Lecture 8
Contents
• Momentum and Fluid Flow
What Did We Do In Last Lecture?
• Model Testing requires…
Dynamic Similarity
This implies here for object in
homogeneous fluid…
(CD)Model=(CD)Prototype
and hence need …
ReModel=RePrototype
What Did We Do In Last Lecture?
Leads to …
PROBLEMS
Need for …
Large Windtunnels
Free-Surface Flows..
BIGGER PROBLEMS
Need …
ReModel=RePrototype
FrModel=FrPrototype
Froude’s Assumption …
C D = C DWave ( Fr ) + C DSkinFr . (Re)
What Did We Do In Last Lecture?
•Applied Froude’s Assumption to
Example of Greek Trireme
Calculate Power and number of
oarsmen required to propel ship at 5
m/s through water.
MOMENTUM AND FLUID FLOW
•
Analysis of fluid-flow phenomena fundamentally
depends upon application of Newton’s laws of
motion, together with recognition of special
properties of fluids in motion.
•
Momentum equation relates sum of forces acting
on a fluid element to its acceleration or rate of
change of momentum in direction of resultant
force.
•
Will now introduce application of momentum
equation to a few very simple problems.
•
Will look at forces exerted upon and by a fluid
as a result of changes in direction and impact
upon surfaces.
Continued...
• Definition of momentum M for object with mass
m and velocity v:
M =mv
• Consider streamtube for steady flow below...
D
A1
v1
ρ1
A2
A
v2
ρ2
C
B
•
• Mass flow rate at entry (station 1): m1 = ρ1 A1 v1
•
• Mass flow rate at exit (station 2): m 2 = ρ 2 A2 v2
• Mass conservation must hold, hence ...
•
•
•
m1 = m 2 = m (A)
Continued...
• Since
momentum, M = m v, the momentum
flow rate at entry (1) and exit (2) are ...
•
m1 v1 = ρ1 A1 v1 v1
•
m 2 v2 = ρ 2 A2 v2 v2
•
•
m1
m2
• Thus,
rate of change of momentum across
control volume is ...
•
•
m 2 v2 − m1 v1
= ρ 2 A2 v2 v2 − ρ1 A1 v1 v1
•
= m (v2 − v1 ) (B)
Continued...
• Thus, the increase of momentum per unit time
(or Momentum flow rate) in direction of
motion is ...
•
•
M = m (v2 − v1 )
(C)
= Mass flow per unit time x Change of velocity
• According to Newton’s second law Eq. (C) will
be caused by a force such that ...
•
F = m (v2 − v1 )
(D)
This is the …
One-Dimensional Momentum
Equation
Continued...
• Eq. (D) from previous slide,
•
F = m (v2 − v1 ) ,
is force acting on fluid element ABCD in direction
of motion.
D
A1
v1
ρ1
A
A2
v2
ρ2
B
C
• By Newton’s third law, the fluid will exert
an equal and opposite reaction on its
surroundings.
Momentum Equation for Two- and
Three-Dimensional Flow Along a
Streamline
• Consider
flow as sketched below to generalize
result of previous section ...
Angle φ
C
v2
vy2
D
vx 2
y
A
v1
Area A1
v y1
vx1
Angle θ
B
• Similar to Eq. B/C/D from previous section write
down, the rate of change of momentum across
control volume but consider x, y component
separately...
x
• Write down velocity components
for use later ...
Angle φ
C
v2
vy2
D
vx 2
y
A
v1
Area A1
v y1
x
vx1
Angle θ
B
vx1 = v1 cosθ
vx 2 = v2 cos φ
v y1 = v1 sin θ
v y 2 = v2 sin φ
(1a-d)
Continued...
• Eq.
B/C/D from previous section gave in
summary ...
•
F = m (v2 − v1 )
• Apply
strategy from previous section in xdirection. This yields ...
Fx = Rate of change of momentum in x-direction
= Mass flow per unit time x Change of velocity
in x-direction
•
(
Fx = m v x2 − vx1
)
(2)
• Now substitute Eqs. 1a,b (i.e. flow components)
into Eq. (2) to get ...
•
Fx = m (v2 cos φ − v1 cosθ )
(3)
Continued...
• Equivalent strategy for y-component yields ...
•
Fy = m (v2 sin φ − v1 sin θ )
NOW….
• Force components Fx and Fy can be combined
to give total force ...
F = Fx2 + Fy2
Finally….
• In three-dimensional flow additional component
Fz from considering velocities vz1 and vz2 in zdirection yielding ...
•
(
Fz = m vz2 − v z1
)
Continued...
• In summary we have in general ...
Total force exerted
on fluid in CV in a
given direction
=
Rate of change of
momentum in given
direction of fluid
passing through CV
•
F = m (vout − vin )
Note …
F is positive in the direction in
which v is assumed to be positive
QUESTION:
What are the forces which act
upon control volume?
Continued...
ANSWER:
For any control volume total force, F, which
acts upon it in a given direction will be made
up of three different components
Components are …
F1 =
Force exerted in given direction on fluid in
CV by any solid body within CV or
coinciding with boundaries of CV.
F2 =
Force exerted in given direction on fluid in
CV by body forces such as gravity.
F3 =
Force exerted in given direction on fluid in
CV by fluid outside the CV (e.g. pressure
forces at exit and entry of stream tube).
•
F = F1 + F2 + F3 = m (vout − vin )
•
Force, R, exerted by the fluid on solid body inside or
coinciding with CV in given direction will be equal and
opposite to F1 so that R=-F1.
Example 1:
Water flows through pipeline 60 m long at
velocity 1.8 m/s. Pressure difference between inlet
and out let ends is 25 kN/m2.
Q: What increase of pressure difference is
required to accelerate water in pipe at rate
0.02 m/s2?
l = 60 m
v = 1.8 m/s
ρ = 1000 kg/m 3
•
Use ...
A=
δp =
Cross-sectional area of pipeline
Increase in pressure at inlet required
required to produce acceleration a.
Solution:
•
Note: THIS IS NOT A STEADY FLOW PROBLEM!
•
Use a control mass comprising whole of water in pipe.
• Start from Newton’s 2nd law:
Force due to δp in
direction of motion
=
Rate of change of
momentum of water in whole pipe
=
=
= A × δp
Mass of water in
pipe x Acceler.
A δp = ρ A l × a
δp = ρ ⋅ l ⋅ a = 1000
kg
m
kN
⋅
⋅
=
60
m
0
.
02
1200
m3
s2
m2
Continued...
• Note:
Here pressure difference is small because
acceleration is small, but very large pressures can
be developed by sudden acceleration or
decelerations, such as may occur when valves are
shut. The elasticity of the fluid and of the pipe must
then be taken in to account.
Example 2:
Force Exerted By A Jet Striking a Plate
(Four possible scenarios)
θ
v
v
v
vn
θ
v
u
• In
θ
u
EACH case the velocity component normal to
plate reduces to zero on impact, hence ...
vn = (v − u ) cosθ
Continued ...
•
Mass flow entering CV given by ...
•
m = ρ A vn = ρ A (v − u ) cosθ
•
For stationary plate perpendicular to jet reduces to ...
•
m = ρ A vn = ρ A v
•
(A)
(B)
For Eq.(A) rate of change of momentum normal to plane
is...
•
dM
= m (v − u ) = ρ A (v − u ) (v − u ) cosθ (A2)
dt
• For a stationary plate this becomes ...
•
dM
= m v = ρ A v 2 cosθ
dt
(A3)
Continued...
•
If plate also perpendicular to jet [equivalent to Eq.(B)]
this becomes ...
•
dM
= m v = ρ A v2
dt
•
(A4) or (B2)
There will be a force (normal) exerted on the plate equal
to the rate of momentum destroyed normal to the plate,
given in the general case by an expression of the form of
Eq. (A2)
Fnormal = ρ A (v − u ) (v − u ) cosθ
(5)
•
There will be an equal an opposite force exerted on the
jet by the plate.
•
In direction parallel to plate, force exerted will depend
upon shear stress between fluid and surface. For an ideal
fluid there would be no shear stress and, hence, no force
parallel to plate.
Worked example for jet striking a plate...
•
A water jet from fixed nozzle has diameter of 25 mm and
strikes a flat plate. The velocity of the jet is v=5 m/s and
the surface of the plate is assumed frictionless. Calculate
the force normal to plate surface for the following cases
(for density of water use 1000 kg/m3):
(I)
u = 0 , θ = 0o
(II)
u = 0 , θ = 30o
(III) u = −2m/s , θ = 30o
•
In each case we have to apply Eq. (5) from previous slide
with appropriate values.
Fnormal = ρ A (v − u ) (v − u ) cos θ
•
First calculate area, A, of jet so we have it when we need
it later...
A = π r 2 = π (0.0125 m )2 = 4.9 × 10− 4 m 2
Continued ...
•
Eq. (5) was ...
Fnormal = ρ A (v − u ) (v − u ) cosθ
• And so we get for the three cases ...
(I): u = 0 , θ = 0o
Fnormal = 1000
kg
m
−4
⋅
×
4
.
9
10
(5
3
m
⎛ m
⎞
− 0) ⎜ 5 − 0 ⎟ cos 0o
s
⎠
⎝ s
=12.25 N
o
(II): u = 0 , θ = 30
Fnormal = 1000
kg
m
−4
4
.
9
10
(5
⋅
×
3
m
⎛ m
⎞
− 0) ⎜ 5 − 0 ⎟ cos 30o
s
⎠
⎝ s
=10.61 N
o
(III): u = −2m/s , θ = 30
kg
Fnormal = 1000 3 ⋅ 4.9 × 10− 4
m
m ⎞⎞
m⎞ ⎛ m ⎛
m ⎛
× (5 − ⎜ − 2 ⎟) ⎜ 5 − ⎜ − 2 ⎟ ⎟ cos 30o
s ⎠⎠
s⎠ ⎝ s ⎝
s ⎝
= 20.79 N
Example 3:
A jet of water from a nozzle is deflected through an
angle of 60 degrees from its original direction by a
curved vane which it enters tangentially without shock
with a mean velocity of 30 m/s and leaves it with a mean
velocity of 25 m/s. Mass discharge from the nozzle is 0.8
kg/s.
Q: What is magnitude and direction of resultant
force on vane if vane is stationary?
Summary of given data:
v1 = 30
•
m
s
m = 0.8
kg
s
m
v2 = 25
s
θ = 60o
Chosen
Control
Volume (CV)
Solution:
•
Resultant force, R, exerted by fluid on vane found
by determining component forces Rx and Ry in xand y-directions.
•
With equation from just before Example 1 which
•
was ...
F = F1 + F2 + F3 = m (vout − vin )
F1 =
Force exerted in given direction on fluid in
CV by any solid body within CV or
coinciding with boundaries of CV.
•
Rx = − F1 = F2 + F3 − m (vout − vin )x
•
Recall ...
F2 = Body Forces ...
F3 = Forces exerted by fluid outside CV ...
(A)
Continued ...
•
Eq. (A) from previous slide was ...
•
Rx = − F1 = F2 + F3 − m (vout − vin )x
NOW ...
•
Neglect body force, F2 , due to gravity and assume
that for free jet pressure is constant everywhere
such that F3 =0 . Then Eq. (A) becomes ...
•
Rx = − F1 = − m (vout − vin )x
•
Rx = m (vin − vout )x
(B)
And similarly ...
•
R y = m (vin − vout ) y
(C)
Continued ...
•
Need to plug appropriate values into (B) and (C) ...
•
Rx = m (vin − vout )x (B)
•
R y = m (vin − vout ) y (C)
v1
•
m = 0.8
θ = 60o
kg
s
v2
x-Direction
vin = (v1 )x = 30
m
s
vout = (v2 )x = v2 cosθ
= 25
Rx = 0.8
m
m
⋅ 0.5 = 12.5
s
s
kg ⎛ m
m⎞
30
12
.
5
−
⎜
⎟
s ⎝ s
s⎠
= 14 N
y-Direction
vin = (v1 )y = 0
m
s
vout = (v2 )y = −v2 sin θ
= −25
R y = 0.8
m
m
⋅ 0.866 = −21.65
s
s
kg ⎛ m ⎛
m⎞ ⎞
⎜ 0 − ⎜ − 21.65 ⎟ ⎟
s ⎝ s ⎝
s ⎠⎠
= 17.32 N
Continued ...
•
Combining the two results for Rx and Ry yields
R = Rx2 + R y2 =
•
(14 N )2 + (17.32 N )2 = 22.27 N
This resultant force, R, will be inclined to the x-direction
at an angle ...
⎛ Ry ⎞
−1 ⎛ 17.32 N ⎞
⎟⎟ = 51o.3'
φ = tan ⎜⎜ ⎟⎟ = tan ⎜⎜
⎝ Rx ⎠
⎝ 14 N x ⎠
−1
φ
Rx
Ry