AE3450
Rayleigh Flow -1
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
with
reversible heat transfer
(heating or cooling)
• Conserved quantities (mass, momentum eqs.)
– since A=const: mass flux=
ρρρρ
v=constant
≡≡≡≡
G
– since no forces but pressure:
p+
ρρρρ
v
2
=constant
– combining:
p+G
2
/
ρρρρ
=constant
• On h-s diagram, can draw this
Rayleigh Line
p
T
v
ρ
M
?
Q
School of Aerospace Engineering
Rayleigh Line
•
p+
ρρρρ
v
2
=constant
,
Þ
high p results in low v
h
s
p high,
v low
M<1
(
ρ
v
)
↑
(
ρ
v
)
↓
s
max
M=1
• ds=
δ
Q/T (reversible)
– heating increases s
– cooling decreases s
• Change mass flux
new Rayleigh line
– M=1 at maximum s
p
M>1
p low,
v high
heating
cooling
AE3450
Rayleigh Flow -3
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
Rayleigh Line - Choking
• Heat addition can only
increase entropy
• For
enough heating
,
M
e
=1 (
Q
in
=
Q
max
)
– heat addition can lead to
choked flow
• What happens if
Q
in
>
Q
max
–
subsonic flow
: must move to
different Rayleigh line (
– – –
), i.e., lower mass flux
–
supersonic flow
: get a shock (
– – –
)
M
1
M
e
m
Q
q
=
h
s
M<1
(
ρ
v
)
↓
s
max
M=1
M>1
heating
stays on same Rayleigh line:
ρρρρ
v and p+
ρρρρ
v
2
also const. for normal shock
School of Aerospace Engineering
(
)
ïþ
ï
ý
ü
ïî
ï
í
ì
−
γ
+
γ
+
δ
−
−
γ
+
=
A
dA
D
dx
f
M
T
c
M
M
M
M
dM
o
p
2
1
q
1
2
1
1
2
2
2
2
2
2
Differential Mach Equations
• Simplify (X.4-5) for
f
=dA=0
•
sign changes
2
2
2
2
M
dM
M
1
M
p
dp
γ
+
γ
−
=
(X.22)
(
)
o
p
2
2
2
2
2
T
c
q
M
1
M
2
1
1
M
1
M
dM
δ
−
÷
ø
ö
ç
è
æ
+
γ
−
γ
+
=
(X.21)
2
2
2
2
p
M
dM
M
1
M
1
c
ds
γ
+
−
=
(X.27)
(X.26)
o
o
2
o
o
T
dT
M
2
p
dp
γ
−
=
2
2
2
2
M
dM
M
1
M
1
h
dh
T
dT
γ
+
γ
−
=
=
(X.23)
2
2
2
dM
1
d
v
dv
γ
+
−
=
ρ
ρ
−
=
(X.25)
o
p
o
o
T
c
q
T
dT
δ
=
AE3450
Rayleigh Flow -5
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
Property Variations
• What can change sign in previous equations
– (1-M
2
) and
δ
q
– (1-
γ
M
2
) in dh,dT for M<1
• h,T same as p,ρ
unless M
2
<1/γ
(low speed flow: T oppos. p,ρ)
• p, ρ
opposite of v (p+ ρv
2
=const)
• Heating: M→1; cooling opposite
• s, T
o
just like q
• p
o
opposite of T
o
and s
M
2
<
γ
-1
M<1
M>1
s, T
o
p
o
M, v
p, ρ
h, T
δδδδ
q
<0
>0
<0
>0
M
2
>
γ
-1
δδδδ
q<0
Þ
Þ
Þ
Þ
cooling
δδδδ
q>0
Þ
Þ
Þ
Þ
heating
School of Aerospace Engineering
Rayleigh Line Extremum
s
max
M=1
• Combine
X.23
and
X.27
2
2
1
1
M
M
c
h
ds
dh
p
−
γ
−
=
γ
=
=
0
for
M
1
ds
dh
1
for
=
∞
=
M
ds
dh
M=
γ
-0.5
h
max
h
s
M<1
M>1
heat
cool
AE3450
Rayleigh Flow -7
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
Integrated Mach Equations
• Integrating (X.22-27)
Þ
δ
=
Þ
δ
=
ò
ò
p
o
o
p
o
o
c
q
dT
T
c
q
T
dT
2
2
1
2
1
2
2
1
2
2
1
1
1
v
v
÷÷ø
ö
ççè
æ
γ
+
γ
+
=
ρ
ρ
=
M
M
M
M
(X.31)
2
2
2
1
1
2
1
1
M
M
p
p
γ
+
γ
+
=
(X.29)
1
2
1
2
2
2
2
2
1
1
1
1
2
2
2
1
2
2
1
1
2
1
1
1
1
−
γ
γ
÷÷
÷
÷
ø
ö
çç
ç
ç
è
æ
−
γ
+
−
γ
+
γ
+
γ
+
=
=
M
M
M
M
p
p
p
p
p
p
p
p
o
o
o
o
(X.32)
2
1
2
2
2
2
2
1
1
2
1
1
÷÷ø
ö
ççè
æ
÷÷ø
ö
ççè
æ
γ
+
γ
+
=
M
M
M
M
T
T
(X.30)
(X.28)
(
)
p
o
o
c
q
T
T
2
−
1
=
12
(X.33)
÷÷
÷
÷
ø
ö
çç
ç
ç
è
æ
−
γ
+
−
γ
+
÷÷ø
ö
ççè
æ
÷÷ø
ö
ççè
æ
γ
+
γ
+
=
2
1
2
2
2
1
2
2
2
2
2
1
1
2
2
1
1
2
1
1
1
1
M
M
M
M
M
M
T
T
o
o
( )
ú
ú
û
ù
ê
ê
ë
é
÷÷ø
ö
ççè
æ
γ
+
γ
+
÷÷ø
ö
ççè
æ
=
−
γ
+
1
γ
2
2
2
1
2
1
2
1
2
1
1
ln
M
M
M
M
c
s
s
p
(X.34)
(energy)
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; M
2
=1, T
o2
= T
o
*
AE3450
Rayleigh Flow -9
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
Direct Solution
• Use two equations to find M change
– e.g., if state 1 and
q
12
known use
X.28 and X.33
doesn’t matter how q
changes along length,
only total q required
M
1
M
2
m
Q
q
=
1
12
1
2
1
o
p
o
o
T
c
q
T
T
+
=
1
2
2
1
2
2
2
1
2
2
2
2
2
1
2
1
1
2
1
1
1
1
o
o
T
T
M
M
M
M
M
M
=
÷÷
÷
÷
ø
ö
çç
ç
ç
è
æ
−
γ
+
−
γ
+
÷÷ø
ö
ççè
æ
÷÷ø
ö
ççè
æ
γ
+
γ
+
a) get
T
o
ratio from X.28
b) get
M
2
from X.33 (must know
M
1
)
requires iterative
solution for M
2
c) get rest of property changes
M
2
from X.29-34, e.g., from X.29
2
2
2
1
1
2
1
1
M
M
p
p
γ
+
γ
+
=
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. M
2
2. q
(including direction)
•
Assume:
N
2
is tpg/cpg,
γ
=1.4, steady, reversible, no work
T
2
=
200K
m
Q
q
=
M
1
=3.0
T
1
=250K
p
1
=50 kPa
AE3450
Rayleigh Flow -11
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
Cooled Duct - Solution
•
Analysis:
–
M
2
since started M>1, can’t be subsonic
–
q
(energy: ins = outs)
(
o
1
o
2
)
p
T
T
c
q
=
−
41
.
3
,
21
.
0
2
=
Þ
M
T
2
=200K
m
Q
q
=
M
1
=3.0
T
1
=250K
p
1
=50 kPa
1
2
2
1
2
2
2
2
2
1
1
1
T
T
M
M
M
M
=
÷÷ø
ö
ççè
æ
÷÷ø
ö
ççè
æ
γ
+
γ
+
( )
80
.
0
250
200
3
4
.
1
1
3
4
.
1
1
2
2
2
2
2
2
=
=
÷÷ø
ö
ççè
æ
÷
÷
ø
ö
ç
ç
è
æ
+
+
M
M
÷
ø
ö
ç
è
æ
+
γ
−
=
2
2
1
1
2
1
1
M
T
T
o
K
700
=
( )
(
)
K
K
T
o
2
=
200
1
+
0
.
2
3
.
41
2
=
665
(
)
K
kmol
kg
kmolK
J
665
700
28
8314
1
4
.
1
4
.
1
−
−
=
kg
kJ
36
=
÷
ø
ö
ç
è
æ
+
−
=
9
2
2
1
4
.
1
1
250
K
h
s
M<1
M>1
heat
cool
M=1
(from X.30)
School of Aerospace Engineering
2
2
2
1
1
2
1
1
M
M
p
p
γ
+
γ
+
=
Reference Solution Method
• Reference state has
M
=1,
T
≡
T
*,
p
≡
p
*,...
2
2
2
1
1
÷÷ø
ö
ççè
æ
γ
+
γ
+
=
M
M
T
T
*
(X.36)
(X.35)
2
1
1
M
p
p
*
+
γ
γ
+
=
2
2
1
1
v
v
M
M
*
*
+
γ
γ
+
=
ρ
ρ
=
(X.37)
1
2
2
2
1
1
2
1
1
1
1
−
γ
γ
÷÷
÷
÷
ø
ö
çç
ç
ç
è
æ
−
γ
+
−
γ
+
γ
+
γ
+
=
M
M
p
p
*
o
o
(X.38)
2
1
1
2
1
1
1
1
2
2
2
2
γ
−
+
−
γ
+
÷÷ø
ö
ççè
æ
γ
+
γ
+
=
M
M
M
T
T
*
o
o
(X.39)
ï
þ
ï
ý
ü
ï
î
ï
í
ì
÷÷ø
ö
ççè
æ
γ
+
γ
+
=
−
γ
γ
+
1
2
2
1
1
M
M
ln
c
s
s
p
*
(X.40)
•
Note:
prop
/
prop
* = f(
M
) only
*
Þ
sonic, but not same
state as A* or Fanno
• State 2
→
sonic in
integ-rated equations (X.29-34)
AE3450
Rayleigh Flow -13
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
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
*
o
o
M
*
o
o
*
o
o
*
o
o
o
o
T
T
T
T
T
T
T
T
T
T
=
=
M
1
M
2
m
Q
q
=
1
12
1
2
1
o
p
o
o
T
c
q
T
T
+
=
1) again get
T
o
ratio from X.28
2) look up
T
o
/
T
o
*
at M
1
• For example, if state 1 and
q
12
known
3) calculate
T
o
/
T
o
*
at M
2
*
o
o
o
o
*
o
o
T
T
T
T
T
T
1
1
2
2
=
×
4) look up
M
2
that corresponds to calculated
T
o
/
T
o
*
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. M
e
,
p
oe
,
T
e
2. q
max
for
M
e
=1
•
Assume:
air is tpg/cpg,
γ
=1.4, steady, reversible, no work
M
e
m
Q
q
=
M
i
=0.2
T
oi
=300K
p
oi
=600 kPa
AE3450
Rayleigh Flow -15
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
Heated Duct - Solution
•
Analysis:
M
e
kg
kJ
q
=
765
M
i
=0.2
T
oi
=300K
p
oi
=600 kPa
–
M
e
another solution for M=3.53, but since
started with M<1, can’t be supersonic
(X.28, energy)
oi
p
oi
oe
T
c
q
T
T
+
=
1
(
0
.
17355
)
0
.
614
54
.
3
=
=
–
p
oe
(
300
)
3
54
1004
10
765
1
3
.
K
kgK
J
kg
/
J
=
×
+
=
2
0.
M
*
o
oi
i
o
e
o
*
o
oe
T
T
T
T
T
T
=
=
i
o
i
o
*
o
*
o
oe
oe
p
p
p
p
p
p
=
kPa
kPa
.
.
600
552
2346
1
1
1351
1
=
=
45
0
.
M
e
=
Þ
(Appendix F)
M
To/To*
p
o
////
p
οοοο
∗∗∗∗
0.19 0.15814 1.23765
0.20 0.17355 1.23460
0.21 0.18943 1.23142
0.44 0.59748 1.13936
0.45 0.61393 1.13508
0.46 0.63007 1.13082
3.53 0.61393 5.47054
School of Aerospace Engineering
Heated Duct Solution (con’t)
–
T
e
–
q
max
oi
oi
oe
oe
e
e
T
T
T
T
T
T
=
(
K
)
K
.
.
.
T
e
3
54
300
1021
45
0
2
4
0
1
1
2
=
+
=
(
)
÷
÷
ø
ö
ç
ç
è
æ
−
=
−
=
1
1
1
o
*
o
o
p
o
*
o
p
max
T
T
T
c
T
T
c
q
kg
MJ
.
.
K
kgK
J
43
1
1
17355
0
1
300
1004
÷
=
ø
ö
ç
è
æ
−
=
M
e
kg
kJ
q
=
765
M
i
=0.2
T
oi
=300K
p
oi
=600 kPa
requires choked exit,
T
oe
=
T
*
K
.
K
T
o
*
1728
17355
0
300
=
=
K
T
T
*
o
*
1440
2
1
1
÷
=
ø
ö
ç
è
æ
+
γ
−
=
AE3450
Rayleigh Flow -17
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
Combustor Example
•
Find:
1. T
e
and compare to value you
would calculate if neglected
kinetic energy change
2. p
o
loss
M
e
air
m
M
i
=0.2
T
i
=500K
air
fuel
0
.
0413
m
m
=
•
Given:
Air entering constant combustor, enters at
M
=0.2
and given inlet conditions.
q
=heating value of fuel (45.5 MJ/kg
fuel
) ×
air
fuel
m
m
•
Assume:
air is tpg/cpg,
γ
=1.4, steady, reversible, no work,
)
(
1
m
m
q
use
just
addition,
mass
neglect
can
air
fuel
<<
School of Aerospace Engineering
Combustor Solution
•
Analysis:
–
T
e
(energy)
oi
p
oi
oe
T
c
q
1
T
T
+
=
M
e
air
m
M
i
=0.2
T
i
=500K
air
fuel
0
.
0413
m
m
=
air
fuel
air
fuel
kg
MJ
88
.
1
kg
MJ
5
.
45
kg
kg
0413
.
0
q
=
÷÷ø
ö
ççè
æ
=
(
1
.
008
)
504
K
K
500
M
2
1
1
T
T
oi
i
2
÷
=
=
ø
ö
ç
è
æ
+
γ
−
=
(
4
.
715
)
2376
K
K
504
)
504
(
1004
10
88
.
1
1
T
T
6
oi
oe
÷÷ø
=
=
ö
ççè
æ
+
×
=
60
.
0
M
818
.
0
)
1736
.
0
(
715
.
4
T
T
T
T
T
T
e
*
o
oi
oi
oe
*
o
oe
=
=
=
Þ
=
(Appendix F)
(
)
1
0
.
2
( )
0
.
6
2216
K
K
2376
2
1
M
1
T
T
e
2
oe
2
=
+
=
−
γ
+
=
even for this subsonic
flow, kinetic energy
reduces final T by 160K
?
combustion just
another kind of
heat addition
AE3450
Rayleigh Flow -19
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
