CP502 Advanced Fluid Mechanics

1
CP502 Advanced Fluid Mechanics
Compressible Flow
Part 01_Set 02 Steady, quasi one-dimensional,
isothermal, compressible flow of an ideal gas in
a constant area duct with wall
friction (continued)
2
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)
3
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
4
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?
5
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?
6
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 ?
7
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.
8
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
9
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 ?
10
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
11
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 ?
12
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.
13
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
14
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.
15
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.
16
Differential equation of Problem 2
(1.2)
can be rearranged to give
In flows where p decreases along the flow
direction
(1.5)
17
Differential equation of Problem 2
(1.2)
can be rearranged to give
In flows where p increases along the flow
direction
(1.6)
18
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
19
Air ? 1.4 molecular mass 29
40 atm 97oC 500 m/s
p?
L
Limiting pressure
20
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
21
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.
22
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
23
In flows where p decreases along the flow
direction
(1.5)
Since
we get
(1.7)
24
In flows where p increases along the flow
direction
(1.6)
Since
we get
(1.8)
25
Summary
x
and
and
Limiting pressure
Limiting Mach number
26
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.
27
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.
28
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)
29
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)