cme-535- chap 4 gas dynamics

GAS DYNAMICSGAS DYNAMICS

M.S Process & Mechanical M.S Process & Mechanical EngineeringEngineering33rdrd Semester Semester

Compressible Flow with Area Compressible Flow with Area ChangeChange

Steady 1-D Isentropic Flow With Area Steady 1-D Isentropic Flow With Area ChangeChange

SO 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. First we consider the effects of area change for adiabatic frictionless flows

Revise development of

equations, given in Table 3.1 and 3.2

Steady, 1-D Isentropic Flow With Area Steady, 1-D Isentropic Flow With Area Change, Change,

Governing Eqs.Governing Eqs.Governing equation for Steady 1-D Flow With Area Change;Assumptions:1. No body forces, gdz = 02. No friction, Ff = 03. No heat Transfer, adiabatic flow, Q = 04. No drag force, D = 05. No work done, W = 0

Continuity Equation

constantm AV

Momentum Equation

0dp VdV

Energy Equation

2

constant2Vh H

Entropy Equation constants

Steady, 1-D Isentropic Flow Process With Area Steady, 1-D Isentropic Flow Process With Area ChangeChange

For An Isentropic Process 1. all the thermodynamic states lie on a vertical

line on h-s plot2. Upper limit of enthalpy is point a, showing

zero KE and h=H3. Lower limit of enthalpy is point b, showing h=0

but maximum isentropic speed4. All the stagnation properties H, P, T, o and so

are constant.5. All this is due to only one driving potential i.e

area change

maxV

Comparison Of Adiabatic, Frictionless And Adiabatic, Frictional Flow 1. Differential equations are valid for both processes2. Difference lies in the integration of these equations3. For given set of initial conditions both processes have different final

conditions

Comparison, Isentropic Comparison, Isentropic && Non-isentropic Non-isentropic Processes, contd.Processes, contd.

Graphic representation

Expansion Compression

Comparison contd.4. Integrate energy eq. for isentropic flow

5. Integrate energy equation for adiabatic flow with friction

6. Intuition tells that for frictional flow the maximum velocity thee fluid can attain will be less than that for isentropic flow

7. Consequently one can say that the maximum enthalpy for frictional flow will always be greater than that for the isentropic flow

Note: point 6 and 7 is valid for positive or negative area changes, i.e.Expansion or Compression

Comparison, Isentropic Comparison, Isentropic && Non-isentropic Non-isentropic ProcessesProcesses

' '2 2

1 1

h V

h V

dh VdV

2 2

1 1

h V

h V

dh VdV

2 2V V

2 2h h


Mathematical Representation of Specific Enthalpy Change

2 22 1

1 2 2e

V Vh h h

2 22 1

1 2 2e

V Vh h h

2 21 2

2 1 2e

V Vh h h

2 21 2

2 1 2e

V Vh h h

Isentropic Flow, expansion

Isentropic Flow, compression

Adiabatic Flow with friction, expansion

Adiabatic Flow with friction, compression

Difference in specific enthalpy change for isentropic and adiabatic flow with friction = heat energy expanded to overcome friction

For Expansion

For Compression

1 2 1 2 1 2f e fh h h h E h Q

2 1 2 1 1 2f e fh h h h E h Q


Effect of Area Change on Flow Effect of Area Change on Flow PropertiesProperties

Continuity Eq.

constantm AV

Differentiate logarithmically

0d dA dVA V

Momentum Eq.

0dp VdV

Can be written as

2 0dp dVVV

Speed of Sound

2

s

pa

For isentropic flow

2a d dp

Combine the three equations and use the definition of Mach Number to get

2 1dA dVMA V

221dA p dpM

A V p

These equations suggest the manner in which area should change to accomplish the required expansion or compression of a compressible fluid

Effect of Area Change on Flow Properties, Effect of Area Change on Flow Properties, contd. contd.

(a) Nozzle flow/action

(b) Diffuser flow/action

The area change which shows the effect of increase in velocity with decrease in pressure

The area change which shows the effect of decrease in velocity with increase in pressure

Relationship between dA & dM for Steady 1-D Isentropic flow

dA M< 1.0 > 1.0

dA < 0 dM > 0 dM < 0dA > 0 dM < 0 dM > 0

Tabl

e 2 1dA dVMA V

Effect of Area Change on Flow Properties, Effect of Area Change on Flow Properties, ChokingChoking

WHAT WOULD HAPPEN ONCE THE

SONIC CONDITIONS ARE REACHED & AREA CHANGE CONTINUES

For subsonic/supersonic flowOnce M=1 at a particular section of converging section then What would be M2, whereas dA < 0

There can be two possibilities/assumptions1. M2 < M1

2. M2 > M1

For: M2 < M1

Table suggests, that for dA < 0 dM > 0

Contradiction with basic assumption, M2 < M1

dA M< 1.0 > 1.0

dA < 0 dM > 0 dM < 0

For: M2 > M1

Table suggests, that for dA < 0 dM > 0

Contradiction with basic assumption, M2 > M1

Effect of Area Change, Choking, Effect of Area Change, Choking, contd. -2contd. -2

The Table Is In Fact A Representation Of The Conservation Laws Under Isentropic Conditions

THESE CONSERVATIONS ARE NEVER VOILATED BY NATURE

dA M< 1.0 > 1.0

dA < 0 dM > 0 dM < 0


WHAT ABOUT HAVING THE FOLLOWING CONFIGURATION AND;ONCE THE SONIC CONDITIONS ARE REACHED

Now consider the possibilities

ORM2 < M1 M2 > M1

Table suggests, that for dA > 0, dM < 0 : dA > 0, dM > 0

Table suggests, that for Both the possibilities the conservation laws are not violated

Once M=1 is achieved at the throat, then how the downstream M will change in the diverging section. We need to answer this query

dA M< 1.0 > 1.0

dA > 0 dM < 0 dM > 0


Ans: The downstream flow can be either subsonic or supersonicWhich one of this is possibleThe subsonic or supersonic flow in the downstream section depends on the downstream physical boundary conditions

at the exit section of the flow passage

IMPORTANT INFERENCES

For supersonic flow starting fromrest

One needs a C-D geometry

For subsonic flow starting from supersonic flow

This CD geometry is named asDe Laval Nozzle

Throat will give the max mass flow rate corresponding to throat velocity.

' * ' * *maxtV a m m m at A

Effect of Downstream Physical Boundary Effect of Downstream Physical Boundary ConditionsConditions

Properties and Property RatiosProperties and Property RatiosProperty Ratios for isentropic flow witharea change

Limiting Values of the Property Ratios for the Steady One-D Isentropic Flow of a Perfect Gas

Properties and Property RatiosProperties and Property RatiosFor CHOKED flow where M=1

The property ratios for isentropic flow leads to t*, p* and *

1 12 212* * * 2 2=a 1 1oRTV Rt a

12

* ** A A=

o

P PmaRT

12 12= 1

Critical Temperature Critical Pressure Critical Density

Critical / Max Mass flow rate

Critical Velocity

* 2= 1t T

1* 2= 1p P

1 1* 2= 1 o

Effect of Area Change On Flow Effect of Area Change On Flow PropertiesProperties

The Governing Equations For Isentropic Flow With Area Change

Eq. of Continuity 0d dA dVA V

Eq. of Motion

Eq. of Energy2

const2p pVc t c T 2-1 =0

p

dt VdV dt dVMt c t t V

Eq. of State p Rt 0dp d dtp t

Eq. of Speed of Sound 2a Rt1 02

da dta t

Eq. of Mach NumberVMa

0dM dV daM V a

Eq. of Stream Thrust

2 2 0dp dV dp V dp dVV dV MV p p p V

21pA M F2

2

2 01

d dp dA M dMp A M M

FF

Effect of Area Change On Flow Properties, Effect of Area Change On Flow Properties, contd. contd.

1. 7 ODE relating, dp/p. dt/t. d/ , dV/V, da/a, dM/M, dF/F to dA/A2. All the 7 equations are linear in derivative form3. Hence if one of the 8 properties is specified all the other can be found4. Choose dA/A as independent variable and write in matrix form5. Thus you get all the dependent variables as function of dA/Ae

2

2

12

2

2

0 0 1 1 0 0 01 0 0 0 0 0

00 1 0 1 0 0 0

01 1 1 0 0 0 0

00 0 0 1 0 0

00 0 0 1 1 1 0

021 0 0 0 0 1

1

dp p dA AM

dt tM

ddV Vda adM M

M d dA AM

F F

Solve to get the solution

Effect of Area Change On Flow Properties, Effect of Area Change On Flow Properties, contd. contd.

2

2

112

1

MdM dAM M A

2

21dp M dAp M A

2

21d M dA

M A

2

2

11

Mdt dAt M A

2

11

dV dAV M A

2

2

12 1

Mda dAa AM

2

11

d dAM A

FF

The coeff. of dA/A are called influence coefficients

Each coefficients of dA/A is the partial derivative of aParticular flow property w.r.t the driving potential dA/A

++--dF/ F-++-da/a+--+dV/V-++-d/-++-dt/t-++-dp/p+--+dM/M

M > 1M < 1M > 1M < 1dA > 0dA < 0

Property Ratio

The Case of Converging NozzleThe Case of Converging NozzleFor given areas of nozzle and back pressure po, we are interested in1.The Mach No at throat as function of pressure ratio2.The pressure ratio that will cause the nozzle to choke

Since flow is assumed to be isentropic, write the energy equation for section 1 and 2'

'constant2t

tV

H h

12' '2t tV H h

1 2 1 21 1' '' 2 21 1

1 1t t

t op pRTV aP P

1 21' ''

' '

2 11

t o tt

t t

V a pMa a P

1 21

''

2 11t

t

PMp

If the gas is perfect

1. If po is decreased for a constant P then there will be increasing flow in nozzle and pt

’ = po until pt’=po=p*

2. Once pt’ = p* then any further reduction in po will have no effect on pt

’ and pt’

will remain p*.3. At this condition the fluid will be flowing at its maximum speed, Vt

’ =a*

4. The critical pressure ratio p*/P does not get reduced to po/P but rather remains constant at p*/P, no matter how much po is reduced to.

5. Even if P is increased it will raise the p* for the same constant p*/P ratio6. Thus the throat Mach number will remain same equal to unity7. Mass flow rate m* will also remain critical for the corresponding P & T8. This applies to the converging nozzle and to throat section of CD nozzle

9. So for max isentropic throat speed and

, complete nozzling & flow choked

The Case of Converging Nozzle, The Case of Converging Nozzle, contd. -2contd. -2

IMPORTANT INFERENCES

1* 2= 1p P

1 21 1 2'' * 2 21

1 1t

t opRTV a aP


Isentropic Mass Flow Rate:

Apply continuity equation

For perfect gas density can be found as

Hence isentropic throat mass flow rate is

12' ' ' '2t t t t t tm A V A H h

1 1' '' = t tt o

p pPP RT P

1212

' '2' tA P 2 1

1RTt tp p

mP P

' tA PRT

m


Effect of decreasing po with constant P

There can be 3 possibilities1. po = pb = p*

2. po > p*

3. po < p*

Cases of po p*

Jet issues from nozzle in parallel cylindrical stream

Case of po < p*

1. Gas expands outside the nozzle2. This expansion is explosive3. Gas particles are accelerated radially4. Cause low pressure in jet core5. Making particles move back6. Thus giving a periodic thin & thick sections

The Case of Converging Nozzle, contd. -5The Case of Converging Nozzle, contd. -5Effect of increasing P with constant po

Once P in slightly more than po flow starts as subsonicFurther increase in P increases flow rate till pt = p*

Now further increase in P has no effect on the pressure ratio po/P

With = 1.4 , the critical pressure ratio remains constant at 0.52828

112oP p

1 12 2

1 21'' *2 2 21

1 1 1t

t opRT RTV a aP

Choked nozzle gas speed for at throat

Increase in P for constant T, increases the local density causing increase in mass flow rate. Increase in mass flow rate is linear with increase in P

However for critical conditions at throat volumetric flow rate is independent of P, as Q* = m* / *. If T is also constant then Q also remains constant

MULTIDIMENSIONAL FLOW EFFECTS1. Assumption of 1-D flow is not valid for large angled nozzles, giving non uniform flow2. Large angles cause the inward radial momentum to form vena contracta outside the

nozzle, Vena contract is also called aerodynamic throat3. Vena contracta shape is affected by the surrounding atmospheric conditions. Even for

pressure ratios larger than the critical ratio

The Case of Converging Nozzle, contd. -6The Case of Converging Nozzle, contd. -6

Line of constant M for 40o c-nozzle and r=4Expt. Sonic lines for 25o c-nozzle

The Case of Converging Nozzle, contd. -7The Case of Converging Nozzle, contd. -7Flow non-uniformities and vena contacta causes reduced flow rates

giving rise to discharge coefficients

1 D isentropic

actualD

mCm

Experimental discharge coefficients for conical converging nozzles

Only CD nozzles can generate supersonic flowsThe converging part makes the flow criticalThe diverging part then makes the flow supersonicWe need to specify property ratios at exit of nozzle as function of pressure ratio

The Case of C-D NozzleThe Case of C-D Nozzle

The Case of C-D Nozzle, contd. -2The Case of C-D Nozzle, contd. -2We earlier calculated the isentropic discharge speed of gas as

1 21'' 2 1

1pRTVP

12

1 2 1 21 1 1' ' ''

max2 21 1 1

1 1e e e

e op p pRTV a VP P P

Similarly the exit Mach number1 21 2 11' '

'

2 21 11 1

e o ee

e e e

V a p PMa a P p

Similarly the exit Mass flow rate

* * *t t e em m A a A V ' tA P

RTm

*

tA P*RT

m

OR

Plot of various ratios of exit to stagnation conditions as function of pressure ratios

The Case of C-D Nozzle, contd. -3The Case of C-D Nozzle, contd. -3

Area Ratio for Complete Expansion in terms of pressure ratio


11

1 2

11 21

' *

* *t t

A P 2 1RT 1 1A AA P* A A

1RT

mm p p

P P

Area Ratio for Complete Expansion in terms of exit Mach number

1

2 1

2*

t

A A 1 2 11A A 1 2

MM

Similarly Velocity ratio of exit to the throat or critical conditions1 21' '

**

1 11

V pMa P


1. When mass flow rate is less than choked mass flow rate then throat area is greater than critical throat area and throat M <1

2. In order to find the exit area put A = Ae and useBy putting p/ =pe and M = Me

3. To get max isentropic discharge speed put pe = po. This is done by designing the nozzle in such a way that Ae/A* gives pe = po

Such a nozzle has optimum area ratioOR

The area ratio gives complete expansion

eq.

Effect of Back Pressure on C-D NozzleEffect of Back Pressure on C-D Nozzle

Under-expansion in CD NozzlesUnder-expansion in CD Nozzles

If the nozzle expands the gas passing through it such that pe = po then the velocities in the diverging section are supersonic through out provided the nozzle is passing maximum mass flow rate through its throat.

If for maximum flow rate condition and given fixed geometry, pe > po, then the gas expands outside the nozzle to po as in the case of only converging nozzle.

Under such conditions the K.E of gas coming out of the nozzle < the K.E of gas expanding isentropically and completely

Thrust produced by such nozzles is less than corresponding to complete expansion

Such CD nozzles are called under-expanding nozzles

Over-expansion in CD NozzlesOver-expansion in CD Nozzles• Similarly if the back pressure is greater than the exit pressure (po > pe), the gas is over-

expanded and the CD nozzle is called over-expanding nozzles• Thrust produced by such nozzles is also less than that corresponding to complete

isentropic expansion• In order to meet with po ( which is not transmitted backward into the nozzle), the gas has

to be compressed through series of shock waves, as discussed earlier

1. There are regions of subsonic flow in the Boundary Layers

2. Information of po gets transmitted through this region

3. If po is slightly greater than pe then oblique shock waves are formed at the edge of nozzle

4. As po is raised further then these oblique shock waves get stronger and cause the jet to separate. Reason being the adverse pressure gradient that exists between the edge of nozzle and inside the nozzle

5. Separation of jet needs regulation for rocket motors, as they operate over an extremely large range of altitude

Over-expansion in CD Nozzles, contd. Over-expansion in CD Nozzles, contd. -2-26. Plot of experimental data of over-expanding nozzles

7. Only one isentropic complete expansion “M” and if p0 is increased from that corresponding to “M” then sharp increase in pressures / shock waves are observed.

8. Shock waves become stronger & even move inside nozzle if po is raise sufficiently raised9. The separation pressure ps 0.4 po and depends on BL character & divergence angle 10. For a given , ps depends on pressure ratio r

Jet Propulsion EnginesJet Propulsion Engines1. Based on the Newton third Law.2. Examples, all types of sea ships or aircrafts etc.3. How to get required reaction force. By giving sufficient momentum to a mass of fluid4. Reaction to time rate of increase in momentum of fluid produces a force called thrust5. What are the available means for propelling a body in a fluid medium differ only in the

manner increase in momentum is imparted to the working fluid6. The working fluid to receive the increase in momentum, water, steam, air, gases

produced by chemical reactions, charged particles or their combinations7. Selection of the working fluid depends on the nature of propulsion problem8. All the propulsion systems are classified by the way the working fluid is handled

1. Propeller propulsion: where the momentum is imparted to the working fluid when it passes through the propeller. The working fluid moves around the body to be propelled

2. Jet Propulsion: where the momentum is imparted to the working fluid by ejecting it as a high speed jet from with in the body to be propelled

9. The jet propulsion devices are further classified into two classes1. The Air Breathing Engines2. The Rocket Engines

Jet Propulsion Engines, contd. -2Jet Propulsion Engines, contd. -2THE AIR BREATHING ENGINES1. They involve the normal combustion process. 2. Fuel is mixed with air drawn into the engine from the atmosphere3. The air fuel mixture is burnt and the temperature of the gases produced after

combustion is increased to the desired value.4. These hot gases are then ejected out of the engine through nozzles to get the

required propulsion and hence thrust5. Turbojet, turbofan, ramjet and SCRAM jet are typical examples of this class

Choose control volume given by dashed line1. DA = Additive Drag2. F =Net force acting on Engine surface3. Pressure forces will be acting on the inlet

and outlet areas of nozzles4. Momentum will be coming in and going out

from inlet and outlet of nozzle

Jet Propulsion Engines, contd. -3Jet Propulsion Engines, contd. -3

a o o e e e e a oF D p A p A mV m V

Net force acting on propulsion device

Pressure force acting on propulsion device inlet/outlet areas

Additive drag force if stream lines are curved

Momentum leaving / enteringthe propulsion device

Net force acting on propulsion device = nozzle e e e o eF mV p p A

Mass of fluid at outlet = e a fm m m Mass of fuel is generally 2-5% of air

Thrust per unit mass flow rate air = aa

FIm

Another performance parameter, thrust specific fuel consumption =fmTSFCF

TSFC should be as low as possible for economical reasons

Jet Propulsion Engines, contd. -3Jet Propulsion Engines, contd. -3The following assumptions are very much valid

1. Mass of fuel is negligible compared to the mass of air used2. DA is generally only few percent of the total thrust

3. For a well designed nozzle pe = po hence the pressure thrust should be zero

Hence the net thrust produced can be approximated as

a e oF m V V

Rocket EnginesRocket EnginesThey differ from the Air Breathing Engines in the following ways

1. They do not draw air from the atmosphere2. Total mass ejected out of the engine to produce thrust is carried with in the engine

Basic Features of A rocker Engine

A similar analysis on a control volume gives

o e e e e eF p A p A m V

e e e o eF mV p p A Pressure Thrust

Jet ThrustNet surface force

acting on the rocket

Similarly Specific thrust = spFIm

Similarly for Max Thrust Nozzle Force = eF mV

Performance of Propulsive NozzlesPerformance of Propulsive NozzlesNet thrust developed by a propulsive nozzle depends on1. Mass flow rate through the nozzle2. The exit area3. The exit static pressure4. The ambient pressure5. The velocity of fluid at exit

Essential features of propulsive nozzles

The converging propulsive nozzles

Performance of Propulsive Nozzles, Performance of Propulsive Nozzles, contd. -2contd. -2

The converging propulsive nozzles

Two Commonly Used Performance Criteria for Propulsive Nozzles

e o esp e

p p AFI Vm m

Specific Impulse

Thrust Coefficiente e o e

Ft t t

mV p p AFCPA PA P P A

One can use the isentropic flow relations and represent CF in terms of m* and Ve


Doing the necessary algebra leads to1

21* 2 1

1t e e o e

F oo t t

PA p p p AC aa PA P P P A

1

1 2112 2 1

1 1e e o e

Ft

p p p AC

P P P A

1

12

1

2

2

*

1211 12 1 1

2

211

e e eF

te e

e e

t

M p AC

P AM M

p AP A

FF


Plot of CF as function of e (area ratio) for = 1.2

IMPORTANT INFERENCES 1. For a given po/P there is a unique value of

area ratio that gives max thrust2. Maximum thrust occurs when po = 0, for

given e

3. Clearly max thrust occurs when such an area

ratio is chooses which gives pe = po

4. Around maximum thrust CF curve is quite flat

5. Thus slight adjustment in area ratios are possible without losing CF.

6. Thus one can design for reduced area and thus less weight and size

7. One can use the correction coefficients to account for the multidimensional effects

Thrust Reduction by Flow Divergence in Thrust Reduction by Flow Divergence in NozzlesNozzlesAxial thrust is first calculated on a differential exit area and then integrating over the total area

Integration and algebraic manipulation gives

21 cos2 e e e e o sF A V p p F

Design Criteria for Propulsive Nozzles for Design Criteria for Propulsive Nozzles for Max ThrustMax Thrust

Thrust developed by a propulsive nozzle is given as e e e o eF mV p p A

We need to find the nozzle exit area which should yield max thrust. Differentiating we get

e e e e e e o e

e e e e e e o e

dF m dV A dp p dA p dA

A dp V dV p p dA

Zero: Bernoulli’s equation

e oe

dF p pdA

Zero: For maximum thrusti.e pe = p0

Hence Area must be chosen so that e op pP P

AssignmentAssignmentChapter 1: 3, 4, 5, 10, 11, 13, 16, 17, 18, 19, 20, 21, 22, 23

Chapter 3: 1, 5, 6, 8, 9, 13, 14, 15, 16, 17, 18, 19

Chapter 4: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 17, 19,20

cme-535- chap 4 gas dynamics

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