CP502 Advanced Fluid Mechanics

1
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
2
Incompressible flow assumption is not valid if
Mach number gt 0.3
What is a Mach number?
Definition of Mach number (M)
For an ideal gas,
specific heat ratio
specific gas constant (in J/kg.K)
absolute temperature of the flow at the point
concerned (in K)
3
For an ideal gas,
u
u

M
c
Unit of u m/s
Unit of c (J/kg.K)(K)0.5

kg.(m/s2).m/kg0.5
J/kg0.5
(N.m/kg)0.5
m2/s20.5
m/s
4
constant area duct quasi one-dimensional
flow compressible flow steady flow isothermal
flow ideal gas wall friction
is a constant
Diameter (D)
speed (u)
u varies only in x-direction
x
Density (?) is NOT a constant
Mass flow rate is a constant
Temperature (T) is a constant
Obeys the Ideal Gas equation
is the shear stress acting on the wall
where is the average Fanning friction factor
5
Friction factor For laminar flow in circular
pipes           where Re is the Reynolds
number of the flow defined as follows For
lamina flow in a square channel
               For the turbulent flow regime
                                                  
           
Quasi one-dimensional flow is closer to turbulent
velocity profile than to laminar velocity profile.
6
Ideal Gas equation of state
temperature
pressure
specific gas constant (not universal gas constant)
volume
mass
Ideal Gas equation of state can be rearranged to
give
K
Pa N/m2
kg/m3
J/(kg.K)
7
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 shall be written as
follows where p, ? and u are the
respective pressure, density and velocity at
distance x from the entrance of the pipe.
(1.1)
8
p
pdp
D
u
udu
dx
x
Write the momentum balance over the differential
volume chosen.
(1)
steady mass flow rate
cross-sectional area
shear stress acting on the wall
is the wetted area on which shear is acting
9
p
pdp
D
u
udu
dx
x
Equation (1) can be reduced to
Substituting
Since ,
and , we get
(1.1)
10
Problem 2 from Problem Set 1 in Compressible
Fluid Flow
Show that the differential equation of
Problem (1) can be converted into which
in turn can be integrated to yield the following
design equation 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 temperature of the
gas, is the mass flow rate of the gas flowing
through the pipe, and A is the cross-sectional
area of the pipe.
(1.2)
(1.3)
11
The differential equation of problem (1) is in
which the variables ? and u must be replaced by
the variable p.
(1.1)
Let us use the mass flow rate equation
and the ideal gas equation to obtain the
following
and
and therefore
It is a constant for steady, isothermal flow in a
constant area duct
12
,
Using
and
in
(1.1)
we get
(1.2)
13
p
pL
L
Integrating (1.2) from 0 to L, we get
which becomes
(1.3)
14
Problem 3 from Problem Set 1 in Compressible
Fluid Flow
Show that the design equation of Problem (2) is
equivalent to where M is the Mach number at
the entry and ML is the Mach number at length L
from the entry.
(1.4)
15
Design equation of Problem (2) is which
should be shown to be equivalent to 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.
(1.3)
(1.4)
We need to relate p to M!
16
We need to relate p to M!
which gives
constant for steady, isothermal flow in a
constant
area duct
Substituting the above in (1.3), we get
(1.4)
17
Summary Design equations for steady, quasi
one-dimensional, isothermal,compressible flow of
an ideal gas in a constant area duct with wall
friction
(1.1)
(1.2)
(1.3)
(1.4)
18
Problem 4 from Problem Set 1 in Compressible
Fluid Flow
Nitrogen (? 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.
1.5 mol/s
0.0066
? 1.4 molecular mass 28
pL ?
D 15 mm
p 600 kPa
T 300 K
L 11.5 m
19
1.5 mol/s
0.0066
? 1.4 molecular mass 28
pL ?
D 15 mm
p 600 kPa
T 300 K
L 11.5 m
Design equation
(1.3)
0.0066
20.240
L 11.5 m
D 15 mm 0.015 m
unit?
20
1.5 mol/s
0.0066
? 1.4 molecular mass 28
pL ?
D 15 mm
p 600 kPa
T 300 K
L 11.5 m
Design equation
(1.3)
p 600 kPa 600,000 Pa
T 300 K
R 8.314 kJ/kmol.K 8.314/28 kJ/kg.K 8314/28
J/kg.K
1.5 mol/s 1.5 x 28 g/s 1.5 x 28/1000
kg/s
A pD2/4 p(15 mm)2/4 p(0.015 m)2/4
71.544
unit?
21
1.5 mol/s
0.0066
? 1.4 molecular mass 28
pL ?
D 15 mm
p 600 kPa
T 300 K
L 11.5 m
Design equation
(1.3)
71.544
20.240
p 600 kPa 600,000 Pa
Solve the nonlinear equation above to determine pL
pL ?
22
71.544
20.240
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.
23
Now, solve the nonlinear equation for pL values
close to 508.1 kPa
71.544
20.240
p 600 kPa 600,000 Pa
pL kPa LHS of the above equation RHS of the above equation
510 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
24
Problem 4 continued
Rework the problem in terms of Mach number
and determine ML.
1.5 mol/s
0.0066
? 1.4 molecular mass 28
ML ?
D 15 mm
p 600 kPa
T 300 K
L 11.5 m
Design equation
(1.4)
20.240 (already calculated in Problem 4)
M ?
25
1.5 mol/s
0.0066
? 1.4 molecular mass 28
ML ?
D 15 mm
p 600 kPa
T 300 K
L 11.5 m
u
u
1
1
RT
M



c
A
p
(
)
0.5
4 (1.5x 28/1000 kg/s)
(8314/28)(300) J/kg

1.4
p
(15/1000 m)2 (600,000 Pa)
0.1
26
1.5 mol/s
0.0066
? 1.4 molecular mass 28
ML ?
D 15 mm
p 600 kPa
T 300 K
L 11.5 m
Design equation
(1.4)
20.240
Solve the nonlinear equation above to determine ML
ML ?
27
20.240
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.
28
Now, solve the nonlinear equation for ML values
close to 0.118
20.240
pL kPa LHS of the above equation RHS of the above equation
0.116 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
29
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.
30
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
(1.5)
in flows where p decreases along the flow
direction, and
(1.6)
in flows where p increases along the flow
direction.
31
Differential equation of Problem 2
(1.2)
can be rearranged to give
In flows where p decreases along the flow
direction
(1.5)
32
Differential equation of Problem 2
(1.2)
can be rearranged to give
In flows where p increases along the flow
direction
(1.6)
33
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.
Air ? 1.4 molecular mass 29
40 atm 97oC 500 m/s
p?
L
34
Air ? 1.4 molecular mass 29
40 atm 97oC 500 m/s
p?
L
Limiting pressure
35
Air ? 1.4 molecular mass 29
40 atm 97oC 500 m/s
p?
L
(40 atm) (500 m/s)

61.4 atm
0.5
(8314/29)(27397) J/kg
36
Air ? 1.4 molecular mass 29
40 atm 97oC 500 m/s
p61.4 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.