Question 8.13: Air at Mach number 1.5 pressure 300 kN/m^2, and temperature ......

Given: M_1=1.5 ; p_1=300  kN / m ^2 ; T_1=288  K ; V_2=a_2

Refer Figure 8.27. Using isentropic table (γ = 1.4) values, corresponding to M_1 = 1.5, the stagnation temperature at the inlet section T_{01} can be calculated.

\begin{aligned}\frac{T_1}{T_{01}}&=\left(\frac{T}{T_0}\right)_{M=1.5}=0.6896 \\T_{01}&=\frac{288}{0.6896}\\&=418  K\end{aligned}

Since the velocity of flow at the exit section is sonic, the exit Mach number is unity and the properties at this section are critical properties. Hence, the flow properties at exit can be directly obtained from property ratios available in the Rayleigh flow tables (Appendix A3).

\begin{aligned}\frac{p_1}{p_2}&=\frac{p_1}{p^*}=\left(\frac{p}{p^*}\right)_{M=1.5}=0.578 \\p_2 &=p^*=\frac{300}{0.578}\\&=519.03    kN / m ^2 \\\frac{T_1}{T_2}&=\frac{T_1}{T^*}=\left(\frac{T}{T^*}\right)_{M=1.5}=0.753 \\T_2 &=T^*=\frac{288}{0.753}\\&=382.47 K \\\frac{T_{01}}{T_{02}}&=\frac{T_{01}}{T_0^*}=\left(\frac{T_0}{T_0^*}\right)_{M=1.5}=0.909 \\T_{02}&=T_0^*=\frac{418}{0.909}\\&=459.84  K\end{aligned}

Heat added/kg of air,

\begin{aligned}Q &=c_p\left(T_{02}-T_{01}\right) \\&=1.005(459.84-418) \\&=42.05   kJ / kg .\end{aligned}