cp502 advanced fluid mechanics compressible flow part 01_set 02: steady, quasi one-dimensional,...
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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)
R. Shanthini 09 Feb 2012
SummaryDesign equations for steady, quasi one-dimensional,
isothermal,compressible flow of an ideal gas in a constant area duct with wall friction
0224
2 du
udp
udx
D
f
(1.1)
2
2
2
2
2
2
ln1)/(
4
p
p
p
p
AmRT
p
D
Lf LL
(1.2)
(1.3)
dpp
pdpAmRT
dxD
f 2
)/(
242
2
2
2
2
2ln1
1 4
LL M
M
M
M
MD
Lf
(1.4)
R. Shanthini 09 Feb 2012
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.
Problem 4 from Problem Set 1 in Compressible Fluid Flow:
p = 600 kPa
pL = ?
L = 11.5 m
D = 15 mm
m = 1.5 mol/s;
T = 300 K
f = 0.0066γ = 1.4; molecular mass = 28;
R. Shanthini 09 Feb 2012
2
2
2
2
2
2
ln1)/(
4
p
p
p
p
AmRT
p
D
Lf LL
(1.3)
Design equation:
f = 0.0066;
L = 11.5 m;
D = 15 mm = 0.015 m;
D
Lf 4= 20.240
unit?
p = 600 kPa
pL = ?
L = 11.5 m
D = 15 mm T = 300 K
m = 1.5 mol/s; f = 0.0066γ = 1.4; molecular mass = 28;
R. Shanthini 09 Feb 2012
p = 600 kPa = 600,000 Pa;
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;
T = 300 K;
A = πD2/4 = π(15 mm)2/4 = π(0.015 m)2/4;
2
2
)/( AmRT
p
= 71.544
m
unit?
2
2
2
2
2
2
ln1)/(
4
p
p
p
p
AmRT
p
D
Lf LL
(1.3)
Design equation:
p = 600 kPa
pL = ?
L = 11.5 m
D = 15 mm T = 300 K
m = 1.5 mol/s; f = 0.0066γ = 1.4; molecular mass = 28;
R. Shanthini 09 Feb 2012
= 71.54420.240
2
2
2
2
ln1p
p
p
p LL
p = 600 kPa = 600,000 Pa
pL = ?
2
2
2
2
2
2
ln1)/(
4
p
p
p
p
AmRT
p
D
Lf LL
(1.3)
Design equation:
Solve the nonlinear equation above to determine pL
p = 600 kPa
pL = ?
L = 11.5 m
D = 15 mm T = 300 K
m = 1.5 mol/s; f = 0.0066γ = 1.4; molecular mass = 28;
R. Shanthini 09 Feb 2012
= 71.54420.240
2
2
2
2
ln1p
p
p
p LL
p = 600 kPa = 600,000 Pa
This value is small when compared to 20.240. And therefore pL = 508.1 kPa is a good first approximation.
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
R. Shanthini 09 Feb 2012
= 71.54420.240
2
2
2
2
ln1p
p
p
p LL
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
Now, solve the nonlinear equation for pL values close to 508.1 kPa:
R. Shanthini 09 Feb 2012
Rework the problem in terms of Mach number and determine ML.
Problem 4 continued:
Design equation:
2
2
2
2
2ln1
1 4
LL M
M
M
M
MD
Lf
(1.4)
D
Lf 4= 20.240 (already calculated in Problem 4)
M = ?
p = 600 kPa
ML = ?
L = 11.5 m
D = 15 mm T = 300 K
m = 1.5 mol/s; f = 0.0066γ = 1.4; molecular mass = 28;
R. Shanthini 09 Feb 2012
M = uc = u
RT= 1
RTAm
=m 1
RTA pRT
RT
Ap
m
p = 600 kPa
ML = ?
L = 11.5 m
D = 15 mm T = 300 K
RT
pD
mM
2
4
=4 (1.5x 28/1000 kg/s)
π (15/1000 m)2 (600,000 Pa) ( )(8314/28)(300) J/kg1.4
0.5
= 0.1
m = 1.5 mol/s; f = 0.0066γ = 1.4; molecular mass = 28;
R. Shanthini 09 Feb 2012
Design equation:
2
2
2
2
2ln1
1 4
LL M
M
M
M
MD
Lf
(1.4)
ML = ?
p = 600 kPa
L = 11.5 m
D = 15 mm T = 300 K
Solve the nonlinear equation above to determine ML
ML = ?
20.240
2
2
2
2
2
)1.0(ln
)1.0(1
1.4)(0.1)(
1
LL MM
m = 1.5 mol/s; f = 0.0066γ = 1.4; molecular mass = 28;
R. Shanthini 09 Feb 2012
20.240
2
2
2
2
2
)1.0(ln
)1.0(1
1.4)(0.1)(
1
LL MM
This value is small when compared to 20.240. And therefore ML = 0.118 is a good first approximation.
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
R. Shanthini 09 Feb 2012
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
Now, solve the nonlinear equation for ML values close to 0.118:
20.240
2
2
2
2
2
)1.0(ln
)1.0(1
1.4)(0.1)(
1
LL MM
R. Shanthini 09 Feb 2012
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:
RTAmp )/(
in flows where p decreases along the flow direction, and
in flows where p increases along the flow direction.
RTAmp )/(
(1.5)
(1.6)
R. Shanthini 09 Feb 2012
Differential equation of Problem 2:
(1.2)dpp
pdpAmRT
dxD
f 2
)/(
242
22
2
2
)/(
)/()/2(2
)/(
2/4
pAmRT
AmpRTDf
pp
AmRT
Df
dx
dp
In flows where p decreases along the flow direction
0dx
dp0)/( 22 pAmRT
RTAmp )/(
can be rearranged to give
(1.5)
R. Shanthini 09 Feb 2012
Differential equation of Problem 2:
(1.2)dpp
pdpAmRT
dxD
f 2
)/(
242
22
2
2
)/(
)/()/2(2
)/(
2/4
pAmRT
AmpRTDf
pp
AmRT
Df
dx
dp
In flows where p increases along the flow direction
can be rearranged to give
0dx
dp0)/( 22 pAmRT
RTAmp )/( (1.6)
R. Shanthini 09 Feb 2012
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.
L
40 atm97oC
500 m/sp*=?
Air: γ = 1.4; molecular mass = 29;
R. Shanthini 09 Feb 2012
0dx
dp RTAmp )/(
L
0dx
dpRTAmp )/(
Limiting pressure: RTAmp )/(*
40 atm97oC
500 m/sp*=?
Air: γ = 1.4; molecular mass = 29;
R. Shanthini 09 Feb 2012
L
RTAmp )/(*
40 atm97oC
500 m/sp*=?
RT
puu
RT
puAuAAm entrance
entranceentranceentrance
)(/)()/(
RT
puRT
RT
pup entranceentrance *
=(40 atm) (500 m/s)
[(8314/29)(273+97) J/kg]0.5 = 61.4 atm
Air: γ = 1.4; molecular mass = 29;
R. Shanthini 09 Feb 2012
L
40 atm97oC
500 m/sp*=61.4 atm
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.
0 wwdAdumAdp
Air: γ = 1.4; molecular mass = 29;
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:
(1.7)/1M
/1M (1.8)
in flows where p decreases along the flow direction
in flows where p increases along the flow direction
R. Shanthini 09 Feb 2012
In flows where p decreases along the flow direction
(1.5)
RT
AM
mp
Since we get
RTAmRT
AM
m)/(
(1.7)/1M
0dx
dpRTAmp )/(
R. Shanthini 09 Feb 2012
In flows where p increases along the flow direction
(1.6)
RT
AM
mp
Since we get
(1.8)
RTAmRT
AM
m)/(
/1M
RTAmp )/( 0dx
dp
R. Shanthini 09 Feb 2012
0dx
dpRTAmp )/(
x
0dx
dpRTAmp )/( /1Mand
and /1M
Limiting pressure:
Limiting Mach number:
RTAmp )/(*
/1* M
Summary:
R. Shanthini 09 Feb 2012
For air, γ = 1.4 4.1/1* M = 0.845
If M < 0.845 then pressure decreases in the flow direction.That is, the pressure gradient causes the flow.
0dx
dp /1M
x
is associated with
is associated with 0dx
dp /1M
If M > 0.845 then pressure increases in the flow direction. That is, momentum causes the flow working against the pressure gradient.
Limiting Mach number for air:
R. Shanthini 09 Feb 2012
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:
RTAmp )/(*
/1* M
22
2
2
2
2
2max ln
1*ln1
*
4 M
M
M
p
p
p
p
D
Lf
where pressure p and Mach number M are the conditions of the flow at the entrance of the pipe.
R. Shanthini 09 Feb 2012
Lmax
p; M p*; M*
2
2
2
2
2
2
ln1)/(
4
p
p
p
p
AmRT
p
D
Lf LL
(1.3)
Start with the following:
Substitute L = Lmax and pL = in (1.3) to getRTAmp )/(*
2
2
2
2max *
ln1*
4
p
p
p
p
D
Lf(part of 1.9)