gas dynamics-rayleigh flow

RAYLEIGH FLOW

GDJP Anna University

Constant area Duct flow with Heat Transfer and negligible friction is called Rayleigh flow (Simple diabatic flow)

•Many compressible flow problems encountered in practice involvechemical reactions such as combustion, nuclear reactions, evaporation,and condensation as well as heat gain or heat loss through the duct wall

•Such problems are difficult to analyze

•Essential features of such complex flows can be captured by a simpleanalysis method where generation/absorption is modeled as heat transferthrough the wall at the same rate

•Still too complicated for introductory treatment since flow may involvefriction, geometry changes, 3D effects

•We will focus on 1D flow in a duct of constant cross-sectional area withnegligible frictional effects

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RAYLEIGH FLOW





RAYLEIGH FLOW- Assumptions


To isolate the effects heat transfer we make the following assumptions:

1. The area of the flow passage or duct is constant.

2. The flow is steady and one-dimensional.

3. There is no work, body forces are negligible, and

the effects of friction are negligible.

4. Heat transfer is the only driving potential.




RAYLEIGH FLOW – Fundamental Equations


Continuity Equation

X-Momentum equation




RAYLEIGH FLOW – Fundamental Equations


Energy equationCV involves no shear, shaft, or other forms of work, and potential energy change is negligible.

For and ideal gas with constant cp, ∆h = cp∆T

Entropy change

In absence of irreversibility's such as friction, entropy changes by heat transfer only




RAYLEIGH LINE


•Infinite number of downstreamstates 2 for a given upstream state 1

•Practical approach is to assume various values for T2, and calculate all other properties as well as q.

•Plot results on T-s diagramCalled a Rayleigh line

•This line is the locus of allphysically attainabledownstream states

•S increases with heat gain topoint a which is the point of maximum entropy (Ma =1)




RAYLEIGH LINE


Various flow parameters on Rayleigh Line




Variation of fluid Properties





Rayleigh Flow Relations


Pressure Impulse Function

12 FF =

Stagnation Pressure Temperature




Cont..


Stagnation Temperature Density & Velocity

Change of Entropy Heat Transfer




Derivation Proof


Momentum

)1()1(

)1()1(

Therefore

that know We

)(

22

21

1

2

2221

21

1212

2221

222

211

22221

211

22221

1221

MM

pp

pMpM

pMpMpp

pMRTMRTpc

cc)p(pAcAc)Ap(p

ccm)Ap(p

γ

γ

γγ

γγ

γγρ

ρρρρ

+

+=

+=+

−=−

==

−=−⇒−=−

−=−•




Derivation Proof


Stagnation Pressure

1

212

11

222

11

211

211

0102

1

212

11

222

11

12

0102

122

110

−

−+

−+

×+

+=

−

−+

−+

=

−

−+=

γγ

γ

γ

γ

γ

γγ

γ

γ

γγ

γ

M

M

M

MPP

M

M

pp

PP

MpP




Derivation Proof


Temperature

(4) eqn 12

12

1p2p

222p , 11p that, know We

(3) eqn 2/1

21

2M1M

12

2c1c

2 & 1 eqn From

(2) eqn 2/1

21

12

11

221122

1M2M

number Mach From

(1) eqn 12

2c1c

2211m Continuity

1

TTSo

RTRTTT

TT

cc

RTc

RTcacac

AcAcFrom

×=

==

==

×===

=⇒==•

ρρ

ρρ

ρρ

γ

γ

ρρ

ρρ




Derivation Proof-Cont..


2

12

221

211

1T2T

eqn above the in 12

2

12

12

1T2T

2/1

12

21

1p2p

(4) in (3) eqn from gSubstituin

×

+

+=

×=⇒

×=

MM

M

M

ppsubstitute

MM

pp

TT

MM

γ

γ




SLOPE OF RAYLEIGH LINE


Slope of S=cons. Line

22222

2

2

21

21

22

212

1

2

)(tan

points, state two anyFor constant ..

MacGdvdp

AmGpp

GpGp

GpTKW

Rρρθ

νν

νν

ν

−=−==−=

−=−=

−−

+=+

=+

•

(1)




Cont..


(2)

1 2




CONS. ENTHALPY LINE


vp

dvdp

vdppdvRTpv

−=

=+===

0 , yields atingDifferentilineconstant Tfor ;constant Note:




Cont..





Problem 1


In your opinion, which assumption(s) in the Rayleigh flow analysis may be potential source(s) of error in solving a real life problem?Answer: •Heat transfer causes the total temperature to change significantly in the flow,which leads to a large variation of static temperature. The perfect gasassumption (constant specific heats) may not be appropriate. At higher andhigher temperatures, more and more energy modes are activated within themolecules. In general, this causes the specific heats to rise with temperature.

•In cases where combustion occurs, chemical composition of the constituentgases changes significantly. Reactant species will be consumed and productspecies will be produced. Their relative ratio changes as combustion proceeds.Values like gas constant R, and specific heat ratio, will no longer be constantbut depend on the extent of combustion.




Problem 2


Unchoked case: Air at 250 K and 1.0 bar is moving at 100 m/sec towards the entrance of a combustion chamber. Determine the exit conditions if 300 kJ/kg is added to the flow during the combustion process.

1 2 P= 1.0 barT= 250 KV= 100 m/s

Heat addition

Solution:

K T whichFromTT obtain table weflow isentropicthe From

M of number Mach inlet an gives This m/s. V K, T For

01

01

1

1

11

255

9805.0

3156.0100250

=

=

=

==




Problem 2- Cont..


bar 7588.0106.2598.1

1PP

K 5174427.09152.0

250TT

is state

598.1PP

,9152.0TT

and ,(subsonic) 5985.0M

of no. Machexit an to scorrespond this table, flow rayleigh the From

8173.03763.0172.2T

TT

TT

(2), stationAt

1TT

)(C q have we energy, of onconservati From

2.106 PP

4427.0TT

3763.0TT

obtain we table, flow Rayleigh the From

22

*11

*22

1

2

22

*11

*22

1

2

*2

2*2

22

*1

01

01

02*

02

02*

0101

*0202

01

02

0101

020102p

*1

1*1

1*01

01

=⇒=⇒=

=⇒=⇒=⇒

===

=×=×=⇒=

+=⇒−=

===

PP

PPPP

TT

TTTT

exitThe

TTTTTTT

TCqTT

o

p




Problem 2- Cont..


Just choked case: How much more heat that can be added without changing the conditions at the entrance to the cumbustor?Solution:

K 678)255(3763.01

is (2) stationat etemperatur stagnation the condition, choked thisAt thermally.it choke we before flow the into kJ/kg 124 extra an add can we Hence,

/ 42413763.012551004

1)TT(C q Therfore,

1M where TTTT additionheat max.for but,

)(C q have we equation, energy the FromTTT implies This .1M case, chokedjust theFor

0101

*01*

01*

0202

01

*01

0101*0p

*02

*01

*0max 0

0102p

*02

*01022

=

====

=

−×=

−=−=

====

−=

===

TTT

TTT

kgkJ

TT

TC

TT

p




Problem 2- Cont..


Choked case: Let us add sufficient fuel to the system so that the exit stagnation temperature is raised to 1500 K now. Assume that the receiver pressure is very low. What do you expect to happen in the system? Describe the flow both qualitatively and quantitatively.Solution: • In this case T02 = 1500 K > 678 K (choking condition)

• The original flow cannot accommodate this large amount ofheat. Something has to happen in order to take in so much heataddition. In other words, it cannot stay on the same Rayleighline.

•Recall that the upstream state can always communicate withthe downstream states in a subsonic flow by means of pressurewaves.




Problem 2- Cont..


• “Sensing” the super-critical heat addition downstream, the flow deceleratesfrom the free stream to the inlet. Spillage occurs ahead of the inlet. It is shownschematically as follows:

Heat addition∞

M=0.3156P=1.0 barT=250 K 1 2V=100 m/s

Spillage

•With a smaller flow rate in the combustion chamber, the flow moves to adifferent Rayleigh line with a smaller mass flow rate/A value.

•Since the receiver (back) pressure is very low, we can assume that the flow ischoked at the station (2), i.e. M2 = 1




Problem 2- Cont..


With M2 = 1 we conclude that

bar 04.1P 0.1Pthat know We932.0

9731.0PP

253T 250Tthat know We9805.09922.0

TT

is stateinlet The

9731.0PP

9922.0TT

table flow isentropic the from obtain we 1977.0M Withinlet theat 1977.0M to 3156.0 from sdecelerate flow The

3156.01977.0Mobtain we table, Rayleigh refering by

, ratio etemperatur above the From

17.0 1500255

T

T to leads This ; 1500

10

0111

10

0111

01

1

01

1

1

1

1

*01

01*02

*012

=∴===

=∴===

==

=

==

=<=

=====

∞∞∞∞

∞∞∞∞

∞

∞

barPPPP

KKTTTT

MM

KTTTo




Problem 2- Cont..


exit

exitupstream

inlet

Amsmaller•

higher ‘s’ due to moreheat addition

K 1250

2532024.01

1T1T

*1T *

1T*2T2T

be, to conditionexit the conclude can we choking, to due state reference the is

stateexit thethat Recall

275.2*1P1P&2024.0*

1T1T ; 1977.01M @

conditionexit determine To

=

×=×===

===

bar 0.45704.1275.21

1P1P

*1P*

1P*2P2P =×=×===




gas dynamics-rayleigh flow

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