Transcript CompressibleFlow_Par..

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)
Summary
Design equations for steady, quasi one-dimensional,
isothermal,compressible flow of an ideal gas in a constant area
duct with wall friction
4f
dx 
D
4f
dx  
D
4f L
D
u
2
dp 
RT ( m / A )
2
du  0
(1.1)
u
2
2
pdp 
2
dp
p
(1.2)
2

 p L2 
pL 
 1  2   ln  2 

2 
RT ( m / A ) 
p 
 p 
p
4f L
R. Shanthini
09 Feb 2012
2
D

2
1
 M
2

M
 1 
M

2
2
L

M
  ln 

M
2
2
L



(1.3)
(1.4)
Problem 4 from Problem Set 1 in Compressible Fluid Flow:
Nitrogen (γ = 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.
γ = 1.4; molecular mass = 28;
p = 600 kPa
m = 1.5 mol/s;
D = 15 mm
= 0.0066
T = 300 K
L = 11.5 m
R. Shanthini
09 Feb 2012
f
pL = ?
γ = 1.4; molecular mass = 28;
p = 600 kPa
m = 1.5 mol/s;
D = 15 mm
f
= 0.0066
T = 300 K
pL = ?
L = 11.5 m
Design equation:
4f L
D
2

 p L2 
pL 
 1  2   ln  2 

2 
RT ( m / A ) 
p 
 p 
p
2
(1.3)
f = 0.0066;
4f L
L = 11.5 m;
D = 15 mm = 0.015 m;
R. Shanthini
09 Feb 2012
D
= 20.240
unit?
γ = 1.4; molecular mass = 28;
p = 600 kPa
m = 1.5 mol/s;
D = 15 mm
f
= 0.0066
T = 300 K
pL = ?
L = 11.5 m
Design equation:
4f L
D
2

 p L2 
pL 
 1  2   ln  2 

2 
RT ( m / A ) 
p 
 p 
p
2
p = 600 kPa = 600,000 Pa;
(1.3)
T = 300 K;
R = 8.314 kJ/kmol.K = 8.314/28 kJ/kg.K = 8314/28 J/kg.K;
m = 1.5 mol/s = 1.5 x 28 g/s = 1.5 x 28/1000 kg/s;
A = πD2/4 = π(15 mm)2/4 = π(0.015 m)2/4;
p
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09 Feb 2012
2
RT ( m / A )
2
= 71.544
unit?
γ = 1.4; molecular mass = 28;
p = 600 kPa
m = 1.5 mol/s;
D = 15 mm
f
= 0.0066
T = 300 K
pL = ?
L = 11.5 m
Design equation:
4f L
D
2

 p L2 
pL 
 1  2   ln  2 

2 
RT ( m / A ) 
p 
 p 
p
2
(1.3)
2

 p L2 
pL 
20.240 = 71.544  1  2   ln  2 
p 

 p 
p = 600 kPa = 600,000 Pa
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09 Feb 2012
pL = ?
Solve the nonlinear equation
above to determine pL
2

 p L2 
pL 
20.240 = 71.544  1  2   ln  2 
p 

 p 
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.
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09 Feb 2012
Now, solve the nonlinear equation for pL values close to 508.1 kPa:
2

 p L2 
pL 
20.240 = 71.544  1  2   ln  2 
p 

 p 
p = 600 kPa = 600,000 Pa
pL kPa
510
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09 Feb 2012
LHS of the above RHS of the above
equation
equation
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
Problem 4 continued:
Rework the problem in terms of Mach number and determine ML.
γ = 1.4; molecular mass = 28;
p = 600 kPa
m = 1.5 mol/s;
D = 15 mm
f
= 0.0066
T = 300 K
ML = ?
L = 11.5 m
Design equation:
4f L
D
4f L
D

1
 M
2
2
L

M
  ln 

M
2
2
L



(1.4)
= 20.240 (already calculated in Problem 4)
M=?
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09 Feb 2012
2

M
 1 
M

γ = 1.4; molecular mass = 28;
m = 1.5 mol/s;
D = 15 mm
p = 600 kPa
f
= 0.0066
ML = ?
T = 300 K
L = 11.5 m
u
M= c =
M 
u
 RT
=
4 m
RT
D p

2
m
A
1
 RT
m
= A
4 (1.5x 28/1000 kg/s)
=
π (15/1000 m)2 (600,000 Pa)
= 0.1
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09 Feb 2012
RT
p
(
1
 RT

m
RT
Ap

(8314/28)(300) J/kg
1.4
0.5
)
γ = 1.4; molecular mass = 28;
p = 600 kPa
m = 1.5 mol/s;
D = 15 mm
f
= 0.0066
ML = ?
T = 300 K
L = 11.5 m
Design equation:
4f L
D
20.240 

1
 M
2

M
 1 
M

1
(1.4)(0.1)
2
2
L

M
  ln 

M
2

( 0 . 1)
 1 
2
M
L

ML = ?
R. Shanthini
09 Feb 2012
2
2
2
L



(1.4)

 ( 0 . 1) 2
  ln 
2
M
L





Solve the nonlinear equation
above to determine ML
20.240 
1
(1.4)(0.1)
2
2

( 0 . 1)
 1 
2
M
L


 ( 0 . 1) 2
  ln 
2
M
L





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.
R. Shanthini
09 Feb 2012
Now, solve the nonlinear equation for ML values close to 0.118:
20.240 
1
(1.4)(0.1)
pL kPa
0.116
R. Shanthini
09 Feb 2012
2
2

( 0 . 1)
 1 
2
M
L


 ( 0 . 1) 2
  ln 
2
M
L





LHS of the above RHS of the above
equation
equation
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
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.
R. Shanthini
09 Feb 2012
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:
p  ( m / A ) RT
(1.5)
in flows where p decreases along the flow direction, and
p  ( m / A ) RT
(1.6)
in flows where p increases along the flow direction.
R. Shanthini
09 Feb 2012
Differential equation of Problem 2:
4f
dx  
D
2
RT ( m / A )
2
pdp 
2
dp
p
(1.2)
can be rearranged to give
dp
4f /D

dx

2
RT ( m / A )
2
p
2

( 2 f / D ) pRT ( m / A )
2
2
RT ( m / A )  p
p
In flows where p decreases along the flow direction
dp
dx
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09 Feb 2012
0
2
2

RT ( m / A )  p  0
p  ( m / A ) RT
(1.5)
2
Differential equation of Problem 2:
4f
dx  
D
2
RT ( m / A )
2
pdp 
2
dp
p
(1.2)
can be rearranged to give
dp
4f /D

dx

2
RT ( m / A )
2
p
2

( 2 f / D ) pRT ( m / A )
2
2
RT ( m / A )  p
p
In flows where p increases along the flow direction
dp
0
dx
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09 Feb 2012
RT ( m / A )  p  0
2
2
p  ( m / A ) RT
(1.6)
2
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.
40 atm
97oC
500 m/s
Air: γ = 1.4; molecular mass = 29;
p*=?
L
R. Shanthini
09 Feb 2012
Air: γ = 1.4; molecular mass = 29;
40 atm
97oC
500 m/s
p*=?
L
dp
0
p  ( m / A ) RT
0
p  ( m / A ) RT
dx
dp
dx
Limiting pressure: p *  ( m / A ) RT
R. Shanthini
09 Feb 2012
Air: γ = 1.4; molecular mass = 29;
40 atm
97oC
500 m/s
p*=?
L
p *  ( m / A ) RT
( m / A )  ( A  u ) entrance / A  (  u ) entrance
p* 
 pu entrance
RT
=
R. Shanthini
09 Feb 2012
RT 
 p

 
u

 RT  entrance
 pu entrance
RT
(40 atm) (500 m/s)
[(8314/29)(273+97) J/kg]0.5
= 61.4 atm
 pu  entrance
RT
40 atm
97oC
500 m/s
Air: γ = 1.4; molecular mass = 29;
p*=61.4 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.
Adp  m du   w dA w  0
R. Shanthini
09 Feb 2012
Problem 8 from Problem Set 1 in Compressible Fluid Flow:
Show that the equations in Problem 6 are equivalent to the
following:
M  1/

(1.7)
in flows where p decreases along the flow direction
M  1/

(1.8)
in flows where p increases along the flow direction
R. Shanthini
09 Feb 2012
In flows where p decreases along the flow direction
dp
p  ( m / A ) RT
0
(1.5)
dx
Since p 
m
RT
AM

we get
m
RT
AM

M  1/
R. Shanthini
09 Feb 2012
 ( m / A ) RT

(1.7)
In flows where p increases along the flow direction
dp
0
p  ( m / A ) RT
dx
Since p 
m
RT
AM

we get
m
RT
AM

 ( m / A ) RT
M  1/
R. Shanthini
09 Feb 2012
(1.6)

(1.8)
Summary:
x
dp
0
p  ( m / A ) RT
and
M  1/

0
p  ( m / A ) RT
and
M  1/

Limiting Mach number:
M *  1/

dx
dp
dx
Limiting pressure: p *  ( m / A ) RT
R. Shanthini
09 Feb 2012
Limiting Mach number for air:
x
For air, γ = 1.4
dp
M *  1 / 1 .4
 0 is associated with
= 0.845
M  1/

dx
If M < 0.845 then pressure decreases in the flow direction.
That is, the pressure gradient causes the flow.
dp
 0 is associated with
M  1/

dx
If M > 0.845 then pressure increases in the flow direction.
That is, momentum causes the flow working against the
R. Shanthini
pressure
gradient.
09
Feb 2012
Problem 9 from Problem Set 1 in Compressible Fluid Flow:
Show that when the flow has reached the limiting pressure
p *  ( m / A ) RT
or the limiting Mach number

the length of the pipe across which such conditions are reached,
denoted by Lmax, shall satisfy the following equation:
M *  1/
4 f L max
D
 p2

 p *2
 
 1   ln 
2
2
p
*
p



 1  M 2
 
 ln  M
2
 M

2

where pressure p and Mach number M are the conditions of the
flow at the entrance of the pipe.
R. Shanthini
09 Feb 2012
p; M
p*; M*
Lmax
Start with the following:
4f L
D
2

 p L2 
pL 
 1  2   ln  2 

2 
RT ( m / A ) 
p 
 p 
p
2
(1.3)
Substitute L = Lmax and pL = p *  ( m / A ) RT in (1.3) to get
4 f L max
D
R. Shanthini
09 Feb 2012
 p2

 p *2
 
 1   ln 
2
2
p
*
p






(part of 1.9)
p; M
p*; M*
Lmax
Start with the following:
4f L
D

1
 M
2

M
 1 
M

2
2
L

M
  ln 

M
2
2
L



(1.4)
Substitute L = Lmax and ML = M *  1 /  in (1.4) to get
4 f L max
D

1  M
 M
2
2
 ln  M
2

(part of 1.9)
Therefore,
4 f L max
R. Shanthini
09 Feb 2012
D
 p2

 p *2
 
 1   ln 
2
2
p
*
p



 1  M 2
 
 ln  M
2
 M

2

(1.9)