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)
R. Shanthini
16 Feb 2012
Problem 10
from Problem Set 1 in Compressible Fluid Flow:
Determine the isothermal mass flow rate of air in a pipe of 10

mm

i.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.
p
= 1 MPa
L
= 1 m
D
= 10 mm
m
= ?;
T
= 300 K
f
= 0.0075
flow is choked
At choking condition,
p
L
= p*,
M
L
=
and
L = L
max
Air:
γ
=
1.4; molecular mass = 29;
/
1
*
M
R. Shanthini
16 Feb 2012
p
= 1 MPa
L
=
L
max
= 1 m
D
= 10 mm
m
= ?;
T
= 300 K
f
= 0.0075
Air:
γ
=
1.4; molecular mass = 29;
/
1
*
M
RT
pM
D
RT
M
RT
p
D
u
A
m
4
4
2
2
=
(
π
) (10/1000 m)
2
(1000,000 Pa)
M
(8314/29)(300) J/kg
0.5
4
(
)
1.4
= 0.317
M
M
at the entrance could be determined using (1.9)
R. Shanthini
16 Feb 2012
p
= 1 MPa
L
=
L
max
= 1 m
D
= 10 mm
m
= ?;
T
= 300 K
f
= 0.0075
Air:
γ
=
1.4; molecular mass = 29;
/
1
*
M
2
2
2
max
ln
1
4
M
M
M
D
L
f
Use
(part of 1.9)
2
2
2
4
.
1
ln
4
.
1
4
.
1
1
)
m
1000
/
10
(
)
m
1
)(
0075
.
0
(
4
M
M
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
p
= 1 MPa
L
=
L
max
= 1 m
D
= 10 mm
m
= ?;
T
= 300 K
f
= 0.0075
Air:
γ
=
1.4; molecular mass = 29;
/
1
*
M
RT
pM
D
RT
M
RT
p
D
u
A
m
4
4
2
2
Determine the exit pressure.
Since
(
pM
)
entrance
= (
pM
)
exit
(1 MPa) (0.352)
=
p
exit
(
)
/
1
p
exit
= (1 MPa) (0.352) (1.4)
0.5
= 0.417 MPa
R. Shanthini
16 Feb 2012
p
= 1 MPa
L
=
L
max
= 1 m
D
= 10 mm
m
= ?;
T
= 300 K
f
= 0.0075
Air:
γ
=
1.4; molecular mass = 29;
/
1
*
M
Reynolds Number:
D
m
D
D
m
D
A
m
uD
4
4
Re
2
=
4 (0.1116 kg/s)
π
(10/1000 m) (2.17 x 10

5
kg/m.s)
= 6.5 x 10
5
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
m
2
. 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
u
A
and
u
B
(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
p
A
= 6.5
bar
A
= 1.5x10

3
m
2
m
= 9.0 kg/s;
T
= 300 K
Air:
γ
=
1.4; molecular mass = 29;
p
B
= 8.5 bar
(i) Velocities
u
A
and
u
B
= ?
A
A
A
Ap
RT
m
A
m
u
=
(9 kg/s) (8314/29 J/kg.K) (300 K)
(1.5x10

3
m
2
) (6.5 bar) (100,000 Pa/bar)
= 794 m/s
B
B
A
A
p
u
A
RT
m
p
u
A
B
A
B
u
p
p
u
=
6.5 bar
8.5 bar
794 m/s
= 607 m/s
R. Shanthini
16 Feb 2012
p
A
= 6.5
bar
A
= 1.5x10

3
m
2
m
= 9.0 kg/s;
T
= 300 K
Air:
γ
=
1.4; molecular mass = 29;
p
B
= 8.5 bar
(ii)
Force acting on the duct wall = ?
Force balance on the entire duct gives the following:
p
A
A + u
A
= p
B
A + u
B
+ Force acting on the duct wall
m
m
Force acting on the duct wall
=
(
p
A
–
p
B
)
A +
(
u
A
–
u
B
)
m
=
(6.5
–
8.5) bar x 100,000 Pa/bar x 1.5 x 10

3
m
2
+
(9.0 kg/s)
(794
–
607) m/s
=

300 Pa.m
2
+ 1683 kg.m/s
2
=
1383 N
=

300 N + 1683 N
R. Shanthini
16 Feb 2012
p
A
= 6.5
bar
A
= 1.5x10

3
m
2
m
= 9.0 kg/s;
T
= 300 K
Air:
γ
=
1.4; molecular mass = 29;
p
B
= 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
+
h
A
+
u
A
2
/
2
= h
B
+
u
B
2
/2
Enthalpy at A
Enthalpy at B
Kinetic energy at A
Kinetic energy at B
R. Shanthini
16 Feb 2012
p
A
= 6.5
bar
A
= 1.5x10

3
m
2
m
= 9.0 kg/s;
T
= 300 K
Air:
γ
=
1.4; molecular mass = 29;
p
B
= 8.5 bar
Rate of heat transfer through the duct wall from the surroundings
=
(
h
B
–
h
A
) +
(
u
B
2
–
u
A
2
)/2
Since (
h
B
–
h
A
) =
c
p
(
T
B
–
T
A
) = 0 for isothermal flow of an ideal gas
Rate of heat transfer through the duct wall from the surroundings
=
(
u
B
2
–
u
A
2
)/2
=
(607
2
–
794
2
)/2 m
2
/s
2
=
(

130993.5 m
2
/s
2
)
=
(

130993.5 J/kg)
=
(

130993.5 J/kg) (9.0 kg/s)
=

1178942 J/s
=

1179 kW
Heat is lost to the surroundings
R. Shanthini
16 Feb 2012
p
A
= 6.5
bar
A
= 1.5x10

3
m
2
m
= 9.0 kg/s;
T
= 300 K
Air:
γ
=
1.4; molecular mass = 29;
p
B
= 8.5 bar
Direction of the gas flow:
RT
A
m
p
)
/
(
*
Determine first the limiting pressure as follows:
= (9.0/1.5x10

3
) kg/m
2
.s (8314*300/29 J/kg)
0.5
= (9.0/1.5x10

3
) kg/m
2
.s (8314*300/29 J/kg)
0.5
= 17.6 bar
Since
p
A
and
p
B
are lower than the limiting pressure,
p
increases along the flow direction (see Problem 6).
Therefore, gas is flowing from A to B.
R. Shanthini
16 Feb 2012
p
A
= 6.5
bar
m
= 9.0 kg/s;
T
= 300 K
p
B
= 8.5 bar
u
A
= 794 m/s
u
B
= 607 m/s
B
B
A
A
p
u
A
RT
m
p
u
Summarizing the results of Problem 11:
Pressure increases in the flow direction and therefore velocity decreases
according to the following equation:
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:
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:
)]
/
7
.
3
log(
4
/[
1
D
f
Additional data:
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 87
o
C. Estimate the
flow rate of coal gas through the pipe?
R. Shanthini
16 Feb 2012
p
= 1 MPa
p
L
= 500 kPa
L
= 100 m
D
= 15 cm
T
= (273+87) K
Properties of gas produced:
Molecular weight = 0.013 kg/mol;
μ
= 10

5
kg/m.s;
γ
= 1.36
What is the flow rate through the pipe?
R. Shanthini
16 Feb 2012
)]
/
7
.
3
log(
4
/[
1
D
f
= 1/[4 log(3.7x15x10/0.046)] = 0.0613
f
= 0.0038
D
L
f
4
= 4 x 0.0038 x 100 m / (15 cm) = 10.1333
= (500/1000)
2
= 0.25
2
2
p
p
L
Using the above in (1.3), we get
2
2
2
2
2
2
ln
1
)
/
(
4
p
p
p
p
A
m
RT
p
D
L
f
L
L
(1.3)
Design equation to be used:
2
2
)
/
(
A
m
RT
p
10.1333
–
ln(0.25)
=
1
–
0.25
= 15.3595
R. Shanthini
16 Feb 2012
2
2
)
/
(
A
m
RT
p
= 15.3595
p
= 1 MPa = 1,000,000 Pa;
R
= 8.314 J/mol.K = 8.314/0.013 J/kg.K;
= 9.4 kg/s;
T
= 360 K;
A
=
π
D
2
/4
=
π
(15 cm)
2
/4 =
π
(0.15 m)
2
/4;
3595
.
15
1
RT
pA
m
Therefore,
Check the Reynolds number:
Re =
uD
ρ
/
μ
= D/
μ
m
= 1.4x10
5
=
(9.4 kg/s)(15/100 m)/(
10

5
kg/m.s)
Therefore, flow is turbulent
R. Shanthini
16 Feb 2012
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
Governing equation for
incompressible
flow:
Density (
ρ
) is a constant
u
A
m
Mass flow rate is a constant
Incompressible flow
Steady flow
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
p
p+dp
u
u+du
x
dx
D
Write the momentum balance over the differential volume chosen.
w
w
dA
du
u
m
A
dp
p
u
m
pA
)
(
)
(
w
0
w
w
dA
du
m
Adp
Ddx
dA
w
2
/
2
u
f
w
4
/
2
D
A
Since , and , we get
0
4
2
2
dx
D
f
u
udu
dp
R. Shanthini
16 Feb 2012
Therefore, we get
0
2
2
4
2
du
u
dp
u
dx
D
f
Rearranging
(2) gives
(2)
Since
ρ
and
u
are constants, integrating the above gives
It means
p
decreases in the flow direction.
L
D
u
f
p
p
L
2
2
0
2
/
2
/
4
2
2
D
u
f
u
D
f
dx
dp
pressure at the entrance
pressure at the exit
R. Shanthini
16 Feb 2012
Rework Problem 12 assuming
incompressible
flow:
p
= 1 MPa
p
L
= 500 kPa
L
= 100 m
D
= 15 cm
T
= (273+87) K
Molecular weight = 0.013 kg/mol;
2
2
2
2
2
2
ln
1
)
/
(
4
p
p
p
p
A
m
RT
p
D
L
f
L
L
(1.3)
Design equation used for
compressible
flow
Design equation to be used with
incompressible
flow
L
D
u
f
p
p
L
2
2
What is the flow rate through the pipe?
f
= 0.0038
R. Shanthini
16 Feb 2012
L
D
u
f
p
p
L
2
2
p
= 1 MPa
p
L
= 500 kPa
L
= 100 m
D
= 15 cm
T
= (273+87) K
Substitute in the above, we get
A
m
u
L
D
A
m
f
p
p
L
2
)
/
(
2
L
f
D
p
p
A
m
L
2
)
(
Molecular weight = 0.013 kg/mol;
f
= 0.0038
R. Shanthini
16 Feb 2012
L
f
D
p
p
D
L
f
D
p
p
A
m
L
L
2
)
(
4
2
)
(
2
What is
ρ
?
= (1,000,000 + 500,000) Pa / [2 x (8.314/0.013) x 300 J/kg]
ρ
=
(
ρ
entrance
+
ρ
exit
) / 2
=
[
(
p/RT
)
entrance
+ (
p/RT
)
exit
)] / 2
=
(
p
entrance
+
p
exit
) / 2
RT
= 1.9545 kg/m
3
p
= 1 MPa
p
L
= 500 kPa
L
= 100 m
D
= 15 cm
T
= (273+87) K
m
Therefore,
= 7.76 kg/s
Molecular weight = 0.013 kg/mol;
f
= 0.0038
Compare 7.76 kg/s with the 9.4 kg/s
obtained considering the flow to be
compressible.
R. Shanthini
16 Feb 2012
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.
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