Question 12.12: 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 onedimensional flow of an ideal gas with constant properties through a constant crosssectional area duct with negligible frictional effects) are valid.
Analysis   The differential forms of the continuity (𝜌V = constant), momentum [rearranged as P + (𝜌V )V = constant], ideal gas (P = 𝜌RT ), and enthalpy change ($Δh = c_p ΔT$ ) equations are 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. 12–40) of an ideal gas with constant specific heats is

$s_2-s_1=c_P \ln \frac{T_2}{T_1}-R \ln \frac{P_2}{P_1}$                (12.40)

$d s=c_p \frac{d T}{T}-R \frac{d P}{P}$              (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

$c_p-R=c_v \quad \rightarrow \quad k c_v-R=c_v \quad \rightarrow \quad c_v=R /(k-1)$

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,

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

Setting ds/dT = 0 and solving the resulting equation R(kRT − V²) = 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 $dT /ds = (ds /dT )^{−1} = 0$ and solving the resulting equation $T (k  −  1) × (RT  −  V^2) = 0$ 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.