GAS DYNAMICSGAS DYNAMICS

M.S. Process Engineering and Mechanical M.S. Process Engineering and Mechanical EngineeringEngineering33rdrd Semester Semester

INTRODUCTIONINTRODUCTIONThe fluid properties of concern in Gas Dynamics are;1. Enthalpy2. Internal Energy3. Entropy4. Temperature5. Pressure6. DensityAll these properties can be referred to some reference state ??

What this reference state should be?A unique state, can always be achieved, if the path is definedIn Fluid Analysis this is referred as

TOTAL OR STAGNATION STATETOTAL OR STAGNATION STATEFluid properties referred to this state are

THE TOTAL OR STAGNATION PROPERTIESTHE TOTAL OR STAGNATION PROPERTIES

INTRODUCTION, contd,-2INTRODUCTION, contd,-2TOTAL OR STAGNATION STATETOTAL OR STAGNATION STATE

A State Where the Flow Velocity Is ZeroOR

A State Attained By The Fluid When It Is Decelerated To Zero Velocity

ORIt Is The Constant Static State From Which The Fluid Can Be Accelerated To The Actual State For A Given Flow.

Fig.1

Effect of PathEffect of PathIf Deceleration is done adiabatically then,Stagnation Enthalpy and Internal Energy are uniquely determinedIf Deceleration is done isentropically then,Stagnation Temperature, Pressure and Density are uniquely determined

INTRODUCTION, contd.-3INTRODUCTION, contd.-3Concept Of Reference State Is Independent Of 1. The Working Fluid2. The process undergoing investigation

1. The actual flow process can involve1. Work2. Heat3. Body Forces4. Friction

3. This reference state exists for every point along actual flow path, hence stagnation conditions are point functions

4. Stagnation properties do change from point to point. This change can be due to friction, area change, heat transfer or work.

5. Hence stagnation properties can be related to these flow driving potentials.

6. This reference state is in fact obtained from a pure imaginary or hypothetical deceleration of fluid

STAGNATION ENTHALPYSTAGNATION ENTHALPY

Steady 1-D energy equation

in differential form

Assumptions;1. Adiabatic flow= Q=02. no work done= W=03. No body

forces=gdz=0

2 221 2

1 202 2 2V VVdh d h h

Let the fluid at condition one decelerates to zero velocity at condition 2, then

2

2 21

2 10 2 2V

V VH h h h

Hence: Stagnation enthalpy at any point = static enthalpy + K.E at that point

Eq.1

202

VW Q dh d gdz

STAGNATION TEMPERATURESTAGNATION TEMPERATURETemperature attained by the fluid when decelerated to zero

velocitySteady 1-D energy equation (Q, W and gdz=0) in differential form

202

Vdh d

Fluid=Perfect Gas

202p

Vc dt d

For constant cp integration between points 1 and 2

2 21 22 2

V Vand Energy due to directed motion of molecules / bulk motion

1 2p pc t and c t Energy due to random motion of molecules

2 2 21 2

1 2 constant2 2 2p p pV V Vc t c t c t Eq.2

STAGNATION TEMPERATURE, contd.-2STAGNATION TEMPERATURE, contd.-2AN IMPORTANT INTERPRETATION OF ENERGY EQUATION

Energy equation (Q, W and gdz=0) shows interconvertibility of energies associated with directed motion and random motion

Mach number is in fact ratio of these two energies

2 2 2 2 2 21 1 2 2 constant2 1 2 1 2 1

V a V a V a

Using Cp = R/(-1) and a2 = RT , Eq.2 becomes

OR2 2 2 2 2 21 1 2 2

1 1 11+ = 1+ = 1+ = const2 2 2a M a M a M

OR

2 2 21 1 2 2

1 1 11+ = 1+ = 1+ = const=T2 2 2t M t M t M

For adiabatic flow process involving a perfect gas

the stagnation temperature

remains constant

STAGNATION TEMPERATURE, contd.-3STAGNATION TEMPERATURE, contd.-3Similarly if the deceleration process is isentropic, the final

temperature attained by the gas (V = 0, M = 0)the corresponding unique Stagnation Temperature

2 2 21 1 2 2

1 1 11+ = 1+ = 1+ = constant=T2 2 2t M t M t M

Values of t/T vs M for ’s, given gas tables undergoing isentropic processes

For a Perfect Gas, h=h(t)Eq.1

2

22 V 0

1= 1+ 2T t t M

Eq.3

2= 2p p

H VT tc c

Eq.4

2 2 22

21 1 1

2 2 2impp

V V Vt t M tc a R

Compare eqs. 3 and 4

T=t+timp= static temperature + impact temperature rise

STAGNATION TEMPERATURE, contd.-4STAGNATION TEMPERATURE, contd.-4

Hence, T is the Static Temperature that the gas will attain when decelerated isentropically to zero speed

For Perfect Gases T=H/cp , Hence any deceleration process leads to a unique H corresponding to a unique T. For other

gases where t=f(h,s) we need further restrictions of isentropic flow to define a unique stagnation temperature

Relative change in stagnation temperature Differentiate eq.3

2

22 V 0

1= 1+ 2T t t M

22

22

12

11 2

MdT dt dMT t MM

An Example: A airplane flies at constant speed of 900km/hr at 10000 ft altitude. The air bought to rest somewhere on fuselage. Find the temperature of air in the stagnation region and impact temperature rise. Assume air = 1.4

At 10000 m altitude t = 223.25 K so a=299.53 m/s M=0.8346.

So T = 457.8K and timp = 31.1K

STAGNATION PRESSURESTAGNATION PRESSUREIt is the static pressure corresponding to the stagnation temperature

Stagnation temperature comes through isentropic process Thus for perfect gas stagnation pressure can be calculated from

stagnation temperature through isentropic relations

1P Tp t

1

2

1

1

21

22

1 1 22

112

112

M

M

Mp Ppp p P

M

Similarly

Isentropic relationship for a Perfect Gas

Hence stagnation pressure is

One can relate two static pressures in isentropic flow of perfect gas

2 2

22

211 2

dP dp M dMP p MM

Value of p/P vs M for

’s In Gas Tables

1211 2

P Mp

STAGNATION DENSITYSTAGNATION DENSITYo the static density corresponding to stagnation state of the gas

oP

RT

Mathematically

Where P and T are at stagnation state

1 1211 2

o M

Values of o/ vs M for ’sreported in Gas Tables

STAGNATION ACOUSTIC SPEEDSTAGNATION ACOUSTIC SPEEDao stagnation acoustic speed corresponding to stagnation state of

gas

2oa RT

Mathematically

Remains constant for adiabatic flow as T is constant

2 1211

2o

a t Ma T

22 2

2 1 1oaV a

ENTROPY CHANGEENTROPY CHANGEWhen a gas changes its state under static conditions from t to T Its entropy should change also This change can be calculated by differentiating the equation of

entropy change for reversible process

1ln lnpp pR c

t ts c const c constpp

1lnptds c d

p

Entropy of fluid in a static state = stagnation entropy corresponding to that state Hence static entropy change between two states = stagnation entropy change between those two states Fig.2

1lno ptds ds c d

p

ENTROPY CHANGE, contd. -2ENTROPY CHANGE, contd. -2Differential of the equation of entropy change for reversible process

1

p

ds dT dPc T P

If the perfect gas is also calorically perfect i.e cp constantthen integration yields

2 22 1

1 1

ln lnpT Ps s s c RT P

For steady adiabatic flow, T is constant

22 1

1ln Ps s s R P

For steady adiabatic frictionless flow Both T and P are constant

You have isentropic flow, ∆s = 0

Hence for any frictional flow, ∆s > 0, so P2 < P1

2

1

expP sRP

CHARACTERISTIC SPEEDSCHARACTERISTIC SPEEDSMAXIMUM ISENTROPIC SPEED

Isentropic Discharge Speed of a Perfect GasWhen it expands isentropically from an infinite

reservoir where V=0, p=P, t=T to static pressure p=p/

1 212 11RT pV

P

V/

max when p/01 2 1 2

2 21 1oRTV a

This speed corresponds to the complete transformation of the KE associated with the random motion of molecules into directed KE

At this complete conversion of KE the static temperature of gas would be zero. A real gas would in fact liquefy before that speed is

attainedConcept applies only to the flow of gas, not to the motion of a body in gas. Bodies motion in gas is limited by the thrust available. For a body at rest and gas moving, the gas cannot attain speed more than V/ Useful relation between V/

max, V and a

22 2max

2 1 2VV a

CHARACTERISTIC SPEEDS, contd. -2CHARACTERISTIC SPEEDS, contd. -2CRITICAL SPEED OF SOUND

One of the forms of energy equation

2 2 2 2 2 21 1 2 2

1 1 11+ = 1+ = 1+ = const2 2 2a M a M a M

Isentropic expansion of a gas

• Its speed increases• Its temperature

decreases• Somewhere V2=a2

• At 2 M2 =1This a2 is called the critical speed of sound, a*

22 21 1 *1+ =2 2a M a

22 2 *1=2 1 2 1aV a 22 1 *= 2oa a

Thermodynamic Properties and Critical Thermodynamic Properties and Critical SpeedSpeed

All the required thermodynamic properties at a*, can be found

* 2T 1t

Similarly,

-1 -1* * 21

p tP T

1 1-1* * 2

1o

pP

Values of ratios of t*/T, p*/P, */ as functions of

Put t = t*

* 2 T1t

For air,p*/P = 0.5283t*/T = 0.833

*/ o =0.6339a*/ao = 0.9129

2

22 V 0

1= 1+ 2T t t M

FaheemAleem

Thermodynamic Properties and Critical Thermodynamic Properties and Critical Speed, contd.Speed, contd.

Critical speed of sound & stagnation acoustic speed can be relatedWhen a gas is expanded isentropically, its discharge speed is;

Substitute for stagnation acoustic speed '2 *2max

11

V a

Similarly one can relate the Mach Number as M* = V/a*

M* is called the dimensionless velocity, a flow characteristic

*22

*2 22

*2

21 1

12 1 11

MMM or M

M M

An interesting thing; 1. M0, M* 02. M1, M* 13. M, M* finite value

12

* 1lim1M

M

Similarly one can relate the stagnation property ratios in terms of M*

22max 1

2oaV

Different Forms of Continuity eq. for Perfect Different Forms of Continuity eq. for Perfect GasGas

Equation of continuity in different forms, if R & are constant

constantm AV pAV Rt

1 2 1 2

1 2 =ApM =constantRt Rt

pAVm pAV RtRt

Substituting for t and p in terms of T and P1 2

21ApM 1 =constantRT 2

m M

1 2 - 1 2 121APM 1 =constant

RT 2m M

1 2 12

APM T =constant11

2

mM

If a perfect gas is expanded isentropically then it will attain critical speed while flowing through an area A* called critical throat area

Critical Flow AreaCritical Flow Area

Writing continuity equation for an infinite reservoir and critical area

* * * constantm AV A a

* ** *

*

RtA aA V V

Writing *, t* in terms of local and stagnation property ratios

12

2

1 1

* 1

211 for isentropic flow

21

AA p p

P P

Critical Flow Area, contd. - 2Critical Flow Area, contd. - 2

Graphical Representation of this equation

Critical area ratio as a function of pressure ratio for = 1.4

1 2 1

2*

1 2 111 2

A MA M

Critical area ratio as a function of M reported in gas tables for isentropic flow

Critical Flow Area, contd. - 3Critical Flow Area, contd. - 3

*

)1(2/)1(2/1

/1

12

AARTPAm

At critical conditions, Where M = 1 )1(2/)1(2/1

**

12

RT

PAm

oaPA

RTPAm

*)1(2/)1(**

12

Where

)1(2/)1(

12

Effect of Compressibility on the Mass Flow Effect of Compressibility on the Mass Flow Rate Rate Change in the density of the fluid affects the mass flow rate

of the fluid

Mass flux is given as

There exist two extreme situations1.p0 / P = 12.p0 / P = 0 Mass flux is zero for both extremes It has a maximum value for a particular p/P ratio call critical

pressure ratio

What is the condition for that???

Condition for Maximum Isentropic Mass Condition for Maximum Isentropic Mass Flow Rate Flow Rate For a given mass flow rate

maximum value of mass flux occurs at throat where area is minimum

Differentiating w.r.t. x logarithmically

For throat dA/dx = 0

Hence

Now Neglecting body forces and dividing by dx

comparing