Transcript CompressibleFlow_Par..

CP502
Advanced Fluid Mechanics
Compressible Flow
Part 01_Set 01:
Steady, quasi one-dimensional, isothermal,
compressible flow of an ideal gas in a
constant area duct with wall friction
Incompressible flow assumption is not valid
if Mach number > 0.3
What is a Mach number?
Definition of Mach number (M):
M≡
For an ideal gas,
Speed of the flow (u)
Speed of sound (c) in the fluid
at the flow temperature
c  RT
specific heat ratio
specific gas constant (in J/kg.K)
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absolute temperature of the flow at the point concerned (in K)
For an ideal gas,
M=
u
c
=
u
RT
Unit of u = m/s
Unit of c = [(J/kg.K)(K)]0.5
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= [J/kg]0.5
= (N.m/kg)0.5
= [m2/s2]0.5
= m/s
= [kg.(m/s2).m/kg]0.5
constant area duct
Diameter (D)
quasi one-dimensional flow
speed (u)
A  D 2 / 4 is a constant
u varies only in x-direction
x
compressible flow
  Au is a constant
Mass flow rate m
steady flow
isothermal flow
ideal gas
wall friction
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Density (ρ) is NOT a constant
Temperature (T) is a constant
Obeys the Ideal Gas equation
 w  fu2 / 2
is the shear stress acting on the wall
where f is the average Fanning friction factor
Friction factor:
f  16/ Re
For laminar flow in circular pipes:
where Re is the Reynolds number of the flow defined as follows:
uD
m D
Re 


A
4m D

D 2 
For lamina flow in a square channel:
For the turbulent flow regime:
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4m

D
f  14.227/ Re
 3.7 D 
 4.0 log10 

  
f
1
Quasi one-dimensional flow is closer to turbulent
velocity profile than to laminar velocity profile.
Ideal Gas equation of state:
pV  mRT
temperature
pressure
specific gas constant
(not universal gas constant)
volume
mass
Ideal Gas equation of state can be rearranged to give
m
p  RT
V
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p  RT
K
Pa = N/m2
kg/m3
J/(kg.K)
Problem 1 from Problem Set 1 in Compressible Fluid Flow:
Starting from the mass and momentum balances, show that the
differential equation describing the quasi one-dimensional,
compressible, isothermal, steady flow of an ideal gas through a
constant area pipe of diameter D and average Fanning friction
factor f shall be written as follows:
4f
2
2
dx  2 dp  du  0
D
u
u
(1.1)
where p, ρ and u are the respective pressure, density and velocity
at distance x from the entrance of the pipe.
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w
x
p
p+dp
u
u+du
D
dx
Write the momentum balance over the differential volume chosen.
 u  ( p  dp) A  m
 (u  du)   wdAw
pA  m
(1)
steady mass flow rate
cross-sectional area
shear stress acting on the wall
dAw  Ddx is the wetted area on which shear is acting
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w
p
p+dp
u
u+du
dx
x
Equation (1) can be reduced to
  Au
Substituting m
Since A  D 2 / 4 ,
 du   wdAw  0
Adp  m
A(dp  udu)   w dAw  0
 w  fu / 2
2
dp  udu 
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D
and dAw  Ddx , we get
u 4 f
2
2
D
dx  0
4f
2
2
dx  2 dp  du  0
D
u
u
(1.1)
Problem 2 from Problem Set 1 in Compressible Fluid Flow:
Show that the differential equation of Problem (1) can be converted
into
4f
2
2
dx  
pdp  dp
2
D
RT (m / A)
p
(1.2)
which in turn can be integrated to yield the following design equation:
4f L
p2

D
RT (m / A) 2

 pL2 
pL2 
1  2   ln 2 
p 

p 
(1.3)
where p is the pressure at the entrance of the pipe, pL is the pressure at
length L from the entrance of the pipe, R is the gas constant, T is the
 is the mass flow rate of the gas flowing
temperature of the gas, m
through the pipe, and A is the cross-sectional area of the pipe.
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The differential equation of problem (1) is
4f
2
2
dx  2 dp  du  0
D
u
u
(1.1)
in which the variables ρ and u must be replaced by the variable p.
Let us use the mass flow rate equation m
  Au and the ideal gas
equation p  RT to obtain the following:
p

RT
m
m RT

and u 
A
Ap
m RT
and therefore pu 
A
It is a constant for steady, isothermal flow in a
constant area duct
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d ( pu)  pdu  udp  0
du
dp

u
p
p
m
m RT
, u

Using  
RT
A
Ap
and
du
dp

u
p
4
f
2
2
in
dx  2 dp  du  0
D
u
u
(1.1)
we get
4f
2
2
dx  
pdp  dp
2
D
RT (m / A)
p
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(1.2)
p
pL
L
Integrating (1.2) from 0 to L, we get
4f
D
L
2
0 dx   RT (m / A) 2
pL
pL
p
p
 pdp  
2
dp
p
which becomes
4f L
p2

D
RT (m / A) 2
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
 pL2 
pL2 
1  2   ln 2 
p 

p 
(1.3)
Problem 3 from Problem Set 1 in Compressible Fluid Flow:
Show that the design equation of Problem (2) is equivalent to
4f L
1

D
 M2

M2 
M2 
1  2   ln 2 
ML 

 ML 
(1.4)
where M is the Mach number at the entry and ML is the Mach number
at length L from the entry.
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Design equation of Problem (2) is
4f L
p2

D
RT (m / A) 2

 pL2 
pL2 
1  2   ln 2 
p 

p 
(1.3)
which should be shown to be equivalent to
4f L
1

D
 M2

M2 
M2 
1  2   ln 2 
ML 

 ML 
(1.4)
where p and M are the pressure and Mach number at the entry and pL
and ML are the pressure and Mach number at length L from the entry.
We need to relate p to M!
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We need to relate p to M!
m
m RT 1 m RT
1
m
p  RT 
RT 


Au
A u
A M RT AM
RT

which gives
m RT
pM 
A 
= constant for steady, isothermal flow in a
constant area duct
Substituting the above in (1.3), we get
4f L
1

D
 M2
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
M2 
M2 
1  2   ln 2 
ML 

 ML 
(1.4)
Summary
Design equations for steady, quasi one-dimensional,
isothermal,compressible flow of an ideal gas in a constant area
duct with wall friction
4f
2
2
dx  2 dp  du  0
D
u
u
4f
2
2
dx  
pdp  dp
2
D
RT (m / A)
p
4f L
p2

D
RT (m / A) 2
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02 Feb 2012
4f L
1

D
 M2
(1.1)
(1.2)

 pL2 
pL2 
1  2   ln 2 
p 

p 

M2 
M2 
1  2   ln 2 
ML 

 ML 
(1.3)
(1.4)