57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
1
Chapter 5 Finite Control Volume Analysis
5.1 Continuity Equation
RTT can be used to obtain an integral relationship expressing
conservation of mass by defining the extensive property B = M
such that  = 1.
B = M = mass
 = dB/dM = 1
General Form of Continuity Equation
dM
d
0
 dV   V  dA
dt
dt CV
CS
or
d

V

dA



 dV
dt CV
CS
net rate of outflow
of mass across CS
rate of decrease of
mass within CV
Simplifications:
d
 dV  0
dt CV
2. V = constant over discrete dA (flow sections):
1. Steady flow: 
 V  dA   V  A
CS
CS
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
2
3. Incompressible fluid ( = constant)
 V  dA  
CS
d
 dV
dt CV
conservation of volume
4. Steady One-Dimensional Flow in a Conduit:
 V  A  0
CS
1V1A1 + 2V2A2 = 0
for  = constant
Q1 = Q2
Some useful definitions:
Mass flux
   V  dA
m
A
Volume flux
Q   V  dA
A
Average Velocity
V  Q/A
Average Density

1
 dA
A
Note: m  Q unless  = constant
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
3
Example
*Steady flow
*V1,2,3 = 50 fps
*At , V varies linearly
from zero at wall to
Vmax at pipe center
 4 , Q4, Vmax
*find m
*water, w = 1.94 slug/ft3
0
 V  dA  0  
CS
d
 dV
dt CV
4
m
i.e., -1V1A1 - 2V2A2 + 3V3A3 +   V4 dA 4 = 0
A4
 = const. = 1.94 lb-s2 /ft4 = 1.94 slug/ft3
 4   V4 dA 4 = V(A1 + A2 – A3)
m

V1=V2=V3=V=50f/s
1.94

 50  12  2 2  1.52
144
4
= 1.45 slugs/s
=

57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
4
 4   .75 ft3/s
Q4 = m
=  V4 dA 4
A4
velocity profile


Q4 =   Vmax 1  r rddr
ro 2 

0 0
ro 
dA4
V4  V4()

r
 2  Vmax 1  rdr
0
 ro 
ro
ro 
r2 
 2Vmax  r  dr
ro 
0
r2
 2Vmax 
 2
r0
0
r3

3ro


0 

ro
1 1 1
 2Vmax ro 2     ro 2 Vmax
 2 3 3
Vmax =
Q4
1 2
ro
3
 2.86 fps
1 2
r V
Q 3 o max
V4  
A
ro2
1
= Vmax
3
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
5
5.2 Momentum Equation
Derivation of the Momentum Equation
Newton’s second law of motion for a system is
time rate of change
of the momentum
of the system
= sum of external
forces acting on
the system
Since momentum is mass times velocity, the momentum of a
small particle of mass 𝜌𝑑𝑉 is 𝑉𝜌𝑑𝑉 and the momentum of the
entire system is ∫𝑠𝑦𝑠 𝑉𝜌𝑑𝑉. Thus,
𝐷
∫ 𝑉𝜌𝑑𝑉 = ∑ 𝐹𝑠𝑦𝑠
𝐷𝑡 𝑠𝑦𝑠
Recall RTT:
𝐷𝐵𝑠𝑦𝑠
𝜕
= ∫ 𝛽𝜌𝑑𝑉 + ∫ 𝛽𝜌𝑉𝑅 ⋅ 𝑑𝐴
𝐷𝑡
𝜕𝑡 𝐶𝑉
𝐶𝑆
With 𝐵𝑠𝑦𝑠 = 𝑀𝑉 and 𝛽 =
𝑑𝐵𝑠𝑦𝑠
𝑑𝑀
= 𝑉,
𝐷
𝜕
∫ 𝑉𝜌𝑑𝑉 = ∫ 𝑉𝜌𝑑𝑉 + ∫ 𝑉𝜌𝑉𝑅 ⋅ 𝑑𝐴
𝐷𝑡 𝑠𝑦𝑠
𝜕𝑡 𝐶𝑉
𝐶𝑆
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
6
Thus, the Newton’s second law becomes
𝜕
∫ 𝑉𝜌𝑑𝑉 + ∫ 𝑉𝜌𝑉𝑅 ⋅ 𝑑𝐴 =
𝜕𝑡
⏟ 𝐶𝑉
⏟𝐶𝑆
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒
𝑜𝑓 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚
𝑖𝑛 𝑡ℎ𝑒 𝐶𝑉
𝑛𝑒𝑡 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑓𝑙𝑜𝑤
𝑜𝑓 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚
𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑡ℎ𝑒 𝐶𝑆
∑ 𝐹𝐶𝑉
⏟
𝑛𝑒𝑡 𝑒𝑥𝑡𝑒𝑟𝑛𝑎𝑙
𝑓𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔
𝑜𝑛 𝑡ℎ𝑒 𝐶𝑉
where,
𝑉 is fluid velocity referenced to an inertial frame (nonaccelerating)
𝑉𝑆 is the velocity of CS referenced to the inertial frame
𝑉𝑅 = 𝑉 − 𝑉𝑆 is the relative velocity referenced to CV
∑ 𝐹𝐶𝑉 = ∑ 𝐹𝐵 + ∑ 𝐹𝑆 is vector sum of all forces acting on
the CV
𝐹𝐵 is body force such as gravity that acts on the entire
mass/volume of CV
𝐹𝑆 is surface force such as normal (pressure and
viscous) and tangential (viscous) stresses acting on the
CS
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
7
Note that, when CS cuts through solids, 𝐹𝑆 may also include
reaction force, 𝐹𝑅
(e.g., the reaction force required to hold nozzle or bend when CS
cuts through the bolts that are holding the nozzle/bend in place)
∑ 𝐹𝑥 = 𝑝1 𝐴1 − 𝑝2 𝐴2 + 𝑅𝑥
∑ 𝐹𝑦 = −𝑊 + 𝑅𝑦
𝑅 = 𝑅𝑥 𝑖̂ + 𝑅𝑦 𝑗̂ = resultant
force on fluid in CV due to 𝑝𝑤
and 𝜏𝑤 , i.e. reaction force on
fluid
Free body diagram
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
8
Important Features (to be remembered)
1) Vector equation to get component in any direction must use
dot product
x equation
d
 Fx 
 udV   u V R  dA
dt CV
CS
y equation
d
 Fy 
 vdV   vV R  dA
dt CV
CS
Carefully define coordinate
system with forces positive in
positive direction of
coordinate axes
z equation
d
 Fz 
 wdV   w V R  dA
dt CV
CS
2) Carefully define control volume and be sure to include all
external body and surface faces acting on it.
For example,
(Rx,Ry) = reaction
force on fluid
(Rx,Ry) = reaction
force on nozzle
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
9
3) Velocity V and Vs must be referenced to a non-accelerating
inertial reference frame. Sometimes it is advantageous to use
a moving (at constant velocity) reference frame: relative
inertial coordinate. Note VR = V – Vs is always relative to
CS.
4) Steady vs. Unsteady Flow
Steady flow 
d
 VdV  0
dt CV
5) Uniform vs. Nonuniform Flow
 VV R  dA = change in flow of momentum across CS
CS
= VVRA
uniform flow across A
6) Fpres =   pndA
 fdV   f nds
V
S
f = constant, f = 0
= 0 for p = constant and for a closed surface
i.e., always use gage pressure
7) Pressure condition at a jet exit
at an exit into the atmosphere jet
pressure must be pa
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
10
Applications of the Momentum Equation
Initial Setup and Signs
1. Jet deflected by a plate or a vane
2. Flow through a nozzle
3. Forces on bends
4. Problems involving non-uniform velocity distribution
5. Motion of a rocket
6. Force on rectangular sluice gate
7. Water hammer
8. Steady and unsteady developing and fully developed pipe
flow
9. Empting and filling tanks
10. Forces on transitions
11. Hydraulic jump
12. Boundary layer and bluff body drag
13. Rocket or jet propulsion
14. Propeller
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
11
1. Jet deflected by a plate or vane
Consider a jet of water turned through a horizontal angle
CV and CS are
for jet so that Fx
and Fy are vane
reactions forces
on fluid
x-equation:
d
 udV   u V  dA
dt
CS
steady flow
Fx   u V  A
 Fx  Fx 
CS
= V1x (V1A1 )  V2 x (V2 A 2 )
continuity equation:
A1V1 = A2V2 = Q
Fx = Q(V2x – V1x)
y-equation:
for A1 = A2
V1 = V2
 Fy  Fy   v V  A
CS
Fy = V1y(– A1V1) + V2y(– A2V2)
= Q(V2y – V1y)
for above geometry only
where:
note:
V1x = V1 V2x = -V2cos V2y = -V2sin V1y = 0
Fx and Fy are force on fluid
- Fx and -Fy are force on vane due to fluid
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
12
If the vane is moving with velocity Vv, then it is convenient to
choose CV moving with the vane
i.e., VR = V - Vv and V used for B also moving with vane
x-equation:
Fx   u V R  dA
CS
Fx = V1x[-(V – Vv)1A1] + V2x[(V – Vv)2A2]
Continuity:
0 =  V R  dA
i.e., (V-Vv)1A1 = (V-Vv)2A2 = (V-Vv)A
Qrel
Fx = (V-Vv)A[V2x – V1x]
Qrel
on fluid
V2x = (V – Vv)2x
V1x = (V – Vv)1x
Power = -FxVv
Fy = Qrel(V2y – V1y)
For coordinate system
moving with vane
i.e., = 0 for Vv = 0
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
13
2. Flow through a nozzle
Consider a nozzle at the end of a pipe (or hose). What force is
required to hold the nozzle in place?
CV = nozzle
and fluid
 (Rx, Ry) =
force required
to hold nozzle
in place the
Then,
Assume either the pipe velocity or pressure is known.
unknown (velocity or pressure) and the exit velocity V2 can be
obtained from combined use of the continuity and Bernoulli
equations.
Bernoulli:
1
1
p1  z1  V12  p 2  z 2  V22
2
2
1
1
p1  V12  V22
2
2
Continuity:
z1=z2
A1V1 = A2V2 = Q
2
A1
 D
V2 
V1    V1
A2
d
4
1 2   D  
p1  V1 1     0
 d 
2


1/ 2




 2p1
V1  
Say p1 known:
4 

D
 1  d  

 
To obtain the reaction force Rx apply momentum equation in xdirection
 
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
14
d
 udV   u V  dA
dt CV
CS
steady flow and uniform
=  u V  A
CS
flow over CS
 Fx 
Rx + p1A1 – p2A2 = V1(-V1A1) + V2(V2A2)
= Q(V2 - V1)
Rx = Q(V2 - V1) - p1A1
To obtain the reaction force Ry apply momentum equation in ydirection
 Fy   v V  A  0
since no flow in y-direction
CS
Ry – Wf  WN = 0 i.e., Ry = Wf + WN
Numerical Example: Oil with S = .85 flows in pipe under
pressure of 100 psi. Pipe diameter is 3” and nozzle tip diameter
is 1”
S

 1.65
g
V1 = 14.59 ft/s
D/d = 3
V2 = 131.3 ft/s
2
 1 
Q =   V2
4  12 
Rx = 141.48 – 706.86 = 569 lbf
= .716 ft3/s
Rz = 10 lbf
This is force on nozzle
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
15
3. Forces on Bends
Consider the flow through a bend in a pipe. The flow is
considered steady and uniform across the inlet and outlet
sections. Of primary concern is the force required to hold the
bend in place, i.e., the reaction forces Rx and Ry which can be
determined by application of the momentum equation.
Rx, Ry = reaction force on
bend i.e., force
required to hold
bend in place
Continuity:
0   V  A  V1A1  V2 A 2
i.e., Q = constant = V1A1  V2 A 2
x-momentum:  Fx   u V  A
p1A1  p 2 A 2 cos   R x  V1x  V1A1   V2 x V2 A 2 
= QV2 x  V1x 
y-momentum:  Fy   vV  A
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
16
p 2 A 2 sin   R y  w f  w b  V1y  V1A1   V2 y V2 A 2 
= QV2 y  V1y 
4. Force on a rectangular sluice gate
The force on the fluid due to the gate is calculated from the xmomentum equation:
 Fx   u V  A
F1  FGW  Fvisc  F2  V1  V1A1   V2 V2 A 2 
FGW  F2  F1  QV2  V1   Fvisc
y
y
=  2  y 2 b   1  y1b  QV2  V1 
2
2
1
FGW  b y 22  y12  QV2  V1 
2
usually can be neglected


V1 
Q
y1b
V2 
Q
y2b
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
17
Q 2  1 1 
  
b  y 2 y1 
5. Application of relative inertial coordinates for a moving but
non-deforming control volume (CV)
The CV moves at a constant velocity V CS with respect to
the absolute inertial coordinates. If V R represents the
velocity in the relative inertial coordinates that move
together with the CV, then:
VR  V  VCS
Reynolds transport theorem for an arbitrary moving deforming
CV:
dBSYS d

  d   CS VR  n dA
dt
dt CV
For a non-deforming CV moving at constant velocity, RTT for
incompressible flow:
dBsyst


d      VR  ndA
dt
CV
t
CS
1) Conservation of mass
Bsyst  M , and   1 :
dM
   VR  ndA
dt
CS
For steady flow:
V
R
CS
 ndA  0
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
18
2) Conservation of momentum
Bsyst  M VR  V CS  and   dBsyst dM  VR  VCS


d [ M VR  V CS ]
dt
 F  


 VR  V CS
t
CV
 d  
 V
R

 V CS VR  ndA
CS
For steady flow with the use of continuity:
 F    V
R

 V CS VR  ndA
CS
   VR VR  ndA  V CS  VR  ndA
CS
F
0
CS
   VR VR  ndA
CS
Example (use relative inertial coordinates):
Ex) A jet strikes a vane which moves to the right at constant velocity 𝑉𝐶 on a
frictionless cart. Compute (a) the force 𝐹𝑥 required to restrain the cart and (b)
the power 𝑃 delivered to the cart. Also find the cart velocity for which (c) the
force 𝐹𝑥 is a maximum and (d) the power 𝑃 is a maximum.
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
19
Solution:
Assume relative inertial coordinates with non-deforming CV i.e. CV moves
at constant translational non-accelerating
𝑉𝐶𝑆 = 𝑢𝐶𝑆 𝑖̂ + 𝑣𝐶𝑆 𝑗̂ + 𝑤𝐶𝑆 𝑘̂ = 𝑉𝐶 𝑖̂
then V R  V  V CS . Also assume steady flow 𝑉 ≠ 𝑉(𝑡) with 𝜌 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 and neglect
gravity effect.
Continuity:
V
R
 ndA  0
CS
Bernoulli without gravity:
0
1
1
VR21  p2  VR22
2
2
VR1  VR 2
VR1 A1  VR 2 A2
A1  A2  A j
0
p1 
Since
Momentum:
∑𝐹 = 𝜌 ∫ 𝑉𝑅 𝑉𝑅 ⋅ 𝑛𝑑𝐴
𝐶𝑆
∑ 𝐹𝑥 = −𝐹𝑥 = 𝜌 ∫ 𝑉𝑅𝑥 𝑉𝑅 ⋅ 𝑛𝑑𝐴
𝐶𝑆
−𝐹𝑥 = 𝜌𝑉𝑅𝑥 1 (−𝑉𝑅1 𝐴1 ) + 𝜌𝑉𝑅𝑥 2 (𝑉𝑅2 𝐴2 )
0 = 𝜌 ∫ 𝑉𝑅 ⋅ 𝑛𝑑𝐴
𝐶𝑆
−𝜌𝑉𝑅1 𝐴1 + 𝜌𝑉𝑅2 𝐴2 = 0
𝑉𝑅1 𝐴1 = 𝑉𝑅2 𝐴2 = ⏟
(𝑉𝑗 − 𝑉𝐶 ) 𝐴𝑗
𝑉𝑅1 =𝑉𝑅𝑥 1 =𝑉𝑗 −𝑉𝐶
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
20
−𝐹𝑥 = 𝜌(𝑉𝑗 − 𝑉𝐶 )[−(𝑉𝑗 − 𝑉𝐶 )𝐴𝑗 ] + 𝜌(𝑉𝑗 − 𝑉𝐶 ) cos 𝜃 (𝑉𝑗 − 𝑉𝐶 )𝐴𝑗
2
𝐹𝑥 = 𝜌(𝑉𝑗 − 𝑉𝐶 ) 𝐴𝑗 [1 − cos 𝜃]
2
𝑃𝑜𝑤𝑒𝑟 = 𝑉𝐶 𝐹𝑥 = 𝑉𝐶 𝜌(𝑉𝑗 − 𝑉𝐶 ) 𝐴𝑗 (1 − cos 𝜃)
𝐹𝑥 𝑚𝑎𝑥 = 𝜌𝑉𝑗2 𝐴𝑗 (1 − cos 𝜃),
𝑃𝑚𝑎𝑥 ⇒
𝑉𝐶 = 0
𝑑𝑃
=0
𝑑𝑉𝐶
𝑃 = 𝑉𝐶 𝜌(𝑉𝑗2 − 2𝑉𝐶 𝑉𝑗 + 𝑉𝐶2 )𝐴𝑗 (1 − cos 𝜃)
= 𝜌(𝑉𝑗2 𝑉𝐶 − 2𝑉𝐶2 𝑉𝑗 + 𝑉𝐶3 )𝐴𝑗 (1 − cos 𝜃)
𝑑𝑃
= 𝜌(𝑉𝑗2 − 4𝑉𝐶 𝑉𝑗 + 3𝑉𝐶2 )𝐴𝑗 (1 − cos 𝜃) = 0
𝑑𝑉𝐶
3𝑉𝐶2 − 4𝑉𝑗 𝑉𝐶 + 𝑉𝑗2 = 0
+4𝑉𝑗 ± √16𝑉𝑗2 − 12𝑉𝑗2
𝑉𝐶 =
=
6
𝑉𝐶 =
4𝑉𝑗 ± 2𝑉𝑗
6
𝑉𝑗
3
𝑉𝑗 2𝑉𝑗 2
𝑃𝑚𝑎𝑥 = 𝜌 ( ) 𝐴𝑗 (1 − cos 𝜃)
3
3
4 3
=
𝑉 𝜌𝐴𝑗 (1 − cos 𝜃)
27 𝑗
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
21
5.3 Energy Equation
Derivation of the Energy Equation
The First Law of Thermodynamics
The difference between the heat added to a system and the work
done by a system depends only on the initial and final states of
the system; that is, depends only on the change in energy E:
principle of conservation of energy
E = Q – W
E = change in energy
Q = heat added to the system
W = work done by the system
E = Eu + Ek + Ep = total energy of the system
potential energy
kinetic energy
Internal energy due to molecular motion
The differential form of the first law of thermodynamics
expresses the rate of change of E with respect to time
dE  
QW
dt
rate of work being done by system
rate of heat transfer to system
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
22
Energy Equation for Fluid Flow
The energy equation for fluid flow is derived from Reynolds
transport theorem with
Bsystem = E = total energy of the system (extensive property)
 = E/mass = e = energy per unit mass (intensive property)
= û + ek + ep
dE d
  edV  CS eV  dA
dt dt CV
d
Q  W    (uˆ  ek  e p )dV    (uˆ  ek  e p )V  dA
CS
dt CV
This can be put in a more useable form by noting the following:
Total KE of mass with velocity V MV 2 / 2 V 2
ek 


V2  V
mass
M
2
E p Vz
(for Ep due to gravity only)
ep 

 gz
M V
V 2

V 2

d
Q W    
 gz  uˆ  dV    
 gz  uˆ V  dA
Cs
dt CV  2

 2

rate of work
done by system
rate of heat
transfer to sysem
rate of change
of energy in CV
flux of energy
out of CV
(ie, across CS)
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
23
 W
s W
f
Rate of Work Components: W
For convenience of analysis, work is divided into shaft work Ws
and flow work Wf
Wf = net work done on the surroundings as a result of
normal and tangential stresses acting at the control
surfaces
= Wf pressure + Wf shear
Ws = any other work transferred to the surroundings
usually in the form of a shaft which either takes
energy out of the system (turbine) or puts energy into
the system (pump)
Flow work due to pressure forces Wf p (for system)
Note: here V uniform over A
CS
System at time t + t
CV
System at time t
Work = force  distance
at 2 W2 = p2A2  V2t (on surroundings)
 2  p 2 A 2 V2  p 2 V 2  A 2
rate of work W
neg. sign since pressure
force on surrounding
fluid acts in a direction
opposite to the motion
of the system boundary
at 1 W1 = p1A1  V1t
 1  p1 V1  A1
W
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
24
In general,
 fp  p V  A
W
for more than one control surface and V not necessarily uniform
over A:
 fp   pV  dA    p V  dA
W
CS
CS

f W
 fp  W
 fshear
W
Basic form of energy equation
 p
Q  Ws  W fshear     V  dA
CS

V 2

V 2

d
  
 gz  uˆ  dV    
 gz  uˆ  V  dA
CS
dt CV  2

 2

Q  Ws  W fshear
V 2

d
  
 gz  uˆ  dV
dt CV  2

Usually this term can be
eliminated by proper choice of
CV, i.e. CS normal to flow lines.
Also, at fixed boundaries the
velocity is zero (no slip
condition) and no shear stress
flow work is done. Not included
or discussed in text!
V 2
p
ˆ
 
 gz  u   V  dA
CS
2


h=enthalpy
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
25
Simplified Forms of the Energy Equation
Energy Equation for Steady One-Dimensional Pipe Flow
Consider flow through the pipe system as shown
Energy Equation (steady flow)
V 2

p
Q  Ws    
 gz   uˆ V  dA
CS

 2

 p1

1V13
Q  Ws     gz1  uˆ1  1V1 A1  
dA1
A1
A
1
2


 p2

2V23
    gz2  uˆ2  2V2 A2  
dA2
A2
A
2
2
 

*Although the velocity varies across the flow sections the
streamlines are assumed to be straight and parallel;
consequently, there is no acceleration normal to the streamlines
and the pressure is hydrostatically distributed, i.e., p/ +gz =
constant.
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
26
*Furthermore, the internal energy u can be considered as
constant across the flow sections, i.e. T = constant. These
quantities can then be taken outside the integral sign to yield
 p1

V13
Q  Ws    gz1  uˆ1    V1dA1   
dA1
A
A
1
1
2


 p2

V23
   gz2  uˆ2    V2 dA2   
dA2
A2
A2 2



Recall that
So that
Q   V  dA  VA

 V  dA  VA  m
Define:
V3
V A
V

  dA  

m
2
2
2
A
3
K.E. flux
mass flow rate
2
K.E. flux for V= V =constant across pipe
3
i.e.,
1 V
     dA = kinetic energy correction factor
A A V 
2
2
p

p

V
V
1
2
1
2
 m    gz2  uˆ2   2
m
Q  W    gz1  uˆ1  1


 

2
2





2

2
p
1
V 1 p2
V2
Q  W  1  gz1  uˆ1  1

 gz2  uˆ2   2
m

2

2
Note that:
 = 1 if V is constant across the flow section
 > 1 if V is nonuniform
laminar flow  = 2
turbulent flow  = 1.05  1 may be used
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
27
Shaft Work
Shaft work is usually the result of a turbine or a pump in the
flow system. When a fluid passes through a turbine, the fluid is
doing shaft work on the surroundings; on the other hand, a pump
does work on the fluid
 t and W
s W
 t W
p
 p are
W
where W
 work 


 time 
Using this result in the energy equation and deviding by g
results in
magnitudes of power
V12 Wt p2
V22 uˆ2  uˆ1 Q
  z1  1


 z2   2


mg 
2 mg 
2
g
mg
Wp
p1
mechanical part
Note: each term has dimensions of length
Define the following:
hp 
p
W
g
m

p
W
Qg

p
W
Q
t
W
ht 
g
m
hL 
uˆ2  uˆ1 Q

 head loss
g
mg
thermal part
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
28
Head Loss
In a general fluid system a certain amount of mechanical energy
is converted to thermal energy due to viscous action. This effect
results in an increase in the fluid internal energy. Also, some
heat will be generated through energy dissipation and be lost
 ). Therefore the term
(i.e. - Q
from 2nd law
uˆ  uˆ
Q
hL  2 1 
0
g
gm
represents a loss in
mechanical energy due
to viscous stresses
 to system will not make hL = 0 since this
Note that adding Q
also increases u. It can be shown from 2nd law of
thermodynamics that hL > 0.
Drop  over V and understand that V in energy equation refers
to average velocity.
Using the above definitions in the energy equation results in
(steady 1-D incompressible flow)
p1
V12
p2
V22
 1
 z1  h p 
 2
 z2  h t  h L

2g

2g
form of energy equation used for this course!
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
29
Comparison of Energy Equation and Bernoulli Equation
Apply energy equation to a stream tube without any shaft work
Infinitesimal stream tube 
Energy eq :
1=2=1
p1 V12
p 2 V22

 z1 

 z2  hL
 2g
 2g
If hL = 0 (i.e.,  = 0) we get Bernoulli equation and
conservation of mechanical energy along a streamline
Therefore, energy equation for steady 1-D pipe flow can be
interpreted as a modified Bernoulli equation to include viscous
effects (hL) and shaft work (hp or ht)
Summary of the Energy Equation
The energy equation is derived from RTT with
B = E = total energy of the system
 = e = E/M = energy per unit mass
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
= û +
internal
Chapter 5
30
1 2
V +gz
2
KE
PE
dE d
 W


 edV   eV  dA  Q
dt dt CV
CS
heat
add
from 1st Law of
Thermodynamics
work
done
Neglected in text presentation
ed
 W
s W
 p W
v
W
shaft work
done on or
by system
(pump or
turbine)
pressure
work done
on CS
Viscous stress
work on CS
 p   p V  dA   p V  dA
W
CV
CS
s W
 t W
p
W
 W
 t W
 p  d  edV   e  p e V  dA
Q
dt CV
CS
1
e  uˆ  V 2  gz
2
For steady 1-D pipe flow (one inlet and one outlet):
1) Streamlines are straight and parallel
 p/ +gz = constant across CS
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
31
2) T = constant  u = constant across CS
3
3)
1 V
     dA = KE correction factor
A CS  V 
define
3

2
 3
V
V

V
dA


A


m

2
2
2
mechanical energy
p1
V12
p2
V22
 1
 z1  h p 
 2
 z2  h t  h L

2g

2g
p m
g
hp  W
t m
g
ht  W
hL 
uˆ2  uˆ1 Q

 head loss
g
mg
Thermal
energy
Note: each term
has
units of length
V is average velocity
(vector dropped) and
corrected by 
> 0 represents loss in mechanical energy due to viscosity
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
32
Concept of Hydraulic and Energy Grade Lines
p1
V12
p2
V22
 1
 z1  h p 
 2
 z2  h t  h L

2g

2g
p
point-by-point
Define HGL =  z

application is
graphically
p
V2
EGL =  z  
displayed

2g
HGL corresponds to pressure tap measurement + z
EGL corresponds to stagnation tube measurement + z
EGL = HGL if V = 0
EGL1 = EGL2 + hL
for hp = ht = 0
L V2
D 2g
i.e., linear variation in L for D,
V, and f constant
hL = f
f = friction factor
f = f(Re)
p2
h

p2
V22
stagnation tube:

h

2g
pressure tap:
h = height of fluid in
tap/tube
EGL1 + hp = EGL2 + ht + hL
EGL2 = EGL1 + hp  ht  hL
abrupt
change due
to hp or ht
L V2
f
D 2g
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
Helpful hints for drawing HGL and EGL
1. EGL = HGL + V2/2g = HGL for V = 0
L V2
2.&3. h L  f
in pipe means EGL and HGL will slope
D 2g
downward, except for abrupt changes due to ht or hp
p1
 z1 
2
V1

p2
 z2 
2
V2

2g

2g
HGL2 = EGL1 - hL
2
V
hL 
for abrupt expansion
2g
 hL
33
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
34
4. p = 0  HGL = z
L V2
5. for h L  f
= constant  L
D 2g
i.e., linearly increased for
f V2
increasing L with slope
D 2g
EGL/HGL slope downward
6. for change in D  change in V
i.e.
V1A1 = V2A2
D12
D 22
V1
 V2
4
4
2
2
V1D1  V1D 2
change in distance between
 HGL & EGL and slope
change due to change in hL
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
7. If HGL < z then p/ < 0
Chapter 5
35
i.e., cavitation possible
condition for cavitation:
p  p va  2000
N
m2
gage pressure p va ,g  p A  p atm   p atm  100,000
p va ,g

 10m
9810 N/m3
N
m2
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
36
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
37
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
38
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
Energy:
p1
V12
p2
V22
 1
 z1  h p 
 2
 z2  h t  h L

2g

2g
Momentum:
2
2
 Fs  V2 A 2  V1 A1  QV2  V1 
Continuity:
A1V1 = A2V2 = Q = constant
one inlet and
one outlet
 = constant
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
39
Abrupt Expansion
Consider the flow from a small pipe to a larger pipe. Would like
to know hL = hL(V1,V2). Analytic solution to exact problem is
extremely difficult due
to the occurrence of
flow separations and
turbulence. However, if
the assumption is made
that the pressure in the
separation region
remains approximately
constant and at the
value at the point of
separation, i.e, p1, an approximate solution for hL is possible:
Apply Energy Eq from 1-2 (1 = 2 = 1)
p1
V12 p 2
V22
 z1 

 z2 
 hL

2g

2g
Momentum eq. For CV shown (shear stress neglected)
 Fs  p1A 2  p 2 A 2  W sin    u V  A
= V1 (V1A1 )  V2 (V2 A 2 )
= V22 A 2  V12 A1
z
L
W sin 
next divide momentum equation by A2
A 2 L
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
÷ A2
Chapter 5
40
p1 p 2
V22 V12 A1 V12 A1  A1 

  z1  z 2  


 1


g
g A2
g A2  A2 
from energy equation

V22 V12
V22 V12 A1

 hL 

2g 2g
g
g A2
V22 V12  2A1 
1 

hL 

2g 2g 
A 2 
hL 
1  2
2
2 A1 
V

V

2
V
2
1
1
2g 
A 2 
2V1V2
hL 
1
V2  V1 2
2g
If V2 << V1,
1 2
hL =
V1
2g
continutity eq.
V1A1 = V2A2
A1 V2

A 2 V1
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
41
Forces on Transitions
Example 7-6
Q = .707 m3/s
V22
head loss = .1
2g
(empirical equation)
Fluid = water
p1 = 250 kPa
D1 = 30 cm
D2 = 20 cm
Fx = ?
First apply momentum theorem
 Fx   u V  A
Fx + p1A1  p2A2 = V1(V1A1) + V2(V2A2)
Fx = Q(V2  V1)  p1A1 + p2A2
force required to hold transition in place
57:020 Mechanics of Fluids and Transport Processes
Professor Fred Stern Fall 2012
Chapter 5
42
The only unknown in this equation is p2, which can be obtained
from the energy equation.
p1 V12 p 2 V22



 hL
 2g
 2g
note: z1 = z2 and  = 1
 V22 V12

p 2  p1   

 hL 
 2g 2 g

drop in pressure

 V22 V12

 Fx  QV2  V1   A 2 p1  

 h L   p1A1
 2g 2g


p2
In this equation,
continuity
V1 = Q/A1 = 10 m/s
V2 = Q/A2 = 22.5 m/s
V22
h L  .1
 2.58m
2g
Fx = 8.15 kN
(note: if p2 = 0 same as nozzle)
A1V1 = A2V2
A
V2  1 V1
A2
i.e. V2 > V1
is negative x direction to hold
transition in place