Question 8.15: A gaseous mixure of air and fuel enters a ramjet combustion ......

Given: V_1=60  m / s ; T_1=323  K ; p_1=35  kPa ; Q=1160  kJ / kg

Refer Figure 8.29. At inlet to the combustion chamber,

\begin{aligned}M_1 &=\frac{V_1}{a_1}=\frac{V_1}{\sqrt{\gamma R T_1}}\\&=\frac{60}{\sqrt{1.4 \times 287 \times 323}}\\&=0.167\end{aligned}

Using isentropic tables, the stagnation temperature at inlet is:

\begin{aligned}\frac{T_1}{T_{01}}&=\left(\frac{T}{T_0}\right)_{M_1=0.167}=0.994 \\T_{01}&=\frac{T_1}{0.994}=\frac{323}{0.994}\\&=324.95  K\end{aligned}

In the combustion chamber, heat of reaction is added to the fuel air mixure (i.e., the fluid flowing through the chamber)

Q=1160  kJ / kg  \text { of fuel air mixture }

From the equation for heat added in a Rayleigh flow:

\begin{aligned}Q &=c_p\left(T_{02}-T_{01}\right) \\1160 &=1.005\left(T_{02}-324.95\right) \\T_{02}&=1479.18  K\end{aligned}

The Mach number at section 2 can be obtained from the stagnation temperature ratio:

\begin{gathered}\frac{T_{02}}{T_{01}}=\frac{1479.18}{324.95}=\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=0.167}}=\frac{\left(T_0 / T_0{}^*\right)_{M_2}}{0.1244}\\\left(\frac{T_0}{T_0{}^*}\right)_{M_2}=0.5663\end{gathered}

Corresponding to this, the Mach number is:

M_2=0.42

Using Rayleigh, tables, the properties at the exit section can be obtained

\begin{aligned}\frac{T_2}{T_1}&=\frac{T_2 / T^*}{T_1 / T^*}=\frac{\left(T / T^*\right)_{M=0.42}}{\left(T / T^*\right)_{M=0.167}}=\frac{0.65345}{0.1508}=4.333 \\T_2 &=323 \times 4.333 \\&=1399.56  K \\\frac{p_2}{p_1}&=\frac{p_2 / p^*}{p_1 / p^*}=\frac{\left(p / p^*\right)_{M=0.42}}{\left(p / p^*\right)_{M=0.167}}=\frac{1.9247}{8.3091}\end{aligned}

\begin{aligned}&=0.8335 \\p_2 &=35 \times 0.8335 \\&=29.1725    kPa\end{aligned}