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Develop impulse - momentum equation, the third of three basic equations of fluid mechanics, added to continuity and work-energy principles.
Typology: Schemes and Mind Maps
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6.1 The Linear Impulse-Momentum Equation
6.2 Pipe Flow Applications
6.3 Open Channel Flow Applications
6.4 The Angular Impulse-Momentum Principle
Objectives:
mechanics, added to continuity and work-energy principles
6.0 Introduction
- Three basic tools for the solution of fluid flow problems
Impulse - momentum equation
Continuity principle
Work-energy principle
~ derived from Newton's 2nd law in vector form
F ma
Multifly by dt
( (^) c )
d F mv dt
where vc
velocity of the center of mass of the system of mass
mv c
linear momentum
6.1 The Linear Impulse – Momentum Equation
Use the same control volume previously employed for conservation of mass and work-energy.
For the individual fluid system in the control volume,
d d F ma mv vdV dt dt
^ ^ ^
(a)
Sum them all
ext^ ^ ^ ^ sys sys
d d F vdV vdV dt dt
^
Use Reynolds Transport Theorem to evaluate RHS
(^) sys c s.. c s out.. c s in..
d dE vdV i v dA v v dA v v dA dt dt
^ ^ ^ ^ ^ ^
(b)
where E = momentum of fluid system in the control volume
i v
for momentum/mass
i v
momentum per unit mass
Because the streamlines are straight and parallel at Sections 1 and 2, velocity is constant over
the cross sections. The cross-sectional area is normal to the velocity vector over the entire
cross section.
∴In Eq. (b)
c s out..^ ^ ^ c s out.. ^ c s out.. 2 2 2 v Q
c s in..^ ^ ^ c s in.. 1 1 1 v
∴ R. H. S. of (b) Q V 2 (^) V 1
(c)
Substitute (c) into (a)
F^ ^ Q^^ ^ ^ V 2^ V 1
(6.1)
In 2-D flow,
Fx^ ^ Q^^ ^ V^2^ x V 1 x (6.2a)
Fz^ ^ Q^^ ^ ^ V 2^ z V 1 z (6.2b)
6.2 Pipe Flow Applications
Forces exerted by a flowing fluid on a pipe bend, enlargement, or contraction in a pipeline
may be computed by a application of the impulse-momentum principle.
Known: flowrate, Q ; pressures, p 1 (^) , p 2 ; velocities, v 1 (^) , v 2
Find: F (equal & opposite of the force exerted by the fluid on the bend)
= force exerted by the bend on the fluid
For streamlines essentially straight and parallel at section 1 and 2, the forces F 1 , and F 2 result
from hydrostatic pressure distributions.
If mean pressure p 1 and p 2 are large, and the pipe areas are small, then F 1 (^) p A 1 1 and
F 2 (^) p A 2 2 , and assumed to act at the centerline of the pipe instead of the center of pressure.
= total weight of fluid, W
= resultant of the pressure distribution over the entire interior of the bend between sections
1&2.
~ distribution is unknown in detail
~ resultant can be predicted by Impulse-momentum Eq.
Now apply Impulse-momentum equation, Eq. (6.2)
(i) x -direction:
[IP 6.1] 300 l/s of water flow through the vertical reducing pipe bend. Calculate the force
exerted by the fluid on the bend if the volume of the bend is 0.085 m
3 .
Given:
3 Q 300 l s 0.3 m s;
3 Vol. of bend 0.085 m
2 2 1 (0.3)^ 0.071 m 4
2 2 2 (0.2)^ 0.031 m 4
3 2 p 1 (^) 70 kPa 70 10 N m
1
4.24 m/s
590.6 N
2
9.55 m/s
2 2 1 1 2 2 1 2 2 2
p V p V z z g g
3 2 2 70 10 (4.24) 0 2 (9.55) 1.
9,800 2(9.8) 9,800 2(9.8)
p
p 2 (^) 18.8 kPa
Apply Eqs. 6.4a and 6.4b
F 1 (^) p A 1 1 4948 N
3 F 2 (^) p A 2 2 18.8 10 0.031 590.6 N
W (volume) 9800 0.085 833 N
The impulse-momentum principle can be employed to predict the fall of the energy line
(energy loss due to a rise in the internal energy of the fluid caused by viscous dissipation) at
an abrupt axisymmetric enlargement in a passage.
Consider the control surface ABCD assuming a one-dimensional flow
i) Continuity
ii) Momentum
Result from hydrostatic pressure distribution over the area
→ For area AB it is an approximation because of the dynamics of eddies in the “dead water” zone.
Energy loss
2 2 ( 1 2 ) 2 ( 2 1 )
p p A V V g
1 2 2 ( 2 1 )
p p V V V g
(a)
iii) Bernoulli equation
2 2 1 1 2 2 2 2
p V p V H g g
2 2 1 2 2 1 2 2
p p V V H g g
(b)
where H Borda-Carnot Head loss
Combine (a) and (b)
2 2 2 (^2 1 ) 2 1 2 2
g g g
2 2 2 2 2 2 2 1 2 2 1 ( 1 2 )
2 2 2 2
g g g g
Fx F 1 (^) F 2 (^) Fx Q ( V 2 (^) (^) x V 1 (^) (^) x ) q V 2 (^) V 1
where
q W
discharge per unit width y V 1 1 (^) y V 2 2
Assume that the pressure distribution is hydrostatic at sections 1 and 2, replace V with q/y
2 2 1 2 2
2 1
x
y y F q y y
[Re] Hydrostatic pressure distribution
2 1 1 1 (^1 1) 2 2
c
y y F h A y
3 1
1 1 1
c p c c
y I l l y l A y y
1 1 1
C (^) p y y y
Discharge per unit width
For ideal fluid (to a good approximation, for a real fluid), the force tnagent to the gate is zero.
→ shear stress is neglected.
→ Hence, the resultant force is normal to the gate.
We don’t need to apply the impulse-momentum equation in the z -direction.
[Re] The impulse-momentum equation in the z -direction
F z W FOB
Non-uniform pressure distribution
2
1
9.8 122.5 kN m 2 2
c
y
2
2
9.8 19.6 kN m 2
3
Fx 122,500 Fx 19,600 (1000 16.65)(8.33 3.33)
Fx 19.65 kN m
[Cf] What is the force if the gate is closed?
Jamshil submerged weir (Seo, 1999)
Jamshil submerged weir with gate opened (Q = 200 m^3 /s)
