CP502
Advanced Fluid Mechanics
Compressible Flow
Lectures 1 & 2
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)
R. Shanthini
08 Dec 2010
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
R. Shanthini
08 Dec 2010
= [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
R. Shanthini
08 Dec 2010
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:
R. Shanthini
08 Dec 2010
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
R. Shanthini
08 Dec 2010
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.
R. Shanthini
08 Dec 2010
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
R. Shanthini
08 Dec 2010
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 
R. Shanthini
08 Dec 2010
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.
R. Shanthini
08 Dec 2010
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
R. Shanthini
08 Dec 2010
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
R. Shanthini
08 Dec 2010
(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
R. Shanthini
08 Dec 2010

 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.
R. Shanthini
08 Dec 2010
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!
R. Shanthini
08 Dec 2010
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
R. Shanthini
08 Dec 2010

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
R. Shanthini
08 Dec 2010
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)
Problem 4 from Problem Set 1 in Compressible Fluid Flow:
Nitrogen (γ = 1.4; molecular mass = 28) is to be fed through a 15
mm-id commercial steel pipe 11.5 m long to a synthetic ammonia
plant. Calculate the downstream pressure in the line for a flow rate of
1.5 mol/s, an upstream pressure of 600 kPa, and a temperature of
27oC throughout. The average Fanning friction factor may be taken as
0.0066.
γ = 1.4; molecular mass = 28;
p = 600 kPa
m = 1.5 mol/s;
D = 15 mm
T = 300 K
L = 11.5 m
R. Shanthini
08 Dec 2010
f = 0.0066
pL = ?
γ = 1.4; molecular mass = 28;
p = 600 kPa
m = 1.5 mol/s;
D = 15 mm
f = 0.0066
T = 300 K
pL = ?
L = 11.5 m
Design equation:
4f L
p2

D
RT (m / A) 2

 pL2 
pL2 
1  2   ln 2 
p 

p 
(1.3)
f = 0.0066;
L = 11.5 m;
D = 15 mm = 0.015 m;
R. Shanthini
08 Dec 2010
4f L
= 20.240
D
unit?
γ = 1.4; molecular mass = 28;
p = 600 kPa
m = 1.5 mol/s;
D = 15 mm
f = 0.0066
T = 300 K
pL = ?
L = 11.5 m
Design equation:
4f L
p2

D
RT (m / A) 2
p = 600 kPa = 600,000 Pa;

 pL2 
pL2 
1  2   ln 2 
p 

p 
(1.3)
T = 300 K;
R = 8.314 kJ/kmol.K = 8.314/28 kJ/kg.K = 8314/28 J/kg.K;
m = 1.5 mol/s = 1.5 x 28 g/s = 1.5 x 28/1000 kg/s;
A = πD2/4 = π(15 mm)2/4 = π(0.015 m)2/4;
R. Shanthini
08 Dec 2010
p2
= 71.544
2
RT (m / A)
unit?
m = 1.5 mol/s;
γ = 1.4; molecular mass = 28;
p = 600 kPa
D = 15 mm
f = 0.0066
T = 300 K
pL = ?
L = 11.5 m
Design equation:
4f L
p2

D
RT (m / A) 2

 pL2 
pL2 
1  2   ln 2 
p 

p 
(1.3)

 p L2 
p L2 
20.240 = 71.544 1  2   ln 2 
p 

p 
p = 600 kPa = 600,000 Pa
R. Shanthini
08 Dec 2010
pL = ?
Solve the nonlinear equation
above to determine pL

 p L2 
p L2 
20.240 = 71.544 1  2   ln 2 
p 

p 
p = 600 kPa = 600,000 Pa
Determine the approximate solution by ignoring the ln-term:
pL = p (1-20.240/71.544)0.5 = 508.1 kPa
Check the value of the ln-term using pL = 508.1 kPa:
ln[(pL /p)2] = ln[(508.1 /600)2] = -0.3325
This value is small when compared to
20.240. And therefore pL = 508.1 kPa
is a good first approximation.
R. Shanthini
08 Dec 2010
Now, solve the nonlinear equation for pL values close to 508.1 kPa:

 p L2 
p L2 
20.240 = 71.544 1  2   ln 2 
p 

p 
p = 600 kPa = 600,000 Pa
pL kPa
510
R. Shanthini
08 Dec 2010
LHS of the above RHS of the above
equation
equation
20.240
19.528
509
20.240
19.727
508.1
20.240
19.905
507
20.240
20.123
506.5
20.240
20.222
506
20.240
20.320
Problem 4 continued:
Rework the problem in terms of Mach number and determine ML.
γ = 1.4; molecular mass = 28;
p = 600 kPa
m = 1.5 mol/s;
D = 15 mm
f = 0.0066
T = 300 K
ML = ?
L = 11.5 m
Design equation:
4f L
1

D
 M2

M2 
M2 
1  2   ln 2 
ML 

 ML 
(1.4)
4f L
= 20.240 (already calculated in Problem 4)
D
M=?
R. Shanthini
08 Dec 2010
γ = 1.4; molecular mass = 28;
m = 1.5 mol/s;
D = 15 mm
p = 600 kPa
f = 0.0066
T = 300 K
ML = ?
L = 11.5 m
m
u
u
M= c =
= A
RT
M
4m
D 2 p

1 = m
A
RT
R. Shanthini
08 Dec 2010
 RT
1  m
RT
Ap 
RT

4 (1.5x 28/1000 kg/s)
=
π (15/1000 m)2 (600,000 Pa)
= 0.1
RT
p
(
(8314/28)(300) J/kg
1.4
0.5
)
γ = 1.4; molecular mass = 28;
p = 600 kPa
m = 1.5 mol/s;
D = 15 mm
f = 0.0066
ML = ?
T = 300 K
L = 11.5 m
Design equation:
4f L
1

D
 M2

M2 
M2 
1  2   ln 2 
ML 

 ML 
(1.4)
 (0.1) 2 
 (0.1) 2 
1
1 
  ln

20.240 
2 
2 
2 
(1.4)(0.1) 
ML 
 ML 
ML = ?
R. Shanthini
08 Dec 2010
Solve the nonlinear equation
above to determine ML
 (0.1) 2 
 (0.1) 2 
1
1 
  ln

20.240 
2 
2 
2 
(1.4)(0.1) 
ML 
 ML 
Determine the approximate solution by ignoring the ln-term:
ML = 0.1 / (1-20.240 x 1.4 x 0.12)0.5
= 0.118
Check the value of the ln-term using ML = 0.118:
ln[(0.1/ML)2] = ln[(0.1 /0.118)2] = -0.3310
This value is small when compared to
20.240. And therefore ML = 0.118 is a
good first approximation.
R. Shanthini
08 Dec 2010
Now, solve the nonlinear equation for ML values close to 0.118:
 (0.1) 2 
 (0.1) 2 
1
20.240 
1 
  ln

2 
2 
2 
(1.4)(0.1) 
ML 
 ML 
pL kPa
0.116
R. Shanthini
08 Dec 2010
LHS of the above RHS of the above
equation
equation
20.240
18.049
0.117
20.240
0.118
20.240
19.798
0.1185
20.240
20.222
0.119
20.240
20.64
Problem 5 from Problem Set 1 in Compressible Fluid Flow:
Explain why the design equations of Problems (1), (2) and (3)
are valid only for fully turbulent flow and not for laminar flow.
R. Shanthini
08 Dec 2010
Problem 6 from Problem Set 1 in Compressible Fluid Flow:
Starting from the differential equation of Problem (2), or otherwise,
prove that p, the pressure, in a quasi one-dimensional, compressible,
isothermal, steady flow of an ideal gas in a pipe with wall friction
should always satisfies the following condition:
 / A) RT
p  (m
(1.5)
in flows where p decreases along the flow direction, and
 / A) RT
p  (m
(1.6)
in flows where p increases along the flow direction.
R. Shanthini
08 Dec 2010
Differential equation of Problem 2:
4f
2
2
dx  
pdp  dp
2
D
RT (m / A)
p
(1.2)
can be rearranged to give
dp

dx 
4f /D
(2 f / D ) pRT(m / A) 2

2
2
 / A) 2  p 2
RT
(
m
p
2
RT (m / A)
p
In flows where p decreases along the flow direction
dp
0
dx
R. Shanthini
08 Dec 2010
2
2

RT (m / A)  p  0
 / A) RT
p  (m
(1.5)
Differential equation of Problem 2:
4f
2
2
dx  
pdp  dp
2
D
RT (m / A)
p
(1.2)
can be rearranged to give
dp

dx 
4f /D
(2 f / D ) pRT(m / A) 2

2
2
 / A) 2  p 2
RT
(
m
p
2
RT (m / A)
p
In flows where p increases along the flow direction
dp
0
dx
R. Shanthini
08 Dec 2010
 / A) 2  p 2  0
RT (m
 / A) RT
p  (m
(1.6)
Problem 7 from Problem Set 1 in Compressible Fluid Flow:
Air enters a horizontal constant-area pipe at 40 atm and 97oC with a
velocity of 500 m/s. What is the limiting pressure for isothermal flow?
It can be observed that in the above case pressure increases in the
direction of flow. Is such flow physically realizable?
If yes, explain how the flow is driven along the pipe.
40 atm
97oC
500 m/s
Air: γ = 1.4; molecular mass = 29;
p*=?
L
R. Shanthini
08 Dec 2010
40 atm
97oC
500 m/s
Air: γ = 1.4; molecular mass = 29;
p*=?
L
dp
0
dx
 / A) RT
p  (m
dp
0
dx
 / A) RT
p  (m
Limiting pressure:
R. Shanthini
08 Dec 2010
 / A) RT
p*  (m
Air: γ = 1.4; molecular mass = 29;
40 atm
97oC
500 m/s
p*=?
L
 / A) RT
p*  (m
 / A)  ( Au) entrance / A  ( u) entrance
(m

puentrance
p* 
RT
=
R. Shanthini
08 Dec 2010
RT

puentrance
 p 

u

RT
 RT  entrance

puentrance

RT
(40 atm) (500 m/s)
[(8314/29)(273+97) J/kg]0.5
= 61.4 atm
40 atm
97oC
500 m/s
Air: γ = 1.4; molecular mass = 29;
p*=61.4 atm
L
Pressure increases in the direction of flow. Is such flow physically
realizable?
YES
If yes, explain how the flow is driven along the pipe.
Use the momentum balance over a differential element of the flow
(given below) to explain.
 du   wdAw  0
Adp  m
R. Shanthini
08 Dec 2010