CP502 Advanced Fluid Mechanics Compressible Flow Part 01_Set 02: Steady, quasi one-dimensional, isothermal, compressible flow of an ideal gas in a constant.

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)


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;


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;


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



= 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



= 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


= 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:


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 = 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



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



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


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


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.



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)


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)


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)



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;


0dx

dp RTAmp )/(

L

0dx

dpRTAmp )/(

Limiting pressure: RTAmp )/(*

40 atm97oC

500 m/sp*=?



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



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




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


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 )/(


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


0dx

dpRTAmp )/(

x

0dx

dpRTAmp )/( /1Mand

and /1M

Limiting pressure:

Limiting Mach number:

RTAmp )/(*

/1* M

Summary:


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:


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.


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)


Lmax

p; M p*; M*

Start with the following:

Substitute L = Lmax and ML = in (1.4) to get

(part of 1.9)

2

2

2

2

2ln1

1 4

LL M

M

M

M

MD

Lf

(1.4)

/1* M

22

2max ln

14 M

M

M

D

Lf

Therefore,

22

2

2

2

2

2max ln

1*ln1

*

4 M

M

M

p

p

p

p

D

Lf

(1.9)

CP502 Advanced Fluid Mechanics Compressible Flow Part 01_Set 02: Steady, quasi one-dimensional, isothermal, compressible flow of an ideal gas in a constant.

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