CP502 Advanced Fluid Mechanics Compressible Flow Lectures 1 & 2 Steady, quasi one-dimensional,...

CP502 Advanced Fluid Mechanics

Compressible Flow

Lectures 1 & 2Steady, quasi one-dimensional, isothermal,

compressible flow of an ideal gas in a constant area duct with wall friction

R. Shanthini 08 Dec 2010

What is a Mach number?

Definition of Mach number (M):

M ≡ Speed of the flow (u)

Speed of sound (c) in the fluid at the flow temperature

Incompressible flow assumption is not valid if Mach number > 0.3

RTc 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)


For an ideal gas,

Unit of c = [(J/kg.K)(K)]0.5

= [m2/s2]0.5 = m/s

= [kg.(m/s2).m/kg]0.5

M = u u

RTc=

Unit of u = m/s

= [J/kg]0.5 = (N.m/kg)0.5


constant area duct

quasi one-dimensional flow

compressible flow

steady flow

isothermal flow

ideal gas

wall friction

Diameter (D) 4/2DA is a constant

speed (u)

x

u varies only in x-direction

Density (ρ) is NOT a constant

Temperature (T) is a constant

Obeys the Ideal Gas equation

is the shear stress acting on the wall2/2ufw where is the average Fanning friction factorf

uAm Mass flow rate is a constant


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:

Re/16f

uD

ReD

A

m

D

D

m2

4

D

m4

Re/227.14f

D

f

7.3log0.4

110

Quasi one-dimensional flow is closer to turbulent velocity profile than to laminar velocity profile.


Ideal Gas equation of state:

mRTpV pressure

volume

mass

specific gas constant(not universal gas constant)

temperature

RTV

mp

Ideal Gas equation of state can be rearranged to give

RTp

kg/m3 J/(kg.K)

K

Pa = N/m2


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, ρ and u are the respective pressure, density and velocity at distance x from the entrance of the pipe.

0224

2 du

udp

udx

D

f

f

Problem 1 from Problem Set 1 in Compressible Fluid Flow:

(1.1)


p p+dp

u u+du

x dx

D

Write the momentum balance over the differential volume chosen.

cross-sectional area

is the wetted area on which shear is actingDdxdAw

steady mass flow rate

shear stress acting on the wall

(1)wwdAduumAdppumpA )()(

w


DdxdAw

Substituting

2/2ufw 4/2DA

uAm

Equation (1) can be reduced to 0 wwdAdumAdp

Since , and , we get

0)( wwdAududpA

04

2

2

dxD

fuududp

0224

2 du

udp

udx

D

f

(1.1)

p p+dp

u u+du

x dx

D

w


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.

m

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


The differential equation of problem (1) is

in which the variables ρ and u must be replaced by the variable p.

0224

2 du

udp

udx

D

f

(1.1)

uAm RTp

Let us use the mass flow rate equation and the ideal gas equation to obtain the following:

RT

p

Ap

RTm

A

mu

and and therefore

A

RTmpu

It is a constant for steady, isothermal flow in a constant area duct

0)( udppdupudp

dp

u

du


0224

2 du

udp

udx

D

f

(1.1)

RT

p

Ap

RTm

A

mu

,Using and

(1.2)dpp

pdpAmRT

dxD

f 2

)/(

242

in

we get

p

dp

u

du


(1.3)

p pL

LIntegrating (1.2) from 0 to L, we get

which becomes

2

2

2

2

2

2

ln1)/(

4

p

p

p

p

AmRT

p

D

Lf LL

dpp

dppAmRT

dxD

f LL p

p

p

p

L

2

)/(

242

0


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 numberat length L from the entry.


(1.4)

2

2

2

2

2ln1

1 4

LL M

M

M

M

MD

Lf


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.

(1.4)

2

2

2

2

2ln1

1 4

LL M

M

M

M

MD

Lf

(1.3)

2

2

2

2

2

2

ln1)/(

4

p

p

p

p

AmRT

p

D

Lf LL

We need to relate p to M!


We need to relate p to M!

which gives

= constant for steady, isothermal flow in a constant area duct

RT

AM

m

RTMA

RTm

uA

RTmRT

Au

mRTp

11

RT

A

mpM

2

2

2

2

2ln1

1 4

LL M

M

M

M

MD

Lf

Substituting the above in (1.3), we get

(1.4)


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.


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


CP502 Advanced Fluid Mechanics Compressible Flow Lectures 1 & 2 Steady, quasi one-dimensional,...

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Transcript of CP502 Advanced Fluid Mechanics Compressible Flow Lectures 1 & 2 Steady, quasi one-dimensional,...