Question 8.14: Air enters a constant area pipe with velocity 150 m/s, tempe......

Given: V_1=150  m / s ; T_1=333  K ; p_1=0.5  MN / m ^2 ; Q=180  kJ / kg

Refer Figure 8.28. From the definition of Mach number, the Mach number at the inlet section is:

\begin{aligned}M_1 &=\frac{V_1}{a_1}=\frac{V_1}{\sqrt{\gamma R T_1}}=\frac{150}{\sqrt{1.4 \times 287 \times 333}}\\&=0.41\end{aligned}

The stagnation properties at the inlet section can be obtained from the isentropic stagnation property ratios available in the isentropic tables (γ = 1.4). Corresponding to M_1 = 0.41.

\begin{aligned}& \frac{T_1}{T_{01}}=\left(\frac{T}{T_0}\right)_{M=0.41}=0.9676 \\& T_{01}=\frac{333}{0.9676}\\&=344.1  K \\& \frac{p_1}{p_{01}}=\left(\frac{p}{p_0}\right)_{M=0.41}=0.8907 \\& p_{01}=\frac{0.5}{0.8907}\\&=0.5614   MN / m ^2 \\&\end{aligned}

The stagnation temperature at the exit section can be obtained from the equation for heat added.

\begin{aligned}Q &=h_{02}-h_{01}=c_p\left(T_{02}-T_{01}\right) \\180 &=1.005\left(T_{02}-344.1\right) \\T_{02}&=523.2  K\end{aligned}

Mach number at the exit section can be obtained from the stagnation temperature ratio between the inlet and exit section.

\frac{T_{02}}{T_{01}}=\frac{T_{02} / T_0^*}{T_{01} / T_0^*}=\frac{\left(T_0 / T_0^*\right)_{M 2}}{\left(T_0 / T_0^*\right)_{M=0.41}}

From Rayleigh tables, corresponding to M_1=0.41 .\left(T_0 / T_0^*\right)_{M=0.41}=0.542

\begin{aligned}\frac{523.2}{344.1}&=\frac{\left(T_0 / T_0^*\right)_{M_2}}{0.542}\\\left(\frac{T_0}{T_0^*}\right)_{M_2}&=0.8488\end{aligned}

Corresponding to this value of stagnation temperature ratio, from Rayleigh tables (Appendix A3), the Mach number at the second section is:

M_2=0.63

Properties at the second section can be obtained from the Rayleigh property ratios available in Rayleigh table (γ = 1.4)

\begin{aligned}\frac{p_2}{p_1}&=\frac{p_2 / p^*}{p_1 / p^*}=\frac{\left(p / p^*\right)_{M=0.63}}{\left(p / p^*\right)_{M=0.41}}=\frac{1.5415}{1.941}=0.7942 \\p_2 &=0.5 \times 0.7942 \\&=0.3971   MN / m ^2 \\\frac{p_{02}}{p_{01}}&=\frac{p_{02}/ p^*}{p_{01}/ p_0 *}=\frac{\left(p_0 / p_0{}^*\right)_{M=0.63}}{\left(p_0 / p_0{}^*\right)_{M=0.41}}=\frac{1.0648}{1.1521}=0.924 \\p_{02}&=0.5614 \times 0.924 \\&=0.5187   MN / m ^2\end{aligned}

Change in stagnation pressure,

p_{01}-p_{02}=0.5614-0.5187=0.0427   MN / m ^2

The static temperature ratio between the two section is:

\frac{T_2}{T_1}=\frac{T_2 / T^*}{T_1 / T^*}=\frac{\left(T / T^*\right)_{M=0.63}}{\left(T / T^*\right)_{M=0.41}}=\frac{0.9443}{0.6343}=1.489

Change in entropy, can be obtained from the general equation for entropy change,

\begin{aligned}s_2-s_1 &=c_p \ln \frac{T_2}{T_1}-R \ln \frac{p_2}{p_1}\\&=1.005 \ln (1.489)-0.287 \ln (0.7942) \\&=0.4662   kJ / kg K .\end{aligned}