Answer
$\dot{Q}=-1050 KW$
Work Step by Step
Knowing stagnation properties, the static properties are determined to be $$
\begin{aligned}
& T_1=T_{01}\left(1+\frac{k-1}{2} \mathrm{Ma}_1^2\right)^{-1}=(350 \mathrm{~K})\left(1+\frac{1.4-1}{2} 1.2^2\right)^{-1}=271.7 \mathrm{~K} \\
& P_1=P_{01}\left(1+\frac{k-1}{2} \mathrm{Ma}_1^2\right)^{-k /(k-1)}=(240 \mathrm{kPa})\left(1+\frac{1.4-1}{2} 1.2^2\right)^{-1.4 / 0.4}=98.97 \mathrm{kPa} \\
& \rho_1=\frac{P_1}{R T_1}=\frac{98.97 \mathrm{kPa}}{(0.287 \mathrm{~kJ} / \mathrm{kgK})(271.7 \mathrm{~K})}=1.269 \mathrm{~kg} / \mathrm{m}^3
\end{aligned}
$$ Then the inlet velocity and the mass flow rate become
$$
\begin{aligned}
& c_1=\sqrt{k R T_1}=\sqrt{(1.4)(0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(271.7 \mathrm{~K})\left(\frac{1000 \mathrm{~m}^2 / \mathrm{s}^2}{1 \mathrm{~kJ} / \mathrm{kg}}\right)}=330.4 \mathrm{~m} / \mathrm{s} \\
& V_1=\mathrm{Ma}_1 c_1=1.2(330.4 \mathrm{~m} / \mathrm{s})=396.5 \mathrm{~m} / \mathrm{s} \\
& \dot{m}_{\text {air }}=\rho_1 A_{c 1} V_1=\left(1.269 \mathrm{~kg} / \mathrm{m}^3\right)\left[\pi(0.20 \mathrm{~m})^2 / 4\right](396.5 \mathrm{~m} / \mathrm{s})=15.81 \mathrm{~kg} / \mathrm{s}
\end{aligned}
$$ The Rayleigh flow functions $T_d T_0{ }^*$ corresponding to the inlet and exit Mach numbers are (Table A-34):
$$
\begin{array}{ll}
\mathrm{Ma}_1=1.8: & T_{01} / T_0{ }^*=0.9787 \\
\mathrm{Ma}_2=2: & T_{02} / T_0{ }^*=0.7934
\end{array}
$$ Then the exit stagnation temperature is determined to be
$$
\frac{T_{02}}{T_{01}}=\frac{T_{02} / T_0^*}{T_{01} / T_0^*}=\frac{0.7934}{0.9787}=0.8107 \quad \rightarrow \quad T_{02}=0.8107 T_{01}=0.8107(350 \mathrm{~K})=283.7 \mathrm{~K}
$$ Finally, the rate of heat transfer is
$$
\dot{Q}=\dot{m}_{\mathrm{irr}} c_p\left(T_{02}-T_{01}\right)=(15.81 \mathrm{~kg} / \mathrm{s})(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(283.7-350) \mathrm{K}=-1053 \mathrm{~kW} \cong-1050 \mathrm{~kW}
$$
