GAS DYNAMICSGAS DYNAMICS

MS Process & Mechanical MS Process & Mechanical EngineeringEngineering33rdrd Semester Semester

Chapter 5Chapter 5Compressible Flow with FrictionCompressible Flow with Friction

Steady 1-D Compressible Flow With FrictionSteady 1-D Compressible Flow With FrictionSO FAR WE HAVE DEVELOPED1. General equations governing steady 1-D flow2. These equations have driving potential terms for,

1. Area change2. Friction3. Heat Transfer4. Body forces

We Need To Quantify The Effects Of These Driving Potentials1. Generally all these potentials can be present 2. In most of the practical cases one of them is more important and rest can be

neglected3. Previously we considered the effects of area change for adiabatic frictionless

flows4. Now we consider the effects of friction in constant area ducts

Steady 1-D Compressible Flow With Friction Steady 1-D Compressible Flow With Friction Governing Eqs.Governing Eqs.

Governing equation for Steady 1-D Flow With Friction;Assumptions:1. No body forces, gdz = 02. Constant Area dA = 03. No heat Transfer, adiabatic flow, Q = 04. No drag force, D = 05. No work done, W = 06. Wall friction is the only driving force

Continuity Equation

=constantm VG VA v

Momentum Equation

2 4fdx 02Vdp VdV

D

Energy Equation

2

02Vdh d

Entropy Equation constants

Thermodynamics of Steady 1-D Comp. Flow Thermodynamics of Steady 1-D Comp. Flow With FrictionWith Friction

First consider only the thermodynamics of the problem.

Let us find out all the thermodynamic states that are possible for this process

Combine the continuity and energy equations

2 2 2

contant2 2V G vh h H

2 2 2 21 2

1 2 contant2 2G v G vh h H

Important comments1. Relationship of enthalpy and specific volume of a gas for a fixed G2. No explicit information about friction effects3. However, all the effects are hidden in the thermodynamic states that the fluid attains

during this frictional process4. Thus for a frictional process this equation should give a curve on h-s diagram not a

vertical straight line5. This curve is called FANNO LINE/CURVE

Thermodynamics of St. 1-D Comp. Flow With Thermodynamics of St. 1-D Comp. Flow With Friction, -2Friction, -2

1. To plot a Fanno Curve assume different values of specific volume(Specific volume increases as pressure decreases)(Since the process is not isentropic the new thermodynamic state will be at higher entropy, thus incorporating the effect of friction)

2. Calculate ‘h’ from above(Energy + continuity equation)

3. Find the corresponding entropy ‘s’ from gas tables

For a steady state process G = constant

A typical such Fanno curve for steam

Thermodynamics of St. 1-D Comp. Flow Thermodynamics of St. 1-D Comp. Flow With Friction, -3With Friction, -3

IMPORTANT INFERENCES FROM FANNO CURVE1. For a given fixed G as the length of the duct increases, the pressure at the far

end decreases. 2. With increased length the enthalpy and pressure decreases, entropy and flow

velocity/ specific volume increases as H is constant. i.e flow expansion3. Curve passes through a maximum entropy corresponding to a pressure pL

4. Thus if back pressure is reduced below pL then according to the equation the pressure and entropy should decrease further as one sees in the plot in its first glance

5. However, 2nd law of thermodynamics gets violated as entropy cannot be decreased, Hence the process lying on the lower portion are not possible if one starts from state 1

6. Lower portion is very much valid if from a lower enthalpy and lower pressure one moves towards higher enthalpy and pressure i.e Flow compression

7. Thus either you move from higher enthalpy and pressure to lower enthalpy and pressure or VICE VERSA for a given G you reach a limiting condition corresponding to maximum enthalpy and pressure pL.

8. This condition corresponds to choking condition, and limiting velocity

Limiting Velocity For a Fanno LineLimiting Velocity For a Fanno Line1. Around point of maximum entropy the Fanno Curve is more or less flat.2. Any infinitesimal change in the state of gas around this point can be assumed to

occur at constant entropy. Hence one may write

3. Similarly one can rewrite the Fanno curve equation for the maximum entropy point as

4. Solve the two equations to get

5. Rewrite this equation as

0L L L L Lt ds dh v dp

2 2L contant

2LG vh H 2

L Lv v 0Ldh G d

2L L L 0Lv dp G v dv

1 L

2 2L LV

2L Ld

2 L

L L

0LL

dp dV

22 2 *

L L

L LL

s

dp pV a ad

Limiting Velocity For a Fanno Line, contd. -Limiting Velocity For a Fanno Line, contd. -22

1. Hence for adiabatic 1-D compressible flow with friction in a constant area duct the entropy of the fluid is at its maximum when the fluid is moving with the local speed of sound OR ML = 1

2. The upper portion of Fanno curve has V < a* and M < 1 and the lower portion has V > a* and M > 1

3. The effect of friction addition is that 1. if the flow is initially subsonic it will be accelerated to sonic conditions2. If the fluid is initially supersonic then it will decelerate to the sonic conditions

4. On the upper portion of Fanno Line the velocity increases and pressure decreases5. On the lower portion of the Fanno Line the velocity decreases and pressure

increases

IMPORTANT INFERENCES

Limiting Velocity For a Fanno Line, contd. -Limiting Velocity For a Fanno Line, contd. -33

Limiting conditions at the maximum entropy point on a Fanno line.

Effect of increasing the flow resistance on flow along a Fanno line.

Limiting Velocity For a Fanno Line, contd. -Limiting Velocity For a Fanno Line, contd. -44

How would the flow will behave if length is increased i.e. friction is added• Flow starts from some static condition 1 and by addition of friction is accelerated to a*.• This change of state corresponds to a critical condition say L1.

• This would in turn give rise to an increase in entropy = sL – s1

• In length of duct is further increased then due to an increased loss of energy, there should be an increase in the entropy change.

• This entropy change cannot occur on the same F-Line starting from condition1• Hence the flow will shift to another F-Line corresponding to a new lesser G.• If you wish to maintain the G and still add length then the increase in entropy change can

only be accounted for if the initial starting conditions are changed.

Effect of decreasing the back pressureon flow along a Fanno line.

Inlet Properties for Fanno FlowInlet Properties for Fanno Flow

Dynamics of Steady 1-D Fanno FlowDynamics of Steady 1-D Fanno FlowThe momentum equation for Fanno Flow is rearranged to give the following form

2 1 4f dx 02

dp dvGv v

D

Either an Average/constant value of ‘f’ is needed or the functional dependence of f with respect to velocity, as it varies with increasing or decreasing velocity, only then we can integrate the above equation. For a mean ‘f’

22 2

11

1 4f Lln 02

vdp Gv v

D

2

2 11

L dx x x 2

1

1f fx

x

dxL

If relationship of p & is known then the dynamic equation can be finally integratedThis relationship comes from the equation of states for a perfect or imperfect gas

Can you get the Darcy Equation form 2

2 2

11

1 4f Lln 02

vdp Gv v

D

2 4fdx 02Vdp VdV

D

Fanno Line Equations for Perfect Gases in Fanno Line Equations for Perfect Gases in Different FormsDifferent FormsSince the Fanno Flow is an adiabatic process, H & T remains constant, Hence

As the Fanno Flow is not a frictionless process so p, P, and o will not be constant,

To relate the pressure & densities at two locations in Fanno Flow use constant G

Hence

21T= 1+ 2t M

21

2

212

11+ 211+ 2

Mtt M

1 1

2 2

1 1 11 1 1 11 1 1 1 2 2

1 1 1 1 2

= = = Constantp M apV pVmG V p M p M

A Rt Rt Rt Rt Rt

12

21

2 1

21 22

112

112

Mp Mp M M

12

22

2 1 2 1 1 1

21 2 1 2 2 21

112

112

Mv p t V Mv p t V M M

Fanno Line Equations for Perfect Gases in Fanno Line Equations for Perfect Gases in Different FormsDifferent Forms

The stagnation pressures are given as

One can also represent the entropy rises as

The impulse function

1211 2

P Mp

1-1 2 -1

2 22 2

2 2 1

2 21 1 21 1

1 11 12 2

1 11 12 2

M MP p MP p MM M

12 -1

21

2 1 1 2

22 12

112ln ln 112

Ms s P MR P M M

12

2211 22

221 2 12

111 211 1

2

MM M

M M M

FF

Friction Parameter for Fanno FlowFriction Parameter for Fanno FlowThe dynamic equation for Fanno Flow

2 4fdx 02Vdp VdV

D

22 4fdx 0

2dp dV MMp V

D

Friction Parameter a Function of M

After doing some necessary algebra 2

3 2

2 14fdx11

2

M dM

M M

D

Integrate from x=0 for M=M to x=L* for M=1, with average friction factor

* 22x=L *

220

11 14fdx 4f L 2ln 12 1

2x

MM

M M

D DValue of

Friction Parameter vs M for ’s

In Gas Tables

Length Vs Mach Number RelationshipLength Vs Mach Number RelationshipUltimately Our interest is to find a relationship of variation of M as length changes

One can reason that the following relation holds

OR

OR

Similarly one can also relate the property ratios in terms of Initial M and Initial and final velocities

1 2

* *4f L 4f L 4f L

M M

D D D

2 2

2 2 1 22 1

2 22 21 22 1

1114f L 1 2ln12 1

2

M MM M

M M M M

D

2 2

2 1 112

1 2 2

14f L 1 11 1 ln2 2

V VMM V V

D

Influence Coefficients for Fanno FlowInfluence Coefficients for Fanno FlowSimilar Method as specified in isentropic flow with area change

2 2

2

11 4fdx22 1

M MdMM M

D

2 2

2

1 1 4fdx2 1

M Mdpp M

D

2

2

4fdx2 1

d MM

D

4

2

1 4fdx2 1

Mdtt M

D

2

2

4fdx2 1

dV MV M

D

2 4fdx2

dP MP

D

2

2

4fdx2 1

d MM

D

FF

2 4fdx2

ds MR

D

Effect of Fanno Flow on Flow Properties in Effect of Fanno Flow on Flow Properties in Constant AreaConstant Area

Property Ratio M < 1 M > 1dp/p - +dt/t - +

d/ - +dP/P - -dV/V + -dM/M + -dF/F - -ds/cp + +

Fanno Line Tables for Perfect GasFanno Line Tables for Perfect GasFor calculation purpose limiting condition of M = 1 is employed

Steady Isothermal Fanno Flow in Constant Steady Isothermal Fanno Flow in Constant Area DuctArea Duct

Energy Equation for flowing fluids with work and heat transfer

With no work, isothermal flow and no body forces

Isothermal flow, no work & no body forces, the dynamic eq for isothermal Fanno flow

From continuity equation we have

From equation of state for perfect gas we have

2

02pVW Q c dt d gdz

2

2 pVQ d dH c dT

2 2

2

4fdx 4fdx 2 4fdx2 02 2V V v dVdp VdV vdp VdV dp

V V

D D D

= constant m V dv dVG VA v v V

dp dv dVp v V

Steady Isothermal Fanno Flow in Constant Steady Isothermal Fanno Flow in Constant Area DuctArea DuctCombine the equations to get relation between friction parameter and static pressure

Now integrate between the two stations

Similarly Stagnation Temperature, Pressure Ratios and Impulse functions ratios

2

2 2 4fdx 0v dppdppV p

D

2

2 2

2 2 constantvpvV G Rt

2 2 L1

21 1 1 2 0

2 4f2 dx 0v dppdppV p

D

2

2 21 21 22

1 1 1

4f L lnv pp ppV p

D

222 2

2 21 1 1

4f L 1 1 lnp pM p p

D

22

2

211

11+2

11+2

MTT M

-122

2 2

21 11

112

112

MP pP p M

22 22

21 1 1

1

1

p M

p M

FF

Steady Isothermal Fanno Flow in Constant Steady Isothermal Fanno Flow in Constant Area DuctArea Duct

2

2 1 04

dx vdp f V p

DWe can get

2

2 2 4fdx 0v dppdppV p

D

For point of chocking 0dxdp

1L L LV v p Rt

1L

aV

Flow chocked at subsonic speed

1LM

For1

LM Fluid is receiving

heat and vice versa

As

Integrating above eq for 1 to L

Putting in above result in the eq.

We get

Or in the form of M

Steady Isothermal Fanno Flow in Constant Steady Isothermal Fanno Flow in Constant Area DuctArea DuctIf the flow is incompressible

0 0 0Integrating, we get

so

Property Ratios for Steady Isothermal Fanno Property Ratios for Steady Isothermal Fanno Flow in Constant Area DuctFlow in Constant Area Duct