Question 17.13: Extrema of Rayleigh Line Consider the T-s diagram of Rayleig...

It is to be shown that Ma _{a}=1 at the point of maximum entropy and Ma _{b}=1 / \sqrt{ k } at the point of maximum temperature on the Rayleigh line.
Assumptions The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional-area duct with negligible frictional effects) are valid.
Analysis The differential forms of the mass (ρV = constant), momentum [rearranged as P + (ρV)V = constant], ideal gas (P = ρRT), and enthalpy change \left(\Delta h=c_{p} \Delta T\right) equations can be expressed as

\rho V=\text { constant } \rightarrow \rho d V+V d \rho=0 \rightarrow \frac{d \rho}{\rho}=-\frac{d V}{V}           (1)

P+(\rho V) V=\text { constant } \rightarrow d P+(\rho V) d V=0 \rightarrow \frac{d P}{d V}=-\rho V                 (2)

P=\rho R T \rightarrow d P=\rho R d T+R T d \rho \rightarrow \frac{d P}{P}=\frac{d T}{T}+\frac{d \rho}{\rho}                       (3)

The differential form of the entropy change relation (Eq. 17–40) of an ideal gas with constant specific heats is

s_{2}-s_{1}=c_{p} \ln \frac{T_{2}}{T_{1}}-R \ln \frac{P_{2}}{P_{1}}

d s=c_{p} \frac{d T}{T}-R \frac{d P}{P}                                     (4)

Substituting Eq. 3 into Eq. 4 gives

d s=c_{p} \frac{d T}{T}-R\left(\frac{d T}{T}+\frac{d \rho}{\rho}\right)=\left(c_{p}-R\right) \frac{d T}{T}-R \frac{d \rho}{\rho}=\frac{R}{k-1} \frac{d T}{T}-R \frac{d \rho}{\rho}                       (5)

since

c_{p}-R=c_{v} \rightarrow k c_{v}-R=c_{v} \rightarrow c_{v}=R /(k-1)

Dividing both sides of Eq. 5 by dT and combining with Eq. 1,

\frac{d s}{d T}=\frac{R}{T(k-1)}+\frac{R}{V} \frac{d V}{d T}                               (6)

Dividing Eq. 3 by dV and combining it with Eqs. 1 and 2 give, after rearranging,

\frac{d T}{d V}=\frac{T}{V}-\frac{V}{R}                   (7)

Substituting Eq. 7 into Eq. 6 and rearranging,

\frac{d s}{d T}=\frac{R}{T(k-1)}+\frac{R}{T-V^{2} / R}=\frac{R^{2}\left(k R T-V^{2}\right)}{T(k-1)\left(R T-V^{2}\right)}                                 (8)

Setting ds/dT = 0 and solving the resulting equation R^{2}\left(k R T-V^{2}\right)=0 for V give the velocity at point a to be

V_{a}=\sqrt{k R T_{a}} \quad \text { and } \quad Ma _{a}=\frac{V_{a}}{c_{a}}=\frac{\sqrt{k R T_{a}}}{\sqrt{k R T_{a}}}=1                             (9)

Therefore, sonic conditions exist at point a, and thus the Mach number is 1. Setting d T / d s=(d s / d T)^{-1}=0 and solving the resulting equation T(k-1)\left(R T-V^{2}\right)=0  for velocity at point b give

V_{b}=\sqrt{R T_{b}} \quad \text { and } \quad Ma _{b}=\frac{V_{b}}{c_{b}}=\frac{\sqrt{R T_{b}}}{\sqrt{k R T_{b}}}=\frac{1}{\sqrt{k}}                                   (10)

Therefore, the Mach number at point b is Ma _{b}=1 / \sqrt{k} . For air, k = 1.4 and thus Ma _{b}=0.845 .
Discussion Note that in Rayleigh flow, sonic conditions are reached as the entropy reaches its maximum value, and maximum temperature occurs during subsonic flow.