School of Aerospace Engineering
Rayleigh Flow - Thermodynamics
p
• Steady, 1-d, constant area, inviscid
T
flow with no external work but with v
ρ
reversible heat transfer
M
(heating or cooling)
Q
• Conserved quantities (mass, momentum eqs.)
?
– since A=const: mass flux=ρ
ρv=constant≡
≡G
– since no forces but pressure: p+ρ
ρv2=constant
ρ=constant
– combining: p+G2/ρ
• On h-s diagram, can draw this Rayleigh Line
AE3450
Rayleigh Flow -1
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
School of Aerospace Engineering
Rayleigh Line
p high,
v low
• p+ρ
ρv2=constant,
h
Þhigh p results in low v
M<1
• ds=δQ/T (reversible)
heating
– heating increases s
– cooling decreases s
– M=1 at maximum s
• Change mass flux
new Rayleigh line
Rayleigh Flow -2
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
M=1
(ρv)↓
cooling
p
(ρv)↑ M>1
p low,
v high
smax
s
AE3450
1
School of Aerospace Engineering
Rayleigh Line - Choking
• Heat addition can only
increase entropy
• For enough heating,
Me=1 (Qin=Qmax)
h
M1
m
q=Q
Me
M<1
M=1
heating
(ρv)↓
– heat addition can lead to
choked flow
M>1
• What happens if Qin>Qmax
– subsonic flow: must move to
smax
s
different Rayleigh line (– – –), i.e., lower mass flux
– supersonic flow: get a shock (– – –)
stays on same Rayleigh line: ρv and p+ρ
ρv2 also const. for normal shock
AE3450
Rayleigh Flow -3
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
School of Aerospace Engineering
Differential Mach Equations
• Simplify (X.4-5) for f=dA=0
(
)
γ −21æ 2 γ − 1 2 ö
1+ ì M ÷ 2
22 11++ γ M çM
dM
f dx 2 dA üï
ï2δq 1 + øγMδ q
dM
2 è
+ γ M 2 dTo−= δqý
=
=
í
2
2
c pToc p To
DTo
1 − M1 − M 2ïî
M2
cAp Tï
M
þo (X.25)
(
dp
− γM 2 dM 2
=
p 1 + γM 2 M 2 (X.22)
)
(X.21)
dT dh 1 − γ M 2 dM 2
=
=
T
h 1 + γ M 2 M 2 (X.23)
dv
dρ
− 1 dM 2
=−
=
v
ρ 1 + γ M 2 M 2 (X.24)
Rayleigh Flow -4
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
γ
dp o
dT
= − M2 o
po
2
To (X.26)
ds 1 − M 2 dM 2
=
c p 1 + γM 2 M 2 (X.27)
• sign changes
AE3450
2
School of Aerospace Engineering
Property Variations
• What can change sign in previous equations
– (1-M2) and δq
– (1-γM2) in dh,dT for M<1
M<1
<0 >0
δq
• s, To just like q
M>1
<0 >0
s, To
po
M, v
p, ρ
h, T
• po opposite of To and s
• Heating: M→1; cooling opposite
• p, ρ opposite of v (p+ ρv2=const)
• h,T same as p,ρ unless M2<1/γ
(low speed flow: T oppos. p,ρ)
M2<γ-1
δq<0Þ
Þcooling
Þheating
δq>0Þ
M2>γ-1
AE3450
Rayleigh Flow -5
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
School of Aerospace Engineering
Rayleigh Line Extremum
• Combine X.23 and X.27
dh
h 1 − γM 2
=
ds c p 1 − M 2
1
dh
= 0 for M =
ds
γ
dh
= ∞ for M = 1
ds
h
hmax
M=γ-0.5
M<1
M=1
heat
cool
M>1
smax
Rayleigh Flow -6
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
s
AE3450
3
School of Aerospace Engineering
Integrated Mach Equations
• Integrating (X.22-27)
δq
δq
dTo
q (X.28)
=
Þ ò dTo = ò
Þ (To 2 − To1 ) = 12 (energy)
To
c pTo
cp
cp
p 2 1 + γM 12 (X.29)
=
p1 1 + γM 22
T2 æ 1 + γ M 12 ö
÷
=ç
T1 çè 1 + γ M 22 ÷ø
2
æ M2
ç
çM
è 1
po 2
p o1
(X.30)
ö
÷
÷
ø
2
v1 ρ 2 1 + γM 22 æ M 1 ö
÷
ç
=
=
v 2 ρ1 1 + γM 12 çè M 2 ÷ø
2
(X.31)
po 2
æ
1+
1 + γ M 12 ç
p2
ç
=
=
p o1 1 + γ M 22 ç
p1
ç1+
p1
è
p2
To 2 æ 1 + γM 12 ö
÷
=ç
To1 çè 1 + γM 22 ÷ø
2
æ M2 ö
çç
÷÷
è M1 ø
(X.32)
γ
γ − 1 2 ö γ −1
M2 ÷
2
÷
γ −1 2 ÷
M1 ÷
2
ø
γ −1 2 ö
M2 ÷
2
ç
÷ (X.33)
çç 1 + γ − 1 M 2 ÷÷
1
è
ø
2
2æ
ç1+
éæ M ö 2 æ 1 + γ M 2 ö (γ +1) γ ù
s 2 − s1
1 ÷
ú (X.34)
= ln êçç 2 ÷÷ çç
cp
êè M 1 ø è 1 + γ M 22 ÷ø
ú
ë
û
Rayleigh Flow -7
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
AE3450
School of Aerospace Engineering
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*
Rayleigh Flow -8
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
AE3450
4
School of Aerospace Engineering
Direct Solution
• Use two equations to find M change
– e.g., if state 1 and q12 known use X.28 and X.33
a) get To ratio from X.28
To 2
q
= 1 + 12
To1
c pTo1
doesn’t matter how q
changes along length,
only total q required
b) get M2 from X.33 (must know M1)
æ 1 + γM 12 ö
ç
÷
ç 1 + γM 2 ÷
2 ø
è
2
æ M2 ö
çç
÷÷
è M1 ø
M1
m
q=Q
M2
γ −1 2 ö
M2 ÷ T
requires iterative
2
ç
÷ = o2
çç 1 + γ − 1 M 2 ÷÷ To1 solution for M2
1
è
ø
2
2æ
ç1 +
c) get rest of property changes M2 from X.29-34, e.g., from X.29
p 2 1 + γM 12
=
p1 1 + γM 22
AE3450
Rayleigh Flow -9
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
School of Aerospace Engineering
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)
M1=3.0
T1=250K
p1=50 kPa
q=Q m
T2=
200K
• Assume: N2 is tpg/cpg, γ=1.4, steady, reversible, no work
Rayleigh Flow -10
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
AE3450
5
School of Aerospace Engineering
Cooled Duct - Solution
• Analysis:
– M2
–q
(from X.30)
æ1 +
ç
ç1 +
è
2
2
M1=3.0
T1=250K
T p1=50 kPa
= 2
T1
ö æ M2 ö
÷
÷ ç
÷ çM ÷
ø è 1 ø
2
æ 1 + 1 .4 (3 )2 ö æ M ö 2 200
ç
÷ ç 2÷ =
= 0 .80
ç 1 + 1 .4 M 22 ÷ çè 3 ÷ø
250
è
ø
γM 12
γM 22
Þ M 2 = 0.21, 3.41
q=Q m
T2=200K
h
M<1
heat M=1
cool
M>1
s
since started M>1, can’t be subsonic
(energy: ins = outs)
æ γ −1 2 ö
To1 = T1ç1 +
M2 ÷
2
è
ø
1.4 8314 J kmolK
æ 1.4 − 1 2 ö = 700 K
9 ÷
= 250K ç1 +
=
(700 − 665)K
2
1.4 − 1 28 kg kmol
è
ø
q = c p (To1 − To 2 )
= 36 kJ kg
(
)
To 2 = 200 K 1 + 0.2(3.41) = 665K
2
AE3450
Rayleigh Flow -11
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
School of Aerospace Engineering
Reference Solution Method
but not same
• Reference state has M=1, T≡T*, p≡p*,... *Þsonic,
state as A* or Fanno
γ
• State 2→sonic in integ(X.38)
γ − 1 2 ö γ −1
æ
M ÷
1+
rated equations (X.29-34) po
1+ γ ç
2
=
ç
÷
1 + γM 2 çç 1 + γ − 1 ÷÷
(X.35)
è
ø
2
1
−
(X.39)
γ
2
1+
M2
2
To æç 1 + γ ö÷
2
2
æ
ö
T
+
γ
1
(X.36)
=
M
÷
= M 2 çç
γ −1
To* çè 1 + γM 2 ÷ø
2÷
1+
T*
M
1
+
γ
è
ø
2
1+ γ ü
(X.40)
v ρ*
1+ γ
(X.37)
ì
*
=
=
M2
γ ï
ö
æ
ï
s
s
1
−
+
γ
*
2
ρ 1 + γM
÷
v
= ln í M 2 çç
ý
2
÷
cp
• Note: prop/prop* = f(M) only
è 1 + γM ø
ï
ï
þ
î
• Values for γ=1.4 in Appendix F, John
pp2 1 +1 +
γ Mγ 12
=
pp*1 1 + γM 222
Rayleigh Flow -12
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
p*o
AE3450
6
School of Aerospace Engineering
Use of Tables
• To get change in M, use change in property ratio (like
using A/A*), e.g.,
To To*
To 2 To 2 To*
M2
=
=
To1 To1 To* To To*
M1
M2
m
q=Q
M1
• For example, if state 1 and q12 known
1) again get To ratio from X.28
2) look up To/To* at M1
3) calculate To/To* at M2
To 2
q
= 1 + 12
To1
c pTo1
To 2
To*
=
To 2 To1
×
To1 To*
4) look up M2 that corresponds to calculated To/To*
AE3450
Rayleigh Flow -13
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
School of Aerospace Engineering
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
Mi=0.2
Toi=300K
poi=600 kPa
m
q=Q
Me
• Assume: air is tpg/cpg, γ=1.4, steady, reversible, no work
Rayleigh Flow -14
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
AE3450
7
School of Aerospace Engineering
Heated Duct - Solution
Mi=0.2
Toi=300K
poi=600 kPa
• Analysis:
q = 765 kJ kg
Me
Toe
q
=1+
Toi
c pToi
(X.28, energy)
765 × 103 J / kg
=1+
= 3.54
J
1004
(300 K )
kgK
– Me
M
0.19
0.20
0.21
0.44
0.45
0.46
3.53
To/To*
0.15814
0.17355
0.18943
0.59748
0.61393
0.63007
0.61393
p o / p ο∗
1.23765
1.23460
1.23142
1.13936
1.13508
1.13082
5.47054
(Appendix F)
Toe Toi
= 3.54(0.17355 ) = 0.614 Þ M e = 0.45
=
To* Toi To* M =0.2
another solution for M=3.53, but since
Toe
started with M<1, can’t be supersonic
– poe
poe =
poe p*o
p*o poi
1
poi = 1.1351
600kPa = 552kPa
1.2346
AE3450
Rayleigh Flow -15
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
School of Aerospace Engineering
Heated Duct Solution (con’t)
– Te
Mi=0.2
Toi=300K
poi=600 kPa
q = 765 kJ kg
Te =
Te Toe
Toi
Toe Toi
Te =
1
3.54(300 K ) = 1021K
0.4
1+
0.452
2
– qmax
qmax
requires choked exit, Toe=T*
(
)
To* =
Me
300 K
= 1728 K
0.17355
æ T*
ö
= c p To* − To = c pTo1ç o − 1÷
ç To1 ÷ T* = To* æç1 + γ − 1ö÷ = 1440K
è
ø
2 ø
è
J
1
MJ
ö
æ
300 K ç
= 1004
− 1÷ = 1.43
kgK
kg
è 0.17355 ø
Rayleigh Flow -16
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
AE3450
8
School of Aerospace Engineering
Combustor Example
• Given: Air entering constant combustor, enters at M=0.2
and given inlet conditions.
m
q=heating value of fuel (45.5 MJ/kgfuel) × fuel
air
m
• Find:
1. Te and compare to value you Mi=0.2
Me
would calculate if neglected Ti=500K
air
m
kinetic energy change
fuel = 0.0413 m
air
m
2. po loss
• Assume: air is tpg/cpg, γ=1.4, steady, reversible, no work,
fuel
m
<< 1 (can neglect mass addition, just use q)
air
m
Rayleigh Flow -17
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
AE3450
School of Aerospace Engineering
Combustor Solution
Mi=0.2
Me
? combustion just Ti=500K
q
Toe
another kind of
air
=1+
m
– Te
heat addition
Toi
cp Toi
fuel = 0.0413 m
air
m
kg fuel æ
MJ ö
MJ
ç 45.5
÷ = 1.88
q = 0.0413
kg air çè
kg fuel ÷ø
kg air
even for this subsonic
γ
−
1
æ
ö
flow, kinetic energy
Toi = Ti ç1 +
M 2 ÷ = 500K (1.008) = 504K
reduces final T by 160K
2
è
ø
æ 1.88 × 106 ö
÷ = 504K (4.715) = 2376K
Toe = Toi çç1 +
÷
(Appendix F)
è 1004(504) ø
Toe Toe Toi
=
= 4.715(0.1736) = 0.818 Þ M e = 0.60
To* Toi To*
Toe
2376K
Te =
=
= 2216K
2
1 + M (γ − 1) 2 1 + 0.2(0.6 )2
• Analysis:
(energy)
Rayleigh Flow -18
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
AE3450
9
School of Aerospace Engineering
Combustor Solution (con’t)
Mi=0.2
Ti=500K
air
m
*
p po
1
= oe
=
1
.
0753
=
0
.
871
fuel = 0.0413 m
air
m
1.2346
p*o poi
– poe/poi
poe
poi
Me
(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
AE3450
Rayleigh Flow -19
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
School of Aerospace Engineering
Choking
• Heating of a flow can lead
to choking just like friction
• For supersonic flow,
depending on back
pressure, can get
– underexpanded flow
– overexpanded flow
– shock inside duct
Rayleigh Flow -20
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
Q
po1
M1>1
p/po1
1
p*/po
pb
Me
pe
shock
inside
shock
at exit
O
pe,sh
pe,R
U
x
AE3450
10