Share this document with a friend

Description:

Gas Dynamics and Propulsion / BY Dr.G.KUMARESAN, / PROFESSOR, / ANNA UNIVERSITY

of 14
/14

Transcript

Isothermal flow with friction

GDJP Anna University

IntroductionØ The Fanno flow of an ideal gas through a constant area duct

under adiabatic condition is achievable in practice when theduct is not very long.

Ø Avoiding heat transfer to the environment is not convenientwhen a gas line is taken over long lengths.

Ø The supply of natural or by-product gases over long pipelines from an industrial area to a consuming city.

The pipe is exposed to the atmosphere and heattransfer through the pipe wall is a reality. The pipe gasattains the environment temperature by heat exchange.

Ø The study of isothermal flow of an ideal gas through aconstant area pipe is applicable to long pipe lines. Friction isaccounted for.

PDF created with pdfFactory trial version www.pdffactory.com

Cont..

GDJP Anna University

Ø The Reynolds number may be written in the form

where, G - mass velocity or mass flow densityD - duct diameterμ - viscosity of the fluid

For a constant temperature the viscosity of the flowingfluid μ is a constant. Since G is constant, the Reynoldsnumber is constant at all points in the flow.

Friction coefficient for a given pipe surface is a functionof the Reynolds number alone, it follows that in isothermalflow friction coefficient is invariant along the pipe.

µµρ GDVD

==Re

PDF created with pdfFactory trial version www.pdffactory.com

Isothermal flow process- T-s diagram

GDJP Anna University

Isothermal line

Stagnation temperature line

γ

1>M

γ

1=M

γ

1<M

PDF created with pdfFactory trial version www.pdffactory.com

Isothermal flow – Assumptions

GDJP Anna University

The following assumptions are used to derive

equations for T=constant flow :

1. Perfect gas

2. Constant diameter duct

3. Absence of body forces

4. Steady, one dimensional flow

5. Simple Diabatic, frictional flow at T=cons.

PDF created with pdfFactory trial version www.pdffactory.com

Isothermal flow – Fundamental equations

GDJP Anna University

To describe isothermal flow process, the following

governing relationships are generally used :

Ø Continuity equation

Ø Equation of state

Ø Energy equation

Ø Momentum equation

PDF created with pdfFactory trial version www.pdffactory.com

Cont..

GDJP Anna University

From continuity G =ρc = constantc

dcd−=

ρρ

From first law of thermodynamics 21 ; ohohodhQ ≠=δ

From equation of state p = ρRTρρd

pdp

=

From momentum equation

Ddxf

M

Mp

dp 4 212

2

−

−=γ

γ

PDF created with pdfFactory trial version www.pdffactory.com

Direction of isothermal flow process

GDJP Anna University

Ddxf

M

Mp

dp 4 212

2

−

−=γ

γ

Case (a) veM +=− 21 γ

γ

γ1M

12M

<∴

<givesThis

γ

γ

γ

112

2M-1 (b) Case

>∴

>

−=

M

MgivesThis

ve

γ

1*M

Case Limitingt

=

PDF created with pdfFactory trial version www.pdffactory.com

Variation of flow parameters- Isothermal flow

GDJP Anna University

Parameter M < 1/√γ M > 1/√γ

Pressure

Density

Velocity

Mach number

Stagnation Temperature

Stagnation pressure

Decreases

Decreases

Increases

Increases

Increases

Decreases

Increases

Increases

Decreases

Decreases

Decreases

12M

12M

+>

+<

γ

γ

forDecreases

forIncreases

(Cooling)(Heating)

PDF created with pdfFactory trial version www.pdffactory.com

Property Variations- Isothermal Flow

GDJP Anna University

0.845 0.912

PDF created with pdfFactory trial version www.pdffactory.com

Isothermal Flow – Property ratios

GDJP Anna University

Velocity and Density

12

21

12

MM

CC

==ρρ

Stagnation Pressure

−+

−+

=222

11

212

11

0201

M

M

TT

γ

γ

Stagnation Temperature

1

22

211

21

211

12

02P01P

−

−+

−+

=

γγ

γ

γ

M

M

MM Impulse Function

221

211

12

21

M

MMM

FF

γ

γ

+

+×=

Change of Entropy

1M2M lnR

2p1p ln 12

=

=− RSS

Duct Length

21

max4max44MD

LfMD

LfDLf

−

=

PDF created with pdfFactory trial version www.pdffactory.com

Limiting Conditions

GDJP Anna University

Choking velocityØ As the gas flows along the pipe, its static pressure

decreases. Eventually a limiting condition is reached wherethe pipe cannot be increased in length without altering theupstream conditions; that is the flow has become choked.

Ø Hence the limiting or choking velocity for the isothermalflow of a perfect gas in a constant-area duct in thepresence of wall friction alone is

LT1T 1 L 1

1

==⇒=

=

Qγγ

γ

aorLaCLaLc

LM

PDF created with pdfFactory trial version www.pdffactory.com

Isothermal flow- Problem

GDJP Anna University

Air flows in a long pipe (diameter=0.150 m) under isothermal conditions. At the pipe inlet, the static temperature and pressure are 300 K and 3.5 bar, respectively, and the velocity is 175 m/s. The friction coefficient is 0.005

Calculate (a) the length of pipe required to choke the flow, (b) the limiting velocity and pressure, and (c) the length of pipe at the station where the Mach number is 0.60

Solution:(a) The inlet speed of sound and Mach number are

50.02.347

175

111

sec/ 2.3473002874.11*

1

===

=××====

acM

mRTaaa γ

PDF created with pdfFactory trial version www.pdffactory.com

Cont..

GDJP Anna University

21

max4max44MD

LfMD

LfDLf

−

=

We know that

mf

DD

Lat

05.6005.04

150.0807.04

807.0maxL 807.0maxf4

table flow isothermalrefer 50.01M

=×

×=

×=⇒=

=

(b) Limiting velocity and pressure

barMMpp

MMthatknowWe

sm

aMFor

07.284515.0

50.05.3211

*2p

21

1p2p

/ 44.2932 . 347 84515.0

***c 84515.01*M flow isothermal

=×=×==⇒=

=×=

=⇒==γ

PDF created with pdfFactory trial version www.pdffactory.com

Cont..

GDJP Anna University

(c) the length of pipe at the station where the Mach no. is 0.60

21

max4max44MD

LfMD

LfDLf

−

=

mL

LDLat

DLat

81.3005.04

150.0508.0

508.0299.0807.0150.0

4(0.005)

299.0f4 table isothermal from 60.02M

and 807.0f4 table isothermal from 50.01M

=×

×=∴

=−=

==

==

PDF created with pdfFactory trial version www.pdffactory.com

Recommended