CP502
Advanced Fluid Mechanics
Compressible Flow
Part 01_Set 03:
Steady, quasi one-dimensional, isothermal,
compressible flow of an ideal gas in a
constant area duct with wall friction
(continued)
Problem 10 from Problem Set 1 in Compressible Fluid Flow:
Determine the isothermal mass flow rate of air in a pipe of 10-mmi.d. and 1m long with upstream condition of 1 MPa and 300 K with a
exit pressure low enough to choke the flow in the pipe assuming an
average Fanning friction factor of 0.0075.
Determine also the exit pressure.
Given μ = 2.17 x 10-5 kg/m.s, calculate the Reynolds number of
the flow to check if the given flow were turbulent.
m = ?;
p = 1 MPa
Air: γ = 1.4; molecular mass = 29;
D = 10 mm
T = 300 K
f = 0.0075
flow is choked
L=1m
At choking condition, pL = p*, ML = M *  1 /  and L = Lmax
R. Shanthini
16 Feb 2012
m = ?;
p = 1 MPa
Air: γ = 1.4; molecular mass = 29;
D = 10 mm
f = 0.0075
M*  1/ 
T = 300 K
L = Lmax = 1 m
m  Au 
=
D 2 p
4 RT
(π) (10/1000
M RT 
m)2 (1000,000
4
D 2
4
pM

RT
0.5
Pa) M
(
1.4
(8314/29)(300) J/kg
)
= 0.317 M
M at the entrance could be determined using (1.9)
R. Shanthini
16 Feb 2012
m = ?;
p = 1 MPa
Air: γ = 1.4; molecular mass = 29;
D = 10 mm
f = 0.0075
T = 300 K
M*  1/ 
L = Lmax = 1 m
4 fLmax 1  M 2
2
Use




ln

M
2
D
M
(part of 1.9)
4(0.0075)(1 m) 1  1.4 M 2
2




ln
1
.
4
M
2
(10 / 1000m)
1.4 M
Solving the nonlinear equation above gives M = 0.352 at the entrance
m
= 0.317 M = 0.317 x 0.352 = 0.1116 kg/s
R. Shanthini
16 Feb 2012
m = ?;
p = 1 MPa
Air: γ = 1.4; molecular mass = 29;
D = 10 mm
f = 0.0075
M*  1/ 
T = 300 K
L = Lmax = 1 m
Determine the exit pressure.
  Au 
Since m
D 2 p
4 RT
M RT 
D 2
pM
4

RT
(pM)entrance = (pM)exit
(1 MPa) (0.352) = pexit (1 / 
0.5
pexit = (1 MPa) (0.352) (1.4)
R. Shanthini
16 Feb 2012
)
= 0.417 MPa
m = ?;
p = 1 MPa
Air: γ = 1.4; molecular mass = 29;
D = 10 mm
T = 300 K
f = 0.0075
M*  1/ 
L = Lmax = 1 m
Reynolds Number:
uD m D 4m D 4m
Re 



2

A  D  D
4 (0.1116 kg/s)
=
π (10/1000 m) (2.17 x 10-5 kg/m.s)
= 6.5 x 105
Therefore, flow is turbulent
R. Shanthini
16 Feb 2012
Problem 11 from Problem Set 1 in Compressible Fluid Flow:
Air flows at a mass flow rate of 9.0 kg/s isothermally at 300 K
through a straight rough duct of constant cross-sectional area
1.5 x 10-3 m2. At one end A the pressure is 6.5 bar and at the other
end B the pressure is 8.5 bar.
Determine the following:
(i) Velocities uA and uB
(ii) Force acting on the duct wall
(iii) Rate of heat transfer through the duct wall
In which direction is the gas flowing?
R. Shanthini
16 Feb 2012
Air: γ = 1.4; molecular mass = 29;
pA = 6.5
bar
A = 1.5x10-3 m2
m = 9.0 kg/s;
T = 300 K
pB = 8.5 bar
(i) Velocities uA and uB = ?
m
m RT
uA 

A A
ApA
=
(9 kg/s) (8314/29 J/kg.K) (300 K)
(1.5x10-3 m2) (6.5 bar) (100,000 Pa/bar)
= 794 m/s
m RT
uA pA 
 uB pB
A
pA
uB 
uA
pB
6.5 bar
= 8.5 bar 794 m/s
R. Shanthini
16 Feb 2012
= 607 m/s
Air: γ = 1.4; molecular mass = 29;
pA = 6.5
bar
A = 1.5x10-3 m2
m = 9.0 kg/s;
T = 300 K
pB = 8.5 bar
(ii) Force acting on the duct wall = ?
Force balance on the entire duct gives the following:
pA A + m
 uA = pB A + m uB + Force acting on the duct wall
Force acting on the duct wall = (pA – pB ) A + m
 (uA – uB )
= (6.5 – 8.5) bar x 100,000 Pa/bar x 1.5 x 10-3 m2
+ (9.0 kg/s) (794 – 607) m/s
= -300 Pa.m2 + 1683 kg.m/s2
R. Shanthini
16 Feb 2012
= -300 N + 1683 N
= 1383 N
m = 9.0 kg/s;
Air: γ = 1.4; molecular mass = 29;
pA = 6.5
bar
A = 1.5x10-3 m2
T = 300 K
pB = 8.5 bar
(iii) Rate of heat transfer through the duct wall = ?
Energy balance on the entire duct gives the following:
Rate of heat transfer through the duct wall from the surroundings
+ hA + uA2/2
Enthalpy at A
=
Enthalpy at B
Kinetic energy at A
R. Shanthini
16 Feb 2012
hB + uB2/2
Kinetic energy at B
Air: γ = 1.4; molecular mass = 29;
pA = 6.5
bar
A = 1.5x10-3 m2
m = 9.0 kg/s;
T = 300 K
pB = 8.5 bar
Rate of heat transfer through the duct wall from the surroundings
= (hB – hA) + (uB 2 – uA 2)/2
Since (hB – hA) = cp (TB – TA) = 0 for isothermal flow of an ideal gas
Rate of heat transfer through the duct wall from the surroundings
= (uB 2 – uA 2)/2
= (607 2 – 794 2)/2 m2/s2
= (-130993.5 m2/s2) = (-130993.5 J/kg)
= (-130993.5 J/kg) (9.0 kg/s)
= -1178942 J/s
= -1179 kW
R. Shanthini
16 Feb 2012
Heat is lost to the surroundings
Air: γ = 1.4; molecular mass = 29;
pA = 6.5
bar
A = 1.5x10-3 m2
m = 9.0 kg/s;
T = 300 K
pB = 8.5 bar
Direction of the gas flow:
Determine first the limiting pressure as follows:
 / A) RT
p*  (m
= (9.0/1.5x10-3) kg/m2.s (8314*300/29 J/kg)0.5
= (9.0/1.5x10-3) kg/m2.s (8314*300/29 J/kg)0.5
= 17.6 bar
Since pA and pB are lower than the limiting pressure,
p increases along the flow direction (see Problem 6).
R. Shanthini
16 Feb 2012
Therefore, gas is flowing from A to B.
Summarizing the results of Problem 11:
m = 9.0 kg/s;
pA = 6.5
ubar
A = 794 m/s
T = 300 K
pB = 8.5 bar
uB = 607 m/s
Pressure increases in the flow direction and therefore velocity decreases
according to the following equation:
m RT
uA pA 
 uB pB
A
Force acting on the entire duct wall is 1383 N
Velocity decreases and therefore kinetic energy is lost across the duct.
The lost energy is transferred from the duct to the surroundings through
the duct wall.
R. Shanthini
16 Feb 2012
Problem 12 from Problem Set 1 in Compressible Fluid Flow:
Gas produced in a coal gasification plant (molecular weight = 0.013
kg/mol, μ = 10-5 kg/m.s, γ = 1.36) is sent to neighbouring industrial
users through a bare 15-cm-i.d. commercial steel pipe 100 m long.
The pressure gauge at one end of the pipe reads 1 MPa absolute. At
the other end it reads 500 kPa. The temperature is 87oC. Estimate the
flow rate of coal gas through the pipe?
Additional data:
ε = 0.046 mm for commercial steel. For fully developed turbulent
flow in rough pipes, the average Fanning friction factor can be
found by use of the following:
f  1 /[4 log(3.7 D /  )]
R. Shanthini
16 Feb 2012
Properties of gas produced:
Molecular weight = 0.013 kg/mol; μ = 10-5 kg/m.s; γ = 1.36
p = 1 MPa
D = 15 cm
T = (273+87) K
L = 100 m
What is the flow rate through the pipe?
R. Shanthini
16 Feb 2012
pL = 500 kPa
Design equation to be used:
4f L
p2

D
RT (m / A) 2

 pL2 
pL2 
1  2   ln 2 
p 

p 
(1.3)
f  1 /[4 log(3.7 D /  )] = 1/[4 log(3.7x15x10/0.046)] = 0.0613
f = 0.0038
4f L
= 4 x 0.0038 x 100 m / (15 cm) = 10.1333
D
 p L2 
 2  = (500/1000)2 = 0.25
p 
Using the above in (1.3), we get
R. Shanthini
16 Feb 2012
p2
2 =
RT (m / A)
10.1333 – ln(0.25)
1 – 0.25
= 15.3595
p2
= 15.3595
2
RT (m / A)
p = 1 MPa = 1,000,000 Pa;
R = 8.314 J/mol.K = 8.314/0.013 J/kg.K;
T = 360 K;
A = πD2/4 = π(15 cm)2/4 = π(0.15 m)2/4;
Therefore,
m 
pA
RT
1
= 9.4 kg/s;
15.3595
Check the Reynolds number:
 D/μ
Re = uDρ/μ = m
= (9.4 kg/s)(15/100 m)/(10-5 kg/m.s) = 1.4x105
R. Shanthini
16 Feb 2012
Therefore, flow is turbulent
Governing equation for incompressible flow:
Starting from the mass and momentum balances, obtain the
differential equation describing the quasi one-dimensional,
incompressible, isothermal, steady flow of an ideal gas through a
constant area pipe of diameter D and average Fanning friction
factor. f
Incompressible flow
Steady flow
Density (ρ) is a constant
  Au is a constant
Mass flow rate m
Constant area pipe
A is a constant
Therefore, u is a constant for a steady, quasi one-dimensional,
compressible flow in a constant area pipe.
R. Shanthini
16 Feb 2012
w
p
p+dp
u
u+du
D
dx
x
Write the momentum balance over the differential volume chosen.
 u  ( p  dp) A  m
 (u  du)   wdAw
pA  m
 du   wdAw  0
Adp  m
Since A  D 2 / 4 ,
R. Shanthini
16 Feb 2012
 w  fu / 2
dp  udu 
2
and dAw  Ddx , we get
u 2 4 f
2
D
dx  0
Therefore, we get
4f
2
2
dx  2 dp  du  0
D
u
u
(2)
dp
4f /D
2 fu 2
Rearranging (2) gives


0
2
dx
2 / u
D
It means p decreases in the flow direction.
Since ρ and u are constants, integrating the above gives
2 fu 2
p  pL 
L
D
R. Shanthini
16 Feb 2012
pressure at the exit
pressure at the entrance
Rework Problem 12 assuming incompressible flow:
f = 0.0038
Molecular weight = 0.013 kg/mol;
p = 1 MPa
D = 15 cm
T = (273+87) K
pL = 500 kPa
L = 100 m
What is the flow rate through the pipe?
Design equation used for compressible flow
4f L
p2

D
RT (m / A) 2

 pL2 
pL2 
1  2   ln 2 
p 

p 
Design equation to be used with incompressible flow
R. Shanthini
16 Feb 2012
2 fu 2
p  pL 
L
D
(1.3)
Molecular weight = 0.013 kg/mol;
p = 1 MPa
D = 15 cm
f = 0.0038
T = (273+87) K
pL = 500 kPa
L = 100 m
2 fu 2
p  pL 
L
D
m
Substitute u 
A
in the above, we get
2 f (m / A) 2
p  pL 
L
D
R. Shanthini
16 Feb 2012
( p  pL ) D
m  A
2 fL
Molecular weight = 0.013 kg/mol;
p = 1 MPa
D = 15 cm
f = 0.0038
T = (273+87) K
pL = 500 kPa
L = 100 m
( p  pL ) D D 2
m  A

2 fL
4
What is ρ?
( p  pL ) D
2 fL
ρ = (ρentrance + ρexit ) / 2 = [(p/RT)entrance + (p/RT)exit )] / 2
= (pentrance + pexit ) / 2RT
= (1,000,000 + 500,000) Pa / [2 x (8.314/0.013) x 300 J/kg]
= 1.9545 kg/m3
 = 7.76 kg/s
Therefore, m
R. Shanthini
16 Feb 2012
Compare 7.76 kg/s with the 9.4 kg/s
obtained considering the flow to be
compressible.
Important Note:
Problems (13) and (14) from Problem Set 1 in
Compressible Fluid Flow are assignments to be
worked out by the students themselves in
preparation to the mid-semester examination.
R. Shanthini
16 Feb 2012