rayleigh flow - thermodynamics (georgia tech)
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AE3450Rayleigh Flow -1
School of Aerospace Engineering
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
• Steady, 1-d, constant area, inviscidflow 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�
AE3450Rayleigh Flow -2
School of Aerospace Engineering
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
• 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
AE3450Rayleigh Flow -3
School of Aerospace Engineering
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
• Heat addition can onlyincrease 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
AE3450Rayleigh Flow -4
School of Aerospace Engineering
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
( )��
���
��
���
−γ+γ+δ−
−γ+=
AdA
DdxfM
TcM
M
M
MdM
op
21q1
211
22
2
2
2
2
• 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=
AE3450Rayleigh Flow -5
School of Aerospace Engineering
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
• 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
AE3450Rayleigh Flow -6
School of Aerospace Engineering
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
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
AE3450Rayleigh Flow -7
School of Aerospace Engineering
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
• Integrating (X.22-27)�
δ=�δ= ��
po
opo
oc
qdTTcq
TdT
2
2
12
1
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)
AE3450Rayleigh Flow -8
School of Aerospace Engineering
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
• 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*
AE3450Rayleigh Flow -9
School of Aerospace Engineering
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
• 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+=
AE3450Rayleigh Flow -10
School of Aerospace Engineering
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
• 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
AE3450Rayleigh Flow -11
School of Aerospace Engineering
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
• 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)
AE3450Rayleigh Flow -12
School of Aerospace Engineering
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
22
21
1
2
11
MM
pp
γ+γ+=
• 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)
AE3450Rayleigh Flow -13
School of Aerospace Engineering
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
• 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*
AE3450Rayleigh Flow -14
School of Aerospace Engineering
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
• 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
AE3450Rayleigh Flow -15
School of Aerospace Engineering
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
• 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
AE3450Rayleigh Flow -16
School of Aerospace Engineering
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
– 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+=
AE3450Rayleigh Flow -17
School of Aerospace Engineering
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
• 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 <<�
�
AE3450Rayleigh Flow -18
School of Aerospace Engineering
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
• Analysis:
– Te
(energy)
oipoi
oe
Tcq1
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
AE3450Rayleigh Flow -19
School of Aerospace Engineering
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
– 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
AE3450Rayleigh Flow -20
School of Aerospace Engineering
Copyright © 2001-2002 by Jerry M. Seitzman. All rights reserved.
• 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�
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