Rayleigh Flow -1 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Rayleigh Flow - Thermodynamics Steady, 1-d, constant area, inviscid flow with no external work but with reversible heat transfer (heating or cooling) Conserved quantities –since A=const: mass flux=  v=constant –since no forces but pressure: p+  v 2 =constant –combining: p+(  v) 2 /  =constant On h-s diagram, can draw this Rayleigh Line pTvMpTvM ?

Rayleigh Flow -2 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Rayleigh Line p+  v 2 =constant,  high p results in low v h s p high, v low M<1 (v)(v) (v)(v) s max M=1 ds=  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

Rayleigh Flow -3 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Rayleigh Line - Choking Heat addition can only increase entropy For enough heating, M e =1 (Q in =Q max ) –heat addition leads 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 ( – – – ) M1M1 MeMe h s M<1 (v)(v) s max M=1 M>1 heating

Rayleigh Flow -4 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Rayleigh Line – Mach Equations Simplify (IX.4-5) for f=dA=0 sign changes (X.21) (X.20) (X.26) (X.25) (X.24) (X.22) (X.23)

Rayleigh Flow -5 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Property Variations What can change sign in previous equations –(1-M 2 ) and  q –(1-  M 2 ) in dh,dT for M<1 T same as p,  unless M 2 <1/  (low speed flow: T oppos. p,  ) p,  opposite of v (p+  v 2 =const) Heating: M  1; cooling opposite s, T o just like q p o opposite of T o M 2 <  -1 M<1M>1 s, T o p o M, v p,  h, T qq <0 >0 M 2 >  -1

Rayleigh Flow -6 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Rayleigh Line Extremum s max M=1 Combine X.22 and X.26 M=  -0.5 h max h s M<1 M>1 heat cool

Rayleigh Flow -7 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 A Solution Method Need to integrate ( X ) to find how properties change due to q –can integrate or use tables of integrated values doesn’t matter how q changes along length Mach number variation M1M1 MeMe to tabularize solution, use reference condition: M 2 =1, T o = T o *

Rayleigh Flow -8 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Use of Tables To get change in M, use change in T o /T o * (like using A/A*) Find values in Appendix F in John so if you know q, T o1 and M 1, 1) T o2 /T o1 = 1 + q/(c p T o1 ) 2) look up T o /T o * at M 1 3) calculate T o /T o * at M 2 4) look up corresponding M 2 (X.27) M1M1 MeMe

Rayleigh Flow -9 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Rayleigh Flow Property Changes To get changes in T, p, p o,... can also use M=1 condition as reference condition (*) (X.28) (X.29) (X.30) (X.32) (X.33) (X.31) Integrate X (also tabulated in Appendix F)

Rayleigh Flow -10 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Example #1 Given: Air flowing in constant area duct, enters at M=0.2 and given inlet conditions. For a heating rate of 765 J/kg, Find: - M e, p oe, T e - q max for M e =1 Assume: air is tpg/cpg,  =1.4, steady, reversible, no work MeMe M i =0.2 T oi =300K p oi =600 kPa

Rayleigh Flow -11 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Solution #1 Analysis: MeMe M i =0.2 T oi =300K p oi =600 kPa –Me–Me another solution for M=3.5, but since started M<1, can’t be supersonic (energy) (Appendix F) –p oe

Rayleigh Flow -12 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Solution #1 (con’t) MeMe M i =0.2 T oi =300K p oi =600 kPa –Te–Te –T oe (choked)

Rayleigh Flow -13 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Example #2 Find: - T e and compare to value you would calculate if neglected kinetic energy change - p o loss MeMe M i =0.2 T i =500K Given: Air entering constant combustor, enters at M=0.2 and given inlet conditions. q=heating value of fuel (45.5 MJ/kg fuel ) × Assume: air is tpg/cpg,  =1.4, steady, reversible, no work,

Rayleigh Flow -14 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Solution #2 Analysis: –Te–Te (energy) MeMe M i =0.2 T i =500K (Appendix F) even for this subsonic flow, kinetic energy reduces final T by 160K ?

Rayleigh Flow -15 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Solution #2 (con’t) –p oe /p oi MeMe M i =0.2 T i =500K (Appendix F) So heat addition (burning fuel) in an engine gives p o loss as was case with friction As we lose p o (~13% loss here) –less thrust –less work output

Rayleigh Flow -16 School of Aerospace Engineering Copyright © 2001 by Jerry M. Seitzman. All rights reserved. AE3450 Choking Heating of a flow can lead to choking just like friction shock inside p e,sh shock at exit O U p*/p o x 1 p e,R p/p o1 For supersonic flow, depending on back pressure, can get –underexpanded flow –overexpanded flow –shock inside duct M 1 >1 MeMe pepe pbpb p o1