Ch. 6 The Impulse-Momentum Principle
Chapter 6 The Impulse-Momentum Principle
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:
-
Develop impulse - momentum equation, the third of three basic equations of fluid
mechanics, added to continuity and work-energy principles
-
Develop linear and angular momentum (moment of momentum) equations
-
6-1
Ch. 6 The Impulse-Momentum Principle
6.0 Introduction
• Three basic tools for the solution of fluid flow problems
Impulse - momentum equation
Continuity principle
Work-energy principle
•Impulse - momentum equation
~ derived from Newton's 2nd law in vector form


 F  ma
Multifly by dt



 F dt  madt  d (mvc )
 
 d

 F  ( mvc )
dt
where

vc  velocity of the center of mass of the system of mass

mvc  linear momentum
6-2
Ch. 6 The Impulse-Momentum Principle
m   dm
sys
vc 
1
vdm
m sys
( F )dt  impulse in time dt
-
Define the fluid system to include all the fluid in a specified control volume whereas
the Euler equations was developed for a small fluid system
-
Restrict the analysis to steady flow
-
Shear stress is not explicitly included
-
This equation will apply equally well to real fluids as well as ideal fluids.
-
Develop linear and angular momentum (moment of momentum) equations
-
Linear momentum equation:
-
Angular momentum equation: calculate line of action of the resultant forces, rotating
calculate magnitude and direction of resultant forces
fluid machinery (pump, turbine)
6-3
Ch. 6 The Impulse-Momentum Principle
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  dt mv  dt  vdV
(a)
Sum them all
F
ext
 
sys
d 
d

  vdV      vdV 
dt
dt sys

i  v for momentum/mass
Use Reynolds Transport Theorem to evaluate RHS

 
  
  
d
dE
vdV
i
v
dA
v
v
dA
v
v  dA (b)















dt 
dt
sys
c.s.
c. s .out
c . s .in

where E = momentum of fluid system in the control volume
6-4



Ch. 6 The Impulse-Momentum Principle

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)

 
   






v

v
dA

v
v
n
dA

v
vdA

V


2 2Q2
  c. s.out 
 c.s.out
 c.s.out  
v
Q


  
   





v

v
dA

v
v
n
dA

V
1 1Q1

 c.s.in
 c.s.in  
v



By Continuity eq: Q11  Q2  2  Q 



∴ R. H. S. of (b)  Q  V2  V1

(c)
Substitute (c) into (a)

 
F

Q

V

2  V1


(6.1)
In 2-D flow,
F
 Q  V2 x  V1 x 
(6.2a)
F
 Q  V2 z  V1z 
(6.2b)
x
z
6-5
Ch. 6 The Impulse-Momentum Principle
General form in case momentum enters and leaves the control volume at more than one
location:



F
    Q v out    Q v in
(6.3)
- The external forces include both normal (pressure) and tangential (shear) forces on the fluid
in the control volume, as well as the weight of the fluid inside the control volume at a given
time.
- Advantages of impulse-momentum principle
~ Only flow conditions at inlets and exits of the control volume are needed for successful
application.
~ Detailed flow processes within the control volume need not be known to apply the principle.
6-6
Ch. 6 The Impulse-Momentum Principle
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.
•
The reducing pipe bend
Known: flowrate, Q; pressures, p1 , p2 ; velocities, v1 , v2
Find:
F (equal & opposite of the force exerted by the fluid on the bend)
= force exerted by the bend on the fluid
6-7
Ch. 6 The Impulse-Momentum Principle
• Pressures:
For streamlines essentially straight and parallel at section 1 and 2, the forces F1, and F2 result
from hydrostatic pressure distributions.
If mean pressure p1 and p2 are large, and the pipe areas are small, then F1  p1 A1 and
F2  p2 A2 , and assumed to act at the centerline of the pipe instead of the center of pressure.
• Body forces:
= total weight of fluid, W
• Force exerted by the bend on the fluid, F
= 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:
6-8
Ch. 6 The Impulse-Momentum Principle
 Fx  p1 A1  p2 A2 cos   Fx
Q  V2 x  V1x   Q  V2 cos   V1 
Combining the two equations to develop an expression for Fx
Fx  p1 A1  p2 A2 cos   Q  (V1  V2 cos  )
(6.4.a)
(ii) z-direction
 Fz  W  p2 A2 sin   Fz
Q  V2 z  V1z   Q  V2 sin   0 
Fz  W  p2 A2 sin   Q V2 sin 
(6.4.b)
If the bend is relatively sharp, the weight may be negligible, depending on the magnitudes of
the pressure and velocities.
6-9
Ch. 6 The Impulse-Momentum Principle
[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 m3.
590.6 N
Given:
Q  300 l s  0.3 m3 s ; Vol. of bend  0.085 m3
A1 

4
(0.3)2  0.071 m2 ; A2 

4
(0.2)2  0.031 m 2
p1  70 kPa  70  103 N m2
1) Continuity Eq.
Q  AV
1 1  A2V2
V1 
(4.5)
0.3
 4.24 m/s
0.071
6-10
Ch. 6 The Impulse-Momentum Principle
V2 
0.3
 9.55 m/s
0.031
2) Bernoulli Eq. between 1 and 2
p1 V12
p2 V22

 z1 

 z2
 2g
 2g
70  103 (4.24) 2
p2
(9.55) 2

0

 1.5
9,800
2(9.8)
9,800 2(9.8)
p2  18.8 kPa
3) Momentum Eq.
Apply Eqs. 6.4a and 6.4b
Fx  p1 A1  p2 A2 cos   Q  (V1  V2 cos  )
Fz  W  p2 A2 sin   Q V2 sin 
F1  p1 A1  4948 N
F2  p2 A2  18.8  103  0.031  590.6 N
W   (volume)  9800  0.085  833 N
6-11
Ch. 6 The Impulse-Momentum Principle
Fx  4,948  (590.6) cos120  (998  0.3)(4.24  9.55cos120 )  7,942 N
Fz  833  (590.6)sin120  (998  0.3)(9.55sin120  0)  3,820 N
F  Fx2  Fz2  8,813 N
  tan 1
Fz
 25.7
Fx
6-12
Ch. 6 The Impulse-Momentum Principle
• Abrupt enlargement in a closed passage ~ Real fluid flow
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.
Energy loss
Consider the control surface ABCD assuming a one-dimensional flow
i) Continuity
Q  AV
1 1  A2V2
Result from hydrostatic pressure distribution over
the area
ii) Momentum
→ For area AB it is an approximation because of
the dynamics of eddies in the “dead water” zone.
 Fx  p1 A2  p2 A2  Q  (V2  V1 )
6-13
Ch. 6 The Impulse-Momentum Principle
( p1  p2 ) A2 

V2 A2
 (V2  V1 )
g
p1  p2 V2
 (V2  V1 )

g
(a)
iii) Bernoulli equation
p1 V12 p2 V22



 H
 2g
 2g
p1  p2 V22 V12


 H

2g 2g
(b)
where H  Borda-Carnot Head loss
Combine (a) and (b)
V2 (V2  V1 ) V22 V12


 H
g
2g 2g
2V22  2V1V2 V22 V12 (V1  V2 ) 2
H 



2g
2g 2g
2g
6-14
Ch. 6 The Impulse-Momentum Principle
6.3 Open Channel Flow Applications
• Applications impulse-momentum principle for Open Channel Flow
-
Computation of forces exerted by flowing water on overflow or underflow structures
(weirs or gates)
-
Hydraulic jump
-
Wave propagation
[Case 1] Sluice gate
Shear force
neglected
is
Consider a control volume that has uniform flow and straight and parallel streamlines at the
entrance and exit
Apply first Bernoulli and continuity equations to find values of depths y1 and y2 and flowrate
per unit width q
Then, apply the impulse-momentum equation to find the force the water exerts on the sluice
gate
6-15
Ch. 6 The Impulse-Momentum Principle
 Fx  Q  (V2  V1 )
Discharge
width
per
 Fx  F1  F2  Fx  Q  (V2 x  V1x )  q V2  V1 
where q 
Q
 discharge per unit width  y1V1  y2V2
W
Assume that the pressure distribution is hydrostatic at sections 1 and 2, replace V with q/y
y12 y22
1 1

 Fx  q 2  (  )
2
2
y2 y1
(6.6)
[Re] Hydrostatic pressure distribution
y1
 y12
F1   hc A   ( y1  1) 
2
2
1( y1 )3
I
1
12
l p  lc  c 
 y1
lc A y1 ( y  1) 6
1
2
Cp 
1
1
1
y1  y1  y1
2
6
3
6-16
unit
Ch. 6 The Impulse-Momentum Principle
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.
F  Fx cos 
We don’t need to apply the impulse-momentum equation in the z-direction.
[Re] The impulse-momentum equation in the z-direction
 Fz  Q  (V2 z  V1z )
 Fz  FOB  W  Fz  Q  (0  0)
Fz  W  FOB
Non-uniform
distribution
6-17
pressure
Ch. 6 The Impulse-Momentum Principle
[IP 6.2]
For the two-dimensional overflow structure, calculate the horizontal component of
the resultant force the fluid exerts on the structure
Lift gate
Ideal fluid
• Continuity Eq.
q  5V1  2V2
(4.7)
• Bernoulli's equation between (1) and (2)
V12
V22
0  5m 
 0 2m 
2g
2g
(5.7)
Combine two equations
V1  3.33 m s
V2  8.33 m s
q  5(3.33)  16.65 m3 s  m
6-18
Ch. 6 The Impulse-Momentum Principle
• Hydrostatic pressure principle
   9.8 kN
m3 
y
(5)2
 122.5 kN m
F1   hc A   y  9.8
2
2
(2)2
F2  9.8
 19.6 kN m
2
• Impulse-Momentum Eq. (   1000 kg m )
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?
6-19
Ch. 6 The Impulse-Momentum Principle
Jamshil submerged weir (Seo, 1999)
Jamshil submerged weir with gate opened (Q = 200 m3/s)
6-20
Ch. 6 The Impulse-Momentum Principle
Bucket roller
Jamshil submerged weir Model Test; Q = 200 m3/s (Seo, 1999)
Jamshil submerged weir Model Test (Q = 5,000 m3/s)
6-21
Ch. 6 The Impulse-Momentum Principle
[Case 2] Hydraulic Jump
When liquid at high velocity discharges into a zone of lower velocity, a rather abrupt rise (a
standing wave) occurs in water surface and is accompanied by violent turbulence, eddying,
air entrainment, surface undulation.
→ such as a wave is known as a hydraulic jump
Head loss due to
hydraulic jump
→ large head loss (energy dissipation)
turbulence, eddying, air
entrainment, surface
undulation in the
hydraulic jump
Neglect shear
force
Apply impulse-momentum equation to find the relation between the depths for a given
flowrate
Construct a control volume enclosing the hydraulic jump between two sections 1 and 2 where
the streamlines are straight and parallel
 Fx  F1  F2 
 y12
2

 y22
2
 q (V2  V1 )
where q  flowrate per unit width
6-22
Ch. 6 The Impulse-Momentum Principle
Substitute the continuity relations
V1 
q
q
; V2 
y1
y2
Rearrange (divide by
)
q 2 y12 q 2
y2


 2
gy1 2 gy2 2
Solve for y2 y1
8q 2  1 
8V12 
y2 1 
  1  1  3    1  1 

y1 2 
gy1  2 
gy1 
Set Fr1 
V1
gy1
Fr2 
V2
gy2
William
Froude
(1810~1879)
where Fr  Froude number 

Inertia Force
Gravity Force
V
gy
6-23
(6.7)
Ch. 6 The Impulse-Momentum Principle
Then, we have
y2 1 
 1  1  8Fr12 

y1 2 
(a) Fr1  1 :

(b) Fr1 1 :

(c) Fr1 1:

Jump Equation
critical flow
y2 1 
 1  1  8   1
y1 2 
y1  y2  No jump
super-critical flow
y2
1
y1
y2  y1
 Hydraulic jump
y2  y1
 physically impossible
sub-critical flow
y2
1
y1
(∵ rise of energy line through the jump is impossible)
Conclusion:
For a hydraulic jump to occur, the upstream conditions must be such that
V12 gy1  1 .
6-24
Ch. 6 The Impulse-Momentum Principle
[IP 6.3] p. 199 ; Water flows in a horizontal open channel.
y1  0.6 m
q  3.7 m3 s  m
Find y2 , and power dissipated in hydraulic jump.
[Sol]
(i) Continuity
q  y1V1  y2V2
V1 
3.7
 6.17 m s
0.6
Fr1 
V1
6.17

 2.54 1  hydraulic jump occurs
9.8(0.6)
gy1
(ii) Jump Eq.
y2 

y1 
1  1  8Fr12 

2
0.6 
1  1  8(2.54)2 

2 
 1.88 m
6-25
Ch. 6 The Impulse-Momentum Principle
V2 
3.7
 1.97 m s
1.88
(iii) Bernoulli Eq. (Work-Energy Eq.)
V12
V22
y1 
 y2 
 E
2g
2g
0.6 
(6.17) 2
(1.97) 2
 1.88 
 E
2(9.8)
2(9.8)
E  0.46 m
Power   QE  9800  3.7  0.46   16.7 kW meter of width
→ The hydraulic jump is excellent energy dissipator (used in the spillway).
6-26
Ch. 6 The Impulse-Momentum Principle
Pulsating
jump
E/E ~ 85%
6-27
Ch. 6 The Impulse-Momentum Principle
[Case 3] Wave Propagation
The velocity (celerity) of small gravity waves in a body of water can be calculated by the
impulse-momentum equation.
•small gravity waves
~ appears as a small localized rise in the liquid surface which propagate at a velocity a
~ extends over the full depth of the flow
[Cf] small surface disturbance (ripple)
~ liquid movement is restricted to a region near the surface
For the steady flow, assign the velocity under the wave as a’
From continuity
6-28
Ch. 6 The Impulse-Momentum Principle
ay  a '  y  dy 
From impulse-momentum
 y2
2

  y  dy 
2
2
  ay    a '  a 
(6.2a)
Combining these two equations gives
a 2  g  y  dy 
Letting dy approach zero results in
a  gy
(6.8)
→ The celerity of the samll gravity wave depends only on the depth of flow.
6-29
Ch. 6 The Impulse-Momentum Principle
6.4 The Angular Impulse-Momentum Principle
The angular impulse-momentum equation can be developed using moments of the force and
momentum vectors
V2z
V2
V2x
(x1, z1)
(x2, z2)
Fig. 6.8
Take a moment of forces and momentum vectors for the small individual fluid system about
0
  d 


d 
 r  F  ( r  mv )  ( r   d Vol v )
dt
dt
Sum this for control volume
  d
 
 r  Fext   ( r  v )  d Vol.
dt sys
6-30
(a)
Ch. 6 The Impulse-Momentum Principle
Use Reynolds Transport Theorem to evaluate the integral
 
i  rv
 
 
dE d
  ( r  v )  d Vol.   i  v  dA
C.S .
dt dt sys
 
C . S .out
   
(r  v)  v  dA  
C . S .in
   
(r  v)  v  dA
(b)
E  moment of momentum of fluid system
where
 
i  r  v = moment of momentum per unit mass
Restrict to control volume where the fluid enters and leaves at sections where the streamlines
are straight and parallel and with the velocity normal to the cross-sectional area
 
 
 
d
r
v
d
Vol
r
v
dQ
r
(

)

.

(

)


(
C.S .out
C.S .in  v)  dQ
dt sys
Because velocity is uniform over the flow cross sections
 
 
 
d
r
v
dVol
Q
r
V
Q
r
(

)

.


(

)


(
out
out
in  Vin )
dt sys
 
 
 Q  ( r  V )out  ( r  V )in 

(c)
where r  position vector from the moment center to the centroid of entering or leaving flow
cross section of the control volume
6-31
Ch. 6 The Impulse-Momentum Principle
Substitute (c) into (a)
 

 
 
( r  Fext )   M 0  Q  ( r V )out  ( r  V )in 
(6.13)
In 2-D flow,
 M 0  Q  ( r2V2 t  rV
1 1t )
(6.14)
where Vt  component of velocity normal to the moment arm r.
In rectangular components, assuming V is directed with positive components in both x and zdirection, and with the moment center at the origin of the x-z coordinate system, for
clockwise positive moments,
 M 0  Q  ( z2V2 x  x2V2 z )  ( z1V1 x  x1V1z )
(6.15)
where x1 , z1  coordinates of centroid of the entering cross section
x2 , z2  coordinates of centroid of the leaving cross section
For the fluid that enter or leave the control volume at more than one cross-section,
 M 0  (  Q  rVt )out  (  Q  rVt )in
6-32
(6.16)
Ch. 6 The Impulse-Momentum Principle
[IP 6.6] Compute the location of the resultant force exerted by the water on the pipe bend.
590 N
8,813 N
Assume that center of gravity of the fluid is 0.525 m to the right of section 1, and the forces
F1 and F2 act at the centroid of the sections rather than at the center of pressure.
Take moments about the center of section 1
 M 0  Q   ( z2v2 x  x2v2 x )  ( z1v1x  x1v1x )
For this case, x1  0, z1  0, x2  0.6, z2  1.5
 T   r (8,813)  0.525(833)  1.5(590cos60 )  0.6( 590sin 60 )
 (0.3  998) 1.5(9.55  cos60 )  0.6(9.55  sin 60 ) 
 r  0.59 m
6-33
Ch. 6 The Impulse-Momentum Principle
[Re] Torque for rotating system

 

d 
T  (r  F )  (r  mv c )
dt

Where T  torque

T dt  torque impulse


r  mvc  angular momentum (moment of momentum)

r  radius vector from the origin 0 to the point of application of a force
[Re] Vector product (cross product)
  
V  F G
-Magnitude:
  
V  F G sin 


-Direction: perpendicular to the plane of F and G (right-hand rule)
If
 

F , G are in the plane of x and y , then the V is in the z plane.
6-34
Ch. 6 The Impulse-Momentum Principle
Homework Assignment # 6
Due: 1 week from today
Prob. 6.1
Prob. 6.6
Prob. 6.14
Prob. 6.16
Prob. 6.30
Prob. 6.34
Prob. 6.36
Prob. 6.40
Prob. 6.55
Prob. 6.60
6-35