CP502 Advanced Fluid Mechanics

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CP502 Advanced Fluid Mechanics
Compressible Flow
Lectures 3 Steady, quasi one-dimensional,
isothermal, compressible flow of an ideal gas in
a constant area duct with wall
friction (continued)
2
Problem 8 from Problem Set 1 in Compressible
Fluid Flow
Show that the equations in Problem 6 are
equivalent to the following
(1.7)
in flows where p decreases along the flow
direction
(1.8)
in flows where p increases along the flow
direction
3
In flows where p decreases along the flow
direction
(1.5)
Since
we get
(1.7)
4
In flows where p increases along the flow
direction
(1.6)
Since
we get
(1.8)
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Summary
x
and
and
Limiting pressure
Limiting Mach number
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Limiting Mach number for air
x
For air, ? 1.4
0.845
is associated with
If M lt 0.845 then pressure decreases in the flow
direction. That is, the pressure gradient causes
the flow.
is associated with
If M gt 0.845 then pressure increases in the flow
direction. That is, momentum causes the flow
working against the pressure gradient.
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Problem 9 from Problem Set 1 in Compressible
Fluid Flow
Show that when the flow has reached the limiting
pressure or the limiting Mach number the
length of the pipe across which such conditions
are reached, denoted by Lmax, shall satisfy the
following equation
where pressure p and Mach number M are the
conditions of the flow at the entrance of the
pipe.
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p M
p M
Lmax
Start with the following
(1.3)
Substitute L Lmax and pL in (1.3) to
get
(part of 1.9)
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p M
p M
Lmax
Start with the following
(1.4)
Substitute L Lmax and ML in (1.4)
to get
(part of 1.9)
Therefore,
(1.9)
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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.
Air ? 1.4 molecular mass 29
?
0.0075
D 10 mm
flow is choked
p 1 MPa
T 300 K
L 1 m
At choking condition, pL p, ML
and L Lmax
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Air ? 1.4 molecular mass 29
?
0.0075
D 10 mm
p 1 MPa
T 300 K
L Lmax 1 m
0.5
(
)
(p) (10/1000 m)2 (1000,000 Pa) M
1.4

(8314/29)(300) J/kg
4
0.317 M
M at the entrance could be determined using (1.9)
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Air ? 1.4 molecular mass 29
?
0.0075
D 10 mm
p 1 MPa
T 300 K
L Lmax 1 m
Use
(part of 1.9)
Solving the nonlinear equation above gives M
0.352 at the entrance
0.317 M
0.317 x 0.352 0.1116 kg/s
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Air ? 1.4 molecular mass 29
?
0.0075
D 10 mm
p 1 MPa
T 300 K
L Lmax 1 m
Determine the exit pressure.
Since
(pM)entrance (pM)exit
(1 MPa) (0.352) pexit ( )
0.5
pexit (1 MPa) (0.352) (1.4)
0.417 MPa
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Air ? 1.4 molecular mass 29
?
0.0075
D 10 mm
p 1 MPa
T 300 K
L Lmax 1 m
Reynolds Number
4 (0.1116 kg/s)

p (10/1000 m) (2.17 x 10-5 kg/m.s)
6.5 x 105
Therefore, flow is turbulent
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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?
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Air ? 1.4 molecular mass 29
9.0 kg/s
A 1.5x10-3 m2
pA 6.5 bar
T 300 K
pB 8.5 bar
(i) Velocities uA and uB ?
(9 kg/s) (8314/29 J/kg.K) (300 K)

(1.5x10-3 m2) (6.5 bar) (100,000 Pa/bar)
794 m/s
6.5 bar

794 m/s
607 m/s
8.5 bar
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Air ? 1.4 molecular mass 29
9.0 kg/s
A 1.5x10-3 m2
pA 6.5 bar
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 uA pB A uB Force acting on
the duct wall
Force acting on the duct wall (pA pB ) A
(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
-300 N 1683 N
1383 N
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Air ? 1.4 molecular mass 29
9.0 kg/s
A 1.5x10-3 m2
pA 6.5 bar
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
hB uB2/2
Enthalpy at A
Enthalpy at B
Kinetic energy at B
Kinetic energy at A
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Air ? 1.4 molecular mass 29
9.0 kg/s
A 1.5x10-3 m2
pA 6.5 bar
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
Heat is lost to the surroundings
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Air ? 1.4 molecular mass 29
9.0 kg/s
A 1.5x10-3 m2
pA 6.5 bar
T 300 K
pB 8.5 bar
Direction of the gas flow
Determine first the limiting pressure as follows
(9.0/1.5x10-3) kg/m2.s (8314300/29 J/kg)0.5
(9.0/1.5x10-3) kg/m2.s (8314300/29 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). Therefore, gas is flowing from
A to B.
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Summarizing the results of Problem 11
9.0 kg/s
pB 8.5 bar
pA 6.5 bar
T 300 K
uB 607 m/s
uA 794 m/s
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.
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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, ? 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
e 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
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Properties of gas produced Molecular weight
0.013 kg/mol µ 10-5 kg/m.s ? 1.36
pL 500 kPa
D 15 cm
p 1 MPa
T (27387) K
L 100 m
What is the flow rate through the pipe?
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Design equation to be used
(1.3)
1/4 log(3.7x15x10/0.046) 0.0613
0.0038
4 x 0.0038 x 100 m / (15 cm) 10.1333
(500/1000)2 0.25
Using the above in (1.3), we get
10.1333 ln(0.25)

15.3595
1 0.25
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15.3595
p 1 MPa 1,000,000 Pa
R 8.314 J/mol.K 8.314/0.013 J/kg.K
T 360 K
A pD2/4 p(15 cm)2/4 p(0.15 m)2/4
Therefore,
9.4 kg/s
Check the Reynolds number
Re uD?/µ D/µ
1.4x105
(9.4 kg/s)(15/100 m)/(10-5 kg/m.s)
Therefore, flow is turbulent
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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.
Density (?) is a constant
Incompressible flow
Mass flow rate is a constant
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.
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p
pdp
D
u
udu
dx
x
Write the momentum balance over the differential
volume chosen.
Since ,
and , we get
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Therefore, we get
(2)
Rearranging (2) gives
It means p decreases in the flow direction.
Since ? and u are constants, integrating the
above gives
pressure at the exit
pressure at the entrance
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Rework Problem 12 assuming incompressible flow
Molecular weight 0.013 kg/mol
0.0038
pL 500 kPa
D 15 cm
p 1 MPa
T (27387) K
L 100 m
What is the flow rate through the pipe?
Design equation used for compressible flow
(1.3)
Design equation to be used with incompressible
flow
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Molecular weight 0.013 kg/mol
0.0038
pL 500 kPa
D 15 cm
p 1 MPa
T (27387) K
L 100 m
Substitute in the above, we
get
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Molecular weight 0.013 kg/mol
0.0038
pL 500 kPa
D 15 cm
p 1 MPa
T (27387) K
L 100 m
What is ??
? (?entrance ?exit ) / 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
Compare 7.76 kg/s with the 9.4 kg/s obtained
considering the flow to be compressible.
Therefore,
7.76 kg/s
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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 final
examination.