Momentum and Fluid Flow

What Did We Do In Last Lecture?
Dynamic Similarity
This implies here for object in homogeneous fluid…
D
Model
D
Prototype and hence need …
Model
Prototype

What Did We Do In Last Lecture?
Leads to …
Need for …
Free-Surface Flows..
Need …
Model
Prototype
Model
Prototype
C
D
=
Froude’s Assumption …
C
D
Wave
( Fr )
+
C
D
SkinFr .
(Re)

What Did We Do In Last Lecture?
Calculate Power and number of oarsmen required to propel ship at 5 m/s through water.

• 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
= m v
•
Consider streamtube for steady flow below...
D
A
1 v
1
ρ
1
A
B C
A
2 v
2
ρ
2
•
•
Mass flow rate at entry (station 1):
Mass flow rate at exit (station 2):
• m
1
= ρ
1
A
1 v
1
• m
2
= ρ
2
A
2 v
2
•
Mass conservation must hold, hence ...
• m
1
=
• m
2
=
• m (A)

Continued...
•
Since momentum, M = m v, the momentum flow rate at entry (1) and exit (2) are ...
• m
1 v
1
= ρ
1
A
1 v
1 v
1
• m
2 v
2
= ρ
2
A
2 v
2 v
2
• m
1
• m
2
•
Thus, rate of change of momentum across control volume is ...
• m
2 v
2
−
• m
1 v
1
= ρ
2
A
2 v
2 v
2
− ρ
1
A
1 v
1 v
1
=
• m
( v
2
− v
1
)
(B)

Continued...
•
Thus, the increase of momentum per unit time
(or Momentum flow rate) in direction of motion is ...
•
M
=
• m
( v
2
− v
1
)
(C)
= Mass flow per unit time x Change of velocity
•
According to Newton’s second law Eq. (C) wil be caused by a force such that ...
F
=
• m
( v
2
− v
1
)
(D)
This is the …

Continued...
•
•
Eq. (D) from previous slide, , is force acting on fluid element ABCD in direction of motion.
D
A
1 v
1
ρ
1
A
B C
A
2 v
2
ρ
2
•
By Newton’s third law, the fluid will exert an equal and opposite reaction on its surroundings.

•
Consider flow as sketched below to generalize result of previous section ...
Angle
φ
C v
2 v y 2
D v x 2 y
A
Area A
1 v x 1 v
1 v y 1
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

A
Area A
1 v x 1 v
1
B
Angle
φ
C v
2 v y 2
D v x 2 y v y 1 x
Angle
θ v x 1
= v
1 cos
θ v y 1
= v
1 sin
θ v x 2
= v
2 cos
φ v y 2
= v
2 sin
φ
(1a-d)

Continued...
•
Eq. B/C/D from previous section gave in summary ...
F
=
• m
( v
2
− v
1
)
•
Apply strategy from previous section in xdirection. This yields ...
F x
= Rate of change of momentum in x-direction
= Mass flow per unit time x Change of velocity in x-direction
F x
=
• m
( v x
2
− v x
1
)
(2)
•
Now substitute Eqs. 1a,b (i.e. flow components) into Eq. (2) to get ...
F x
=
• m
( v
2 cos
φ − v
1 cos
θ )
(3)

Continued...
•
Equivalent strategy for y-component yields ...
F y
=
• m
( v
2 sin
φ − v
1 cos
θ )
NOW….
•
Force components F x to give total force ...
and F y can be combined
F
=
F x
2 +
F y
2
Finally….
•
In three-dimensional flow additional component
F z from considering velocities v direction yielding ...
z1 and v z2 in z-
F z
=
• m
( v z
2
− v z
1
)

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 the fluid passing through the
CV,
F
=
• m
( v out
− v in
)
Note …
F is positive in the direction in which v is assumed to be positive

Continued...
For any control volume total force, F, which acts upon it in a given direction will me made up of three different components
Components are …
F
1
=
Force exerted in given direction on fluid in
CV by any solid body within CV or coinciding with boundaries of CV .
F
2
=
F
3
=
Force exerted in given direction on fluid in
CV by body forces such as gravity .
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
=
F
1
+
F
2
+
F
3
=
• m
( v out
− v in
)
•
Force, R, excerted by the fluid on solid body inside or coinciding with CV in given direction will be equal and opposite to F1 so that R=-F.

Example 1:
Water flows through pipeline 60m long at velocity 1.8m/s . Pressure difference between inlet and out let ends is 25kN/m 2 .
Q: What increase of pressure difference is reuired to accelerate water in pipe at rate 0.02m/s 2 ?
l
=
60 m v
=
1 .
8 m/s
ρ =
1000 kg/m
3
•
Use ...
A
=
Cross-sectional area of pipeline
p
= 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:
δ direction of motion
=
Rate of change of momentum of water in whole pipe
=
A
×
=
δ p
=
Mass of water in pipe x Acceler.
A
δ p
= ρ
A l
× a
δ p
= ρ ⋅ l
⋅ a
=
1000 kg m
3
⋅
60 m
⋅ m
0 .
02 s
2
=
1200 kN m
2

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:
θ v n v v v
θ u v θ u
•
In EACH case the velocity component normal to plate reduces to zero on impact with ...
v n
=
( v
− u ) cos
θ

Continued ...
•
Mass flow entering CV given by ...
• m
= ρ
A v n
= ρ
A ( v
− u ) cos
θ
(A)
•
For stationary plate perpendicular to jet reduces to ...
• m
= ρ
A v n
= ρ
A v (B)
•
For Eq.(A) rate of change of momentum normal to plane is...
dM dt
=
• m
( v
− u
) = ρ
A ( v
− u )
( v
− u
) cos
θ
(A2)
•
For a stationary plate this becomes ...
dM dt
=
• m v
= ρ
A v
2 cos
θ
(A3)

Continued...
•
If plate also perpendicular to jet [equivalent to Eq.(B)] this becomes ...
dM dt
=
• m v
= ρ
A v
2 (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)
F normal
= ρ
A ( v
− u )
( v
− u
) cos
θ
(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 25mm 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/m 3 ):
(I)
(II) u u
=
0 ,
θ =
0 o
=
0 ,
θ =
30 o
(III) u
= −
2 m/s ,
θ =
30 o
•
In each case we have to apply Eq. (5) from previous slide with appropriate values.
F normal
= 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 ...
F normal
=
A ( v
− u )
( v
− u
) cos
•
And so we get for the three cases ...
(I): u
=
0 ,
θ =
0 o
F normal
= kg
1000 m
3
⋅
4 .
9
×
10
−
4
( 5 m s
−
0 )
⎝
⎜
⎛
5 m s
−
0
⎠
⎟
⎞ cos 0 o
=
12 .
25 N
(II): u
=
0 ,
θ =
30 o
F normal
=
1000 kg m
3
⋅
4 .
9
×
10
−
4
( 5 m s
−
0 ) 5 m s
−
0 cos 30 o
=
10 .
61 N
(III): u
= −
2 m/s ,
θ =
30 o
F normal
=
1000 kg m
3
⋅
4 .
9
×
×
( 5 m s
10
−
−
4
2 m s
⎞
⎠
) 5 m s
−
2 m s cos 30 o
=
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. 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: v
1
=
30 m s v
2
=
25 m s
• m
=
0 .
8 kg s
=
60 o

Chosen
Control
Volume (CV)
Solution:
•
Resultant force, R, exerted by fluid on vane found by determining component forces R x and y-directions.
and R y in x-
•
With equation from just before Example 1 which was ...
F
=
F
1
+
F
2
+
F
3
=
• m
( v out
− v in
)
F
1
=
Force exerted in given direction on fluid in
CV by any solid body within CV or coinciding with boundaries of CV .
R x
= −
F
1
=
F
2
+
F
3
−
• m
( v out
− v in
)
(A)
•
Recall ...
F
2
= Body Forces ...
F
3
= Forces exerted by fluid outside CV ...

Continued ...
•
Eq. (A) from previous slide was ...
R x
= −
F
1
=
F
2
+
F
3
−
• m
( v out
− v in
)
•
Neglect body force, F
2
, due to gravity and assume that for free jet pressure is constant everywhere such that F
3
R x
=0 . Then Eq. (A) becomes ...
•
= −
F
1
= − m
( v out
− v in
)
R x
=
• m
( v out
− v in
) x
And similarly ...
R y
=
• m
( v out
− v in
) y
(B)
(C)

Continued ...
•
Need to plug appropriate values into (B) and (C) ...
R x
=
• m
( v out
− v in
) x
(B) R y
=
• m
( v out
− v in
) y
(C) v
1
• m
=
0 .
8 kg s
θ =
60 o
x-Direction v in
=
( )
1 x
=
30 m s v
2
y-Direction v in
=
( )
1 y
=
0 m s v out
=
( )
2 x
= v
2 cos
θ
=
25 m s
⋅
0 .
5
=
12 .
5 m s v out
=
( )
2 y
= v
2 sin
θ
=
25 m s
⋅
0 .
866
=
21 .
65 m s
R x
=
0 .
8 kg s
⎝
⎜
⎛
30
=
14 N m s
−
12 .
5 m s
⎠
⎟
⎞
R y
=
0 .
8 kg s
⎝
⎜
⎛
21 .
65 m s
=
17 .
32 N
−
0 m s
⎠
⎟
⎞

Continued ...
•
Combining the two results for R x and R y yields
R
=
R x
2 +
R
2 y
= (
14 N
) (
17 .
32 N
)
2 =
22 .
27 N
•
This resultant force, R, will be inclined to the x-direction at an angle ...
= tan
−
1
⎛
⎜⎜
R y
R x
⎞
⎟⎟
= tan
−
1
⎛
⎜⎜
17 .
32
14 N x
N
⎞
⎟⎟
=
51 o
.
3
'
R x
R y