Ch 5 Momentum and forces in fluid flow
Outline
¾ Momentum equation
¾ Applications of the momentum equation
Objectives
•
After completing this chapter, you should be able to
•
•
•
Identify the various kinds of forces and moments acting on a
control volume.
Use control volume analysis to determine the forces associated
with fluid flow.
Use control volume analysis to determine the moments caused
by fluid flow.
6.1 Development of the Momentum Principle
Start with a modified-form of Newton’s 2nd law:
( )
r
r d mV
ΣF =
dt
r
∑F
= sum of forces on fluid system
r
mV = linear momentum of system
r
r
Also, ΣFdt = d (mV ) s-------Impulse-momentum principle
6.1 Development of the Momentum Principle
Momentum entering = ρ A1u 1δ tu 1
Momentum leaving = ρ A 2 u 2 δ tu 2
Force = rate of change of momentum
( ρA2u 2δ tu 2 − ρA1u1δ tu1 )
F=
δt
Q= A1 u1 = A2 u2,
F = ρQ (u 2 − u1 )
Fig 6.1 Momentum in a flowing fluid
Momentum equation for twoand three-dimensional flow
The force in the x-direction
Fig 6.2 Two dimensional
flow in a streamtube
Momentum equation for twoand three-dimensional flow
The force in the y-direction
Fig 6.2 Two dimensional flow in a streamtube
Momentum equation for twoand three-dimensional flow
Total force exerted on the fluid =
Rate of change of momentum
through the control volume
r
r r
F = ρQ(u2 − u1 )
For steady flow with one inlet and one outlet, the momentum equation is
r
r
r
F = ρQ( β 2u2 − β1u1 )
β—— momentum correction factor
Momentum correction factor
True momentum per unit time = β × Mass per unit time × Mean velocity
β = ∫ u 2 dA / V 2 A
The momentum equation
r
r
r
F = ρQ( β 2V2 − β1V1 )
the resultant force
F = Fx2 + Fy2
the angle which this force acts at is given by
What are the forces acting on the
fluid in the control volume?
the total force, FT, is given by the sum of these forces
r
r
r
r
FT = FR + FB + FP
Step in Analysis with Momentum Equation
1. Draw a control volume: Based on the problem, selecting the
stream between two gradually varied flow sections as the control
volume;
2. Decide on co-ordinate axis system: Determining the
directions of co-ordinate axis, magnitudes and directions of
components of all forces and velocities on each axis.
3. Plotting diagram for computation : Analyzing the forces on
control volume and plotting the directions of all forces on the
control volume.
4. Writing momentum equation and solving it: Substituting
components of all forces and velocities on axes into
momentum equation and solving it. All the pressures are
relative to the relative pressure.
Application of the Momentum Equation
1. Force due to the flow of fluid round a pipe bend.
2. Force on a nozzle at the outlet of a pipe.
3. Impact of a jet on a plane surface.
4. Force due to flow round a curved vane.
Force due to Jet Striking Surface
Force by Flow Round a Pipe-bend
Example 1
Find the horizontal thrust of the water on each meter of
width of the sluice gate shown in the Fig., given y1=2.2
m, y2=0.4 m, and y3=0.5 m. Neglect friction.
( 6.4.2)
Solution
6.5.5
Solution
The flow rate
Momentum equation
6.8
A reducing right-angled bend lies in a horizontal plane.
Water enters from the west with a velocity of 3 m/s and a
pressure of 30 kPa, and it leaves toward the north. The
diameter at the entrance is 500 mm and at the exit it is
400 mm. Neglecting any friction loss, find the magnitude
and direction of the resultant force on the bend.
Solution
Energy
Momentum
Example
Water flows through a reducing 180°bend. The bend is shown in
plan. Determine the magnitude of the force exerted on the bend in
the x-direction. Assume energy losses to be negligible.
Solution:
Example
T=15 oC
•
•
•
Given: Figure
Find: Horizontal force required to
hold plate in position
Solution:
Q=0.4 m3/s
F
B
pA=75 kPa
V A2 p B
VB2
+ zA +
=
+ zB +
γ
γ
2g
2g
pA
VB2
=
γ
2g
pA
r
r
V = Vi
− F = ρQ(V2 − V1 )
p
VB = 2 A = 2 * 75000 / 999 = 12.3 m / s
ρ
F = ρQV = 999 * 0.4 *12.3 = 4.9 kN
question
If the value of force calculated from
momentum equation is negative, what
does that mean? Does the magnitude of
the unknown force has anything or
nothing to do with that of the control
volume? How to select control volume in
the application?
If the value of force calculated from momentum
equation is negative, what does that mean? Does the
magnitude of the unknown force has anything or
nothing to do with that of the control volume? How to
select control volume in the application?
directions are inverse；independency（when
there is no gravitation）；calculated crosssection and solid wall
Sluice Gate
Find: Force due to pressure on
face of gate
Solution:
Assume: v1 and v2 are uniform
(so pressure is hydrostatic)
Application of the Energy, Momentum, and
Continuity Equations in Combination
In general, when solving fluid mechanics problems,
one should use all available equations in order to
derive as much information as possible about the
flow. For example, consistent with the approximation
of the energy equation we can also apply the
momentum and continuity equations
Forces on Transitions
Example
Example: Energy Equation
(energy loss)
An irrigation pump lifts 50 L/s of water from a reservoir and
discharges it into a farmer’s irrigation channel. The pump
supplies a total head of 10 m. How much mechanical energy
is lost? What is hL?
cs2
4m
2.4 m
2m
cs1
datum
Why can’t I draw the cs at the end of the pipe?
p in
g
+ z in + a
h p = z out + hL
in
V in2
p
+ hP = out + z out + a
2g
g
hL = h p - z out
out
V o2u t
+ hT + h L
2g
Example: Energy Equation
(pressure at pump outlet)
The total pipe length is 50 m and is 20 cm in diameter. The
pipe length to the pump is 12 m. What is the pressure in the
pipe at the pump outlet? You may assume (for now) that the
only losses are frictional losses in the pipeline.
50 L/s
hP = 10 m
4m
cs2
2.4 m
2m
cs1
datum
p in
g
+ z in + a
in
V in2
p
+ hP = out + z out + a
2g
g
out
V o2u t
+ hT + h L
2g
We need _______ in the pipe, __, and ____ ____.
Example: Energy Equation
(pressure at pump outlet)
¾How do we get the velocity in the pipe?
¾How do we get the frictional losses?
¾What about α?
Kinetic Energy Correction Term:
α
¾α is a function of the velocity distribution in
the pipe.
¾For a uniform velocity distribution ____
¾For laminar flow ______
¾For turbulent flow _____________
¾Often neglected in calculations because it is so
close to 1
Example: Energy Equation
(pressure at pump outlet)
V = 1.6 m/s
α = 1.05
hL = 1.44 m
2.4 m
2m
hP =
p out
g
+ z out + a
out
V o2u t
+ hL
2g
50 L/s
hP = 10 m
4m
datum
2
⎡
⎤
(1.6
m/s)
3
p2 = (9810N/m )⎢(10m)− (2.4m)− (1.05)
− (1.44m)⎥
2
2(9.81m/s )
⎣
⎦
Example: Energy Equation
(Hydraulic Grade Line - HGL)
¾We would like to know if there are any
places in the pipeline where the pressure is
too high (_________) or too low (water
might boil - cavitation).
¾Plot the pressure as piezometric head
(height water would rise to in a piezometer)
¾How?
Example: Energy Equation
(Energy Grade Line - EGL)
p = 59 kPa
Loss due to shear
HP = 10 m
Entrance loss
Exit loss
2
p
V
+a
2g
g
50 L/s
4m
2.4 m
2m
datum
What is the pressure at the pump intake?
p in
g
+ z in + a
in
V in2
p out
+ hP =
+ z out + a
2g
g
out
V o2u t
+ hT + h L
2g
EGL (or TEL) and HGL
EGL =
p
γ
+ z +α
V
2
2g
HGL =
p
γ
+ z
What is the difference between EGL defined by Bernoulli
and EGL defined here?
EGL (or TEL) and HGL
¾The energy grade line may never be
horizontal or slope upward (in direction of
flow) unless energy is added (______)
¾The decrease in total energy represents the
head loss or energy dissipation per unit
weight
¾EGL and HGL are ____________and lie at
the free surface for water at rest (reservoir)
¾Whenever the HGL falls below the point in
the system for which it is plotted the local
Example HGL and EGL
velocity head α
V
2
2g
pressure head
p
γ
energy grade line
hydraulic grade line
z elevation
pump
z=0
p in
g
+ z in + a
datum
2
in
in
V
p
+ hP = out + z out + a
2g
g
out
V o2u t
+ hT + h L
2g
See you next time.