizontal situation, we have
𝑝2 = 3208
𝑙𝑙𝑙
𝑙𝑙𝑙
𝑙𝑙𝑙 ∙ 𝑠2
𝑓𝑓
)
=
3208
+
𝜌𝜌(𝑧
−
𝑧
+
1.938
× 32.2 2 × (40 − 0)𝑓𝑓
1
2
4
𝑓𝑓 2
𝑓𝑓 2
𝑠
𝑓𝑓
𝑝2 = 5704
which is the same as for the flow downward.
𝑙𝑙𝑙
= 39.6 𝑝𝑝𝑝
𝑓𝑓 2
The pressures for the vertical orientation are greater than for the horizontal orientation due to the
hydrostatic pressure. The pressures for the vertical directions are the same since the hydrostatic pressure
difference is the same regardless of flow direction.
Problem 6.31
(Difficulty 2)
6.31 In a pipe 0.3 𝑚 in diameter, 0.3
𝑚3
𝑠
of water are pumped up a hill. On the hilltop (elevation 48),
the line reduces to 0.2 𝑚 diameter. If the pump maintains a pressure of 690 𝑘𝑘𝑘 at elevation 21,
calculate the pressure in the pipe on the hilltop.
Find: Calculate the pressure 𝑝2 .
Assumption: The flow is steady, incompressible, uniform, and frictionless.
Solution: Apply the continuity and Bernoulli equations to find the pressure
The continuity equation is:
𝑄 = 𝑉1 𝐴1 = 𝑉2 𝐴2
The Bernoulli equation along a streamline is
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝜌 2
The velocity can be calculated using the continuity equation:
𝐴1 =
The velocity at location 1 is
𝐴2 =
𝜋𝐷12
4
𝜋𝐷22
4
𝑚3
0.3
𝑄
𝑚
𝑠
=
= 4.24
𝑉1 =
2
(0.3
𝑚)
𝐴1 𝜋 ×
𝑠
4
And at location 2
𝑚3
0.3
𝑄
𝑚
𝑠
=
= 9.55
𝑉2 =
2
𝐴2 𝜋 × (0.2 𝑚)
𝑠
4
Applying the Bernoulli equation from inlet and outlet, we have:
𝑝2 𝑉22
𝑝1 𝑉12
+
+ 𝑔𝑧1 = +
+ 𝑔𝑧2
𝜌
2
𝜌
2
Or the pressure is
𝑝2 = 690 𝑘𝑘𝑘 +
𝜌
𝑝2 = 𝑝1 + (𝑉12 − 𝑉22 ) + 𝜌𝜌(𝑧1 − 𝑧2 )
2
𝑘𝑘
2
2
𝑚3 × ��4.24 𝑚� − �9.55 𝑚� � + 998 𝑘𝑘 × 9.81 𝑚 × (−27 𝑚)
𝑠
𝑠
𝑚3
𝑠2
2
998
𝑝2 = 389 𝑘𝑘𝑘
Problem 6.32
Problem
6.42
[Difficulty: 2]
6.32
Given:
Air jet hitting wall generating pressures
Find:
Speed of air at two locations
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
⎜
3
2
s⎠
2.65 kg
1000⋅ mm
⎝
m
s
V = 44.8
m
s
Problem 6.33
Problem
6.44
[Difficulty: 2]
6.33
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
2
⋅ ρair⋅ U
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 6.34
Problem
6.45
6.34
[Difficulty: 2]
Problem 6.35
Problem
6.46
6.35
[Difficulty: 2]
Problem 6.36
(Difficulty 3)
6.36 Water is flowing. Calculate 𝐻 (𝑚) and 𝑝 (𝑘𝑘𝑘).
Find: The manometer reading and the pressure
Assumption: The flow is steady, incompressible, uniform, and frictionless.
Solution: Apply the continuity and Bernoulli equations to find the pressure
The continuity equation is:
𝑄 = 𝑉1 𝐴1 = 𝑉2 𝐴2
The Bernoulli equation along a streamline is
From the continuity equation we have:
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝜌 2
𝑄 = 𝑉1 𝐴1 = 𝑉2 𝐴2 = 𝑉3 𝐴3
𝑉12 𝐴23 𝐷34
=
=
𝑉32 𝐴12 𝐷14
𝑉22 𝐴23 𝐷34
=
=
𝑉32 𝐴22 𝐷24
The hydrostatic pressure is determined from the manometer reading:
𝑝1 + 𝛾𝐻2 𝑜 𝑧1 = 𝑝2 + 𝛾𝐻2 𝑜 (𝑧2 − 0.175) + 𝛾𝐻𝐻 (0.175)
�
𝛾𝐻𝐻
𝑝1
𝑝2
(0.175)
+ 𝑧1 � − �
+ 𝑧2 � = −0.175 +
𝛾𝐻2𝑜
𝛾𝐻2 𝑜
𝛾𝐻2𝑜
Appling the Bernoulli equation between points 1 and 2 as:
Or
�
1
𝑝2
1
𝑝1
+ 𝑔𝑧1 + 𝑉12 =
+ 𝑔𝑧2 + 𝑉22
2
2
𝜌𝐻2 𝑜
𝜌𝐻2 𝑜
𝑝1
1 2
𝑝2
1 2
+ 𝑧1 +
𝑉 =
+ 𝑧2 +
𝑉
2𝑔 1
2𝑔 2
𝛾𝐻2 𝑜
𝛾𝐻2 𝑜
𝑝2
𝑉22 𝑉12
𝐷34 𝐷34 𝑉32
𝑝1
+ 𝑧1 � − �
+ 𝑧2 � =
−
= � 4 − 4�
𝛾𝐻2𝑜
𝛾𝐻2 𝑜
2𝑔 2𝑔
𝐷2 𝐷1 2𝑔
Combining the equations we have:
�
𝛾𝐻𝐻
𝐷34 𝐷34 𝑉32
(0.175)
4 − 4 � 2𝑔 = −0.175 + 𝛾
𝐷2 𝐷1
𝐻2 𝑜
𝛾𝐻𝑔
= 13.57
𝛾𝐻2 𝑜
𝑉32 −0.175 𝑚 + 13.57 × (0.175 𝑚)
=
= 11.78 𝑚
(0.075 𝑚)4 (0.075 𝑚)4
2𝑔
�
−
�
(0.1 𝑚)4
(0.125 𝑚)4
Appling the Bernoulli equation from the water surface to the out let we have:
𝐻=
𝑉32
= 11.78 𝑚
2𝑔
Appling the Bernoulli equation from section 2 to section 3
𝑉22 𝑉32
𝑝2
+
=
𝛾𝐻2 𝑜 2𝑔 2𝑔
Thus the pressure at location 2 is
𝑝2 = 9810
𝑉32
𝐷34
𝑝2
=
�1 − 4 �
𝛾𝐻2 𝑜 2𝑔
𝐷2
𝛾𝐻2 𝑜 𝑉32
𝐷34
𝑝2 =
�1 − 4 �
2𝑔
𝐷2
(0.075 𝑚)4
𝑁
×
11.78
𝑚
×
�1
−
� = 78.9 𝑘𝑘𝑘
(0.1 𝑚)4
𝑚3
Problem 6.37
(Difficulty 2)
6.37 If each gage shows the same reading for a flow rate of 1.00
constriction?
𝑓𝑓 3
.
𝑠
What is the diameter of the
Find: Calculate the diameter of the constriction 𝐷𝐴 .
Assumption: The flow is steady, incompressible, uniform, and frictionless.
Solution: Apply the continuity and Bernoulli equations to find the pressure
The continuity equation is:
𝑄 = 𝑉1 𝐴1 = 𝑉2 𝐴2
The Bernoulli equation along a streamline is
From the continuity equation:
The velocity at B can be calculated by:
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝜌 2
𝑄 = 𝑉𝐴 𝐴𝐴 = 𝑉𝐵 𝐴𝐵
𝜋𝐷𝐴2
𝐴𝐴 =
4
𝐴𝐵 =
𝜋𝐷𝐵2
4
𝑄
𝑉𝐵 =
=
𝐴𝐵
1.00
𝜋×�
𝑓𝑓 3
𝑠
3
𝑓𝑓�
12
4
2
= 20.4
𝑓𝑓
𝑠
Apply the Bernoulli equation for the streamline from A to B we have:
As each gage shows the same reading:
𝑝𝐵 𝑉𝐵2
𝑝𝐴 𝑉𝐴2
+
+ 𝑔𝑧𝐴 =
+
+ 𝑔𝑧𝐵
𝜌
2
𝜌
2
𝑝𝐴 = 𝑝𝐵
The velocity at A is then
𝑉𝐴 = �𝑉𝐵2 + 2𝑔(𝑧𝐵 − 𝑧𝐴 ) = ��20.4
𝑓𝑓 2
𝑓𝑓
𝑓𝑓
� + 2 × 32.2 2 × (108 𝑓𝑓 − 90 𝑓𝑓) = 39.7
𝑠
𝑠
𝑠
So the constriction diameter is from the continuity equation:
𝐴𝐴 =
𝜋𝐷𝐴2 𝑄
=
𝑉𝐴
4
𝑓𝑓 3
4 × 1.00
4𝑄
𝑠 = 0.179 𝑓𝑓 = 2.15 𝑖𝑖
𝐷𝐴 = �
=�
𝑓𝑓
𝜋𝑉𝐴
𝜋 × 39.7
𝑠
Problem 6.38
(Difficulty 2)
𝑚3
6.38 Derive a relation between 𝐴1 and 𝐴2 so that for a flow rate of 0.28
the static pressure will be
𝑠
the same at sections 1 and 2. Also calculate the manometer reading for this condition and state which
leg has the higher mercury column.
Assumption: The flow is steady and incompressible
Solution: Use the continuity and Bernoulli equations together with manometer relations to find the height
of the mercury column.
From the continuity equation:
𝑄 = 𝑉1 𝐴1 = 𝑉2 𝐴2
Appling the Bernoulli equation from section 1 to section 2 we have:
𝑝1
1
𝑝2
1
+ 𝑔𝑧1 + 𝑉12 =
+ 𝑔𝑧2 + 𝑉22
2
2
𝜌𝑔𝑔𝑔
𝜌𝑔𝑔𝑔
We also have the pressures at location 1 and 2 as specified as equal:
From the Bernoulli equation
𝑝1 = 𝑝2
1 2 1 2
𝑉 − 𝑉 = 𝑔(𝑧2 − 𝑧1 )
2 1 2 2
Using the continuity equation to replace the velocities with the volume flow rate and areas
𝑉12 − 𝑉22 =
The relation between the areas is then
𝑄2 𝑄2
−
= 2𝑔(𝑧2 − 𝑧1 )
𝐴12 𝐴22
𝑚
1
2𝑔(𝑧2 − 𝑧1 ) 2 × 9.81 𝑠2 × (−1.5 𝑚)
1
1
=
= −375 4
2− 2 =
2
2
3
𝑚
𝑄
𝐴1 𝐴2
𝑚
�
�0.28
𝑠
For the static pressure equation of the manometer we have:
𝑝1 + 𝛾𝑔𝑔𝑔 (1.5 + ℎ) = 𝑝2 +𝛾𝐻𝐻 ℎ
ℎ=
𝛾𝑔𝑔𝑔 (1.5 + ℎ) = 𝛾𝐻𝐻 ℎ
𝛾𝑔𝑔𝑔 × 1.5 𝑚 0.85 × 1.5 𝑚
=
= 0.1002 𝑚 = 100.2 𝑚𝑚
13.57 − 0.85
𝛾𝐻𝐻 − 𝛾𝑔𝑔𝑔
The mercury column is higher in the left hand leg of the manometer.
Problem 6.39
Problem
6.48
[Difficulty: 2]
6.39
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
⎞
1
2
p 1 = ρ⋅ ⎜
+ g ⋅ z2
2
⎝
⎠
When it is inverted
z2 = −10⋅ ft
⎛⎜ V 2 − V 2
⎞
1
2
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 6.40
Problem
6.50
[Difficulty: 2]
6.40
Given:
Siphoning of gasoline
Find:
Flow rate
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 6.41
Problem
6.52
[Difficulty: 2]
6.41
Given:
Ruptured pipe
Find:
Height benzene rises from tank
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 6.42
Problem
6.54
[Difficulty: 3]
6.42
Given:
Flow rate through siphon
Find:
Maximum height h to avoid cavitation
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
p atm − p v
2
Hence
V
p A = p v = p atm − ρ⋅ g ⋅ h − ρ⋅
2
Hence
m
s
h = ( 101 − 2.358 ) × 10 ⋅
×
⋅
×
×
− × ⎛⎜ 10.2 ⎞ ×
h = 4.76 m
2
9.81⋅ m
2
9.81⋅ m
999 kg
2 ⎝
s⎠
m
N⋅ s
3 N
h=
or
1
3
m
2
s
ρ⋅ g
kg⋅ m
−
V
2⋅ g
1
2
2
Problem 6.43
Problem
6.56
[Difficulty: 2]
6.43
Given:
Flow through tank-pipe system
Find:
Velocity in pipe; Rate of discharge
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
Combining equations
ρ⋅ g ⋅ H − ρ⋅
Note that we have water on one side and mercury on
the other of the manometer
2
Hence
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 6.44
Problem
6.58
[Difficulty: 3]
6.44
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
⎜
⎜
⎜
0.861 ⋅ kg
2
2
hr ⎠
s
⎝
⎝ 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 6.45
(Difficulty 1)
6.45 A closed tank contains water with air above it. The air is maintained at a gage pressure of 150 𝑘𝑘𝑘
and 3 𝑚 below the water surface a nozzle discharges into the atmosphere. At what velocity will water
emerge from the nozzle?
Find: Calculate the manometer reading for this condition.
Assumption: The flow is steady, incompressible, uniform, and frictionless.
Solution: Apply the Bernoulli equation to find the pressure
The Bernoulli equation along a streamline is
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝜌 2
Appling the Bernoulli equation for the streamline from the interface to outlet we have:
𝑝2 𝑉22
𝑝1 𝑉12
+
+ 𝑔𝑧1 = +
+ 𝑔𝑧2
𝜌
2
𝜌
2
The pressure at the nozzle is atmosphere pressure:
And we also have:
𝑝2 = 0
𝑧2 = 0
𝑉1 = 0
So we obtain:
The exit velocity is
𝑉22
𝑝1
+ 𝑔𝑧1 =
𝜌
2
2𝑝1
𝑉2 = �
+ 2𝑔𝑧1
𝜌
𝑚
𝑚
2 × 150 × 103 𝑃𝑃
𝑉2 = �
+ 2 × 9.81 2 × 3𝑚 = 19.0
𝑘𝑘
𝑠
𝑠
998 3
𝑚
Problem 6.46
(Difficulty 2)
6.46 Water jets upward through a 3 𝑖𝑖 diameter nozzle under a head of 10 𝑓𝑓. At what height ℎ will be
the liquid stand in the pitot tube? What is the cross-sectional area of jet at section B?
Find: The height ℎ and cross-sectional area 𝐴𝐵 .
Assumption: The flow is steady, incompressible, uniform, and frictionless.
Solution: Apply the continuity and Bernoulli equations to find the pressure
The continuity equation is:
𝑄 = 𝑉1 𝐴1 = 𝑉2 𝐴2
The Bernoulli equation along a streamline is
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝜌 2
The pitot tube reading will be identical to total head at the nozzle outlet:
ℎ = 10 𝑓𝑓
Appling the Bernoulli equation of the streamline from top to the bottom across B we have:
𝑝𝑜 𝑉𝑜2
𝑝𝑡 𝑉𝑡2
𝑝𝐵 𝑉𝐵2
+
+ 𝑔𝑧𝐵 =
+
+ 𝑔𝑧𝑜 = +
+ 𝑔𝑧𝑡
𝜌
2
𝜌
2
𝜌
2
Where we have for this situation
𝑝𝑡 = 0, 𝑉𝑡 = 0, 𝑧𝑡 = ℎ, 𝑝𝑜 = 0, 𝑧𝑜 = 0, 𝑝𝐵 = 0
The Bernoulli equation reduces to
0+
𝑉𝑜2
𝑉𝐵2
+ 𝑔𝑧𝐵 = 0 +
+ 0 = 0 + 0 + 𝑔ℎ
2
2
The velocity for the outlet can be found as:
𝑉𝑜 = �2𝑔ℎ = �2 × 32.2
The cross-sectional area at outlet is:
The volumetric flow rate is:
𝑓𝑓
𝑓𝑓
×
10
𝑓𝑓
=
25.4
𝑠2
𝑠
2
3
𝜋𝐷𝑜2 𝜋 × �12 𝑓𝑓�
=
= 0.0491 𝑓𝑓 2
𝐴𝑜 =
4
4
𝑄 = 𝑉𝑜 𝐴0 = 25.4
𝑓𝑓
𝑓𝑓 3
× 0.0491 𝑓𝑓 2 = 1.247
𝑠
𝑠
For the jet cross-section B, from the Bernoulli equation:
The outlet area is then
𝑉𝐵 = �2𝑔(𝑧𝑡 − 𝑧𝐵 ) = �2 × 32.2
𝑓𝑓
𝑓𝑓
× 5 𝑓𝑓 = 17.94
2
𝑠
𝑠
𝑓𝑓 3
1.247
𝑄
𝑠 = 0.0695 𝑓𝑓 2
=
𝐴𝐵 =
𝑓𝑓
𝑉𝐵
17.94
𝑠
Problem 6.47
(Difficulty 2)
6.47 Calculate the rate of flow through this pipeline and the pressures at 𝐴, 𝐵, 𝐶 and 𝐷. Sketch the 𝐸𝐸
and 𝐻𝐻𝐻 showing vertical distances.
Find: The flow rate and the pressures
Assumption: The flow is steady, incompressible, uniform, and frictionless.
Solution: Apply the continuity and Bernoulli equations to find the pressure
The continuity equation is:
𝑄 = 𝑉1 𝐴1 = 𝑉2 𝐴2
The Bernoulli equation along a streamline is
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝜌 2
Applying the Bernoulli equation from the water surface 1 to the outlet 2, we have:
Where
𝑝2 𝑉22
𝑝1 𝑉12
+
+ 𝑔𝑧1 = +
+ 𝑔𝑧2
𝜌
2
𝜌
2
𝑝1 = 0, 𝑉1 = 0, 𝑧1 = 𝐻, 𝑝2 = 0, 𝑧2 = 0,
The Bernoulli equation then becomes
0 + 0 + 𝑔𝑔 = 0 +
𝑉22
+0
2
Or, solving for the outlet velocity
𝑉22
=𝐻
2𝑔
𝑉2 = �2𝑔𝐻
The volumetric flow rate is the
= �2 × 32.2
𝑓𝑓
𝑓𝑓
× 12 𝑓𝑓 = 27.8
2
𝑠
𝑠
2
𝑓𝑓 𝜋
2
𝑓𝑓 3
𝑄 = 𝑉𝐴2 = 27.8
× × � 𝑓𝑓� = 0.606
𝑠 4
12
𝑠
The velocities at 𝐴, 𝐵, 𝐶 𝑎𝑎𝑎 𝐷 are the same and using the continuity equation are:
𝑓𝑓 3
0.606
𝑄
𝑓𝑓
𝑠
= 3.09
𝑉𝐴 = 𝑉𝐵 = 𝑉𝐶 = 𝑉𝐷 = =
2
𝐴 𝜋
𝑠
6
× � 𝑓𝑓�
12
4
The velocity head in pipe is the same at these locations:
𝑓𝑓 2
�
𝑠
=
= 0.148 𝑓𝑓
𝑓𝑓
2𝑔
2 × 32.2 2
𝑠
𝑉2
�3.09
Apply Bernoulli equation from water surface to A we have:
𝑝𝐴 = 𝛾𝛾 − 𝛾
From the surface to B
𝑝𝐴 𝑉𝐴2
+
+ 𝑧𝐴
𝐻=
𝛾 2𝑔
𝑉𝐴2
𝑙𝑙𝑙
𝑙𝑙𝑙
𝑙𝑙𝑙
= 62.4 3 × 20 𝑓𝑓 − 62.4 3 × 0.148 𝑓𝑓 = 1238 2 = 8.59 𝑝𝑝𝑝
𝑓𝑓
𝑓𝑓
𝑓𝑓
2𝑔
𝐻=
𝑝𝐵 = −𝛾
From the surface to C
𝑝𝐵 𝑉𝐵2
+
+𝐻
𝛾 2𝑔
𝑉𝐵2
𝑙𝑙𝑙
𝑙𝑙𝑙
= −62.4 3 × 0.148 𝑓𝑓 = −9.2352 2 = −0.0641 𝑝𝑝𝑝
𝑓𝑓
𝑓𝑓
2𝑔
𝐻=
𝑝𝑐 𝑉𝑐2
+
+𝐻+5
𝛾 2𝑔
𝑝𝑐
= −5 𝑓𝑓 − 0.148 𝑓𝑓 = −5.148 𝑓𝑓
𝛾
From the surface to D
𝑝𝑐 = −62.4
𝑙𝑙𝑙
𝑙𝑙𝑙
× 5.148 𝑓𝑓 = −321 2 = −2.23 𝑝𝑝𝑝
3
𝑓𝑓
𝑓𝑓
𝐻=
𝑝𝐷 𝑉𝐷2
+
+ 𝑧𝐷
𝛾 2𝑔
𝑝𝐷
= 20 𝑓𝑓 − 8𝑓𝑓 − 0.148 𝑓𝑓 = 11.852 𝑓𝑓
𝛾
𝑝𝐷 = 62.4
𝑙𝑙𝑙
𝑙𝑙𝑙
× 11.852 𝑓𝑓 = 740 2 = 5.14 𝑝𝑝𝑝
3
𝑓𝑓
𝑓𝑓
The energy and hydraulic grade lines are sketched below
Problem 6.48
Problem
6.60
6.48
[Difficulty: 3]
Problem 6.49
Problem
6.62
[Difficulty: 2]
6.49
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 ⎞
⎜
⎝ V1 ⎠
⎡
⎢
∆p = ⋅ ρ⋅ V1 ⋅ 1 −
⎢
2
⎣
The freestream dynamic pressure is
Then
p atm
2
V2
2
2⎤
⎥
⎥
⎦
⎛ V2 ⎞
⎜
⎝ V1 ⎠
p 2 = 15.6 psi
2⎤
⎥
⎥
⎦
∆p = 0.917 ⋅ psi
2
q = 0.980 ⋅ psi
q =
1
∆p
= 93.5⋅ %
q
2
⋅ ρ⋅ V1
Note that at this speed the flow is borderline compressible!
Problem 6.50
Problem
6.64
[Difficulty: 3]
6.50
6.6.
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
v→0
V→0
v=0
)
2
V=
q⋅ x
(2
π⋅ x + h
)
2
2
q⋅ x
⎡
⎤
=
+ ⋅
ρ
2
2 ⎥
ρ
2 ⎢
⎣ π⋅ x + h ⎦
p atm
Hence the Bernoulli equation becomes
p
1
(
)
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
(x2 + h2)
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 ⎞
1
10
N⋅ s
F = −
× 1000⋅
× ⎜ 2⋅
×
× ⎛⎜ −
+ atan( 10) ⎞ ×
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 6.51
Problem
6.66
6.51
[Difficulty: 3]
Problem 6.52
Problem
6.68
6.52
[Difficulty: 3]
Problem 6.53
Problem
6.70
[Difficulty: 3]
6.53
Given:
Flow nozzle
Find:
Mass flow rate in terms of ∆p, 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
2⋅ (p2 − p1)
⎛ D2 ⎞
V2 − V2 ⋅ ⎜
=
ρ
⎝ D1 ⎠
(
4
Hence
Then the mass flow rate is
Using
For a flow nozzle
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 ⎞
⎜
⎝ D1 ⎠
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 ⎞
⎜
⎝ D1 ⎠
4⎤
⎥
⎥
⎦
∆p⋅ ρ
⎡
⎢
⎢1 −
⎣
⎛ D2 ⎞
⎜
⎝ D1 ⎠
4⎤
⎥
⎥
⎦
∆p⋅ p 1
⎡
⎢
R ⋅ T1 ⋅ 1 −
⎢
⎣
⎛ D2 ⎞
⎜
⎝ D1 ⎠
4⎤
⎥
⎥
⎦
p1
⎡
⎢
R ⋅ T1 ⋅ 1 −
⎢
⎣
4⎤
⎛ D2 ⎞ ⎥
⎜
⎥
⎝ D1 ⎠ ⎦
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 6.54
(Difficulty 3)
6.54 The head of water on a 50 𝑚𝑚 diameter smooth nozzle is 3 𝑚. If the nozzle is directed upward at
angles of (a) 30°, (b) 45°, (c) 60°, and (d) 90°, how high above the nozzle will the jet rise, and how far
from the nozzle will the jet pass through the horizontal plane in which the nozzle lies? What is the
diameter of the jet at the top of the trajectory?
Find: The height ℎ, the distance 𝑥 and the diameter of jet on the top.
Assumption: The flow is steady, incompressible, uniform, and frictionless.
Solution: Apply the continuity and Bernoulli equations to find the pressure
The continuity equation is:
𝑄 = 𝑉1 𝐴1 = 𝑉2 𝐴2
The Bernoulli equation along a streamline is
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝜌 2
The total head is the 3 m. The velocity leaving the nozzle is then:
𝑉 = �2𝑔𝑔 = �2 × 9.81
The horizontal velocity at the outlet is:
𝑚
𝑚
×
3𝑚
=
7.67
𝑠2
𝑠
The vertical velocity at the outlet is:
𝑉𝑥 = 𝑉 cos 𝜃
The maximum height can be calculated by:
𝑉𝑦 = 𝑉 sin 𝜃
ℎ=
𝑉𝑦2
2𝑔
or
2
𝑚
𝑉𝑦2 �7.67 𝑠 × sin 90°�
ℎ=
=
= 3.00 𝑚 = 𝐻
𝑚
2𝑔
2 × 9.81 2
𝑠
The maximum height is the total head.
To calculate the jet trajectory, we follow a particle of water from the time it leaves the nozzle. Applying
Newton’s second law
𝐹 = 𝑚𝑚 = 𝑚
𝑑𝑉𝑦
𝑑𝑑
The force on the particle is its weight W, and acts downward. The mass is the weight divided by g.
Newton’s law is then
Or, separating variables
−𝑊 =
𝑊 𝑑𝑉𝑦
𝑔 𝑑𝑑
𝑑𝑉𝑦 = −𝑔 𝑑𝑑
Integrating from the initial velocity Vy at time equal 0, the y-component of velocity at any time is
𝑉𝑦 (𝑡) = 𝑉𝑦 − 𝑔𝑔
The time for jet to return to the outlet elevation is:
The distance x can be calculated by:
At the top of the jet,
𝑡=2
𝑉𝑦
𝑔
𝑥 = 𝑉𝑥 𝑡 = 𝑉𝑥 2
𝑉𝑦
𝑔
𝑉𝑦 = 0
The volumetric flow rate is calculated from the velocity leaving the nozzle, and equals the volume flow
rate at the top of the trajectory:
𝑄 = 𝑉𝑉 = 𝑉
𝜋𝑑𝑜2
𝜋𝑑𝑡2
= 𝑉𝑥
4
4
The diameter of the jet at the top of the trajectory is related to the nozzle diameter as
𝑉
𝑑𝑡 = 𝑑 𝑜 �
𝑉𝑥
𝑑𝑜 = 0.05 𝑚
(a) For 𝜃 = 30°, the height is
and the distance is
2
𝑚
𝑉𝑦2 �7.67 𝑠 × sin 30°�
ℎ=
=
= 0.75 𝑚
𝑚
2𝑔
2 × 9.81 2
𝑠
𝑚
𝑚
𝑉𝑦 2 × �7.67 𝑠 × sin 30°� × �7.67 𝑠 × cos 30°�
= 5.19 𝑚
𝑥 = 2𝑉𝑥 =
𝑚
𝑔
9.81 2
𝑠
The jet diameter is
𝑚
7.67
𝑉
𝑠
𝑑𝑡 = 𝑑𝑜 � = 0.05 𝑚 × �
= 53.7 𝑚𝑚
𝑚
𝑉𝑥
7.67 × cos 30°
𝑠
(b) For 𝜃 = 45°, the height is
and the distance is
2
𝑚
𝑉𝑦2 �7.67 𝑠 × sin 45°�
ℎ=
=
= 1.50 𝑚
𝑚
2𝑔
2 × 9.81 2
𝑠
𝑚
𝑚
𝑉𝑦 2 × �7.67 𝑠 × sin 45°� × �7.67 𝑠 × cos 45°�
=6𝑚
𝑥 = 2𝑉𝑥 =
𝑚
𝑔
9.81 2
𝑠
The jet diameter is
𝑚
7.67
𝑉
𝑠
𝑑𝑡 = 𝑑𝑜 � = 0.05 𝑚 × �
= 59.5 𝑚𝑚
𝑚
𝑉𝑥
7.67 × cos 45°
𝑠
(c) For 𝜃 = 60°, the height is
and the distance is
2
𝑚
𝑉𝑦2 �7.67 𝑠 × sin 60°�
ℎ=
=
= 2.25 𝑚
𝑚
2𝑔
2 × 9.81 2
𝑠
𝑚
𝑚
𝑉𝑦 2 × �7.67 𝑠 × sin 60°� × �7.67 𝑠 × cos 60°�
= 5.19 𝑚
𝑥 = 2𝑉𝑥 =
𝑚
𝑔
9.81 2
𝑠
The jet diameter is
𝑚
7.67
𝑉
𝑠
𝑑𝑡 = 𝑑𝑜 � = 0.05 𝑚 × �
= 70.7 𝑚𝑚
𝑚
𝑉𝑥
7.67 × cos 60°
𝑠
(d) For 𝜃 = 90° the distance is
𝑚
𝑚
𝑉𝑦 2 × �7.67 𝑠 × sin 90°� × �7.67 𝑠 × cos 90°�
𝑥 = 2𝑉𝑥 =
=0𝑚
𝑚
𝑔
9.81 2
𝑠
and the diameter is
𝑚
7.67
𝑉
𝑠
𝑑𝑡 = 𝑑𝑜 � = 0.05 𝑚 × �
= 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖
𝑚
𝑉𝑥
7.67 × cos 90°
𝑠
Problem 6.55
(Difficulty 3)
6.55 Water flows from one reservoir in a 200 𝑚𝑚 pipe, while water flows from a second reservoir in a
150 𝑚𝑚 pipe. The two pipes meet in a “tee” junction with a 300 𝑚𝑚 pipe that discharges to the
atmosphere at an elevation of 20 𝑚. If the water surface in the reservoirs is at 30 𝑚 elevation, what is
the total flow rate?
Find: The total flow rate.
Assumption: The flow is steady, incompressible, uniform, and frictionless.
Solution: Apply the continuity and Bernoulli equations to find the pressure
The continuity equation is:
The Bernoulli equation along a streamline is
𝑄=𝑉𝐴
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝜌 2
We will apply the Bernoulli equation along a streamline from water surface of the left hand reservoir to
the discharge of the pipe at z4. This assumes that the flow is frictionless without any irreversibilities such
as mixing of the fluid at the junction of pipe 1 and pipe 2. We will discuss this assumption at the end of
the problem.
The Bernoulli equation becomes
𝑝4 𝑉42
𝑝1 𝑉12
+
+ 𝑔𝑧1 = +
+ 𝑔𝑔4
𝜌
2
𝜌
2
For this situation, we have p1 = 0, V1 = 0, and p4 = 0. The Bernoulli equation becomes
𝑉42
+ 𝑔𝑔4
𝑔𝑔1 =
2
Or, V4 is given by
The area at 4 is
𝑉4 = �2𝑔(𝑧1 − 𝑧4 ) = �2 × 9.81
And the total flow rate is
𝑚
𝑚
× (30 𝑚 − 20 𝑚) = 14
2
𝑠
𝑠
1
1
𝐴4 = 𝜋𝐷42 = × 𝜋 × (0.3 𝑚)2 = 0.0707 𝑚2
4
4
𝑚
𝑚3
2
𝑄 = 𝑉4 𝐴4 = 14 × 0.0707 𝑚 = 0.99
𝑠
𝑠
With the assumption that the flow is frictionless from the surface of either reservoir to the outlet, the
diameters of pipe 1 and pipe 2 do not matter. In reality they would matter. The combined flow area of
these two pipes is 0.049 m2, which is about 70 % of that of pipe 3. Therefore the velocities in pipes 1 and
2 would be 45% greater than that in pipe 3. The deceleration of the flows would create mixing and the
frictionless flow assumption would not strictly valid.
The velocities of the flows in pipes 1 and 2 are equal since the heads are equal. There would then be no
mixing between these two streams.
Another factor in the flow might be whether the pressure at the junction was low enough for cavitation to
occur.
Problem 6.56
(Difficulty 2)
6.56 Barometric pressure is 14.0 𝑝𝑝𝑝. What is the maximum flow rate that can be obtained by opening
the valve if (a) cavitation is not a consideration and (b) cavitation needs to be prevented?
Assumption: The flow is steady, incompressible, uniform, and frictionless.
Solution: Apply the Bernoulli equations to find the minimum pressure in the system and check for
cavitation.
The Bernoulli equation along a streamline is
For 𝐶𝐶𝐶4 we have:
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝜌 2
𝑝𝑣 = 1.9 𝑝𝑝𝑝 𝑎𝑎𝑎 𝑆𝑆 = 1.59
𝑝𝑎𝑎𝑎 = 14 𝑝𝑝𝑝
Assume that the lowest pressure will occur at the 4 𝑖𝑖 constriction where the velocity is highest. We
apply the Bernoulli equation from the water surface where the velocity is zero to the 4 𝑖𝑖 constriction.
We take the height datum as the constriction.
(a) If cavitation is not a problem, we apply the Bernoulli equation from the water surface to the valve:
𝑉42
=ℎ
2𝑔
𝑉4 = �2𝑔ℎ = �2 × 32.2
𝑓𝑓
𝑓𝑓
× 45 𝑓𝑓 = 53.8
2
𝑠
𝑠
2
𝑓𝑓 𝜋
6
𝑓𝑓 3
𝑄 = 𝑉4 𝐴4 = 53.8
× ×�
𝑓𝑡� = 10.56
𝑠 4
12
𝑠
(b) If we wish to prevent cavitation, the minimum pressure at the constriction will be the vapor pressure
pv:
ℎ+
The velocity at the constriction is
𝑝𝑎𝑎𝑎 𝑝𝑣 𝑉42
= +
𝛾
𝛾 2𝑔
𝑉42 = 2𝑔 �ℎ +
𝑝𝑎𝑎𝑎 𝑝𝑣
− �
𝛾
𝛾
𝑝𝑎𝑎𝑎 𝑝𝑣
𝑓𝑓
− � = �2 × 32.2 2 × �20 𝑓𝑓 +
𝑉4 = �2𝑔 �ℎ +
𝑠
𝛾
𝛾
The volumetric flow rate is:
𝑄 = 𝑉4 𝐴4 = 49.2
𝑉4 = 49.2
𝑓𝑓
𝑠
𝑙𝑙𝑙
𝑓𝑓 2
�
𝑙𝑙𝑙
1.59 × 62.4 3
𝑓𝑓
(14 − 1.9) × 144
2
𝑓𝑓 𝜋
4
𝑓𝑓 3
× × � 𝑓𝑓� = 4.29
𝑠 4
12
𝑠
This is the maximum flow rate which can occur without cavitation.
Problem 6.57
Problem
6.72
6.57
[Difficulty: 3] Part 1/2
Problem 6.72
[Difficulty: 3] Part 2/2
Problem 6.58
Problem
6.74
[Difficulty: 3]
6.58
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 6.59
Problem
6.76
[Difficulty: 4]
6.59
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 ⋅
⎝
Rx = −0.002377⋅
2
s⎠
slug
3
× ⎛⎜ 225 ⋅
The force of the jet on the plate is then F = −Rx
1
2
p 0 = p atm + ⋅ ρair⋅ V
2
⎝
ft ⎞
s⎠
1
×
)
ft
The stagnation pressure is
ft ⎞
2 π⋅ d
Rx = V⋅ −ρair⋅ A⋅ V = −ρair⋅ V ⋅
4
(
2
∆h =
so
2 ⋅ 1.75
×
2
⋅ ρair⋅ V
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
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
⎛ 7.5 ⋅ ft⎞ × lbf ⋅ s F = 18.5⋅ lbf
⎜
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 +
⋅⎛
− ⎞
⎝
⎠
⎜
128
2
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 6.60
Problem
6.78
6.60
[Difficulty: 4] Part 1/2
Problem 6.78
[Difficulty: 4] Part 2/2
Problem 6.61
Problem
6.80
[Difficulty: 3]
6.61
Given:
Air flow over "bubble" structure
Find:
Net vertical force
Solution:
The net force is given by
L = 50⋅ ft
Available data
→ ⌠
→
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
ft
(pi − p)⋅ sin(θ)⋅ R⋅ L dθ
0
⌠
FV = ⎮
⎮
⌡
slug
∆p = 187 Pa
Through symmetry only the vertical component of force is no-zero
Hence
3
ρair = 0.00238 ⋅
(
(
)
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 6.62
Problem
6.82
[Difficulty: 4]
6.62
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
V2
+
ρ
2
⋅ ⎛ V − V1
2 ⎝ 2
ρ
2
V2
=
2
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⋅ ⋅ ⎜
2
s ⎜ 2
⎝ 2 − 1.5 ⎠
2
p1 =
Hence
The x mometum is
1
2
× 1.94⋅
slug
ft
where we work in gage pressure
2
ft
V2 = 45.7⋅
s
2
2 ft
× ( 45.7 − 20 ) ⋅ ⎛⎜ ⎞
s
2
⎝ ⎠
3
(
)
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 ⎛
ft
lbf ⋅ s
π⋅ ⎡⎣( 2 ⋅ in) − ( 1.5⋅ in) ⎤⎦ ⎤ ⎛ 1 ⋅ ft ⎞
⎢⎜ 20⋅
⎥×⎜
×
− ⎜ 45.7⋅ ⎞ ×
×
s⎠
4
4
slug⋅ ft
⎣⎝ s ⎠
⎝
⎦ ⎝ 12⋅ in ⎠
in the direction shown
Problem 6.63
Problem
6.84
[Difficulty: 5]
6.63
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 6.64
Problem
6.86
[Difficulty: 5]
6.64
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 6.65
(Difficulty: 2)
6.65 Carefully sketch the energy grade lines (EGL) and hydraulic grade lines (HGL) for the system
shown in Fig. 6.6 if the pipe is horizontal (i.e., the outlet is at the base of the reservoir), and a water
turbine (extracting energy) is located at point 2, or at point 3. In chapter 8 we will investigate the effects
of friction on internal flows. Can you anticipate and sketch the effect of the friction on the EGL and HGL
for the two cases?
Solution:
(a) For the turbine located at point 2, the EGL and HGL would be:
(b) For the turbine located at point 3, the RGL and HGL would be
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.
Problem 6.66
(Difficulty: 2)
6.66 Carefully sketch the energy grade lines (EGL) and hydraulic grade lines (HGL) for the system
shown in Fig. 6.6 if a pump adding energy to the fluid is located at point (2), such that flow is into the
reservoir. In Chapter 8 we will investigate the effects of friction on internal flows. Can you anticipate and
sketch the effect of friction on the EGL and HGL for the two cases?
Solution:
(a) The EGL and HGL for the a pump at point 2 is:
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 EGL in a manner similar to that shown.
(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.
Problem 6.67
(Difficulty 2)
6.67 Water is being pumped from the lower reservoir through a nozzle into the upper reservoir. If the
vacuum gage at 𝐴 reads 2.4 𝑝𝑝𝑝 vacuum,
(a) find the flow velocity through the nozzle.
(b) find the horsepower the pump must add to the water.
(c) draw the energy line and the hydraulic grade line.
Assumption: The flow is steady, incompressible, uniform, and frictionless.
Solution: Apply the continuity and Bernoulli equations to find the velocity and power.
The continuity equation is:
𝑄 = 𝑉1 𝐴1 = 𝑉2 𝐴2
The Bernoulli equation along a streamline is
The gage pressure at A is:
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝜌 2
𝑝𝐴 = −2.4 𝑝𝑝𝑝 = −345.6
𝑙𝑙𝑙
𝑓𝑓 2
Applying the Bernoulli equation from station 1 to station A:
𝑝𝐴 𝑉𝐴2
𝑝1 𝑉12
+
+ 𝑔𝑧1 =
+
+ 𝑔𝑧𝐴
𝜌
2
𝜌
2
Where p1 = 0 and V1 = 0
𝑔𝑧1 =
The water density is
𝑝𝐴 𝑉𝐴2
+
+ 𝑔𝑧𝐴
𝜌
2
𝜌 = 1.938
The velocity at A is then
𝑙𝑙𝑙 ∙ 𝑠2
𝑠𝑠𝑠𝑠
=
1.938
𝑓𝑓 3
𝑓𝑓 4
𝑙𝑙𝑙
2 × −345.6 2
2𝑝𝐴
𝑓𝑓
𝑓𝑓
𝑓𝑓
− 2𝑔𝑧𝐴 = �2 × 32.2 2 × (20 𝑓𝑓 − 25 𝑓𝑓) −
= 5.90
𝑉𝐴 = �2𝑔𝑧1 −
2
𝑙𝑙𝑙 ∙ 𝑠
𝑠
𝑠
𝜌
1.938
𝑓𝑓 4
The volumetric flow rate is:
𝑄 = 𝑉𝐴 𝐴𝐴 = 5.90
2
𝑓𝑓 𝜋
12
𝑓𝑓 3
× × � 𝑓𝑓� = 4.63
𝑠 4
12
𝑠
Applying the continuity equation, the volumetric flow rate at location 2 is the same as at A:
𝑓𝑓 3
4.63
𝑄
𝑓𝑓
𝑠
=
= 53.1
𝑉2 =
2
𝐴2 𝜋
𝑠
4
× � 𝑓𝑓�
12
4
Applying the energy equation from station 1 to station 2:
𝑝2 𝑉22
𝑝1 𝑉12
+
+ 𝑧1 + 𝐸𝑝 =
+
+ 𝑧2
𝜌𝜌 2𝑔
𝜌𝜌 2𝑔
Where Ep is the head provided by the pump
The pump power is then
𝐸𝑝 = (𝑧2 − 𝑧1 ) +
𝑊̇𝑝 =
𝑉22
= 148.8 𝑓𝑓
2𝑔
𝑄𝑄𝑄𝐸𝑝
= 70 ℎ𝑝
550
The energy line and hydraulic grade lines are
Problem 6.68
(Difficulty: 2)
6.68 The turbine extracts power from the water flowing from the reservoir. Find the horsepower
extracted if the flow through the system is 1000
𝑓𝑓 3
.
𝑠
Draw the energy line and the hydraulic grade line.
Find: The power produced by the turbine
Assumption: The flow is steady, incompressible, uniform, and frictionless.
Solution: Apply the continuity and Bernoulli equations to find the power.
The continuity equation is:
𝑄 = 𝑉1 𝐴1 = 𝑉2 𝐴2
The Bernoulli equation along a streamline is
The volumetric flow rate 𝑄 is:
𝑝 𝑉2
+
+ 𝑔𝑔 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝜌 2
𝑄 = 1000
From the continuity equation, the velocity at station 2:
𝑓𝑓 3
𝑠
𝑓𝑓 3
𝑄
𝑠 = 8.84 𝑓𝑓
=𝜋
𝑉2 =
𝐴2
𝑠
× (12 𝑓𝑓)2
4
1000
Applying the energy equation from station 1 to station 2, where ET is the turbine head:
𝑝2 𝑉22
𝑝1 𝑉12
+
+ 𝑧1 = 𝐸𝑇 +
+
+ 𝑧2
𝜌𝜌 2𝑔
𝜌𝜌 2𝑔
𝐸𝑇 = 499 𝑓𝑓
The power is be calculated by:
𝑊̇ 𝑇 =
𝑄𝑄𝑄𝐸𝑇
= 56600 ℎ𝑝
550
The energy line and hydraulic grade line are shown in the figure.
Problem 6.69
(Difficulty: 2)
6.69 Consider a two-dimensional fluid flow: 𝑢 = 𝑎𝑎 + 𝑏𝑏 and 𝑣 = 𝑐𝑐 + 𝑑𝑑, where a, b, c, and d are
constant. If the flow is incompressible and irrotational, find the relationships among a, b, c, and d. Find
the stream function and velocity potential function of this flow.
Given: 2D incompressible, inviscid flow field
Find: Relationships among constants; stream function; velocity potential.
Solution:
Basic equations
For incompressible flow the velocity components are
𝑢=
For irrotational flow
Check to be certain the flow is incompressibile:
This requires that:
𝜕𝜕
𝜕𝜕
𝑣=−
𝜕𝜕
𝜕𝜕
𝑢=−
𝜕𝜕
𝜕𝜕
𝑣=−
𝜕𝜕
𝜕𝜕
𝜕𝜕(𝑥, 𝑦) 𝜕𝜕(𝑥, 𝑦)
+
= 𝑎+𝑑 =0
𝜕𝜕
𝜕𝜕
Check to see if the flow is irrotational
𝑑 = −𝑎
𝜕𝜕(𝑥, 𝑦) 𝜕𝜕(𝑥, 𝑦)
−
=𝑐−𝑏
𝜕𝜕
𝜕𝜕
This requires that:
𝑐=𝑏
Solving for the stream function:
𝛹(𝑥, 𝑦) = � 𝑢(𝑥, 𝑦)𝑑𝑑 =
𝑏𝑏 2
+ 𝑎𝑎𝑎 + 𝑓(𝑥)
2
𝑑𝑑
𝜕𝜕
= 𝑎𝑎 +
= −𝑣 = −𝑐𝑐 − 𝑑𝑑
𝑑𝑑
𝜕𝜕
𝑑𝑑
= −𝑎𝑎 − 𝑑𝑑 − 𝑐𝑐 = −𝑐𝑐
𝑑𝑑
𝑐𝑐 2
𝑓(𝑥) = −
2
So the stream function is:
𝛹(𝑥, 𝑦) =
Solving for the velocity potential:
𝑏𝑏 2
𝑐𝑐 2
𝑏
+ 𝑎𝑎𝑎 −
= 𝑎𝑎𝑎 + (𝑦 2 − 𝑥 2 )
2
2
2
𝑎𝑎 2
𝜙(𝑥, 𝑦) = − � 𝑢(𝑥, 𝑦)𝑑𝑑 = −
− 𝑏𝑏𝑏 + 𝑔(𝑦)
2
𝑑𝑑
𝜕𝜕(𝑥, 𝑦)
= −𝑏𝑏 +
= −𝑐𝑐 − 𝑑𝑑
𝑑𝑦
𝜕𝜕
𝑑𝑑
= 𝑏𝑏 − 𝑐𝑐 − 𝑑𝑑 = −𝑑𝑑
𝑑𝑑
So the velocity potential is:
𝑔(𝑦) = −
𝜙(𝑥, 𝑦) = −
𝑑𝑦 2
2
𝑎𝑎 2
𝑑𝑦 2
− 𝑏𝑏𝑏 −
2
2
Problem 6.100
6.70
Problem
6.70
[Difficulty: 2]
Problem 6.71
(Difficulty: 2)
6.71 A flow field is characterized by the stream function 𝛹 = 𝐴𝐴𝐴 where 𝐴 = 2 𝑠 −1 and the coordinates
are measured in feet. Verify that the flow is irrotational and determine the velocity potential 𝜙. Plot the
streamlines and potential lines and visually verify that they are orthogonal.
Find: The velocity potential 𝜙 and plot the streamline and potential lines.
Assumption: The flow is incompressible, steady, and frictionless
Solution: Apply the definitions of the stream function and the velocity potential
The flow is irrotational if the vorticity is zero. The vorticity is defined as:
As we have for the stream function:
�⃗
𝜉⃗ = ∇ × 𝑉
The velocities u and v are
𝛹 = 2𝑥𝑥
𝑢=
Thus the vorticity is
This flow is irrotational.
𝑣=−
𝜕𝜕
= 2𝑥
𝜕𝜕
𝜕𝜕
= −2𝑦
𝜕𝜕
𝜕𝜕 𝜕𝜕
�⃗ = � − � 𝑘� = (0 − 0)𝑘� = 0
𝜉⃗ = ∇ × 𝑉
𝜕𝜕 𝜕𝜕
For the velocity potential, u and v are given by:
𝑢=−
So we have:
𝑣=−
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
= −𝑢 = −2𝑥
𝜕𝜕
𝜕𝜕
= −𝑣 = 2𝑦
𝜕𝜕
𝜙(𝑥, 𝑦) = −𝑥 2 + 𝑓(𝑦)
𝜕𝜕 𝜕𝜕
=
= 2𝑦
𝜕𝜕 𝜕𝜕
𝑓(𝑦) = 𝑦 2 + 𝑐
𝜙(𝑥, 𝑦) = −𝑥 2 + 𝑦 2 + 𝑐
Plotting the streamlines and potential lines:
Problem 6.102
6.72
Problem
[Difficulty: 3]
6.72
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 6.104
6.73
Problem
[Difficulty: 2]
6.73
Given:
Stream function
Find:
Velocity potential
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 6.74
(Difficulty: 2)
6.74 A flow field is characterized by the stream function
𝛹 = 2𝑦 +
1
𝑦−𝑎
𝑦+𝑎
�tan−1
− tan−1
�
2𝜋
𝑥
𝑥
Derive an expression for the location of the stagnation points. Sketch the flow field.
Find: The stagnation points in the flow
Assumption: The flow is incompressible, steady, and frictionless
Solution: Apply the definitions of the stream function
The stream function is given by
𝛹 = 2𝑦 +
1
𝑦−𝑎
𝑦+𝑎
�tan−1
− tan−1
�
2𝜋
𝑥
𝑥
The velocity components u and v can be calculated as:
𝑢=
𝜕𝜕
1
𝑥
𝑥
=2+
� 2
−
�
𝜕𝜕
2𝜋 𝑥 + (𝑦 − 𝑎)2 𝑥 2 + (𝑦 + 𝑎)2
𝑣=−
The stagnation points are where:
1
𝑦+𝑎
𝑦−𝑎
𝜕𝜕
=− � 2
− 2
�
2
2𝜋 𝑥 + (𝑦 + 𝑎)
𝑥 + (𝑦 − 𝑎)2
𝜕𝜕
𝑢=0
For the stagnation point we have for u = 0:
And for v = 0
[𝑥 2
+ (𝑦 −
𝑣=0
𝑥𝑥
𝑎)2 ][𝑥 2
+ (𝑦 +
𝑥 2 + 𝑎2 = 𝑦 2
𝑎)2 ]
=−
𝜋
𝑎
The above equations govern all of the stagnation points in the flow field. We plot the flow field as
follows:
For 𝑎 = 0.1,
For 𝑎 = 0.5,
For 𝑎 = 1,
Problem 6.75
(Difficulty: 2)
6.75 A flow field is characterized by the stream function
𝛹 = 𝑥𝑦 2 + 𝐵𝑥 3
What does the value of B need to be for the flow to be irrotational? For that value of B, determine the
velocity potential 𝜙. Sketch the streamlines and potential lines.
Find: The value of B for irrotational flow and the velocity potential
Assumption: The flow is incompressible, steady, and frictionless
Solution: Apply the definitions of the stream function
The flow is irrotational if the vorticity is zero. The vorticity is defined as:
�⃗ =
𝜉⃗ = ∇ × 𝑉
𝜕𝜕 𝜕𝜕
−
=0
𝜕𝜕 𝜕𝜕
The velocity field is calculated in terms of the stream function as:
𝑢=
𝜕𝜕
= 2𝑥𝑥
𝜕𝜕
𝜕𝜕
= −𝑦 2 − 3𝐵𝑥 2
𝜕𝜕
Thus
𝑣=−
The vorticity will be zero if
∇ × 𝑽 = −6𝐵𝐵 − 2𝑥 = 0
𝐵=−
To find the velocity potential we use the definitions:
𝑢=−
For the expression for u we have
𝑣=−
1
3
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
𝜕𝜕
= −𝑢 = −2𝑥𝑥
𝜕𝜕
Integrating
From the expression for v we have
The function f is then
𝜙 = −𝑥 2 𝑦 + 𝑓(𝑦)
𝜕𝜕
𝜕𝜕
= −𝑣 = 𝑦 2 − 𝑥 2 = −𝑥 2 +
𝜕𝜕
𝜕𝜕
𝜕𝜕
= 𝑦2
𝜕𝜕
Where 𝐶 is a constant.
1
𝑓(𝑦) = 𝑦 3 + 𝐶
3
Thus the velocity potential is
And the stream function is:
The streamlines are shown as:
1
𝜙 = −𝑥 2 𝑦 + 𝑦 3 + 𝐶
3
1
𝛹 = 𝑥𝑦 2 − 𝑥 3
3
The potential lines are shown as:
We can find that the streamlines and potential lines are orthogonal .
Problem 6.76
Problem
6.106
[Difficulty: 2]
6.76
Given:
Stream function
Find:
Velocity potential
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
Problem6.108
6.77
Problem
[Difficulty: 2]
6.77
Given:
Stream function
Find:
Velocity field; Show flow is irrotational; Velocity potential
Solution:
Basic equations: Incompressibility because ψ exists
∂
Irrotationality
We have
6
v −
∂y
4 2
∂
∂y
∂
∂y
∂
v=− ψ
∂x
ψ
∂
u=− φ
∂x
2 4
6
2 3
ψ( x , y )
4
u ( x , y ) = 60⋅ x ⋅ y − 30⋅ x ⋅ y − 6 ⋅ y
∂
v ( x , y ) = − ψ( x , y )
∂x
v ( x , y ) = 60⋅ x ⋅ y − 6 ⋅ x − 30⋅ x ⋅ y
∂
Hence flow is IRROTATIONAL
∂x
v( x , y) −
∂
∂y
3 2
u( x , y) = 0
∂
v=− φ
∂y
u =0
ψ( x , y ) = x − 15⋅ x ⋅ y + 15⋅ x ⋅ y − y
u( x , y) =
Hence
∂x
∂
u=
5
5
4
∂
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 6.110
6.78
Problem
[Difficulty: 2]
6.78
Given:
Velocity potential
Find:
Show flow is incompressible; Stream function
Solution:
u=
Basic equations: Irrotationality because φ exists
∂
Incompressibility
We have
Hence
Hence
5
∂x
u +
3 2
∂
∂y
∂
∂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
∂
Hence flow is INCOMPRESSIBLE
∂x
u( x , y) +
u=
∂
∂y
ψ
∂
v=− ψ
∂x
∂
∂y
2 2
4
3
v( x , y) = 0
∂
v=− φ
∂y
4
3
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 6.79
Problem
6.112
[Difficulty: 4]
6.79
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:
u=
Basic equations: Irrotationality because φ exists
∂
Incompressibility
6
f ( z) = z = ( x + i ⋅ y )
6
u +
∂x
∂
∂y
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
Hence we see
∂
∂x
∂
∂x
v( x , y) −
u( x , y) +
df
dz
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⎛⎜
⎞
dz
⎝ ⎠
df
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
Problem 6.114
6.80
6.80
[Difficulty: 2]
Problem 6.116
6.81
Problem
6.81
[Difficulty: 3]
Problem 6.82
(Difficulty: 2)
6.82 Consider an air flow over a flat wall with an upstream velocity of 6
which air is drawn in at a flow rate of 0.2
𝑚3
𝑠
𝑚
.
𝑠
There is a narrow slit through
per meter of width. Represent the flow as a combination of
a uniform flow and a sink. Determine the location of the stagnation point. Sketch the dividing line
between the air that enters slit and the air that continues downstream.
y
r
𝜃
x
Find: The stagnation point and sketch the streamlines
Assumption: The flow is incompressible, steady, and frictionless
Solution: Apply the potential flow methods. We set the location of the sink as the origin of the
coordinate. For this flow, we can apply the superposition of the uniform flow and sink.
The velocity for the uniform flow is:
𝑢=𝑈
The velocity field for the sink flow is:
𝑣=0
𝑢=−
The combined flow velocity is:
𝑣=−
𝑞
cos 𝜃
2𝜋𝜋
𝑞
sin 𝜃
2𝜋𝜋
𝑢=𝑈−
𝑣=−
𝑞
cos 𝜃
2𝜋𝜋
𝑞
sin 𝜃
2𝜋𝜋
The velocity vector is
𝑽 = �𝑈 −
𝑞
𝑞
cos 𝜃� 𝑖 + �−
sin 𝜃� 𝑗
2𝜋𝜋
2𝜋𝜋
𝑥 = 𝑟 cos 𝜃
𝑦 = 𝑟 sin 𝜃
For the stagnation point we have:
𝑢=𝑈−
𝑣=−
So we have:
𝑞
cos 𝜃 = 0
2𝜋𝜋
𝑞
sin 𝜃 = 0
2𝜋𝜋
𝜃=0
𝑈−
The value of the radius is
𝑞
=0
2𝜋𝜋
𝑚2
0.2
𝑞
𝑠 = 0.0053 𝑚
=
𝑟=𝑥=
2𝜋𝜋 2𝜋 × 6 𝑚
𝑠
The stagnation point is at the (0.0053 𝑚, 0 𝑚).
The streamlines are sketched below
Problem 6.83
(Difficulty: 2)
6.83 A source with a strength of 𝑞 = 3𝜋
𝑚2
𝑠
and a sink with a strength of 𝑞 = 𝜋
𝑚2
𝑠
are located on the
𝑥 = −1 𝑚 and 𝑥 = 1 𝑚 respectively. Determine the stream function and velocity potential for the
combined flow and sketch the streamlines.
Find: The stream function and velocity potential.
Assumption: The flow is incompressible, steady, and frictionless
Solution: Apply the potential flow methods.
We set the location of the sink as the origin. Then:
The stream function for the combined flow is:
Or in terms of x and y
𝛹=
𝑎 = ±1
𝑞2
3
1
𝑞1
𝜃1 −
𝜃2 = 𝜃1 − 𝜃2
2
2
2𝜋
2𝜋
3
𝑦
1
𝑦
𝛹 = tan−1 �
� − tan−1 �
�
2
𝑥+1
2
𝑥−1
The velocity potential for the combined flow:
𝜙=−
𝑞2
3
1
𝑞1
ln 𝑟1 +
ln 𝑟2 = − ln 𝑟1 + ln 𝑟2
2
2
2𝜋
2𝜋
Or in terms of x and y
3
1
𝜙 = − ln�(𝑥 + 1)2 + 𝑦 2 + ln�(𝑥 − 1)2 + 𝑦 2
2
2
The streamlines are shown as:
10
9
8
7
y
6
5
4
3
2
1
0
-10
-8
-6
-4
-2
0
x
2
4
6
8
10
Problem 6.84
Problem
6.118
6.84
[Difficulty: 3] Part 1/2
Problem 6.118
[Difficulty: 3] Part 2/2
Problem 6.85
Problem
6.120
6.85
[Difficulty: 2]
Problem 6.86
(Difficulty: 2)
6.86 The flow in a corner with an angle 𝛼 can be described in radial coordinates by the stream function
𝜋
as 𝛹 = 𝐴𝑟 𝑎 sin
𝜋𝜋
.
𝑎
Determine the velocity potential for the flow and plot streamlines for flow for
𝑎 = 60 degrees.
Find: The velocity potential.
Assumption: The flow is incompressible, steady, and frictionless
Solution: Apply the potential flow methods.
We have the radial and tangential velocities in terms of the stream function and velocity potential as:
𝑉𝑟 =
1 𝜕𝜕
𝜕𝜕
=−
𝑟 𝜕𝜕
𝜕𝜕
𝑉𝜃 = −
𝜕𝜕
1 𝜕𝜕
=−
𝜕𝜕
𝑟 𝜕𝜕
So we have for this problem using the stream function:
𝑉𝑟 =
1 𝜕𝜕 1 𝜋 𝜋
𝜋𝜋 𝐴𝐴
𝜋𝜋 𝜋−𝑎
= 𝐴𝑟 𝑎 cos
=
cos
𝑟 𝑎
𝑟 𝜕𝜕 𝑟
𝑎
𝑎
𝑎
𝑎
The derivative of the velocity potential with respect to r is then
𝐴𝐴
𝜋𝜋 𝜋−𝑎
𝜕𝜕
=−
cos
𝑟 𝑎
𝑎
𝑎
𝜕𝜕
Integrating with respect to r
𝜙=−
𝐴𝐴 𝑎
𝜋𝜋 𝜋
𝜋𝜋 𝜋
cos
𝑟 𝑎 + 𝑓(𝜃) = −𝐴 cos
𝑟 𝑎 + 𝑓(𝜃)
𝑎 𝜋
𝑎
𝑎
The derivative of the velocity potential with respect to θ is then
𝜕𝜕 𝐴𝜋 𝜋
𝜋𝜋 𝑑𝑑(𝜃)
=
𝑟 𝑎 sin
+
𝜕𝜕
𝑎
𝑎
𝑑𝑑
We also have for the tangential velocity:
𝑉𝜃 = −
And the derivative with respect to θ is
𝜕𝜕
𝐴𝜋 𝜋−𝑎
𝜋𝜋
=−
𝑟 𝑎 sin
𝜕𝜕
𝑎
𝑎
𝜋𝜋
1 𝜕𝜕 𝐴𝜋 𝜋−𝑎
=
𝑟 𝑎 sin
𝑎
𝑎
𝑟 𝜕𝜕
Or
𝜋𝜋
𝜕𝜕 𝐴𝜋 𝜋
=
𝑟 𝑎 sin
𝑎
𝑎
𝜕𝜕
Equating the two expressions for the derivative:
𝜋𝜋 𝑑𝑑(𝜃) 𝐴𝜋 𝜋
𝜋𝜋
𝐴𝜋 𝜋
𝑟 𝑎 sin
+
=
𝑟 𝑎 sin
𝑎
𝑎
𝑎
𝑎
𝑑𝑑
Therefore
𝑑𝑑(𝜃)
=0
𝑑𝑑
Which yields
𝑓(𝜃) = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
The velocity potential is then:
𝜋
3
𝜙 = −𝐴 cos
The velocity field for 𝑎 = 60 = degrees is:
𝑉𝑟 =
𝛹 = 𝐴𝑟 𝑎 sin
𝐴𝐴
𝜋𝜋 𝜋−𝑎
cos
𝑟 𝑎
The streamlines are shown as:
𝑎
𝜋
𝜋𝜋 𝜋
𝑟 𝑎 + 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝑎
𝑎
𝐴𝜋
𝑉𝜃 = − 𝜋
3
𝜋𝜋
= 𝐴𝑟 3 sin3θ
𝑎
𝐴𝐴
𝜋𝜋
= 𝜋 cos 𝜋
3
3
𝜋
𝜋− 3
𝜋
𝑟 3
𝜋
𝜋− 3
𝜋
𝑟 3
= 3𝐴𝑟 2 cos3θ
𝜋𝜋
sin 𝜋 = −3𝐴𝑟 2 sin3θ
3
Problem 6.87
(Difficulty: 2)
6.87 Consider the two-dimension flow against a flat plate that is characterized by the stream function
𝛹 = 𝐴𝐴𝐴. Superimpose a plane source of strength B placed in the origin. Determine the relation
between the height of the stagnation point h, the constant A, and the strength B. Sketch streamlines for
the flow and identify the streamline that divides the two flows.
Find: The stagnation point properties.
Assumption: The flow is incompressible, steady, and frictionless
Solution: Apply the potential flow methods.
For the plane source of strength B we have:
The stream function for the combined flow is:
𝛹 = 𝐴𝐴𝐴 +
𝛹=
𝐵
𝜃
2𝜋
𝐵
𝑦
𝐵
𝜃 = 𝐴𝐴𝐴 +
tan−1 � �
2𝜋
𝑥
2𝜋
The velocity field is calculated using the stream function as:
𝑢=
𝑣=−
𝜕𝜕
𝐵
𝑥
= 𝐴𝐴 +
�
�
𝜕𝜕
2𝜋 𝑥 2 + 𝑦 2
𝜕𝜕
𝐵
𝑦
= −𝐴𝐴 +
� 2
�
𝜕𝜕
2𝜋 𝑥 + 𝑦 2
For the stagnation point we have:
𝑢=𝑣=0
For 𝑥 = 0 we have:
𝑢=0
Then
𝑣 = −𝐴ℎ +
𝐵 ℎ
� �=0
2𝜋 ℎ2
So the relation between the height of the stagnation point and strength B and constant A is:
𝐴=
𝐵 1
� �
2𝜋 ℎ2
For the streamline pass the stagnation point we have:
𝛹 = 𝐴𝐴𝐴 +
𝐵
𝐵
𝑦
𝐵
tan−1 � � =
tan−1 (∞) =
4
2𝜋
𝑥
2𝜋
The streamline is shown as (assume A=B=0.1):
10
9
8
7
y
6
5
4
3
2
1
0
-10
-8
-6
-4
-2
0
x
2
The streamline divides the two flows can be seen from the figure.
4
6
8
10
Problem 6.88
Problem
6.122
6.88
[Difficulty: 3]
Problem 6.89
Problem
6.124
6.89
[Difficulty: 3] Part 1/2
Problem 6.124
[Difficulty: 3] Part 2/2
Problem 6.90
Problem
6.126
6.90
[Difficulty: 3] Part 1/2
Problem 6.126
[Difficulty: 3] Part 2/2
Problem 7.1
Problem
7.3
[Difficulty: 2]
7.1
Given:
Find:
Equation describing the slope of a steady wave in a shallow 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 V o. Obtain the dimensionless
groups that characterize the flow.
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 7.2
Problem
7.4
[Difficulty: 2]
7.2
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.
*
gL h *
u *
* u
u



t *
x *
V02 x *
Problem 7.3
Problem
7.6
[Difficulty: 2]
7.3
Given:
Equations for modeling atmospheric motion
Find:
Non-dimensionalized equation; Dimensionless groups
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 t *
L
 L
The final dimensionless equation is
The dimensionless groups are



p
V * 
 L  
p *
 V *  * V * 2 
  * 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 7.4
(Difficulty 2)
7.4 Fluid fills the space between two parallel plates. The differential equation that describes the
instantaneous fluid velocity for unsteady flow with the fluid moving parallel to the walls is
𝜌
𝜕 2𝑢
𝜕𝜕
=𝜇 2
𝜕𝜕
𝜕𝜕
The lower plate is stationary and the upper plate oscillates in the x-direction with a frequency 𝜔 and an
amplitude in the plate velocity of 𝑈. Use the characteristic dimensions to normalize the differential
equation and obtain the dimensionless groups that characterize the flow.
Find: Use the characteristic dimensions to normalize the equation and obtain the dimensionless groups.
Solution: This is unidirectional flow with oscillating boundary with characteristic dimensions U and H.
We will normalize the variables using these characteristic dimensions as:
𝑢∗ =
𝑦∗ =
𝑡∗ =
𝑢
𝑈
𝑦
𝐻
𝑡
𝐻
𝑈
Substituting these dimensions into the differential equation we have:
𝜌
𝜕 2 (𝑈𝑈∗ )
𝜕(𝑈𝑈∗ )
=𝜇
𝐻
𝜕(𝐻𝐻 ∗ )2
𝜕 � 𝑡 ∗�
𝑈
𝜇 𝜕 2 𝑢∗
𝜕𝑢∗
=
𝜕𝑡 ∗ 𝜌𝜌𝜌 𝜕𝜕 ∗2
On the boundary we have:
At 𝑦 = 0, 𝑢 = 0 so we have:
At 𝑦 = 𝐻, 𝑢 = 𝑈 cos 𝜔𝜔 so we have:
We define the Reynolds number as:
The differential equation becomes:
𝑢∗ = 0 at 𝑦 ∗ = 0
𝑢∗ = cos 𝜔𝜔 at 𝑦 ∗ = 1
𝑅𝑅 =
𝜌𝜌𝜌
𝜇
1 𝜕 2𝑢∗
𝜕𝑢∗
=
𝜕𝑡 ∗ 𝑅𝑅 𝜕𝜕 ∗2
The fluid flow is governed by the following two dimensionless group:
𝑅𝑅 =
It can be written as:
𝜌𝜌𝜌
𝜇
П = cos 𝜔𝜔
𝜌𝜌𝜌
𝑢
= 𝑓�
, cos 𝜔𝜔�
𝜇
𝑈
Problem 7.5
Problem
7.7
[Difficulty: 4]
7.5
Given:
Find:
The Prandtl boundary-layer equations for steady, incompressible, two-dimensional flow neglecting gravity
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 V 0. Obtain the dimensionless
groups that characterize the flow.
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
*
u V0
 v V0


  x* L
 x* L
 y*L
 y*L  y*L
*
u*
 
Simplifying this expression yields:
*
1 p
  2 u * Now every term in this equation has been non-dimensionalized except the
u *
* u
v




V02 x * V0 L y * 2 pressure gradient. We define a dimensionless pressure as:
y *
x *
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 7.6
(Difficulty 2)
7.6 Consider a disk of radius 𝑅 rotating in an incompressible fluid at a speed 𝜔. The equations that
describe the boundary layer on the disk are:
𝜌 �𝑣𝑟
𝜕𝑣𝑧
1 𝜕(𝑟𝑣𝑟 )
�
�+
=0
𝑟
𝜕𝜕
𝜕𝜕
𝜕 2 𝑣𝑟
𝜕𝑣𝑟 𝑣𝜃2
𝜕𝑣𝑟
−
+ 𝑣𝑧
�=𝜇
𝜕𝜕
𝑟
𝜕𝜕
𝜕𝑧 2
Use the characteristic dimensions to normalize the differential equation and obtain the dimensionless
groups that characterize the flow.
Find: Use the characteristic dimensions to normalize the equation and obtain the dimensionless groups.
Solution: This is unidirectional flow with oscillating boundary. We will normalize the variable with
the following characteristic dimensions. Because there is no characteristic velocity in the problem, we
will use the product 𝜔𝜔 as the characteristic velocity:
𝑣𝑟∗ =
𝑣𝑟
𝜔𝜔
𝑣𝑧∗ =
𝑣𝑧
𝜔𝜔
𝑣𝜃∗ =
𝑣𝜃
𝜔𝜔
𝑟∗ =
𝑧∗ =
𝑟
𝑅
𝑧
𝑅
Substitute these equations in to the differential equation we have:
For the continuity equation:
𝜕𝜔𝜔𝑣𝑧∗
1 1 𝜕(𝑅𝑟 ∗ 𝜔𝜔𝑣𝑟∗ )
�
�
+
=0
𝑅 𝑟∗
𝜕𝑅𝑧 ∗
𝜕𝜕𝑟 ∗
Thus
𝜔
Or
1 𝜕(𝑟 ∗ 𝑣𝑟∗ )
𝜕𝑣𝑧∗
�
�
+
𝜔
=0
𝑟∗
𝜕𝑧 ∗
𝜕𝑟 ∗
𝜕𝑣𝑧∗
1 𝜕(𝑟 ∗ 𝑣𝑟∗ )
�
�
+
=0
𝑟∗
𝜕𝑧 ∗
𝜕𝑟 ∗
For the r-momentum equation:
𝜌 �𝜔𝜔𝑣𝑟∗
𝜕 2 𝜔𝜔𝑣𝑟∗
𝜕𝜔𝜔𝑣𝑟∗ (𝜔𝜔𝑣𝜃∗ )2
𝜕𝜔𝜔𝑣𝑟∗
∗
−
+
𝜔𝜔𝑣
�
=
𝜇
𝑧
𝜕𝜕𝑟 ∗
𝑅𝑟 ∗
𝜕𝑅𝑧 ∗
𝜕𝑅 2 𝑧 ∗2
𝜌 �𝜔2 𝑅𝑣𝑟∗
(𝑣𝜃∗ )2
𝜇𝜔 𝜕 2 𝑣𝑟∗
𝜕𝑣𝑟∗
𝜕𝑣𝑟∗
2𝑅
2 𝑅𝑣 ∗
−
𝜔
+
𝜔
�
=
𝑧
𝑅 𝜕𝑧 ∗2
𝜕𝑟 ∗
𝑟∗
𝜕𝑧 ∗
𝜌𝜔2 𝑅 �𝑣𝑟∗
�𝑣𝑟∗
Defining the Reynolds number as:
𝜇𝜔 𝜕 2 𝑣𝑟∗
𝜕𝑣𝑟∗ (𝑣𝜃∗ )2
𝜕𝑣𝑟∗
∗
−
+
𝑣
�
=
𝑧
𝑅 𝜕𝑧 ∗2
𝜕𝑟 ∗
𝑟∗
𝜕𝑧 ∗
𝜇 𝜕 2 𝑣𝑟∗
𝜕𝑣𝑟∗ (𝑣𝜃∗ )2
𝜕𝑣𝑟∗
∗
−
+
𝑣
�
=
𝑧
𝜌𝜔𝑅 2 𝜕𝑧 ∗2
𝜕𝑟 ∗
𝑟∗
𝜕𝑧 ∗
The differential equation becomes:
�𝑣𝑟∗
𝑅𝑅 =
𝜌𝜔𝑅 2
𝜇
1 𝜕 2 𝑣𝑟∗
𝜕𝑣𝑟∗ (𝑣𝜃∗ )2
𝜕𝑣𝑟∗
∗
−
+
𝑣
�
=
𝑧
𝑅𝑅 𝜕𝑧 ∗2
𝜕𝑟 ∗
𝑟∗
𝜕𝑧 ∗
The Reynolds number characterizes the fluid flow as:
�⃗
𝜌𝜔𝑅 2
𝑉
= 𝑓�
�
𝜔𝜔
𝜇
Problem 7.7
Problem
7.8
[Difficulty: 2]
7.7
Given:
Equation for unsteady, 2D compressible, inviscid flow
Find:
Dimensionless groups
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






*
*
*
*
2
*
*
0
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 *
y *2
x *2
t
t *2
No dimensionless group is needed for this equation!
Problem 7.10
7.8
Problem
[Difficulty: 2]
7.8
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
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   Lc  M0 L0 t0
 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   Lc  M0 L0 t0
   3  
 L t   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 7.9
Problem
7.12
[Difficulty: 2]
7.9
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   Lc  M0 L0 t0
 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 7.10
Problem
7.14
[Difficulty: 2]
7.10
Functional relationship between buoyant force of a fluid and physical parameters
Given:
Find:
Solution:
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
V
γ
F
3
L
4
5
V
n = 3 parameters
γ
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 7.11
(Difficulty 1)
7.11 Assume that the velocity acquired by a body falling from rest (without resistance) depends on
weight of body, acceleration due to gravity, and distance of fall. Prove by dimensional analysis that
𝑉 = 𝐶�𝑔𝑛 ℎ and is thus independent of the weight of the body.
Find: Prove 𝑉 is independent of the weight of the body.
Solution:
(1) There are four dimensional parameters: 𝑉, 𝑊, 𝑔𝑛 , ℎ so 𝑛 = 4
(2) Select primary dimensions 𝑀, 𝐿 𝑎𝑎𝑎 𝑡.
(3) We have the following relations:
𝑉
𝑀
0
𝐿
1
𝑡
−1
𝐶
0
0
0
𝑊
1
1
−2
𝑔𝑛
0
1
−2
ℎ
0
1
0
All three primary dimensions are represented so r = 3. The number of repeating variables will
then be m = r = 3.
The number of groups will be n – r = 4 – 3 = 1. All of the dimensional parameters will be
combined into one group.
The group is then assumed to be of the form
So we get:
Π = 𝑉 𝑊 𝑎 𝑔𝑛𝑏 ℎ𝑐
𝐿 𝑀𝑀 𝑎 𝐿 𝑏
Π = � 2 � � 2 � (𝐿)𝑐 = 𝑀0 𝐿0 𝑡 0
𝑡
𝑡 𝑡
We evaluate the coefficients by setting the coefficients of each dimension on the left hand side
equal to the coefficient for the dimension on the right hand side (which is 0).
Solving for the values of b and c
𝑎=0
1+𝑎+𝑏+𝑐 =0
−1 − 2𝑎 − 2𝑏 = 0
1
2
1
𝑐=−
2
𝑏=−
Thus
Π = 𝑉𝑔𝑛−0.5 ℎ−0.5 =
𝑉
�𝑔𝑛 ℎ
Because there is only one group, it is a constant and we then have
𝑉 = 𝐶�𝑔𝑛 ℎ
The coefficient a for the only variable that contains mass (the weight W) is zero, so it is clear that
𝑉 is independent of the weight of the body.
Problem 7.12
(Difficulty 2)
7.12 Derive by dimensional analysis an expression for the local velocity in established pipe flow through
a smooth pipe if this velocity depends only on mean velocity, pipe diameter, distance from pipe wall,
and density and viscosity of the fluid.
Find: The appropriate dimensionless groups.
Solution:
(1) There are 6 dimensional parameters : 𝑉, 𝑉𝑚 , 𝑑, 𝑦, 𝜌, 𝜇 so 𝑛 = 6
(2) Select primary dimensions 𝑀, 𝐿 𝑎𝑎𝑎 𝑡.
(3) We have the following relation:
𝑉
𝑉𝑚
𝑀
0
0
𝐿
1
1
𝑡
−1
−1
𝑑
0
1
0
𝑦
0
1
0
𝜌
1
−3
0
𝜇
1
−1
−1
We have all three dimensions so r = m = 3. We need 3 repeating parameters that include all of
the dimensions and we pick the following three repeating parameters:
𝑉𝑚 𝑑 𝜌
The number of dimensionless groups will be n – m = 6 – 3 = 3 dimensionless groups: For the
first group we have combine the repeating variables with the dimensional parameter V
П1 = 𝑉𝑚
𝑎 𝑏 𝑐
𝑑 𝜌 𝑉
𝐿 𝑎
𝑀 𝑐 𝐿
𝑏
= � � (𝐿) � 3 � � � = 𝑀0 𝐿0 𝑡 0
𝑡
𝐿
𝑡
Equating the exponents for each of the dimensions
Solving for the exponents
So the first group is
𝑐=0
𝑎 + 𝑏 − 3𝑐 + 1 = 0
−𝑎 − 1 = 0
𝑎 = −1
𝑏=0
П1 =
𝑉
𝑉𝑚
For the second group we combine with the dimensional parameter y:
П2 = 𝑉𝑚
𝐿 𝑑
𝑀 𝑓
𝑒
= � � (𝐿) � 3 � (𝐿) = 𝑀0 𝐿0 𝑡 0
𝑡
𝐿
𝑓=0
𝑑 + 𝑒 − 3𝑓 + 1 = 0
𝑑 𝑒 𝑓
𝑑 𝜌 𝑦
Solving for the exponents
−𝑑 = 0
𝑒 = −1
The second group is
П2 =
𝑦
𝑑
Similarly, for the third group, combining with the dimensional parameter µ
𝐿 𝑔
𝑀 𝑖 𝑀
П3 = 𝑉𝑚 𝑔 𝑑 ℎ 𝜌 𝑖 𝜇 = � � (𝐿)ℎ � 3 � � � = 𝑀0 𝐿0 𝑡 0
𝑡
𝐿
𝐿𝑡
The exponents are then related as
Solving for the exponents
The third group is then
𝑖+1=0
𝑔 + ℎ − 3𝑖 − 1 = 0
−𝑔 − 1 = 0
𝑖 = −1
𝑔 = −1
ℎ = −1
П3 =
𝜇
𝑉𝑚 𝑑𝑑
The functional relation among the groups is then
𝑉
𝑦 𝜇
= 𝑓� ,
�
𝑉𝑚
𝑑 𝑉𝑚 𝑑𝑑
Problem 7.13
(Difficulty 2)
7.13 The speed of shallow water waves in the ocean (𝑒. 𝑔. 𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠 𝑤𝑤𝑤𝑤𝑤 𝑜𝑜 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡) depends
only on the still water depth and the acceleration due to gravity. Derive an expression for wave speed.
Find: The appropriate dimensionless groups.
Solution:
(1) There are three dimensional parameters 𝑉, 𝑑, 𝑔 so 𝑛 = 3
(2) Select primary dimensions 𝑀, 𝐿 𝑎𝑎𝑎 𝑡.
(3) We have the following relation:
𝑉
𝑑
𝑀
0
0
𝐿
1
1
𝑡
−1
0
𝑔
0
1
−2
Mass does not appear in any of the parameters so we need only 2 repeating variables
𝑟=2
We pick up the following two repeating parameters that include all of the dimensions:
𝑑 𝑔
The number of dimensionless groups is n – m = n – r = 3 – 2 = 1
This group is then
𝐿 𝑏 𝐿
П1 = 𝑑 𝑎 𝑔𝑏 𝑉 = 𝐿𝑎 � 2 � � � = 𝑀0 𝐿0 𝑡 0
𝑡
𝑡
Equating the coefficients of the dimensions. There is no exponent for the mass
Solving for the values
𝑎+𝑏+1=0
−2𝑏 − 1 = 0
1
2
1
𝑎=−
2
𝑏=−
The group is
П1 =
𝑉
�𝑑𝑑
There is only one group and so it is a constant. The velocity is then
𝑉 = 𝐶 �𝑑 𝑔
Problem 7.14
Problem
7.16
[Difficulty: 2]
7.14
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
Summing exponents,
L : a  3b  c  1  0 c  
Hence
2
1
t:
 2a  1  0
a
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
Summing exponents,
Hence
L : a  3b  c  0 c  2
 2 a  2  0 a  1
t:
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 7.15
Problem
7.18
[Difficulty: 2]
7.15
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

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   Lb  L   M0 L0 t0
   3
 
 L t   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 7.16
Problem
7.20
[Difficulty: 2]
7.16
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
ρ
g
M
L
3
2
L
ρ
n = 5 parameters
g
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 7.17
(Difficulty 2)
7.17 Derive an expression for the velocity of very small ripples on the surface of a liquid if this velocity
depends only on ripple length and density and surface tension of the liquid.
Find: The appropriate dimensionless groups.
Solution:
(1) The dimensional parameters are 𝑉, 𝐿, 𝜌, 𝜎 so 𝑛 = 4
(2) Select primary dimensions 𝑀, 𝐿 𝑎𝑎𝑎 𝑡.
(3) We have the following relation:
𝑉
𝑀
0
𝐿
1
𝑡
−1
𝐿
0
1
0
𝜌
1
−3
0
All three dimensions appear so 𝑟 = 3.
𝜎
1
0
−2
We need 3 repeating parameters that include all of the dimensions. We pick up the following
three repeating parameters:
𝐿 𝜌 𝜎
The number of groups is n – m = n – r = 4 – 3 = 1
Thus this group is
𝑀 𝑏 𝑀 𝑐 𝐿
П1 = 𝐿𝑎 𝜌 𝑏 𝜎 𝑐 𝑉 = 𝐿𝑎 � 3 � � 2 � � � = 𝑀0 𝐿0 𝑡 0
𝐿
𝑡
𝑡
Equating the coefficients of the dimensions
Solving for the coefficients
The group is then
𝑏+𝑐 = 0
𝑎 − 3𝑏 + 1 = 0
−2𝑐 − 1 = 0
1
𝑐=−
2
1
𝑏=
2
1
𝑎=
2
П1 =
𝑉
𝜎
�𝜌𝜌
Since there is only one group, it is a constant and then the velocity is given by
𝜎
𝑉 = 𝐶�
𝜌𝜌
Problem 7.18
(Difficulty 3)
7.18 Derive an expression for the axial thrust exerted by a propeller if the thrust depends only on
forward speed, angular speed, size, and viscosity and density of the fluid. How would the expression
change if gravity were a relevant variable in the case of a ship propeller?
Find: The appropriate dimensionless groups.
Solution:
(1) There are 6 dimensional parameters 𝐹, 𝑉, 𝜔, 𝑑, 𝜇, 𝜌, g so 𝑛 = 7
(2) Select primary dimensions 𝑀, 𝐿 𝑎𝑎𝑎 𝑡.
(3) We have the following relation:
𝐹
𝑉
𝑀
1
0
𝐿
1
1
𝑡
−2
−1
𝜔
0
0
−1
𝑑
0
1
0
𝜇
1
−1
−1
𝜌
1
−3
0
𝑔
0
1
−2
All dimensions are present so the number of repeating variables is 𝑟 = 3. We pick the following
three repeating parameters that include all of the dimensions:
𝑉 𝑑 𝜌
We will have n – m = n – r = 7 – 3 = 4 nondimensional groups if we include gravity. If we do
not include gravity, we will have only three groups. For the first group we will combine the
force F with the repeating variables
П1 =
𝑉 𝑎 𝑑𝑏 𝜌𝑐 𝐹
𝐿 𝑎 𝑏 𝑀 𝑐 𝑀𝑀
= � � 𝐿 � 3 � 2 = 𝑀0 𝐿0 𝑡 0
𝑡
𝐿
𝑡
Equating the exponents of the dimensions
𝑐+1=0
𝑎 + 𝑏 − 3𝑐 + 1 = 0
Solving for the values of the exponents
𝑐 = −1
−𝑎 − 2 = 0
𝑎 = −2
𝑏 = −2
The first group is then
𝐹
П1 =
𝜌𝜌 2 𝑑 2
We combine the viscosity with the repeating variables for the second group:
П2 =
Solving for the exponents
𝑉𝑑 𝑑𝑒 𝜌 𝑓 𝜇
𝐿 𝑑 𝑒 𝑀 𝑓 𝑀
= � � 𝐿 � 3 � � � = 𝑀0 𝐿0 𝑡 0
𝑡
𝐿
𝐿𝐿
𝑓+1=0
𝑓 = −1
𝑑 + 𝑒 − 3𝑓 − 1 = 0
−𝑑 − 1 = 0
𝑑 = −1
𝑒 = −1
П2 =
Similarly for the third group that combines ω
𝜇
𝜌𝜌𝜌
𝐿 𝑔
𝑀 𝑖1
П3 = 𝑉𝑔 𝑑 ℎ 𝜌 𝑖 𝜔 = � � 𝐿ℎ � 3 � = 𝑀0 𝐿0 𝑡 0
𝑡
𝐿 𝑡
Solving for the exponents
𝑖=0
𝑔 + ℎ − 3𝑖 = 0
−𝑔 − 1 = 0
𝑔 = −1
ℎ=1
𝜔𝜔
П3 =
𝑉
For the three groups that do not include gravity, we have the relation
𝜔𝜔 𝜇
𝐹
= 𝑓�
,
�
2
2
𝑉 𝜌𝜌𝜌
𝜌𝜌 𝑑
For the fourth group with the addition of gravity 𝑔, we have:
The exponents are then
𝐿 𝑗
𝑀 𝑙 𝐿
П3 = 𝑉𝑗 𝑑 𝑘 𝜌 𝑙 𝑔 = � � 𝐿𝑘 � 3 � � 2 � = 𝑀0 𝐿0 𝑡 0
𝑡
𝐿
𝑡
𝑙=0
𝑗 + 𝑘 − 3𝑙 + 1 = 0
−𝑗 − 2 = 0
𝑗 = −2
𝑘=1
𝑔𝑔
П4 = 2
𝑉
So if 𝑔 is also the variable, the expression for thrust becomes:
𝜔𝜔 𝜇 𝑔𝑔
𝐹
= 𝑓�
,
, �
2
2
𝑉 𝜌𝜌𝜌 𝑉 2
𝜌𝜌 𝑑
Problem 7.19
(Difficulty 2)
7.19 Derive an expression for drag force on a smooth submerged object moving through incompressible
fluid if this force depends only on speed and size of object and viscosity and density of the fluid.
Find: The appropriate dimensionless groups.
Solution:
(1) We have the dimensional parameters 𝐷, 𝑉, 𝐿, 𝜇, 𝜌 so 𝑛 = 5 dimensional parameters
(2) Select primary dimensions 𝑀, 𝐿 𝑎𝑎𝑎 𝑡.
(3) We have the following relation:
𝐷
𝑉
𝑀
1
0
𝐿
1
1
𝑡
−2
−1
𝐿
0
1
0
𝜇
1
−1
−1
𝜌
1
−3
0
All three dimensions are present and r = 3 so we need 3 repeating parameters that include all of
the dimensions. We pick up the following three repeating parameters:
𝑉 𝐿 𝜌
We will have n – m = n – r = 5 – 3 = 2 dimensionless groups. For the first group we will
combine the repeating variables with the parameter D:
𝐿 𝑎
𝑀 𝑐
П1 = 𝑉 𝑎 𝐿𝑏 𝜌 𝑐 𝐷 = � � 𝐿𝑏 � 3 � 𝐿 = 𝑀0 𝐿0 𝑡 0
𝑡
𝐿
Equating the coefficients for each dimension and solving for the values
The first group is
𝑐+1=0
𝑐 = −1
𝑎 + 𝑏 − 3𝑐 + 1 = 0
−𝑎 − 2 = 0
𝑎 = −2
𝑏 = −2
П1 =
𝐷
𝜌𝜌 2 𝐿2
For the second group we combine the repeating variables with the viscosity:
The exponents are then
П2 =
𝑉 𝑑 𝐿𝑒 𝜌 𝑓 𝜇
The second group is
The relation between the groups is
𝐿 𝑑 𝑒 𝑀 𝑓𝑀
= � � 𝐿 � 3�
= 𝑀0 𝐿0 𝑡 0
𝑡
𝐿
𝐿𝐿
𝑓+1=0
𝑓 = −1
𝑑 + 𝑒 − 3𝑓 − 1 = 0
−𝑑 − 1 = 0
𝑑 = −1
𝑒 = −1
П2 =
𝜇
𝜌𝜌𝜌
𝜇
𝐷
= 𝑓�
�
2
2
𝜌𝜌𝜌
𝜌𝜌 𝐿
Problem 7.20
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).)
7.20
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
2
t
Summing exponents:
2  3 a  c  0
t:
2  b  0
a b
c
Thus:
M
t:
2  b  0
2 1
t 
F t
2

M
L
5
c  5
E t
2
ρ R
5
1
a
The solution to this system is:
a  1
Check using F, L, t dimensions:
F
L
The functional relationship is:
4
b2
Π1 
b c
0 0 0
 3  t L  M L t
L t  L 
2
M: 1  a  0
1  3  a  c  0
a
b c
0 0 0
 3  t L  M L t
L 
L
F L
Summing exponents:
L:
M
a  1
Check using F, L, t dimensions:
Π2  p  ρ  t  R

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
c  2
Π2 
p t
2
ρ R
2
1
 p t2 

 f
5

2
ρ R
 ρ R 
E t
2
E
ρ R
t
2
5
 p t2 

f 
 ρ R2 


Problem 7.24
7.21
Problem
[Difficulty: 2]
7.21
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
h
5
g
3
h
L
L
t
4
g
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 7.22
Problem
7.27
[Difficulty: 2]
7.22
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
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
0 0 0
 F t 
  2   F L t
 t  L 
F L  
b  1
L t
M L 1
  t
1
2
2
M
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 7.23
(Difficulty 2)
7.23 Derive an expression for the drag force on a smooth object moving through compressible fluid if
this force depends only on speed and size of object, and viscosity, density and modulus of elasticity of
the fluid.
Find: The appropriate dimensionless groups.
Solution:
(1) We have six dimensional parameters 𝐷, 𝑉, 𝐿, 𝜇, 𝜌, 𝐸 so 𝑛 = 6 dimensional parameters
(2) Select primary dimensions 𝑀, 𝐿 𝑎𝑎𝑎 𝑡.
(3) We have the following relation:
𝐷
𝑉
𝑀
1
0
𝐿
1
1
𝑡
−2
−1
𝐿
0
1
0
𝜇
1
−1
−1
𝜌
1
−3
0
𝐸
1
−1
−2
All three dimensions are present so 𝑟 = 3. We need 3 repeating parameters that include all of
the dimensions and we will pick the following three repeating parameters:
𝑉 𝐿 𝜌
We will have n – m = n – r = 6 – 3 = 3 dimensionless variables. The first group will combine the
dimension D
𝐿 𝑎 𝑏 𝑀 𝑐
𝑎
𝑏
𝑐
П1 = 𝑉 𝐿 𝜌 𝐷 = � � 𝐿 � 3 � 𝐿 = 𝑀0 𝐿0 𝑡 0
𝑡
𝐿
Equating the exponents of the dimensions and solving for their values
The first group is then
𝑐+1=0
𝑐 = −1
𝑎 + 𝑏 − 3𝑐 + 1 = 0
−𝑎 − 2 = 0
𝑎 = −2
𝑏 = −2
П1 =
𝐷
𝜌𝜌 2 𝐿2
For the second group we will combine the viscosity:
П2 =
The exponents are related as
𝑉 𝑑 𝐿𝑒 𝜌 𝑓 𝜇
𝐿 𝑑 𝑒 𝑀 𝑓𝑀
= � � 𝐿 � 3�
= 𝑀0 𝐿0 𝑡 0
𝑡
𝐿
𝐿𝐿
𝑓+1=0
𝑓 = −1
𝑑 + 𝑒 − 3𝑓 − 1 = 0
−𝑑 − 1 = 0
𝑑 = −1
𝑒 = −1
The second group is then
П2 =
𝜇
𝜌𝜌𝜌
The third group combines the modulus of elasticity
𝐿 𝑔
𝑀 𝑖 𝑀
П3 = 𝑉𝑔 𝐿ℎ 𝜌 𝑖 𝐸 = � � 𝐿ℎ � 3 � 2 = 𝑀0 𝐿0 𝑡 0
𝑡
𝐿 𝐿𝑡
Solving for the exponents
𝑖+1=0
𝑖 = −1
𝑔 + ℎ − 3𝑖 − 1 = 0
−𝑔 − 2 = 0
𝑔 = −2
ℎ=0
The third group is
The relation among the groups is then
П3 =
𝐸
𝜌𝑉 2
𝜇
𝐸
𝐷
= 𝑓�
, 2�
2
2
𝜌𝜌𝜌 𝜌𝑉
𝜌𝜌 𝐿
Problem 7.24
(Difficulty 2)
7.24 A circular disk of diameter 𝑑 and of negligible thickness is rotated at a constant angular speed, 𝜔, in
a cylindrical casing filled with a liquid of viscosity 𝜇 and density 𝜌. The casing has an internal diameter 𝐷,
and there is a clearance 𝑦 between the surfaces of disk and casing. Derive an expression for the torque
required to maintain this speed if it depends only on the foregoing variables.
Find: The appropriate dimensionless groups.
Solution:
(1) There are seven dimensional parameters 𝑇,𝑑, 𝜔, 𝜇, 𝜌, 𝐷, 𝑦 so 𝑛 = 7 dimensional parameters
(2) Select primary dimensions 𝑀, 𝐿 𝑎𝑎𝑎 𝑡.
(3) We have the following relation:
𝑇
𝑑
𝑀
1
0
𝐿
2
1
𝑡
−2
0
𝜔
0
0
−1
𝜇
1
−1
−1
𝜌
1
−3
0
𝐷
0
1
0
All three dimensions are present so r = 3. We need 3 repeating parameters that include all of the
dimensions and we will pick the following three repeating parameters:
𝜔 𝑑 𝜌
We will have n – m = n – r = 7 – 3 = 4 dimensionless groups. For the first group we will
combine the repeating variables with the torque T
П1 =
𝜔𝑎 𝑑 𝑏 𝜌 𝑐 𝑇
1 𝑎 𝑏 𝑀 𝑐 𝑀𝐿2
= � � 𝐿 � 3�
= 𝑀0 𝐿0 𝑡 0
𝑡
𝐿
𝑡2
Equating the exponents of the dimensions and solving for the exponents
𝑐+1=0
𝑐 = −1
𝑏 − 3𝑐 + 2 = 0
𝑏 = −5
−𝑎 − 2 = 0
𝑎 = −2
𝑇
П1 =
𝜌𝜌 2 𝑑 5
𝑦
0
1
0
For the second group, we will combine the repeating variable with the viscosity
1 𝑎
𝑀 𝑐𝑀
П2 = 𝜔𝑑 𝑑 𝑒 𝜌 𝑓 𝜇 = � � 𝐿𝑏 � 3 �
= 𝑀0 𝐿0 𝑡 0
𝑡
𝐿
𝐿𝐿
Equating the exponents of the dimensions
𝑓+1=0
𝑓 = −1
𝑒 − 3𝑓 − 1 = 0
𝑒 = −2
−𝑑 − 1 = 0
𝑑 = −1
The second group is
П2 =
𝜇
𝜌𝜌𝑑 2
For the third group we combine the internal diameter
1 𝑎
𝑀 𝑐
П3 = 𝜔 𝑔 𝑑 ℎ 𝜌 𝑖 𝐷 = � � 𝐿𝑏 � 3 � 𝐿 = 𝑀0 𝐿0 𝑡 0
𝑡
𝐿
Solving for the exponents
𝑔=0
𝑖=0
ℎ=1
The third group is
The last group combines the clearance y:
The exponents are
П3 =
𝐷
𝑑
1 𝑎
𝑀 𝑐
П4 = 𝜔 𝑗 𝑑 𝑘 𝜌 𝑙 𝑦 = � � 𝐿𝑏 � 3 � 𝐿 = 𝑀0 𝐿0 𝑡 0
𝑡
𝐿
𝑗=0
𝑙=0
𝑘=1
𝑦
П4 =
𝑑
The relation among the groups is
𝜇 𝐷 𝑦
𝑇
= 𝑓�
, , �
2
5
𝜌𝜌𝑑 2 𝑑 𝑑
𝜌𝜌 𝑑
Problem 7.25
(Difficulty 2)
7.25 Two cylinders are concentric, the outer one fixed and the inner one movable. A viscous
incompressible fluid fills the gap between them. Derive an expression for the torque required to
maintain constant-speed rotation of the inner cylinder if this torque depends only on the diameters and
lengths of the cylinders, the viscosity and density of the fluid, and the angular speed of the inner
cylinder.
Find: The appropriate dimensionless groups.
Solution:
(1) There are eight dimensionless parameters 𝑇,𝑑1 , 𝑑2 , 𝐿1 , 𝐿2 , 𝜇, 𝜌 𝜔 so 𝑛 = 8
(2) Select primary dimensions 𝑀, 𝐿 𝑎𝑎𝑎 𝑡.
(3) We have the following relation:
𝑇
𝑑1 𝑑2
𝑀
1
0
𝐿
2
1
𝑡
−2
0
𝐿1 𝐿2
0
1
0
𝜇
1
−1
−1
𝜌
1
−3
0
𝜔
0
0
−1
All of the primary dimensions appear and we have r= 3. We need 3 repeating parameters that
include all of the dimensions and we pick the following three repeating parameters:
𝜔 𝑑1 𝜌
There will be n – m = n – r = 8 – 3 = 5 dimensionless groups. For the first group we will
combine the torque T:
П1 =
𝜔𝑎 𝑑1𝑏 𝜌 𝑐 𝑇
1 𝑎 𝑏 𝑀 𝑐 𝑀𝐿2
= � � 𝐿 � 3�
= 𝑀0 𝐿0 𝑡 0
𝑡
𝐿
𝑡2
Equating the exponents of the dimensions and solving for the value of the exponents
𝑐+1=0
𝑐 = −1
𝑏 − 3𝑐 + 2 = 0
𝑏 = −5
−𝑎 − 2 = 0
𝑎 = −2
The first group is then
П1 =
𝑇
𝜌𝜔 2 𝑑15
For the second group, we will combine the repeating variables with the viscosity:
1 𝑑
𝑀 𝑓𝑀
= 𝑀0 𝐿0 𝑡 0
П2 = 𝜔𝑑 𝑑1𝑒 𝜌 𝑓 𝜇 = � � 𝐿𝑒 � 3 �
𝑡
𝐿
𝐿𝐿
Equating the exponents of the dimensions
The second group is
𝑓+1=0
𝑓 = −1
𝑒 − 3𝑓 − 1 = 0
𝑒 = −2
−𝑑 − 1 = 0
𝑑 = −1
The third group combines the diameter d2:
П2 =
𝜇
𝜌𝜌𝑑12
1 𝑔
𝑀 𝑖
П3 = 𝜔 𝑔 𝑑1ℎ 𝜌 𝑖 𝑑2 = � � 𝐿ℎ � 3 � 𝐿 = 𝑀0 𝐿0 𝑡 0
𝑡
𝐿
Equating the exponents and solving for their value
So the third group is
𝑔=0
𝑖=0
ℎ = −1
П3 =
𝑑2
𝑑1
The fourth group is similar to the third with the parameter L1:
1 𝑗
𝑀 𝑙
П4 = 𝜔 𝑗 𝑑1𝑘 𝜌 𝑙 𝐿1 = � � 𝐿𝑘 � 3 � 𝐿 = 𝑀0 𝐿0 𝑡 0
𝑡
𝐿
The solution is the same as for the third group
𝑗=0
𝑙=0
𝑘 = −1
And the fourth group is
П4 =
𝐿1
𝑑1
The fifth group is similar to the previous two groups, and could be obtained by inspection
1 𝑚
𝑀 𝑜
П5 = 𝜔𝑚 𝑑1𝑛 𝜌 𝑜 𝐿2 = � � 𝐿𝑛 � 3 � 𝐿 = 𝑀0 𝐿0 𝑡 0
𝑡
𝐿
Solving for the exponents
The fifth group is
𝑚=0
𝑜=0
𝑛 = −1
П5 =
The functional relation between the groups is
𝐿2
𝑑1
𝜇 𝑑2 𝐿1 𝐿2
𝑇
5 = 𝑓 �𝜌𝜌𝑑 2 , 𝑑 , 𝑑 , 𝑑 �
2
𝜌𝜌 𝑑1
1
1
1
1
Problem
Problem7.26
7.28 (In Excel)
[Difficulty: 2]
7.26
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 7.27
Problem
7.30
[Difficulty: 2]
7.27
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
g
μ
g
ρ
μ
L
M
M
2
3
L t
t
4
5
ρ
d
n = 7 parameters
ρ
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
2
2
3
1
L F t
L
2
L
2
Π1 
μ
3
1
2
2
ρ d  g
Problem 7.28
Problem
7.31
[Difficulty: 3]
7.28
Given:
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

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
M
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



3
t
L 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 7.29
(Difficulty 2)
7.29 Derive an expression for the frictional torque exerted on the journal of a bearing if this torque
depends only on the diameters of journal and bearing, their axial lengths (these are the same), viscosity
of the lubricant, angular speed of the journal, and the transverse load (force) on the bearing.
Find: The appropriate dimensionless groups.
Solution:
(1) There are seven dimensional parameters 𝑇,𝑑1 , 𝑑2 , 𝐻, 𝜇, 𝜔, 𝐹 so 𝑛 = 7s
(2) Select primary dimensions 𝑀, 𝐿 𝑎𝑎𝑎 𝑡.
(3) We have the following relation:
𝑇
𝑑1
𝑑2
𝑀
1
0
0
𝐿
2
1
1
𝑡
−2
0
0
𝐻
0
1
0
𝜇
1
−1
−1
𝜔
0
0
−1
𝐹
1
1
−2
All of the dimensions are present so 𝑟 = 3. We will have 3 repeating parameters and we pick the
following three repeating parameters that include all of the dimensions:
𝑑1 𝜔 𝐹
There will be n – m = n – r = 7 – 3 = 4 dimensionless groups. For the first group we will
combine the repeating variables with the torque:
1 𝑏 𝑀𝑀 𝑐 𝑀𝐿2
= 𝑀0 𝐿0 𝑡 0
П1 = 𝑑1𝑎 𝜔𝑏 𝐹 𝑐 𝑇 = 𝐿𝑎 � � � 2 �
𝑡
𝑡
𝑡2
We equate the exponents of the dimensions and solve for their values
The first group is then
𝑐+1=0
𝑐 = −1
𝑎+𝑐+2=0
𝑎 = −1
−𝑏 − 2𝑐 − 2 = 0
𝑏=0
П1 =
𝑇
𝐹𝑑1
For the second group we will combine the repeating variables with the diameter d2:
П2 =
𝑑1𝑑 𝜔𝑒 𝐹 𝑓 𝑑2
=
𝐿𝑑
1 𝑒 𝑀𝑀 𝑓
� � � 2 � 𝐿 = 𝑀0 𝐿0 𝑡 0 𝑓 = 0
𝑡
𝑡
Equating the exponents of the dimensions and solving for their values
The second group is
𝑑+𝑓+1=0
𝑑 = −1
−𝑒 − 2𝑓 = 0
𝑒=0
П2 =
Similarly, for the third group with the length H
𝑑2
𝑑1
1 ℎ 𝑀𝑀 𝑖
𝑔
П3 = 𝑑1 𝜔ℎ 𝐹 𝑖 𝐻 = 𝐿𝑔 � � � 2 � 𝐿 = 𝑀0 𝐿0 𝑡 0
𝑡
𝑡
𝑔 = −1
ℎ=0
𝑖=0
The third group is similar to the second group and could have been determined by inspection
П3 =
𝐿
𝑑1
For the fourth group we combine the viscosity with the repeating variables:
1 𝑏 𝑀𝑀 𝑐 𝑀
𝑗
= 𝑀0 𝐿0 𝑡 0
П4 = 𝑑1 𝜔𝑘 𝐹 𝑙 𝜇 = 𝐿𝑎 � � � 2 �
𝑡
𝑡
𝑡𝑡
Equating the exponents of the dimensions and solving for them
The fourth group is
𝑙+1=0
𝑙 = −1
𝑗+𝑙−1= 0
𝑗=2
−𝑘 − 2𝑙 − 1 = 0
𝑘=1
П4 =
𝜇𝜔𝜔12
𝐹
The relation among the groups is
𝑑2 𝐿 𝜇𝜔𝜔12
𝑇
= 𝑓� , ,
�
𝐹𝑑1
𝑑1 𝑑1 𝐹
Problem 7.30
(Difficulty 3)
7.30 Tests on the established flow of six different liquids in smooth pipes of various sizes yield the
following data: Make a dimensional analysis of this problem and a plot of the resulting dimensionless
numbers as ordinate and abscissa. What conclusion may be drawn from the plot?
Diameter
mm
300
250
150
100
50
25
Velocity
m/s
2.26
2.47
1.22
1.39
0.20
0.36
Viscosity
mPa s
862
431
84.3
44.0
1.5
1.0
Density
Kg/m3
1247
1031
907
938
861
1000
Wall shear stress
Pa
51.2
33.5
5.41
9.67
0.162
0.517
Find: Make a dimensional analysis and draw a conclusion from the plot.
Solution: Use dimensional analysis to determine the relation.
parameters of the problem
First, determine the dimensionless
(1) There five variables in the table, so there are 5 dimensional parameters: 𝜏, 𝑉, 𝑑, 𝜌, 𝜇. 𝑛 = 5
(2) Select primary dimensions 𝑀, 𝐿 𝑎𝑎𝑎 𝑡.
(3) We have the following relations:
𝜏
𝑉
𝑀
1
0
𝐿
−1
1
𝑡
−2
−1
𝑑
0
1
0
𝜌
1
−3
0
𝜇
1
−1
−1
All of the dimensions are present and so 𝑟 = 3. We need 3 repeating parameters that include all
of the dimensions and we pick the following three repeating parameters:
𝑉 𝑑 𝜌
There will be n – m = n – r = 5 – 3 = 2 dimensionless groups. The first group will combine the
repeating variables with the wall shear stress:
𝐿 𝑎
𝑀 𝑐 𝑀
П1 = 𝑉 𝑎 𝑑 𝑏 𝜌 𝑐 𝜏 = � � 𝐿𝑏 � 3 �
= 𝑀0 𝐿0 𝑡 0
𝑡
𝐿
𝐿 𝑡2
Equating the exponents of the dimensions and solving for the values
The first group is
𝑐+1=0
𝑐 = −1
𝑎 + 𝑏 − 3𝑐 − 1 = 0
−𝑎 − 2 = 0
𝑎 = −2
𝑏=0
П1 =
𝜏
𝜌𝑉 2
For the second group we combine the repeating variables with the viscosity:
𝐿 𝑑
𝑀 𝑓𝑀
= 𝑀0 𝐿0 𝑡 0
П2 = 𝑉 𝑑 𝑑 𝑒 𝜌 𝑓 𝜇 = � � 𝐿𝑒 � 3 �
𝑡
𝐿
𝐿𝐿
Equating the exponents of the dimensions and solving for them
The second group is
𝑓+1=0
𝑓 = −1
𝑑 + 𝑒 − 3𝑓 − 1 = 0
−𝑑 − 1 = 0
𝑑 = −1
𝑒 = −1
The relation between the two groups is:
П2 =
𝜇
𝜌𝜌𝜌
𝜇
𝜏
= 𝑓�
�
2
𝜌𝜌𝜌
𝜌𝑉
The data in the table are combined into the two dimensionless groups in the table below. The
second group is recognized as the inverse of the Reynolds number, so the Reynolds number is
calculated:
𝜏
𝜌𝑉 2
0.00804
0.00533
0.00401
0.00534
0.00470
0.00399
𝜌𝜌𝜌
𝜇
981
1477
1969
2963
5740
9000
The data are plotted in the following figure
The plot demonstrates a universal relationship for all pipes and all fluid between
established flow in smooth cylindrical pipes.
The discontinuity at
𝜌𝜌𝜌
𝜇
𝜏
𝜌𝑉 2
and
𝜇
𝜌𝜌𝜌
for
= 2100 delineates
laminar and turbulent zones, and the different trends of the curves suggest a large difference
between the physical laws governing the two regimes.
Note too that you could not draw such a conclusion from the data alone. It needs to be put in
nondimensional form for us to draw conclusions.
Problem 7.31
Problem
7.32
[Difficulty: 2]
7.31
Given:
Functional relationship between the power required to drive a fan and other physical parameters
Find:
Expression for P in terms of the other variables
Solution:
We will use the Buckingham pi-theorem.
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
2  3 a  b  0
t:
3  c  0
M
 3
L 
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:

3
t
Summing exponents:
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
2 5
t
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 7.32
Problem
7.36
[Difficulty: 3]
7.32
Functional relationship between the height of a ball suported by a vertical air jet and other physical parameters
Given:
Find:
Solution:
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
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:
1  3 a  b  c  0
t:
b  0
b c
Π2  D ρ  V  d
 
L
b
c
0 0 0
 3   t   L  M  L  t
L 
b0
Thus:
M: a  0
L 
M
a
 
L
h
Π1 
d
c  1
Check using F, L, t dimensions:
Summing exponents:
t:
a
a0
L:
L:
M
The solution to this system is:
M: a  0
a
L 
L
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
1  3 a  b  c  0
b  0
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
Thus:
Summing exponents:
M: 1  a  0
μ
Π3 
ρ V d
c  1
4
Check using F, L, t dimensions:
1  b  0
a
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:

M L
t
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 7.33
Problem
7.38 (In Excel)
[Difficulty: 3]
7.33
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 7.34
Problem
7.42
[Difficulty: 3]
7.34
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.
Given:
1
m
2
Select primary dimensions M, L, t:
3
m
A
A
M
L
t
4
5
p
p
T
p
T
M
2
L t
A
T
n = 5 parameters
R
L
T
2
(Mathcad can't render dots!)
R
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
2 
b
M 
2
c L 
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
7.44
Problem 7.35
[Difficulty: 3]
7.35
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.
Given:
1
m
2
Select primary dimensions M, L, t:
3
m
A
A
M
L
t
4
5
ρ
h
ρ
M
2
L
A
h
ρ
L
3
Δp
g
M
L
L t
n = 6 parameters
g
Δp
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:
Summing exponents:
M
t

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
2
5
1
L
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  Δp 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

Δp
2


A ρ g  A 

L
4

F t
t
Δp
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

Δp


A ρ g  A 
Problem 7.36
Problem
7.45
[Difficulty: 3]
7.36
Functional relationship between the aerodynamic torque on a spinning ball and other physical parameters
Given:
Find:
Solution:
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
V
M L
t
4
5
ρ
L
M
M
t
3
L t
2
V
μ
V
2
ρ
ρ
L
μ
D
d
ω
d
D
ω
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



3
t
L 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:
t 1
F t L

  1
2
2 L L
L F t
a
b
c
Π3  ω ρ  V  D
Thus:
Summing exponents:

M
a
 
L
a0
L:
3  a  b  c  0
t:
1  b  0
b
c
Π4  d  ρ  V  D
b
c
0 0 0
L  M L t



3
t
t
L   
ω D
Π3 
V
The solution to this system is:
M: a  0
a
1
b  1
c1
Check using F, L, t dimensions:
Thus:
L 
t:
b  0
The functional relationship is:
 
L
t
L
1
b
The solution to this system is:
a0
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
1
b0
c  1
Check using F, L, t dimensions:

Π1  f Π2 Π3 Π4

d
Π4 
D
1
L
L  1
T
2
3
ρ V  D
ω D d 
 
 ρ V D V D 
 f 
μ

Problem
7.48
Problem 7.37
[Difficulty: 3]
7.37
Functional relationship between the power loss in a journal bearing and other physical parameters
Given:
Find:
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
D
D
c
l
D
c
L
L
L
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
ω
ω
We have n - m = 4 dimensionless groups. Setting up dimensional equations:
a
b c
Π1  P D  ω  p
Thus:
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
t
a
1
a  3
b c
Summing exponents:
F:
c0
L:
1  a  2 c  0
t:
b  0
c
Thus:
1
L L   
t
a
b  1
b
c  1
L L  
a
1
P
3
D  ω p
c
The solution to this system is:
Thus:
Π1 
F
0 0 0
   F L t
 2
L 
a  1
Π3  c D  ω  p
b
0 0 0
F
  2   F L t
 t  L 
 L  
The solution to this system is:
Summing exponents:
F:
F L
b0
b
c0
l
Π2 
D
c
0 0 0
F
  2   F L t
 t  L 
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
1
b
c
0 0 0
F
  2   F L t
 t  L 
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 7.38
Problem
7.50
[Difficulty: 3]
7.38
Functional relationship between the thrust of a marine propeller and other physical parameters
Given:
Find:
Solution:
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
2
ρ
D
ρ
L
V
D
L
3
V
g
ω
p
μ
V
g
ω
p
μ
L
L
1
2
t
t
t
M
L 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
c
Π2  g  ρ  V  D
Thus:
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
b  2
a
a0
b
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



3
t
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
 
L
M
c
0 0 0
       L  M  L  t


2
3
t
t L   
L
Summing exponents:
Π3  ω ρ  V  D
a
c
0 0 0
 3   t   L  M  L  t
L 
a  1
L:
b
M
The solution to this system is:
M: 1  a  0
a
2

ω D
Π3 
V
a
b
c
Π4  p  ρ  V  D
a  1
L:
1  3  a  b  c  0
t:
2  b  0
b
c
M
b
L
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
The solution to this system is:
M: 1  a  0
L:

b  2
Π4 
c
0 0 0
L  M L t



3
t
L t
L   
Thus:
Summing exponents:
6
 
The solution to this system is:
M: 1  a  0
a
a
M
c
0 0 0
 3   t   L  M  L  t
L t  L 
2
Summing exponents:
Π5  μ ρ  V  D

M
Thus:
a  1
L
4
F t
2

t
2
L
2

1
L
2
b  1
1
L
t
2
μ
Π5 
ρ V D
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 7.39
Problem
7.51
[Difficulty: 3]
7.39
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 7.40
Problem
7.54
[Difficulty: 4]
7.40
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
Given:
Find:
We will use the Buckingham pi-theorem.
Solution:
1
p max
ρ
2
U0
EV
Select primary dimensions M, L, t:
3
p max
ρ
U0
M
L
3
t
M
L t
4
5
ρ
2
L
n = 4 parameters
EV
M
L t
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
Summing exponents:
1  3  a  b  0
t:
2  b  0
a
Π2  Ev  ρ  U0
b
M
6
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:
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 7.41
Problem
7.56
[Difficulty: 3]
7.41
Given:
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
2
2
20
2
Fp  250  N 
    ( 20)
5.333  75 
1
2
 Vp   Lp 
Fp  Fm

  
ρm Vm
   Lm 
ρp
Fp  1.333  kN
Problem 7.42
(Difficulty 2)
7.42 An airplane wing of 3 𝑚 chord length moves through still air at 15 ℃ and 101.3 𝑘𝑘𝑘 at a speed of
320
𝑘𝑘
.
ℎ
A 1: 20 scale model of this wing is placed in a wind tunnel, and dynamic similarity between a
model and prototype is desired. (a) What velocity is necessary in a tunnel where the air has the same
pressure and temperature as that in flight? (b) What velocity is necessary in a variable-density wind
tunnel where absolute pressure is 1400 𝑘𝑘𝑘 and temperature is 15 ℃? (c) At what speed must the
model move through water (15 ℃) for dynamic similarity?
Find: The wind velocity for dynamic similarity between a model and prototype
Solution:
(a) For the dynamic similarity we have the Reynolds number as:
𝜌𝜌𝜌
𝜌𝜌𝜌
�
� =�
�
𝜇 𝑚
𝜇 𝑝
As the air has same temperature and pressure, so the air properties are the same. For dynamic
similarity we then have
The length scale is
The prototype velocity is
Thus the model velocity must be
(𝑉𝑉)𝑚 = (𝑉𝑉)𝑝
𝐿𝑝
= 20
𝐿𝑚
𝑉𝑝 = 320
𝑘𝑘
𝑚
= 88.9
ℎ
𝑠
𝑉𝑚 = 1778
𝑚
𝑠
This is a supersonic velocity and in a different regime from that for the prototype
(b) In this case we have the same viscosity for the model and prototype:
𝜇𝑚 = 𝜇𝑝
Using the ideal gas law we have:
𝜌
𝜌
� � =� �
𝑝 𝑚
𝑝 𝑝
With the specified pressures, the density ratio is
𝜌𝑚 =
Thus, for dynamic similarity
𝑝𝑚
1400
𝜌𝑝 =
𝜌
101.3 𝑝
𝑝𝑝
Or
𝜌𝜌𝜌
𝜌𝜌𝜌
�
� =�
�
𝜇 𝑚
𝜇 𝑝
The model velocity would then be
(𝜌𝜌𝜌)𝑚 = (𝜌𝜌𝜌)𝑝
𝑉𝑚 = 𝑉𝑝
𝜌𝑝 𝐿𝑝
𝑚 101.3
𝑚
= 88.9 ×
× 20 = 128.7
𝑠 1400
𝑠
𝜌𝑚 𝐿𝑚
This is a reasonable velocity, but the wind tunnel pressure is about 15 atmospheres and very high.
(c) The properties for the model of water at 15 ℃ are:
We have for the prototype:
𝜇𝑚
𝑘𝑘
𝑚3
= 1.139 × 10−3 𝑃𝑃 ∙ 𝑠
𝜌𝑚 = 998
𝑘𝑘
𝑚3
−5
𝜇𝑝 = 1.789 × 10 𝑃𝑃 ∙ 𝑠
𝜌𝑝 = 1.225
Using the similarity for Reynolds number:
𝑘𝑘
𝜌𝑝 𝐿𝑝 𝜇𝑚
𝑚 1.225 𝑚3
1.139 × 10−3 𝑃𝑃 ∙ 𝑠
𝑉𝑚 = 𝑉𝑝
= 88.9 ×
× 20 ×
𝑘𝑘
𝑠
1.789 × 10−5 𝑃𝑃 ∙ 𝑠
𝜌𝑚 𝐿𝑚 𝜇𝑝
998 3
𝑚
𝑚
𝑉𝑚 = 138.8
𝑠
This is a very high velocity for water flow.
Problem 7.43
(Difficulty 2)
7.43 A flat plate 1.5 𝑚 long and 0.3 𝑚 wide is towed at 3
𝑚
𝑠
in a towing basin containing water at 20 ℃,
and the drag force is observed to be 14 𝑁. Calculate the dimensions of similar plate which will yield
𝑚
dynamically similar conditions in an airstream (101.4 𝑘𝑘𝑘 𝑎𝑎𝑎 15 ℃) having a velocity of 18 . What
𝑠
drag force may be expected on this plate?
Find: The dimensions of a model and the drag force 𝐷𝑝 .
Solution:
For the dynamic similarity we have equal Reynolds numbers for the model and protype:
𝜌𝑚 𝑉𝑚 𝐿𝑚 𝜌𝑝 𝑉𝑝 𝐿𝑝
=
𝜇𝑚
𝜇𝑝
For the prototype in water:
𝜌𝑝 = 998
𝑘𝑘
𝑚3
𝜇𝑝 = 1.002 × 10−3 𝑃𝑃 ∙ 𝑠
𝐿𝑝 = 1.5 𝑚
For the model in air the properties are
𝑉𝑝 = 3
𝑚
𝑠
𝜌𝑚 = 1.225
𝑘𝑘
𝑚3
𝜇𝑚 = 1.789 × 10−5 𝑃𝑃 ∙ 𝑠
𝑉𝑚 = 18
The length of the model for dynamic similarity is then
𝐿𝑚 =
𝐿𝑚
𝑚
𝑠
𝜇𝑚 𝜌𝑝 𝑉𝑝 𝐿𝑝
𝜇𝑚 𝜌𝑝 𝑉𝑝
= 𝐿𝑝
𝜌𝑚 𝑉𝑚 𝜇𝑝
𝜇𝑝 𝜌𝑚 𝑉𝑚
𝑘𝑔
𝑚
998 3
3
1.789 × 10−5 𝑃𝑃 𝑠
𝑠
𝑚
= 1.5 𝑚 ×
×
×
1.002 × 10−3 𝑃𝑃 ∙ 𝑠 1.225 𝑘𝑘 18 𝑚
𝑠
𝑚3
𝐿𝑝 = 3.64 𝑚
Under this dynamical similar condition, we have:
𝜌𝜌𝜌
𝐷
= 𝑓�
�
2
2
𝜇
𝜌𝑉 𝐴
From the geometric similarity,
𝑏𝑝 =
We also have:
The drag force will be:
𝐿𝑚 𝐿𝑝
=
𝑏𝑚 𝑏𝑝
𝑏𝑚
0.3 𝑚
𝐿𝑝 =
× 3.64 𝑚 = 0.73 𝑚
1.5 𝑚
𝐿𝑚
𝐷
𝐷
� 2 � =� 2 �
𝜌𝑉 𝐴 𝑚
𝜌𝑉 𝐴 𝑝
𝑘𝑘
1.225 3 18𝑚/𝑠2 3.64𝑚 × 0.73𝑚
𝜌𝑚 𝑉𝑚2 𝐴𝑚
𝑚
𝐷𝑚 = 𝐷𝑝
= 14𝑁 ×
×
×
𝑘𝑘
1.5𝑚 × 0.3𝑚
𝜌𝑝 𝑉𝑝2 𝐴𝑝
3𝑚/𝑠2
998 3
𝑚
𝐷𝑚 = 3.65 𝑁
Problem 7.44
(Difficulty 3)
7.44 This 1: 12 pump model (𝑢𝑢𝑢𝑢𝑢 𝑤𝑤𝑤𝑤𝑤 𝑎𝑎 15 ℃) simulates a prototype for pumping oil of specific
gravity 0.9. The input to the model is 0.522 𝑘𝑘. Calculate the viscosity of the oil and the prototype
power for complete dynamic similarity between model and prototype.
Find: The viscosity of the oil 𝜇𝑝 and the prototype power.
Solution: Use similitude to determine the parameters of the prototype.
Using the continuity relation, the velocity at the outlet is
𝑚3
0.014
𝑄
𝑚
𝑠
= 3.2
𝑉𝑚 = =
𝐴 1 × 𝜋 × (0.075 𝑚)2
𝑠
4
We have the following properties for the model:
𝜌𝑚 = 998
𝑘𝑘
𝑚3
𝜇𝑚 = 1.139 × 10−3 𝑃𝑃 ∙ 𝑠
𝜌𝑝 = 0.9 × 998
𝑘𝑘
𝑘𝑘
=
898
𝑚3
𝑚3
For dynamic similarity, we need to have equal Reynolds number and Froude numbers. The Reynolds
number dynamic similarity means that:
𝑉𝑉𝑉
𝑉𝑉𝑉
�
� =�
�
𝜇 𝑚
𝜇 𝑝
The Froude number dynamic similarity means that:
�
𝑉2
𝑉2
� =� �
𝑑𝑑 𝑚
𝑑𝑑 𝑝
From the Reynolds number similarity
3.2
𝑘𝑘
𝑘𝑘
𝑚
× 0.075 𝑚 × 998 3 𝑉𝑝 × (0.075 𝑚 × 12) × �998 𝑚3 �
𝑠
𝑚 =
𝜇𝑝
1.139 × 10−3 𝑃𝑃 ∙ 𝑠
And from the Froude number similarity
𝑚 2
2
�
�𝑉𝑝 �
𝑠
𝑚 =
𝑚
�0.075 𝑚 × 9.81 2 � �0.075 𝑚 × 12 × 9.81 2 �
𝑠
𝑠
�3.2
Using the Froude number similarity we can find the prototype velocity
𝑉𝑝 = 11.09
𝑚
𝑠
With this value of velocity, we can find the prototype viscosity
𝜇𝑝 = 0.043 𝑃𝑃 ∙ 𝑠
We now apply the First Law of Thermodynamics equation for the model from the pipe at location 1
(100 𝑚𝑚 𝑑𝑑𝑑) to the pipe at location 2 (75 𝑚𝑚). The velocity at the outlet is the model velocity used in
the Reynolds number. For the ideal flow without losses (pump inefficiency) we have
𝑉12 𝑊𝑚̇ 𝚤
𝑉𝑚2
+
+ 𝑧1 =
+ 𝑧2
𝑚̇
2𝑔
2𝑔
The velocity at location 1 is
𝑉1 = 𝑉𝑚
𝑚 (0.075 𝑚)2
𝑚
𝑑22
=
3.2
×
= 1.8
2
2
(0. 1 𝑚)
𝑠
𝑠
𝑑1
The difference in elevation is given by the manometers. These measure the pressure, but also reflect the
height difference
𝑧𝑚 − 𝑧1 = 3 𝑚
The ideal pump power per unit mass flow is then
𝑊𝑚̇ 𝚤
𝑚̇
=
�3.2
𝑚 2
𝑚 2
� − �1.8 �
𝑠
𝑠 + 3 𝑚 = 3.35 𝑚
𝑚
2 × 9.81 2
𝑠
The total ideal power for the model is then
𝑊𝑚̇ 𝚤 = 𝑄𝑚 𝛾
𝑊𝑚̇ 𝚤
𝑚3
𝑁
= 0.014
× 9810 3 × 3.35 𝑚 = 460 𝑊
𝑚
𝑚̇
𝑠
The pump efficiency is the ratio of the ideal model power to the input:
𝜂𝑚 =
𝑊𝑚 𝑖 460 𝑊
=
= 0.881
522 𝑊
𝑊𝑚
We now compute the ideal power for the prototype. Because of dynamic similarity, the power per unit
mass flow will be the same. The prototype is 12 times as large as the model, so:
𝑊𝑝̇ 𝚤 𝑊𝑚̇ 𝚤
=
× 12 = 3.35 𝑚 × 12 = 40.2 𝑚
𝑚̇
𝑚̇
The outlet diameter is also 12 time larger, so the volume flow rate for the prototype is
𝑄𝑝 = 11.09
𝑚 𝜋
𝑚3
× × (0.075 𝑚 × 12)2 = 7.05
𝑠 4
𝑠
The ideal prototype power is then
𝑊𝑝 𝑖 = 0.9𝑄𝑝
𝑊𝑝̇ 𝚤
𝑚3
𝑁
𝛾 = 0.9 × 7.05
× 40.2 𝑚 × 9810 3 = 2.50 𝑀𝑀
𝑚
𝑠
𝑚̇
The efficiency is assumed to be the same. The prototype power is then
𝑊𝑝 =
𝑊𝑝 𝑖 2.50 𝑀𝑀
=
= 2.84 𝑀𝑀
0.881
𝜂
Problem 7.45
Problem
7.58
[Difficulty: 5]
7.45
Given:
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



3
t
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
1
F L L
F t L
3 1

t 
1

 t
1
2
5
2
2
2
t
L
L
L F t
F t

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 7.46
Problem
7.60
[Difficulty: 2]
7.46
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 7.47
Problem
7.61
[Difficulty: 3]
7.47
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
m
μm  1.81  10
 5 N s


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
3
3
110 1
m
m
1.00  10
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

2
ρp  Vp  Dp
2
 28    5 

  
 110   1 
2
Thus:
 Vp 
Fp  Fm


ρm Vm
 
ρp
2
 Dp 


 Dm 
2
2
Fp  43.4 kN
Problem 7.48
(Difficulty 3)
7.48 A flow rate of 0.18
𝑚3
𝑠
of water 20 ℃ discharges a 0.3 𝑚 pipe through a 0.15 𝑚 nozzle into the
atmosphere. The axial force component exerted by water on the nozzle is 3 𝑘𝑘. If frictional effects may
be ignored, what corresponding force will be exerted on a 4: 1 prototype of nozzle and pipe discharging
1.13
𝑚3
𝑠
of air (101.4 𝑘𝑘𝑘 𝑎𝑎𝑎 15 ℃) to the atmosphere? If frictional effects are included, the axial
component is 3.56 𝑘𝑘. What flow rate of air is then required for dynamic similarity? What is the
corresponding force on the nozzle discharging air?
Find: (a) The force 𝐹𝑎𝑎𝑎 . (b) The flow rate 𝑄 and 𝐹𝑎𝑎𝑎 with frictional effects.
Solution:
(a) For dynamic similarity for pressure force of frictionless flow we have the equality of the pressure
coefficient:
∆𝑝
∆𝑝
� 2� = � 2�
𝜌𝑉 𝑚
𝜌𝑉 𝑝
Using the continuity equation to replace the velocity with the volume flow rate
𝑄=𝐴𝑉
So the dynamic similarity becomes
�
∆𝑝𝐴2
∆𝑝𝐴2
�
=
�
�
𝜌𝑄 2 𝑚
𝜌𝑄 2 𝑝
The product of the pressure difference and the area is the force
So for dynamic similarity we have
𝐹 = ∆𝑝 𝐴
F𝐴
F𝐴
� 2� = � 2�
𝜌𝑄 𝑚
𝜌𝑄 𝑝
We have the following properties for the force on model. We take the water flow as the model and the
air flow as the prototype:
𝐴𝑚 =
𝜌𝑚 = 998
𝑘𝑘
𝑚3
𝑄𝑚 = 0.18
𝑚3
𝑠
𝜌𝑝 = 1.225
𝑘𝑘
𝑚3
𝜋
× (0.15 𝑚)2 = 0.0177 𝑚2
4
𝐹𝑚 = 3 𝑘𝑘
And for the prototype
𝐴𝑝 =
𝜋
× (0.15 𝑚 × 4)2 = 0.283 𝑚2
4
Thus we have for the force on the prototype
𝑄𝑝 = 1.13
𝑚3
𝑠
𝑘𝑘 �1.13
F𝐴
0.0177 𝑚2 1.225 𝑚3
𝐴𝑚 𝜌𝑝 𝑄𝑝2
� 2 � 𝐹𝑝 = 𝐹𝑚
×
2 = 3𝑘𝑘 × 0.283 𝑚2 ×
𝑘𝑘
𝜌𝑄 𝑚
𝐴𝑝 𝜌𝑚 𝑄𝑚
998 3
�0.18
𝑚
𝐹𝑝 = 𝐹𝑎𝑎𝑎 = 9.08 𝑁
2
𝑚3
�
𝑠
2
𝑚3
�
𝑠
(b) With friction, we need to have the dynamic similarity for Reynolds number and Froude number.
For the Reynolds number:
ρVD
ρVD
�
� =�
�
𝜇 𝑚
𝜇 𝑝
And for the Froude number
ρQD
ρQD
�
� =�
�
𝐴𝐴 𝑚
𝐴𝐴 𝑝
We have the properties
𝜇𝑚
𝐷𝑚 = 0.15 𝑚
= 1.002 × 10−3 𝑃𝑃 ∙ 𝑠
𝐷𝑝 = 0.6 𝑚
𝜇𝑝 = 1.789 × 10−5 𝑃𝑃 ∙ 𝑠
We have the velocity for the model as
𝑄𝑚
𝑉𝑚 =
𝐴𝑚
𝑚3
𝑚
𝑠
=𝜋
= 10.9
𝑠
× (0.15 𝑚)2
4
0.18
The prototype model velocity is then
𝑘𝑘
−5
998 3
𝑚
𝜌𝑚 𝜇𝑝 𝐿𝑚
𝑚 × 1.789 × 10 𝑃𝑃 ∙ 𝑠 × 1
= 10.9 ×
𝑉𝑝 = 𝑉𝑚
𝑠 1.225 𝑘𝑘 1.002 × 10−3 𝑃𝑃 ∙ 𝑠 4
𝜌𝑝 𝜇𝑚 𝐿𝑝
𝑚3
𝑚
𝑉𝑝 = 39.6
𝑠
Thus the prototype flow rate is
𝑚
𝑚3
× 0.283 𝑚2 = 10.47
𝑠
𝑠
Finally, we have for the force for dynamic similarity:
F𝐴
F𝐴
� 2� = � 2�
𝜌𝑄 𝑚
𝜌𝑄 𝑝
𝑄𝑝 = 𝑉𝑝 𝐴𝑝 = 39.6
Or
where
𝐹𝑝 = 𝐹𝑚 ×
𝜌𝑝
𝑄𝑝 2 𝐴𝑚
×� � ×
𝜌𝑚
𝑄𝑚
𝐴𝑝
𝐹𝑚 = 3.56 𝑘𝑘
The force on the prototype is then
2
𝑘𝑘
𝑚3
2
10.47
3
𝑠 � × 0.0177 𝑚
𝑚 ×�
𝐹𝑝 = 3.56 𝑘𝑘 ×
𝑘𝑘
𝑚3
0.283 𝑚2
998 3
0.18
𝑠
𝑚
1.225
𝐹𝑝 = 𝐹𝑎𝑎𝑎 = 925 𝑁
Problem 7.49
(Difficulty 2)
7.49 A force of 9 𝑁 is required to tow a 1: 50 ship model at 4.8
𝑘𝑘
.
ℎ
Assuming the same water in towing
basin and sea, calculate the corresponding speed and force in the prototype if the flow is dominated by:
(a) density and gravity (b) density and surface tension (c) density and viscosity
Find: The corresponding speed 𝑉𝑝 and 𝐹𝑝 .
Solution:
(a) If the flow is dominated by density and gravity, for dynamic similarity we must have equal
Froude numbers. It is easier to work with the square of the Froude number:
𝑉2
𝑉2
� � =� �
𝑔𝑔 𝑚
𝑔𝑔 𝑝
Or, since it is the same gravity for the model and prototype
𝐿𝑝 1/2
𝑘𝑘
50 1/2
� �
𝑉𝑝 = 𝑉𝑚 × � � = 4.8
ℎ
1
𝐿𝑚
Thus the velocity is
𝑘𝑘
𝑉𝑝 = 33.9
ℎ
We also have that for dynamic similarity that the drag coefficients are equal
𝐹
𝐹
� 2 2 � = � 2 2�
𝜌𝑉 𝐿 𝑚
𝜌𝑉 𝐿 𝑝
The density is the same for both the model and prototype and so we have
𝑘𝑘 2
33.9
𝐿𝑝
𝑉𝑝
50
ℎ �
𝐹𝑝 = 𝐹𝑚 × � � × � � = 9𝑁 × � � × �
𝑘𝑘
1
𝐿𝑚
𝑉𝑚
4.8
ℎ
2
Or the force is:
2
2
𝐹𝑝 = 1122 𝑘𝑘
(b) If the flow is governed by the density and surface tension we have equal Weber numbers for
dynamic similarity:
𝜌𝜌 2 𝐿
𝜌𝜌 2 𝐿
�
� =�
�
𝜎 𝑚
𝜎 𝑝
The surface tension and density are the same for both model and prototype so we have for the
velocity
1/2
𝑘𝑘
1 1/2
� �
ℎ
50
𝑘𝑘
𝑉𝑝 = 0.68
ℎ
which is much lower than the speed for Froude number similarity
𝐿𝑚
𝑉𝑝 = 𝑉𝑚 × � �
𝐿𝑝
= 4.8
For the force, we again use that the force coefficients are the same for dynamic similarity:
𝐹
𝐹
� 2 2 � = � 2 2�
𝜌𝑉 𝐿 𝑝
𝜌𝑉 𝐿 𝑚
So we get:
𝑘𝑘 2
𝐿𝑝
𝑉𝑝
50
ℎ �
𝐹𝑝 = 𝐹𝑚 × � � × � � = 9𝑁 × � � × �
𝑘𝑘
1
𝐿𝑚
𝑉𝑚
4.8
ℎ
𝐹𝑝 = 452 𝑁
2
2
2
0.68
(c) If the flow is dominated by the density and viscosity, the Reynolds numbers are the same for
dynamic similarity:
𝜌𝜌𝜌
𝜌𝜌𝜌
�
� =�
�
𝜇 𝑚
𝜇 𝑝
Or, since the density and viscosity are the same for both model and prototype
𝐿𝑚
𝑘𝑘
1
� = 4.8
� �
ℎ
50
𝐿𝑝
𝑘𝑘
𝑉𝑝 = 0.096
ℎ
Again, because the force coefficients are then equal:
𝐹
𝐹
� 2 2 � = � 2 2�
𝜌𝑉 𝐿 𝑝
𝜌𝑉 𝐿 𝑚
𝑉𝑝 = 𝑉𝑚 × �
Or
or
𝑘𝑘 2
0.096
𝐿𝑝 2
𝑉𝑝 2
50 2
ℎ �
𝐹𝑝 = 𝐹𝑚 × � � × � � = 9𝑁 × � � × �
𝑘𝑘
1
𝐿𝑚
𝑉𝑚
4.8
ℎ
𝐹𝑝 = 9 𝑁
We see that the force we predict on the model depends very heavily on what we assume the flow is
dominated by. In this case the forces differ by 106!
Problem 7.50
Problem
7.64
[Difficulty: 2]
7.50
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
m
 1.21 kg 

3
m 
m 
Vm  7.5 

kg 

s
 998  3 
m 

Then
Fm
2
ρm Vm  lm sm

 1.01  10 3 N s 

2 
m 
 10   
  
 5 N s 
1
 1.81  10  2 
m 

Fp
2
ρp  Vp  lp  sp
2
μp  1.81  10

2
m
m
Vm  5.07
s
ρm Vm lm sm
998





Fp
ρp
2 l p  sp
1.21
Vp
Fm
 5 N s
2
 5.07   1  1  3.77


10 10
 7.5 
Problem 7.51
Problem
7.66
[Difficulty: 3]
7.51
Given:
Model of water pump
Find:
Model flow rate for dynamic similarity (ignoring Re); Power of prototype
Solution:
Q
From Buckingham Π
Hence
3
Qp

ωm Dm
ωp  Dp
3
3
 Dm 
Qm  Qp 


ωp Dp
 
ωm
3
Then
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
ft
Qm  15

s
From Table A.8 at 68 oF
3
ω D
Qm
For dynamic similarity
P
and
3
ρp  1.94
 2400  


 750 
slug
ft
3
5
ρm ωm  Dm
3
ft
Qm  0.750 
s
ρm  0.00234 
slug
ft
3
Pp

3
ρp  ωp  Dp
3
5
 ωp   Dp 
Pp  Pm

  
ρm ωm
   Dm 
ρp
3
From Table A.9 at 68 oF
3
Pm
1
 
4
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 7.52
Problem
7.68
[Difficulty: 3]
7.52
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 7.53
(Difficulty 3)
7.53 A ship 120 𝑚 long moves through freshwater at 15 ℃ at 32
𝑘𝑘
.
ℎ
A 1: 100 model of this ship is to be
tested in a towing basin containing a liquid of specific gravity 0.92. What viscosity must this liquid have
for both Reynolds’ and Froude’s laws to be satisfied? At what velocity must the model be towed? What
propulsive force on this ship corresponds to a towing force of 9𝑁 in the model?
Find: The liquid viscosity of the model 𝜇𝑚 . The velocity of the model 𝑉𝑚 . The force on the ship 𝐹𝑝 .
Solution:
For dynamic similarity we must have equal Froude and Reynolds numbers. It is easier to work with the
square of the Froude number. For the Froude number
�
For the Reynolds number
𝑉2
𝑉2
� =� �
𝑔𝑔 𝑚
𝑔𝑔 𝑝
𝜌𝜌𝜌
𝜌𝜌𝜌
�
� =�
�
𝜇 𝑚
𝜇 𝑝
From Froude number similarity, since gravity is the same for both model and prototype
𝑉𝑚 = 𝑉𝑝 × �
1/2
𝐿𝑚
�
𝐿𝑝
= 32
𝑉𝑚 = 3.2
𝑘𝑘
1 1/2
�
�
ℎ𝑟
100
𝑘𝑘
ℎ𝑟
For Reynolds number similarity, we need to account for the difference in density. For dynamic similarity
we have for the viscosity
with
𝜇𝑚 = 𝜇𝑝 × �
𝜌𝑚
𝑉𝑚
𝐿𝑚
� � �� �
𝜌𝑝
𝑉𝑝
𝐿𝑝
𝐿𝑝 = 120 𝑚
𝜌𝑝 = 998
𝑘𝑘
𝑚3
𝜇𝑝 = 1.139 × 10−3 𝑃𝑃 ∙ 𝑠
𝐿𝑚 = 1.2 𝑚
𝜌𝑚 = 0.92 × 998
The viscosity of the liquid for the model is
𝜇𝑚 = 1.139
The viscosity must be
× 10−3 𝑃𝑃
𝑘𝑘
𝑘𝑘
=
918
𝑚3
𝑚3
𝑘𝑘
𝑘𝑘
918 3
3.2
ℎ𝑟 � × � 1 �
𝑚
∙𝑠×�
� �
𝑘𝑘
𝑘𝑘
100
998 3
32
ℎ𝑟
𝑚
𝜇𝑚 = 1.05 × 10−6 𝑃𝑃 ∙ 𝑠
This is a very low viscosity and the fluid must be as “thin” as air.
For the force, we have that the drag coefficients are equal:
𝐹
𝐹
� 2 2 � = � 2 2�
𝜌𝑉 𝐿 𝑝
𝜌𝑉 𝐿 𝑚
𝑘𝑘 2
32
𝐿𝑝 2
𝑉𝑝 2
100 2
ℎ𝑟 �
� �
𝐹𝑝 = 𝐹𝑚 × � � × � � = 9𝑁 × �
𝑘𝑘
1
𝐿𝑚
𝑉𝑚
3.2
ℎ𝑟
Thus
𝐹𝑚 = 9 𝑁
𝐹𝑝 = 9 × 106 𝑁
Problem 7.54
(Difficulty 3)
7.54 A 1: 30 scale model of a cavitating overflow structure is to be tested in a vacuum tank wherein the
pressure is maintained at 2.0 𝑝𝑝𝑝𝑝. The prototype liquid is water at 70 ℉. The barometric pressure on
the prototype is 14.5 𝑝𝑝𝑝𝑝. If the liquid to be used in the model has vapor pressure of 1.5 𝑝𝑝𝑝𝑝, what
values of density, viscosity , and surface tension must it have for complete dynamic similarity between
model and prototype?
Find: The parameters for completing the dynamic similarity.
Solution:
For the dynamic similarity we have dynamic similarity with the Froude number, Reynolds number and
Euler number. The Froude number squared is easier to work with, and is
�
𝑉2
𝑉2
� =� �
𝑔𝑔 𝑚
𝑔𝑔 𝑝
The Reynolds number relation is written in terms of the kinematic viscosity
𝑉𝑉
𝑉𝑉
� � =� �
𝜈 𝑚
𝜈 𝑝
The Euler number relation is
And the Weber number is
∆𝑝
∆𝑝
�
� =�
�
1 2
1 2
𝜌𝑉
𝜌𝑉
2
2
𝑚
𝑝
�
𝜌𝜌 2 𝐿
𝜌𝜌 2 𝐿
� =�
�
𝜎 𝑚
𝜎 𝑝
The prototype is water at 70 ℉. The properties are
Vapor pressure
𝑝𝑣 = 0.36 𝑝𝑝𝑝a
𝜌𝑝 = 1.934
𝑠𝑠𝑠𝑠
𝑓𝑓 3
𝑣𝑝 = 1.059 × 10−5
𝜎𝑝 = 0.00498
𝑓𝑓 2
𝑠
𝑙𝑙𝑙
𝑓𝑓
From the Euler number, we have the pressure difference for the prototype
∆𝑝𝑝 = 14.5 𝑝𝑝𝑝𝑝 − 0.36 𝑝𝑝𝑝𝑝 = 14.14 𝑝𝑝𝑝
And for the model the pressure difference is:
∆𝑝𝑚 = 2.0 𝑝𝑝𝑝𝑝 − 1.5 𝑝𝑝𝑝𝑝 = 0.5 𝑝𝑝𝑝
From Euler number similarity we have for the density we have:
𝜌𝑚 = 𝜌𝑝 × �
And from the Froude number similarity we have
𝑉𝑝 2
∆𝑝𝑚
� � �
∆𝑝𝑝
𝑉𝑚
𝑉𝑝
𝐿𝑝 1/2
=� �
𝑉𝑚
𝐿𝑚
So the density for the model can be calculated as
𝜌𝑚 = 𝜌𝑝 × �
𝐿𝑝
∆𝑝𝑚
𝑠𝑠𝑠𝑠
0.5 𝑝𝑝𝑝
30
� × � � = 1.934
×
×
𝑓𝑓 3 14.14 𝑝𝑝𝑝 1
∆𝑝𝑝
𝐿𝑚
𝜌𝑚 = 2.05
𝑠𝑠𝑠𝑠
𝑓𝑓 3
From Reynolds number similarity we have for the kinematic viscosity, where we use the relation for the
velocities in terms of lengths from the Froude number
𝑉𝑚
𝐿𝑚
𝐿𝑚
𝜈𝑚 = 𝜈𝑝 × � � × � � = � �
𝑉𝑝
𝐿𝑝
𝐿𝑝
3/2
= 1.059
𝑣𝑚 = 6.44 × 10−8
𝑓𝑓 2
𝑠
× 10−5
𝑓𝑓 2
1 3/2
� �
30
𝑠
With the density of the model fluid, we have the viscosity
𝜇𝑚 = 𝜈 𝜌 = 6.44 × 10−8
𝑓𝑓 2
𝑠𝑠𝑠𝑠
𝑙𝑙𝑙 ∙ 𝑠
× 2.05
= 1.32 × 10−7
3
𝑓𝑓
𝑓𝑓 2
𝑠
From the Weber number similarity we will get the surface tension:
�
𝜌𝜌 2 𝐿
𝜌𝜌 2 𝐿
� =�
�
𝜎 𝑚
𝜎 𝑝
Thus
2
𝜌𝑚
𝑉𝑚
𝐿𝑚
𝜌𝑚
𝐿𝑚
𝜎𝑚 = 𝜎𝑝 × � � × � � × � � = 𝜎𝑝 × � � × � �
𝜌𝑝
𝑉𝑝
𝐿𝑝
𝜌𝑝
𝐿𝑝
𝜎𝑚 == 0.00498
𝑙𝑙𝑙
1 2
𝑙𝑙𝑙
× � � = 5.86 × 10−6
𝑓𝑓
30
𝑓𝑓
2
Problem 7.55
Problem
7.70
[Difficulty: 3]
7.55
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
t
4
5
d
ρ
ρ
L
V
d
3
L
V
V
n = 5 parameters
μ
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



3
t
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



3
t
L 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
t 1
1 t
F t L
 L  1

  1
2
2 L L
t 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

 
d 1 V2
2 2
f1
f2


1
2
1
4
Problem 7.56
Problem
7.71
[Difficulty: 3]
7.56
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
m
3.23 kg
Am  0.1452 m
2

1
kg m
 s  



2
2
 75.0 m 
0.1452 m
N s
CDm  0.0970
Assuming standard conditions and equal viscosities:

3600 s 
1
 11
hr  8
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 
2
2 N s
FDp 
 0.0970  1.23
 ( 55  10)



 0.1452 m  8 

3 
2
hr
ft
3600 s
mi
kg m
m
FDp  468 N
Problem 7.57
Problem
7.72
[Difficulty: 2]
7.57
Given:
Flow around cruise ship smoke stack
Find:
Range of wind tunnel speeds
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 7.58
(Difficulty 3)
7.58 When a sphere of 0.25 𝑚𝑚 diameter and specific gravity 5.54 is dropped in water at 25 ℃ it will
attain a constant velocity of 0.07
𝑚
.
𝑠
What specific gravity must a 2.5 𝑚𝑚 sphere have so that when it is
dropped in crude oil (25 ℃) the two flows will be dynamically similar when the terminal velocity is
attained ?
Find: The specific gravity for the prototype 𝑆𝑆𝑝 .
Solution:
For the dynamic similarity with a falling sphere, we will have dynamic similarity if the Reynolds numbers
are equal:
𝜌𝜌𝜌
𝜌𝜌𝜌
� =�
�
�
𝜇 𝑝
𝜇 𝑚
The model falls in water and the properties are
𝜌𝑚 = 998
The properties of the sphere are
𝑘𝑘
𝑚3
𝜇𝑚 = 0.89 × 10−3 𝑃𝑃 ∙ 𝑠
𝑑𝑚 = 0.25 𝑚𝑚
And the terminal velocity is
𝑆𝑆𝑚 = 5.54
𝑉𝑚 = 0.07
𝑚
𝑠
𝜌𝑝 = 858
𝑘𝑘
𝑚3
The prototype falls through crude oil and the properties are
The prototype sphere diameter is
𝜇𝑝 = 6.8 × 10−3 𝑃𝑃 ∙ 𝑠
𝑑𝑝 = 2.5 𝑚𝑚
The prototype sphere velocity is then computed using the equality of the Reynolds numbers
𝑘𝑘
998 3
𝜇𝑝
𝜌𝑚
𝐿𝑚
𝑚
0.25 𝑚𝑚
6.8 × 10−3 𝑃𝑃 ∙ 𝑠
𝑚
𝑉𝑝 = 𝑉𝑚 × � � × � � × � � = 𝑉𝑚 = 0.07 × �
��
��
�
𝑘𝑘
𝑠
2.5 𝑚𝑚
0.89 × 10−3 𝑃𝑃 ∙ 𝑠
𝜌𝑝
𝐿𝑝
𝜇𝑚
858 3
𝑚
𝑉𝑝 = 0.062
𝑚
𝑠
At the terminal velocity condition, the buoyance force and drag force will balance the weight:
𝑊 − 𝐹𝐵 = 𝐷
We don’t know what the drag force is, but we know how the difference between the weight and the
buoyancy force for the model is related to the properties of the sphere and the water:
3
(𝑊 − 𝐹𝐵 )𝑚 ∝ (𝑆𝑆𝑚 𝜌𝑤𝑤𝑤𝑤𝑤 − 𝜌𝑤𝑤𝑤𝑤𝑤 )𝑔𝑑𝑚
And for the prototype, the same realtion is true:
(𝑊 − 𝐹𝐵 )𝑝 = �𝑆𝑆𝑝 𝜌𝑤𝑤𝑤𝑤𝑤 − 𝜌𝑜𝑜𝑜 �𝑔𝑑𝑝3
For dynamic similarity, the crag coefficients are equal
𝐷
𝐷
� 2 2 � = � 2 2�
𝜌𝑉 𝐿 𝑚
𝜌𝑉 𝐿 𝑝
Or, with the relation for the drag in terms of the sphere and fluid properties:
3
(𝑆𝑆𝑚 𝜌𝑤𝑤𝑤𝑤𝑤 − 𝜌𝑤𝑎𝑎𝑎𝑎 )𝑔𝑑𝑚
𝜌𝑚 𝑉𝑚2 𝑑𝑚
The specific gravity of the sphere is then:
𝑆𝑆𝑝 =
1
𝜌𝑤𝑤𝑤𝑤𝑤
=
�𝑆𝑆𝑝 𝜌𝑤𝑤𝑤𝑤𝑤 − 𝜌𝑜𝑜𝑜 �𝑔𝑑𝑝3
𝜌𝑝 𝑉𝑝2 𝑑𝑝
�𝜌𝑜𝑜𝑜 + (𝑆𝑆𝑚 𝜌𝑤𝑤𝑤𝑤𝑤 − 𝜌𝑤𝑤𝑤𝑤𝑤 )
Which yields the value for specific gravity as
𝑆𝑆𝑝 = 1.166
2 𝜌 𝑉2
𝑑𝑚
𝑝 𝑝
�
2
𝑑𝑝 𝜌𝑚 𝑉𝑚2
Problem 7.59
(Difficulty 2)
7.59 The flow about a 150 𝑚𝑚 artillery projectile which travels at 600
𝑚
𝑠
through still air at 30 ℃ and
absolute pressure 101.4 𝑘𝑘𝑘 is to be modeled in a high-speed wind tunnel with a 1: 6 model. If the wind
tunnel air has a temperature of −18 ℃ and absolute pressure of 68.9 𝑘𝑘𝑘, what velocity is required? If
the drag force on the model is 35 𝑁, what is the drag force on the prototype if skin friction may be
neglected?
Find: The velocity 𝑉𝑚 and the drag force 𝐷𝑝 .
For dynamic similarity for high speed flow, the Mach numbers are equal:
𝑉𝑚 𝑉𝑝
=
𝑐𝑚 𝑐𝑝
We calculate the sound speed from the relation for ideal gases.
𝑐 = √𝑘𝑘𝑘
For the model, the speed of sound is
𝑇𝑚 = −18 ℃
And for the prototype
𝑐𝑚 = �1.4 × (286.8) × (273 − 18)
𝑇𝑝 = 30 ℃
𝑐𝑝 = �1.4 × (286.8) × (273 + 30)
Under Mach number dynamic similarity, the model velocity is
𝑚
𝑚
= 320
𝑠
𝑠
𝑚
𝑚
= 349
𝑠
𝑠
𝑚
𝑚 320 𝑠
𝑐𝑚
= 600 ×
𝑉𝑚 = 𝑉𝑝
𝑠 349 𝑚
𝑐𝑝
𝑠
𝑉𝑚 = 550
For dynamic similarity, the drag coefficients are equal:
Or, the drag for the model is given by
𝑚
𝑠
𝐷
𝐷
� 2 2� = � 2 2�
𝜌𝑉 𝑑 𝑝
𝜌𝑉 𝑑 𝑚
2
𝑑𝑝
𝑉𝑝 2
𝜌𝑝
𝐷𝑝 = 𝐷𝑚 × � � × � � × � �
𝑑𝑚
𝑉𝑚
𝜌𝑚
The model is a 1:6 scale model so the model diameter is
𝑑𝑚 = 𝑑𝑝
1
1
= 150 𝑚𝑚 × = 25 𝑚𝑚
6
6
The densities of the air for the model an prototype depend on the pressure and temperature according to
the ideal gas law
𝜌𝑚 =
𝜌𝑝 =
𝜌=
𝑝
𝑅𝑅
68900
𝑘𝑘
𝑘𝑘
= 0.942 3
3
(−18
286.8 ×
+ 273) 𝑚
𝑚
101400
𝑘𝑘
𝑘𝑘
= 1.167 3
3
286.8 × (30 + 273) 𝑚
𝑚
Thus the drag force on the prototype is
Thus
𝑚 2
1.167
600
150 𝑚𝑚 2
𝑠� ×�
� �
𝐷𝑝 = 35𝑁 × �
𝑚
25 𝑚𝑚
550
0.942
𝑠
𝐷𝑝 = 1858 𝑁
𝑘𝑘
𝑚3 �
𝑘𝑘
𝑚3
Problem 7.60
Problem
7.76
[Difficulty: 4]
7.60
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
t
4
5
ρ
g
ρ
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
M
 

3
t
L 
L
a
b
L
c
0 0 0
  L  M L t
 2
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




3
2
L t
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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
2
2
1
3
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
a
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
Lp
Lp Vm
Lm Lp Vm
1
day
τp 



 τm

τp  9.6 hr 
 0.00900 
Vp
Lm Vp
5
Vm Lm Vp
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
7.78
Problem 7.61
[Difficulty: 3]
7.61
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:
Vm  24 knot 
Thus for 1:50 scale:
Vm
g  Lm

Lm
or: Vm  Vp 
Lp
g  Lp
Vp
1
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

ρp  Vp  Lp

ρp Lp μm
or: Vm  Vp 


ρm Lm μp
μp
 3 N s
at 20oC. Thus for 1:50 scale:
2
m
3
3
FDm
1

2
ρ V A
2 m m m
For surface travel:
FDp
FDm
For submerged travel:

2
FDp
1
2
 ρ  V  Ap
2 p p
2
 Vp  Ap ρp  Vp Lp 





 
 Substituting in known values:
FDm ρm Vm
Am
ρm Vm Lm
 


FDp
1.025
0.998
FDp
m
Vm  9.99
s
Vm  19.41  knot or
Under dynamically similar conditions, the drag coefficients will match:
Solving for the ratio of forces:
m
Vm  1.75
s
Vm  3.39 knot or
ρp
2

 24  50   1.29  105


 3.39 1 
2
0.35
50 

 

  0.835
FDm 0.998  19.41
1 
1.025
FDp
FDm
 1.29  10
FDp
FDm
5
 0.835
(on surface)
(submerged)
Problem 7.62
Problem
7.80
[Difficulty: 3]
7.62
Given:
Find:
Solution:
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
 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 
2
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
7.82
Problem 7.63
[Difficulty: 3]
7.63
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
m
CD  2  350  N 
1.21 kg

2
 s   1  N s


2
kg m
 75 m 
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
km 1000 m
1  hr 
N s
2

Fp 
 1.21
  90


  10 m  1.028 
3 
2
hr
1  km
3600 s 
kg m
m
1
For dynamic similarity
kg
ρm Vm Lm
μm

ρp  Vp  Lp
km 1000 m
1  hr
10
Vm  90



hr
1  km
3600 s
1
c 
Hence we have
M
1.40  286.9 
Vm
c

250
343
c
N m
kg K
 0.729
Fp  3.89 kN
ρp Lp μm
Lp
Vm  Vp 


 Vp 
ρm Lm μp
Lm
μp
For air at standard conditions, the speed of sound is
2
Ap  10 m
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
7.84
Problem 7.64
[Difficulty: 4]
7.64
Power to drive a fan is a function of ρ, Q, D, and ω.
Given:
3
P
2
Select primary dimensions M, L, t:
3
P
4
5
Q
ρ
Q
M
L
3
t
2
L
D
D
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
M
 3
L 
a  1
L:
2  3 a  b  0
t:
3  c  0
b

a
 L  
b
1
c
0 0 0
  M L t
t
The solution to this system is:
M: 1  a  0
a
2
3
t
Summing exponents:
c
Π2  Q ρ  D  ω
Thus:
Summing exponents:
L
3
t

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 


 15 1800 
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

1 3
1
F L L
L

 t  1
 t
1
2 5
t
3
t
L
F t L
3
P
c
The solution to this system is:
M: a  0
6
ω2  1800 rpm
We will use the Buckingham pi-theorem.
1
t
ft
Condition 2: Q2  88
s
Fan diameter for condition 2 to insure dynamic similarity
Find:
Solution:
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 7.65
Problem
7.85 (In Excel)
[Difficulty: 3]
7.65
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 7.66
Problem
7.86
[Difficulty: 2]
7.66
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.
n = 5 parameters
1
Δp
2
Select primary dimensions M, L, t:
3
Δp
Q
M
L
L t
4
5
ρ
ρ
Q
3
t
2
ω
ω
ρ
ω
M
1
3
t
L
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  Δp ρ  ω  D
t:
2  b  0
a
b
c
Π2  Q ρ  ω  D
a  1
6
t:
1  b  0
1
b
Check using F, L, t dimensions:
3
b  2
a
c  2
Π1 
Δp
2
2
ρ ω  D
b
The solution to this system is:
M: a  0
3  3 a  c  0
 
1
M
c
0 0 0
       L  M  L  t


t
3
t
L   
L
Thus:
Summing exponents:
L:
a
The solution to this system is:
M: 1  a  0
1  3  a  c  0
M
c
0 0 0
L  M L t



2
3
t
L t  L   
Summing exponents:
L:

M
Thus:
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:
Δp
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
Δpp

ρm ωm  Dm
3
m
183
Qp  0.0928


min 367
 ωm Dm 
Δpm  Δpp 



ρp ωp Dp


ρm
Δpm
2
ρp  ωp  Dp
 150 


 50 
3
3
m
Qp  1.249 
min
2
Δpm  52.5 kPa 
2
999
800

 367  50 


 183 150 
2
Δpm  29.3 kPa
Problem 7.67
Problem
7.89
[Difficulty: 3]
7.67
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
ωp  Dp
3
hm
then
3
2
3
ωm
3
hp

2
ωp  Dp
Pm
and
2
3
5
ωm  Dm

Pp
3
ωp  Dp
5
3
(1)
then
2
2
2
Dm
1000  Dm




  2  4 2
hp
2
2
500 

ωp Dp
Dp
Dp
and
Pm
We can find Pp from
kg
m
J
Pp  ρ Q h  1000
 0.75
 15
 11.25 kW
3
s
kg
m
hm
2
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

P
and
2
Dm
ωm
3
(2)
5
5
5
3
Dm
1000  Dm




  5  8 5
Pp
3
5
500 

ωp Dp
Dp
Dp
Dm
ωm
(3)
3
1
From Eq 3
From Eq 1
From Eq 2
Pm
Pp
5
 8
Dm
Dp
5
 Dm 
 2 

Qp
 Dp 
Qm
 Dm 
 4 

hp
 Dp 
hm
so
 1 Pm 
Dm  Dp   

 8 Pp 
so
 Dm 
Qm  Qp  2  

 Dp 
so
 Dm 
hm  hp 4 

 Dp 
3
2
1
5
3
Dm  0.25 m 
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 7.68
Problem
7.90
[Difficulty: 3]
7.68
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
[Difficulty: 3]
Problem 7.69
Problem
7.91
7.69
7.38.
7.5
For a marine propeller (Problem 7.38) the thrust force is: FT
Given:
 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
5
ρ
F L
t
ω
D
ρ
T
F L
F
4
T
D
ρ
F t
L
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
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
 F t2   1  b c 0 0 0

F 
L  F L t
 4   t 
L


The solution to this system is:
a  1
Thus:
b  2
c  2
Π1 
FT
2
4
ρ ω  D
a
 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

F L 
L  F L t
 4   t 
L


Thus:
Summing exponents:
F:
1a0
L:
1  4 a  c  0
t:
2 a  b  0
a
b
c
Π4  V ρ  ω  D
The solution to this system is:
a  1
a0
L:
1  4 a  c  0
t:
1  2  a  b  0
a
c
Π5  g  ρ  ω  D
Thus:
The solution to this system is:
 F t2 


t  4 
L 
L
Thus:
Summing exponents:
6
F:
a0
L:
1  4 a  c  0
t:
1  2  a  b  0
Check using M, L, t dimensions:
2
5
ρ ω  D
a
a0
b
c  5
T
 F t2   1  b c 0 0 0
   L  F L t

t  4  t
L 
L
Summing exponents:
F:
b  2
Π3 
b  1
a
 
1
t
V
Π4 
ω D
c  1
b
c
0 0 0
 L  F L t

The solution to this system is:
a0
3
b  1
M L L 2 1
 t 
1
2 M
4
t
L
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
t
L
Based on the dependent and independent variables, the "scaling laws" are:
FT
g 
V
 f1 


ω D 2 
2 4
ω D 
ρ ω  D

P
g 
V
 f2 


ω D 2 
3 5
ρ ω  D
ω D 

T
g 
V
 f3 


ω D 2 
2 5
ρ ω  D
ω D 

Problem 7.70
Problem
7.93
[Difficulty: 2]
7.70
267.
Given:
Find:
Solution:
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
1
hp s
lbf  s
KEratio 
 0.00238 
 ( 40 ft  80 ft)   507   


3
s
125000 hp 550  ft lbf
2
slug ft

ft
KEratio  7.22
Problem 7.71
Problem
7.96
[Difficulty: 3]
7.71
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 
3
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
Substituting in values:
2
 ρ V  Am
2
2
s 
1
1000 mm 
kg m
CDm  2  5.82 N 
 

 



1.23 kg  26.5 m 
2 
2
m

30400  mm
N s
m
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
1
kg 
km 1000 m
hr
  7.782  m2  0.443  N s
FDp 
 1.23
  100 



3 
2
hr
km 3600  s 
kg m
m
FDp  1.636  kN
(The rolling resistance must also be included to obtain the total tractive effort needed to propel the vehicle.)
Problem 7.72
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 




2 π 
ρ λ
 λ 
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
Problem 9.1
9.1
[Difficulty: 2]
9.1
Given:
Minivan traveling at various speeds
Find:
Plot of boundary layer length as function of speed
Solution:
Governing equations:
The critical Reynolds number for transition to turbulence is
VL crit/ =500000
Re crit =
The critical length is then
L crit = 500000/V 
Tabulated or graphical data:
=
3.79E-07
=
0.00234
lbf.s/ft
2
3
slug/ft
(Table A.9, 68oF)
Computed results:
V (mph)
L crit (ft)
10
13
15
18
20
30
40
50
60
70
80
90
5.52
4.42
3.68
3.16
2.76
1.84
1.38
1.10
0.920
0.789
0.690
0.614
Length of Laminar Boundary Layer
on the Roof of a Minivan
6
5
4
L crit (ft)
3
2
1
0
0
10
20
30
40
50
V (mph)
60
70
80
90
100
Problem 9.2
Problem
9.2
[Difficulty: 2]
9.2
Given:
Model of riverboat
Find:
Distance at which transition occurs
Solution:
Basic equation
For water at 10oC
Hence
For the model
Rex 
ρ U x
μ

ν  1.30  10
xp 
xm 
ν Rex
U
xp
18
U x
2
6 m

5
and transition occurs at about
Rex  5  10
(Table A.8)
and we are given
ν
s
x p  0.186 m
x p  18.6 cm
x m  0.0103 m
x m  10.3 mm
U  3.5
m
s
Problem 9.3
(Difficulty 2)
9.3 For flow over a smooth plate, what approximately is the maximum length of the laminar boundary
layer if 𝑉0 = 9.0
𝑚
𝑠
in the irrotational uniform flow and the fluid is air? Water?
Find: The maximum length 𝐿 for laminar boundary layer.
Solution: Use the critical Reynolds number to determine the maximum length.
The Reynolds number is defined as:
𝑅𝑅 =
𝑉0 𝐿
𝑣
For the flow over a smooth plate, the transition is considered to occur at:
For air, from the figure in A-3 we have:
𝑅𝑅 = 500000
𝑣𝑎𝑎𝑎 = 1.461 × 10−5
So the maximum length can be calculated as:
𝑚2
𝑠
𝑅𝑅 ∙ 𝑣𝑎𝑎𝑎 500000 × 1.461 × 10
𝐿=
=
𝑚
𝑉0
9.0
𝑠
−5
𝑚2
𝑠 = 0.812 𝑚
For water, from the figure in A-3 we have:
𝑣𝑤𝑤𝑤𝑤𝑤 = 1.0 × 10−6
So the maximum length can be calculated as:
𝑚2
𝑠
𝑅𝑅 ∙ 𝑣𝑤𝑤𝑤𝑤𝑤 500000 × 1.0 × 10
𝐿=
=
𝑚
𝑉0
9.0
𝑠
−6
𝑚2
𝑠 = 0.056 𝑚
Problem 9.4
(Difficulty 1)
9.4 A model of a thin streamlined body is placed in a flow for testing. The body is 0.9 𝑚 long and the
flow velocity is 0.6
𝑚
.
𝑠
What 𝑣 is needed to ensure that the boundary layer on the body is laminar?
Find: The viscosity 𝑣 to ensure the boundary layer is laminar.
Assumption: The flow is steady
Solution: Use the critical Reynolds number to determine the viscosity.
The Reynolds number is defined as:
𝑅𝑅 =
𝑉0 𝐿
𝑣
For the flow over a smooth plate, the transition from laminar to turbulent is considered to occur at:
𝑅𝑅 = 500000
For this particular case, to ensure the boundary layer is laminar, we should have the viscosity as:
𝑚
𝑚2
𝑉0 𝐿 0.6 𝑠 × 0.9 𝑚
=
= 1.08 × 10−6
𝑣=
𝑅𝑅
500000
𝑠
Problem 9.5
Problem
9.4
[Difficulty: 2]
9.5
Given:
Experiment with 1 cm diameter sphere in SAE 10 oil
Find:
Reasonableness of two flow extremes
Solution:
Basic equation
ReD 
ρ U D
μ

U D
ν  1.1  10
For
ReD  1
For
ReD  2.5  10
Note that for
ReD  2.5  10
For water
ν  1.01  10
For
ReD  2.5  10
ReD  2.5  10
(Fig. A.3 at 20 oC)
and
ν
2
4 m
For SAE 10

s
we find
5
ν ReD
D
ν ReD
D
D  1  cm
U  0.011 
U  2750
m
s
m
s
U  1.10
cm
s
which is reasonable
which is much too high!
we need to increase the sphere diameter D by a factor of about 1000, or reduce the
viscosity ν by the same factor, or some combination of these. One possible solution is
2
6 m

5
U 
U 
5
5
and transition occurs at about
s
(Table A.8 at 20 oC)
we find
U 
D  10 cm
and
ν ReD
D
Hence one solution is to use a 10 cm diameter sphere in a water tank.
U  2.52
m
s
which is reasonable
Problem 9.6
Problem
9.6
[Difficulty: 2]
9.6
Given:
Sheet of plywood attached to the roof of a car
Find:
Speed at which boundary layer becomes turbulent; Speed at which 90% is turbulent
Solution:
Rex 
Basic equation
ρ U x
μ

ν  1.50  10
For air
U x
Rex  5  10
and transition occurs at about
ν
2
5 m

5
(Table A.10)
s
Now if we assume that we orient the plywood such that the longer dimension is parallel to the motion of the car, we can say:
Hence
U 
ν Rex
x
U  3.8
m
s
When 90% of the boundary layer is turbulent
x  0.1  2  m
Hence
U 
ν Rex
x
U  37.5
m
s
x  2 m
U  13.50 
km
U  135.0 
km
hr
hr
Problem 9.7
Problem
9.8
9.7
Given:
Aircraft or missile at various altitudes
Find:
Plot of boundary layer length as function of altitude
Solution:
Governing equations:
The critical Reynolds number for transition to turbulence is
Re crit =
UL crit/ = 500000
The critical length is then
L crit = 500000/U 
Let L 0 be the length at sea level (density 0 and viscosity 0). Then
L crit/L 0 = (/0)/(/0)
The viscosity of air increases with temperature so generally decreases with elevation;
the density also decreases with elevation, but much more rapidly.
Hence we expect that the length ratio increases with elevation
For the density , we use data from Table A.3.
For the viscosity , we use the Sutherland correlation (Eq. A.1)
 = bT 1/2/(1+S /T )
b =
S =
1.46E-06
110.4
kg/m.s.K1/2
K
[Difficulty: 2]
Computed results:
z (km)
T (K)
/0
/0
L crit/L 0
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
6.0
288.2
284.9
281.7
278.4
275.2
271.9
268.7
265.4
262.2
258.9
255.7
249.2
1.0000
0.9529
0.9075
0.8638
0.8217
0.7812
0.7423
0.7048
0.6689
0.6343
0.6012
0.5389
1.000
0.991
0.982
0.973
0.965
0.955
0.947
0.937
0.928
0.919
0.910
0.891
1.000
1.04
1.08
1.13
1.17
1.22
1.28
1.33
1.39
1.45
1.51
1.65
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
242.7
236.2
229.7
223.3
216.8
216.7
216.7
216.7
216.7
216.7
216.7
216.7
216.7
216.7
218.6
220.6
222.5
224.5
226.5
0.4817
0.4292
0.3813
0.3376
0.2978
0.2546
0.2176
0.1860
0.1590
0.1359
0.1162
0.0993
0.0849
0.0726
0.0527
0.0383
0.0280
0.0205
0.0150
0.872
0.853
0.834
0.815
0.795
0.795
0.795
0.795
0.795
0.795
0.795
0.795
0.795
0.795
0.800
0.806
0.812
0.818
0.824
1.81
1.99
2.19
2.41
2.67
3.12
3.65
4.27
5.00
5.85
6.84
8.00
9.36
10.9
15.2
21.0
29.0
40.0
54.8
Length of Laminar Boundary Layer
versus Elevation
60
50
40
L/L 0
30
20
10
0
0
10
20
z (m)
30
Problem 9.8
Problem
9.10
[Difficulty: 2]
9.8
Given:
Find:
Linear, sinusoidal, and parabolic velocity profiles
Solution:
Here are the profiles:
Plots of y/δ vs u/U for all three profiles
Laminar Boundary Layer Velocity Profiles
Linear
Sinusoidal
Parabolic
Dimensionless Distance y/δ
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
Dimensionless Velocity u/U
0.8
Problem 9.9
Problem
9.12
[Difficulty: 2]
9.9
Given:
Laminar boundary layer profile
Find:
If it satisfies BC’s; Evaluate */ and /
Solution:
3
4
The boundary layer equation is
u
y
 y  y
 2  2     for which u = U at y = 
U

   
The BC’s are
u 0  0


0
y 
u
3
4
 20  20  0  0
U
 1
 1
2
3 
du
y2
y3 
 U  2  6 3  4 4 
 U  2  6 3  4 4   0

  y 

 
dy
 
 
At y = 0
At y = 
du
dy



0

u
u

dy   1  dy
U
U
0
For *:
 *   1 
Then
1
1
* 1 u
u   y
u


  1  dy   1  d     1  d
  0 U
U    0  U 
0
with
u
 2  2 3   4
U
*
u
1
1 
3


  1  d   1  2  2 3   4 d     2   4   5  
 0 .3

U
2
5
10




0
0
0
1
Hence
1
1


For :
u
u
u
u
   1  dy   1  dy
U U
U U
0
0
Then
1
1
 1 u u
u
u   y
u
u
  1  dy   1  d     1  d
  0U  U
U  U    0 U  U 
0
Hence

u
u
  1  d   2   3   4 1  2   3   4 d   2  4 2  2 3  9 4  4 5  4 6  4 7   8 d
 0U  U 
0
0
1
1
1
  2 4 3 1 4 9 5 4 7 1 8 1 9
37
               
 0.117
 
3
2
5
7
2
9  0 315
1
Problem
9.15
Problem 9.10
[Difficulty: 2]
9.10
9.8
Given:
Find:
Solution:
Linear, sinusoidal, and parabolic velocity profiles
The momentum thickness expressed as θ/δ for each profile
We will apply the definition of the momentum thickness to each profile.

θ 


Governing
Equation:
δ
u
U
  1 
0

u
 dy
(Definition of momentum thickness)
U
1 
 
δ
δ 

δ
θ
If we divide both sides of the equation by δ, we get:
u
U
0
the variable of integration to η = y/δ, resulting in:
For the linear profile:
u
U
dη 
1
δ
 dy
1

u
For the sinusoidal profile:
U

 sin
π
2
 η
0
1
0
θ
δ




δ 

u
U

2
π

0

θ
δ
1

2

2
1
3

u
U
  1 

u
 dη
U
1
θ
6
δ
 0.1667
2
 π
π
sin  η   sin  η   dη
  2    2  
2 π

π 4
 2 η  η
1
θ
Into the momentum thickness:


1

π 
π 
θ 




sin  η   1  sin  η  dη  
δ 

2  
 2 


1
 dy However, we can change
U
Therefore:
1


2
  η ( 1  η) dη   η  η dη Evaluating this integral:

δ 0
0
For the parabolic profile:

u
 η Into the momentum thickness:
θ
Evaluating this integral:
  1 
θ
δ
 0.1366
Into the momentum thickness:
1





2
2
2
3
4
  2  η  η  1  2  η  η  dη   2  η  5  η  4  η  η dη

δ 0
0
θ
Evaluating this integral:
θ
δ
1
5
3
1
1
5

2
θ
15
δ
 0.1333
Problem 9.11
(Difficulty 2)
9.11 Evaluate the displacement thickness 𝛿 ∗ and the momentum thickness 𝜃 for a velocity profile
given by
𝑢
𝑈
𝑦
𝛿
= . Plot the non-dimensional velocity profile and show the thickness
𝛿∗
𝛿
𝜃
𝛿
and on the
plot. Does this expression satisfy the boundary conditions applicable to a laminar boundary layer?
Assumption: The flow is steady
Solution: Use the definitions of displacement and momentum thickness.
The linear velocity profile is given by:
𝑢 𝑦
=
𝑈 𝛿
Using the definition of the displacement thickness, we have for this profile:
𝛿
𝛿
𝑢
𝑦
1
𝛿 ∗ = � �1 − � 𝑑𝑑 = � �1 − � 𝑑𝑑 = 𝛿
𝑈
𝛿
2
0
0
And for the momentum thickness for this profile we have:
𝛿
𝜃=�
0
𝛿
𝑢
𝑢
𝑦
𝑦
1
�1 − � 𝑑𝑑 = � �1 − � 𝑑𝑑 = 𝛿
𝑈
𝑈
𝛿
6
0 𝛿
The plot of the non-dimensional velocity profile is shown as:
1
0.9
Dimensionless u/U
Thickness δ*/δ
0.8
Thickness θ/δ
0.7
y/ δ
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
u/U, δ*/δ and θ*/δ
0.7
0.8
0.9
1
As we know for this linear velocity approximation:
𝛽=
𝛿∗ 1
=
2
𝛿
𝜃 1
= = 0.167
𝛿 6
For a laminar boundary layer, the exact results for these parameters are:
𝛿∗
= 0.344
𝛿
𝛽=
𝜃
= 0.133
𝛿
This expression satisfies the boundary conditions applicable to a laminar boundary layer. However,
the displacement and momentum thicknesses are not too close to the exact values.
Problem 9.12
(Difficulty 2)
9.12 Evaluate the displacement thickness 𝛿 ∗ and the momentum thickness 𝜃 for a power law
velocity profile given by
𝛿∗
𝛿
𝜃
𝛿
𝑢
𝑈
=
1
𝑦 7
� �.
𝛿
Plot the non-dimensional velocity profile and show the thickness
and on the plot. Does this expression satisfy the boundary conditions applicable to a laminar
boundary layer?
Assumption: The flow is steady
Solution: Use the definitions of displacement and momentum thickness.
We have the one-seventh power profile as:
1
𝑦 7
𝑢
=� �
𝛿
𝑈
Using the definition of the displacement thickness we have:
𝛿∗
1
𝛿
𝛿
𝑢
𝑦 7
1
= � �1 − � 𝑑𝑑 = � �1 − � � � 𝑑𝑑 = 𝛿
𝑈
𝛿
8
0
0
For the momentum thickness we have:
𝛿
1
1
𝛿
𝑢
𝑢
𝑦 7
𝑦 7
7
𝜃=�
�1 − � 𝑑𝑑 = � � � �1 − � � � 𝑑𝑑 =
𝛿
𝑈
𝛿
72
0 𝑈
0 𝛿
The plot of the non-dimensional velocity profile is shown as:
1
0.9
Dimensionless u/U
Thickness δ*/δ
0.8
Thickness θ/δ
0.7
y/ δ
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
u/U, δ*/δ and θ*/δ
0.7
0.8
0.9
1
For the one-seventh power velocity profile approximation:
𝛿∗ 1
= = 0.125
8
𝛿
𝛽=
7
𝜃
=
= 0.0972
𝛿 72
The exact results for these for a laminar boundary layer are:
𝛿∗
= 0.344
𝛿
𝛽=
𝜃
= 0.133
𝛿
So this expression does not satisfy the conditions applicable to a laminar boundary layer. This power
law actually models the turbulent boundary layer.
Problem
9.18
Problem 9.13
[Difficulty: 3]
9.13
Given:
Data on fluid and boundary layer geometry
Find:
Mass flow rate across ab; Drag
CV
Solution:
The given data is
ρ  1.5
slug
ft
Governing
equations:
U  10
3
ft
s
d
L  10 ft δ  1  in b  3  ft
c
Rx
Mass
Momentum
Assumptions:
(1) Steady flow (2) No pressure force (3) No body force in x direction (4) Uniform flow at a
Applying these to the CV abcd
δ
Mass

( ρ U b  δ)   ρ u  b dy  mab  0

0
For the boundary layer
u
U

y
δ
dy
η
δ
 dη
1
Hence

1
mab  ρ U b  δ   ρ U η δ dy  ρ U b  δ   ρ U b  δ

2
0
1
mab   ρ U b  δ
2
slug
mab  1.875 
s
δ
Momentum

Rx  U ( ρ U δ)  mab u ab   u  ρ u  b dy

0
u ab  U
Note that
and
1
δ


2
2
 u  ρ u  b dy   ρ U  b  δ η dη


0
2
Rx  ρ U  b  δ 
0
1

2
2
 ρ U b  δ U   ρ U  b  δ η dy

2
1
0
2
Rx  ρ U  b  δ 
1
2
2
 ρ U  δ 
1
3
2
 ρ U  δ
1
2
Rx    ρ U  b  δ
6
Rx  6.25 lbf
We are able to compute the boundary layer drag even though we do not know the viscosity because it is the viscosity
that creates the boundary layer in the first place
Problem 9.14
Problem
9.20
9.14
[Difficulty: 3]
9.13
9.8.
δ = 1 in
Flow over a flat plate with parabolic laminar boundary layer profile
Given:
Find:
(a) Mass flow rate across ab
(b) x component (and direction) of force needed to hold the plate in place
We will apply the continuity and x-momentum equations to this system.
Solution:
Governing
 

(Continuity)

d
V


V
 dA  0
Equations:


CV
CS
t
 

(x- Momentum)
udV   uV  dA  Fsx  Fbx

CS
t CV
(1) Steady flow
Assumptions:
(2) No net pressure forces
(3) No body forces in the x-direction
(4) Uniform flow at da
CV
d

ρ U b  δ   ρ u  b dy  mab  0

From the assumptions, the continuity equation becomes:
c
Rx
δ
The integral can be written as:
0
δ
δ



 ρ u  b dy  ρ b   u dy  ρ U b  δ 



0
0
1
2η  η2 dη
where η 
0
y
δ
This integral is equal to: ρ U b  δ  1 

1
2
   ρ U b  δ
3 3
2
1
mab  ρ U b  δ   ρ U b  δ   ρ U b  δ Substituting known values:
3
3
Solving continuity for the mass flux through ab we get:
1
slug
ft
ft
mab 
 1.5
 10  3.0 ft  1  in 
3
s
12 in
3
ft
slug
mab  1.250 
s
δ
From the assumptions, the momentum equation becomes:

Rx  u da ( ρ U b  δ)  u ab mab   u  ρ u  b dy where u da  u ab  U

0
1
2
2
Thus: Rx  ρ U  b  δ   ρ U  b  δ 
3
δ

2
2
 u  ρ u  b dy    ρ U  b  δ 

3
0

δ
δ
 2


2
 u  ρ u  b dy  ρ b   u dy  ρ U  b  δ 



0
0
ρ U  b  δ 
2
4
3
1
1
 2 η  η 
2
0
2
δ

 u  ρ u  b dy The integral can be written as:

0

dη  ρ U  b  δ 

2
1
4η2  4η3  η4 dη
This integral is equal to:
0
1
8
2
2
2
 8 2
   ρ U b δ Therefore the force on the plate is: Rx      ρ U  b  δ    ρ U  b δ
5  15
15
 15 3 
Substituting known values:
Rx  
2
15
 1.5
slug
ft
3
  10

This force must be applied to the control volume by the plate.
ft 

s
2
 3.0 ft  1  in 
ft
12 in
2

lbf  s
slug ft
Rx  5.00 lbf
(to the left)
Problem 9.15
Problem
9.22
[Difficulty: 2]
9.15
Given:
Data on boundary layer in a cylindrical duct
Find:
Velocity U2 in the inviscid core at location 2; Pressure drop
Solution:
The solution involves using mass conservation in the inviscid core, allowing for the fact that as the boundary layer grows it
reduces the size of the core. One approach would be to integrate the 1/7 law velocity profile to compute the mass flow in the
boundary layer; an easier approach is to simply use the displacement thickness!
The given or available data (from Appendix A) is
ρ  1.23
kg
m
U1  12.5
s
3
m
D  100  mm
δ1  5.25 mm
δ2  24 mm
Governing
equations: Mass
p
Bernoulli
ρ
2

V
2
 g  z  constant
(4.24)
The displacement thicknesses can be computed from boundary layer thicknesses using Eq. 9.1
1




u


δdisp 
1   dy  δ 


 
U
0

1


δ
7
 1  η  dη 
δ
8
0
Hence at locations 1 and 2
δ1
δdisp1 
8
Applying mass conservation at locations 1 and 2
δdisp1  0.656  mm
δ2
δdisp2 
8
δdisp2  3  mm
ρ U1 A1  ρ U2 A2  0
A1
U2  U1 
A2
or
The two areas are given by the duct cross section area minus the displacement boundary layer


π
2
A1   D  2  δdisp1
4
Hence
A1  7.65  10
3
2
m
A1
U2  U1 
A2
For the pressure drop we can apply Bernoulli to locations 1 and 2 to find


π
2
A2   D  2  δdisp2
4
A2  6.94  10
m
U2  13.8
s
ρ
2
2
p 1  p 2  Δp    U2  U1  Δp  20.6 Pa


2
3 2
m
Problem 9.16
(Difficulty 2)
9.16 Evaluate the displacement thickness 𝛿 ∗ and the momentum thickness 𝜃 for a velocity profile
given by
𝜃
𝛿
𝑢
𝑈
𝑦 2
𝛿
𝑦
𝛿
= 2 � � − � � . Plot the non-dimensional velocity profile and show the thickness
𝛿∗
𝛿
and
on the plot. Does this expansion satisfy the boundary conditions applicable to a laminar boundary
layer?
Assumption: The flow is steady
Solution: Use the definitions of displacement and momentum thickness.
We have the parabolic velocity profile as:
𝑢
𝑦
𝑦 2
= 2� � − � �
𝑈
𝛿
𝛿
Using the definition of the displacement thickness we have:
𝛿∗
𝛿
𝛿
𝑢
𝑦
𝑦 2
1
= � �1 − � 𝑑𝑑 = � �1 − 2 � � + � � � 𝑑𝑑 = 𝛿
𝑈
𝛿
𝛿
3
0
0
For the momentum thickness we have:
𝛿
𝜃=�
0
𝛿
𝑢
𝑢
𝑦
𝑦 2
𝑦
𝑦 2
2
�1 − � 𝑑𝑑 = � �2 � � − � � � �1 − 2 � � + � � � 𝑑𝑑 =
𝛿
𝑈
𝑈
𝛿
𝛿
𝛿
𝛿
15
0
The plot of the non-dimensional velocity profile is shown as:
1
0.9
Dimensionless u/U
Thickness δ*/δ
0.8
Thickness θ/δ
0.7
y/ δ
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
u/U, δ*/δ and θ*/δ
0.7
0.8
0.9
1
For the parabolic velocity profile approximation we have:
𝛿∗ 1
= = 0.333
3
𝛿
𝛽=
2
𝜃
=
= 0.133
𝛿 15
The exact results for the laminar boundary layer profile are:
𝛿∗
= 0.344
𝛿
𝛽=
𝜃
= 0.133
𝛿
So this expression satisfies the boundary conditions applicable to a laminar boundary layer.
exact velocity profile for a laminar boundary is closely approximated by the parabolic profile.
The
Problem 9.17
(Difficulty 2)
9.17 Evaluate the displacement thickness 𝛿 ∗ and the momentum thickness 𝜃 for a velocity profile
given by
𝑢
𝑈
𝜋𝜋
2𝛿
= sin � �. Plot the non-dimensional velocity profile and show the thickness
𝛿∗
𝛿
𝜃
𝛿
and on
the plot. Does this expression satisfy the boundary conditions applicable to a laminar boundary
layer?
Assumption: The flow is steady
Solution: Use the definitions of displacement and momentum thickness.
We have the sinusoidal velocity profile as:
𝜋𝜋
𝑢
= sin � �
2𝛿
𝑈
Using the definition of the displacement thickness we have:
𝛿
𝛿
𝑢
𝜋𝜋
2
𝛿 ∗ = � �1 − � 𝑑𝑑 = � �1 − sin � �� 𝑑𝑑 = �1 − � 𝛿
𝑈
2𝛿
𝜋
0
0
For the momentum thickness we have:
𝛿
𝛿
𝑢
𝑢
𝜋𝜋
𝜋𝜋
4−𝜋
𝜃=�
�1 − � 𝑑𝑑 = � sin � � �1 − sin � �� 𝑑𝑑 = �
�𝛿
𝑈
2𝛿
2𝛿
2𝜋
0 𝑈
0
The plot of the non-dimensional velocity profile is shown as:
1
0.9
Dimensionless u/U
Thickness δ*/δ
0.8
Thickness θ/δ
0.7
y/ δ
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
u/U, δ*/δ and θ*/δ
0.7
0.8
0.9
1
For this sinusoidal velocity profile approximation:
𝛿∗
2
= 1 − = 0.363
𝜋
𝛿
𝛽=
𝜃 4−𝜋
=
= 0.1366
𝛿
2𝜋
The exact results for the laminar boundary layer are:
𝛿∗
= 0.344
𝛿
𝛽=
𝜃
= 0.133
𝛿
So this expression satisfies the boundary conditions applicable to a laminar boundary layer.
values are quite close to the exact and the sinusoidal profile is a good approximation.
The
Problem 9.19
Problem
9.25
[Difficulty: 2]
9.19
Given:
Data on wind tunnel and boundary layers
Find:
Pressure change between points 1 and 2
Solution:
Basic
equations
(4.12)
p
ρ
2

V
2
 g  z  const
Assumptions: 1) Steady flow 2) Incompressible 3) No friction outside boundary layer 4) Flow along streamline 5) Horizontal
For this flow
ρ U A  const
The given data is
ft
U0  100 
s
We also have
δdisp2  0.035  in
Hence at the Point 2
A2  h  2  δdisp2
U1  U0

h  3  in
2

p1
ρ
Hence
Δp 
  U  U2
2  1
ρ
2
2
A1  9  in
2
 
The pressure change is found from Bernoulli
2
A2  8.58 in
Applying mass conservation between Points 1
and 2
ρ U1  A1  ρ U2  A2  0

A1  h

o
r
U1
2
2
2

The pressure drops by a small amount as the air accelerates

p2
ρ

U2
2
A1
U2  U1 
A2
ft
U2  105 
s
wit
h
ρ  0.00234 
2
slug
ft
3
Δp  8.05  10
 psi
Δp  1.16
lbf
ft
2
3
Problem9.26
9.20
Problem
[Difficulty: 3]
9.20
Given:
Find:
Developing flow of air in flat horizontal duct. Assume 1/7-power law velocity profile in boundary layer.
Solution:
We will apply the continuity and x-momentum equations to this problem.
(a) Displacement thickness is 1/8 times boundary layer thickness
(b) Static gage pressure at section 1.
(c) Average wall shear stress between entrance and section 2.
Governing
Equations:

δdisp  


infinity
δ
 1  u  dy  




U



0
0
 
dV   V  dA  0
 1  u  dy


U

(Definition of displacement thickness)

CS
t CV
 


ud
V

u

V
CS  dA  Fsx  Fbx
t CV
Assumptions:
(Continuity)
(x- Momentum)
(1) Steady, incompressible flow
(2) No body forces in the x-direction
(3) No viscous forces outside boundary layer
(4) Boundary layers only grow on horizontal walls
L = 20 ft
H = 1 ft
V1 = 40 ft/s
δ 2 = 4 in
δdisp
If we divide both sides of the displacement thickness definition by δ, we get:
δ
1 
 
δ 

δ
0
However, we can change the variable of integration to η = y/δ, resulting in:
dη 
1
δ
 dy
 1  u  dy


U

Therefore:
δdisp
δ




1
0
1
1
For the power law profile:
u
U
η
7
Into the displacement thickness:


δdisp 


δ
0
Evaluating this integral:
δdisp
δ
 1  u  dη


U

1


7
 1  η  dη
1
7
8

1
δdisp
8
δ

1
8
V1  A1  V2  A2 or
After applying the assumptions from above, continuity reduces to:
Solving for the velocity at 2:
H
V2  V1 
 V1 
H  2  δdisp2
H
H
p0
ρ
1
2
p 1g  p 1  p 0    ρ V1
2
p 1g  
1
2
p 2g  p 2  p 0    ρ V2
2
p 2g  
1
2
1
2

Substituting known values:
4
2
p
V

ρ
 0.00234 
2
slug
ft
 0.00234 
3
slug
ft
3

δ2
ft
V2  40  1  ft 
s
From Bernoulli equation, since z = constant:

V1  w H  V2  w H  2  δdisp2
1

 1

1
4  ft
1   
4 12 

ft
V2  43.6
s
along a streamline. Therefore:
  40

  43.6

2
2
lbf  s
ft 
 
 

s
slug ft  12 in 
ft 
2
2
2
lbf  s
ft 
 
 

s
slug ft  12 in 
ft 
p 1g  0.01300  psi
2
Now if we apply the momentum equation to the control volume (considering the assumptions shown):
p 2g  0.01545  psi
 
Fsx   uV  dA
CS
H
p1  p2 w 2

H
 τ w L  V1  ρ V1   w  
2  

δ2
0
H
u  ρ u  w dy  V2  ρ V2    δ2  w
2


 
1

2



2
2
2 7
7
The integral is equal to: ρ w  u dy  ρ V2  δ2  w  η dη  ρ V2  δ2  w


9
0
0
δ2
H
p1  p2 w 2
τ
1
L
Therefore the momentum equation becomes:
2
2 H
 w  ρ V2     δ2  w Simplifying and solving for the shear stress we get:
2
2
9 

2 H
 τ w L  ρ V1 
H
2
2 H
2 H
  p 1  p 2   ρ V1   V2     δ2 Substituting in known values we get:
2
2
2
9 





2
2
2
2

lbf 1  ft
slug 
ft  1  ft 
ft   1
2 4   lbf  s  ft  



τ 
 [ ( 0.01328 )  ( 0.01578 ) ] 

 0.00234 
  40  
  43.6        ft 


2 2
3 
20 ft 
s 2
s  2
9 12   slug ft  12 in  

in
ft


1
5
τ  5.46  10
 psi
Problem 9.21
Problem
9.28
[Difficulty: 3]
9.21
Given:
Data on fluid and boundary layer geometry
Find:
Gage pressure at location 2; average wall stress
Solution:
The solution involves using mass conservation in the inviscid core, allowing for the fact that as the boundary layer grows it
reduces the size of the core. One approach would be to integrate the 1/7 law velocity profile to compute the mass flow in the
boundary layer; an easier approach is to simply use the displacement thickness!
The average wall stress can be estimated using the momentum equation for a CV
The given and available (from Appendix A) data is
ρ  0.00234 
slug
ft
3
ft
U1  50
s
L  20 ft
D  15 in
δ2  4  in
Governing equations:
Mass
Momentum
Bernoulli
p
ρ
2

V
 g  z  constant
2
(4.24)
Assumptions: (1) Steady flow (2) No pressure force (3) No body force in x direction
The displacement thickness at location 2 can be computed from boundary layer thickness using Eq. 9.1
1

δdisp2  


δ2
0
Hence


u
 1   dy  δ  


2

U

0
δ2
δdisp2 
8
Applying mass conservation at locations 1 and 2
1

δ2

7
 1  η  dη 
8
δdisp2  0.500  in
ρ U1 A1  ρ U2 A2  0
π 2
A1   D
4
A1
U2  U1 
A2
or
A1  1.227  ft
2
The area at location 2 is given by the duct cross section area minus the displacement boundary layer


π
2
A2   D  2  δdisp2
4
A2  1.069  ft
2
Hence
A1
U2  U1 
A2
ft
U2  57.4
s
For the pressure change we can apply Bernoulli to locations 1 and 2 to find
Hence
ρ
2
2
p 1  p 2  Δp    U2  U1 

2 
Δp  6.46  10
p 2 ( gage )  p 1 ( gage )  Δp
p 2  6.46  10
3
 psi
3
p 2  Δp
 psi
For the average wall shear stress we use the momentum equation, simplified for this problem
D
2

2
2 π
2
Δp A1  τ π D L  ρ U1  A1  ρ U2   D  2  δ2  ρ 
4

D


2
2
2  π r u dr
 δ2
1
where
y
u ( r)  U2   
 δ2 
7
r
and
D
2
y
dr  dy
0


2


2
2
ρ 
2  π r u dr  2  π ρ U2 


D  δ

δ
2
D
The integral is
2
2
7
 D  y    y  dy

 
2
  δ2 
2
D
2

ρ 

D
2
Hence
τ 
 δ2
 D δ2 
2
2
2  π r u dr  7  π ρ U2  δ2   

8 
9
 D δ2 
2
2 π
2
2
Δp A1  ρ U1  A1  ρ U2   D  2  δ2  7  π ρ U2  δ2   

8 
4
9
τ  6.767  10

π D L
5
 psi

Problem 9.22
Problem
9.30
[Difficulty: 2]
Given:
Find:
Blasius exact solution for laminar boundary layer flow
Solution:
The Blasius solution is given in Table 9.1; it is plotted below.
u
Plot and compare to parabolic velocity profile:
U
 2 
y

δ

y
 
δ
2
Parabolic
Blasius
Dimensionless Height y/δ
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
Dimensionless Velocity u/U
0.8
Problem 9.23
Problem
9.32
[Difficulty: 3]
9.23
9.8.
Blasius (Table 9.1 on the web) exact solution for laminar boundary layer flow
Given:
Find:
(a) Evaluate shear stress distribution
(b) Plot τ/τw versus y/δ
(c) Compare with results from sinusoidal velocity profile: u  2  y 
U
δ
2
We will apply the shear stress definition to both velocity profiles.
Solution:
Governing
Equation:
τ  μ

y
τ
2
ρ U

μ
ρ U
From the above equation:
 f''( η) 
τ
τw
For the parabolic profile: τ 

(Shear stress in Newtonian fluid)
u
U
For Blasius: u  U f'( η) and η  y 
Therefore:
y
δ
 
ν x
U
ν x

f''( η)
f''( 0 )
 
U
The shear stress is: τ  μ  ( U f'( η) )  U μ ( f''( η) )    η   U μ f''( η) 
ν
x
y
y 
f''( η)
τ is proportional to f''(η)
Rex

f''( η)
0.33206
Since y  δ at η  5 it follows that
μ U d
 u   μ U   2  2  y 

 


δ   y   U 
δ 
δ
d 

 δ

τw 
μ U
δ
2
y
δ
Thus:

η
5
τ
τw
1
y
δ
Both profiles are plotted here:
Dimensionless Height y/δ
0.8
0.6
0.4
0.2
Parabolic
Blasius
0
0
0.2
0.4
0.6
Dimensionless Shear Stress τ/τw
0.8
Problem 9.24
Problem
9.33
[Difficulty: 3]
9.24
Given:
Blasius (Table 9.1 on the web) exact solution for laminar boundary layer flow
Find:
Solution:
Plot v/U versus y/δ for Rex  10
We will apply the stream function definition to the Blasius solution.
5
For Blasius: u  U f'( η) and η  y 
U
ν x
The stream function is:
 1 ν U
From the stream function: v   ψ   
 f ( η) 
x
2 x
Thus
 1 ν U  f ( η) 
v   
2 x
ψ
U ν x  f ( η)
 d f     η  But  η   1  y  U   1  η
  
2 x ν x
2 x
x
 dη  x 
ν U x  
 d f     1  η   1  ν U  ( η f'( η)  f ( η) ) and


 dη   2 x  2 x
ν U x  
v
U

1
2

ν
U x
 ( η f'( η)  f ( η) )
v
U
Since y  δ at η  5 it follows that
y
δ

η
5

η f'( η)  f ( η)
2 Rex
Plotting v/U as a function of y/δ:
Dimensionless Height y/δ
0.8
0.6
0.4
0.2
0
0
3
1 10
3
2 10
Dimensionless flow Velocity v/U
3
3 10
Problem 9.25
(Difficulty 2)
9.25 A smooth flat plate 2.4 𝑚 long and 0.6 𝑚 wide is placed in an airstream (101.3 𝑘𝑘𝑘 𝑎𝑎𝑎 15℃) of
velocity 9
𝑚
.
𝑠
Calculate the total drag force on this plate (2 𝑠𝑠𝑠𝑠𝑠) if the boundary layer at the trailing
edge is (a) laminar, (b) transition, and (c) turbulent.
Find: The drag force on this plate.
Solution: Use the boundary layer relations for drag to determine forces.
Laminar flow:
𝐶𝐷 =
Turbulent flow:
𝐶𝐷 =
1.33
�𝑅𝑅𝐿
= 0.0011
0.0742
1
𝑅𝑅 5
= 0.0043
The kinematic viscosity and density of air at (101.3 𝑘𝑘𝑘 𝑎𝑎𝑎 15℃) are:
𝑣 = 1.46 × 10−5
The Reynolds number at the end of the plate is:
𝑚2
,
𝑠
𝜌 = 1.225
𝑘𝑘
𝑚3
The drag force is calculated by:
𝑚
× 2.4 𝑚
𝑠
6
2 = 1.48 × 10
𝑚
1.46 × 10−5
𝑠
or
1
𝐹𝐷 = 2𝐶𝐷 𝐴 𝜌𝑈 2
2
𝑈𝑈
=
𝑅𝑅 =
𝑣
9
𝐹𝐷 = 𝐶𝐷 × 2.4 𝑚 × 0.6 𝑚 × 1.225
For laminar flow we have (Eq 9.33):
𝐶𝐷 =
1.33
�𝑅𝑅𝐿
=
𝑘𝑘
𝑚 2
×
�9
� = 142.9𝐶𝐷 𝑁
𝑚3
𝑠
1.33
√1.48 × 106
= 0.0011
The force is then
𝐹𝐷 = 142.9 × 0.0011 𝑁 = 0.1572 𝑁
For turbulent flow we have (Eq.9.34):
𝐶𝐷 =
0.0742
1
𝑅𝑅 5
The force is
= 0.0043
𝐹𝐷 = 142.9 × 0.0043 𝑁 = 0.615 𝑁
For transitional flow, we need to calculate the transition location. For a transitional Reynolds
number of 5x106, the transition occurs at:
𝑥𝑐𝑐𝑐𝑐
−5
1.46 × 10
𝜈
=
𝑅𝑅𝑐𝑐𝑐𝑐 =
𝑚
𝑈
9
𝑠
𝑚2
𝑠 × 5 × 106 = 0.912 𝑚
The drag coefficient for the laminar section is
𝐶𝐷 =
The drag on the laminar portion is
1.33
�𝑅𝑅𝑐𝑐𝑐𝑐
=
1.33
√5.0 × 105
𝐹𝐷 = 0.00188 × 0.912 𝑚 × 0.6 𝑚 × 1.225
= 0.00188
𝑘𝑘
𝑚 2
×
�9
� = 0.102 𝑁
𝑚3
𝑠
We need to subtract from the total drag for the plate with a turbulent boundary layer the drag of the
section that is now laminar. The drag coefficient for turbulent flow at the transition location is
𝐶𝐷 =
0.0742
1
𝑅𝑅 5
=
The turbulent drag force for this section is then:
0.0742
1
5000005
= 0.00538
𝐹𝐷 = 0.00538 × 0.912 𝑚 × 0.6 𝑚 × 1.225
The total drag force is then
𝑘𝑘
𝑚 2
×
�9
� = 0.292 𝑁
𝑚3
𝑠
𝐹𝐷 = 𝐹𝐷,𝐿 + 𝐹𝐷,𝑇 = 0.102 𝑁 + (0.615 − 0.292) 𝑁 = 0.425 𝑁
Problem 9.26
Problem
9.36
[Difficulty: 2]
9.26
Given:
Data on flow over flat plate
Find:
Plot of laminar thickness at various speeds
Solution:
Given or available data:
Governing
Equations:
δ
x

5.48
2
5 m
ν  1.5  10
(9.21)

and
Rex
The critical Reynolds number is
(from Table A.10 at 20oC)
s
U x
Rex 
ν
δ  5.48
so
ν x
U
Recrit  500000
Hence, for velocity U the critical length xcrit is
x crit  500000
ν
U
The calculations and plot were generated in Excel and are shown below:
U (m/s)
x c rit (m)
1
7.5
2
3.8
3
2.5
4
1.9
5
1.5
10
0.75
x (m)
δ (mm)
δ (mm)
δ (mm)
δ (mm)
δ (mm)
δ (mm)
0.000
0.025
0.050
0.075
0.100
0.2
0.5
1.5
1.9
2.5
3.8
5.0
0.00
3.36
4.75
5.81
6.71
9.49
15.01
25.99
29.26
33.56
41.37
47.46
0.00
2.37
3.36
4.11
4.75
6.71
10.61
18.38
20.69
23.73
29.26
0.00
1.94
2.74
3.36
3.87
5.48
8.66
15.01
16.89
19.37
0.00
1.68
2.37
2.91
3.36
4.75
7.50
13.00
14.63
0.00
1.50
2.12
2.60
3.00
4.24
6.71
11.62
0.00
1.06
1.50
1.84
6.0
7.5
51.99
58.12
Laminar Boundary Layer Profiles
δ (mm)
70
60
U = 1 m/s
50
U = 2 m/s
U = 3 m/s
40
U = 4 m/s
30
U = 5 m/s
20
U = 10 m/s
10
0
0
2
4
x (m)
6
8
Problem
9.38
Problem 9.27
[Difficulty: 2]
9.27
Given:
Parabolic solution for laminar boundary layer
Find:
Plot of δ, δ*, and τ w versus x/L
Solution:
Given or available data:
Basic
equations:
u
U
 2  
y
ν  1.08  10
y
 
δ δ
2
 5 ft

δ
x
2
s
(From Table A.8 at 68 oF) L  9  in
5.48

cf 
Rex

τw
1
2
2
 ρ U

U  5
ft
s
0.730
Rex
1
1
1
1 3 

 u   y
 u
2
2
Hence:  *   1  dy    1  d      1  2   d        
U
U   
3 0 3

0
0
0


The computed results are from Excel, shown below:
Laminar Boundary Layer Profiles
δ (in)
0.000
0.019
0.026
0.032
0.037
0.042
0.046
0.050
0.053
0.056
0.059
0.062
0.065
0.067
0.070
0.072
0.075
0.077
0.079
0.082
0.084
δ * (in) τ w (psi)
0.000
0.006 0.1344
0.009 0.0950
0.011 0.0776
0.012 0.0672
0.014 0.0601
0.015 0.0548
0.017 0.0508
0.018 0.0475
0.019 0.0448
0.020 0.0425
0.021 0.0405
0.022 0.0388
0.022 0.0373
0.023 0.0359
0.024 0.0347
0.025 0.0336
0.026 0.0326
0.026 0.0317
0.027 0.0308
0.028 0.0300
0.09
0.16
0.08
0.14
0.07
0.12
δ
0.06
0.10
0.05
τw (psi)
0.00
0.45
0.90
1.35
1.80
2.25
2.70
3.15
3.60
4.05
4.50
4.95
5.40
5.85
6.30
6.75
7.20
7.65
8.10
8.55
9.00
Re x
0.00.E+00
1.74.E+04
3.47.E+04
5.21.E+04
6.94.E+04
8.68.E+04
1.04.E+05
1.22.E+05
1.39.E+05
1.56.E+05
1.74.E+05
1.91.E+05
2.08.E+05
2.26.E+05
2.43.E+05
2.60.E+05
2.78.E+05
2.95.E+05
3.13.E+05
3.30.E+05
3.47.E+05
δ and δ * (in)
x (in)
0.08
0.04
τw
0.03
0.02
0.06
0.04
δ*
0.02
0.01
0.00
0.00
0
3
6
x (in)
9
Problem 9.28
(Difficulty 3)
9.28 For a laminar boundary layer on a flat plate, evaluate the kinetic energy lost between free
stream and any point in the boundary layer. Assume that the boundary layer is linear (see Problem
9.8) and use a control volume so that the flow rate for the oncoming flow and boundary layer are
equal How may this loss of kinetic energy be accounted for?
Assumption: the flow is steady
Solution: Evaluate the kinetic energy from the definition, where the energy needs to be integrated
over the region as the velocity varies
𝐾𝐾 =
𝜌
𝛾
� 𝑢3 𝑑𝑑 =
� 𝑢3 𝑑𝑑
2 𝐴
2𝑔 𝐴
At section (1), the velocity is uniform and the kinetic energy per unit width can be evaluated. The
flow area is chosen so the flow rate is the same entering and in the boundary layer. The kinetic
energy is:
𝐾𝐾1 =
The linear velocity profile is given by
𝛾 ℎ 3
𝛾𝑉03 ℎ
� 𝑉0 𝑑𝑑 =
2𝑔 0
2𝑔
𝑢 𝑦
=
𝑉0 𝛿
At section (2), we have the kinetic energy in the boundary layer as:
𝐾𝐾2 =
𝛾 𝛿 3
𝛾 𝛿 𝑉0 3 3
𝛾 𝑉0 3 𝛿 4 𝛾𝑉03 𝛿
� 𝑢 𝑑𝑑 =
� � � 𝑦 𝑑𝑑 =
� �
=
2𝑔 0
2𝑔 0 𝛿
2𝑔 𝛿
8𝑔
4
The relation between h and d is found using the continuity relation. The flow rate per unit area q is
the same for the oncoming flow and that in the boundary layer.
For the oncoming flow
ℎ
𝑞 = � 𝑉0 𝑑𝑑 = 𝑉0 ℎ
0
And in the boundary layer
𝛿
The height h is then
𝛿
𝑞 = � 𝑢 𝑑𝑑 = �
0
0
𝑉0
𝑉0 𝛿 2
𝛿
𝑦 𝑑𝑑 =
= 𝑉0
2
𝛿
𝛿 2
ℎ=
The oncoming kinetic energy is then
𝐾𝐾1 =
𝛿
2
𝛾𝑉03 ℎ 𝛾𝑉03 𝛿
=
2𝑔
4𝑔
Comparing the values of the oncoming kinetic energy and that in the boundary layer, we see that the
kinetic energy in the boundary layer is 50 % of that in the oncoming flow. There is a 50 % loss of
kinetic energy. If the exact laminar boundary layer profile had been used, the kinetic energy loss
would have been about 32 %.
The loss in kinetic energy is due to the viscous friction. The kinetic energy is converted to thermal
energy due to the viscous work done.
Problem 9.29
(Difficulty 2)
9.29 Air at atmospheric pressure and 20 ℃ flows over both sides of a flat plate that is 0.8 𝑚 long and
0.3 𝑚 wide at a velocity of 5
𝑚
.
𝑠
Determine the total drag force on the plate. If the single plate is
replaced by two plates each 0.4 𝑚 long and 0.3 𝑚 wide, what is the total drag force? Explain why there
is a difference in the total drag force even though the total surface area is the same.
Find: The drag force
Assumption: the flow is steady and incompressible
Solution: Use the boundary layer relations for drag on a flat plate.
The drag force is calculated by:
1
𝐹𝐷 = 𝐶𝐷 𝜌𝑉 2 𝐴
2
Where the drag coefficient CD is a function of the Reynolds number.
We first determine the Reynolds number. For air at atmospheric pressure and 20 ℃ from table A.9:
𝜌 = 1.21
For the single plate we have:
𝑘𝑘
𝑚3
𝑎𝑎𝑎 𝜇 = 1.81 × 10−5 𝑃𝑃 ∙ 𝑠
𝑘𝑘
𝑚
𝜌𝜌𝜌 1.21 𝑚3 × 5 𝑠 × 0.8 𝑚
𝑅𝑅𝐿 =
=
= 2.67 × 105
𝜇
1.81 × 10−5 𝑃𝑃 ∙ 𝑠
We assume the flow is laminar. The drag coefficient is given by
𝐶𝐷 =
1.33
�𝑅𝑅𝐿
=
1.33
√2.67 × 105
The drag force on both sides of the plate is calculated by:
= 0.0026
1
1
𝑘𝑘
𝑚 2
𝐹𝐷 = 𝐶𝐷 𝜌𝑉 2 𝐴 = × 0.0026 × 1.21 3 × �5 � × 2 × 0.8 𝑚 × 0.3 𝑚 = 0.0189 𝑁
2
2
𝑚
𝑠
For the case of two plates we have:
𝑘𝑘
𝑚
𝜌𝜌𝜌 1.21 𝑚3 × 5 𝑠 × 0.4 𝑚
𝑅𝑅𝐿 =
=
= 1.34 × 105
𝜇
1.81 × 10−5 𝑃𝑃 ∙ 𝑠
𝐶𝐷 =
1.33
�𝑅𝑅𝐿
=
1.33
√1.34 × 105
= 0.00363
The drag force for both sides of both plates is calculated by:
1
𝑘𝑘
𝑚 2
𝐹𝐷 = 2 � 𝐶𝐷 𝜌𝑉 2 𝐴� = 0.00363 × 1.21 3 × �5 � × 3 × 0.4 𝑚 × 0.3 𝑚 = 0.0264 𝑁
2
𝑚
𝑠
We have the different total drag force although the area is the same because the length of the
boundary layer for these two cases is different. It gives us different drag coefficient 𝐶𝐷 .
Problem
9.40
Problem 9.30
[Difficulty: 3]
9.30
Thin flat plate installed in a water tunnel. Laminar BL's with parabolic profiles form on both sides of the plate.
Given:
L  0.3 m
b  1 m
U  1.6
m
s
ν  1  10
2
6 m

u
s
U
 2  
y
y
 
δ δ
2
Total viscous drag force acting on the plate.
Find:
Solution:
We will determine the drag force from the shear stress at the wall
U L
First we will check the Reynolds number of the flow: ReL 
ν
5
 4.8  10
Therefore the flow is laminar throughout.
L
The viscous drag for the two sides of the plate is:

FD  2   τw b dx

The wall shear stress τw is:
0
2
   at y = 0, which for the parabolic profile yields:
2  0  2  μ U
u
τw  μ U  


2 
δ
δ
 y 
δ


τw  μ 
1
δ  5.48
The BL thickness δ is:
L

2
 x Therefore: FD  2  b  

U




L
ν

  1
4
2  μ U
U 
2
 x
dx
dx 
 b  μ U
1
5.48
ν 0
ν 2
5.48
x
U
0
Evaluating this integral:
FD 
8  b  μ U
5.48

U L
ν
FD  1.617 N
Problem 9.31
Problem
9.42
[Difficulty: 3]
9.31
Given:
Triangular plate
Find:
Drag
Solution:
Basic
equations:
cf 
τw
1
2
cf 
2
 ρ U
L  2  ft
3
0.730
Rex
L  1.732  ft
2
W  2  ft
U  15
ft
s
Assumptions:
(1) Parabolic boundary layer profile
(2) Boundary layer thickness is based on distance from leading edge (the "point" of the triangle).
From Table A.9 at 70 oF
ν  1.63  10
 4 ft

2
ρ  0.00233 
s
ft
ReL 
First determine the nature of the boundary layer
The drag (one side) is
We also have
slug
U L
ν
3
ReL  2  10
5
so definitely laminar
L

FD   τw dA



FD   τw w( x ) dx

w( x )  W
0
x
L
1
2 1
2 0.730
τw  cf   ρ U   ρ U 
2
2
Rex
L
Hence
1
2 W 

FD   ρ U  
2
L 



L

1


0.730
0.730  x
W
2
2
dx 
 ρ U   ν  x dx

2
L
U x
0
3
ν
0
L
The integral is

1
3


2
2
2
 x dx   L

3
0
so
3
3
FD  0.243  ρ W ν L U
FD  1.11  10
 lbf
Note: For two-sided solution
2  FD  2.21  10
3
 lbf
Problem 9.32
Problem
9.44
[Difficulty: 3]
9.32
Given:
Parabolic plate
Find:
Drag
Solution:
Basic equations:
cf 
τw
1
2
0.730
cf 
2
 ρ U
Rex
 W
 
2
L 
W  1  ft
2
L  0.25 ft
1  ft
U  15
ft
s
Note: "y" is the equation of the upper and lower surfaces, so y = W/2 at x = L
From Table A.9 at 70 oF
ν  1.63  10
 4 ft

2
ρ  0.00233 
s
ft
ReL 
First determine the nature of the boundary layer
The drag (one side) is
slug
3
U L
4
ReL  2.3  10
ν
so just laminar
L

FD   τw dA



FD   τw w( x ) dx

w( x )  W
0
We also have
1
2 1
2 0.730
τw  cf   ρ U   ρ U 
2
2
Rex
Hence



1
2
FD   ρ U  W 
2



x
L
L
0.730 
U x
x
L
3
dx 
0.730
2
L
ν 
 ρ U  W
  1 dx
L 
2
0
ν
0
3
FD  0.365  ρ W ν L U
FD  3.15  10
Note: For two-sided solution
4
 lbf
4
2  FD  6.31  10
 lbf
Problem 9.33
Problem
9.46
[Difficulty: 3]
9.33
Given:
Pattern of flat plates
Find:
Drag on separate and composite plates
Solution:
Basic
equations:
cf 
1
0.730
cf 
2
 ρ U
Rex
2
Parabolic boundary layer profile
Assumption:
For separate plates
τw
L  3  in
W  3  in
U  3
We also have
Hence
From Table A.7 at 70 oF ν  1.06  10
s
ReL 
First determine the nature of the boundary layer
The drag (one side) is
ft
U L
ReL  7.08  10
ν
4
 5 ft

2
ρ  1.93
slug
s
so definitely laminar
L

FD   τw dA



FD   τw W dx

0
1
2 1
2 0.730
τw  cf   ρ U   ρ U 
2
2
Rex

1
2

FD   ρ U  W

2



L
L

  1

0.730
0.730
2
2
dx 
dx
 ρ U  W ν  x

2
U x
0
3
ν
0
L
The integral is

1
  1

2
2
dx  2  L
 x

so
0
This is the drag on one plate. The total drag is then
3
FD  0.730  ρ W ν L U
FD  0.0030 lbf
FTotal  4  FD
FTotal  0.0119 lbf
For both sides:
For the composite plate
L  4  3  in
L  1.00 ft
ReL 
U L
ν
 2.83  10
5
2  FTotal  0.0238 lbf
so still laminar
3
FComposite  0.730  ρ W ν L U
FComposite  0.0060 lbf
For both sides:
2  FComposite  0.0119 lbf
The drag is much lower on the composite compared to the separate plates. This is because τ w is largest near the
leading edges and falls off rapidly; in this problem the separate plates experience leading edges four times!
ft
3
Problem 9.34
Problem
9.48
[Difficulty: 2]
9.34
Horizontal surface immersed in a stream of standard air. Laminar BL with sinusoidal profile forms.
Given:
L  1.8 m b  0.9 m
Find:
Solution:
Plot δ,
δ*,
U  3.2
m
s
ν  1.46  10
2
5 m

s




δ
1
0


δ 

1
θ
u
U
0
 1  u  dη


U

 dη

(Wall shear stress)
(Displacement thickness)
u
  1 
(Momentum thickness)
U

1

π 
π 
θ 




sin  η   1  sin  η  dη  
δ 

2  
 2 


For the sinusoidal velocity profile:
0
Evaluating this integral:
θ
δ

4π
Separating variables yields:

2 π
2 4 π
 dx  δ dδ or
δ dδ 
2  ρ U




1
0
  π    π   2
sin  η   sin  η   dη
  2    2  
  4  π  
d
d
θ  θ  δ 
  δ
2  π  x 
dx
dδ  x 
π 0 π
π μ U
2 4  π  
τw  μ U cos   

 ρ U 
  δ
2 δ
2  π  x 
 2 δ  2 δ
π μ U
Solving this expression for δ/x:
1
0
 0.1366 Therefore it follows that
2 π
To determine the wall shear stress:
δ
π y
 
 2 δ
We will determine the drag force from the shear stress at the wall
δdisp
Also,
U
 sin
and τw versus x/L for the plate
Governing τw  ρ U2  d θ   μ   u  at y = 0
 dx 
y 
Equations:
δdisp
u
δ
x

π
2
4π
 1  sin π  η  dη




 2 
The Reynolds number is related to x through:

π
2

μ
4  π ρ U
2
 dx Integrating yields:
δ
2

π
2
μ

x
4  π ρ U
μ
δ
ρ U x
x
Evaluating this integral:
δdisp
δ
5
Rex  2.19  10  x
Plots of δ, δdisp and τ w as functions of x are shown on the next page.
1
2
π
 0.363
where x is measured in meters.
δdisp
δ

4.80
Rex
 0.363
BL Thickness
Disp. Thickness
Wall Shear
0.04
10
0.03
0.02
5
0.01
0
0
0.5
1
x (m)
1.5
0
Wall Shear Stress (Pa)
Boundary Layer and Displacement Thicknesses (mm)
15
Problem
9.49
Problem 9.35
[Difficulty: 3]
9.35
Given:
Water flow over flat plate
Find:
Drag on plate for linear boundary layer
Solution:
Basic
equations:

FD  2   τw dA


du
τw  μ
dy
L  0.35 m
From Table A.8 at 10 oC ν  1.30  10
6 m

ρ  1000
s
ReL 
First determine the nature of the boundary
layer
y
The velocity profile is
u  U  U η
δ
du
U
Hence
τw  μ
 μ
dy
δ
We also have
The integral is
2 dδ 
τw  ρ U   




1
0

u
 dη
U
s
ReL  2.15  10
5
so laminar
but we need δ(x)
1
so
1
2 dδ
2 dδ
τw  ρ U 
  ρ U 
dx
6
dx
(2)
U
1
2 dδ
τw  μ   ρ U 
δ
6
dx
δ dδ 
Hence
δ
6 μ
 dx
δ
x
or
δ
ρ U
12 μ
ρ U

FD  2 


2
or
2

x

L


U


τw dA  2 W 
μ  dx  2 W 


δ


0

L
0
6 μ
ρ U
x  c
12
Rex

but δ(0) = 0 so c = 0
3.46
Rex

  1

2
dx  2  L
 x

so
0
FD 
2
3
3
 ρ W ν L U
L

  1

ρ U
μWU U 
2
2
 x
dx
μ  U
x
dx 

12 μ
ν 0
3
1
L
The integral is
U
  1 
3
m
U L
ν

m
u
kg
u
η  η2 dx  16
Separating variables
Then
1
(1)
U  0.8
dx
1
u
2 dδ 
  1   dη  ρ U    η ( 1  η) dη
dx
U 
U
dx 0

0
0
Comparing Eqs 1 and 2
at y = 0, and also
W  1 m
2
2 dδ 
τw  ρ U   

FD 
2 μ  W  U
3
FD  0.557N

U L
ν
Problem 9.36
(Difficulty 2)
9.36 Use the momentum integral equation to derive expressions for the displacement thickness 𝛿 ∗ , the
momentum thickness 𝜃, and the friction coefficient 𝐶𝑓 for the linear velocity profile
𝑢
𝑈
𝑦
𝛿
= . Compare
your results to that in Table 9.2. What is the percent error in the total drag on a plate if the linear
approximation is used?
Find: Displacement thickness, momentum thickness, and friction factor.
Solution: Use the momentum integral equation:
𝜏𝑤
𝑑
𝑑𝑑
(𝑈 2 𝜃) + 𝛿 ∗ 𝑈
=
𝑑𝑑
𝑑𝑑
𝜌
For the constant free stream velocity U the momentum integral equation becomes:
We have the linear velocity profile as:
𝜏𝑤
𝑑𝑑 1 𝑑𝑑
=
=
2
𝑑𝑑 6 𝑑𝑑
𝜌𝑈
For the displacement thickness we then have:
𝑢 𝑦
=
𝑈 𝛿
𝛿
𝛿
𝑢
𝑦
1
𝛿 ∗ = � �1 − � 𝑑𝑑 = � �1 − � 𝑑𝑑 = 𝛿
𝑈
𝛿
2
0
0
And for the momentum thickness we have:
𝛿
𝜃=�
0
𝛿
𝑢
𝑦
𝑦
1
𝑢
�1 − � 𝑑𝑑 = � �1 − � 𝑑𝑑 = 𝛿
𝑈
𝛿
6
𝑈
0 𝛿
As the friction coefficient is defined as:
𝐶𝑓 =
We also have for the wall shear stress
𝜏𝑤
𝜏𝑤
𝑑𝑑 1 𝑑𝑑
=2 2=2
=
1 2
𝑑𝑑 3 𝑑𝑑
𝜌𝑈
𝜌𝑈
2
𝜏𝑤 = 𝜇
𝜕𝜕
𝑈
=𝜇
𝜕𝜕𝑦=0
𝛿
For the constant U we have:
Or, rearranging
Integrating we have:
𝑈
𝜇
1 𝑑𝑑
𝜏𝑤
= 𝛿2 =
2
6 𝑑𝑑
𝜌𝑈
𝜌𝑈
6𝜇
𝑑𝑑 = 𝛿𝛿𝛿
𝜌𝜌
1
6𝜇
𝑥 = 𝛿2
2
𝜌𝜌
The friction factor is then
𝛿
12𝜇 3.46
=�
=
𝑥
𝜌𝜌𝜌 √𝑅𝑅
12𝜇𝜇
1 𝑑𝑑 1 𝑑� 𝜌𝜌
1 12𝜇 0.577
=
= �
=
𝐶𝑓 =
3 𝑑𝑑 3 𝑑𝑑
6 𝜌𝜌𝜌
√𝑅𝑅
We get the same result as the linear approximation in the Table 9.2 Compared with the exact result
for friction coefficient as:
𝐶𝑓 =
0.664
√𝑅𝑅
The error of the total drag by using linear approximation is about 10.4%.
Problem 9.37
(Difficulty 2)
9.37 A smooth flat plate 1.6 𝑓𝑓 long is immersed in 68 ℉ water flowing at 1.2
plate is a small 1 𝑖𝑖𝑖ℎ square sensor. What is the friction force on this sensor?
𝑓𝑓
.
𝑠
In the center of the
Find: Friction force
Solution: Use the expression for drag force to find the friction force.
Then density and the viscosity are:
𝜌 = 1.937
𝑠𝑠𝑠𝑠
𝑙𝑙𝑙 ∙ 𝑠
𝑎𝑎𝑎 𝜇 = 2.104 × 10−5
𝑓𝑓 3
𝑓𝑓 2
The Reynolds number at the end of the plate is:
𝑓𝑓
𝑠𝑠𝑠𝑠
𝜌𝜌𝜌 1.937 𝑓𝑓 3 × 1.2 𝑠 × 1.6 𝑓𝑓
=
= 1.76 × 105
𝑅𝑅 =
𝑙𝑙𝑙 ∙ 𝑠
𝜇
−5
2.104 × 10
𝑓𝑓 2
So this is laminar flow over the plate.
The shear stress at any location is given by:
The friction force can be calculated as:
1
0.664
𝜏𝑤 = 𝜌𝑈 2
2
𝜌𝜌𝜌
�
𝜇
𝐴
𝑥2
𝐹 = � 𝜏𝑤 𝑑𝑑 = 𝑏 � 𝜏𝑤 𝑑𝑑
0
𝑥1
Because the length of the plate in the flow direction is only 1 inch, we can assume the shear stress is
constant and equal to the value at the center. Thus:
𝐹 = 𝑏 2 𝜏𝑤𝑤
𝑓𝑓
𝑠𝑠𝑠𝑠
× 1.2
× 0.8 𝑓𝑓
1.937
𝜌𝜌𝑥𝑐
𝑠
𝑓𝑓 3
=�
= 297
�𝑅𝑒𝑒𝑐 = �
𝑙𝑙𝑙 ∙ 𝑠
𝜇
2.104 × 10−5
𝑓𝑓 2
Thus
𝜏𝑤𝑤
1
𝑙𝑙𝑙 ∙ 𝑠2
𝑓𝑓 2 0.664
𝑙𝑙𝑙
= × 1.937
×
�1.2
�
×
=
0.0031
2
𝑠
297
𝑓𝑓 2
𝑓𝑓 4
2
1
𝑙𝑙𝑙
𝐹 = 𝑏 2 𝜏𝑤𝑤 = � 𝑓𝑓� × 0.0031 2 = 2.15 × 10−5 𝑙𝑙𝑙
12
𝑓𝑓
Problem 9.38
Problem
9.50
[Difficulty: 2]
9.38
Horizontal surface immersed in a stream of standard air. Laminar BL with linear profile forms.
Given:
L  0.8 m b  1.9 m
Find:
Solution:
Plot δ,
δ*,
U  5.3
m
s
ν  1.46  10
2
5 m

u
s
U

y
δ
and τw versus x/L for the plate
We will determine the drag force from the shear stress at the wall
Governing τw  ρ U2  d θ   μ   u  at y = 0
 dx 
y 
Equations:




δdisp
δ
1
0


δ 

1
θ
u
U
0
 1  u  dη


U

  1 

For the linear velocity profile:
(Wall shear stress)
(Displacement thickness)
u
 dη
(Momentum thickness)
U
1
1


2
  η  ( 1  η ) dη   η  η dη

δ 0
0
θ


δ
  1  
Therefore it follows that d θ  d θ    δ      δ  To determine the wall shear stress:
dx
dδ  x  6  x 
Separating variables yields:
Also,
δdisp
δ
6 μ
ρ U
2
 dx  δ dδ
1

  ( 1  η ) dη

Evaluating this integral:
0
The Reynolds number is related to x through:
δ
Integrating yields:

2
δdisp
δ
5

Rex  3.63  10  x
Plots of δ, δdisp and τ w as functions of x are shown on the next page.
6 μ
ρ U
θ
Evaluating this integral:
x
τw 
μ U
δ
1

6
2

ρ U
6
 0.1667
 
δ
x 

Solving this expression for δ/x:
δ
x
1
δdisp
2
δ
where x is measured in meters.


3.46
Rex
1
2
BL Thickness
Disp. Thickness
Wall Shear
0.04
4
0.03
2
0.02
0
0
0.2
0.4
x (m)
0.6
0.01
0.8
Wall Shear Stress (Pa)
Boundary Layer and Displacement Thicknesses (mm)
6
Problem 9.39
Problem
9.54
[Difficulty: 3]
9.39
P9.13.
9.13 only
Note: Figure data applies to problem 9.18
Given:
Data on fluid and turbulent boundary layer
Find:
Solution:
Mass flow rate across ab; Momentum flux across bc; Distance at which turbulence occurs
CV
Mass
Basic
equations:
d
Momentum
c
Rx
Assumptions: 1) Steady flow 2) No pressure force 3) No body force in x direction 4) Uniform flow at ab
The given or available data (Table A.10) is
U  50
m
s
δ  19 mm
Consider CV abcd
b  3 m
ρ  1.23
kg
ν  1.50  10
3
2
5 m

m
kg
mad  3.51
s
mad  ρ U b  δ
(Note: Software cannot render a dot)
1
δ

mad   ρ u  b dy  mab  0

Mass
s
and in the boundary layer
u
U
0

y
1
7
7
  η
δ
 
dy  dη  δ
1
Hence

1


7
7
mab  ρ U b  δ   ρ U η  δ dη  ρ U b  δ   ρ U b  δ

8
0
1
mab   ρ U b  δ
8
kg
mab  0.438 
s
1
δ
The momentum flux
across bc is

  δ
mfbc   u  ρ V dA  


0
0

2


7
2
2
7
u  ρ u  b dy   ρ U  b  δ η dη  ρ U  b  δ

9
0
7
2
mfbc   ρ U  b  δ
9
mfbc  136.3 
2
s
From momentum
Rx  U ( ρ U δ)  mab u ab  mfbc
Transition occurs at
Rex  5  10
5
kg m
and
2
Rx  ρ U  b  δ  mab U  mfbc
U x
Rex 
ν
x trans 
Rx  17.04  N
Rex  ν
U
x trans  0.1500 m
Problem 9.40
Problem
9.56
9.40
9.3.
Turbulent boundary layer flow of water
L  1 m
Find:
Solution:
Governing
Equations:
Plot δ,
δ*,
U  1
1
2
6 m
m
ν  1.00  10
s

u
s
U

y
 
δ
7
and τw versus x/L for the plate
We will determine the drag force from the shear stress at the wall
δ
x

0.382
(Boundary layer thickness)
1
Rex
δdisp
δ
Cf 

5
1
(Displacement thickness)
8
τw
1
2
2

0.0594
1
 ρ U
Rex
(Skin friction factor)
5
Assumption: Boundary layer is turbulent from x = 0
For the conditions given:
ReL 
U L
ν
6
 1.0  10
q 
1
2
2
 ρ U  500 Pa
τw 
0.0594
1
Rex
30
Boundary Layer and Displacement Thicknesses (mm)
Here is the plot of boundary layer thickness
and wall shear stress:

 q  29.7 Pa Rex
1
5
5
3
BL Thickness
Disp. Thickness
Wall Shear
20
2
10
1
0
0
0.5
x (m)
0
1
Wall Shear Stress (Pa)
Given:
[Difficulty: 2]
Problem 9.41
(Difficulty 2)
9.41 A flat-bottomed barge having a 150 𝑓𝑓 by 20 𝑓𝑓 bottom is towed through still water (60 ℉) at
10 𝑚𝑚ℎ. What is the frictional drag force exerted by the water on the bottom of the barge? How long
could the laminar portion of the boundary layer be, using critical Reynolds number of 537000? What is
the thickness of the laminar layer at its downstream end? What is the approximate thickness of the
boundary layer at the rear end of the bottom of the barge?
Find: Frictional force on the bottom of the boat
Solution: Use the boundary layer relations to find the force and thhicknesses
The velocity is:
𝑈 = 10 𝑚𝑚ℎ = 14.67
The density and viscosity of water are:
𝜌 = 1.938
𝑓𝑓
𝑠
𝑙𝑙𝑓 ∙ 𝑠
𝑠𝑠𝑠𝑠
𝑎𝑎𝑎 𝜇 = 2.34 × 10−5
3
𝑓𝑓 2
𝑓𝑓
The Reynolds number at the end of the barge is:
𝑅𝑅 =
𝜌𝜌𝜌
= 537000
𝜇
The length of the laminar portion boundary layer is:
−5 𝑙𝑙𝑙 ∙ 𝑠 × 537000
𝜇𝜇𝜇 2.34 × 10
𝑓𝑓 2
=
= 0.442 𝑓𝑓
𝐿=
𝑓𝑓
𝑙𝑙𝑙 ∙ 𝑠2
𝜌𝜌
×
14.67
1.938
𝑠
𝑓𝑓 4
This is a very short distance and negligible compared to the length of the barge. For the thickness of
the laminar boundary layer we have:
𝛿
30
=�
𝑥
𝑅𝑅
At the downstream end we have:
30
30
𝛿 = 𝐿� = 0.442 𝑓𝑓 × �
= 0.0033 𝑓𝑓
𝑅𝑅
537000
The drag force can be calculated assuming that the boundary is turbulent over the entire bottom:
𝐶𝐷 =
1
𝐹𝐷 = 𝐶𝐷 × 𝜌𝜌𝑈 2
2
0.0742
1
𝑅𝑅 5
−
1740
= 0.0021
𝑅𝑅
𝑙𝑙𝑙 ∙ 𝑠2
𝑓𝑓 2
×
20
𝑓𝑓
×
150
𝑓𝑓
×
�14.67
� = 1313 𝑙𝑙𝑙
𝑠
𝑓𝑓 4
𝐹𝐷 = 0.0021 × 0.5 × 1.938
The Reynolds number at the rear end of the barge is:
𝑅𝑅 =
𝜌𝜌𝜌
=
𝜇
1.938
𝑓𝑓
𝑙𝑙𝑙 ∙ 𝑠2
× 14.67
× 150 𝑓𝑓
𝑠
𝑓𝑓 4
= 1.823 × 108
𝑙𝑙𝑙
∙
𝑠
2.34 × 10−5
𝑓𝑓 2
𝛿 0.382
=
1
𝑥
𝑅𝑅 5
So we have the boundary layer thickness at the end is:
𝛿=𝑥
0.382
1
𝑅𝑅 5
= 150 𝑓𝑓 ×
0.382
1
(1.823 × 108 )5
= 1.276 𝑓𝑓
Problem 9.42
(Difficulty 2)
9.42 European Intercity Express trains operate at speeds of up to 280
𝑘𝑘
.
ℎ𝑟
Suppose that a train is 120 𝑚
long. Treat the sides and top of the train as a smooth flat plate 9 𝑚 wide. When the train moves through
still air at sea level, calculate the possible length of the laminar boundary layer and thickness of this
layer at its down-stream end. What is the thickness of the boundary layer at the rear end of the train?
What is the viscous drag force on the train and what power must be expended to overcome this
resistance at maximum speed? At 50% of maximum?
Find The drag force and other boundary layer parameters
Solution: Use the boundary layer relations to find the force and other parameters
The velocity is:
𝑈 = 280
The density and the viscosity of the air are:
𝜌 = 1.225
The Reynolds number is defined as:
𝑘𝑘
𝑎𝑎𝑎 𝜇 = 1.789 × 10−5 𝑃𝑃 ∙ 𝑠
𝑚3
The length of the laminar boundary layer is:
𝐿=
𝑘𝑘
𝑚
= 77.8
ℎ𝑟
𝑠
𝑅𝑅 =
𝜌𝜌𝜌
𝜇
𝜇𝜇𝜇 1.789 × 10−5 𝑃𝑃 ∙ 𝑠 × 500000
=
= 0.0939𝑚
𝑘𝑘
𝑚
𝜌𝜌
1.225 3 × 77.8
𝑠
𝑚
For the thickness of the laminar boundary layer we have:
𝛿
30
=�
𝑥
𝑅𝑅
At the downstream end we have:
30
30
𝛿 = 𝐿 � = 0.0939 𝑚 × �
= 7.27 × 10−4 𝑚
𝑅𝑅
500000
At the end of the train we have for the Reynolds number:
𝑘𝑘
𝑚
𝜌𝜌𝜌 1.225 𝑚3 × 77.8 𝑠 × 120 𝑚
=
= 6.39 × 108
𝑅𝑅 =
𝜇
1.789 × 10−5 𝑃𝑃 ∙ 𝑠
The boundary layer thickness at the end is:
𝛿=𝑥
0.382
1
𝑅𝑅 5
𝛿 0.382
=
1
𝑥
𝑅𝑅 5
= 120 𝑚 ×
The drag force can be calculated by:
The drag coefficient for turbulent flow is
1
(6.39 × 108 )5
= 0.795 𝑚
1
𝐹𝐷 = 𝐶𝐷 × 𝜌𝜌𝑈 2
2
𝐶𝐷 =
And the force is
𝐹𝐷 = 0.0013 × 0.5 × 1.225
The power to overcome this drag is:
0.382
0.0742
1
𝑅𝑅 5
= 0.0013
𝑘𝑘
𝑚 2
× 120𝑚 × 9 𝑚 × �77.8 � = 5.21 𝑘𝑘
3
𝑚
𝑠
𝑃 = 𝐹𝐷 𝑈 = 5.21 𝑘𝑘 × 77.8
For 50% of the maximum velocity we have:
𝑚
= 405 𝑘𝑘
𝑠
1
𝐹𝐷 = 𝐶𝐷 × 𝜌𝜌𝑈 2
2
The Reynolds number is
𝑘𝑘
𝑚
𝜌𝜌𝜌 1.225 𝑚3 × 0.5 × 77.8 𝑠 × 120 𝑚
=
= 3.195 × 108
𝑅𝑅 =
𝜇
1.789 × 10−5 𝑃𝑃 ∙ 𝑠
And the drag force is
𝐶𝐷 =
The force is
𝐹𝐷 = 0.0015 × 0.5 × 1.225
The power to overcome this drag is:
0.0742
1
𝑅𝑅 5
= 0.0015
𝑘𝑘
𝑚 2
×
120𝑚
×
9
𝑚
×
�0.5
×
77.8
� = 1.502 𝑘𝑘
𝑚3
𝑠
𝑃 = 𝐹𝐷 𝑈 = 1.502 𝑘𝑘 × 0.5 × 77.8
𝑚
= 58.4 𝑘𝑘
𝑠
Problem 9.43
(Difficulty 2)
9.43 Grumman Corp has proposed (Mechanical Engineering, 115, 8, August 1993, p.74f) to build a
magnetic levitation train to operate at a top speed of 300 𝑚𝑚ℎ. The vehicle is 114 𝑓𝑓 long. Assuming
that the sides and top can be treated approximately as a smooth flat plate of 30 𝑓𝑓 width, with a
turbulent boundary layer on it, calculate the drag force and the power expended to overcome the drag
at the maximum speed.
Find: The drag force and power
Solution: Use the boundary layer relation for drag
The velocity is:
𝑈 = 300 𝑚𝑚ℎ = 440
The density and the viscosity of the air are:
𝜌 = 2.377 × 10−3
𝑓𝑓
𝑠
𝑠𝑠𝑠𝑠
𝑙𝑙𝑙 ∙ 𝑠
𝑎𝑎𝑎 𝜇 = 3.74 × 10−7
3
𝑓𝑓
𝑓𝑓 2
At the end of the train we have the Reynolds number:
−3 𝑠𝑠𝑠𝑠 × 440 𝑓𝑓 × 114 𝑓𝑓
𝜌𝜌𝜌 2.377 × 10
𝑠
𝑓𝑓 3
=
= 3.19 × 108 > 500000
𝑅𝑅 =
𝑙𝑙𝑙
∙
𝑠
𝜇
−7
3.74 × 10
𝑓𝑓 2
So the boundary is turbulent essentially over the entire length of the surface.
The drag force can be calculated by:
1
𝐹𝐷 = 𝐶𝐷 × 𝜌𝜌𝑈 2
2
For a wholly turbulent boundary layer, the drag coefficient is
𝐶𝐷 =
0.0742
And the force is
1
𝑅𝑅 5
𝐹𝐷 = 0.0015 × 0.5 × 2.377 × 10−3
The power to overcome this drag is:
=
0.0742
1
(3.19 × 108 )5
= 0.0015
𝑙𝑙𝑙 ∙ 𝑠2
𝑓𝑓 2
×
114
𝑓𝑓
×
30
𝑓𝑓
×
�440
� = 1180 𝑙𝑙𝑙
𝑠
𝑓𝑓 4
𝑃 = 𝐹𝐷 𝑈 = 1180 𝑙𝑙𝑙 × 440
𝑓𝑓
𝑙𝑙𝑙 ∙ 𝑓𝑓
= 519200
= 944 ℎ𝑝
𝑠
𝑠
Problem 9.44
Problem
9.58
[Difficulty: 3]
9.44
Given:
Parabolic plate
Find:
Drag
Solution:
Basic
equations:
cf 
τw
1
2
cf 
2
 ρ U
0.0594
1
Rex
5
W
 
2
L
W  1  ft
2
L  3 in
1 ft
U  80
ft
s
Note: "y" is the equation of the upper and lower surfaces, so y = W/2 at x = L
From Table A.9 at 70 oF
ν  1.63  10
 4 ft

2
ρ  0.00233 
s
ft
ReL 
First determine the nature of the boundary layer
The drag (one side) is
We also have
3
U L
ReL  1.23  10
ν
5
so still laminar, but we are
told to assume turbulent!
L

FD   τw dA



FD   τw w( x ) dx

w( x )  W
0
x
L
1
2 1
2 0.0594
τw  cf   ρ U   ρ U 
1
2
2
Rex
Hence
slug



1
2
FD   ρ U  W 
2





5
L
x
0.0594
L
1
 U x 


 ν 
9
dx 
0.0594
2

1
1 

L
3
5
2 5 
10
 ρ U  W L
 ν   x dx

0
5
0
1

4

9
FD  0.0228 ρ W ν L  U
5
FD  0.00816  lbf
Note: For two-sided solution
2  FD  0.01632  lbf
Problem 9.45
Problem
9.60
[Difficulty: 3]
9.45
9.4
1
u
6

1
y  η 6
 
δ
Given:
Turbulent boundary layer flow with 1/6 power velocity profile:
Find:
Expressions for δ/x and Cf using the momentum integral equation; compare to 1/7-power rule results.
Solution:
We will apply the momentum integral equation
τw
Governing
Equations:
ρ



2
d
d 
U  θ  δdisp U  U 
dx
 dx 
τw
Cf 
1
2
U
(Momentum integral equation)
(Skin friction coefficient)
2
 ρ U
(1) Zero pressure gradient, so U is constant and dp/dx = 0
(2) δ is a function of x only, and δ = 0 at x = 0
0.25
(3) Incompressible flow
2 ν 
τ

0.0233

ρ

U

 
(4) Wall shear stress is:
w
U δ
Assumptions:



  1
u 
u  
2 d  


τw  ρ U   θ   ρ U 
δ
  1   dη
U 
U  
dx  
 dx 

  0

  1

2  
    1

2  
2 6
6
6  
Substituting for the velocity profile: τw  ρ U  d δ   η  η  dη  ρ U    d δ 
Setting our two τ w's equal:
dx  

56
d
x


  0

Applying the assumptions to the momentum integral equation yields:
0.0233 ρ U  
2


U

δ
 
ν
0.25
2 d
1
1
 d δ

56  dx 
2 6
 ρ U 

Simplifying and separating variables:
4
δ  dδ  0.0233
56
6
 
ν
4
  dx
 U
4
1 


4
4 
5
56  ν 
56  ν 
4 4
 δ  0.0233     x  C but C = 0 since δ = 0 at x = 0. Therefore: δ    0.0233     x
6  U
6  U
5
4

5
Integrating both sides:
1
5
In terms of the Reynolds number:
δ
x

0.353
1
Rex
5
For the skin friction factor:
1
0.0233 ρ U  
2
Cf 
τw
1
2
2

 ρ U
1
2


 U δ 
ν
2
 ρ U
4
1
4
1


1 
  Re 5 
4
4
ν  x
x
4
 0.0466 
     0.0466 Rex   0.353  Upon simplification:
 U x   δ 


1
1
Cf 
0.0605
1
Rex
These results compare to
δ
x

0.353
1
Rex
5
and
Cf 
0.0605
1
Rex
5
for the 1/7-power profile.
5
Problem 9.46
(Difficulty 2)
9.46 The U.S. Navy has built the Sea Shadow, which is a small waterplane twin-hull (SWATH) ship whose
object is to achieve the same reduced radar profile as the STEALTH aircraft. This catamaran is 160 𝑓𝑓
long and its twin hulls have a draft of 14 𝑓𝑓. Assume that the ocean turbulence triggers a fully turbulent
boundary layer on the sides of each hull. Treat these as flat plate boundary layers and calculate the drag
on the ship and the power required to overcome it as a function of speed. Plot the results for speeds
from 5 to 13 knots.
Find: The drag force and power as a function of speed
Solution: Use the relation for boundary layer drag.
The velocity is:
1 𝑘𝑘𝑘𝑘 = 1.69
The density and the viscosity of sea water are:
𝜌 = 1.99
𝑓𝑓
𝑠
𝑠𝑠𝑠𝑠
𝑙𝑙𝑙 ∙ 𝑠
𝑎𝑎𝑎 𝜇 = 2.25 × 10−5
3
𝑓𝑓
𝑓𝑓 2
At the end of the train we have for the Reynolds number as a function of speed:
𝑅𝑅 =
𝜌𝜌𝜌
=
𝜇
𝑓𝑓
1.69
𝑠𝑠𝑠𝑠
𝑠 � × 160 𝑓𝑓
× 𝑈(𝑘𝑘𝑘𝑘𝑘) × �
1.99
𝑘𝑘𝑘𝑘
𝑓𝑓 3
2.25 × 10−5
𝑙𝑙𝑙 ∙ 𝑠
𝑓𝑓 2
= 2.39 × 107 𝑈(𝑘𝑘𝑘𝑘𝑘)
Since the boundary layer is turbulent we have the drag force can be calculated by: (note there are
four surfaces on the catamaran).
𝐶𝐷 =
𝐹𝐷 = 4 ×
0.00248
1
(𝑈)5
1
𝐹𝐷 = 4𝐶𝐷 × 𝜌𝜌𝑈 2
2
0.0742
1
𝑅𝑅 5
× 0.5 × 1.99
=
0.0742
(2.39 ×
1
107 𝑈)5
1.8
𝑃 = 𝐹𝐷 𝑈 = 63.1�𝑈(𝑘𝑘𝑘𝑘𝑘)�
× 𝑈(𝑘𝑘𝑘𝑘𝑘) × 1.69
For the velocity 𝑈 from 5 to 13 knots:
0.00248
1
(𝑈)5
𝑙𝑙𝑙 ∙ 𝑠2
𝑓𝑓 2
×
160
𝑓𝑓
×
14
𝑓𝑓
×
�𝑈(𝑘𝑘𝑘𝑘𝑘)
×
1.69
�
𝑠
𝑓𝑓 4
1.8
𝐹𝐷 = 63.1�𝑈(𝑘𝑘𝑘𝑘𝑘)�
The power can be calculated as:
=
𝑙𝑙𝑙
𝑓𝑓
×
𝑠
ℎ𝑝
2.8
= 0.1938�𝑈(𝑘𝑘𝑘𝑘𝑘)� ℎ𝑝
𝑓𝑓 − 𝑙𝑙𝑙
550
𝑠
Problem 9.47
(Difficulty 2)
9.47 The two rectangular smooth flat plates are to have the same drag in the same fluid stream.
Calculate the required x. If the two plates are combined into the T-shape indicated, what ratio exists
between the drag of the combination and that of either one? Assume laminar boundary layers in all
calculations.
Find: Drag forces on plates
Solution: Use the boundary layer relations to find the drag force
For laminar boundary layer, the drag coefficient is calculated as:
𝐶𝐷 =
𝑅𝑅 =
And the force is
Thus the force for the first plate is
And for the second plate
1.33
√𝑅𝑅
𝜌𝜌𝜌
𝜇
1
𝐹𝐷 = 𝐶𝐷 × 𝜌𝜌𝑉 2
2
1
1.33
1
× 𝜌 × 3𝑥 × 𝑉 2 𝑁
𝐹𝐷1 = 𝐶𝐷1 × 𝜌𝐴1 𝑉 2 =
2
𝜌𝜌𝜌 2
�
𝜇
1
𝐹𝐷2 = 𝐶𝐷2 × 𝜌𝐴2 𝑉 2 =
2
1.33
1
× 𝜌 × 24 × 𝑉 2 𝑁
𝜌𝜌 × 4 2
�
𝜇
For equal forces
𝐹𝐷1 = 𝐹𝐷2
1
1.33
1
× 𝜌 × 3𝑥 × 𝑉 2 =
× 𝜌 × 24 × 𝑉 2
𝜌𝜌 × 4 2
𝜌𝜌𝜌 2
�
�
𝜇
𝜇
1.33
1
√𝑥
Thus the length x is
The drag force for the T-shape is:
𝐹𝐷3 = 2 ×
𝐹𝐷3 =
1.33
× 3𝑥 = 12
𝑥 = 16 𝑚
1.33
1
1.33
1
× 𝜌 × 4 × 1.5 × 𝑉 2 +
× 𝜌 × 20 × 3 × 𝑉 2
2
𝜌𝜌 × 4
𝜌𝜌 × 20 2
�
�
𝜇
𝜇
1
1 1.33 1
1.33 1 2
× 𝜌 × 6 × 𝑉2 +
× 𝜌 × 60 × 𝑉 2 = 19.42
× 𝜌𝑉
𝜌𝜌 2
𝜌𝜌 2
√20 𝜌𝜌 2
�
�
�
𝜇
𝜇
𝜇
𝐹𝐷3 𝐹𝐷3
=
𝐹𝐷1 𝐹𝐷2
1.33 1 2
× 𝜌𝑉
𝜌𝜌 2
�
𝜇
=
= 1.618
1.33 1 2
12
× 𝜌𝑉
𝜌𝜌 2
�
𝜇
19.42
Problem 9.48
Problem
9.63
[Difficulty: 3]
9.48
Turbulent boundary layer flow of water, 1/7-power profile
Given:
The given or available data (Table A.9) is
U  20
m
s
L  1.5 m
b  0.8 m ν  1.46  10
2
5 m

s
ρ  1.23
kg
3
m
x 1  0.5 m
(a) δ at x = L
(b) τw at x = L
(c) Drag force on the portion 0.5 m < x < L
Find:
Solution:
Basic
equations:
δ
x

0.382
Rex
Cf 
(Boundary Layer Thickness)
1
5
τw
1
2
2
0.0594

(Skin friction factor)
1
 ρ U
Rex
5
Assumptions: 1) Steady flow
2) No pressure force
3) No body force in x direction
At the trailing edge of the plate:
ReL 
U L
ν
 2.05  10
6
δL  L
Therefore
δL  31.3 mm
1
ReL
1
2 0.0594
Similarly, the wall shear stress is: τwL   ρ U 
1
2
ReL
0.382
5
τwL  0.798  Pa
5
L
To find the drag:

L
1



1
1



L

5


1
U
2
5
5
dx where c is defined:
FD   τw b dx   0.0594   ρ U    
x
 b dx  c  b   x



2
ν




x
0
x
1
1

1
U
2
c  0.0594   ρ U    
2
  ν
1
4
5
Therefore the drag is:
5
FD   c b  L
4
5

5 1
2
  ρ U  b  L CfL  x 1  Cfx1
4 2


At x = x1:
Rex1 
U x 1
ν
 6.849  10
5
Cfx1 
0.0594
1
Rex1
3
 4.043  10
and at x = L CfL 
5
0.0594
1
ReL
3
 3.245  10
5
FD  0.700 N
Therefore the drag is:
Alternately, we could solve for the drag using the momentum thickness:
At x = L
δL  31.304 mm
7
θL 
 δ  3.043  mm At x = x1:
72 L

2
FD  ρ U  b  θL  θx1
δx1  x 1 
0.382
1
Rex1

where θ 
 12.999 mm θx1 
7
72
δ
7
 δ  1.264  mm
72 x1
5
Therefore the drag is:
FD  0.700 N
Problem 9.49
Problem
9.64
[Difficulty: 3]
9.49
Air at standard conditions flowing over a flat plate
Given:
The given or available data (Table A.10) is
U  30
ft
s
Find:
x  3  ft
ν  1.57  10
 4 ft

2
ρ  0.00238
s
slug
ft
3
δ and τw at x assuming:
(a) completely laminar flow (parabolic velocity profile)
(b) completely turbulent flow (1/7-power velocity profile)
Solution:
(Laminar Flow)
Basic
equations:
δ
x

(Turbulent Flow)
5.48
δ
Rex
x

0.382
Rex
Cf 
τw
1
2
2
 ρ U

0.730
Rex 
The Reynolds number is:
ν
δlam  x 
For laminar flow:
1
2
U x
 5.73  10
δturb  x 
5.48
0.382
Comparing results:
δlam
1
 ρ U
Rex
(Skin friction factor)
5
τwlam  7.17  10
5
The turbulent boundary layer has a much larger skin friction, which causes it to
grow more rapidly than the laminar boundary layer.
 4.34
5
τwturb  3.12  10
 3.72
τwlam
 psi
5
Rex
τwturb
6
δturb  0.970  in
1
1
2 0.0594
τwturb   ρ U 
1
2
δturb
0.0594
δlam  0.261  in
Rex
Rex
2

5
1
2 0.730
τwlam   ρ U 
2
Rex
For turbulent flow:
5
τw
Cf 
Rex
(Boundary Layer Thickness)
1
 psi
Problem
9.65
Problem 9.50
[Difficulty: 3]
9.50
Given:
Air at standard conditions flowing through a plane-wall diffuser with negligible
BL thickness. Walls diverge slightly to accomodate BL growth, so p = constant.
The given or available data (Table A.9) is
U  60
m
s
Find:
L  1.2 m
W1  75 mm
2
5 m
ν  1.46  10

s
ρ  1.23
kg
3
m
(a) why Bernoulli is applicable to this flow.
(b) diffuser width W2 at x = L
Solution:
p1
Basic
equations:
ρ

V1
2
2
p2
 g  z1 

ρ
V2
2
 g  z2
2
 


 ρ dV   ρ V dA  0

t 
Assumptions:
(Bernoulli Equation)
(Continuity)
(1) Steady flow
(2) Turbulent, 1/7-power velocity profile in boundary layer
(3) z = constant
(4) p = constant
The Bernoulli equation may be applied along a streamline in any steady, incompressible flow in the absence of
friction. The given flow is steady and incompressible. Frictional effects are confined to the thin wall boundary
layers. Therefore, the Bernoulli equation may be applied along any streamline in the core flow outside the boundary
layers. (In addition, since there is no streamline curvature, the pressure is uniform across sections 1 and 2.
From the assumptions, Bernoulli reduces to: V1  V2 and from continuity: ρ V1  A1  ρ V2  A2eff  0


or A2eff  W2  2  δdisp2  b  W1  b
The Reynolds number is:
ReL 
U L
ν
Therefore: W2  W1  2  δdisp2
6
 4.932  10
From turbulent BL theory:
δ2  L
0.382
1
ReL
5
 21.02  mm
The displacement thickness is determined from:

δdisp2  δ2  


1
1
u
1 

 dη
U

where
u
U
η
7
η
y
δ
0
Substituting the velocity profile and valuating the integral:



δdisp2  δ2  

1
1

δ2

7
 1  η  dη 
0
Therefore:
W2  W1  2  δdisp2
8
δdisp2  2.628  mm
W2  80.3 mm
Problem
9.66
Problem 9.51
[Difficulty: 3]
9.51
Given:
Laboratory wind tunnel has flexible wall to accomodate BL growth. BL's are well
represented by 1/7-power profile. Information at two stations are known:
The given or available data (Table A.9) is
U  90
ft
s
Find:
H1  1  ft
W1  1  ft
δ1  0.5 in
δ6  0.65 in
ν  1.57  10
 4 ft

2
ρ  0.00238 
s
slug
ft
3
(a) Height of tunnel walls at section 6.
(b) Equivalent length of flat plate that would produce the inlet BL
(c) Estimate length of tunnel between stations 1 and 6.
Solution:
Basic
equations:
Assumptions:
 


 ρ dV   ρ V dA  0

t 
(Continuity)
(1) Steady flow
(2) Turbulent, 1/7-power velocity profile in boundary layer
(3) z = constant
(4) p = constant
Applying continuity between 1 and 6:
A1  U1  A6  U6
where A is the effective flow area. The velocities at 1 and 6 must be
equal since pressure is constant. In terms of the duct dimensions:
W1  2 δdisp1H1  2 δdisp1  W1  2 δdisp6 H6  2 δdisp6
solving for the height at 6:
H6 
W1  2 δdisp1H1  2 δdisp1
W1  2 δdisp6
The displacement thickness is determined from:

δdisp  δ 


1
 2  δdisp6
1
 1  u  dη


U

u
where
U
η
7
η
y
δ
0
Substituting the velocity profile and valuating the integral:



δdisp  δ 

1
0
1


δ
7
 1  η  dη 
δdisp1  0.0625 in
Therefore:
8
We may now determine the height at 6:
δdisp6  0.0813 in
H6  1.006  ft
1
For a flat plate turbulent boundary layer with 1/7-power law profile: δ1  L1 
5
1
4
 δ1   U  4
L1  
  
 0.382   ν 
0.382
1
Re1
 ν
 0.382  
4
5
5
  L1 Solving for L1:
 U
5
L1  1.725  ft
To estimate the length between 1 and 6, we determine length necessary to build the BL at section 6:
5
1
4
L6 
 δ6   U  4

     2.394  ft
 0.382   ν 
Therefore, the distance between 1 and 6 is:
L  L6  L1
L  0.669  ft
Problem 9.52
Problem
9.68
[Difficulty: 3]
9.52
Given:
Data on flow in a duct
Find:
Velocity at location 2; pressure drop; length of duct; position at which boundary layer is 20 mm
Solution:
The given data is
D  6  in
δ1  0.4 in
Table A.9
ρ  0.00234 
slug
ft
Governing
equations
Mass
In the boundary layer
δ
x

ν  1.56  10
3
ft
U1  80
s
δ2  1.2 in
0.382
 4 ft

2
s
(9.26)
1
5
Rex
In the the inviscid core, the Bernoulli equation holds
p
ρ
2

V
2
 g  z  constant
(4.24)
Assumptions: (1) Steady flow
(2) No body force (gravity) in x direction
For a 1/7-power law profile, from Example 9.4 the displacement thickness is
Hence
δ
δdisp 
8
δ1
δdisp1 
8
δdisp1  0.0500 in
δ2
δdisp2 
8
δdisp2  0.1500 in
From the definition of the displacement thickness, to compute the flow rate, the uniform flow at locations 1
and 2 is assumed to take place in the entire duct, minus the displacement thicknesses


π
2
A1   D  2  δdisp1
4
A1  0.1899 ft
2


π
2
A2   D  2  δdisp2
4
2
A2  0.1772 ft
Mass conservation (Eq. 4.12) leads to U2
ρ U1 A1  ρ U2 A2  0
or
A1
U2  U1 
A2
ft
U2  85.7
s
The Bernoulli equation applied between locations 1 and 2 is
p1
ρ
or the pressure drop is

U1
2

2
p2
ρ

U2
2
2
ρ
2
2
p 1  p 2  Δp    U2  U1 


2
Δp  7.69  10
3
 psi (Depends on ρ value selected)
The static pressure falls continuously in the entrance region as the fluid in the central core accelerates into a decreasing core.
If we assume the stagnation pressure is atmospheric, a change in pressure of about 0.008 psi is not significant; in addition, the
velocity changes by about 5%, again not a large change to within engineering accuracy
To compute distances corresponding to boundary layer thicknesses, rearrange Eq.9.26
1
δ
x

0.382
1
Rex


U

x


 0.382  
ν
5
5
so
x
 δ 


 0.382 
4
1
 
U

 ν
5
Applying this equation to locations 1 and 2 (using U = U1 or U2 as approximations)
5
1
4
4
 δ1   U1 
x1  
  
 0.382   ν 
For location 3
x 1  1.269  ft
5
1
 δ2   U2 
x2  
  
 0.382   ν 
4
4
x 2  x 1  3.83 ft
(Depends on ν value selected)
δ3  0.6 in
δ3
δdisp3 
8

x 2  5.098  ft

δdisp3  0.075  in
π
2
A3   D  2  δdisp3
4
A3  0.187  ft
A1
U3  U1 
A3
ft
U3  81.4
s
5
1
 δ3   U2 
x3  
  
 0.382   ν 
4
4
x 3  x 1  0.874  ft
(Depends on ν value selected)
x 3  2.143  ft
2
4
Problem 9.53
Problem
9.70
[Difficulty: 3]
9.53
9.4
Given:
Data on a large tanker
Find:
Cost effectiveness of tanker; compare to Alaska pipeline
Solution:
The given data is
L  360  m
B  70 m
D  25 m
kg
ρ  1020
U  6.69
3
m
s
m
4
P  1.30  10  hp
(Power consumed by drag)
P  9.7 MW
The power to the propeller is
P
Pprop 
70 %
Pprop  1.86  10  hp
The shaft power is
Ps  120% Pprop
Ps  2.23  10  hp
The efficiency of the engines is
η  40 %
Hence the heat supplied to the engines is
Q 
t 
The journey time is
Ps
4
4
8 BTU
Q  1.42  10 
η
x
hr
t  134  hr
U
10
Qtotal  Q t
The total energy consumed is
x  2000 mi
Qtotal  1.9  10  BTU
From buoyancy the total ship weight equals the displaced seawater volume
M ship g  ρ g  L B D
9
M ship  ρ L B D
M ship  1.42  10  lb
Hence the mass of oil is
M oil  75% M ship
M oil  1.06  10  lb
The chemical energy stored in the petroleum is
q  20000 
E  q  M oil
The total chemical energy is
The equivalent percentage of petroleum cargo used is then
9
BTU
lb
13
E  2.13  10  BTU
Qtotal
E
The Alaska pipeline uses
epipeline  120 
BTU
but for the
ton mi ship
The ship uses only about 15% of the energy of the pipeline!
 0.089  %
eship 
Qtotal
M oil x
eship  17.8
BTU
ton mi
Problem 9.54
Problem
9.72
[Difficulty: 4]
9.54
Given:
Laminar (Blasius) and turbulent (1/7-power) velocity distributions
Find:
Plot of distributions; momentum fluxes
Solution:
δ
The momentum flux is given by
Using the substitutions
the momentum flux becomes

2
mf   ρ u dy

u
U
per unit width of the boundary layer
0
y
 f ( η)
δ
η
1

2
mf  ρ U  δ  f ( η ) dη

2
0
For the Blasius solution a numerical evaluation (a Simpson's rule) of the integral is needed
2 Δη 
2
2
2
2
mflam  ρ U  δ
 f η0  4  f η1  2  f η2  f ηN 


3
 
 
 
 
where Δη is the step size and N the number of steps
The result for the Blasius profile is
2
mflam  0.525  ρ U  δ
1
For a 1/7 power velocity profile

2

2 
7
mfturb  ρ U  δ  η dη

0
7
2
mfturb   ρ U  δ
9
The laminar boundary has less momentum, so will separate first when encountering an adverse pressure gradient. The
computed results were generated in Excel and are shown below:
(Table 9.1) (Simpsons Rule)
η
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
1.00
Laminar Weight Weight x
u/U
0.000
0.166
0.330
0.487
0.630
0.751
0.846
0.913
0.956
0.980
0.992
w
1
4
2
4
2
4
2
4
2
4
1
Simpsons':
y /δ = η
2
(u/U )
0.00
0.11
0.22
0.95
0.79
2.26
1.43
3.33
1.83
3.84
0.98
0.525
0.0
0.0125
0.025
0.050
0.10
0.15
0.2
0.4
0.6
0.8
1.0
t
u/U
0.00
0.53
0.59
0.65
0.72
0.76
0.79
0.88
0.93
0.97
1.00
Laminar and Turbulent Boundary Layer
Velocity Profiles
0.75
y /δ
0.50
Laminar
Turbulent
0.25
0.00
0.00
0.25
0.50
0.75
u/U
1.00
Problem 9.55
(Difficulty 1)
9.55 A fluid flows enters the plane-wall diffuser that has an entrance area of 𝐴0 at a velocity of 𝑈0 . (a)
Assuming the fluid is inviscid, determine the velocity gradient
𝑑𝑑
𝑑𝑑
in terms of 𝑈0 and 𝐴0 for a value of
𝜙 = 0 degrees and 𝜙 = 20 degrees. (b) Assuming a real viscous fluid, what is the effect of the boundary
layers on the pressure gradient? In which assumption (inviscid or viscous) will the computed exit
pressure be highest? Explain your answer.
Find: Velocity gradients
Assumption: flow is uniform at a cross-section, steady, and incompressible
Solution: Use the continuity and boundary layer relations
(a) When the value of 𝜙 = 0 degrees, this is straight channel, the cross section area will be
constant. From the continuity equation we have:
𝑑𝑑
=0
𝑑𝑑
When the value of 𝜙 = 20 degrees, the cross section area increases:(Assume at the inlet x=0)
𝑄 = 𝑈0 𝐴0 = 𝑈𝑈
𝐴0
𝐷0 = �
1
𝜋
4
𝐷 = 𝐷0 + 2𝑥 tan 𝜙
2
Thus
1
1
1
4𝐴0
𝐴 = 𝜋𝐷 2 = 𝜋(𝐷0 + 2𝑥 tan 𝜙)2 = 𝜋 ��
+ 2𝑥 tan 𝜙�
4
4
4
𝜋
𝑈=
𝑑𝑑
=−
𝑑𝑑
𝑈0 𝐴0
𝑈0 𝐴0
4𝑈0 𝐴0
=
=
2
2
𝐴
4𝐴0
4𝐴0
1
�
�
𝜋�
+ 2𝑥 tan 𝜙�
𝜋�
+ 2𝑥 tan 𝜙�
𝜋
𝜋
4
16𝑈0 𝐴0
4𝐴
𝜋 �� 0 + 2𝑥 tan 𝜙�
𝜋
3 tan 𝜙
=−
16𝑈0 𝐴0
4𝐴
𝜋 �� 0 + 2𝑥 tan 20°�
𝜋
3 tan 20°
(a) When the value of 𝜙 = 0 degrees, for viscous flow, the boundary layer will grow and
decrease the effective flow area. As a result, the velocity will increase and the pressure will
decrease.
When the value of 𝜙 = 20 degrees, the growth rate of the boundary layer will be greater
and the pressure will decrease faster.
The inviscid fluid will have the higher exit pressure because the pressure for the viscous
fluid will decrease for all the value of 𝜙 when the boundary layer grows.
Problem
9.74
Problem 9.56
[Difficulty: 3]
9.56
Given:
u
Laminar boundary layer with velocity profile
U
2
 a  b  λ  c λ  d  λ
3
λ
y
δ
Separation occurs when shear stress at the surface becomes zero.
Find:
(a) Boundary conditions on the velocity profile at separation
(b) Appropriate constants a, b, c, d for the profile
(c) Shape factor H at separation
(d) Plot the profile and compare with the parabolic approximate profile
Solution:
Basic
equations:
u
U
y
 
δ δ
 2  
y
2
(Parabolic profile)
The boundary conditions for the separation profile are:
The velocity gradient is defined as:
du
dy
Applying the boundary conditions:

at y  0
u0
τ  μ
du
at y  δ
uU
τ  μ
du


dy
dy
0
Four boundary
conditions for four
coefficients a, b, c, d
0
U d  u  U
2
      b  2  c λ  3  d  λ
δ
δ dλ  U 
y0 λ0
u
U
du
dy
2
3
 a  b  0  c 0  d  0  0

The velocity profile and gradient may now be written as:

δ
U
 b  2  c 0  3  d  0
u
 c λ  d  λ
U
2
0
2
du
3

dy
U
δ
Therefore:
a0
Therefore:
b0

 2  c λ  3  d  λ

2
Applying the other boundary conditions:
yδ λ1
u
U
du
dy
The velocity profile is:
u
U
δdisp
δ



1
0
1

2
 3 λ  2 λ
3
2
3
 c 1  d  1  1

U
δ

 2  c 1  3  d  1
0
2

1

δdisp
θ

c3
d  2
δdisp δ

δ θ



2
3
2
3
  3  λ  2  λ  1  3  λ  2  λ dλ Expanding out the
δ 0
integrand yields:
θ

9
4
9
1
2
3
4
5
6

2

  3  λ  2  λ  9  λ  12 λ  4  λ dλ  1 
5
7
70
δ 0
2
θ
2 c  3 d  0
H
The shape parameter is defined as:
1  3λ2  2λ3 dλ  1  1  12  12
Solving this system
of equations yields:
cd1
Thus
H 
1
2

70
9
H  3.89
The two velocity profiles are plotted here:
Height y/δ
1
0.5
Separated
Parabolic
0
0
0.5
Velocity Distribution u/U
1
Problem 9.57
Problem
9.75
[Difficulty: 4]
9.57
Discussion: Shear stress decreases along the plate because the freestream flow speed remains constant while the
boundary-layer thickness increases.
The momentum flux decreases as the flow proceeds along the plate. Momentum thickness θ (actually proportional
to the defect in momentum within the boundary layer) increases, showing that momentum flux decreases. The
forct that must be applied to hold the plate stationary reduces the momentum flux of the stream and boundary
layer.
The laminar boundary layer has less shear stress than the turbulent boundary layer. Therefore laminar boundary
layer flow from the leading edge produces a thinner boundary layer and less shear stress everywhere along the
plate than a turbulent boundary layer from the leading edge.
Since both boundary layers continue to grow with increasing distance from the leading edge, and the turbulent
boundary layer continues to grow more rapidly because of its higher shear stress, this comparison will be the
same no matter the distance from the leading edge.
Problem 9.58
Problem
9.76
[Difficulty: 5]
9.58
Given:
Laboratory wind tunnel has fixed walls. BL's
are well represented by 1/7-power profile.
Information at two stations are known:
The given or available data (Table A.9) is
ft
U1  80
s
dp
dx
 0.035 
Find:
H1  1  ft
in H2 O
W1  1  ft
L  10 in
in
δ1  0.4 in
ν  1.62  10
 4 ft

2
ρ  0.00234 
s
slug
ft
3
(a) Reduction in effective flow area at section 1
(b) dθ/dx at section 1
(c) θ at section 2
Solution:
Basic
equations:
 


 ρ dV   ρ V dA  0

t 
τw
ρ
Assumptions:


(Continuity)

2
d
d 
U  θ  δdisp U  U 
dx
 dx 
(Momentum integral equation)
(1) Steady flow
(2) Turbulent, 1/7-power velocity profile in boundary layer
(3) z = constant
The percent reduction in flow area at 1 is given as:
The displacement thickness is determined from:
Aeff  A
A

δdisp  δ 


1
0
Substituting the velocity profile and valuating the integral:

W1  2 δdisp H1  2 δdisp  W1 H1
W1  H1
1
 1  u  dη


U




δdisp  δ 

1
0
u
where
U
1


δ
7
 1  η  dη 
8
η
7
η
Therefore:
Thus:
y
δ
δdisp1  0.0500 in
Aeff  A
A
 1.66 %
Solving the momentum integral equation for the momentum thickness gradient:
1
At station 1:
τw1
ρ U1
2
 0.0233 

U1  δ1 


L


1


u
θ  u 

  1   dη  

U
δ  U 
0

ν
4
0.0233 
8
Solving for the velocity gradient:
1
2

τw
2
 ( H  2)
ρ U
θ dU

U dx
4
  2.057  10 3
U1  δ1 


ν
δdisp1
7
Thus: θ1 
 δ1  0.0389 in H 
 1.286
72
θ1
72
9
0
p
dx
1
2
 1
 7
7
7
7
7
 η  η  dη   
Now outside the boundary layer
dθ
2
 ρ U  constant
1 dU


U dx
1

dp
2 dx
ρ U
from the Bernoulli equation. Then:
 0.1458
1
ft
dp
dx
 ρ U
dx
Substituting all of this information into the above expression:
dθ
dx
We approximate the momentum thickness at 2 from:
dU
dθ
θ2  θ1 
L
dx
 4.89  10
4
 0.00587 
in
ft
θ2  0.0438 in
Problem 9.59
Problem
9.78
[Difficulty: 3]
9.59
Given:
Barge pushed upriver
L  80 ft B  35 ft
Find:
2
 5 ft
D  5  ft
From Table A.7: ν  1.321  10

s
ρ  1.94
slug
ft
3
Power required to overcome friction; Plot power versus speed
Solution:
CD 
Basic
equations:
FD
1
2
CD 
(9.32)
2
 ρ U  A
From Eq. 9.32
1
2
FD  CD A  ρ U
2
The power consumed is
P  FD U
0.455
logReL
2.58
A  L ( B  2  D)
and
1
3
P  CD A  ρ U
2

1610
(9.37b)
ReL
A  3600 ft
Re L
CD
P (hp)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
9.70E+06
1.94E+07
2.91E+07
3.88E+07
4.85E+07
5.82E+07
6.79E+07
7.76E+07
8.73E+07
9.70E+07
1.07E+08
1.16E+08
1.26E+08
1.36E+08
1.45E+08
0.00285
0.00262
0.00249
0.00240
0.00233
0.00227
0.00222
0.00219
0.00215
0.00212
0.00209
0.00207
0.00205
0.00203
0.00201
0.0571
0.421
1.35
3.1
5.8
9.8
15
22
31
42
56
72
90
111
136
150
120
P (hp) 90
60
30
0
6
ν
The calculated results and the plot were generated in Excel:
U (mph)
3
U L
2
Power Consumed by Friction on a Barge
0
ReL 
9
U (mph)
12
15
Problem 9.60
Problem
9.80
[Difficulty: 3]
9.60
Given:
Towboat model at 1:13.5 scale to be tested in towing tank.
Lm  3.5 m
Find:
Bm  1  m
d m  0.2 m
m
Up  7  knot  3.601
s
Disp m  5500 N
(a) Estimate average length of wetted surface on the hull
(b) Calculate skin friction drag force on the prototype
Solution:
Basic
equations:
1
2
FD  CD  ρ U  A
2
CD 
(Drag)
0.455
logReL
2.58

1610
(Drag Coefficient)
ReL
We will represent the towboat as a rectangular solid of length L av, with the displacement of the boat. From buoyancy:
W  ρ g  V  ρ g  Lav Bm d m thus:
For the prototype:
Lav 
W
ρ g  Bm d m
Lp  13.5 Lav
Lp  37.9 m
ReL 
The Reynolds number is:
Lav  2.80 m
Up  Lp
ReL  1.36  10
ν
8
This flow is predominantly turbulent, so we will use a turbulent analysis.
The drag coefficient is: CD 
The area is:
2

0.455
logReL
2.58


1610
ReL
 0.00203
2
A  13.5  Lav Bm  2  d m  716 m
The drag force would then be:
1
2
FD  CD  ρ Up  A
2
FD  9.41 kN
This is skin friction only.
Problem 9.61
(Difficulty 2)
𝛿
𝑥
9.61 Plot the local friction coefficient 𝑐𝑓 , the boundary layer thickness ratio , and the drag coefficient 𝐶𝑓
for both laminar and turbulent boundary layer on a flat plate for 𝑅𝑥 from 0 to 500000, assuming in the
turbulent case that the layer is tripped at the leading edge and so is fully turbulent along the length of
the plate. Discuss the ratio of drag force as a function of 𝑅𝑥 .
Find: Friction coefficient, boundary layer thickness, and drag coefficient
Solution: Use the boundary layer relations
For laminar boundary layer, the friction coefficient, boundary layer thickness, and drag coefficient
are given by
𝑐𝑓 =
0.664
𝐶𝑓 =
For turbulent flow, these are given by
𝐶𝑓 =
The Reynolds number is given as
1.33
�𝑅𝑥
𝛿 5.48
=
𝑥 �𝑅𝑥
𝑐𝑓 =
The drag force is calculated as:
�𝑅𝑥
0.0594
1
(𝑅𝑥 )5
0.0742
1
(𝑅𝑥 )5
𝛿 0.382
=
1
𝑥
(𝑅𝑥 )5
1
𝐹𝐷 = 𝐶𝑓 × 𝜌𝜌𝑉 2
2
With all these we have:
𝑅𝑥 =
𝜌𝜌𝜌
𝜇
Drag forces appear to be higher for laminar flow at low 𝑅𝑥 , however the correlations used to
determine 𝐶𝑓 break down at low 𝑅𝑥 (< 5x105) As a general rule, turbulent boundary layers produces
more drag.
Problem 9.62
(Difficulty 2)
9.62 A smooth plate 3 𝑚 long and 0.9 𝑚 wide moves through still sea level air at 4.5
𝑚
.
𝑠
Assuming the
boundary layer to be wholly laminar, calculate (a) the thickness of the layer at 0.5, 1.0, 1.5, 2.0, 2.5, and
3.0 𝑚 from the leading edge of the plate; (b) the shear stress,𝜏0 at those points. (c) the total drag force
on the side of the plate. (d) Calculate the thickness at the above points if the layer is turbulent. (e)
Calculate the total drag for the turbulent boundary layer. (f) What percentage saving in drag is effected
by a laminar boundary layer?
Find: The boundary layer properties
Assumptions: The flow is steady and without a pressure gradient
Solution: Use the relations for laminar and turbulent boundary layers on a flat plate.
For laminar flow, the thickness of layer is given by:
5
𝜈𝑥
𝑥 = 5�
𝑈
�𝑅𝑅𝑥
𝛿=
For the friction coefficient:
𝐶𝑓 =
𝜏0
0.664
=
1 2 �𝑅𝑅
𝑥
𝜌𝑈
2
For turbulent flow the boundary layer thickness is given by:
𝛿=
And for the friction coefficient
𝜈 1�5 4�
𝑥
=
0.382
�
� 𝑥 5
1
𝑈
5
0.382
(𝑅𝑅𝑥 )
𝐶𝑓 =
For sea level air we have:
0.0594
1�
5
𝑅𝑅𝑥
𝜌 = 1.225
𝑘𝑘
𝑚3
𝜇 = 1.789 × 10−5 𝑃𝑃 ∙ 𝑠
The Reynolds number, for either laminar or turbulent flow, is given as:
𝑅𝑅𝑥 =
𝜌𝜌𝜌
1.225 × 4.5
=
𝑥 = 314420𝑥
𝜇
1.789 × 10−5
So we have the results as listed in the table:
𝑅𝑅𝑥
x (m)
157000
0.5
314000
1
471000
1.5
628000
2
785000
2.5
942000
3
Laminar 𝛿
(m)
0.0063
𝜏0 (Pa)
0.0208
Turbulent 𝛿
(m)
0.0175
0.0089
0.0147
0.0304
0.0109
0.012
0.042
0.0126
0.0104
0.0529
0.0141
0.0093
0.0632
0.0155
0.0085
0.0732
For the total shear stress for laminar flow we use the relation for drag coefficient:
𝐶𝐷 =
And for turbulent flow
𝐹𝐷
1 2
𝜌𝑈 𝐴
2
𝐶𝐷 =
The total drag force for laminar flow is:
=
1.33
�𝑅𝑅𝐿
0.0742
1�
5
𝑅𝑅𝐿
1.33
1
𝑘𝑘
𝑚 2
1
× × 1.225 3 × �4.5 � × 3𝑚 × 0.9 𝑚 = 0.0469 𝑁
𝐹𝐷𝐷 = 𝐶𝐷 𝜌𝑈 2 𝐴 =
𝑚
𝑠
2
√942000 2
Similarly for turbulent flow we have the total drag force as:
1
𝐹𝐷𝐷 = 𝐶𝐷 𝜌𝑈 2 𝐴 =
2
The savings is:
0.0742
942000
𝑆𝑆𝑆𝑆𝑆𝑆𝑆 =
1�
5
1
𝑘𝑘
𝑚 2
× × 1.225 3 × �4.5 � × 3𝑚 × 0.9 𝑚 = 0.157 𝑁
2
𝑚
𝑠
𝐹𝐷𝐷 − 𝐹𝐷𝐷 0.157 𝑁 − 0.0469 𝑁
=
= 70.1%
0.157 𝑁
𝐹𝐷𝐷
Problem 9.63
Problem
9.82
[Difficulty: 3]
9.63
Given:
Towboat model at 1:13.5 scale to be tested in towing tank.
Lm  7.00 m
Find:
Bm  1.4 m
d m  0.2 m
Vp  10 knot
(a) Model speed in order to exhibit similar wave drag behavior
(b) Type of boundary layer on the prototype
(c) Where to place BL trips on the model
(d) Estimate skin friction drag on prototype
Solution:
Basic
equations:
1
2
FD  CD  ρ V  A
2
(Drag)
Vm
The test should be conducted to match Froude numbers:
Rep 
The Reynolds number is:
g  Lm
A  L ( B  2  d )
0.0594
ReL
Therefore
CDm 
0.0743
0.2
0.2

0.2
 2.97  10
3
Rem
For the prototype:
CDp 
Vm  2.72 knot
8
Rep  4.85  10
ν
5
Ret  5  10
so
xt
L

Ret
Rep
 0.00155
x t  0.0109 m
We calculate the drag coefficient from turbulent BL theory:
0.0743
ReL
g  Lp
x t  0.00155  Lm
Thus the location of transition would be:
CD  1.25 Cf  1.25 
Lm
Vm  Vp 
Lp
Vp
Vp  Lp
Therefore the boundary layer is turbulent. Transition occurs at
The wetted area is:

0.455
logRep
2.56
For the model: Lm  7 m
Rem 
Vm Lm
ν
6
2
 9.77  10 Am  12.6 m
1
2
and the drag force is: FDm  CDm  ρ Vm  Am
2

1610
Rep
CDp  1.7944  10
3
FDm  36.70 N
3
2
Ap  2.30  10  m
1
2
FDp  CDp  ρ Vp  Ap
2
FDp  54.5 kN
Problem 9.64
Problem
9.84
[Difficulty: 4]
9.64
Given:
Nuclear submarine cruising submerged. Hull approximated by circular cylinder
L  107  m
Find:
D  11.0 m
V  27 knot
(a) Percentage of hull length for which BL is laminar
(b) Skin friction drag on hull
(c) Power consumed
Solution:
Basic
equations:
1
2
FD  CD  ρ V  A
2
5
Transition occurs at Ret  5  10
(Drag)
so the location of transition would be:
xt
L

Ret ν
xt
V L
L
 0.0353%
We will therefore assume that the BL is completely turbulent.
The Reynolds number at x = L is:
The wetted area of the hull is:
ReL 
V L
ν
 1.42  10
9
For this Reynolds number:
CD 
3
0.455
logReL
2.58
 1.50  10
2
A  π D L  3698 m
So the drag force is:
The power consumed is:
1
2
FD  CD  ρ V  A
2
P  FD V
5
FD  5.36  10 N
P  7.45 MW
Problem 9.65
Problem
9.85
[Difficulty: 3]
9.65
Given:
Racing shell for crew approximated by half-cylinder:
L  7.32 m
Find:
D  457  mm
V  6.71
m
s
(a) Location of transition on hull
(b) Thickness of turbulent BL at the rear of the hull
(c) Skin friction drag on hull
Solution:
Basic
equations:
1
2
FD  CD  ρ V  A
2
5
Transition occurs at Ret  5  10
so the location of transition would be:
δ
For the turbulent boundary layer
x
The Reynolds number at x = L is:
The wetted area of the hull is:
A 
(Drag)

0.382
Rex
ReL 
π D
2
0.2
V L
ν
Therefore δ 
x t  0.0745 m
V
0.2
so the BL thickness is:
δ 
0.382  L
ReL
2
 L  5.2547 m
So the drag force is:
7
Ret ν
0.382  L
ReL
 4.91  10
xt 
For this Reynolds number:
1
2
FD  CD  ρ V  A
2
Note that the rowers must produce an average power of
CD 
δ  0.0810 m
0.2
3
0.455
logReL
2.58
 2.36  10
FD  278 N
P  FD V  1.868  kW to move the shell at this speed.
Problem 9.66
(Difficulty 1)
9.66 The drag coefficient of a circular disk when place normal to the flow is 1.12. Calculate the force and
power necessary to derive a 0.3 𝑚 disk at 48
𝑘𝑘
ℎ
through (a) standard air at sea level and (b) water.
Find: The force and power for air and water
Solution: Use the drag coefficient to find the force and power
The drag coefficient is:
𝐶𝐷 = 1.12
The drag force is:
1
𝐹𝐷 = 𝐶𝐷 × 𝜌𝜌𝑉 2
2
The power is:
𝑃 = 𝐹𝐷 𝑉
The velocity is:
𝑉 = 48
The area is:
𝐴=
𝑚
𝑘𝑘
= 13.33
𝑠
ℎ
𝜋 2 𝜋
𝐷 = × (0.3 𝑚)2 = 0.0707 𝑚2
4
4
(a) For standard air at sea level:
𝜌 = 1.225
𝑘𝑘
𝑚3
1
𝑘𝑘
𝑚 2
𝐹𝐷 = 1.12 × × 1.225 3 × 0.0707 𝑚2 × �13.33 � = 8.62 𝑁
2
𝑚
𝑠
(b) For water:
𝑃 = 𝐹𝐷 𝑉 = 8.62 𝑁 × 13.33
𝜌 = 998
𝑚
= 114.9 𝑊
𝑠
𝑘𝑘
𝑚3
1
𝑘𝑘
𝑚 2
𝐹𝐷 = 1.12 × × 998 3 × 0.0707 𝑚2 × �13.33 � = 7021 𝑁
2
𝑚
𝑠
𝑃 = 𝐹𝐷 𝑉 = 7021 𝑁 × 13.33
𝑚
= 93.6 𝑘𝑘
𝑠
Problem 9.67
(Difficulty 1)
9.67 A steel sphere of 0.25 𝑖𝑖 diameter has a velocity of 200
Standard Atmosphere. Calculate the drag force on this sphere.
𝑓𝑓
𝑠
at an altitude of 30000 𝑓𝑓 in the U.S.
Find: The drag force on the sphere
Assumption: The sphere velocity changes slowly enough so that the flow can be considered steady
Solution: Use the relations for drag coefficient
For the sphere, the drag coefficient is defined as
𝐶𝐷 =
𝐹𝐷
1 2
𝜌𝑈 𝐴
2
Where the area is the frontal area of the sphere. The drag coefficient is a function of Reynolds
number, as shown in Figure 9.11. At 30000 𝑓𝑓 attitude we have:
𝜌 = 0.000891
𝜇 = 3.107 × 10−7
The Reynolds number is:
𝑅𝑅 =
𝜌𝜌𝜌
=
𝜇
𝑠𝑠𝑠𝑠
𝑓𝑓 3
0.000891
From the figure 9.11 we have:
The drag force is calculated as:
𝑙𝑙𝑙 ∙ 𝑠
𝑓𝑓 2
𝑓𝑓
𝑙𝑙𝑙 ∙ 𝑠2
× 200 × 0.0208 𝑓𝑓
𝑠
𝑓𝑓 4
= 1.193 × 104
𝑙𝑙𝑙
∙
𝑠
3.107 × 10−7
𝑓𝑓 2
𝐶𝐷 = 0.4
1
𝑠𝑠𝑠𝑠
𝑓𝑓 2 𝜋
𝜌 2
𝐹𝐷 = 𝐶𝐷 𝑈 𝐴 = 0.4 × × 0.000891
× �200 � × × (0.0208 𝑓𝑓)2 = 0.00242 𝑙𝑙𝑙
2
𝑓𝑓 3
𝑠
4
2
Problem 9.68
(Difficulty 2)
9.68 A steel sphere (𝑠. 𝑔. = 7.8) of 13 𝑚𝑚 diameter falls at a constant velocity of 0.06
𝑚
𝑠
through an oil
(𝑠. 𝑔. = 0.90). Calculate the viscosity of the oil, assuming that the fall occurs in a large tank.
Assumption: The velocity is very low
Find: The oil viscosity
Solution: Use the drag coefficient to find the viscosity
For the drag force we have for low Reynolds number:
The force balance in the 𝑦 direction is:
𝐹𝐷 = 3𝜋𝜋𝜋𝜋
𝐹𝐷 − 𝑊 + 𝐹𝐵 = 0
For the buoyant force and gravity we have:
𝐹𝐵 = 𝛾𝑜𝑜𝑜 𝑉𝑉𝑉
𝑊 = 𝛾𝑠𝑠𝑠𝑠𝑠 𝑉𝑉𝑉
𝑉𝑉𝑉 =
Thus
𝜋 3
𝑑
6
𝜋
𝜋
3𝜋𝜋𝜋𝜋 − 𝛾𝑠𝑠𝑠𝑠𝑠 𝑑 3 + 𝛾𝑜𝑜𝑜 𝑑 3 = 0
6
6
𝜋
𝜋
𝜋
𝛾𝑠𝑠𝑠𝑠𝑠 𝑑 3 − 𝛾𝑜𝑜𝑜 𝑑 3 (𝑆𝑆𝑠𝑠𝑠𝑠𝑠 − 𝑆𝑆𝑜𝑜𝑜 )𝛾𝑤𝑤𝑤𝑤𝑤 𝑑 2
6
6
6
𝜇=
=
3𝜋𝜋𝜋
3𝜋𝜋
𝜇=
𝑁 𝜋
× × (0.013 𝑚)2
𝑚3 6
= 10.59 𝑃𝑃 ∙ 𝑠
𝑚
3𝜋 × 0.06
𝑠
(7.8 − 0.9) × 9810
To check the validation for drag force equation, the Reynolds number is:
𝑘𝑘
𝑚
𝜌𝜌𝜌 998 𝑚3 × 0.06 𝑠 × 0.013 𝑚
𝑅𝑅 =
=
= 0.0735 ≪ 1
𝜇
10.59 𝑃𝑃 ∙ 𝑠
So the fluid flow is low Reynolds number flow, the computation is valid.
Problem 9.69
Problem
9.86
[Difficulty: 3]
9.69
Given:
Plastic sheet falling in water
Find:
Terminal speed both ways
Solution:
Basic
equations:
h  0.5 in
ΣFy  0
FD
CD 
for terminal
speed
1
2
W  4  ft
L  2  ft
SG  1.7
2
 ρ V  A
CD 
0.0742
(9.34) (assuming 5 x 105 < ReL < 107)
1
ReL
5
From Table A.8 at 70 oF ν  1.06  10
 5 ft

A  W L
2
ρ  1.94
s
slug
ft
3
A free body diagram of the sheet is shown here. Summing the forces in the vertical (y) direction:
FD  Fb  Wsheet  0
FD  Wsheet  Fb  ρ g  h  A ( SG  1 )
FD
Fb
Also, we can generate an expression for the drag coefficient in terms of the geometry of the sheet
and the water properties:
y
V
x
4
1
9
W sheet
1
0.0742
1
0.0742
2
2
2
5 5
5
FD  2  CD A  ρ V  2 
 A  ρ V 
 W L ρ V  0.0742 W L  ν  ρ V
1
1
2
2
ReL
5
 V L 


 ν 
(Note that we double FD because drag
acts on both sides of the sheet.)
5
5
9

Hence
ρH2O g  h  W ( SG  1 )  0.0742 W L
Check the Reynolds number
Repeating for
ReL 
ReL 
5
1
9
5
5
 ν  ρ V
Solving for V
V L
5
ν
1


 g h  ( SG  1)  L  5
V  
  
 0.0742
ν 
L  4  ft
Check the Reynolds number
1
V L
ν
1


 g h  ( SG  1)  L  5
ft
V  
    V  15.79 
s
 0.0742
ν 
ReL  2.98  10
6
Hence Eq. 9.34 is reasonable
6
Eq. 9.34 is still reasonable
9
V  17.06 
ft
s
ReL  6.44  10
The short side vertical orientation falls more slowly because the largest friction is at the region of the leading edge (τ tails
off as the boundary layer progresses); its leading edge area is larger. Note that neither orientation is likely - the plate will
flip around in a chaotic manner.
Problem 9.70
Problem
9.89
9.70
[Difficulty: 4]
7.5
Given:
"Resistance" data on a ship
Lp  130  m
Lm 
Find:
Lp
80
 1.625 m
ρ  1023
2
Ap  1800 m
Am 
Ap
80
2
kg
3
 3 N s
μ  1.08  10
m

2
m
2
 0.281 m
Plot of wave, viscous and total drag (prototype and model); power required by prototype
Solution:
Basic
equations:
CD 
FD
1
2
From Eq. 9.32
(9.32)
2
Fr 
U
gL
 ρ U  A
1
2
FD  CD A  ρ U
2
This applies to each component of the drag (wave and viscous) as well as to the total
The power consumed is
P  FD U
From the Froude number
U  Fr  gL
1
3
P  CD A  ρ U
2
The solution technique is: For each speed Fr value from the graph, compute U; compute the drag from
the corresponding "resistance" value from the graph. The results were generated in Excel and are shown
below:
Model
Fr
Wave
"Resistance"
Viscous
"Resistance"
0.10
0.20
0.30
0.35
0.40
0.45
0.50
0.60
0.00050
0.00075
0.00120
0.00150
0.00200
0.00300
0.00350
0.00320
0.0052
0.0045
0.0040
0.0038
0.0038
0.0036
0.0035
0.0035
Wave
Total
U (m/s)
Drag (N)
"Resistance"
0.0057
0.0053
0.0052
0.0053
0.0058
0.0066
0.0070
0.0067
0.40
0.80
1.20
1.40
1.60
1.80
2.00
2.40
0.0057
0.0344
0.1238
0.2107
0.3669
0.6966
1.0033
1.3209
Viscous
Drag (N)
0.0596
0.2064
0.4128
0.5337
0.6971
0.8359
1.0033
1.4447
Total
Power (W)
Drag (N)
0.0654
0.2408
0.5366
0.7444
1.0640
1.5324
2.0065
2.7656
Drag on a Model Ship
3.0
2.5
Total
Wave
Viscous
2.0
F (N)
1.5
1.0
0.5
0.0
0.0
0.5
1.0
1.5
2.0
U (m/s)
2.5
3.0
2.5
3.0
Power Requirements for a Model Ship
7.0
6.0
5.0
P (W)
4.0
3.0
2.0
1.0
0.0
0.0
0.5
1.0
1.5
2.0
U (m/s)
0.0261
0.1923
0.6427
1.0403
1.6993
2.7533
4.0057
6.6252
Prototype
Fr
Wave
"Resistance"
Viscous
"Resistance"
0.10
0.20
0.30
0.35
0.40
0.45
0.50
0.60
0.00050
0.00075
0.00120
0.00150
0.00200
0.00300
0.00350
0.00320
0.0017
0.0016
0.0015
0.0015
0.0013
0.0013
0.0013
0.0013
Total
U (m/s)
"Resistance"
0.0022
0.0024
0.0027
0.0030
0.0033
0.0043
0.0048
0.0045
3.6
7.1
10.7
12.5
14.3
16.1
17.9
21.4
Wave
Drag
(MN)
0.0029
0.0176
0.0634
0.1079
0.1879
0.3566
0.5137
0.6763
Viscous
Drag (MN)
0.0100
0.0376
0.0793
0.1079
0.1221
0.1545
0.1908
0.2747
Total
Drag
(MN)
0.0129
0.0552
0.1427
0.2157
0.3100
0.5112
0.7045
0.9510
Drag on a Prototype Ship
1.0
F (MN)
0.8
Total
0.6
Wave
Viscous
0.4
0.2
0.0
0
5
10
15
U (m/s)
20
25
Power Required by a Prototype Ship
25000
20000
P (kW)
15000
10000
5000
0
0
5
10
15
U (m/s)
20
For the prototype wave resistance is a much more significant factor at high speeds! However, note that for
both scales, the primary source of drag changes with speed. At low speeds, viscous effects dominate, and so
the primary source of drag is viscous drag. At higher speeds, inertial effects dominate, and so the wave drag
is the primary source of drag.
25
Power
(kW)
Power
(hp)
46.1
394.1
1528.3
2696.6
4427.7
8214.7
12578.7
20377.5
61.8
528.5
2049.5
3616.1
5937.6
11015.9
16868.1
27326.3
Problem 9.71
Problem
9.90
[Difficulty: 2]
9.71
Given:
Flag mounted vertically
H  194  ft W  367  ft
Find:
V  10 mph  14.67 
ft
s
ρ  0.00234 
slug
ft
3
ν  1.62  10
 4 ft

2
s
Force acting on the flag. Was failure a surprise?
Solution:
Basic
equations:
1
2
FD  CD  ρ V  A
2
(Drag)
We should check the Reynolds number to be sure that the data of Fig. 9.10 are applicable:
ReW 
V W
ν
 3.32  10
7
(We used W as our length scale here since it is the lesser of the two dimensions of the flag.) Since the Reynolds number is less than
1000, we may use Figure 9.10 to find the drag coefficient.
4
The area of the flag is: A  H W  7.12  10  ft
So the drag force is:
2
AR 
W
H
 1.89
1
2
FD  CD  ρ V  A
2
From Fig. 9.10: CD  1.15
4
FD  2.06  10  lbf
This is a large force.
Failure should have
been expected.
Problem 9.72
(Difficulty 2)
9.72 What constant speed will be attained by a lead (𝑠. 𝑔. = 11.4) sphere of 0.5 𝑖𝑖. diameter failing
freely through an oil of kinematic viscosity 0.12
𝑓𝑓 2
𝑠
and (𝑠. 𝑔. = 0.95), if the fall occurs in a large tank?
Assumption: The velocity is very low
Find: The lead sphere speed
Solution: Use the drag coefficient to find the velocity
For the drag force we have for low Reynolds number:
𝐹𝐷 = 3𝜋𝜋𝜋𝜋
For the force balance in the 𝑦 direction we have:
𝐹𝐷 − 𝑊 + 𝐹𝐵 = 0
For the buoyant force and gravity we have:
𝐹𝐵 = 𝛾𝑜𝑜𝑜 𝑉𝑉𝑉
𝑊 = 𝛾𝑠𝑠ℎ𝑒𝑒𝑒 𝑉𝑉𝑉
𝑉𝑉𝑉 =
Thus
𝑉=
𝛾𝑠𝑠ℎ𝑒𝑒𝑒
3𝜋𝜋𝜋𝜋 − 𝛾𝑠𝑠ℎ𝑒𝑒𝑒
𝜋 3
𝑑
6
𝜋 3
𝜋
𝑑 + 𝛾𝑜𝑜𝑜 𝑑 3 = 0
6
6
𝜋 2
𝜋
𝜋
𝜋
𝑑 − 𝛾𝑜𝑜𝑜 𝑑 2 �𝑆𝑆𝑠𝑠ℎ𝑒𝑒𝑒 − 𝑆𝑆𝑜𝑜𝑜 �𝛾𝑤𝑤𝑤𝑤𝑤 𝑑 2 �𝑆𝑆𝑠𝑠ℎ𝑒𝑒𝑒 − 𝑆𝑆𝑜𝑜𝑜 �𝛾𝑤𝑤𝑤𝑤𝑤 𝑑 2
6
6
6
6
=
=
𝛾
3𝜋𝜋
3𝜋𝜋𝜋
3𝜋𝜋 𝑜𝑜𝑜
𝑔
𝑉=
2
𝑙𝑙𝑙 𝜋
0.5
×
×
�
𝑓𝑓�
𝑓𝑓
12
𝑓𝑓 3 6
= 0.27
𝑙𝑙𝑙
𝑠
0.95 × 62.4 3
𝑓𝑓 2
𝑓𝑓
3𝜋 × 0.12
×
𝑠
𝑓𝑓
32.18 2
𝑠
(11.4 − 0.95) × 62.4
To check the validation for drag force equation, the Reynolds number is:
𝑓𝑓 0.5
𝑉𝑉 0.27 𝑠 × 12 𝑓𝑓
𝑅𝑅 =
=
= 0.09375 ≪ 1
𝑓𝑓 2
𝑣
0.12
𝑠
So the fluid flow is low Reynolds number flow, the computation is valid.
Problem 9.73
(Difficulty 2)
9.73 Assuming a critical Reynolds number of 0.1, calculate the approximate diameter of the largest air
bubble which will obey Stoke’s law while rising through a large tank of oil of viscosity 0.19 𝑃𝑃 ∙ 𝑠 and
𝑆. 𝐺. 0.90.
Assumption: The velocity is very low
Find: The air bubble diameter
Solution: Use the drag coefficient to find the diameter
For the drag force we have for low Reynolds number:
𝐹𝐷 = 3𝜋𝜋𝜋𝜋
For the force balance in the 𝑦 direction we have:
−𝐹𝐷 − 𝑊 + 𝐹𝐵 = 0
For the buoyant force and gravity we have:
𝐹𝐵 = 𝛾𝑜𝑜𝑜 𝑉𝑉𝑉
𝑊 = 𝛾𝑎𝑎𝑎 𝑉𝑉𝑉
𝑉𝑉𝑉 =
Thus
−3𝜋𝜋𝜋𝜋 − 𝛾𝑎𝑎𝑎
𝑉=
𝛾𝑜𝑜𝑜
The Reynolds number is defined as:
𝜋 3
𝑑
6
𝜋 3
𝜋
𝑑 + 𝛾𝑜𝑜𝑜 𝑑 3 = 0
6
6
𝜋 2
𝜋
𝑑 − 𝛾𝑎𝑎𝑎 𝑑 2 (𝛾𝑜𝑜𝑜 − 𝛾𝑎𝑎𝑎 )𝑑 2
6
6
=
3𝜋𝜋
18𝜇
𝑅𝑅 =
𝜌𝜌𝜌 𝛾𝑜𝑜𝑜 𝑉𝑉
=
𝜇
𝑔𝑔
𝑉=
𝑔𝑔𝑔𝑔
𝛾𝑜𝑜𝑜 𝑑
Thus
𝑔𝑔𝑔𝑔 (𝛾𝑜𝑜𝑜 − 𝛾𝑎𝑎𝑎 )𝑑 2
=
𝛾𝑜𝑜𝑜 𝑑
18𝜇
18𝜇2 𝑔𝑔𝑔
𝑑 =
𝛾𝑜𝑜𝑜 (𝛾𝑜𝑜𝑜 − 𝛾𝑎𝑎𝑎 )
3
The specific weight of air compared to oil is negligible. The diameter is then.
𝑚
2
18𝜇2 𝑔𝑔𝑔 3 18 × (0.19 𝑃𝑎 ∙ 𝑠) × 9.81 𝑠 2 × 0.1
=�
= 0.002 𝑚
𝑑=�
𝛾𝑜𝑜𝑜 2
𝑁 2
�0.9 × 9810 3 �
𝑚
3
Problem 9.74
(Difficulty 2)
9.74 Glass spheres of 0.1 𝑖𝑖 diameter fall at constant velocities of 0.1 𝑎𝑎𝑎 0.05
𝑓𝑓
𝑠
through two different
oils (of the same specific gravity) in very large tanks. If the viscosity of the first oil is 0.002
the viscosity of the second?
Assumption: (1) Low Reynolds number flow
(2) The same specific gravity for two different oils
Find: The oil viscosity
Solution: Use the drag coefficient to find the viscosity
For the drag force we have for low Reynolds number:
𝐹𝐷 = 3𝜋𝜋𝜋𝜋
For the force balance in the 𝑦 direction we have:
𝐹𝐷 − 𝑊 + 𝐹𝐵 = 0
For the buoyance force and gravity we have:
𝐹𝐵 = 𝛾𝑜𝑜𝑜 𝑉𝑉𝑉
𝑊 = 𝛾𝑔 𝑉𝑉𝑉
Thus
𝑉𝑉𝑉 =
𝜋 3
𝑑
6
𝜋
𝜋
3𝜋𝜋𝜋𝜋 − 𝛾𝑔 𝑑 3 + 𝛾𝑜𝑜𝑜 𝑑 3 = 0
6
6
𝜋
𝜋
𝜋
𝛾𝑔 𝑑 3 − 𝛾𝑜𝑜𝑜 𝑑 3 �𝑆𝑆𝑔 − 𝑆𝑆𝑜𝑜𝑜 �𝛾𝑤𝑤𝑤𝑤𝑤 𝑑 2
6
6
6
𝜇=
=
3𝜋𝜋𝜋
3𝜋𝜋
𝑙𝑙𝑓∙𝑠
,
𝑓𝑓 2
what is
So we have:
𝜋
�𝑆𝑆𝑔 − 𝑆𝑆𝑜𝑜𝑜 �𝛾𝑤𝑤𝑤𝑤𝑤 𝑑 2
𝑓𝑓
6
𝑉2 0.05 𝑠
1
𝜇1
3𝜋𝑉1
=
=
=
=
𝜋
𝑓𝑓
2
𝜇2 �𝑆𝑆𝑔 − 𝑆𝑆𝑜𝑜𝑜 �𝛾𝑤𝑤𝑤𝑤𝑤 𝑑 2 𝑉1
0.1
6
𝑠
3𝜋𝑉2
𝜇2 = 2𝜇1 = 0.004
𝑙𝑙𝑙 ∙ 𝑠
𝑓𝑓 2
Problem 9.75
Problem
9.92
[Difficulty: 2]
9.75
Given:
Rotary mixer rotated in a brine solution
R  0.6 m
ω  60 rpm
d  100  mm SG  1.1
ρ  ρw SG
ρ  1100
kg
3
m
ν  1.05  1.55  10
Find:
2
6 m

s
2
6 m
 1.63  10

s
(a) Torque on mixer
(b) Horsepower required to drive mixer
Solution:
Basic
equations:
1
2
FD  CD  ρ V  A
2
(Drag)
T  2  R FD
(Torque)
P  T ω
(Power)
Assumptions: Drag on rods and motion induced in the brine can be neglected.
The speed of the disks through the brine is:
The area of one disk is:
A 
π
4
2
V  R ω  3.77
m
s
From Table 9.2: CD  1.17 for a disk.
2
 d  0.00785  m
So the drag force is:
1
2
FD  CD  ρ V  A  71.8 N
2
and the torque is: T  2  R FD
The power consumed to run the mixer is:
P  T ω  542 W
T  86.2 N m
P  0.726  hp
Problem 9.76
Problem
9.95
[Difficulty: 3]
9.76
Given:
Data on airplane landing
M  9500 kg
Find:
km
Vi  350 
hr
km
Vf  100 
hr
x f  1200 m CD  1.43 (Table 9.3)
Solution:
1
2
FD  CD  ρ V  A
2
(Drag)
Assumptions: (1) Standard air
(2) Parachute behaves as open hemisphere
(3) Vertical speed is constant
Newton's second law for the aircraft is
M
dV
1
2
 CD  ρ A V
dt
2
where A and CD are the single parachute area and drag coefficient
Separating variables
dV
2

V
Integrating, with IC V = Vi
CD ρ A
2 M
1
Integrating again with respect to t
x ( t) 
Eliminating t from Eqs. 1 and 2
x
 dt
Vi
V( t) 
CD ρ A
2 M
(1)
 Vi t

CD ρ A 
2 M
 ln 1 
CD ρ A
2 M


 Vi t
 Vi 

CD ρ A  V 
2 M
 ln
(2)
(3)
To find the minimum parachute area we must solve Eq 3 for A with x = xf when V = Vf
A
2 M
CD ρ x f
 Vi 

 Vf 
 ln
(4)
For three parachutes, the analysis is the same except A is replaced with 3A, leading to
A
2 M
3  CD ρ x f
 Vi 

 Vf 
 ln
kg
3
m
Single and three-parachute sizes; plot speed against distance and time; maximum "g''s
Basic
equations:
ρ  1.23
(5)
dV
The "g"'s are given by
2
dt
CD ρ A V

g
which has a maximum at the initial instant (V = Vi)
2 M g
The results generated in Excel are shown below:
Single:
A =
D =
Triple:
11.4 m
3.80 m
2
A = 3.8 m2
D = 2.20 m
"g "'s = -1.01 Max
t (s) x (m) V (km/hr)
0.00
2.50
5.00
7.50
10.0
12.5
15.0
17.5
20.0
22.5
24.6
0.0
216.6
393.2
542.2
671.1
784.7
886.3
978.1
1061.9
1138.9
1200.0
350
279
232
199
174
154
139
126
116
107
100
Aircraft Velocity versus Time
350
300
250
V (km/hr)
200
150
100
50
0
0
5
10
15
t (s)
20
25
Aircraft Velocity versus Distance
350
300
250
V (km/hr) 200
150
100
50
0
0
200
400
600
800
x (m)
1000
1200
Problem 9.77
Problem
9.96
[Difficulty: 3]
9.77
Given:
Data on airplane and parachute
Find:
Time and distance to slow down; plot speed against distance and time; maximum "g"'s
Solution:
The given data or available data is
M  8500 kg
km
Vi  400 
hr
km
Vf  100 
hr
π
2
2
Asingle   Dsingle  28.274 m
4
Newton's second law for the aircraft is
CD  1.42
ρ  1.23
kg
3
Dsingle  6  m
Dtriple  3.75 m
m
π
2
2
Atriple   Dtriple  11.045 m
4
M
dV
1
2
 CD  ρ A V
dt
2
where A and C D are the single parachute area and drag coefficient
Separating variables
dV
2

V
Integrating, with IC V = Vi
CD ρ A
2 M
Vi
V( t) 
1
Integrating again with respect to t
x ( t) 
Eliminating t from Eqs. 1 and 2
x
 dt
CD ρ A
2 M
(1)
 Vi t

CD ρ A 
2 M
 ln 1 
CD ρ A
2 M


 Vi t
 Vi 

CD ρ A  V 
2 M
 ln
(2)
(3)
To find the time and distance to slow down to 100 km/hr, Eqs. 1 and 3 are solved with V = 100
km/hr (or use Goal Seek)
dV
The "g"'s are given by
dt
g
2

CD ρ A V
2 M g
which has a maximum at the initial instant (V = Vi)
For three parachutes, the analysis is the same except A is replaced with 3A. leading to
Vi
V( t) 
1
x ( t) 
3  CD ρ A
2 M
 Vi t

3  CD ρ A 
2 M
 ln 1 
3  CD ρ A
2 M


 Vi t
The results generated in Excel are shown here:
t (s) x (m) V (km/hr)
t (s) x (m) V (km/hr)
0.0 0.0
1.0 96.3
2.0 171
3.0 233
4.0 285
5.0 331
6.0 371
7.0 407
8.0 439
9.0 469
9.29 477
0.0 0.0
1.0 94.2
2.0 165
3.0 223
4.0 271
5.0 312
6.0 348
7.0 380
7.93 407
9.0 436
9.3 443
400
302
243
203
175
153
136
123
112
102
100
400
290
228
187
159
138
122
110
100
91
89
"g "'s = -3.66 Max
Aircraft Velocity versus Time
400
350
One Parachute
Three Parachutes
300
V (km/hr)
250
200
150
100
50
0
0
1
2
3
4
5
6
t (s)
7
8
9
10
450
500
Aircraft Velocity versus Distance
400
350
V (km/hr)
300
One Parachute
250
Three Parachutes
200
150
100
50
0
0
50
100
150
200
250
300
350
x (m)
400
Problem 9.78
(Difficulty 2)
9.78 Calculate the drag of a smooth sphere of 0.3 𝑚 diameter in a stream of standard sea level air at
Reynolds numbers of 1, 10, 100 𝑎𝑎𝑎 1000.
Find: The drag force for a sphere
Assumption The flow is steady
Solution: Use the relations for drag force
The drag coefficient of a sphere is defined as:
𝐶𝐷 =
𝐹𝐷
1 2
𝜌𝑉 𝐴
2
Where the area is the frontal area of the sphere. The drag coefficient is a function of Reynolds
number defined as:
𝑅𝑅 =
𝜌𝜌𝜌
𝜇
The density and dynamic viscosity of air at standard conditions are:
𝜇 = 1.79 × 10−5
At 𝑅𝑅 = 1, from Fig. 9.11 we have:
The velocity for this Reynolds number is
The drag force is then
𝜌 = 1.225
𝑘𝑘
𝑚∙𝑠
𝑘𝑘
𝑚3
𝐶𝐷 = 27
−5 𝑘𝑘
𝑅𝑅 𝜇 1 × 1.79 × 10 𝑚 ∙ 𝑠
𝑚
𝑉=
=
= 0.000049
𝑘𝑘
𝜌𝜌
𝑠
1.225 3 × 0.3𝑚
𝑚
1
𝑘𝑘
𝑚 2 𝜋
× 27 × 1.225 3 × �0.000049 � × × (0.3 𝑚)2 = 2.8 × 10−9 𝑁
2
𝑚
𝑠
4
𝐹𝐷 =
At 𝑅𝑅 = 10, from Fig. 9.11 we have:
And a velocity of
And a drag force of
𝐹𝐷 =
−5 𝑘𝑘
𝑚
𝑅𝑅𝑅 10 × 1.79 × 10 𝑚 ∙ 𝑠
=
= 0.00049
𝑉=
𝑘𝑘
𝑠
𝜌𝜌
1.225 3 × 0.3𝑚
𝑚
1
𝑘𝑘
𝑚 2 𝜋
× 4.2 × 1.225 3 × �0.00049 � × × (0.3 𝑚)2 = 4.3 × 10−8 𝑁
2
𝑚
𝑠
4
At 𝑅𝑅 = 100, with Fig. 9.11 we have:
And a velocity of
And a drag force of
𝐹𝐷 =
𝐶𝐷 = 4.2
𝐶𝐷 = 1.05
−5 𝑘𝑘
𝑅𝑅𝑅 100 × 1.79 × 10 𝑚 ∙ 𝑠
𝑚
𝑉=
=
= 0.0049
𝑘𝑘
𝜌𝜌
𝑠
1.225 3 × 0.3𝑚
𝑚
1
𝑘𝑘
𝑚 2 𝜋
× 1.05 × 1.225 3 × �0.0049 � × × (0.3 𝑚)2 = 1.09 × 10−6 𝑁
2
𝑚
𝑠
4
At 𝑅𝑅 = 1000, with Fig. 9.11 we have:
And a velocity of
And a drag force of
𝐶𝐷 = 0.47
−5 𝑘𝑘
𝑅𝑅𝑅 1000 × 1.79 × 10 𝑚 ∙ 𝑠
𝑚
𝑉=
=
= 0.049
𝑘𝑘
𝜌𝜌
𝑠
1.225 3 × 0.3𝑚
𝑚
𝐹𝐷 =
1
𝑘𝑘
𝑚 2 𝜋
× 0.47 × 1.225 3 × �0.049 � × × (0.3 𝑚)2 = 4.8 × 10−5 𝑁
2
𝑚
𝑠
4
Problem 9.79
(Difficulty 2)
9.79 Calculate the drag of a smooth sphere of 0.5 𝑚 diameter when placed in an airstream
(15 ℃ 𝑎𝑎𝑎 101.3 𝑘𝑘𝑘) if the velocity is (a) 6
𝑚
𝑠
(b) 8.4
𝑚
.
𝑠
For the same drag coefficient as at 8.4
what velocity will the sphere attain the same drag which it had at a velocity of 6
Find: The drag force for a sphere
𝑚
?
𝑠
𝑚
,
𝑠
Assumption The flow is steady
Solution: Use the relations for drag force
The drag coefficient of a sphere is defined as:
𝐶𝐷 =
𝐹𝐷
1 2
𝜌𝑉 𝐴
2
Where the area is the frontal area of the sphere. The drag coefficient is a function of Reynolds
number defined as:
𝑅𝑅 =
The air density and viscosity are:
𝜌𝜌𝜌
𝜇
𝜌 = 1.225
𝜇 = 1.789 × 10−5 𝑃𝑃 ∙ 𝑠
The fromtal area is calculated as:
(a) For the case 𝑈 = 6
𝑚
,
𝑠
𝐴=
𝑘𝑘
𝑚3
𝜋 2 𝜋
𝐷 = × (0.5 𝑚)2 = 0.1963 𝑚2
4
4
we have:
𝑘𝑘
𝑚
𝜌𝜌𝜌 1.225 𝑚3 × 6 𝑠 × 0.5 𝑚
𝑅𝑅 =
=
= 205000
𝜇
1.789 × 10−5 𝑃𝑃 ∙ 𝑠
From the figure 9.11, we have:
𝐶𝐷 = 0.42
And the drag force is
1
1
𝑘𝑘
𝑚 2
𝐹𝐷 = 𝐶𝐷 𝜌𝑉 2 𝐴 = × 0.42 × 1.225 3 × �6 � × 0.1963 𝑚2 = 1.82 𝑁
2
2
𝑚
𝑠
at
(b) For the case 𝑈 = 8.4
𝑚
,
𝑠
we have:
𝑘𝑘
𝑚
𝜌𝜌𝜌 1.225 𝑚3 × 8.4 𝑠 × 0.5 𝑚
𝑅𝑅 =
=
= 288000
𝜇
1.789 × 10−5 𝑃𝑃 ∙ 𝑠
From the figure 9.11, the drag coefficient is:
𝐶𝐷 = 0.19
And the drag force is
1
1
𝑘𝑘
𝑚 2
𝐹𝐷 = 𝐶𝐷 𝜌𝑉 2 𝐴 = × 0.19 × 1.225 3 × �8.4 � × 0.1963 𝑚2 = 1.61 𝑁
2
2
𝑚
𝑠
The drag coefficient for 8.4 m/s is 0.19 and the drag force for 6 m/s is 1.82 N. The velocity that
would give the same drag force for a drag coefficient of 0.19 is:
1
𝐹𝐷 = 𝐶𝐷 𝜌𝑉 2 𝐴 = 1.82 𝑁
2
The velocity is
1.82 𝑁
𝐹𝐷
=�
=
𝑉= �
1
𝑘𝑘
1
2
𝐶𝐷 𝜌 𝐴
0.19
×
×
1.225
×
0.1963
𝑚
2
2
𝑚3
𝑉 = 8.3
𝑚
𝑠
Problem 9.80
(Difficulty 1)
9.80 A cylindrical chimney 0.9 𝑚 in diameter and 22.5 𝑚 high is exposed to a 56
𝑘𝑘
ℎ
wind
(15 ℃ 𝑎𝑎𝑎 101.3 𝑘𝑘𝑘); estimate the bending moment at the bottom of the chimney. Neglect end
effects.
Find: Bending moment
Solution: Use the drag coefficient to find the force and moment
The velocity is:
𝑈 = 56
The density and viscosity of the air are:
𝜌 = 1.225
The Reynolds number is defined as:
𝑘𝑘
𝑚
= 15.6
ℎ
𝑠
𝑘𝑘
𝑎𝑎𝑎 𝜇 = 1.789 × 10−5 𝑃𝑃 ∙ 𝑠
𝑚3
𝑘𝑘
𝑚
𝜌𝜌𝜌 1.225 𝑚3 × 15.6 𝑠 × 0.9 𝑚
=
= 960000
𝑅𝑅 =
𝜇
1.789 × 10−5 𝑃𝑃 ∙ 𝑠
From figure 9.13 we have for the drag coefficient for a cylinder:
𝐶𝐷 = 0.34
The drag force is:
The frontal area of the cylinder is:
1
𝐹𝐷 = 𝐶𝐷 𝜌𝜌𝑈 2
2
𝐴 = 𝐷𝐷 = 0.9 𝑚 × 22.5 𝑚 = 20.25 𝑚2
1
𝑘𝑘
𝑚 2
𝐹𝐷 = 0.34 × × 1.225 3 × 20.25 𝑚2 × �15.6 � = 1026 𝑁
2
𝑚
𝑠
The resultant force acts in the midpoint of the chimney. The moment is then:
𝑀0 = 𝐹𝐷
22.5 𝑚
ℎ
= 1026 𝑁 ×
= 11540 𝑁 ∙ 𝑚
2
2
Problem 9.81
Problem
9.98
[Difficulty: 2]
9.81
Given:
Bike and rider at terminal speed on hill with 8% grade.
W  210  lbf A  5  ft
Find:
ft
Vt  50
s
2
CD  1.25
(a) Verify drag coefficient
(b) Estimate distance needed for bike and rider to decelerate to 10 m/s after reaching level road
Solution:
Basic
equations:
1
2
FD  CD  ρ V  A
2
(Drag)
Assumptions: (1) Standard air
(2) Neglect all losses other than aerodynamic drag
θ  atan( 9  %)  5.143  deg Summing forces in the x-direction: ΣFx  FG  FD  0
1
2
Expanding out both force terms: M  g  sin( θ)  CD  ρ Vt  A Solving this expression for the drag coefficient:
2
The angle of incline is:
CD 
2  W sin( θ)
CD  1.26
2
ρ Vt  A
The original estimate for the
drag coefficient was good.
W d  W d 
Once on the flat surface: ΣFx  FD 
  V 
 V  V  Therefore:
g  dt 
g
 ds 
Separating variables:
ds  
2 W

dV
CD ρ g  A V
W
g
Integrating both sides yields:
 d V  C  1  ρ V2 A

D 2
 ds 
 V 
Δs  
 V2 

CD ρ g  A
 V1 
2 W
 ln
Δs  447  ft
Problem 9.82
Problem
9.100
[Difficulty: 2]
9.82
Given:
Ballistic data for .44 magnum revolver bullet
m
m
Vi  250 
Vf  210 
Δx  150  m M  15.6 gm D  11.2 mm
s
s
Average drag coefficient
Find:
Solution:
Basic
1
2
FD  CD  ρ V  A
equations:
2
(Drag)
Assumptions: (1) Standard air
(2) Neglect all losses other than aerodynamic drag
Newton's 2nd law:
Separating variables:
1
2
d 
d 
d 
ΣFx  FD  M   V   M  V  V  Therefore: M  V  V   CD  ρ V  A
2
d
d
d
t
s
x
 
 


dx  
2 M

dV
CD ρ A V
Solving this expression for the drag coefficient:
Integrating both sides yields:
CD  
 Vf 

Δx ρ A
 Vi 
2 M
 ln
Δx  
 Vf 

CD ρ A
 Vi 
2 M
The area is:
 ln
A 
π
4
2
2
 D  98.52  mm
Therefore the drag coefficent is:
CD  0.299
Problem 9.83
Problem
9.102
[Difficulty: 3]
9.83
Given:
Data on cyclist performance on a calm day
Find:
Performance hindered and aided by wind; repeat with high-tech tires; with fairing
Solution:
The given data or available data is
FR  7.5 N
M  65 kg
CD  1.2
ρ  1.23
2
A  0.25 m
kg
V  30
3
m
The governing equation is
1
2
FD   ρ A V  CD
2
Power steady power generated by the cyclist is
P  FD  FR  V
Now, with a headwind we have
km
Vw  10
hr

km
hr
FD  12.8 N

P  169 W
The aerodynamic drag is greater because of the greater effective wind speed


1
2
FD   ρ A V  Vw  CD
2
(1)
The power required is that needed to overcome the total force FD + FR, moving at the cyclist's speed is

P  V FD  FR

(2)
Combining Eqs 1 and 2 we obtain an expression for the cyclist's maximum speed V cycling into a
headwind (where P = 169 W is the cyclist's power)
Cycling into the wind:
1
2
P  FR   ρ A V  Vw  CD  V
2




(3)
This is a cubic equation for V; it can be solved analytically, or by iterating. It is convenient to use Excel's
Goal Seek (or Solver). From the associated Excel workbook
V  24.7
From Solver
km
hr
By a similar reasoning:
Cycling with the wind:
1
2
P  FR   ρ A V  Vw  CD  V
2




(4)
P  0.227  hp
V  35.8
From Solver
km
hr
With improved tires
FR  3.5 N
Maximum speed on a calm day is obtained from
1
2
P   FR   ρ A V  CD  V
2


This is a again a cubic equation for V; it can be solved analytically, or by iterating. It is convenient to use
Excel's Goal Seek (or Solver). From the associated Excel workbook
V  32.6
From Solver
km
hr
Equations 3 and 4 are repeated for the case of improved tires
From Solver
Against the wind
V  26.8
km
V  29.8
km
With the wind
hr
V  39.1
km
V  42.1
km
hr
For improved tires and fairing, from Solver
V  35.7
km
hr
Against the wind
hr
With the wind
hr
Problem 9.84
(Difficulty 2)
9.84 A standard marine torpedo is 0.533 𝑚 in diameter and about 7.2 𝑚 long. Make an engineering
estimate of the power required to derive this torpedo at 80
𝑘𝑘
ℎ
through freshwater at 20 ℃. Assume
hemispherical nose, cylinder body, and flat tail. 𝐶𝐷 for a solid hemisphere (flat side downstream) is
about 0.42?
Assumption: The drag of the torpedo is equal to the sum of the drag of each of the components.
Solution: Use the relations for drag for a sphere and flat plate.
The velocity is:
𝑈 = 80
The density and viscosity of the air are:
𝜌 = 998
The Reynolds number is defined as:
𝑘𝑘
𝑚
= 22.22
ℎ
𝑠
𝑘𝑘
𝑎𝑎𝑎 𝜇 = 1.002 × 10−3 𝑃𝑃 ∙ 𝑠
𝑚3
𝑘𝑘
𝑚
𝜌𝜌𝜌 998 𝑚3 × 22.22 𝑠 × 7.2 𝑚
=
= 1.59 × 108
𝑅𝑅 =
𝜇
1.002 × 10−3 𝑃𝑎 ∙ 𝑠
From the problem we have the drag coefficient of the hemisphere from Figure 9.11:
The area of the hemisphere is:
𝐴1 =
The drag force on the hemisphere:
𝐶𝐷1 = 0.42
𝜋 2 𝜋
𝐷 = × ( 0.533 𝑚)2 = 0.223 𝑚2
4
4
1
𝑘𝑘
𝑚 2
1
𝐹𝐷1 = 𝐶𝐷1 𝜌𝐴1 𝑈 2 = 0.42 × × 998 3 × 0.223 𝑚2 × �22.22 � = 23.1 𝑘𝑘
2
𝑚
𝑠
2
To do the unrolled cylinder, treat it as a flat plate. The drag coefficient is:
𝐶𝐷2 =
The area of the cylinder is:
The drag force on the cylinder is:
0.0742
1
𝑅𝑅 5
= 0.0017
𝐴2 = 𝜋𝜋𝜋 = 12.06 𝑚2
1
𝑘𝑘
𝑚 2
1
𝐹𝐷2 = 𝐶𝐷2 𝜌𝐴2 𝑈 2 = 0.0017 × × 998 3 × 12.06 𝑚2 × �22.22 � = 5.05 𝑘𝑘
2
𝑚
𝑠
2
The total drag force is:
The power can be calculated by:
𝐹𝐷 = 𝐹𝐷1 + 𝐹𝐷2 = 28.2 𝑘𝑘
𝑃 = 𝐹𝐷 𝑈 = 28.2 𝑘𝑘 × 22.22
𝑚
= 627 𝑘𝑘
𝑠
Problem 9.85
(Difficulty 2)
9.85 A large truck has an essentially boxlike body that causes flow separation at the front edges of the
cab at any speed. The drag is mostly profile drag and 𝐶𝐷 = 0.75 . If the projected frontal area of the
truck is 9 𝑚2 , determine and plot as a function of speed between zero and the legal limit the power that
must be delivered to the road to propel the truck.
Solution: Use the drag coefficient to find the power
The drag coefficient is:
𝐶𝐷 = 0.75
The frontal area is:
𝐴 = 9 𝑚2
The density of the air at 15 ℃ 𝑎𝑎𝑎 101.3 𝑘𝑘𝑘 is:
The drag force is calculated by:
𝜌 = 1.225
𝑘𝑘
𝑚3
1
𝐹𝐷 = 𝐶𝐷 𝜌𝜌𝑈 2
2
The power needed to be delivered to the road is:
1
𝑃 = 𝐹𝐷 𝑈 = 𝐶𝐷 𝜌𝜌𝑈 3
2
1
𝑘𝑘
𝑃 = 0.75 × × 1.225 3 × 9 𝑚2 𝑈 3 = 4.13𝑈 3 𝑊
2
𝑚
The maximum legal limit on the highway is taken as:
𝑈𝑙𝑙𝑙𝑙𝑙 = 75
𝑚𝑚𝑚𝑚𝑚
𝑚
= 33.5
ℎ𝑟
𝑠
The plot of power as a function of velocity is shown in the figure:
4
16
x 10
14
12
Power (W)
10
8
6
4
2
0
0
5
10
15
20
Velocity (m/s)
25
30
35
Problem 9.86
Problem
9.103
[Difficulty: 3]
9.86
FBnet
V
FD
y
x

T
Given:
Series of party balloons
Find:
Wind velocity profile; Plot
Wlatex
Solution:
Basic equations:
CD 
FD
1
FB  ρair g  Vol
2
 ρ V  A
2
The above figure applies to each balloon
For the horizontal forces FD  T sin( θ)  0
T cos( θ)  FBnet  Wlatex  0
Here
π D
FBnet  FB  W  ρair  ρHe  g 
6

(2)
3

D  20 cm
M latex  3  gm
RHe  2077
Rair  287 
N m
p He  111  kPa
kg K
N m
p air  101  kPa
kg K


Applying Eqs 1 and 2 to the top balloon, for which
Wlatex  0.02942  N
p He
kg
THe  293  K ρHe 
ρHe  0.1824
RHe THe
3
m
p air
kg
Tair  293  K ρair 
ρair  1.201 
Rair Tair
3
m
FBnet  0.0418 N
θ  65 deg
FBnet  Wlatex
cos( θ)

Wlatex  M latex g
3
π D
FBnet  ρair  ρHe  g 
6
FD  T sin( θ) 
This problem is ideal for computing and
plotting in Excel, but we will go through
the details here.
(1)
For the vertical forces
We have (Table A.6)

ΣF  0
 sin( θ)

Hence
FD  FBnet  Wlatex  tan( θ)
FD  0.0266 N
But we have
1
1
2
2 π D
FD  CD  ρair V  A  CD  ρair V 
2
2
4
2
V 
8  FD
2
CD ρair π D
From Table A.9
ν  1.50  10
2
5 m

s
V  1.88
CD  0.4
with
from Fig. 9.11 (we will
check Re later)
m
s
The Reynolds number is Red 
V D
ν
4
Red  2.51  10
We are okay!
For the next balloon

θ  60 deg
8  FD
V 
2
V  1.69
CD ρair π D
The Reynolds number is Red 
For the next balloon
V D

2
V  1.40
CD ρair π D
The Reynolds number is Red 
For the next balloon
V D

2
V  1.28
CD ρair π D
The Reynolds number is Red 
For the next balloon
V D
2
CD ρair π D
The Reynolds number is Red 
For the next balloon
V D
2
CD ρair π D
The Reynolds number is Red 
For the next balloon
V D
2
CD ρair π D
The Reynolds number is Red 
For the next balloon
V D
2
CD ρair π D
The Reynolds number is Red 
For the next balloon
V D
ν
θ  10 deg
CD  0.4
We are okay!

FD  0.00452 N
with
CD  0.4
with
CD  0.4
m
s
4
Red  1.03  10

with
s

V  0.77
FD  0.00717 N
m
FD  FBnet  Wlatex  tan( θ)
8  FD
V 

4
θ  20 deg
CD  0.4
We are okay!
Red  1.30  10
ν
with
s

V  0.97
FD  0.00870 N
m
FD  FBnet  Wlatex  tan( θ)
8  FD
V 

4
θ  30 deg
CD  0.4
We are okay!
Red  1.43  10
ν
with
s

V  1.07
FD  0.01043 N
m
FD  FBnet  Wlatex  tan( θ)
8  FD
V 

4
θ  35 deg
CD  0.4
We are okay!
Red  1.57  10
ν
with
s

V  1.18
FD  0.01243 N
m
FD  FBnet  Wlatex  tan( θ)
8  FD
V 

4
θ  40 deg
CD  0.4
We are okay!
Red  1.71  10
ν
with
s
FD  FBnet  Wlatex  tan( θ)
8  FD
FD  0.01481 N
m
4
θ  45 deg
V 

Red  1.87  10
ν
CD  0.4
We are okay!
FD  FBnet  Wlatex  tan( θ)
8  FD
with
s
4
θ  50 deg
FD  0.0215 N
m
Red  2.25  10
ν
V 

FD  FBnet  Wlatex  tan( θ)
We are okay!

FD  FBnet  Wlatex  tan( θ)
FD  0.002191 N
8  FD
V 
2
V  0.54
CD ρair π D
The Reynolds number is Red 
V D
m
s
Red  7184.21
ν
We are okay!
V  ( 0.54 0.77 0.97 1.07 1.18 1.28 1.40 1.69 1.88 ) 
In summary we have
m
s
h  ( 1 2 3 4 5 6 7 8 9 )m
10
h (m)
8
6
4
2
0
0.5
1
1.5
2
V (m/s)
This does not seem like an unreasonable profile for the lowest portion of an atmospheric boundary layer - over cities or rough terrain
the atmospheric boundary layer is typically 300-400 meters, so a near-linear profile over a small fraction of that distance is not out of
the question.
Problem 9.104
9.87
Problem
[Difficulty: 2]
9.87
FB
V
FD
y

T
W
x
Given:
Sphere dragged through river
Find:
Relative velocity of sphere
Solution:
CD 
Basic
equations:
FD
1
FB  ρ g  Vol
2
 ρ V  A

ΣF  0
2
The above figure applies to the sphere
For the horizontal forces FD  T sin( θ)  0
For the vertical forces
Here
V  5
m
s
(1)
T cos( θ)  FB  W  0
D  0.5 m
The Reynolds number is Red 
(2)
SG  0.30
V D
6
Red  1.92  10
ν
and from Table A.8 ν  1.30  10
2
6 m

s
ρ  1000
kg
3
m
Therefore we estimate the drag coefficient: CD  0.15 (Fig 9.11)
FB  W
FD  T sin( θ) 
 sin( θ)  ρ g  Vol  ( 1  SG )  tan( θ)
cos( θ)
3
π D
Hence
FD  ρ g 
6
 ( 1  SG )  tan( θ)
Therefore
1
π D
2 π D
CD  ρ V 
 ρ g 
 ( 1  SG)  tan( θ)
2
4
6
2
Solving for θ:
tan( θ) 
3
But we have
1
2
1
2
2 π D
FD  CD  ρ V  A  CD  ρ V 
2
2
4
3
2
CD V

4 g  D ( 1  SG )
2


CD V
3
θ  atan 

 4 g  D ( 1  SG)
The angle with the horizontal is:
α  90 deg  θ
α  50.7 deg
Problem 9.88
(Difficulty 2)
9.88 A simple but effective anemometer to measure wind speed can be made from a thin plate hinged
to deflect in the wind. Consider a thin plate made from brass that is 20 𝑚𝑚 high and 10 𝑚𝑚 wide.
Derive a relationship for wind speed as a function of deflection angle, 𝜃. What thickness of brass should
be used to give 𝜃 = 30 ° at 10
𝑚
?
𝑠
Find: The plate thickness
Assumption: The flow is steady
Solution: Use the relations for drag together with a moment balance on the plate.
The moment of the drag force plus the moment of the weight force balance. The sum of the
moments about the pivot is
� 𝑀 = 𝐹𝑁
ℎ
ℎ
− 𝑚𝑚 sin 𝜃
2
2
Where FN is the drag force. In terms of the drag coefficient and the velocity normal to the plate
1
1
𝐹𝑁 = 𝐶𝐷 𝐴 𝜌𝑉𝑛2 = 𝐶𝐷 𝐴 𝜌𝑉 2 cos 2 𝜃
2
2
The moment of the drag force equals that due to the weight
1
𝐶𝐷 𝐴 𝜌𝑉 2 cos 2 𝜃 = 𝑚𝑚 sin 𝜃
2
The relation between the wind velocity and the deflection angle is
1
2 𝑚 𝑔 sin 𝜃 2
𝑉=�
�
𝐶𝐷 𝐴 𝜌 cos 2 𝜃
From plate geometry, where the specific gravity of brass is given in Table A.1 as 8.55
From the moment relation we have:
The thickness is then
𝑡=
𝑚 = 𝜌 𝑤 ℎ 𝑡 = 𝑆𝑆𝜌𝐻2𝑜 𝑤 ℎ 𝑡
1
𝑆𝑆𝜌𝐻2 𝑜 𝑔𝑔ℎ𝑡 sin 𝜃 = 𝐶𝐷 𝐴 𝜌𝑉 2 cos 2 𝜃
2
𝐶𝐷 𝐴𝐴𝑉 2 cos 2 𝜃
𝐶𝐷 𝑤ℎ𝜌𝑉 2 cos 2 𝜃
𝐶𝐷 𝜌𝑉 2 cos 2 𝜃
=
=
2𝑆𝑆𝜌𝐻2 𝑜 𝑔𝑔ℎ sin 𝜃 2𝑆𝑆𝜌𝐻2 𝑜 𝑔𝑔ℎ sin 𝜃 2𝑆𝑆𝜌𝐻2 𝑜 𝑔 sin 𝜃
From figure 9.10, the drag coefficient at
The thickness is then
𝑏
ℎ
= 2.0, is
𝐶𝐷 = 1.2
1.2 1.23 𝑘𝑘
𝑚3
1
𝑠2
𝑚𝑚
𝑚2
2
2
(30°)
(10)
𝑡=
×
×
×
×
cos
×
×
×
1000
8.55
2 𝑚3
𝑚
𝑠2
999 𝑘𝑘 sin(30°) 9.81 𝑚
𝑡 = 1.30 𝑚𝑚
Problem 9.89
(Difficulty 2)
9.89 The Willis Tower in Chicago is 1454 𝑓𝑓 tall. Assuming that it is a tall rectangle with a square base of
120 𝑓𝑓 sides, calculating the maximum drag force on the building and the force when the wind is along
the diagonal of the structure as a function of wind speed from Beauford Wind Scales of strong breeze
(28 𝑚𝑚ℎ) to hurricane (75 𝑚𝑚ℎ). Assuming that the wind field is uniform, calculate the moment about
the base of the Tower also.
Assumption: The total drag is the sum of the drag on the components
Solution: Use the drag coefficient to find the forces
The density and viscosity of the air is:
The velocity is:
𝑠𝑠𝑠𝑠
𝜌 = 0.00238
𝑓𝑡
3
𝑎𝑎𝑎 𝜇 = 3.75 × 10−7
𝑉1 = 28 𝑚𝑚ℎ = 41.4
The Reynolds number is:
From the table 9.3, we have:
Thus the drag force is:
𝐹𝐷𝐷𝐷𝐷1
𝑉2 = 75 𝑚𝑚ℎ = 110
𝑅𝑅 =
𝑙𝑙𝑙 ∙ 𝑠
𝑓𝑓 2
𝑓𝑓
𝑠
𝑓𝑓
𝑠
𝜌𝑉1 𝑥
= 3.14 × 107 ≫ 1000
𝜇
𝐶𝐷𝐷𝐷𝐷 = 1.5
1
𝐹𝐷𝐷𝐷𝐷 = 𝐶𝐷𝐷𝐷𝐷 𝜌𝑈 2 𝐴
2
1
𝑠𝑠𝑠𝑠
𝑓𝑓 2
= 1.5 × × 0.00238
× �41.4 � × 1454 𝑓𝑓 × 120𝑓𝑓 = 534000 𝑙𝑙𝑙
2
𝑓𝑓 3
𝑠
1
𝑠𝑠𝑠𝑠
𝑓𝑓 2
𝐹𝐷𝐷𝐷𝐷1 = 1.5 × × 0.00238
×
�110
� × 1454 𝑓𝑓 × 120𝑓𝑓 = 3770000 𝑙𝑙𝑙
2
𝑓𝑓 3
𝑠
For the moment we have:
1
ℎ3
𝑀𝑚𝑚𝑚 = 𝐶𝐷𝐷𝐷𝐷 𝜌𝑈 2
2
3
1
𝑠𝑠𝑠𝑠
𝑓𝑓 2 (1454 𝑓𝑓)3
𝑀𝑚𝑚𝑚1 = 1.5 × × 0.00238
×
�41.4
� ×
= 3.13 × 109 𝑙𝑙𝑙 ∙ 𝑓𝑓
2
𝑓𝑓 3
𝑠
3
𝑀𝑚𝑚𝑚2
1
𝑠𝑠𝑠𝑠
𝑓𝑓 2 (1454 𝑓𝑓)3
= 1.5 × × 0.00238
× �110 � ×
= 2.21 × 1010 𝑙𝑙𝑙 ∙ 𝑓𝑓
2
𝑓𝑓 3
𝑠
3
For the diagonal wind we have:
𝐶𝐷𝐷 = 1.05
Thus the drag force is:
𝐹𝐷𝐷1
1
𝐹𝐷𝐷𝐷𝐷 = 𝐶𝐷𝐷 𝜌𝑈 2 𝐴
2
1
𝑠𝑠𝑠𝑠
𝑓𝑓 2
= 1.05 × × 0.00238
× �41.4 � × √2 × 1454 𝑓𝑓 × 120𝑓𝑓 = 529000 𝑙𝑙𝑙
2
𝑓𝑓 3
𝑠
1
𝑠𝑠𝑠𝑠
𝑓𝑓 2
𝐹𝐷𝐷2 = 1.05 × × 0.00238
×
�110
� × √2 × 1454 𝑓𝑓 × 120𝑓𝑓 = 3730000 𝑙𝑙𝑙
2
𝑓𝑓 3
𝑠
For the moment we have:
𝑀𝐷 = 𝐶𝐷𝐷𝐷𝐷
𝑀𝐷𝐷1 = 1.05 ×
𝑀𝐷𝐷2 = 1.05 ×
√2 2 ℎ3
𝜌𝑈
2
3
𝑠𝑠𝑠𝑠
𝑓𝑡 2 (1454 𝑓𝑓)3
√2
× 0.00238
×
�41.4
� ×
= 3.10 × 109 𝑙𝑙𝑙 ∙ 𝑓𝑓
𝑓𝑓 3
𝑠
2
3
𝑠𝑠𝑠𝑠
𝑓𝑓 2 (1454 𝑓𝑓)3
√2
× 0.00238
×
�110
� ×
= 2.19 × 1010 𝑙𝑙𝑙 ∙ 𝑓𝑓
𝑓𝑓 3
𝑠
2
3
Problem 9.90
(Difficulty 2)
9.90 It is proposed to build a pyramidal building with a square base with sides of 160 𝑓𝑓, which has the
same volume as the Willis Tower. Calculate the maximum drag force on this building. Do you expect the
drag force to be greater, the same or less than that for Willis Tower? Why? And compare it to that for
the Willis Tower under hurricane force conditions (75 mph).
Find: The drag force
Solution: Use the drag coefficient to find the force
The density of air is:
𝜌 = 0.00238
For the volume we have:
𝑠𝑠𝑠𝑠
𝑓𝑓 3
ℎ
ℎ 2
𝑉𝑉𝑉 = 𝑉𝑉𝑉𝑠𝑠𝑠𝑠𝑠 𝑡𝑡𝑡𝑡𝑡 = � 𝑥 2 �1 − � 𝑑ℎ = 20937600 𝑓𝑓 3
𝐻
0
20937600 𝑓𝑓 3
𝐻
ℎ 2
= 817.9 𝑓𝑓 =
� �1 − � 𝑑ℎ =
3
𝐻
160 𝑓𝑓 × 160 𝑓𝑓
0
ℎ
𝐻 = 2454 𝑓𝑓
We don’t have values for CD for a triangular shaped object, we will approximate it as a rectangular
object with the same frontal. The height of a rectangular building with the same frontal area would
be one-half the height, or 1227 ft. We can use Figure 9.10. The aspect ratio of the building is
The drag coefficient is about
The maximum velocity is:
The drag force is calculated by:
𝑏 1227
=
= 7.6
ℎ
160
𝐶𝐷 = 1.25
𝑈 = 75 𝑚𝑚ℎ = 110
𝑓𝑓
𝑠
1
𝐹𝐷 = 𝐶𝐷 𝜌𝑈 2 𝐴
2
1
𝐴 = 𝐻𝐻
2
Thus
𝐹𝐷 = 1.25 ×
1
𝑠𝑠𝑠𝑠
𝑓𝑓 2 1
× 0.00238
×
�110
� × × 2453.6 𝑓𝑓 × 160 𝑓𝑓 = 3.53 × 106 𝑙𝑙𝑙
2
𝑓𝑓 3
𝑠
2
The drag coefficient for the Willis Tower in Problem 9.89 is also determined from Figure 9.10. The
aspect ratio of the tower is
This yields a drag coefficient of about
𝑏 1454
=
= 12.1
ℎ
120
𝐶𝐷 = 1.3
The drag force should be slightly smaller than the Willis Tower as computed in Problem 9.89
because of the smaller drag coefficient.
Problem 9.91
(Difficulty 2)
9.91 Calculate the drag forces on a
1
scale
200
model of the Willis Tower that is tested in a large water
flume under conditions corresponding to those in problem 9.89. Ignore any free surface effects and
assume dynamic similarity and that the drag coefficient is unchanged.
Find: The drag force
Solution: Use the drag coefficient to find the force
The density and viscosity of the water is:
𝜌 = 1.936
𝑠𝑠𝑠𝑠
𝑓𝑓 3
𝜇 = 2.344 × 10−5
For the model we have:
ℎ=
With dynamic similarity we have:
𝑙𝑙𝑙 ∙ 𝑠
𝑓𝑓 2
1454 𝑓𝑓
= 7.27 𝑓𝑓
200
𝑥=
120 𝑓𝑓
= 0.6 𝑓𝑓
200
𝐶𝐷 = 1.5
For the Willis Tower we have the Reynolds number as:
𝑅𝑅 =
𝜌𝜌𝜌
= 3.14 × 107 − 8.40 × 107
𝜇
For model we have the Reynolds number is:
𝑅𝑅 =
𝜌𝜌𝜌
= 49500𝑉
𝜇
Thus to match the dynamic similarity we have:
𝑉 = 634
The drag coefficient can be calculated by:
𝑓𝑓
𝑓𝑓
− 1697
𝑠
𝑠
𝐹𝐷1
1
𝐹𝐷 = 𝐶𝐷 𝜌𝑈 2 𝐴
2
1
𝑠𝑠𝑠𝑠
𝑓𝑓 2
= 1.5 × × 1.936
× �634 � × 7.27 𝑓𝑓 × 0.6 𝑓𝑓 = 2.55 × 106 𝑙𝑙𝑙
2
𝑓𝑓 3
𝑠
𝐹𝐷2 = 1.5 ×
1
𝑠𝑠𝑠𝑠
𝑓𝑓 2
× 1.936
×
�1697
� × 7.27 𝑓𝑓 × 0.6 𝑓𝑓 = 1.824 × 107 𝑙𝑙𝑙
2
𝑓𝑓 3
𝑠
The water velocity is much higher than can be produced and the forces are extremely high. Dynamic
similarity is not really possible.
Problem 9.92
Problem
9.107
[Difficulty: 2]
9.92
Given:
Circular disk in wind
Find:
Mass of disk; Plot α versus V
Solution:
CD 
Basic
equations:

ΣM  0
FD
1
2
2
 ρ V  A
Summing moments at the pivotW L sin( α)  Fn  L  0
Hence
M  g  sin( α) 
The data is
ρ  1.225 
1
2
2
2 π D
 ρ ( V cos( α) ) 
kg
3
m
M 
V
Rearranging
4
V  15
2
2
1
2
Fn   ρ Vn  A CD
2
and
 CD
m
D  25 mm
s
M  0.0451 kg
8  g  sin( α)

2
π ρ D  CD
CD  1.17
2
π ρ V  cos( α)  D  CD
8 M g
α  10 deg
tan( α)
V  35.5
cos( α)
m
s

tan( α)
cos( α)
We can plot this by choosing α and computing V
80
V (m/s)
60
40
20
0
10
20
30
40
Angle (deg)
This graph can be easily plotted in Excel
50
60
70
(Table 9.3)
Problem 9.93
Problem
9.109
9.93
[Difficulty: 3]
Problem 9.110
9.94
Problem
9.94
[Difficulty: 3]
9.95
Problem 9.95
Problem
9.111
[Difficulty: 3]
Problem 9.96
Problem
9.114
[Difficulty: 4]
9.96
Given:
Data on a sports car
Find:
Speed for aerodynamic drag to exceed rolling resistance; maximum speed & acceleration at 100
km/h; Redesign change that has greatest effect
Solution:
1
2
Basic equation: FD   ρ A V  CD
2
P  FD V
The given data or available data is
M  1250 kg
2
A  1.72 m
CD  0.31
Pengine  180  hp  134.23 kW FR  0.012  M  g
To find the speed at which aerodynamic drag first equals rolling resistance, set the two forces equal
2  FR
ρ A CD
V  21.2
m
V  76.2
s
V  100 
1
2
FD   ρ V  A CD
2
η 
Pused
1
2
 ρ V  A CD  FR
2
hr
The power consumed by drag and rolling resistance at this speed is
Hence the drive train efficiency is
km
V  27.8
hr
m
s
Pengine  17 hp  12.677 kW
FD  253  N


Pused  FD  FR  V
Pused  11.1 kW
η  87.7 %
Pengine
The acceleration is obtained from Newton's second law
where T is the thrust produced by the engine, given by
M  a  ΣF  T  FR  FD
P
T
V
The maximum acceleration at 100 km/h is when full engine power is used. Pengine  180  hp  134.2  kW
Because of drive train inefficiencies the maximum power at the wheels is Pmax  η Pengine
Hence the maximum thrust is Tmax 
The maximum acceleration is then
Pmax
V
3
km
To find the drive train efficiency we use the data at a speed of
The aerodynamic drag at this speed is
kg
m
FR  147.1  N
The rolling resistance is then
Hence V 
ρ  1.23
Pmax  118  kW
Tmax  4237 N
amax 
Tmax  FD  FR
M
amax  3.07
m
2
s
The maximum speed is obtained when the maximum engine power is just balanced by power consumed by drag and rolling resistance
Pmax 
For maximum speed:
 1  ρ V 2 A C  F   V

max
D
R max
2

This is a cubic equation that can be solved by iteration or by using Excel's Goal Seek or Solver
km
Vmax  248 
hr
We are to evaluate several possible improvements:
For improved drive train
η  η  6 %
η  93.7 %
Pmax 
Pmax  η Pengine
Pmax  126  kW
 1  ρ V 2 A C  F   V

max
D
R max
2

km
Vmax  254 
hr
Solving the cubic (using Solver)
Improved drag coefficient:
CDnew  0.29
Pmax 
Pmax  118  kW

 1  ρ V 2 A C

max
Dnew  FR  Vmax
2


km
This is the
Vmax  254 
hr best option!
Solving the cubic (using Solver)
Reduced rolling resistance:
FRnew  0.91 % M  g
FRnew  111.6 N
1
2
Pmax    ρ Vmax  A CD  FRnew  Vmax
2

Solving the cubic (using Solver)

km
Vmax  250 
hr
The improved drag coefficient is the best option.
Problem 9.97
Problem
9.106
[Difficulty: 3]
9.97
Given:
Data on dimensions of anemometer
Find:
Calibration constant; compare to actual with friction
Solution:
The given data or available data is
D  2  in
R  3  in
ρ  0.00234 
slug
ft
3
The drag coefficients for a cup with open end facing the airflow and a cup with open end facing downstream are, respectively,
from Table 9.3
CDopen  1.42
CDnotopen  0.38
1
2
The equation for computing drag is FD   ρ A V  CD
2
(1)
2
A 
where
π D
A  0.0218 ft
4
2
Assuming steady speed ω at steady wind speed V the sum of moments will be zero. The two cups that are momentarily parallel
to the flow will exert no moment; the two cups with open end facing and not facing the flow will exert a moment beacuse of their
drag forces. For each, the drag is based on Eq. 1 (with the relative velocity used!). In addition, friction of the anemometer is
neglected
1
1
2
2
ΣM  0    ρ A ( V  R ω)  CDopen  R    ρ A ( V  R ω)  CDnotopen  R
2
2

or
2



2
( V  R ω)  CDopen  ( V  R ω)  CDnotopen
This indicates that the anemometer reaches a steady speed even in the abscence of friction because it is the
relative velocity on each cup that matters: the cup that has a higher drag coefficient has a lower relative
velocity
Rearranging for
k
V
ω
2
 V  R  C

 Dopen 
ω

2
 V  R  C

 Dnotopen
ω

Hence
CDnotopen 


1

CDopen 
 R
k 
CDnotopen 


1

CDopen 

k  9.43 in
k  0.0561
mph
rpm
For the actual anemometer (with friction), we first need to determine the torque produced when the anemometer is
stationary but about to rotate
Minimum wind for rotation is
Vmin  0.5 mph
The torque produced at this wind speed is
Tf 
 1  ρ A V 2 C


min Dopen  R 
2

Tf  3.57  10
 1  ρ A V 2 C


min Dnotopen   R
2

6
 ft lbf
A moment balance at wind speed V, including this friction, is
ΣM  0 
or

 1  ρ A ( V  R ω) 2 C

Dopen  R 
2


 1  ρ A ( V  R ω) 2 C

Dnotopen  R  Tf
2

2  Tf
2
2
( V  R ω)  CDopen  ( V  R ω)  CDnotopen 
R ρ A
This quadratic equation is to be solved for ω when
V  20 mph
After considerable calculations
ω  356.20 rpm
This must be compared to the rotation for a frictionless model, given by
V
ωfrictionless 
k
The error in neglecting friction is
ωfrictionless  356.44 rpm
ω  ωfrictionless
ω
 0.07 %
Problem 9.98
Problem
9.118
9.98
[Difficulty: 4]
Problem 9.99
Problem
9.119
9.99
[Difficulty: 4]
Problem
9.120
Problem 9.100
[Difficulty: 2]
9.100
Given:
Data on advertising banner
Find:
Power to tow banner; Compare to flat plate; Explain discrepancy
Solution:
Basic equation:
1
2
FD   ρ A  V  CD
2
P  FD V
V  55 mph
The given data or available data is
1
2
FD   ρ A  V  CD
2
FD  771 lbf
P  FD V


7
0.455

2.58
log ReL
1
2
FD   ρ A V  CD
2

s
A  180 ft
 4 ft
V L
ReL 
ν
ft
A  L h
ν  1.62  10
For a flate plate, check Re
CD 
V  80.7
2
L  45 ft
h  4 ft
slug
ft
CD  0.05
L
CD  0.563
h
4 ft  lbf
P  6.22  10 
s
P  113 hp
2
(Table A.9, 69oF)
s
ReL  2.241  10
so flow is fully turbulent. Hence use Eq 9.37b
1610
CD  0.00258
ReL
ρ  0.00234
FD  3.53 lbf
This is the drag on one side. The total drag is then 2  FD  7.06 lbf . This is VERY much less than the banner drag.
The banner drag allows for banner flutter and other secondary motion which induces significant form drag.
3
Problem 9.101
Problem
9.122
[Difficulty: 1]
9.101
Given:
Data on car antenna
Find:
Bending moment
FD
Solution:
V
1
2
Basic equation:
FD   ρ A V  CD
2
The given or available data is
V  120 
km
 33.333
hr
s
L  1.8 m
D  10 mm
x
A  0.018 m
kg
3
M
2
ν  1.50  10
5 m

m
For a cylinder, check Re
Re 
V D
ν
y
2
A  L D
ρ  1.225 
m
Re  2.22  10
(Table A.10, 20 oC)
s
4
From Fig. 9.13
CD  1.0
1
2
FD   ρ A V  CD
2
The bending moment is then
L
M  FD
2
M  11.0 N m
FD  12.3 N
Problem 9.102
Problem
9.126
9.102
[Difficulty: 2]
Problem 9.103
Problem
9.127
9.103
[Difficulty: 2]
Problem 9.104
Problem
9.130
[Difficulty: 2]
9.104
Given:
3 mm raindrop
Find:
Terminal speed
Solution:
Basic equation:
1
2
FD   ρ A V  CD
2
Given or available data is
D  3  mm
Summing vertical forces
ΣF  0
kg
ρH2O  1000
3
m
kg
ρair  1.225 
3
m
1
2
M  g  FD  M  g   ρair A V  CD  0
2
M  1.41  10
Assume the drag coefficient is in the flat region of Fig. 9.11 and verify Re later
V 
Re 
2 M g
CD ρair A
V D
ν
V  8.95
2
5 m

(Table A.10, 20 oC)
s
Buoyancy is negligible
3
π D
M  ρH2O
6
Check Re
ν  1.50  10
5
2
kg
A 
π D
4
CD  0.4
m
s
3
Re  1.79  10 which does place us in the flat region of the curve
Actual raindrops are not quite spherical, so their speed will only be approximated by this result
6
A  7.07  10
2
m
Problem 9.105
Problem
9.131
9.105
[Difficulty: 3]
Problem 9.106
Problem
9.134
[Difficulty: 3]
9.106
Given:
Data on a tennis ball
Find:
Terminal speed time and distance to reach 95% of terminal speed
Solution:
The given data or available data is
M  57 gm
2
A 
Then
π D
4
2
5 m
D  64 mm
ν  1.45 10
3
A  3.22  10
At terminal speed drag equals weight
FD  M  g
The drag at speed V is given by
1
2
FD   ρ A V  CD
2
Hence the terminal speed is
Vt 
Check the Reynolds number
Re 
s
kg
3
m
2
CD  0.5
(from Fig. 9.11)
M g
2
ρ  1.23
m
Assuming high Reynolds number
1

 ρ A CD
Vt D
m
Vt  23.8
s
Re  1.05  10
ν
5
Check!
For motion before terminal speed Newton's second law applies
M a  M
dV
1
2
 M  g    ρ V  A CD
dt
2
Separating variables
2
d
V  g  k V
dt
or





V
1
2
dV  t
g  k V
0
Hence
Evaluating at V = 0.95Vt
For distance x versus time, integrate
g
V( t) 
k
0.95 Vt 
dx
dt

g
k
k 
where





1
2
ρ A CD
k  0.0174
2 M
dV 
g  k V
1
g k
 k  V

 g 
 atanh
 tanh g  k  t
g
k
 tanh g  k  t
 tanh g  k  t
1
t 
g k

x




 atanh 0.95 Vt
t
0
g
k
 tanh g  k  t dt
k

g
t  4.44 s
1
m
Note that

1
 tanh( a t) dt   ln( cosh( a t) )
a

Hence
x ( t) 
Evaluating at V = 0.95Vt
1
k
t  4.44 s
 ln cosh g  k  t 
so
x ( t)  67.1 m
Problem 9.107
Problem
9.136
9.107
[Difficulty: 3]
Problem 9.108
Problem
9.138
9.108
[Difficulty: 3]
Problem 9.109
Problem
9.140
9.109
[Difficulty: 4]
Problem 9.110
(Difficulty 2)
9.110 A rectangular airfoil of 40 𝑓𝑓 span and 6 𝑓𝑓 chord has lift and drag coefficients of 0.5 and 0.04,
respectively, at an angle of attack of 6°. Calculate the drag and horsepower necessary to drive this airfoil
at 50,100 𝑎𝑎𝑎 150 𝑚𝑚ℎ horizontally through still air (40℉ 𝑎𝑎𝑎 13.5 𝑝𝑝𝑝𝑝) . What lift forces are
obtained at these speeds?
Find: The drag and lift forces and power
Solution: Use the drag and lift coefficients to find the forces and power
The density of the air can be calculated from the ideal gas law as:
For the velocity we have:
𝜌=
𝑃
𝑠𝑠𝑠𝑠
= 0.002265
𝑅𝑅
𝑓𝑓 3
𝑉1 = 50 𝑚𝑚ℎ = 73.3
𝑓𝑓
𝑠
𝑉3 = 150 𝑚𝑚ℎ = 220
𝑓𝑓
𝑠
𝑉2 = 100 𝑚𝑚ℎ = 146.7
The drag force can be calculated as:
Thus
𝑓𝑓
𝑠
1
𝐹𝐷 = 𝐶𝐷 𝐴 𝜌𝑉 2
2
1
1
𝑙𝑙𝑙 ∙ 𝑠2
𝑓𝑓 2
𝐹𝐷1 = 𝐶𝐷 𝐴 𝜌𝑉12 = 0.04 × 40 𝑓𝑓 × 6 𝑓𝑓 × × 0.002265
×
�73.3
� = 58.4 𝑙𝑙𝑙
2
2
𝑠
𝑓𝑓 4
𝐹𝐷2
1 2
1
𝑙𝑙𝑙 ∙ 𝑠2
𝑓𝑓 2
= 𝐶𝐷 𝐴 𝜌𝑉2 = 0.04 × 40 𝑓𝑓 × 6 𝑓𝑓 × × 0.002265
× �146.7 � = 234 𝑙𝑙𝑙
2
2
𝑠
𝑓𝑓 4
1
1
𝑙𝑙𝑓 ∙ 𝑠2
𝑓𝑓 2
𝐹𝐷3 = 𝐶𝐷 𝐴 𝜌𝑉32 = 0.04 × 40 𝑓𝑓 × 6 𝑓𝑓 × × 0.002265
×
�220
� = 526 𝑙𝑙𝑙
2
2
𝑠
𝑓𝑓 4
The power necessary is as:
𝑃 = 𝐹𝐷 𝑉
Thus
𝑃1 = 𝐹𝐷1 𝑉1 = 58.4 𝑙𝑙𝑙 × 73.3
𝑃2 = 𝐹𝐷2 𝑉2 = 234 𝑙𝑙𝑙 × 146.7
𝑃3 = 𝐹𝐷3 𝑉3 = 526 𝑙𝑙𝑙 × 220
The lift force can be calculated as:
𝑓𝑓
𝑙𝑙𝑙 ∙ 𝑓𝑓
= 4281
= 7.78 ℎ𝑝
𝑠
𝑠
𝑓𝑓
𝑙𝑙𝑙 ∙ 𝑓𝑓
= 34328
= 62.4 ℎ𝑝
𝑠
𝑠
𝑓𝑓
𝑙𝑙𝑙 ∙ 𝑓𝑓
= 115700
= 210 ℎ𝑝
𝑠
𝑠
1
𝐹𝐿 = 𝐶𝐿 𝐴 𝜌𝑉 2
2
Thus
1
1
𝑙𝑙𝑙 ∙ 𝑠2
𝑓𝑓 2
𝐹𝐿1 = 𝐶𝐿 𝐴 𝜌𝑉12 = 0.5 × 40 𝑓𝑓 × 6 𝑓𝑓 × × 0.002265
×
�73.3
� = 730 𝑙𝑙𝑙
2
2
𝑠
𝑓𝑓 4
𝐹𝐿2
1 2
1
𝑙𝑙𝑙 ∙ 𝑠2
𝑓𝑓 2
= 𝐶𝐿 𝐴 𝜌𝑉2 = 0.5 × 40 𝑓𝑓 × 6 𝑓𝑓 × × 0.002265
× �146.7 � = 2920 𝑙𝑙𝑙
2
2
𝑠
𝑓𝑓 4
1
1
𝑙𝑙𝑙 ∙ 𝑠2
𝑓𝑓 2
𝐹𝐿3 = 𝐶𝐿 𝐴 𝜌𝑉32 = 0.5 × 40 𝑓𝑓 × 6 𝑓𝑓 × × 0.002265
×
�220
� = 6580 𝑙𝑙𝑙
2
2
𝑠
𝑓𝑓 4
Problem 9.111
(Difficulty 2)
9.111 A rectangular airfoil of 9 𝑚 span and 1.8 𝑚 chord moves horizontally at a certain angle of attack
through still air at 240
𝑘𝑘
.
ℎ
Calculate the lift and drag, and the power necessary to drive the airfoil at this
speed through (a) 101.3 𝑘𝑘𝑘 𝑎𝑎𝑎 15 ℃ and (b) 79.3 𝑘𝑘𝑘 𝑎𝑎𝑎 − 18 ℃ . 𝐶𝐷 = 0.035 ; 𝐶𝐿 = 0.46 .
Calculate the speed and power required for condition (b) to obtain the lift of the condition (a).
Find: Drag and lift
Solution: Use the drag and lift coefficients to find the forces
The velocity is:
𝑉 = 240
𝑚
𝑘𝑘
= 66.7
𝑠
ℎ
(a) For the air at 101.3 𝑘𝑘𝑘 𝑎𝑎𝑎 15 ℃, we have:
𝑝
𝑘𝑘
𝜌=
= 1.226 3
𝑅𝑅
𝑚
The drag force is calculated by:
1
𝐹𝐷 = 𝐶𝐷 𝜌𝜌𝑈 2
2
𝐹𝐷 = 0.035 × 0.5 × 1.226
The lift force is calculated by:
𝑘𝑘
𝑚 2
×
9
𝑚
×
1.8
𝑚
×
�66.7
� = 1546 𝑁
𝑚3
𝑠
1
𝐹𝐿 = 𝐶𝐿 𝜌𝜌𝑈 2
2
𝐹𝐿 = 0.46 × 0.5 × 1.226
𝑘𝑘
𝑚 2
×
9
𝑚
×
1.8
𝑚
×
�66.7
� = 20300 𝑁
𝑚3
𝑠
The power necessary to derive the airfoil is:
𝑝 = 𝐹𝐷 𝑈 = 1546 𝑁 × 66.7
𝑚
= 103 𝑘𝑘
𝑠
(b) For the air at 79.3 𝑘𝑘𝑘 𝑎𝑎𝑎 − 18 ℃, we have:
𝑝
𝑘𝑘
𝜌=
= 1.084 3
𝑅𝑅
𝑚
The drag force is calculated by:
1
𝐹𝐷 = 𝐶𝐷 𝜌𝜌𝑈 2
2
𝐹𝐷 = 0.035 × 0.5 × 1.084
The lift force is calculated by:
𝑘𝑘
𝑚 2
× 9 𝑚 × 1.8 𝑚 × �66.7 � = 1367 𝑁
3
𝑚
𝑠
1
𝐹𝐿 = 𝐶𝐿 𝜌𝜌𝑈 2
2
𝐹𝐿 = 0.46 × 0.5 × 1.084
𝑘𝑘
𝑚 2
×
9
𝑚
×
1.8
𝑚
×
�66.7
� = 17970 𝑁
𝑚3
𝑠
The power necessary to derive the airfoil is:
𝑝 = 𝐹𝐷 𝑈 = 1367 𝑁 × 66.7
𝑚
= 91.2 𝑘𝑘
𝑠
Now we want the lift force for case (b) to be the same as for case (a), or 𝐹𝐿 = 20300 𝑁. The
velocity must then be
𝑈2 =
2𝐹𝐿
𝐶𝐿 𝜌𝜌
2 × 20300 𝑁
𝑚
2𝐹𝐿
𝑈=�
=�
= 70.9
𝑘𝑘
𝑠
𝐶𝐿 𝜌𝜌
0.46 × 1.084 3 × 9 𝑚 × 1.8 𝑚
𝑚
The drag force at this speed is:
𝐹𝐷 = 0.035 × 0.5 × 1.084
𝑘𝑘
𝑚 2
×
9
𝑚
×
1.8
𝑚
×
�70.9
� = 1545 𝑁
𝑚3
𝑠
𝑝 = 𝐹𝐷 𝑈 = 1545 𝑁 × 70.9
𝑚
= 109.5 𝑘𝑘
𝑠
Problem 9.112
Problem
9.142
[Difficulty: 4]
9.112
Given:
Data on an air bubble
Find:
Time to reach surface
Solution:
The given data or available data is
h  100  ft  30.48 m
ρ  1025
kg
CD  0.5 (Fig. 9.11)
(Table A.2)
3
p atm  101  kPa
m
1
dx  V dt where
To find the location we have to integrate the velocity over time:
V
 patm  ρ g h

3 CD p atm  ρ g  ( h 

4 g  d 0
The results (generated in Excel) for each bubble diameter are shown below:
d 0 = 0.3 in
d 0 = 7.62 mm
d0=
t (s) x (m) V (m/s)
t (s) x (m) V (m/s)
0
5
10
15
20
25
30
35
40
45
50
63.4
0
2.23
4.49
6.76
9.1
11.4
13.8
16.1
18.6
21.0
23.6
30.5
0.446
0.451
0.455
0.460
0.466
0.472
0.478
0.486
0.494
0.504
0.516
0.563
5
mm
0
5
10
15
20
25
30
35
40
45
50
55
0
1.81
3.63
5.47
7.32
9.19
11.1
13.0
14.9
16.9
18.8
20.8
0.362
0.364
0.367
0.371
0.374
0.377
0.381
0.386
0.390
0.396
0.401
0.408
60
65
70
75
77.8
22.9
25.0
27.1
29.3
30.5
0.415
0.424
0.435
0.448
0.456
d0 =
15
mm
t (s) x (m) V (m/s)
0.0
5.0
10.0
15.0
20.0
25.0
30.0
35.0
40.0
45.1
0
3.13
6.31
9.53
12.8
16.1
19.5
23.0
26.6
30.5
0.626
0.635
0.644
0.655
0.667
0.682
0.699
0.721
0.749
0.790
Use Goal Seek for the last time step to make x = h !
Depth of Air Bubbles versus Time
30
25
20
x (m)
15
10
Initial Diameter = 5 mm
Initial Diameter = 0.3 in
5
Initial Diameter = 15 mm
0
0
10
20
30
40
50
t (s)
60
70
80


x)

6
Problem 9.113
Problem
9.144
9.113
[Difficulty: 4]
Problem 9.114
(Difficulty 2)
9.114 If 𝐶𝐿 = 1.0 and 𝐶𝐷 = 0.05 for an airfoil, then find the span needed for a rectangular wing of 10 𝑚
chord to lift 3560 𝑘𝑘 at a take-off speed of 282
𝑘𝑘
.
ℎ
What is the wing drag at take-off?
Find: The span necessary to lift a given load
Solution: Use the lift coefficient
The velocity is:
The density of the air is:
𝑈 = 282
𝑘𝑘
𝑚
= 78.3
ℎ
𝑠
𝑘𝑘
𝑚3
For the lift force we have:
𝜌 = 1.225
The plan area is calculated by:
1
𝐹𝐿 = 𝐶𝐿 𝜌𝜌𝑈 2
2
where S is the length of the span.
𝐴=𝐿∙𝑆
So we have for the lift force:
The necessary span is then
1
𝐹𝐿 = 𝐶𝐿 𝜌𝜌𝜌𝑈 2
2
𝑆=
2𝐹𝐿
=
𝐶𝐿 𝜌𝜌𝑈 2
2 × 3560 × 103 𝑁
= 94.8 𝑚
𝑘𝑘
𝑚 2
1.0 × 1.225 3 × 10 𝑚 × �78.3 �
𝑠
𝑚
The drag force can be calculated as:
Thus
1
𝐹𝐷 = 𝐶𝐷 𝜌𝜌𝑈 2
2
1
𝑘𝑘
𝑚 2
𝐹𝐷 = 0.05 × × 1.225 3 × 94.8 𝑚 × 10 𝑚 × �78.3 � = 178 𝑘𝑘
2
𝑚
𝑠
Problem 9.115
(Difficulty 1)
9.115 A wing model of 5 𝑖𝑖 chord and 2.5 𝑓𝑓 span is tested at a certain angle of attack in a wind tunnel
at 60 𝑚𝑚ℎ using air at 14.5 𝑝𝑝𝑝𝑝 and 70℉. The lift and drag are found to be 6.0 𝑙𝑙𝑙 and 0.4 𝑙𝑙𝑙,
respectively. Calculate the lift and drag coefficient for the model at this angle of attack.
Find: The lift and drag coefficients
Solution: Determine the coefficients from measure values of lift and drag
The velocity is:
𝑈 = 60 𝑚𝑚ℎ = 88
𝑓𝑓
𝑠
The density of air at 14.5 𝑝𝑝𝑝 and 70℉ is calculated by ideal gas law as:
The lift force is calculated by:
1
𝐹𝐿 = 𝐶𝐿 𝜌𝜌𝑈 2
2
Thus the lift coefficient is:
𝐶𝐿 =
2𝐹𝐿
=
𝜌𝜌𝑈 2
∙ 𝑠2
𝑙𝑙𝑙
0.0023
𝑓𝑓 4
The drag force is calculated by:
Thus the drag coefficient is:
𝑝
𝑠𝑠𝑠𝑠
= 0.0023
𝑅𝑅
𝑓𝑓 3
𝜌=
2 × 6.0 𝑙𝑙𝑙
𝑓𝑓 2
5
× 2.5 𝑓𝑓 × � 𝑓𝑓� × �88 �
12
𝑠
1
𝐹𝐷 = 𝐶𝐷 𝜌𝜌𝑈 2
2
= 0.647
𝐶𝐷 =
2𝐹𝐷
=
𝜌𝜌𝑈 2
𝑠2
𝑙𝑙𝑙 ∙
0.0023
𝑓𝑓 4
2 × 0.4 𝑙𝑙𝑙
𝑓𝑓 2
5
× 2.5 𝑓𝑓 × � 𝑓𝑓� × �88 �
12
𝑠
= 0.043
Problem 9.116
Problem
9.146
[Difficulty: 4]
9.116
Given:
Data on barge and river current
Find:
Speed and direction of barge
Solution:
Basic
equation:
CD 
FD
1
2
Given or available data is
2
 ρ A V
W  8820 kN w  10 m
kg
CDa  1.3 ρw  998 
3
m
2
6 m
νw  1.01  10

s
L  30 m
h  7 m
kg
ρa  1.21
3
m
m
Vriver  1 
s
νa  1.50  10
h sub 
W
ρw g  w L
 3.00 m
Vsub  w L h sub
CDw  1.3
2
 5 m (Water data from Table A.8, air

s data from Table A.10, 20 oC)
First we need to calculate the amount of the barge submerged in the water. From Archimedes' Principle:
The submerged volume can be expressed as:
m
Vwind  10
s
W  ρw g  Vsub
Combining these expressions and solving for the depth:
h air  h  h sub  4.00 m
Therefore the height of barge exposed to the wind is:
Assuming the barge is floating downstream, the velocities of the water and air relative to the barge is:
Vw  Vriver  Vbarge
Assuming that the barge is rectangular, the areas exposed to the air and water are:
Va  Vwind  Vbarge
2
Aa  L w  2  ( L  w)  h air  620 m
2
Aw  L w  2  ( L  w)  h sub  540 m
In order for the barge to be traveling at a constant speed, the drag forces due to the air and water must match:
1
2
2
 CDw ρw Vw  Aw   CDa ρa Va  Aa
2
2
1
Solving for the speed relative to the water:
2
2
2 ρa Aa
Vw  Va 

ρw Aw
ρa Aa
In terms of the barge speed:
Vw  Va

ρw Aw
So solving for the barge speed:
2
Since the drag coefficients are equal, we can simplify: ρw Vw  Aw  ρa Va  Aa
Since the speeds must be in opposite directions:
ρa Aa
Vriver  Vbarge   Vwind  Vbarge 

ρw Aw
ρa Aa
Vriver  Vwind

ρw Aw
m
Vbarge 
Vbarge  1.426
s
ρa Aa
1

ρw Aw


downstream
Problem 9.117
Problem
9.148
[Difficulty: 4]
9.117
Given:
Data on sonar transducer
Find:
Drag force at required towing speed; minimum depth necessary to avoid cavitation
Solution:
CD 
Basic equation:
FD
1
2
Given or available data is
2
 ρ A V
D  15 in
A 
π
4
2
 D  1.227  ft
2
p  p inf
1
2
V  55
The Reynolds number of the flow is: Re 
The area is:
CP 
V D
ν
ft
s
p min  5  psi ρ  1.93
 6.486  10
h 
6
ρ g
3
ν  1.06  10
 5 ft

2
(Table A.7, 70oF)
s
From Fig. 9.11, we estimate the drag coefficient:
Therefore the drag force is:
p inf  p atm
slug
ft
From Fig. 9.12 the minimum pressure occurs where CP  1.2
Solving for the required depth:
p  p atm  ρ g  h
2
 ρ V
1
2
FD   CD ρ V  A
2
Therefore:
CD  0.18
FD  645  lbf
1
2
p inf  p min   CP ρ V  29.326 psi
2
h  33.9 ft
9.118
Problem 9.118
Problem
9.149
[Difficulty: 4]
Problem 9.119
(Difficulty 2)
9.119 If the mean velocity adjacent to the top of a wing 1.8 𝑚 chord is 40
bottom of the wing 31
span.
𝑚
𝑠
when the wing moves through still air at 33.5
𝑚
,
𝑠
𝑚
𝑠
and that adjacent to the
estimate the lift per meter of
Find: The lift on an airfoil
Assumption: The flow is steady
Solution: Use the Bernoulli relations for flow to compute the pressures and the forces. For air flow
over the top of the airfoil we have:
Or
1
1
𝑝𝑎 + 𝜌𝑈𝑎2 = 𝑝𝑇 + 𝜌𝑈𝑇2
2
2
The pressure difference is then
1
1
𝑝𝑇 − 𝑝𝑎 = 𝜌𝑉𝑎2 − 𝜌𝑉𝑇2
2
2
𝑝𝑇 − 𝑝𝑎 = 0.5 × 1.225
𝑘𝑘
𝑚 2
𝑚 2
×
��33.5
�
−
�40
� � = −292 𝑃𝑃
𝑚3
𝑠
𝑠
On the bottom of the airfoil we have the same relation:
The pressure difference is
1
1
𝑝𝑎 + 𝜌𝑈𝑎2 = 𝑝𝐵 + 𝜌𝑈𝐵2
2
2
𝑝𝐵 − 𝑝𝑎 = 0.5 × 1.225
𝑘𝑘
𝑚 2
𝑚 2
×
��33.5
�
−
�31
� � = 98.8 𝑃𝑃
𝑚3
𝑠
𝑠
The difference in pressure between the top and bottom of the air foil is
𝑝𝐵 − 𝑝𝑇 = 98.8 𝑃𝑃 − (−292 𝑃𝑃) = 390 𝑃𝑃
The lift force per meter of span is then, where C is the airfoil chord:
𝐹𝐿 = (𝑝𝐵 − 𝑝𝑇 )𝐶 = 390 𝑃𝑃 × 1.8 𝑚 = 702
𝑁
𝑚
Problem 9.120
(Difficulty 2)
9.120 The NACA 23015 airfoil is to move at 180 𝑚𝑚ℎ through standard sea level air. Determine the
minimum drag, drag at optimum
𝐿
𝐷
and drag at point of maximum lift. Calculate the lift at these points
and the power that must be expended to obtain these lifts.
Find: Drag at different conditions
Solution: Use the airfoil drag characteristics given in Figure 9.17 to find the drag
The velocity is:
𝑈 = 180 𝑚𝑚ℎ = 264
The density of air is:
𝜌 = 0.00238
The area is:
𝑓𝑓
𝑠
𝑠𝑠𝑠𝑠
𝑓𝑓 3
𝐴 = 𝑐 ∙ 𝑏 = 384 𝑓𝑓 2
The minimum drag coefficient is:
𝐶𝐷 = 0.0086
The minimum drag force is:
1
𝑠𝑠𝑠𝑠
𝑓𝑓 2
1
2
𝐹𝐷 = 𝐶𝐷 𝜌𝜌𝑈 2 = 0.0086 × × 0.00238
× 384 𝑓𝑓 × �264 � = 274 𝑙𝑙𝑙
2
𝑓𝑓 3
𝑠
2
The drag coefficient at optimum
The drag force at optimum
𝐿
𝐷
is:
𝐿
𝐷
is:
𝐶𝐷 = 0.018
1
𝑠𝑠𝑠𝑠
𝑓𝑓 2
1
2
𝐹𝐷 = 𝐶𝐷 𝜌𝜌𝑈 2 = 0.018 × × 0.00238
× 384 𝑓𝑓 × �264 � = 573 𝑙𝑙𝑙
2
𝑓𝑓 3
𝑠
2
The drag coefficient at maximum lift point is:
The drag force at maximum lift point is:
𝐶𝐷 = 0.166
1
𝑠𝑠𝑠𝑠
𝑓𝑓 2
1
2 × �264
𝐹𝐷 = 𝐶𝐷 𝜌𝜌𝑈 2 = 0.166 × × 0.00238
×
384
𝑓𝑓
� = 5290 𝑙𝑙𝑙
2
𝑓𝑓 3
𝑠
2
The lift coefficient at minimum drag force point is:
𝐶𝐿 = 0
The lift at this point force is:
1
𝐹𝐿 = 𝐶𝐿 𝜌𝜌𝑈 2 = 0 𝑙𝑙𝑙
2
The lift coefficient at optimum
𝐿
𝐷
is:
The lift force at this point is:
𝐶𝐿 = 0.4
1
𝑠𝑠𝑠𝑠
𝑓𝑓 2
1
2 × �264
𝐹𝐿 = 𝐶𝐿 𝜌𝜌𝑈 2 = 0.4 × × 0.00238
×
384
𝑓𝑓
� = 12740 𝑙𝑙𝑙
2
𝑓𝑓 3
𝑠
2
The maximum lift coefficient is:
The maximum lift force is:
𝐶𝐿 = 1.51
1
𝑠𝑠𝑠𝑠
𝑓𝑓 2
1
2 × �264
𝐹𝐿 = 𝐶𝐿 𝜌𝜌𝑈 2 = 1.51 × × 0.00238
×
384
𝑓𝑓
� = 48100 𝑙𝑙𝑙
2
𝑓𝑓 3
𝑠
2
The power at minimum drag force point is:
𝑃 = 𝐹𝐷 𝑈 = 274 𝑙𝑙𝑙 × 264
𝑓𝑓
𝑙𝑙𝑙 ∙ 𝑓𝑓
= 72336
= 131.5 ℎ𝑝
𝑠
𝑠
𝑃 = 𝐹𝐷 𝑈 = 573 𝑙𝑙𝑙 × 264
𝑓𝑓
𝑙𝑙𝑙 ∙ 𝑓𝑓
= 151272
= 275 ℎ𝑝
𝑠
𝑠
The power at optimum
𝐿
𝐷
point is:
The power at maximum lift point is:
𝑃 = 𝐹𝐷 𝑈 = 5290 𝑙𝑙𝑙 × 264
𝑓𝑓
𝑙𝑙𝑙 ∙ 𝑓𝑓
= 1396560
= 2540 ℎ𝑝
𝑠
𝑠
Problem 9.121
(Difficulty 2)
9.121 A human-powered aircraft has a gross weight of 240 𝑙𝑙𝑙 including the pilot. Its wing has a lift
coefficient of 1.5 and a lift to drag ratio of 70. Estimate the wing area needed and the pilot power that
must be provided for this craft to cruise at 15 𝑚𝑚ℎ. Assume that the wing profile drag is about 40 % of
the total drag and the propeller efficiency is 80 percent.
Find: The wing area and power
Solution: Use the lift coefficient to find the power
The velocity is:
𝑈 = 15 𝑚𝑚ℎ = 22
The density of air is:
𝜌 = 0.00238
To lift the aircraft we need to have:
𝑠𝑠𝑠𝑠
𝑓𝑓 3
𝐹𝐿 = 𝑊
The lift coefficient is
1
𝐹𝐿 = 𝐶𝐿 𝜌𝜌𝑈 2
2
The required area is then:
𝐴=
𝑓𝑓
𝑠
2𝐹𝐿
2𝑊
=
=
2
𝐶𝐿 𝜌𝑈 2
𝐶𝐿 𝜌𝑈
2 × 240 𝑙𝑙𝑙
𝑙𝑙𝑙 ∙ 𝑠2
𝑓𝑓 2
1.5 × 0.00238
×
�22
�
𝑠
𝑓𝑓 4
The drag force on the wing is calculated from the lift/drag ratio of 70 as:
The total drag force is:
𝐹𝐷𝐷 =
𝐹𝐿 𝑊 240 𝑙𝑙𝑙
=
=
= 3.43 𝑙𝑙𝑙
70
70 70
𝐹𝐷𝐷 =
𝐹𝐷𝐷 3.43 𝑙𝑙𝑙
=
= 8.58 𝑙𝑙𝑙
40%
𝜂
= 278 𝑓𝑓 2
The power needed is computed as:
𝑓𝑓
𝐹𝐷𝐷 𝑈 8.58 𝑙𝑙𝑙 × 22 𝑠
𝑙𝑙𝑙 ∙ 𝑓𝑓
𝑃=
=
= 236
= 0.429 ℎ𝑝
𝑠
𝜂𝑝
0.8
Problem 9.122
Problem
9.152
9.122
[Difficulty: 5]
Problem 9.123
Problem
9.154
9.123
[Difficulty: 5] Part 1/2
Problem 9.154
[Difficulty: 5] Part 2/2
Problem
9.156
Problem 9.124
[Difficulty: 3]
9.124
Given:
Data on airfoil and support in wind tunnel, lift and drag measurements
Find:
Lift and drag coefficients of airfoil
FL
Solution:
V
Basic equations: CD 
1
2
The given or available data is
FD
2
 ρ A V
L  6  in
CL 
FL
1
2
2
 ρ A V
W  30 in
FL  10 lbf
FD
V  100 
y
ft
s
Dcyl  1  in
FD  1.5 lbf
ρ  0.00233 
slug
ft
Re 
V Dcyl
4
Re  5.112  10
ν
3
ν  1.63  10
 4 ft

2
s
FD  FDcyl  FDairfoil
We need to determine the cylindrical support's contribution to the total drag force:
Compute the Reynolds number
x
Lcyl  10 in
Therefore: CDcyl  1
1
2
So the drag force on the support is: FDcyl   CDcyl ρ V  Lcyl Dcyl  0.809  lbf
2
So the airfoil drag is: FDairfoil  FD  FDcyl  0.691  lbf The reference area for the airfoil is: A  L W  1.25 ft
The lift and drag coefficients are:
CL 
FL
1
2
CD 
2
 ρ V  A
2
CL  0.687
FDairfoil
1
2
2
 ρ V  A
CD  0.0474
Problem 9.125
Problem
9.157
[Difficulty: 2]
9.125
Given:
Antique airplane guy wires
Find:
Maximum power saving using optimum streamlining
Solution:
Basic equation:
Given or available data is
CD 
FD
1
2
 ρ A V
2
L  50 m
The Reynolds number is
Hence
Re 
V  175 
km
hr
V  48.6
m
s
A  0.25 m
kg
3
m
V D
ν
D  5  mm
2
A  L D
ρ  1.21
P  FD V
ν  1.50  10
2
5 m
Re  1.62  10

(Table A.10, 20 oC)
s
4
1
2
P   CD  ρ A V   V
2


so from Fig. 9.13
CD  1.0
P  17.4 kW
with standard wires
Figure 9.19 suggests we could reduce the drag coefficient to CD  0.06
Hence
1
2
Pfaired   CD  ρ A V   V
2


Pfaired  1.04 kW
The maximum power saving is then
ΔP  P  Pfaired
ΔP  16.3 kW
Thus
ΔP
P
 94 %
which is a HUGE savings! It's amazing the antique planes flew!
Problem 9.126
Problem
9.159
9.126
[Difficulty: 5]
Problem 9.127
Problem
9.160
9.127
[Difficulty: 1]
Problem 9.128
Problem
9.164
[Difficulty: 3]
9.128
Given:
Data on F-16 fighter
Find:
Minimum speed at which pilot can produce 5g acceleration; flight radius, effect of altitude on results
Solution:
The given data or available data is
ρ  0.00234 
slug
ft
3
A  300  ft
2
CL  1.6
W  26000  lbf
At 5g acceleration, the corresponding force is:
FL  5  W  130000 lbf
The minimum velocity corresponds to the maximum lift coefficient:
Vmin 
2  FL
ρ A CL
 481 
ft
ft
Vmin  481 
s
s
To find the flight radius, we perform a vertical force balance:
β  90 deg  asin 
FL sin ( 90 deg  β)  W  0
Now set the horizontal force equal to the centripetal acceleration:
W
  78.5 deg
 FL 
W
FL cos ( 90 deg  β) 
a
g c
ac  g 
FL
W
 cos ( 90 deg  β)
ac  157.6
ft
s
2
The flight radius corresponding to this acceleration is:
R 
As altitude increases, the density decreases, and both the velocity and radius will increase.
Vmin
ac
R  1469 ft
2
Problem 9.129
Problem
9.166
[Difficulty: 3]
9.129
Given:
Data on a light airplane
Find:
Angle of attack of wing; power required; maximum "g" force
Solution:
The given data or available data is
ρ  1.23
kg
3
2
M  1000 kg
A  10 m
CL  0.72
CD  0.17
W  M  g  FL
T  FD
m
V  63
m
s
The governing equations for steady flight are
where W is the weight T is the engine thrust
The lift coeffcient is given by
1
2
FL   ρ A V  Cd
2
Hence the required lift coefficient is
CL 
M g
1
2
2
 ρ A V
From Fig 9.17, for at this lift coefficient
α  3  deg
and the drag coefficient at this angle of attack is
CD  0.0065
CL  0.402
(Note that this does NOT allow for aspect ratio effects on lift and drag!)
Hence the drag is
1
2
FD   ρ A V  CD
2
FD  159 N
and
T  FD
T  159 N
The power required is then
P  T V
P  10 kW
The maximum "g"'s occur when the angle of attack is suddenly increased to produce the maximum lift
From Fig. 9.17
CL.max  1.72
1
2
FLmax   ρ A V  CL.max
2
The maximum "g"s are given by application of Newton's second law
M  aperp  FLmax
where a perp is the acceleration perpendicular to the flight direction
FLmax  42 kN
Hence
In terms of "g"s
aperp 
aperp
g
FLmax
aperp  42
M
m
2
s
 4.28
Note that this result occurs when the airplane is banking at 90 o, i.e, when the airplane is flying momentarily in a
circular flight path in the horizontal plane. For a straight horizontal flight path Newton's second law is
M  aperp  FLmax  M  g
Hence
In terms of "g"s
aperp 
aperp
g
FLmax
M
 3.28
g
aperp  32.2
m
2
s
Problem 9.131
Problem
9.169
9.131
[Difficulty: 3]
Problem 9.132
Problem
9.172
9.132
[Difficulty: 3]
Problem 9.133
Problem
9.175
[Difficulty: 4]
9.133
Given:
Car spoiler
Find:
Whether they are effective
Solution:
To perform the investigation, consider some typical data
For the spoiler, assume
b  4 ft
c  6 in
ρ  1.23
kg
3
A  b c
m
From Fig. 9.17 a reasonable lift coefficient for a conventional airfoil section is
Assume the car speed is
V  55 mph
Hence the "negative lift" is
1
2
FL   ρ A  V  CL
2
CL  1.4
FL  21.7 lbf
This is a relatively minor negative lift force (about four bags of sugar); it is not likely to produce a noticeable
difference in car traction
The picture gets worse at 30 mph:
FL  6.5 lbf
For a race car, such as that shown on the cover of the text, typical data might be
b  5  ft
In this case:
c  18 in
A  b c
FL  1078 lbf
Hence, for a race car, a spoiler can generate very significant negative lift!
A  7.5 ft
2
V  200  mph
A  2 ft
2
Problem 9.134
Problem
9.176
9.134
[Difficulty: 5]
Problem 9.135
Problem
9.178
9.135
[Difficulty: 2]
Problem 9.136
Problem
9.180
[Difficulty: 2]
9.136
Given:
Data on rotating cylinder
Find:
Lift force on cylinder
Solution:
CL 
Basic equations:
FL
1
2
The given or available data is
2
 ρ A V
ρ  1.21
kg
3
2
5 m
ν  1.50 10

m
The spin ratio is:
The area is
ω D
2 V
 0.419
s
L  30 cm
D  5  cm
ω  240  rpm
V  1.5
m
s
From Fig. 9.29, we can estimate the maximum lift coefficient: CL  1.0
2
A  D L  0.015 m
Therefore, the lift force is:
1
2
FL   CL ρ A V
2
FL  0.0204 N
Problem 9.137
Problem
9.182
[Difficulty: 2]
9.137
Data on original Flettner rotor ship
Given:
Find:
Maximum lift and drag forces, optimal force at same wind speed, power requirement
Solution:
CL 
Basic equations:
FL
1
2
The given or available data is
2
 ρ A V
ρ  0.00234 
slug
ft
ν  1.62  10
The spin ratio is:
The area is
ω D
2 V
 9.52
A  D L  500  ft
L  50 ft
3
 4 ft

D  10 ft
ω  800  rpm
V  30 mph  44
ft
s
2
s
From Fig. 9.29, we can estimate the lift and drag coefficients: CL  9.5 CD  3.5
2
Therefore, the lift force is:
1
2
FL   CL ρ A V
2
FL  1.076  10  lbf
The drag force is:
1
2
FD   CD ρ A V
2
FD  3.964  10  lbf
This appears to be close to the optimum L/D ratio. The total force is:
F 
2
4
3
2
4
FL  FD
To determine the power requirement, we need to estimate the torque on the cylinder.
F  1.147  10  lbf
T  τ A R  τ π L D
D
2
2

π τ D  L
2
In this expression τ is the average wall shear stress. We can estimate this stress using the flat plate approximation:
 V  ω D   D


2
7

Re 
 2.857  10
ν
τ
FD
A
τ 
1
2
For a cylinder at this Reynolds number: CD  0.003 Therefore, the shear stress is:
 ρ V  CD  6.795  10
2
 3 lbf

ft
2
2
So the torque is:
T 
π τ D  L
2
The power is: P  T ω  4471
 53.371 ft lbf
ft lbf
s
P  8.13 hp
Problem 9.138
Problem
9.183
[Difficulty: 4]
9.138
x
R
L

Given:
Baseball pitch
Find:
Spin on the ball
Solution:
Basic equations:
1
2
The given or available data is


Σ F  M  a
FL
CL 
2
 ρ A V
ρ  0.00234 
slug
ft
M  5  oz
Compute the Reynolds number
C  9  in
ν  1.62  10
3
D 

2
L  60 ft
s
2
C
D  2.86 in
π
V D
Re 
 4 ft
Re  1.73  10
ν
A 
π D
2
A  6.45 in
4
V  80 mph
5
This Reynolds number is slightly beyond the range of Fig. 9.27; we use Fig. 9.27 as a rough estimate
The ball follows a trajectory defined by Newton's second law. In the horizontal plane ( x coordinate)
2
V
FL  M  aR  M  ax  M 
R
1
2
FL   ρ A V  CL
2
and
where R is the instantaneous radius of curvature of the trajectory
From Eq 1 we see the ball trajectory has the smallest radius (i.e. it curves the most) when C L is as large as
possible. From Fig. 9.27 we see this is when CL  0.4
Solving for R
Also, from Fig. 9.27
Hence
From the trajectory geometry
R 
ω D
2 V
2 M
(1)
CL A ρ
 1.5
ω  1.5
to
2 V
D
x  R cos( θ)  R
Solving for x
x  R  R 1 
L
 
 R
2 V
 1.8
ω  1.8
where
sin( θ) 
L  R
 
 R
x  R 1 
ω D
ω  14080  rpm
2
Hence
R  463.6  ft
2
x  3.90 ft
2 V
D
L
R
defines the best range
ω  16896  rpm
Problem 8.1
Problem
8.2
8.1
[Difficulty: 2]
Problem 8.2
(Difficulty 2)
8.2 What is the maximum flow rate of air that may occur at laminar condition in a 4 in diameter pipe at
an absolute pressure of 30 𝑝𝑝𝑝𝑝 and 100 ℉ ? If the pressure is raised to 60 𝑝𝑝𝑝𝑝, what is the maximum
flow rate ? If the temperature is raise to 200 ℉, what is the maximum flow rate? Explain the differences
in answers in terms of the physical mechanisms involved.
Find: The maximum flow rate for laminar flow.
Assumption: Air behaves as an ideal gas
Solution: The basic equations are the definition of Reynolds number, the continuity expression, and the
ideal gas law
𝑅𝑅 =
𝑉𝑉𝑉 𝑉𝑉
=
𝜈
𝜇
𝑚 = 𝜌̇ 𝐴 𝑉
𝜌=
We have
𝑝
𝑅𝑅
𝑝 = 30 𝑝𝑝𝑝𝑝 = 4320
𝑅 = 1716
𝑙𝑙𝑙 ∙ 𝑓𝑓
𝑠𝑠𝑠𝑠 ∙ °𝑅
𝑇 = 100 ℉ = 560 °𝑅
Thus the density is
𝑝
𝜌=
=
𝑅𝑅
𝑙𝑙𝑙
𝑓𝑓 2
𝑙𝑙𝑙
𝑠𝑠𝑠𝑠
𝑙𝑙𝑙 ∙ 𝑠2
𝑓𝑓 2
= 0.0045
= 0.0045
𝑙𝑙𝑙 ∙ 𝑓𝑓
𝑓𝑓 3
𝑓𝑓 4
1716
× 560 °𝑅
𝑠𝑠𝑠𝑠 ∙ °𝑅
4320
For the maximum laminar flow we have the Reynolds number at the critical value:
For this situation
𝑅𝑅𝑐𝑐𝑐𝑐 = 2300
𝐷 = 4 𝑖𝑖
𝜇 = 3.94 × 10−7
The maximum velocity is then:
The cross section area is:
𝑙𝑙𝑙 ∙ 𝑠
𝑓𝑓 2
−7 𝑙𝑙𝑙 ∙ 𝑠 × 2300
𝜇𝜇𝜇 3.94 × 10
𝑓𝑓
𝑓𝑓 2
𝑉=
=
= 0.605
2
𝑙𝑙𝑙 ∙ 𝑠
4
𝜌𝜌
𝑠
� 𝑓𝑓� × 0.0045
12
𝑓𝑓 4
𝐴=
The maximum flow rate:
𝑚̇ = 𝜌𝜌𝜌 = 0.0045
𝜋𝐷 2
4
=
𝜋×�
2
4
𝑓𝑓�
12
= 0.0873 𝑓𝑓 2
4
𝑓𝑓
𝑠𝑠𝑠𝑠
𝑠𝑠𝑠𝑠
× 0.605
× 0.0873 𝑓𝑓 2 = 2.38 × 10−4
3
𝑠
𝑠
𝑓𝑓
If the pressure is raised up to 60 𝑝𝑝𝑝 = 8640
𝑙𝑙𝑙
,
𝑓𝑓 2
the density of the air will become:
𝑙𝑙𝑙
𝑠𝑠𝑠𝑠
𝑙𝑏𝑏 ∙ 𝑠2
𝑓𝑓 2
𝜌=
= 0.009
=
0.009
𝑙𝑙𝑙 ∙ 𝑓𝑓
𝑓𝑓 3
𝑓𝑓 4
1716
× 560 °𝑅
𝑠𝑠𝑠𝑠 ∙ °𝑅
8640
The maximum velocity in this case is:
−7 𝑙𝑙𝑙 ∙ 𝑠 × 2300
𝜇𝜇𝜇 3.94 × 10
𝑓𝑓
𝑓𝑓 2
𝑉=
=
= 0.302
2
𝑙𝑙𝑙 ∙ 𝑠
4
𝜌𝜌
𝑠
� 𝑓𝑓� × 0.009
12
𝑓𝑓 4
And the maximum flow rate:
𝑚̇ = 𝜌𝜌𝜌 = 0.009
𝑠𝑠𝑠𝑠
𝑓𝑓
𝑠𝑠𝑠𝑠
× 0.302
× 0.0873 𝑓𝑓 2 = 2.38 × 10−4
3
𝑓𝑓
𝑠
𝑠
The maximum flow rate is the same. The reason is that the density cancels out of the flow rate using
Reynolds number:
𝑚̇ = 𝜌𝜌𝜌 =
𝜇𝜇𝜇𝜇
𝜇𝜇𝜇
𝜌𝜌 =
𝐷
𝜌𝜌
The pressure will not change the viscosity 𝜇 (as an assumption)
When the temperature is raised to 200℉, the viscosity decreases. The density also decreases, but we
have seen that this has no effect. The viscosity is
𝜇 = 4.49 × 10−7
𝑙𝑙𝑙 ∙ 𝑠
𝑓𝑓 2
Thus the flow rate is the ratio of the previous flow rate times the ratio of viscosities
𝑚̇ = 2.38 × 10−4
−7 𝑙𝑙𝑙 ∙ 𝑠
𝑠𝑠𝑠𝑠 4.49 × 10
𝑠𝑠𝑠𝑠
𝑓𝑓 2
×
= 2.71 × 10−4
𝑙𝑙𝑙 ∙ 𝑠
𝑠
𝑠
3.94 × 10−7
𝑓𝑓 2
The reason for the increase is that the viscosity is a function of temperature.
Problem
8.4
Problem 8.3
[Difficulty: 2]
8.3
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
Vw =
2 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
s
4
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 8.4
(Difficulty 2)
8.4 Air flows at 100 𝐹 in a pipe system in which the diameter increases in two stages from 2 𝑖𝑖 to 3 𝑖𝑖
to 4 𝑖𝑖. Each section is 6 𝑓𝑓 long. The initial flow rate is high enough so that the flow is turbulent in all
sections. As the flow rate is decreased, which section will become laminar first? Determine the flow
rates at which one, two and then three sections first become laminar. At each of those flow rates,
determine which , if any, of the sections attain fully developed flow.
Find: Flow rates for laminar flow
Assumptions: The flow is steady
Solution: Use the Reynolds number criteria to determine the transition to laminar flow
The Reynolds number is defined as:
𝑄
𝜋 2𝐷
𝑉𝑉 4 𝐷
4𝑄
𝑅𝑅 =
=
=
𝑣
𝜋𝑣𝐷
𝑣
In terms of Reynolds number, the volume flow rate is
The critical Reynolds number is:
𝑄=
𝑅𝑅 𝜋 𝑣 𝐷
4
𝑅𝑅𝑐 = 2300
From Table A.9 we have the viscosity of air as:
We also have the following dimensions:
𝐿 = 6 𝑓𝑓,
𝐷1 = 2 𝑖𝑖 =
𝑣 = 1.79 × 10−4
1
𝑓𝑓,
6
𝑓𝑓 2
𝑠
𝐷2 = 3 𝑖𝑖 = 0.25 𝑓𝑓,
𝐷3 = 4 𝑖𝑖 =
1
𝑓𝑓
3
As the flow rate is decreased, to achieve the critical Reynolds number, we have in each section:
−4
𝑅𝑅𝑐 𝜋𝜋𝐷1 2300 × 𝜋 × 1.79 × 10
𝑄1 =
=
4
4
𝑅𝑅𝑐 𝜋𝜋𝐷2 2300 × 𝜋 × 1.79 × 10
=
𝑄2 =
4
4
−4
𝑅𝑅𝑐 𝜋𝜋𝐷3 2300 × 𝜋 × 1.79 × 10
=
𝑄3 =
4
4
𝑓𝑓 2 1
× 𝑓𝑓
𝑓𝑓 3
6
𝑠
= 0.0539
𝑠
𝑓𝑓 2
× 0.25 𝑓𝑓
𝑓𝑓 3
𝑠
= 0.0808
𝑠
−4
𝑓𝑓 2 1
× 𝑓𝑓
𝑓𝑓 3
3
𝑠
= 0.1078
𝑠
So the largest pipe becomes laminar flow first (at the highest flow rate).
For the largest pipe transition to laminar (𝑸𝟑 ):
For pipe 3:
𝑅𝑅3 = 2300, 𝐿𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = 0.06𝑅𝑅3 𝐷3 = 46 𝑓𝑓, 𝐿𝑙𝑙𝑙𝑙𝑙𝑙𝑙 > 𝐿: Flow is not fully developed.
Or for turbulent flow:
𝐿𝑚𝑚𝑚 = 25𝐷3 = 8.33 𝑓𝑓, 𝐿𝑚𝑚𝑚 = 40𝐷3 = 13.33 𝑓𝑓, 𝐿𝑚𝑚𝑚/𝑚𝑚𝑚 > 𝐿: Not fully developed.
For pipe 1 and 2:
𝐿1𝑚𝑚𝑚 = 25𝐷1 = 4.16𝑓𝑓, 𝐿2𝑚𝑚𝑚 = 40𝐷1 = 6.67 𝑓𝑓, 𝐿1𝑚𝑚𝑚 < 𝐿 < 𝐿2𝑚𝑚𝑚 , may be fully developed or not.
𝐿2𝑚𝑚𝑚 = 25𝐷2 = 6.25𝑓𝑓, 𝐿2𝑚𝑚𝑚 = 40𝐷1 = 10 𝑓𝑓, 𝐿𝑚𝑚𝑚/𝑚𝑚𝑚 > 𝐿: Not fully developed.
For the middle pipe transition to laminar (𝑸𝟐 ):
For pipe 2
𝑅𝑅2 = 2300, 𝐿𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = 0.06𝑅𝑅2 𝐷2 = 34.5 𝑓𝑓, 𝐿𝑙𝑙𝑙𝑙𝑙𝑙𝑙 > 𝐿: Not fully developed.
Or for turbulent:
𝐿2𝑚𝑚𝑚 = 25𝐷2 = 6.25 𝑓𝑓, 𝐿2𝑚𝑚𝑚 = 40𝐷3 = 10 𝑓𝑓, 𝐿𝑚𝑚𝑚/𝑚𝑚𝑚 > 𝐿: Not fully developed.
For pipe 3
𝑄=
𝑅𝑅3 =
𝑅𝑅𝑅𝑅𝑅
4
𝑅𝑅3 𝐷3 = 𝑅𝑅2 𝐷2
𝑅𝑅2 𝐷2 2300 × 0.25 𝑓𝑓
=
= 1725
1
𝐷3
𝑓𝑓
3
𝐿𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = 0.06𝑅𝑅3 𝐷3 = 34.5 𝑓𝑓, 𝐿𝑙𝑙𝑙𝑙𝑙𝑙𝑙 > 𝐿: Not fully developed.
For pipe 1
𝐿1𝑚𝑚𝑚 = 25𝐷1 = 4.16𝑓𝑓, 𝐿2𝑚𝑚𝑚 = 40𝐷1 = 6.67 𝑓𝑓, 𝐿1𝑚𝑚𝑚 < 𝐿 < 𝐿2𝑚𝑚𝑚 , may be fully developed or not.
For the smallest pipe transition to laminar (𝑸𝟏 ):
For pipe 1
𝑅𝑅1 = 2300, 𝐿𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = 0.06𝑅𝑅1 𝐷1 = 23 𝑓𝑓, 𝐿𝑙𝑙𝑙𝑙𝑙𝑙𝑙 > 𝐿: Not fully developed.
Or for turbulent:
𝐿1𝑚𝑚𝑚 = 25𝐷1 = 4.16𝑓𝑓, 𝐿2𝑚𝑚𝑚 = 40𝐷1 = 6.67 𝑓𝑓, 𝐿1𝑚𝑚𝑚 < 𝐿 < 𝐿2𝑚𝑚𝑚 , may be fully developed or not.
For pipe 3
𝑄=
𝑅𝑅𝑅𝑅𝑅
4
𝑅𝑅3 𝐷3 = 𝑅𝑅1 𝐷1
1
𝑅𝑅1 𝐷1 2300 × 6 𝑓𝑓
=
= 1150
𝑅𝑅3 =
1
𝐷3
𝑓𝑓
3
𝐿𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = 0.06𝑅𝑅3 𝐷3 = 23 𝑓𝑓, 𝐿𝑙𝑙𝑙𝑙𝑙𝑙𝑙 > 𝐿: Not fully developed.
For pipe 2
𝑄=
𝑅𝑅𝑅𝑅𝑅
4
𝑅𝑅2 𝐷2 = 𝑅𝑅1 𝐷1
1
𝑅𝑅1 𝐷1 2300 × 6 𝑓𝑓
=
= 1533
𝑅𝑅3 =
0.25 𝑓𝑓
𝐷3
𝐿𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = 0.06𝑅𝑅2 𝐷2 = 23 𝑓𝑓, 𝐿𝑙𝑙𝑙𝑙𝑙𝑙𝑙 > 𝐿: Not fully developed.
Problem
8.6
Problem 8.5
8.5
[Difficulty: 2]
Problem
8.9
Problem 8.6
[Difficulty: 3]
8.6
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 =
=
2
A
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
3
2
2
3
ft
1
− 5 ft
⎞ × 358 ⋅ lbf × 144 ⋅ in ×
× Q = 1.25 × 10 ⋅
2
2
−3
s
12⋅ in ⎠
2 ⋅ in
in
1 ⋅ ft
1.25 × 10
lbf ⋅ s
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
×
s
π
⎛ 1 × 12⋅ in ⎞
⎜
⎝ 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 8.11
8.7
Problem
[Difficulty: 2]
8.7
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
⋅⎛−
2⋅ y ⎞
⎜ 2
⎝ h ⎠
= −y ⋅
h
2
Basic equation
⋅
⎡
dp
2⋅ μ dx
⋅ ⎢1 −
⎣
b
=−
2
⎛ y ⎞ ⎤⎥
⎜
⎝h⎠ ⎦
(from Eq. 8.7)
dp
dx
1⋅ m
3 N
τyx = −1.5⋅ mm ×
× 1.25 × 10 ⋅
2
1000⋅ mm
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
2
0.5⋅ N⋅ s
1000⋅ mm ⎠
m ⋅m
Q
b
2⋅ h ⋅ b dp
⋅
3⋅ μ dx
= −5.63 × 10
2
−6 m
s
Problem 8.8
(Difficulty 2)
8.8 Calculate 𝛼 for the flow in the two-dimensional passage if 𝑄 is 1.5
𝑚2
.
𝑠
Find: The kinetic energy coefficient 𝛼 for this velocity profile.
Solution: Use the definition of α, the kinetic energy coefficient. The parameter α is defined as:
∫ 𝜌𝑉 3 𝑑𝑑
∫ 𝑉 3 𝑑𝑑
𝛼 = � 2𝐴
= � 2𝐴
𝑉 ∫𝐴 𝜌𝜌𝜌𝜌 𝑉 ∫𝐴 𝑉𝑉𝑉
According to the figure, we can express the velocity as:
𝑦 𝑦2
𝑉 = 𝑉𝑐 �2 − 2 � , 0 < 𝑦 < 𝑎
𝑎 𝑎
𝑉 = 𝑉𝑐 , 𝑎 < 𝑦 < 0.3
As the flow is symmetric, we can integrate one-half of the domain to get the volumetric flow rate is:
𝑎
As we know:
0.3
𝑦 𝑦2
𝑄 = � 𝑉𝑉𝑉 = 2 �� 𝑉𝑐 �2 − 2 � 𝑑𝑑 + � 𝑉𝑐 𝑑𝑑�
𝑎 𝑎
𝐴
0
𝑎
𝑉𝑐 = 3
𝑄 = 1.5
Solving the equation we have the parameter a equal to:
𝑚
𝑠
𝑚2
𝑠
𝑎 = 0.15 𝑚
Thus
�𝑉
𝐴
3 𝑑𝑑
𝑎
3
0.3
𝑦 𝑦2
𝑚4
3
= 2 �� 𝑉𝑐 �2 − 2 � 𝑑𝑑 + � 𝑉𝑐 𝑑𝑑� = 11.82 3
𝑎 𝑎
𝑠
0
𝑎
3
𝑚2
𝑄 1.5 𝑠
𝑚
𝑉� = =
= 2.5
𝐴
𝑠
0.6 𝑚
Thus
∫ 𝑉 3 𝑑𝑑
𝛼 = � 2𝐴
=
𝑉 ∫ 𝑉𝑉𝑉
𝐴
11.82
𝑚4
𝑠3
𝑚 2
𝑚2
�2.5 � × 1.5
𝑠
𝑠
= 1.261
This is a 26 % increase in the actual kinetic energy over that calculated using the average velocity.
Problem 8.9
(Difficulty 2)
8.9 If the velocity file in a two-dimensional open channel may be approximated by the parabola shown,
calculate the flow rate and the kinetic energy coefficient 𝛼.
Find: The flow rate and the parameter 𝛼 for the flow.
Assumption: Fluid is incompressible
Solution: Use the definition of the kinetic energy coefficient α, which is defined as:
∫ 𝜌𝑉 3 𝑑𝑑
∫𝐴 𝑉 3 𝑑𝑑
𝛼 = � 2𝐴
=
𝑉 ∫𝐴 𝜌𝜌𝜌𝜌 𝑉� 2 ∫𝐴 𝑉𝑉𝑉
Choose the coordinate system such that 𝑦 = 0 corresponds to the maximum velocity. Then we can write
our velocity profile as:
𝑉 = −𝑘𝑦 2 + 4
At the bottom surface we have:
𝑦 = −8 𝑓𝑓
Substitute these into the equation we can have:
𝑉 = 2𝑓𝑓
𝑘=
The volumetric flow rate per unit width is:
𝑉=−
2
𝑄 = � 𝑉𝑉𝑉 = � �−
𝐴
−8
1
32
1 2
𝑦 +4
32
1 2
𝑓𝑓 2
𝑦 + 4� 𝑑𝑑 = 34.6
32
𝑠
The average velocity is:
𝑓𝑓 2
𝑓𝑓
𝑄 34.6 𝑠
= 3.46
𝑉� = =
𝑠
ℎ
10 𝑓𝑓
For the kinetic energy coefficient we need the following integral
� 𝑉 3 𝑑𝑑
𝐴
3
1 2
𝑓𝑓 4
= � �− 𝑦 + 4� 𝑑𝑑 = 448 3
32
𝑠
−8
2
Using the definition of the kinetic energy coefficient, we have
𝑓𝑓 4
𝑠3
𝛼=
= 1.082
𝑓𝑓
𝑓𝑓
𝑓𝑓 2
3.46
× 3.46
× 34.6
𝑠
𝑠
𝑠
448
Problem 8.10
Problem
8.12
[Difficulty: 3]
8.10
Given:
Piston-cylinder assembly
Find:
Mass supported by piston
Solution:
Basic equation
Available data
Q
l
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
2
68 °F = 20 °C
− 3 lbf ⋅ s
μ = 2.089 × 10
or
⋅
m
Hence, solving for ∆p
∆p =
12⋅ μ⋅ L⋅ Q
π⋅ D⋅ a
4
∆p = 2.133 × 10 ⋅ psi
3
F = ∆p⋅ A = ∆p⋅
A force balance for the piston involves the net pressure force
2
π⋅ D
M =
Note the following
Q
Vave =
a ⋅ π⋅ D
4
⋅
∆p
Hence
2
ft
π
4
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 8.11
Problem
8.14
8.11
[Difficulty: 3]
Problem 8.12
Problem
8.16
8.12
[Difficulty: 3]
Problem 8.13
(Difficulty 2)
8.13 When a horizontal laminar flow occurs between two parallel plates of infinite extent 0.3 𝑚 apart,
the velocity at the midpoint between the plates is 2.7
𝑚
.
𝑠
Calculate (a) the flow rate through a cross
section 0.9 𝑚 wide, (b) the velocity gradient at the surface of the plate, (c) the wall shearing stress if the
fluid has viscosity 1.44 𝑃𝑃 ∙ 𝑠, (d) the pressure drop in each 30 𝑚 along the flow.
Assumptions Flow is steady, fully established, and incompressible.
Solution: Use the expressions for the velocity profile for laminar flow between parallel plate. For
this laminar flow we have the velocity profile in terms of position and pressure gradient as:
𝑢=
The velocity gradient is
𝑎2 𝜕𝜕
𝑦 2
𝑦
� � �� � − � ��
𝑎
𝑎
2𝜇 𝜕𝜕
2𝑦
1
𝑑𝑑 𝑎2 𝜕𝜕
=
� � �� 2 � − � ��
𝑎
𝑎
𝑑𝑑 2𝜇 𝜕𝜕
In this particular case:
𝑎 = 0.3 𝑚
For the velocity at the midpoint we have:
𝑦 = 0.15 𝑚
(0.3 𝑚)2 𝜕𝜕
𝑎2 𝜕𝜕
𝑦 2
𝑦
0.15 𝑚 2
0.15 𝑚
𝑚
� � �� � − � �� =
� � ��
� −�
�� = 2.7
𝑉𝑐 =
𝑎
𝑎
𝜕𝜕
0.3 𝑚
0.3 𝑚
𝑠
2𝜇 𝜕𝜕
2𝜇
Thus the pressure gradient is
1 𝜕𝜕
� �=
2𝜇 𝜕𝜕
𝑚
1
𝑠
= −120
2
𝑚∙𝑠
0.15 𝑚
0.15 𝑚
(0.3 𝑚)2 × ��
� −�
��
0.3 𝑚
0.3 𝑚
2.7
(a) The average velocity is then:
1
−120
1 𝜕𝜕 2
𝑚
∙ 𝑠 × (0.3 𝑚)2 = 1.8 𝑚
𝑉� = −
� �𝑎 = −
12𝜇 𝜕𝜕
𝑠
6
The volumetric flow rate for width 𝑑 = 0.9 𝑚 is:
𝑄 = 𝑉�𝐴 = 1.8
𝑚
𝑚3
× 0.3 𝑚 × 0.9 𝑚 = 0.486
𝑠
𝑠
(b) The velocity gradient at the surface of the plate 𝑦 = 0 or 𝑦 = 0.3 𝑚.
At 𝑦 = 0:
2𝑦
1
1
1
1
𝑑𝑑 𝑎2 𝜕𝜕
=
� � �� 2 � − � �� = (0.3 𝑚)2 × �−120
� �− �
�� = 36
𝑎
𝑎
𝑚∙𝑠
0.3 𝑚
𝑠
𝑑𝑑 2𝜇 𝜕𝜕
At 𝑦 = 0.3 𝑚
2𝑦
1
1
2 × 0.3𝑚
1
1
𝑑𝑑 𝑎2 𝜕𝜕
=
� � �� 2 � − � �� = (0.3 𝑚)2 × �−120
� ��
�
−
�
��
=
−36
(0.3 𝑚)2
𝑎
𝑎
𝑚∙𝑠
0.3𝑚
𝑠
𝑑𝑑 2𝜇 𝜕𝜕
(c) For the shear stress of the wall we have:
𝑑𝑑
1
𝜏𝑤 = 𝜇
= 1.44 𝑃𝑃 ∙ 𝑠 × 36 = 51.8 𝑃𝑃
𝑑𝑑
𝑠
(d) As the viscosity we have:
𝜇 = 1.44 𝑃𝑃 ∙ 𝑠
Thus
𝜕𝜕
1
𝑃𝑃
� � = �−120
� × 2 × 1.44 𝑃𝑃 ∙ 𝑠 = −346
𝜕𝜕
𝑚∙𝑠
𝑚
For the length we have is:
The pressure drop is:
∆𝑥 = 30 𝑚
𝜕𝜕
𝑃𝑃
∇𝑝 = � � ∆𝑥 = −346
× 30 𝑚 = −10380 𝑃𝑃 = −10.38 𝑘𝑘𝑘
𝜕𝜕
𝑚
Problem 8.14
(Difficulty 2)
8.14 In a laminar flow of water at 0.007
𝑚2
𝑠
between parallel plates spaced 75 𝑚𝑚 apart, the measured
shearing stress at the pipe wall is 47.9 𝑃𝑃. What is the viscosity of the fluid? Is the flow laminar?
Find: The viscosity of the fluid and whether the flow is laminar.
Assumptions: Flow is fully developed, steady, and incompressible
Solution: Use the expressions for the velocity profile for laminar flow between parallel plates. We have
the following equations:
𝑢=
𝑎2 𝜕𝜕
𝑦 2
𝑦
� � �� � − � ��
𝑎
𝑎
2𝜇 𝜕𝜕
𝑑𝑑 𝑎2 𝜕𝜕
2𝑦
1
=
� � �� 2 � − � ��
𝑑𝑑 2𝜇 𝜕𝜕
𝑎
𝑎
𝑄
1 𝜕𝜕 3
𝑚2
=−
� � 𝑎 = 0.007
𝐿
12𝜇 𝜕𝜕
𝑠
From the relation for flow rate
𝜕𝜕 𝑎3
𝜇 = −� �
𝜕𝜕 12 𝑄
𝐿
𝑎 = 0.075 𝑚
The shear stress at the wall is:
𝜏𝑤 = 𝜇
Thus
�
𝑑𝑑
𝑎 𝜕𝜕
= − � � = 48.9 𝑃𝑃
𝑑𝑑
2 𝜕𝜕
𝜕𝜕
48.9 𝑃𝑃 × 2
𝑃𝑃
�=−
= −1304
𝜕𝜕
0.075 𝑚
𝑚
(0.075 𝑚)3
𝜕𝜕 𝑎3
𝑃𝑃
𝜇 = −� �
= − �−1304
�×
= 6.55 𝑃𝑃 ∙ 𝑠
𝑚2
𝜕𝜕 12 𝑄
𝑚
12
×
0.007
𝐿
𝑠
The Reynolds number is defined as:
𝑘𝑘
𝑚2
𝑄
𝜌𝑉�𝑎 𝜌 𝐿 998 𝑚3 × 0.007 𝑠
𝑅𝑅 =
=
=
= 1.06 ≪ 2300
𝜇
𝜇
6.55 𝑃𝑃 ∙ 𝑠
The fluid flow is laminar.
Problem 8.15
Problem
8.18
8.15
[Difficulty: 4]
Problem 8.16
Problem
8.20
8.16
[Difficulty: 2]
Problem 8.17
Problem
8.21
8.17
[Difficulty: 3]
8.15
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
⎜
⎝ k L ⎠ 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 8.18
(Difficulty 2)
8.18 In a laminar flow between parallel plates spaced 12 𝑖𝑖 apart, the shear stress at the wall is 1.0 𝑝𝑝𝑝
and the fluid viscosity 0.002
centerline?
𝑙𝑙𝑙∙𝑠
.
𝑓𝑓 2
What is the centerline velocity and the velocity gradient 1 𝑖𝑖 from the
Find: The centerline velocity and velocity gradient 1 𝑖𝑖 from the centerline.
Assumptions: Flow is fully developed, steady, and incompressible
Solution: Use the expressions for the velocity profile for laminar flow between parallel plates. For this
flow we have the velocity as:
𝑢=
𝑎2 𝜕𝜕
𝑦 2
𝑦
� � �� � − � ��
𝑎
𝑎
2𝜇 𝜕𝜕
𝑑𝑑 𝑎2 𝜕𝜕
2𝑦
1
=
� � �� 2 � − � ��
𝑑𝑑 2𝜇 𝜕𝜕
𝑎
𝑎
In this particular case
The shear stress on the wall at 𝑦 = 0 is:
𝑎 = 12 𝑖𝑖 = 1 𝑓𝑓
𝑑𝑑 𝑎2 𝜕𝜕
2𝑦
1
= � � �� 2 � − � ��
𝜏𝑤 = 𝜇
𝑑𝑑
𝑎
𝑎
2 𝜕𝜕
The pressure gradient is related to the wall shear stress as
𝑙𝑙𝑙
2 × 1.0 2
𝜕𝜕
2𝜏𝑤
𝑙𝑙𝑙
𝑓𝑓
� �=
=
= −2 3
1
[(−1𝑓𝑓)]
𝜕𝜕
𝑓𝑓
𝑎2 �− � ��
𝑎
At the centerline:
𝑢=
𝑦=
𝑎
2
(1𝑓𝑓)2
𝑙𝑙𝑙
1 2
1
𝑓𝑓
× �−2 3 � × �� � − � �� = 125
𝑙𝑙𝑙 ∙ 𝑠
𝑓𝑓
2
2
𝑠
2 × 0.002
𝑓𝑓 2
The velocity gradient 1 𝑖𝑖 from the centerline is at 𝑦 = 5 𝑖𝑖 𝑜𝑜 7𝑖𝑖.
At 𝑦 = 5 𝑖𝑖
2𝑦
1
𝑑𝑑 𝑎2 𝜕𝜕
=
� � �� 2 � − � �� =
𝑎
𝑎
𝑑𝑑 2𝜇 𝜕𝜕
At 𝑦 = 7 𝑖𝑖
5
2×
𝑓𝑓
(1𝑓𝑓)2
𝑙𝑙𝑙
1
1
12
�−2 3 � ��
�−�
�� = 83.3
2
𝑙𝑙𝑙 ∙ 𝑠
(1𝑓𝑓)
𝑓𝑓
1 𝑓𝑓
𝑠
2 × 0.002
𝑓𝑓 2
1
𝑑𝑑
= −83.3
𝑠
𝑑𝑑
The gradient is equal in magnitude and opposite in sign.
Problem 8.19
(Difficulty 2)
8.19 A fluid of specific gravity 0.9 flows at a Reynolds number of 1500 between parallel plates spaced
0.3 𝑚 apart. The velocity 50 𝑚𝑚 from the wall is 3
at the wall.
𝑚
.
𝑠
Calculate the flow rate and the velocity gradient
Find: The flow rate and velocity gradient at the wall.
Assumptions: Flow is fully developed, steady, and incompressible
Solution: Use the expressions for the velocity profile for laminar flow between parallel plate. From the
Reynolds number, we know this is laminar flow. For this flow we have the velocity profile as:
𝑢=
𝑎2 𝜕𝜕
𝑦 2
𝑦
� � �� � − � ��
𝑎
𝑎
2𝜇 𝜕𝜕
𝑑𝑑 𝑎2 𝜕𝜕
2𝑦
1
=
� � �� 2 � − � ��
𝑑𝑑 2𝜇 𝜕𝜕
𝑎
𝑎
In this particular case
𝑎 = 0.3 𝑚
The density of the fluid is:
𝜌 = 0.9𝜌𝐻2 𝑜 = 0.9 × 998
The velocity at 𝑦 = 50 𝑚𝑚 = 0.05 𝑚 is:
𝑢=
𝑘𝑘
𝑘𝑘
= 898 3
3
𝑚
𝑚
(0.3 𝑚)2 𝜕𝜕
0.05 2
0.05
𝑚
� � ��
� −�
�� = 3
𝜕𝜕
0.3
0.3
𝑠
2𝜇
The term with the pressure gradient is then
1 𝜕𝜕
� �=
2𝜇 𝜕𝜕
The average velocity is:
𝑚
1
𝑠
= −72
2
𝑚∙𝑠
0.05
0.05
(0.3 𝑚)2 × ��
� −�
��
0.3
0.3
3
1
−72
1 𝜕𝜕 2
𝑚
∙ 𝑠 × (0.3 𝑚)2 = 1.08 𝑚
� �𝑎 = −
𝑉� = −
𝑠
12𝜇 𝜕𝜕
6
The volumetric flow rate is:
𝑄 = 𝑉� 𝐴 = 1.08
The velocity gradient at the wall 𝑦 = 0 is:
𝑚
𝑚2
× 0.3 𝑚 = 0.324
𝑠
𝑠
1
𝑎 𝜕𝜕
1
1
𝑑𝑑 𝑎2 𝜕𝜕
=
� � �− � �� = − � � = −0.3 𝑚 × �−72
� = 21.6
𝑎
2𝜇 𝜕𝜕
𝑚∙𝑠
𝑠
𝑑𝑑 2𝜇 𝜕𝜕
The velocity gradient at the wall 𝑦 = 0.3 𝑚 is:
2𝑦
1
1
2 × 0.3 𝑚
1
1
𝑑𝑑 𝑎2 𝜕𝜕
=
� � �� 2 � − � �� = −72
× (0.3 𝑚)2 × ��
�−�
�� = −21.6
2
(0.3 𝑚)
𝑎
𝑎
𝑚∙𝑠
0.3 𝑚
𝑠
𝑑𝑑 2𝜇 𝜕𝜕
The gradient is equal in magnitude and opposite in sign.
Problem 8.20
Problem
8.25
[Difficulty: 3]
8.20
Given:
Laminar flow of two fluids between plates
Find:
Velocity at the interface
Solution:
Using the analysis of Section 8.2, the sum of forces in the x direction is
⎡ ∂ dy ⎛ ∂ dy ⎞⎤
⎛ ∂ dx
dx ⎞
∂
⋅ 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 8.21
Problem
8.26
[Difficulty: 2]
8.21
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
⋅m ×
−5
1.45 × 10
m
s
from Table A.10 at 15oC
s
s
−6
× 0.25 × 10
⋅
V = 22.3⋅
60⋅ s
2
−5 m
V⋅ a
=
μ
1⋅ 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
× 22.3⋅
m
s
2
from Table A.10 at 15oC
m
1
×
⋅
0.25 × 10
τ = 1.60⋅ kPa
−6
⋅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 8.22
Problem
8.29
8.22
[Difficulty: 2]
Problem 8.23
(Difficulty 2)
8.23 In a flow of air between parallel plates spaced 0.03 𝑚 apart, the centerline velocity is 1.2
that 5 𝑚𝑚 from the pipe wall 0.8
𝑚
.
𝑠
𝑚
𝑠
and
Assuming laminar flow, determine the wall shear stress using each
of the measurements. Explain whether the flow is laminar or turbulent.
Find: The shear stress.
Assumptions: Air is at atmospheric pressure and temperature, the flow is fully developed, steady, and
incompressible
Solution: Use the expressions for the velocity profile for laminar flow between parallel plates. We have
the following equations:
𝑢=
𝑎2 𝜕𝜕
𝑦 2
𝑦
� � �� � − � ��
𝑎
𝑎
2𝜇 𝜕𝜕
𝑑𝑑 𝑎2 𝜕𝜕
2𝑦
1
=
� � �� 2 � − � ��
𝑑𝑑 2𝜇 𝜕𝜕
𝑎
𝑎
For this problem we have:
𝑎 = 0.03 𝑚
For the case center velocity is known:
Thus
𝜇 = 1.825 × 10−5 𝑃𝑃 ∙ 𝑠
𝑎 2
𝑎
(0.03 𝑚)2 𝜕𝜕 1
𝑎2 𝜕𝜕
𝑎2 𝜕𝜕
1 2
1
𝑚
2
𝑉𝑐 =
� � �� � − � 2 �� =
� � �� � − � �� = −
� � � � = 1.2
2
2
𝜕𝜕 4
𝑠
2𝜇 𝜕𝜕
𝑎
𝑎
2𝜇 𝜕𝜕
2𝜇
The shear stress at the wall is:
𝑚
1.2 × 4
1
1 𝜕𝜕
𝑠
� �=−
=
−5133
(0.03 𝑚)2
𝑚∙𝑠
2𝜇 𝜕𝜕
𝜏𝑤 = 𝜇
𝑑𝑑
𝑎 𝜕𝜕
= −𝜇
� �
𝑑𝑑
2𝜇 𝜕𝜕
𝜏𝑤 = −1.825 × 10−5 𝑃𝑃 ∙ 𝑠 × 0.03 𝑚 × �−5133
1
� = 0.0281𝑃𝑃
𝑚∙𝑠
The Reynolds number is:
𝑅𝑅 =
𝜌𝑉𝑐 𝑎 1.20 × 1.2 × 0.03
=
= 2367
1.002 × 10−3
𝜇
If we use the velocity for the pipe 5 𝑚𝑚 from the wall:
𝑢=
𝑦 2
𝑦
𝑎2 𝜕𝜕
� � �� � − � ��
𝑎
𝑎
2𝜇 𝜕𝜕
(0.03 𝑚)2 𝜕𝜕
0.005 𝑚 2
0.005 𝑚
𝑚
𝑢=
� � ��
� −�
�� = 0.64
𝜕𝜕
0.03 𝑚
0.03 𝑚
𝑠
2𝜇
The shear stress at the wall is:
1
1 𝜕𝜕
� � = −6400
𝑚∙𝑠
2𝜇 𝜕𝜕
𝜏𝑤 = −1.825 × 10−5 𝑃𝑃 ∙ 𝑠 × 0.03 𝑚 × �−6400
1
� = 0.0350𝑃𝑃
𝑚∙𝑠
The flow is probably transitional. The Reynolds number is on the upper edge of the laminar flow range
and the shear stresses computed using the two velocity measurements are about 25 % different.
Problem 8.24
Problem
8.30
[Difficulty: 3]
8.24
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 μ2 − μ1
c4 = 3 ⋅ ρ⋅ g ⋅ sin( θ) ⋅ h ⋅
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 ⋅ μ2
(
2 μ2 − μ1
⋅ ⎢3 ⋅ h ⋅
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 ν2 − ν1
⋅ ⎢3 ⋅ h ⋅
⎣
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 8.25
Problem
8.32
[Difficulty: 3]
8.25
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
a
l
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
⎜
ft
ft
⎝ 0.1⋅ ft ⎠
0.1
y (in)
0.075
0.05
0.025
−4
− 1× 10
0
−4
1× 10
Shear Stress (lbf/ft3)
−4
2× 10
−4
3× 10
⋅ psi
Problem 8.26
Problem
8.34
[Difficulty: 3]
8.26
Given:
Flow between parallel plates
Find:
Pressure gradient for no flow; plot velocity and stress distributions; also plot for u = U/2 at y = a/2
Solution:
U⋅ y
Basic equations
u(y) =
Available data
U = 1.5⋅
From Eq 2 for Q = 0
dp
dx
=
a
+
m
⋅
dp
⋅ ⎢⎛⎜
y⎤
Q
⎥ (1)
a⎦
l
=
U⋅ a
2
−
a
3
dp
⋅
12⋅ μ dx
(2)
τ = μ⋅
From Fig. A.2 for castor oil at 20oC
U
a
μ = 1⋅
+ 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
×
dp
1
( 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)
For u = U at y = a/2 we need to adjust the pressure gradient. From Eq. 1
⎡⎢⎛ a ⎞ 2 a ⎤⎥
2
a dp ⎢⎜ 2
2
2⎥
U=
+
⋅ ⋅ ⎢⎜
−
2 ⋅ μ dx ⎣⎝ a ⎠
a ⎥⎦
a
U⋅
Hence
u(y) =
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 8.27
Problem
8.36
8.27
Using the result for average velocity from Example 8.3
[Difficulty: 3]
Problem
8.38
Problem 8.28
8.28
[Difficulty: 5]
Problem 8.29
Problem
8.40
8.29
[Difficulty: 2]
8.29
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 8.30
Problem
8.42
8.30
[Difficulty: 5] Part 1/2
Problem 8.42
[Difficulty: 5] Part 2/2
Problem 8.31
Problem
8.45
8.31
[Difficulty: 3]
Problem 8.32
Problem
8.46
[Difficulty: 3]
8.32
Given:
Paint flow (Bingham fluid)
Find:
Maximum thickness of paint film before flow occur
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 8.33
Problem
8.47
[Difficulty: 4]
8.33
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
∂u
∂p
∂u ⎞
⎟
⎜
+
u
+
=
−
v
+
+
w
g
ρ⎜
ρ x
µ ⎜⎜ 2 + 2 + 2 ⎟⎟
∂x
∂y
∂x
∂z ⎟⎠
∂y
∂z ⎠
⎝ ∂t
⎝ ∂x
1
4
5
3
4
6
5
3
⎛ ∂ 2v ∂ 2v ∂ 2v ⎞
⎛ ∂v
∂v
∂v ⎞
∂v
∂p
ρ ⎜⎜ + u + v + w ⎟⎟ = ρg y − + µ ⎜⎜ 2 + 2 + 2 ⎟⎟
∂x
∂y
∂z ⎠
∂y
∂y
∂z ⎠
⎝ ∂t
⎝ ∂x
1
3
3
3
3
3
3
(5.27a)
3
(5.27b)
3
⎛∂ w ∂ w ∂ w⎞
⎛ ∂w
∂w
∂w
∂w ⎞
∂p
+u
+v
+w
⎟⎟ = ρg z −
+ µ ⎜⎜ 2 + 2 + 2 ⎟⎟
∂x
∂y
∂z ⎠
∂y
∂z ⎠
∂z
⎝ ∂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
2 µ dx 1
µ
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 = −
µU
1 dp
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
dx 2
L2
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
h1 −
h2 −
=−
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 8.34
Problem
8.48
8.34
[Difficulty: 2]
Problem 8.35
Problem
8.50
[Difficulty: 3]
8.35
Given:
Data on water temperature and tube
Find:
Maximum laminar flow; plot
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
−3
1× 10
−4
1× 10
− 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 8.36
Problem
8.53
8.36
[Difficulty: 3]
Problem 8.37
(Difficulty 1)
8.37 Carbon dioxide flows in a 50 𝑚𝑚 plate at a velocity of 1.5
𝑚
,
𝑠
temperature 66 ℃, and absolute
pressure 50 𝑘𝑘𝑘. Is the flow laminar or turbulent? If the temperature is lowered to 30 ℃, what is the
flow regime? If the pressure is reduced to 20 𝑘𝑘𝑘, what is the flow regime? Explain the differences in
answers in terms of the physical mechanisms involved.
Find: Whether the flow is laminar or turbulent
Assumptions: Flow is fully developed, steady, and incompressible. Carbon dioxide is an ideal gas.
Solution: Use the Reynolds number criteria to determine the flow regime
𝑅𝑅 =
For carbon dioxide we have:
𝜌𝜌𝜌
𝜇
𝑅 = 187.8
𝐽
𝑘𝑘 ∙ 𝐾
𝜇 = 1.47 × 10−5 𝑃𝑃 ∙ 𝑠
Applying the ideal gas law we have the density:
𝜌=
𝑝
=
𝑅𝑅 187.8
The Reynolds number is calculated to be:
50 𝑘𝑘𝑘
𝐽
× (66 + 273) 𝐾
𝑘𝑘 ∙ 𝐾
= 0.79
𝑘𝑘
𝑚3
𝑘𝑘
𝑚
𝜌𝜌𝜌 0.79 𝑚3 × 1.5 𝑠 × 0.05 𝑚
𝑅𝑅 =
=
= 4.03 × 103
𝜇
1.47 × 10−5 𝑃𝑃 ∙ 𝑠
The criteria for laminar flow is that the Reynolds number is less than 2300. The Reynolds number is
greater than the value so the flow is turbulent.
When the temperature is lowered to 30 ℃, we have:
𝜌=
𝑝
=
𝑅𝑅 187.8
The Reynolds number is calculated to be:
50 𝑘𝑘𝑘
𝐽
× (30 + 273) 𝐾
𝑘𝑘 ∙ 𝐾
= 0.88
𝑘𝑘
𝑚3
𝑘𝑘
𝑚
𝜌𝜌𝜌 0.88 𝑚3 × 1.5 𝑠 × 0.05 𝑚
𝑅𝑅 =
=
= 4.49 × 103
𝜇
1.47 × 10−5 𝑃𝑃 ∙ 𝑠
The Reynolds number is greater than 2300 so the flow is turbulent.
When the pressure is decreased to 20 𝑘𝑘𝑘 we have:
𝜌=
𝑝
=
𝑅𝑅 187.8
The Reynolds number is calculated to be:
20 𝑘𝑘𝑘
𝑘𝑘
= 0.31 3
𝐽
𝑚
× (66 + 273) 𝐾
𝑘𝑘 ∙ 𝐾
𝑘𝑘
𝑚
𝜌𝜌𝜌 0.31 𝑚3 × 1.5 𝑠 × 0.05 𝑚
𝑅𝑅 =
=
= 1582
𝜇
1.47 × 10−5 𝑃𝑃 ∙ 𝑠
The Reynolds number is less than 2300 so the flow is laminar.
The Reynolds number of the carbon dioxide flows in a plate will increase when temperature decreases.
The Reynolds number will decrease when pressure decreases. The reason is that lower temperature will
increases the density and lower pressure will decrease the density for a gas.
Problem 8.38
Problem
8.54
8.38
[Difficulty: 3]
Problem 8.39
(Difficulty 1)
8.39 What is the largest diameter of pipeline that may be used to carry 100 𝑔𝑔𝑔 of jet fuel (JP-4) at
59 ℉ if the flow is to be laminar?
Find: The largest diameter of the pipe
Assumptions: Flow is fully developed, steady, and incompressible.
Solution: Use the Reynolds number criteria to determine the largest diameter.
𝑅𝑅 =
For JP-4, we have:
𝜌𝜌𝜌
𝜇
𝑆𝑆 = 0.77 and 𝜇 = 1.817 × 10−5
For the volumetric flow rate we have:
𝑄 = 100 𝑔𝑔𝑔 = 0.223
The velocity is:
𝑉=
𝑙𝑙𝑙∙𝑠
𝑓𝑓 2
𝑓𝑓 3
𝑠
𝑄
𝑄
0.284 𝑓𝑓
=
=
𝐴 𝜋 𝐷2
𝐷2
𝑠
4
For the flow to be laminar, the maximum Reynolds number is 2300:
The density is given by:
𝜌𝜌𝜌 𝜌
𝑅𝑅 =
=
𝜇
𝜌 = 1.94
Thus the largest diameter of the pipeline is:
𝐷=
𝜌0.284
=
2300𝜇
0.284
𝐷
𝐷2
= 2300
𝜇
𝑙𝑙𝑙 ∙ 𝑠2
𝑠𝑠𝑠𝑠
=
1.94
𝑓𝑓 3
𝑓𝑓 4
𝑓𝑓 3
𝑙𝑙𝑙 ∙ 𝑠2
× 0.284
4
𝑠
𝑓𝑓
= 10.15 𝑓𝑓
𝑙𝑙𝑙 ∙ 𝑠
2300 × 1.817 × 10−5
𝑓𝑓 2
0.77 × 1.94
Problem 8.40
(Difficulty 1)
8.40 Consider fully developed laminar flow in the annular space formed by the two concentric cylinders
shown in the diagram for problem 8.36, but with pressure gradient, 𝜕𝜕⁄𝜕𝜕 , and the inner cylinder
stationary. Let 𝑟0 = 𝑅 and 𝑟𝑖 = 𝑘𝑘. Show that the velocity profile is given by:
𝑢=−
𝑟 2
1 − 𝑘2
𝑟
𝑅 2 𝜕𝜕
�1 − � � + �
� ln �
𝑅
𝑅
4𝜇 𝜕𝜕
ln(1⁄𝑘)
Show that the volume flow rate is given by:
𝑄=−
(1 − 𝑘 2 )2
𝜋𝑅 4 𝜕𝜕
�(1 − 𝑘 4 ) −
�
8𝜇 𝜕𝜕
ln(1⁄𝑘)
Compare the volume flow rate for the limiting case, 𝑘 → 0, with the corresponding expression for flow
in a circular pipe.
Find: Show the two equations above.
Assumptions: Flow is fully developed, steady, and incompressible.
Solution:
For the fully developed flow in two concentric cylinders we have:
So we have:
1 𝜕
𝜕𝜕
1 𝜕𝜕
�𝑟 � =
𝑟 𝜕𝜕 𝜕𝜕
𝜇 𝜕𝜕
𝜕𝜕
𝑟 𝜕𝜕
𝜕
�𝑟 � =
𝜇 𝜕𝜕
𝜕𝜕 𝜕𝜕
𝑟
𝜕𝜕 𝑟 2 𝜕𝜕
=
+ 𝐶1
𝜕𝜕 2𝜇 𝜕𝜕
𝑢=
For the boundary condition we have:
𝑟 𝜕𝜕 𝐶1
𝜕𝜕
=
+
𝜕𝜕 2𝜇 𝜕𝜕 𝑟
𝑟 2 𝜕𝜕
+ 𝐶1 ln 𝑟 + 𝐶2
4𝜇 𝜕𝜕
𝑢 = 0 𝑎𝑎 𝑟 = 𝑘𝑘 𝑎𝑎𝑎 𝑟 = 𝑅
(𝑘𝑘)2 𝜕𝜕
+ 𝐶1 ln(𝑘𝑘) + 𝐶2 = 0
4𝜇 𝜕𝜕
(𝑅)2 𝜕𝜕
+ 𝐶1 ln(𝑅) + 𝐶2 = 0
4𝜇 𝜕𝜕
Solving for 𝐶1 and 𝐶2 we have:
𝑅 2 𝜕𝜕 1 − 𝑘 2
𝐶1 = −
4𝜇 𝜕𝜕 ln(1⁄𝑘)
The velocity will be expressed as:
𝑢=
𝐶2 = −
𝑅 2 𝜕𝜕 𝑅 2 𝜕𝜕 1 − 𝑘 2
+
ln 𝑅
4𝜇 𝜕𝜕 4𝜇 𝜕𝜕 ln(1⁄𝑘 )
𝑅 2 𝜕𝜕 𝑅 2 𝜕𝜕 1 − 𝑘 2
𝑟 2 𝜕𝜕 𝑅 2 𝜕𝜕 1 − 𝑘 2
−
ln 𝑟 −
+
ln 𝑅
4𝜇 𝜕𝜕 4𝜇 𝜕𝜕 ln(1⁄𝑘 )
4𝜇 𝜕𝜕 4𝜇 𝜕𝜕 ln(1⁄𝑘 )
𝑢=−
𝑟 2
1 − 𝑘2
𝑟
𝑅 2 𝜕𝜕
�1 − � � + �
� ln � ��
𝑅
𝑅
4𝜇 𝜕𝜕
ln(1⁄𝑘)
The volume flow rate is calculated as:
𝑅
So we have:
𝑅
𝑄 = � 𝑢𝑢𝑢 = � 𝑢2𝜋𝜋𝜋𝜋 = −2𝜋
𝑘𝑘
𝑘𝑘
𝑄 = −2𝜋
When 𝑘 → 0 we have:
𝑅 2 𝜕𝜕 𝑅
𝑟3
1 − 𝑘2
𝑟
� 𝑟− 2+�
� �𝑟 ln � �� 𝑑𝑑
⁄
)
𝑅
4𝜇 𝜕𝜕 𝑘𝑘
𝑅
ln(1 𝑘
𝑅 2 𝜕𝜕 (−1 + 𝑘 2 )𝑅 2 (1 − 𝑘 2 + (−1 − 𝑘 2 ) ln(1⁄𝑘 ))
4 ln(1⁄𝑘 )
4𝜇 𝜕𝜕
𝑄=−
(1 − 𝑘 2 )2
𝜋𝑅 4 𝜕𝜕
�(1 − 𝑘 4 ) −
�
8𝜇 𝜕𝜕
ln(1⁄𝑘)
𝑄=−
𝜋𝑅 4 𝜕𝜕
8𝜇 𝜕𝜕
which is the same volume flow rate in a circular pipe.
Problem 8.41
Problem
8.58
[Difficulty: 2]
8.41
Given:
Tube dimensions and volumetric flow rate
Find:
Pressure difference and hydraulic resistance
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 =
8 10 ×10 −9 m 3
0.01
m
×
×
×
× 4 ×1.0 ×10 −3 Pa.s = 0.00424 Pa
−12
π
60
s 1×10
m
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 8.42
(Difficulty 1)
8.42 In the laminar flow of an oil of viscosity 1 𝑃𝑃 ∙ 𝑠, the velocity at the center of a 0.3 𝑚 pipe is
4.5 𝑚⁄𝑠 and the velocity distribution is parabolic. Calculate the shear stress at the pipe wall and within
the fluid 75 𝑚𝑚 from the pipe wall.
Find: The shear stress at the pipe wall and 75 𝑚𝑚 from the pipe wall.
Assumptions: Flow is steady, and incompressible.
Solution:
As the velocity distribution is parabolic, the fluid flow is fully developed so that we have:
The velocity profile is:
𝑢=−
We have:
𝑟 2
𝑅 2 𝜕𝜕
�1 − � � �
𝑅
4𝜇 𝜕𝜕
𝐷 = 0.3 𝑚
𝑅 = 0.15 𝑚
𝜇 = 1 𝑃𝑃 ∙ 𝑠
For the velocity at the center of the pipe (r=0) we have:
𝑢=−
𝑅 2 𝜕𝜕
4𝜇 𝜕𝜕
𝑚
4 × 1 𝑃𝑃 ∙ 𝑠 × 4.5
4𝜇𝜇
𝜕𝜕
𝑠 = −800 𝑃𝑃
=− 2 =−
2
(0.15 𝑚)
𝑅
𝑚
𝜕𝜕
The shear stress can be calculated as:
𝜏𝑟𝑟 = 𝜇
At the pipe wall we have:
𝜏𝑤𝑤𝑤𝑤 =
𝑑𝑑 𝑟 𝜕𝜕
=
𝑑𝑑 2 𝜕𝜕
𝑅 𝜕𝜕 0.15 𝑚
𝑃𝑃
=
× 800
= 60 𝑃𝑃
2 𝜕𝜕
2
𝑚
In the fluid 75 𝑚𝑚 from the pipe wall we have:
𝜏𝑟𝑟 =
(𝑅 − 0.075 𝑚) 𝜕𝜕 0.15 𝑚 − 0.075
𝑃𝑃
=
× 800
= 30 𝑃𝑃
2
𝑚
𝜕𝜕
2
Problem 8.43
(Difficulty 2)
8.43 In a laminar flow of 0.007
𝑚3
𝑠
in a 75 𝑚𝑚 pipeline the shearing stress at the pipe wall is known to
be 47.9 𝑃𝑃. Calculate the viscosity of the fluid.
Find: The fluid viscosity
Assumptions: Flow is fully developed, steady, and incompressible
Solution: Use the expressions for fully developed laminar flow in a pipe. The shearing stress on the wall
is related to the viscosity through the Newton’s law of viscosity and to the pressure gradient through the
momentum equation.
𝜏=𝜇
𝑑𝑑 𝑟 𝑑𝑑
= � �
𝑑𝑑 2 𝑑𝑑
The wall shear stress equals the negative of the shear on the fluid at the wall and is given in terms of
the velocity gradient as
𝜏𝑤 = −𝜏 = −𝜇 �
𝑑𝑑
𝑅 𝑑𝑑
�
=− � �
𝑑𝑑 𝑟=𝑅
2 𝑑𝑑
The average velocity and flow rate are given in terms of the pressure gradient as by
In this particular case the pipe radius is:
𝑉=
𝑄
𝑅 2 𝑑𝑑
=−
� �
𝐴
8 𝜇 𝑑𝑑
𝑅=
𝐷
= 0.0375 𝑚
2
The pressure gradient can be calculated from the wall shear stress
𝑑𝑑
2
𝑃𝑃
2
� � = −𝜏𝑤 = −47.9 𝑃𝑃 ×
= −2554
𝑑𝑑
0.0375 𝑚
𝑚
𝑅
Using the continuity expression, the velocity is
𝑚3
0.007
𝑚
𝑄
𝑠
= 1.585
𝑉� = =
2
𝑠
𝐴 𝜋 × (0.075 𝑚)
4
The viscosity is then calculated as
𝜇=−
(0.0375𝑚)2
𝑅 2 𝑑𝑑
𝑃𝑃
� �=−
× �−2554
� = 0.283 𝑃𝑃 ∙ 𝑠
𝑚
𝑚
8𝑉 𝑑𝑑
8 × 1.585
𝑠
Problem 8.44
Problem
8.60
[Difficulty: 4]
8.44
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
dr
2 dx
(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
100
Pa / m
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 8.45
Problem
8.62
[Difficulty: 2]
8.45
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
πR 4 ∂p ⎡
l⎤
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 8.46
Problem
8.64
[Difficulty: 2]
8.46
Given:
The expression of hydraulic resistance of straight channels with different cross sectional shapes
Find:
Hydraulic resistance
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
a
(1×10 )
m
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 8.47
(Difficulty 2)
8.47 In a laminar flow in a 12 𝑖𝑖 pipe the shear stress at the wall is 1.0 𝑝𝑝𝑝 and the fluid viscosity
0.002
𝑙𝑙𝑙∙𝑠
.
𝑓𝑓 2
Calculate the velocity gradient 1 𝑖𝑖 from the centerline.
Find : The velocity gradient 1 𝑖𝑖 from the centerline.
Assumptions: Flow is fully developed, steady, and incompressible
Solution: Use the expressions for the velocity profile for laminar flow in a pipe. We have the following
equations for the velocity profile:
𝑢=−
𝑟 𝜕𝜕
𝑑𝑑
=
� �
𝑑𝑑 2𝜇 𝜕𝜕
At the wall we have:
𝑟 = 𝑅 = 6 𝑖𝑖 = 0.5 𝑓𝑓
The shearing stress on the wall is:
Thus
𝑅 2 𝜕𝜕
𝑟 2
� � �1 − � � �
𝑅
4𝜇 𝜕𝜕
𝜏𝑤 = 𝜇
𝑑𝑑 𝑅 𝜕𝜕
𝑙𝑙𝑙
= � � = 1.0 2
𝑑𝑑 2 𝜕𝜕
𝑓𝑓
𝑙𝑙𝑙
2 × 1.0 2
𝜕𝜕
𝑙𝑙𝑙
𝑓𝑓
� �=
=4 3
𝜕𝜕
𝑓𝑓
0.5 𝑓𝑓
For the location 1 𝑖𝑖 from the center line:
𝑟 𝜕𝜕
𝑑𝑑
=
� �=
𝑑𝑑 2𝜇 𝜕𝜕
𝑟 = 1 𝑖𝑖 =
1
𝑓𝑓
12
1
𝑓𝑓
𝑙𝑙𝑙
1
12
× 4 3 = 83.3
𝑙𝑙𝑙 ∙ 𝑠
𝑓𝑓
𝑠
2 × 0.002
.
𝑓𝑓 2
Problem 8.48
(Difficulty 2)
8.48 A fluid of specific gravity 0.9 flows at a Reynolds number of 1500 in a 0.3 𝑚 pipeline. The velocity
50 𝑚𝑚 from the wall is 3
𝑚
.
𝑠
Calculate the flow rate and the velocity gradient at the wall.
Find : The flow rate 𝑄 and velocity gradient
𝑑𝑑
𝑑𝑑
at the wall.
Assumptions: Flow is fully developed, steady, and incompressible
Solution: Use the expressions for the velocity profile for laminar flow in a pipe. We have the following
equations for the velocity profile:
The Reynolds number is:
𝑅𝑒 = 1500 < 2300
And so the flow is laminar. The velocity profile is given by:
𝑢=−
𝑅 2 𝜕𝜕
𝑟 2
� � �1 − � � �
𝑅
4𝜇 𝜕𝜕
𝑟 𝜕𝜕
𝑑𝑑
=
� �
𝑑𝑑 2𝜇 𝜕𝜕
At the location which the velocity is known we have:
𝑅 = 0.15 𝑚
𝑟 = 𝑅 − 50 𝑚𝑚 = 0.1 𝑚
The velocity is:
𝑢=−
Thus
(0.15 𝑚)2 𝜕𝜕
0.1 𝑚 2
𝑚
� � �1 − �
� �=3
𝜕𝜕
0.15 𝑚
𝑠
4𝜇
The flow rate is calculated by:
𝑄=−
1 𝜕𝜕
1
� � = −960
𝜇 𝜕𝜕
𝑚∙𝑠
−𝜋 × (0.15 𝑚)4
1
𝑚3
𝜋𝑅 4 𝜕𝜕
� �=
× �−960
� = 0.1909
𝑚∙𝑠
8𝜇 𝜕𝜕
8
𝑠
The velocity gradient at the wall is:
𝑟 𝜕𝜕
𝑅 𝜕𝜕
𝑑𝑑
=
� �=
� �
2𝜇 𝜕𝜕
𝑑𝑑 2𝜇 𝜕𝜕
1
1
𝑑𝑑 0.15 𝑚
=
× �960
� = 72
2
𝑚∙𝑠
𝑠
𝑑𝑑
Problem 8.49
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
4 ⋅ μ2 L
μ2 ⎝ 2 ⎠
lim
c1
r → 0 μ1 ⋅ r
(4)
2
⎛ D⎞
⎜
⎝ 4 ⎠ ⋅ ∆p + c1 ⋅ ln⎛ D ⎞ + c =
⎜
2
4 ⋅ μ1 L
μ1 ⎝ 4 ⎠
2
⎛ D⎞
⎜
⎝ 4 ⎠ ⋅ ∆p + c3 ⋅ ln⎛ D ⎞ + c
⎜
4
4 ⋅ μ2 L
μ2 ⎝ 4 ⎠
4 ⋅ c1
4 ⋅ c3
D ∆p
D ∆p
⋅
+
= ⋅
+
D
8 L
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 )⎤⎥
⎜
⎥
4 ⋅ μ2
⎝2⎠
⎦
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
2
5 ⋅ N⋅ s 5 ⋅ m
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
= 1.02
max
2
N⋅ s 5 ⋅ m
5 × .5
s
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 8.50
Problem
8.66
[Difficulty: 2]
8.50
Given:
Turbulent pipe flow
Find:
Wall shear stress
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
1
12
τw = − × 750 ⋅ psi ×
15
4
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 8.51
Problem
8.68
[Difficulty: 2]
8.51
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 8.52
(Difficulty 2)
8.52 In a flow of water in a 0.3 m pipe, the centerline velocity is 6 m/s and that 50 mm from the
pipe wall 5.2 m/s. Assuming laminar flow, determine the wall shear stress using each of the
measurements. Explain whether the flow is laminar or turbulent.
Find: The wall shear stress
Assumptions: Flow is fully developed, steady, and incompressible.
Solution: Use the expressions for the shear stress for fully developed laminar flow in a pipe. The
shearing stress on the wall is related to the viscosity through the Newton’s law of viscosity and to the
pressure gradient through the momentum equation.
𝜏=𝜇
𝑑𝑑 𝑟 𝑑𝑑
= � �
𝑑𝑑 2 𝑑𝑑
The wall shear stress equals the negative of the shear on the fluid at the wall and is given in terms of
the velocity gradient as
𝜏𝑤 = −𝜏 = −𝜇 �
𝑑𝑑
𝑅 𝑑𝑑
�
=− � �
𝑑𝑑 𝑟=𝑅
2 𝑑𝑑
The velocity profile for laminar flow in a pipe is given by:
𝑢=−
𝑅 2 𝑑𝑑
𝑟 2
� � �1 − � � �
𝑅
4𝜇 𝑑𝑑
The water viscosity at 20 C is 0.00101 Pa s and the radius of the pipe is 0.15 m. For the centerline
velocity, the pressure gradient is calculated as
𝑑𝑑
� �=−
𝑑𝑑
𝑚
4 × 0.00101 𝑃𝑃 ∙ 𝑠 × 6
4𝜇𝑢
𝑠 = −1.067 𝑃𝑃
=
2
2
𝑟
𝑚
𝑅 2 �1 − � � � (0.15 𝑚)2 × �1 − � 0 � �
𝑅
0.15 𝑚
The wall shear stress for this pressure gradient is
𝑅 𝑑𝑑
0.15𝑚
𝑃𝑃
𝜏𝑤 = − � � = −
× �−1.067 � = 0.0800 𝑃𝑃
2 𝑑𝑑
2
𝑚
Using the velocity at a value of r = 0.1 m, the pressure gradient is calculated as
𝑑𝑑
� �=−
𝑑𝑑
𝑚
4 × 0.00101 𝑃𝑃 ∙ 𝑠 × 5.2
4𝜇𝑢
𝑠 = −1.664 𝑃𝑃
=
2
2
𝑟
𝑚
𝑅 2 �1 − � � � (0.15 𝑚)2 × �1 − � 0.1 𝑚 � �
𝑅
0.15 𝑚
The wall shear stress for this pressure gradient is
𝑅 𝑑𝑑
0.15𝑚
𝑃𝑃
𝜏𝑤 = − � � = −
× �−1.664 � = 0.1248𝑃𝑃
2 𝑑𝑑
2
𝑚
For fully developed laminar flow, the wall stress for the two calculations should agree. They do not,
implying that the flow is not laminar. We can check the Reynolds number criteria to see if the flow is in
the laminar regime.
𝑅𝑅 =
𝑉𝜌𝐷
𝜇
If we assume the flow is laminar, the centerline velocity is 1.5 times the average velocity V. The average
velocity is then 4 m/s and the Reynolds number is then
𝑘𝑘
𝑚
× 1000 3 × 0.3 𝑚
𝑠
𝑚
= 1.2 × 106
𝑅𝑅 =
0.00101 𝑃𝑃 ∙ 𝑠
4
The flow is undoubtedly turbulent.
Problem 8.53
Problem
8.70
8.53
[Difficulty: 3]
Problem8.54
8.73
Problem
[Difficulty: 3]
8.54
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 8.55
Problem
8.74
8.55
[Difficulty: 3]
Problem 8.56
Problem
8.76
[Difficulty: 3]
8.56
Given:
Laminar flow between parallel plates
Find:
Kinetic energy coefficient, α
Solution:
Basic Equation: The kinetic energy coefficient, α is given by
∫
α=
A
From Section 8-2, for flow between parallel plates
ρ V 3dA
(8.26b)
m& V 2
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
31
3
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 8.57
Problem
8.77
8.57
[Difficulty: 3]
Problem 8.58
Problem
8.78
[Difficulty: 3]
8.58
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
⎞
R⎠
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
2.
2
R
3
2 2 3
n
2 ⋅ π⋅ ρ⋅ R ⋅ n ⋅ U
r⎞
⎛
dr =
ρ⋅ 2 ⋅ π⋅ r⋅ U ⋅ ⎜ 1 −
R⎠
( 3 + n) ⋅ ( 3 + 2⋅ n)
⎝
3
0
r
To integrate substitute
m=1−
Then
r = R⋅ ( 1 − m)
⌠
⎮
⎮
⎮
⎮
⌡
3
⌠
3
⎮
⎮
n
r
3
3
ρ⋅ V dA = ⎮ ρ⋅ 2 ⋅ π⋅ r⋅ U ⋅ ⎛⎜ 1 − ⎞ dr
⎮
R⎠
⎝
⌡
⌠
⎮
⎮
⌡
⌠
⎮
⎮
⎮
⎮
⌡
2
R
0
R
dm = −
dr
R
dr = −R⋅ dm
3
0
⌠
3
⎮
n
⎮
r
3
n
ρ⋅ 2 ⋅ π⋅ r⋅ U ⋅ ⎛⎜ 1 − ⎞ dr = −⎮ ρ⋅ 2 ⋅ π⋅ R⋅ ( 1 − m) ⋅ m ⋅ R dm
⌡
R⎠
⎝
1
1
⌠
3 ⎞
⎛ 3
⎮
+1
⎜
⎮
3
n
n
ρ⋅ V dA = ⎮ ρ⋅ 2 ⋅ π⋅ R⋅ ⎝ m − m
⎠ ⋅ R dm
⌡
⌠
⎮
⎮
⌡
Hence
0
2 2
3
⌠
2 ⋅ R ⋅ n ⋅ ρ⋅ π⋅ U
⎮
3
ρ
⋅
V
d
A
=
⎮
( 3 + n) ⋅ ( 3 + 2⋅ n)
⌡
α=
Putting all these results together
⌠
⎮
3
⎮ ρ⋅ V dA
⌡
2
2
=
mrate⋅ Vav
α=
2
3
2⋅ R ⋅ n ⋅ ρ⋅ π⋅ U
( 3+ n) ⋅ ( 3+ 2⋅ n)
2
3
ρ⋅ π⋅ R ⋅ Vav
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 8.59
(Difficulty 3)
8.59 If the turbulent velocity profile for water flow in a pipe 0.6 m in diameter may be
1/ 7
1/ 7
u y
r 

approximated by =   = 1 −  where u is in m/s and y is in m, and the velocity 0.15
U R
 R
m from the pipe wall is 2.7 m/s, estimate the wall shear stress and volume flow rate.
Find: The wall shear stress and the volume flow rate
Assumptions: Flow is fully developed turbulent flow, steady, and incompressible.
Solution: Use the empirical expressions for the turbulent velocity profile.
The empirical expression for the velocity profile in the wholly turbulent region away from the
pipe wall is
𝑢
𝑦 𝑢∗
�
� + 5.0
=
2.5
𝑙𝑙
𝜈
𝑢∗
For water at 20 C, n = 1.01 10-6 m2/s. The velocity is 2.7 m/s at a distance of y = 0.15 m. Using
the profile relation
2.7 𝑚/𝑠
0.15𝑚 𝑢∗
� + 5.0
= 2.5 𝑙𝑙 �
𝑢∗
1.01 10−6 𝑚2 /s
Solving this implicit expression for the shear velocity, we have
𝑢∗ = 0.936 𝑚/𝑠
The relation between the shear velocity and the wall shear stress is
𝜏𝑤 1/2
𝑢 =� �
𝜌
With the value of r = 1000 kg/m3 we have
∗
𝑘𝑘
𝑚 2
⋅ �0.936 � = 8.56 𝑃𝑃
𝑚3
𝑠
For the volume flow rate, we will use the empirical relation for the velocity profile for turbulent
flow to find the centerline velocity from the local velocity
𝜏𝑤 = 𝜌(𝑢∗ )2 = 1000
𝑢
𝑟 1/7
= �1 − �
𝑈
𝑅
Or, for u = 2.7 m at r = 0.6m – 0.15m = 0.45m
𝑚
2.7 𝑠
𝑢
𝑚
𝑈=
=
= 3.29
1/7
1/7
𝑟
𝑠
0.45𝑚
�1 − 𝑅�
�1 − 0.6𝑚 �
The centerline velocity is related to the average velocity using the empirical relation
2𝑛2
𝑉
=
𝑈 (𝑛 + 1)(2𝑛 + 1)
Where for the 1/7th velocity profile n = 7. The average velocity is then
𝑉=𝑈
𝑚
2 ⋅ 72
𝑚
2𝑛2
= 3.29 ⋅
= 2.69
(𝑛 + 1)(2𝑛 + 1)
𝑠 (7 + 1)(2 ⋅ 7 + 1)
𝑠
The volume flow rate is then
𝜋
𝑚
𝑚3
2
𝑄 = 𝐴𝐴 = ⋅ (0.6𝑚) ⋅ 2.69 = 0.19
𝑠
4
𝑠
Problem 8.60
Problem
8.80
[Difficulty: 2]
8.60
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 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 ⋅
(Eq.1)
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 (1) 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 (1) 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 (1) becomes
=z2
h lT
−z1
g
(Gage)
z2 = −28.1 m
Problem 8.82
8.61
Problem
[Difficulty: 2]
8.61
Given:
A given piping system and volume flow rate with two liquid choices.
Find:
Which liquid has greater pressure loss
Solution:
Governing equation:
⎞
⎞ ⎛P
⎛ P1
V2
V2
⎜⎜ + α1 1 + gz1 ⎟⎟ − ⎜⎜ 2 + α 2 2 + gz 2 ⎟⎟ = hlT
2
2
⎠
⎠ ⎝ρ
⎝ρ
2
2
LV
V
+K
hlT = hl + hlm = f
2
D 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 8.84
8.62
Problem
8.62
[Difficulty: 2]
Exam-
ple 8.7
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
⎜
3
1 ⋅ min
⎝ 12 ⎠
1 ⋅ ft
Q = 372 ⋅ gpm
(From Table G.2 1 ft3 = 7.48 gal)
Problem 8.63
Problem
8.87
8.63
[Difficulty: 2]
Problem 8.64
(Difficulty 1)
8.64 When oil (kinematic viscosity 1 × 10−4
𝑚2
,
𝑠
specific gravity 0.92) flows at a mean velocity of 1.5
𝑚
𝑠
through a 50 𝑚𝑚 pipeline, the head lost in 30 𝑚 of pipe is 5.4 𝑚. What will be the head loss when the
𝑚
𝑠
velocity is increased to 3 ?
𝑚
𝑠
Find : The head loss 𝐻𝑙2 when the velocity increased to 3 .
Assumptions: Flow is fully developed, steady, and incompressible
Solution: Use the expressions for the head loss for laminar flow in a pipe.
64 𝐿 𝑉� 2
ℎ𝑙 = � �
𝑅𝑅 𝐷 2
For the two cases, we have the Reynolds number as:
𝑅𝑒1
𝑅𝑒2
𝑚
𝑉�1 𝐷 1.5 𝑠 × 0.05 𝑚
=
=
= 750
𝑚2
𝑣
−4
1 × 10
𝑠
𝑚
𝑉�2 𝐷 3 𝑠 × 0.05 𝑚
=
=
= 1500
𝑚2
𝑣
−4
1 × 10
𝑠
So both of these flows are laminar flow. Thus the ratio of head loss is
ℎ𝑙2
ℎ𝑙1
2
64 𝐿 𝑉�2
1
�
× �3
𝑅𝑒2 𝐷 2
1500
=
2 = 1
64 𝐿 𝑉�1
× �1.5
� �
750
𝑅𝑒1 𝐷 2
�
ℎ𝑙2
𝐻𝑙2
𝑔
=
=2
𝐻𝑙1 ℎ𝑙1
𝑔
𝐻𝑙2 = 2𝐻𝑙1 = 10.8 𝑚
𝑚 2
�
𝑠
=2
𝑚 2
�
𝑠
Problem 8.65
(Difficulty 1)
8.65 When fluid of specific weight 50
𝑙𝑙𝑙
𝑓𝑓 3
flows in a 6 𝑖𝑖 pipeline, the frictional stress is 0.5 𝑝𝑝𝑝.
𝑓𝑓 3
,
𝑠
Calculate the head lost per foot of pipe. If the flow rate is 2.0
pipe?
Find : The head lost per foot
𝐻𝑙
,the
𝐿
how much power is lost per foot of
𝑊̇
𝐿
power lost per foot .
Assumptions: Flow is fully developed, steady, incompressible, and laminar.
Solution: Use the expressions for the head loss and velocity profile for laminar flow in a pipe. The wall
shear stress is related to the pressure drop as:
The head loss is defined as:
Thus
𝜏=𝜇
𝑑𝑑 𝑟 𝜕𝜕
= � �
𝑑𝑑 2 𝜕𝜕
𝜕𝜕
∆𝑝 𝜕𝜕 𝐿
=
= ℎ𝑙 = 𝐻𝑙 𝑔
𝜌
𝜌
𝜕𝜕 𝐻𝑙 𝜌𝜌 𝐻𝑙 𝛾
=
=
𝐿
𝐿
𝜕𝜕
𝜏=
For this situation
The head loss per unit length is
The power lost is calculated by:
𝑟 = 3 𝑖𝑖 =
𝐻𝑙 2𝜏
=
=
𝛾𝛾
𝐿
𝑟 𝐻𝑙 𝛾
2 𝐿
3
𝑓𝑓 = 0.25 𝑓𝑓
12
2 × 0.5
𝑙𝑙𝑙
𝑓𝑓 2
𝑙𝑙𝑙
50 3 × 0.25 𝑓𝑓
𝑓𝑓
= 0.08
𝑄∆𝑝 𝑄𝐻𝑙 𝜌𝜌 𝑄𝐻𝑙 𝛾
𝑊̇
=
=
=
𝐿
𝐿
𝐿
𝐿
𝑄𝐻𝑙 𝛾
𝑓𝑓 3
𝑙𝑙𝑙
𝑙𝑙𝑙 ∙ 𝑓𝑓
ℎ𝑝
𝑊̇
=
= 2.0
× 0.08 × 50 3 = 8
= 0.0145
𝑓𝑓
𝑓𝑓 ∙ 𝑠
𝑓𝑓
𝐿
𝐿
𝑠
Problem 8.66
(Difficulty 1)
8.66 If the head loss in a 30 m of length of a 75-mm-diameter pipe is 7.6 m for a given flow rate of water,
what is the total drag force exerted by the water on this length of pipe?
Find: The drag force on the pipe.
Assumptions: The flow is fully developed, steady and incompressible.
Solution: Use the energy equation with head loss and the momentum equation
�
𝑝2 𝑉�22
𝑝1 𝑉�12
+
+ 𝑧1 � − � +
+ 𝑧2 � = 𝐻𝑙
𝜌𝜌 2𝑔
𝜌𝜌 2𝑔
Momentum equation for steady flow
𝐹𝑠𝑠 = � 𝑢𝑢𝑉� ∙ 𝑑𝐴̅ = 𝑚̇�𝑉�2 − 𝑉�1 �
𝐶𝐶
The pipe is on constant diameter and so the flow area at sections 1 and 2 is the same. From the
continuity equation then, the velocities at 1 and 2 are equal. Assuming that the pipe is horizontal, the
elevations at 1 and 2 are the same. The energy equation with head loss reduces to
𝑝1 𝑝2
� − � = 𝐻𝑙
𝜌𝜌 𝜌𝜌
Or, the pressure differ