AE 341 Fluid Mechanics
2009 Fall
Lecture Note 3. Integral Relations for a Control Volume
Prof. K. D. Kihm
[http://minsfet.utk.edu]
Ch. 3 Integral Relations for a Control Volume
■ Reynolds Transport theorem (RTT)
■ Conservation of Mass
■ Linear Momentum Equation
Homework Assignment #3
Due:
at 12:19 pm on Wednesday of September 30, 2009
E-o-C Problems:
3.28, 3.33, 3.39, 3.49, 3.55, 3.68
■ Linear Momentum Equation under Acceleration
■ Energy Equation
■ Bernoulli Equation
Homework Assignment #4
Due: 12:19 pm on Wednesday of October 7, 2009
E-o-C Problems: 3.70, 3.88, 3.135, 3.160, 3.165, 3.167
Midterm Exam 2 [20%]
Scheduled on Wednesday of October 14, 2009
Coverage: Ch. 3
1
AE 341 Fluid Mechanics
2009 Fall
Lecture Note 3. Integral Relations for a Control Volume
Prof. K. D. Kihm
[http://minsfet.utk.edu]
Ch. 3 INTEGRAL RELATIONS FOR A CONTROL VOLUME
3.1/3.2 The Reynolds Transport Theorem (RTT)
■ System versus Control Volume
2
AE 341 Fluid Mechanics
2009 Fall
Lecture Note 3. Integral Relations for a Control Volume
Prof. K. D. Kihm
[http://minsfet.utk.edu]
■ Rate of flow crossing a boundary
3
AE 341 Fluid Mechanics
2009 Fall
Lecture Note 3. Integral Relations for a Control Volume
Prof. K. D. Kihm
[http://minsfet.utk.edu]
■ RTT
 This theorem converts the Basic Laws discovered for a closed mass SYSTEM to more useful
Equations formulating for an open CONTROL VOLUME.
 The standard derivation procedure shown on pp. 141-145 gives the Reynolds Transport
Theorem as:
d
Bsyst   d
dt
dt
The rate of change of
B of the closed system

CV

dV    Vr  n dA
 
CS
The rate of change
of B within the
control volume
The net rate of flux of
B out through the
control surface
 Extensive and Intensive Properties:
B: Extensive property of a system
 : Intensive property, i.e., the corresponding extensive property per unit mass
Mass:
Bm
Momentum:


B  P  mV
Angular Momentum:

 
B  H  mr  V
  B/m  m/m 1


  mV / m  V


  r V
Energy:
B  E  me
 e
Entropy:
B  S  ms
 s
4
AE 341 Fluid Mechanics
2009 Fall
Lecture Note 3. Integral Relations for a Control Volume
Prof. K. D. Kihm
[http://minsfet.utk.edu]
3.3 Conservation of Mass
RTT:
d
Bsyst   d
dt
dt

CV

dV    Vr  n dA
 
CS
With B  M and   1 ,

d
msyst   0  d ( CV dV )  CS  (Vr  n )dA
dt
dt
Rate of increase of
mass in CV
Net outflux of
mass through CS
0  (
For a fixed control volume:
CV
 

dV )    (Vr  n )dA
CS
t
(3.21)
For a control volume having uniform inlets and outlets:
(
CV

dV )    i AiVi out    i AiVi in  0
t
i
i
* For incompressible fluids with a fixed CV:
  AV 
i i out
   AiVi in  0
i
* For steady flow:
i
  AV 
i
i i out
i
   i AiVi in  0
i
*For one inlet and one outlet flows:
(
CV

dV )  AV out  AV in  0
t
*For steady, incompressible, one inlet and one outlet flows:
AV out  AV in
5
(3.22)
AE 341 Fluid Mechanics
2009 Fall
Lecture Note 3. Integral Relations for a Control Volume
Prof. K. D. Kihm
[http://minsfet.utk.edu]
Example 3.5
6
AE 341 Fluid Mechanics
2009 Fall
Lecture Note 3. Integral Relations for a Control Volume
Prof. K. D. Kihm
[http://minsfet.utk.edu]
3.4 Linear Momentum Equation
Inertial – Zero acceleration, i.e, either stationary or moving at a constant velocity.
 

d
mV
dt
Newton’s second law for a system:
syst
d
Bsyst   d
dt
dt
Combining with the RTT,

 F


dV    Vr  n dA
CV
 
CS


with , B  mV and   V .
The integral form of momentum equation for an inertial CV is
 d
F
  dt

CV


  
VdV   V (Vr  n )dA
or
CS


t

 
udV   uV  dA
y


t

 
vdV   vV  dA
z


t

 
wdV   wV  dA
F
x
F
F
CV
CV
CV
CS
CS
CS
Alternatively,
 F  dt 

d
CV




VdV   m iVi

out


  m iVi
7

in
(Vector equation)
AE 341 Fluid Mechanics
2009 Fall
Lecture Note 3. Integral Relations for a Control Volume
Prof. K. D. Kihm
[http://minsfet.utk.edu]
Example 3.10
8
AE 341 Fluid Mechanics
2009 Fall
Lecture Note 3. Integral Relations for a Control Volume
Prof. K. D. Kihm
[http://minsfet.utk.edu]
Example 3.8
9
AE 341 Fluid Mechanics
2009 Fall
Lecture Note 3. Integral Relations for a Control Volume
Prof. K. D. Kihm
[http://minsfet.utk.edu]
10
AE 341 Fluid Mechanics
2009 Fall
Lecture Note 3. Integral Relations for a Control Volume
Prof. K. D. Kihm
[http://minsfet.utk.edu]
Example 3.9
11
AE 341 Fluid Mechanics
2009 Fall
Lecture Note 3. Integral Relations for a Control Volume
Prof. K. D. Kihm
[http://minsfet.utk.edu]
■ Momentum Flux Correction Factor for Nonuniform Velocity Profiles

Fully-developed axi-symmetric laminar pipe flow:
  r 2 
u  u max 1     ;
  R  
The volume flow rate Q  R 2 
● Momentum Equation:
 F  dt 

d
CV

u max
u
 A  max  AVav
2
2

  
VdV   V (Vr  n )dA
CS
● For 1-D pipe flow, the second momentum flux term is reduced to:
  u 2 dA  AVav2
CS
● The correction factor :
 = 1.33 for fully-developed laminar flows
 ~ 1.0 for fully-developed turbulent flows (see Textbook p. 163)
12
AE 341 Fluid Mechanics
2009 Fall
Lecture Note 3. Integral Relations for a Control Volume
Prof. K. D. Kihm
[http://minsfet.utk.edu]
■ Linearly accelerating CV
The force term must be modified to account for the accelerating forces as:


  

d
a ref dm  (  VdV )   V (Vr  n )dA
CV
CS
dt CV
F  
Example 3.12
13
AE 341 Fluid Mechanics
2009 Fall
Lecture Note 3. Integral Relations for a Control Volume
Prof. K. D. Kihm
[http://minsfet.utk.edu]
3.6 The Energy Equation in the “Head” Expression
For steady flow,
Mass continuity gives,
m 1  m 2  m
Energy conservations states,
1
1




Q  W s  W v  m 1  h1  V12  gz1   m 2  h2  V22  gz 2 
2
2




Combining the two,
1
1
( h1  V12  gz1 )  q  ws  wv  h2  V22  gz 2
2
2
Recalling h  u 
p

,
u1 V12
p2 u2 V22
 
 z1  hq  hs  hv 
 
 z2

g 2g

g 2g
p1
This can be expressed as,
p1


V12
p V2
 z1      2  2  z2
2g
 2g
: added heads from 1 to 2 - hpump
: consumed heads from 1 to 2 - hturbine, hfriction
Therefore, the resulting “head form” of energy equation is given as,
(
p1


V12
p V2
 z1 ) in  h pump  hturbinr  h friction  ( 2  2  z2 ) out
2g
 2g
14
AE 341 Fluid Mechanics
2009 Fall
Lecture Note 3. Integral Relations for a Control Volume
Prof. K. D. Kihm
[http://minsfet.utk.edu]
Example 3.19
15
AE 341 Fluid Mechanics
2009 Fall
Lecture Note 3. Integral Relations for a Control Volume
Prof. K. D. Kihm
[http://minsfet.utk.edu]
3.7 Bernoulli Equation
The head-form energy equation under:
*Steady flow
*Incompressible flow
*Frictionless flow
*Non-tangled streamlines
V12
p2 V22
( 
 z )  h pump  hturbinr  ( 
z )
 2g 1
 2g 2
p1
Or in the absence of pump or turbine,
V12
p2 V22
( 
z )( 
 z )  Const
 2g 1
 2g 2
p1
[Valid/invalid regions for the Bernoulli equation]
16
AE 341 Fluid Mechanics
2009 Fall
Lecture Note 3. Integral Relations for a Control Volume
Prof. K. D. Kihm
[http://minsfet.utk.edu]
■ Kinetic Energy Correction Factor for Nonuniform Velocity Profiles
● The incompressible steady flow energy equation (Bernoulli Equation):
(
p1


V12
p
V2
 z1 ) in  h pump  hturbinr  h friction  ( 2   2  z2 )out
2g

2g
● The correction factor :
 = 2.0 for fully-developed laminar flows
 ~ 1.0 for fully-developed turbulent flows (see Textbook p. 180)
17
AE 341 Fluid Mechanics
2009 Fall
Lecture Note 3. Integral Relations for a Control Volume
Prof. K. D. Kihm
[http://minsfet.utk.edu]
Example 3.22
18
AE 341 Fluid Mechanics
2009 Fall
Lecture Note 3. Integral Relations for a Control Volume
Prof. K. D. Kihm
[http://minsfet.utk.edu]
Example 3.24
19
AE 341 Fluid Mechanics
2009 Fall
Lecture Note 3. Integral Relations for a Control Volume
Prof. K. D. Kihm
[http://minsfet.utk.edu]
SUMMARY
■ Mass Continuity
 
d
0  (  dV )    (Vr  n )dA
CS
dt CV
■ Momentum Equation – Inertial CV
 d
F
  dt

CV




VdV   m iVi





m
V
 ii
out

in
■ Momentum Equation – Linearly Accelerating CV


d
F

a
dm

 CV ref
dt

CV




VdV   m iVi

out



m iVi

in
■ Energy Equation in Head Form
V12
p2 V22
( 
 z1 ) in  h pump  hturbinr  h friction  ( 
 z2 ) out
 2g
 2g
p1
20