Rayleigh Flow - Thermodynamicsseitzman.gatech.edu/classes/ae3450/rayleighflow_two.pdfRayleigh Flow -...

1

AE3450Rayleigh Flow -1

School of Aerospace Engineering

Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.

Rayleigh Flow - Thermodynamics• Steady, 1-d, constant area, inviscid

flow with no external work but withreversible heat transfer(heating or cooling)

• Conserved quantities (mass, momentum eqs.)– since A=const: mass flux=ρρρρv=constant≡≡≡≡G– since no forces but pressure: p+ρρρρv2=constant– combining: p+G2/ρρρρ=constant

• On h-s diagram, can draw this Rayleigh Line

pTvρM

?

Q�




Rayleigh Line

• p+ρρρρv2=constant,�high p results in low v

h

s

p high,v low

M<1

(ρv)↑

(ρv)↓

smax

M=1• ds=δQ/T (reversible)

– heating increases s – cooling decreases s

• Change mass fluxnew Rayleigh line

– M=1 at maximum sp

M>1

p low,v high

heating

cooling

2




Rayleigh Line - Choking• Heat addition can only

increase entropy• For enough heating,

Me=1 (Qin=Qmax)– heat addition can lead to

choked flow• What happens if Qin>Qmax

– subsonic flow: must move todifferent Rayleigh line (– – –), i.e., lower mass flux

– supersonic flow: get a shock (– – –)

M1 Me

mQq ��=h

s

M<1

(ρv)↓

smax

M=1

M>1

heating

stays on same Rayleigh line: ρρρρv and p+ρρρρv2 also const. for normal shock




( )��

��

��

��

−γ+γ+δ−

−γ+=

AdA

DdxfM

TcM

M

M

MdM

op

21q1

211

22

2

2

2

2

Differential Mach Equations• Simplify (X.4-5) for f=dA=0

• sign changes

2

2

2

2

MdM

M1M

pdp

γ+γ−=

(X.22)

( )op

2

22

2

2

Tcq

M1

M2

11M1

MdM δ

−

��

��

� −γ+γ+=

(X.21)

2

2

2

2

p MdM

M1M1

cds

γ+−=

(X.27)

(X.26)o

o2

o

oT

dTM2p

dp γ−=

2

2

2

2

MdM

M1M1

hdh

TdT

γ+γ−==

(X.23)

2

2

2 MdM

M11d

vdv

γ+−=

ρρ−=

(X.24)

(X.25)opo

oTcq

TdT δ=

3




Property Variations• What can change sign in previous equations

– (1-M2) and δq– (1-γM2) in dh,dT for M<1

• h,T same as p,ρ unless M2<1/γ(low speed flow: T oppos. p,ρ)

• p, ρ opposite of v (p+ ρv2=const)• Heating: M→1; cooling opposite

• s, To just like q• po opposite of To and s

M2<γ-1

M<1 M>1

s, To

po

M, vp, ρh, T

δδδδq <0 <0>0 >0

M2>γ-1 δδδδq<0��coolingδδδδq>0��heating




Rayleigh Line Extremum

smax

M=1

• Combine X.23 and X.27

2

2

11

MM

ch

dsdh

p −γ−=

γ== 1for 0 M

dsdh

1for =∞= Mdsdh

M=γ-0.5hmax

h

s

M<1

M>1

heatcool

4




Integrated Mach Equations• Integrating (X.22-27)

�δ=�

δ= ��p

oopo

oc

qdTTcq

TdT

2

2

121

22

1

2

2

1

11

vv

��

��

�

γ+γ+=

ρρ=

MM

MM

(X.31)

22

21

1

2

11

MM

pp

γ+γ+= (X.29) 1

21

22

22

21

1

11

2

22

1

2

211

211

11

−γγ

��

�

�

��

�

�

−γ+

−γ+

γ+γ+==

M

M

MM

ppp

ppp

pp

o

o

o

o

(X.32)

2

1

22

22

21

1

2

11

��

��

��

��

�

γ+γ+=

MM

MM

TT

(X.30)

(X.28)( )p

oo cqTT 12

12 =−

(X.33)��

�

�

��

�

�

−γ+

−γ+��

��

��

��

�

γ+γ+=

21

222

1

22

22

21

1

2

211

211

11

M

M

MM

MM

TT

o

o

( )

��

�

�

��

�

�

��

�

�

γ+γ+

��

�

�=−

γ+γ 1

22

21

2

1

212

11ln

MM

MM

css

p(X.34)

(energy)




Solution Approaches• Two methods to “solve” Rayleigh problems, i.e., for state

change from 1→2• In both, need to “know” one state (including M) and something

about other state (1 TD property or q or M)• Method 1: direct solution of integrated equations (X.28-34)

– requires iterative solution• Method 2: tabular solutions using reference condition

– uses sonic reference condition (similar to A/A*and Fanno methods)

– sonic condition based on heat transfer required for flow to go sonic; M2=1, To2= To*

5




Direct Solution• Use two equations to find M change

– e.g., if state 1 and q12 known use X.28 and X.33

doesn’t matter how q changes along length, only total q required

M1 M2

mQq ��=1

12

1

2 1opo

oTc

qTT +=

1

2

21

222

1

22

22

21

211

211

11

o

oTT

M

M

MM

MM =

��

�

�

��

�

�

−γ+

−γ+��

��

��

��

�

γ+γ+

a) get To ratio from X.28

b) get M2 from X.33 (must know M1)requires iterative solution for M2

c) get rest of property changes M2 from X.29-34, e.g., from X.29

22

21

1

2

11

MM

pp

γ+γ+=




Cooled Duct Example

• Given: Nitrogen flowing in constant area duct, enters at M=3, T=250 K, p=50 kPa and exits with temperature of 200 K

• Find:1. M2

2. q (including direction)

• Assume: N2 is tpg/cpg, γ=1.4, steady, reversible, no work

T2=200K

mQq ��=M1=3.0T1=250Kp1=50 kPa

6




Cooled Duct - Solution• Analysis:

– M2

since started M>1, can’t be subsonic– q (energy: ins = outs)

( )21 oop TTcq −=

41.3,21.02 =� M

T2=200KmQq ��=M1=3.0

T1=250Kp1=50 kPa

1

22

1

22

22

21

11

TT

MM

MM =�

��

��

��

��

�

γ+γ+

( )80.0

250200

34.1134.11 2

2

2

22

2

==��

��

�

��

�

�

��

�

�

+

+ MM

��

��

� −γ+= 2211 2

11 MTTo

K700=

( )( ) KKTo 66541.32.01200 22 =+=

( )Kkmolkg

kmolKJ 66570028

831414.1

4.1 −−

=

kgkJ36=

��

��

� −+= 292

14.11250K

h

s

M<1

M>1

heatcool

M=1

(from X.30)




22

21

1

2

11

MM

pp

γ+γ+=

Reference Solution Method• Reference state has M=1, T≡T*, p≡p*,...

2

22

11

��

��

�

γ+γ+=

MM

TT

*(X.36)

(X.35)21

1Mp

p* γ+

γ+=

221

1vv M

M

*

* γ+γ+=

ρρ= (X.37)

12

2

211

211

11

−γγ

��

�

�

��

�

�

−γ+

−γ+

γ+γ+=

M

Mpp

*o

o

(X.38)

211

211

11

22

2

2 −γ+

−γ+��

��

�

γ+γ+=

MM

MTT

*o

o

(X.39)

��

��

�

��

��

�

��

��

γ+γ+=− γ

γ+1

22

11

MMln

css

p

*(X.40)

• Note: prop/prop* = f(M) only• Values for γ=1.4 in Appendix F, John

*�sonic, but not same state as A* or Fanno

• State 2→sonic in integ-rated equations (X.29-34)

7




Use of Tables• To get change in M, use change in property ratio (like

using A/A*), e.g.,

1

2

1

2

1

2

M*oo

M*oo

*oo

*oo

o

o

TT

TT

TTTT

TT == M1 M2

mQq ��=

1

12

1

2 1opo

oTc

qTT +=1) again get To ratio from X.28

2) look up To/To* at M1

• For example, if state 1 and q12 known

3) calculate To/To* at M2 *

o

o

o

o*o

oTT

TT

TT 1

1

22 ×=

4) look up M2 that corresponds to calculated To/To*




Heated Duct Example

• Given: Air flowing in constant area duct, enters at M=0.2 and given inlet conditions and a heating rate of 765 kJ/kg

• Find:1. Me, poe, Te

2. qmax for Me=1

• Assume: air is tpg/cpg, γ=1.4, steady, reversible, no work

Me

mQq ��=Mi=0.2Toi=300Kpoi=600 kPa

8




Heated Duct - Solution• Analysis: Me

kgkJq 765=Mi=0.2Toi=300Kpoi=600 kPa– Me

another solution for M=3.53, but since started with M<1, can’t be supersonic

(X.28, energy) oipoi

oeTcq

TT +=1

( ) 614.017355.054.3 ==

– poe

( )543

3001004

1076513

.K

kgKJ

kg/J =×+=

20.M*o

oi

io

eo*o

oeTT

TT

TT

=

=

ioio

*o

*o

oeoe p

pp

ppp = kPakPa

.. 552600

23461113511 ==

450.Me =�

(Appendix F)

M To/To* po////pοοοο∗∗∗∗0.19 0.15814 1.237650.20 0.17355 1.234600.21 0.18943 1.231420.44 0.59748 1.139360.45 0.61393 1.135080.46 0.63007 1.130823.53 0.61393 5.47054




Heated Duct Solution (con’t)– Te

– qmax

oioi

oe

oe

ee T

TT

TTT =

( ) KK...Te 1021300543450

2401

12

=+

=

( )��

�

�

��

�

�−=−= 1

11

o

*o

opo*opmax T

TTcTTcq

kgMJ.

.K

kgKJ 4311

17355013001004 =�

�

��

� −=

Me

kgkJq 765=Mi=0.2Toi=300Kpoi=600 kPa

requires choked exit, Toe=T*K

.KT*

o 1728173550

300 ==

KTT *o

* 14402

11 =��

��

� −γ+=

9




Combustor Example

• Find:1. Te and compare to value you

would calculate if neglected kinetic energy change

2. po loss

Me

airm�

Mi=0.2Ti=500K

airfuel m 0413.0m �� =

• Given: Air entering constant combustor, enters at M=0.2 and given inlet conditions.q=heating value of fuel (45.5 MJ/kgfuel) ×

air

fuel

mm�

�

• Assume: air is tpg/cpg, γ=1.4, steady, reversible, no work,)( 1

mm q usejust addition, massneglect can

air

fuel <<�

�




Combustor Solution• Analysis:

– Te

(energy)

oipoi

oeTcq1

TT +=

Me

airm�

Mi=0.2Ti=500K

airfuel m 0413.0m �� =

airfuelair

fuel

kgMJ88.1

kgMJ5.45

kgkg0413.0q =��

�

��

�=

( ) K504008.1K500M2

11TT 2ioi ==�

�

��

� −γ+=

( ) K2376715.4K504)504(1004

1088.11TT6

oioe ==��

��

� ×+=

60.0M818.0)1736.0(715.4TT

TT

TT

e*o

oi

oi

oe*o

oe =�===

(Appendix F)

( ) ( )K2216

6.02.01K2376

21M1TT 22

oee =

+=

−γ+=

even for this subsonic flow, kinetic energy reduces final T by 160K

? combustion just another kind of heat addition

10




Combustor Solution (con’t)– poe/poi Me

airm�

Mi=0.2Ti=500K

airfuel m 0413.0m �� =871.02346.110753.1

pp

pp

pp

oi

*o

*o

oe

oi

oe ===

(Appendix F)

• So heat addition (burning fuel) in an engine gives po loss as was case with friction

• As we lose po (~13% loss here)– less thrust– less work output




Choking• Heating of a flow can lead

to choking just like friction

shock inside

pe,sh

shock at exit

O

Up*/po

x

1

pe,R

p/po1

• For supersonic flow, depending on back pressure, can get– underexpanded flow– overexpanded flow– shock inside duct

M1>1 Mepe

pb

po1

Q�

Rayleigh Flow - Thermodynamicsseitzman.gatech.edu/classes/ae3450/rayleighflow_two.pdfRayleigh Flow -...

Documents

Transcript of Rayleigh Flow - Thermodynamicsseitzman.gatech.edu/classes/ae3450/rayleighflow_two.pdfRayleigh Flow -...