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 involve chemical 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 simple analysis method where generation/absorption is modeled as heat transfer through the wall at the same rate •Still too complicated for introductory treatment since flow may involve friction, geometry changes, 3D effects •We will focus on 1D flow in a duct of constant cross-sectional area with negligible frictional effects PDF created with pdfFactory trial version www.pdffactory.com
Gas Dynamics and Propulsion / BY Dr.G.KUMARESAN, / PROFESSOR, / ANNA UNIVERSITY
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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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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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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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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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• “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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