Thermodynamics

864 | ThermodynamicsTMa 1T maxdT 0ds bba ds 0dT aEXAMPLE 17–13Extrema of Rayleigh LineConsider the T-s diagram of Rayleigh flow, as shown in Fig. 17–54. Usingthe differential forms of the conservation equations and property relations,show that the Mach number is Ma a 1 at the point of maximum entropy(point a), and Ma b 1/ 1k at the point of maximum temperature (point b).Ma 1s maxFIGURE 17–54The T-s diagram of Rayleigh flowconsidered in Example 17–13.sSolution It is to be shown that Ma a 1 at the point of maximum entropyand Ma b 1/ 1k at the point of maximum temperature on the Rayleigh line.Assumptions The assumptions associated with Rayleigh flow (i.e., steadyone-dimensional flow of an ideal gas with constant properties through a constantcross-sectional-area duct with negligible frictional effects) are valid.Analysis The differential forms of the mass (rV constant), momentum[rearranged as P + (rV)V constant], ideal gas (P rRT), and enthalpychange (h c p T) equations can be expressed asrV constant S r dV V dr 0 S dr r dV VP 1rV2V constant S dP 1rV2 dV 0 S dPdV rV(1)(2)P rRT S dP rR dT RT dr S dP P dT T dr r(3)The differential form of the entropy change relation (Eq. 17–40) of anideal gas with constant specific heats isds c p dT T R dP P(4)Substituting Eq. 3 into Eq. 4 givesds c p dT T R adT T dr r b 1c p R2 dT T R dr r sincec p R c v S kc v R c v S c v R/(k 1)Dividing both sides of Eq. 5 by dT and combining with Eq. 1,Dividing Eq. 3 by dV and combining it with Eqs. 1 and 2 give, after rearranging,dT(7)dV T V V RSubstituting Eq. 7 into Eq. 6 and rearranging,dsdT dsdT RT 1k 12 RT 1k 12 R VRT V 2 >R R2 1kRT V 2 2T 1k 12 1RT V 2 2Setting ds/dT 0 and solving the resulting equation R 2 (kRT V 2 ) 0 forV give the velocity at point a to bedVdTRk 1dTT R dr r(5)(6)(8)V a 2kRT a andMa a V ac a 2kRT a2kRT a 1(9)