CP502
Advanced Fluid Mechanics
Compressible Flow
Lectures 5 and 6
Steady, quasi one-dimensional, isentropic
compressible flow of an ideal gas
in a variable area duct
(continued)
Problem 6 from Problem Set #2 in Compressible Fluid Flow:
Show that the steady, one-dimensional, isentropic, compressible flow of an
ideal gas with constant specific heats can be described by the following

equations:

p0  T0   1   0 
     
p T 
  
T0
 1 2
 1
M
T
2
(2.5)
(2.6)

p0    1 2   1
 1 
M 
p 
2

0    1 2 
 1 
M 
 
2

(2.7)
1
 1
(2.8)
where p0, T0 and ρ0 are the stagnation (where fluid is assumed to be at rest)
properties, p, T and ρ are the properties at Mach number M and γ is the specific
heat
ratio assumed to be a constant.
R. Shanthini
22 Dec 2010
T0
p0
u0=0
Stagnation
properties
T
p
u
M
Ideal gas satisfies p  RT
Isentropic flow of an ideal gas satisfies
p  
Using the above two equations, we can easily prove
R. Shanthini
22 Dec 2010
p0  T0 
 
p T 

 1
 
  0 
  

(2.5)
T0
p0
u0=0
Stagnation
properties
T
p
u
M
Start with the energy balance for steady, adiabatic, inviscid, quasi one-dimensional,
compressible flow:
(2.2)
dh + udu = 0
Using dh = cp dT, which is applicable for ideal gas, in the above, we get
cp dT + udu = 0
Integrating the above between the two cross-sections, we get
cp (T – T0) + (u2 – 0)/2 = 0
R. Shanthini
22 Dec 2010
Using the definition of M, we get
cp (T – T0) + M2 γRT/2 = 0
Since cp = γR/(γ -1) for an ideal gas, the above can be written as
γR (T – T0) /(γ -1) + M2 γRT/2 = 0
which can be rearranged to give the following:
T0
 1 2
 1
M
T
2
(2.6)
The above equation relates the stagnation temperature
(at near zero velocity) to a temperature at Mach number
M for steady, isentropic, one-dimensional, compressible
flow of an ideal gas in a variable area duct.
R. Shanthini
22 Dec 2010
Using (2.6) in (2.5), we can easily get the following equations relating the
stagnation properties (at near zero velocity) to properties at Mach number M for
steady, isentropic, one-dimensional, compressible flow of an ideal gas in a variable
area duct:

p0    1 2   1
 1 
M 
p 
2

0    1 2 
 1 
M 
 
2

R. Shanthini
22 Dec 2010
(2.7)
1
 1
(2.8)
Problem 7 from Problem Set #2 in Compressible Fluid Flow:
Show that the mass flow rate in a steady, one-dimensional, isentropic,
compressible flow of an ideal gas with constant specific heats is given
by the following equations:
m  AMp
 AMp0
R. Shanthini
22 Dec 2010

(2.9)
RT



 
1


RT0  1    1 M 2 


2


 1
2 (  1)
(2.10)
Problem 8 from Problem Set #2 in Compressible Fluid Flow:
A large air reservoir contains air at a temperature of 400 K and a pressure of
600 kPa. The air reservoir is connected to a second chamber through a
converging duct whose exit area is 100 mm2. The pressure inside the second
chamber can be regulated independently. Assuming steady, isentropic flow in
the duct, calculate the exit Mach number, exit temperature, and mass flow rate
through the duct when the pressure in the second chamber is (i) 600 kPa, (ii)
500 kPa, (iii) 400 kPa, (iv) 300 kPa and (v) 200 kPa.
Assumptions:
Steady, isentropic flow
Ae = 100 mm2
Air
T0 = 400 K
p0 = 600 kPa
R. Shanthini
22 Dec 2010
pb kPa is given
Determine the following at the
exit of the converging duct:
Mach Number Me = ?
Temperature Te = ?
Mass flow rate = ?
pb is known as the back pressure
(i) For pb = 600 kPa, there will be no flow since p0 = 600 kPa as well
(ii) For pb = 500 kPa, assume pe is the same as pb.
600 
0.4 2 
 1 
Me 
Using (2.7), we get
500 
2

p 
Using (2.5), we get Te  T0  e 
 p0 
 1

1.4
0.4
(6 / 5) 0.4 / 1.4  1
Me 
= 0.517
0.2
 500
 (400 K)

600


0.4
1.4
= 379.7 K
Using (2.9), we get
m  AMp

RT
 Ae M e pe

RTe
(
= (100 x10-6 m2) (0.517) (500,000 Pa)
R. Shanthini
22 Dec 2010
1.4
(8314/29)(379.7) J/kg
0.5
)
= 0.0927 kg/s
Results summarized:
Back pressure, Exit pressure,
pb (in kPa)
pe (in kPa)
Exit Mach Exit temperature,
Mass flow
number, Me
Te (in K)
rate (in kg/s)
600
600
0
400
0
500
500
0.517
379.7
0.0927
400
400
0.784
356.2
0.1161
300
300
1.046
328.1
0.1211
200
200
1.358
292.2
0.1110
Ae = 100 mm2
Air
T0 = 400 K
pR.0Shanthini
= 600 kPa
22 Dec 2010
pb kPa
(given)
Is there a
problem with
the above
results?
Results summarized:
Back pressure, Exit pressure,
pb (in kPa)
pe (in kPa)
Exit Mach Exit temperature,
Mass flow
number, Me
Te (in K)
rate (in kg/s)
600
600
0
400
0
500
500
0.517
379.7
0.0927
400
400
0.784
356.2
0.1161
300
300
1.046 > 1
328.1
0.1211
200
200
1.358 > 1
292.2
0.1110
Ae = 100 mm2
Air
T0 = 400 K
pR.0Shanthini
= 600 kPa
22 Dec 2010
pb kPa
(given)
Yes, there is. Supersonic
Mach numbers have
been reached from
stagnation condition in a
converging duct?
It can not happen!
Maximum Mach number at the exit of a converging duct must be Me = 1.
Corresponding pe (denoted by p*) could be calculated using (2.7) as follows:
600
0.4 2 
p* = 600 1 
Me 
500 
2

1.4
0.4
= 317 kPa
 p *

Using (2.5), we get T *  T0 
 p0 
 1

 317
 (400 K)

600


0.4
1.4
Using (2.9), we get the mass flow rate as 0.1213 kg/s.
R. Shanthini
22 Dec 2010
= 333.3 K
Results summarized:
Back pressure, Exit pressure,
pb (in kPa)
pe (in kPa)
Exit Mach Exit temperature,
Mass flow
number, Me
Te (in K)
rate (in kg/s)
600
600
0
400
0
500
500
0.517
379.7
0.0927
400
400
0.784
356.2
0.1161
300
317
1
333.3
0.1213
200
317
1
333.3
0.1213
Ae = 100 mm2
Air
T0 = 400 K
pR.0Shanthini
= 600 kPa
22 Dec 2010
pb kPa
(given)
Flow chocks at the
throat of the converging
duct at an exit pressure
of 317 kPa at a
maximum flow rate of
0.1213 kg/s?
Ae = 100 mm2
Air
T0 = 400 K
p0 = 600 kPa
pe kPa
(given)
Pressure at the throat cannot be less than the limiting pressure (317 kPa in this
case) even if we keep a lower pressure in the second chamber (known as the back
pressure).
Mass flow rate cannot be increased above the maximum mass flow rate ( 0.1213
kg/s in this case) even if we increase the driving force by decreasing the back
pressure in the second chamber.
R. Shanthini
22 Dec 2010
Problem 9 from Problem Set #2 in Compressible Fluid Flow:
Air at 900 kPa and 400 K enters a converging nozzle with a negligible
velocity. The throat area of the nozzle is 10 cm2. Assuming isentropic flow,
calculate and plot the exit pressure, the exit Mach number, the exit velocity,
and the mass flow rate versus the back pressure pb for 900 ≤ pb ≤ 100 kPa.
R. Shanthini
22 Dec 2010
Results:
900
Exit pressure (in kPa)
800
Limiting pressure = 475.45 kPa
700
600
500
400
300
200
100
100
R. Shanthini
22 Dec 2010
200
300
400
500
600
Back pressure (in kPa)
700
800
900
Results (continued):
1.1
1
Exit Mach number
0.9
0.8
0.7
0.6
0.5
Sonic condition M = 1
0.4
0.3
0.2
0.1
0
100
R. Shanthini
22 Dec 2010
200
300
400
500
600
Back pressure (in kPa)
700
800
900
Results (continued):
400
Exit velocity (in m/s)
350
300
250
200
Maximum velocity 365.77 m/s
150
100
50
0
100
R. Shanthini
22 Dec 2010
200
300
400
500
600
Back pressure (in kPa)
700
800
900
Results (continued):
20
Mass flow rate (in kg/s)
18
16
14
12
10
Maximum mass flow rate 18.1981 kg/s
8
6
4
2
0
100
R. Shanthini
22 Dec 2010
200
300
400
500
600
Back pressure (in kPa)
700
800
900
Problem 10 from Problem Set #2 in Compressible Fluid Flow:
Consider a converging-diverging duct with a circular cross-section for a mass
flow rate of 3 kg/s of air and inlet stagnation conditions of 1400 kPa and
200oC. Assume that the flow is isentropic and the exit pressure is 100 kPa. Plot
the pressure and temperature of the air flow along the duct as a function of M.
Plot also the diameter of the duct as a function of M.
p0 = 1400 kPa
T0 = (273+200) K
M0 ≈ 0
pe = 100 kPa
m = 3 kg/s
R. Shanthini
22 Dec 2010
p0 and T0 are known.
Use the equations below to calculate p and T for different values of M.
T0
 1 2
 1
M
T
2
(2.6)

p0    1 2   1
 1 
M 
p 
2

(2.7)
Use (2.9) to calculate the exit are of the duct as follows:
R. Shanthini
22 Dec 2010
1600
Pressure along the duct (in kPa)
1400
Temperature along the duct (in K)
1200
Diameter of the duct (in mm)
1000
800
600
400
200
0
0
0.5
1
1.5
2
Mach number along the duct
R. Shanthini
22 Dec 2010
2.5
3
Enlarged version
250
Diameter of the duct (in mm)
200
150
100
50
0
0
0.5
1
1.5
2
Mach number along the duct
R. Shanthini
22 Dec 2010
2.5
3
Shape of the converging-diverging duct
250
200
150
100
50
0
-50
-100
-150
-200
-250
0
R. Shanthini
22 Dec 2010
0.5
1
1.5
2
Mach number along the duct
2.5
3
Problem 11 from Problem Set #2 in Compressible Fluid Flow:
Air at approximately zero velocity enters a converging-diverging duct at a
stagnation pressure and a stagnation temperature of 1000 kPa and 480 K,
respectively. Throat area of the duct is 0.002 m2. The flow inside the duct is
isentropic, and the exit pressure is 31.7 kPa. For air, γ = 1.4 and R = 287 J/kg.
Determine (i) the exit Mach number, (ii) the exit temperature, (iii) the exit
area of the duct, and (iii) the mass flow rate through the duct.
p0 = 1000 kPa
T0 = 480 K
M0 ≈ 0
pe = 31.7 kPa
Me = ?
Ae = ?
Athroat = 0.002 m2
R. Shanthini
22 Dec 2010
(i) Rearrange (2.7) to determine Me as follows:
 1


2  p0  

Me 
  1
p
e
  1 



100031.7
0 .4
1 .4

1
= 2.9
0 .2
(ii) Rearrange (2.5) to determine Te as follows:
 pe 
Te  T0  
 p0 
R. Shanthini
22 Dec 2010
 1

 31.7 
 (480 K)

 1000
0.4
1.4
= 179 K
(iii) Using (2.10), the exit area of the duct can be calculated as follows:
m  At M t p0



 
1




1
RT0  1 
M t2 

2


 1
2 (  1)
 Ae M e p0



 
1




1
RT0  1 
M e2 

2


 1
2 (  1)
Since the exit Mach number is 2.9, the flow is supersonic in the diverging
section. It is not possible unless sonic conditions are achieved at the throat.
Therefore Mt = 1 in the above expression.
Therefore, we get


1
 1 

At 
  Ae M e 
2 
 1.2 
 1  0.2 M e 
3
3
Since Me = 2.9 and At = 0.002 m2, the above expression gives
R. Shanthini
22 Dec 2010
At
Ae 
Me
 1  0.2 M

1.2

2
e
3

 = 0.0077 m2

(iv) Use (2.9) to get the mass flow rate through the duct as follows:
m  Ae M e pe

RTe
= 3.695 kg/s
R. Shanthini
22 Dec 2010
 (0.0077m 2 )(2.9)(31.7x1000kPa)
1.4
(287x179 J/kg)