rayleigh flow -1 school of aerospace engineering copyright © 2001 by jerry m. seitzman. all rights...

AE3450Rayleigh Flow -1

School of Aerospace Engineering

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

Rayleigh Flow - Thermodynamics

• Steady, 1-d, constant area, inviscidflow with no external work but withreversible heat transfer(heating or cooling)

• Conserved quantities– since A=const: mass flux=v=constant – since no forces but pressure: p+v2=constant– combining: p+(v)2/=constant

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

pTvM

?

Q




Rayleigh Line

• p+v2=constant,high p results in low v

h

s

p high,v low

M<1

(v)

(v)

smax

M=1• ds=Q/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




Rayleigh Line - Choking

• Heat addition can onlyincrease entropy

• For enough heating,Me=1 (Qin=Qmax)

– heat addition leads tochoked 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

(v)

smax

M=1

M>1

heating




Rayleigh Line – Mach Equations• Simplify (IX.4-5) for f=dA=0

• sign changes

2

2

2

2

M

dM

M1

M

p

dp

(X.21)

op

2

22

2

2

Tc

q

M1

M2

11M1

M

dM

(X.20)

2

2

2

2

p M

dM

M1

M1

c

ds

(X.26)

(X.25)o

o2

o

o

T

dTM

2p

dp

(X.24)opo

o

Tc

q

T

dT

2

2

2

2

M

dM

M1

M1

h

dh

T

dT

(X.22)

2

2

2 M

dM

M1

1d

v

dv

(X.23)




Property Variations• What can change sign in previous equations

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

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

• p, opposite of v (p+ v2=const)

• Heating: M1; cooling opposite

• s, To just like q

• po opposite of To

M2<-1

M<1 M>1

s, To

po

M, v

p, h, T

q <0 <0>0 >0

M2>-1




Rayleigh Line Extremum

smax

M=1

• Combine X.22 and X.26

2

2

p M1

M1

c

h

ds

dh

1Mfor 0

ds

dh

1Mfor ds

dh

M=-0.5

hmax

h

s

M<1

M>1

heatcool




A Solution Method• Need to integrate (X.20-26) to find how properties

change due to q– can integrate or use tables of integrated values

only )M(fT

T*o

o

2o

1o

22

21

T

T o

oq

0 op2

2M

M22

2

T

dT

Tc

q

M

dM

M2

11M1

M1

doesn’t matter how q changes along length

• Mach number variation

*o

o2

21

M22

2

T

Tln

M

dM

M2

11M1

1M

2

M1 Me

mQq

to tabularize solution,use reference condition: M2=1, To= To

*




Use of Tables• To get change in M, use change in To/To

*

(like using A/A*)

• Find values in Appendix F in John

so if you know q, To1 and M1,

1) To2/To1= 1 + q/(cpTo1)

2) look up To/To* at M1

3) calculate To/To* at M2

4) look up corresponding M2

1

M

21

M

2

M

M

2

22

2

1o

2o

21

22

21

22

dMexpdMexp

dMM

21

1M1

M11exp

T

T

(X.27)

12 M*o

o

M*o

o

1o

2o

T

T

T

T

T

T

M1 Me

mQq




Rayleigh Flow Property Changes• To get changes in T, p, po, ... can also use M=1

condition as reference condition (*)2

22

* M1

1M

T

T

(X.28)

2* M1

1

p

p

(X.29)

22

*

*M

M1

1

v

v

(X.30)12

2*o

o

21

1

M2

11

M1

1

p

p

(X.32)

21

1

M2

11

MM1

1

T

T2

22

2*o

o

(X.33)

1

22

p

*

M1

1Mln

c

ss

(X.31)

• Integrate X.21-26(also tabulated in Appendix F)




Example #1

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

• Find:

- Me, poe, Te

- qmax for Me=1

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

Me

mQq

Mi=0.2Toi=300Kpoi=600 kPa




Solution #1• Analysis: Me

mQq

Mi=0.2Toi=300Kpoi=600 kPa– Me

another solution for M=3.5, but since started M<1, can’t be supersonic

(energy)

54.3kg/J)300(1004

kg/J107651

Tc

q1

T

T

3

oipoi

oe

(Appendix F)

2.0M*o

io

oi

oe*o

oe

T

T

T

T

T

T

614.017355.054.3 45.0Me

(Appendix F)

– poe

kPa552kPa6002346.1

11351.1p

p

p

p

pp io

io

*o

*o

oeoe




Solution #1 (con’t)

Me

mQq

Mi=0.2Toi=300Kpoi=600 kPa

– Te

K10210405.1

K30054.3

45.024.0

1

1T

T

TT

M2

11

T

T

2oi

oi

oee

2e

e

oe

– Toe(choked)

kg

MJ43.11

17355.0

1K300

kgK

J1004

1T

TTcTTcq

1o

*o

1op1o*opmax




Example #2

• Find:

- Te and compare to value you would calculate if neglected kinetic energy change

- 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

m

m

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

)( 1m

mq usejust addition, massneglect can

air

fuel




Solution #2• Analysis:

– Te

(energy)

oipoi

oe

Tc

q1

T

T

Me

airm

Mi=0.2Ti=500K

airfuel m 0413.0m

airfuelair

fuel

kg

MJ88.1

kg

MJ5.45

kg

kg0413.0q

K504008.1K500M2

11TT 2

ioi

K2376715.4K504)504(1004

1088.11TT

6

oioe

60.0M818.0)1736.0(715.4T

T

T

T

T

Te*

o

oi

oi

oe*o

oe

(Appendix F)

K2216

6.02.01

K2376

21M1

TT

22oe

e

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

?




Solution #2 (con’t)

– poe/poiMe

airm

Mi=0.2Ti=500K

airfuel m 0413.0m 871.02346.1

10753.1

p

p

p

p

p

p

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

U

p*/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 Me

pe

pb

po1

Q

rayleigh flow -1 school of aerospace engineering copyright © 2001 by jerry m. seitzman. all rights...

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