gas dynamics-rayleigh flow
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RAYLEIGH FLOW
GDJP Anna University
Constant area Duct flow with Heat Transfer and negligible friction is called Rayleigh flow (Simple diabatic flow)
•Many compressible flow problems encountered in practice involvechemical reactions such as combustion, nuclear reactions, evaporation,and condensation as well as heat gain or heat loss through the duct wall
•Such problems are difficult to analyze
•Essential features of such complex flows can be captured by a simpleanalysis method where generation/absorption is modeled as heat transferthrough the wall at the same rate
•Still too complicated for introductory treatment since flow may involvefriction, geometry changes, 3D effects
•We will focus on 1D flow in a duct of constant cross-sectional area withnegligible frictional effects
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RAYLEIGH FLOW
GDJP Anna University
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RAYLEIGH FLOW- Assumptions
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To isolate the effects heat transfer we make the following assumptions:
1. The area of the flow passage or duct is constant.
2. The flow is steady and one-dimensional.
3. There is no work, body forces are negligible, and
the effects of friction are negligible.
4. Heat transfer is the only driving potential.
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RAYLEIGH FLOW – Fundamental Equations
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Continuity Equation
X-Momentum equation
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RAYLEIGH FLOW – Fundamental Equations
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Energy equationCV involves no shear, shaft, or other forms of work, and potential energy change is negligible.
For and ideal gas with constant cp, ∆h = cp∆T
Entropy change
In absence of irreversibility's such as friction, entropy changes by heat transfer only
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RAYLEIGH LINE
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•Infinite number of downstreamstates 2 for a given upstream state 1
•Practical approach is to assume various values for T2, and calculate all other properties as well as q.
•Plot results on T-s diagramCalled a Rayleigh line
•This line is the locus of allphysically attainabledownstream states
•S increases with heat gain topoint a which is the point of maximum entropy (Ma =1)
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RAYLEIGH LINE
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Various flow parameters on Rayleigh Line
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Variation of fluid Properties
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Rayleigh Flow Relations
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Pressure Impulse Function
12 FF =
Stagnation Pressure Temperature
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Cont..
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Stagnation Temperature Density & Velocity
Change of Entropy Heat Transfer
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Derivation Proof
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Momentum
)1()1(
)1()1(
Therefore
that know We
)(
22
21
1
2
2221
21
1212
2221
222
211
22221
211
22221
1221
MM
pp
pMpM
pMpMpp
pMRTMRTpc
cc)p(pAcAc)Ap(p
ccm)Ap(p
γ
γ
γγ
γγ
γγρ
ρρρρ
+
+=
+=+
−=−
==
−=−⇒−=−
−=−•
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Derivation Proof
GDJP Anna University
Stagnation Pressure
1
212
11
222
11
211
211
0102
1
212
11
222
11
12
0102
122
110
−
−+
−+
×+
+=
−
−+
−+
=
−
−+=
γγ
γ
γ
γ
γ
γγ
γ
γ
γγ
γ
M
M
M
MPP
M
M
pp
PP
MpP
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Derivation Proof
GDJP Anna University
Temperature
(4) eqn 12
12
1p2p
222p , 11p that, know We
(3) eqn 2/1
21
2M1M
12
2c1c
2 & 1 eqn From
(2) eqn 2/1
21
12
11
221122
1M2M
number Mach From
(1) eqn 12
2c1c
2211m Continuity
1
TTSo
RTRTTT
TT
cc
RTc
RTcacac
AcAcFrom
×=
==
==
×===
=⇒==•
ρρ
ρρ
ρρ
γ
γ
ρρ
ρρ
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Derivation Proof-Cont..
GDJP Anna University
2
12
221
211
1T2T
eqn above the in 12
2
12
12
1T2T
2/1
12
21
1p2p
(4) in (3) eqn from gSubstituin
×
+
+=
×=⇒
×=
MM
M
M
ppsubstitute
MM
pp
TT
MM
γ
γ
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SLOPE OF RAYLEIGH LINE
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Slope of S=cons. Line
22222
2
2
21
21
22
212
1
2
)(tan
points, state two anyFor constant ..
MacGdvdp
AmGpp
GpGp
GpTKW
Rρρθ
νν
νν
ν
−=−==−=
−=−=
−−
+=+
=+
•
(1)
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Cont..
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(2)
1 2
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CONS. ENTHALPY LINE
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vp
dvdp
vdppdvRTpv
−=
=+===
0 , yields atingDifferentilineconstant Tfor ;constant Note:
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Cont..
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Problem 1
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In your opinion, which assumption(s) in the Rayleigh flow analysis may be potential source(s) of error in solving a real life problem?Answer: •Heat transfer causes the total temperature to change significantly in the flow,which leads to a large variation of static temperature. The perfect gasassumption (constant specific heats) may not be appropriate. At higher andhigher temperatures, more and more energy modes are activated within themolecules. In general, this causes the specific heats to rise with temperature.
•In cases where combustion occurs, chemical composition of the constituentgases changes significantly. Reactant species will be consumed and productspecies will be produced. Their relative ratio changes as combustion proceeds.Values like gas constant R, and specific heat ratio, will no longer be constantbut depend on the extent of combustion.
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Problem 2
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Unchoked case: Air at 250 K and 1.0 bar is moving at 100 m/sec towards the entrance of a combustion chamber. Determine the exit conditions if 300 kJ/kg is added to the flow during the combustion process.
1 2 P= 1.0 barT= 250 KV= 100 m/s
Heat addition
Solution:
K T whichFromTT obtain table weflow isentropicthe From
M of number Mach inlet an gives This m/s. V K, T For
01
01
1
1
11
255
9805.0
3156.0100250
=
=
=
==
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Problem 2- Cont..
GDJP Anna University
bar 7588.0106.2598.1
1PP
K 5174427.09152.0
250TT
is state
598.1PP
,9152.0TT
and ,(subsonic) 5985.0M
of no. Machexit an to scorrespond this table, flow rayleigh the From
8173.03763.0172.2T
TT
TT
(2), stationAt
1TT
)(C q have we energy, of onconservati From
2.106 PP
4427.0TT
3763.0TT
obtain we table, flow Rayleigh the From
22
*11
*22
1
2
22
*11
*22
1
2
*2
2*2
22
*1
01
01
02*
02
02*
0101
*0202
01
02
0101
020102p
*1
1*1
1*01
01
=⇒=⇒=
=⇒=⇒=⇒
===
=×=×=⇒=
+=⇒−=
===
PP
PPPP
TT
TTTT
exitThe
TTTTTTT
TCqTT
o
p
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Problem 2- Cont..
GDJP Anna University
Just choked case: How much more heat that can be added without changing the conditions at the entrance to the cumbustor?Solution:
K 678)255(3763.01
is (2) stationat etemperatur stagnation the condition, choked thisAt thermally.it choke we before flow the into kJ/kg 124 extra an add can we Hence,
/ 42413763.012551004
1)TT(C q Therfore,
1M where TTTT additionheat max.for but,
)(C q have we equation, energy the FromTTT implies This .1M case, chokedjust theFor
0101
*01*
01*
0202
01
*01
0101*0p
*02
*01
*0max 0
0102p
*02
*01022
=
====
=
−×=
−=−=
====
−=
===
TTT
TTT
kgkJ
TT
TC
TT
p
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Problem 2- Cont..
GDJP Anna University
Choked case: Let us add sufficient fuel to the system so that the exit stagnation temperature is raised to 1500 K now. Assume that the receiver pressure is very low. What do you expect to happen in the system? Describe the flow both qualitatively and quantitatively.Solution: • In this case T02 = 1500 K > 678 K (choking condition)
• The original flow cannot accommodate this large amount ofheat. Something has to happen in order to take in so much heataddition. In other words, it cannot stay on the same Rayleighline.
•Recall that the upstream state can always communicate withthe downstream states in a subsonic flow by means of pressurewaves.
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Problem 2- Cont..
GDJP Anna University
• “Sensing” the super-critical heat addition downstream, the flow deceleratesfrom the free stream to the inlet. Spillage occurs ahead of the inlet. It is shownschematically as follows:
Heat addition∞
M=0.3156P=1.0 barT=250 K 1 2V=100 m/s
Spillage
•With a smaller flow rate in the combustion chamber, the flow moves to adifferent Rayleigh line with a smaller mass flow rate/A value.
•Since the receiver (back) pressure is very low, we can assume that the flow ischoked at the station (2), i.e. M2 = 1
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Problem 2- Cont..
GDJP Anna University
With M2 = 1 we conclude that
bar 04.1P 0.1Pthat know We932.0
9731.0PP
253T 250Tthat know We9805.09922.0
TT
is stateinlet The
9731.0PP
9922.0TT
table flow isentropic the from obtain we 1977.0M Withinlet theat 1977.0M to 3156.0 from sdecelerate flow The
3156.01977.0Mobtain we table, Rayleigh refering by
, ratio etemperatur above the From
17.0 1500255
T
T to leads This ; 1500
10
0111
10
0111
01
1
01
1
1
1
1
*01
01*02
*012
=∴===
=∴===
==
=
==
=<=
=====
∞∞∞∞
∞∞∞∞
∞
∞
barPPPP
KKTTTT
MM
KTTTo
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Problem 2- Cont..
GDJP Anna University
exit
exitupstream
inlet
Amsmaller•
higher ‘s’ due to moreheat addition
K 1250
2532024.01
1T1T
*1T *
1T*2T2T
be, to conditionexit the conclude can we choking, to due state reference the is
stateexit thethat Recall
275.2*1P1P&2024.0*
1T1T ; 1977.01M @
conditionexit determine To
=
×=×===
===
bar 0.45704.1275.21
1P1P
*1P*
1P*2P2P =×=×===
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