Question 4.6: A single-spool turbojet engine is powering an aircraft flyin......
A single-spool turbojet engine is powering an aircraft flying at Mach number of 0.85 at altitude where the ambient conditions are T_{\text{a}} = 233 \text{K and } P_\text{a} = 26.4 \text{ kPa}.The nozzle is choked and the gases are leaving the nozzle with velocity of 600 m/s; the nozzle area is 0.2 m² . The maximum temperature is 1200 K and fuel heating value is 43, 000 kJ/kg. It is required to calculate
(a) Compressor pressure ratio
(b) The fuel-to-air ratio
(c) Thrust force
(e) Overall efficiency
(a) Compressor inlet temperature is
T_{02}=T_{\text{a}}\left(1+\frac{\gamma -1}{2}M^2 \right) =255 \text{ K}
Since no losses are assumed, then states (5) and (6) are coincident, or
T_{06}=T_{05}, \quad P_{06}=P_{05}
Since the nozzle is choked
\begin{matrix} \frac{T_{05}}{T_{\text{C}}} &=& \frac{\gamma +1}{2} \\ V_{\text{j}}&=&\sqrt{\gamma RT_{\text{C}}}&=& \sqrt{\frac{2\gamma}{\gamma+1}RT_{05} } \\ T_{05}&=&\frac{\gamma+1}{2 \gamma R}V^2_{\text{j}}&=&1075 \text{ K} \end{matrix}
The temperature drop in the turbine is
(T_{04}-T_{05})=1200-1075=125 \text{ K}
The fuel-to-air ratio is
f=\frac{Cp_{\text{h}}T_{04}-Cp_{\text{c}}T_{03}}{Q_{\text{R}}-Cp_{\text{h}}T_{04}}
Since W_{\text{c}}=W_{\text{t}}
\begin{matrix} Cp_{\text{c}}(T_{03}-T_{02})&=& (1+f)Cp_{\text{h}}(T_{04}-T_{05}) \\ \therefore (T_{03}-T_{02}) &=& \frac{Cp_{\text{h}}}{Cp_{\text{c}}}\left(1+\frac{Cp_{\text{h}}T_{04}-Cp_{\text{c}}T_{03}}{Q_{\text{R}}-Cp_{\text{h}}T_{04}} \right) (T_{04}-T_{05}) \\ T_{03}&=& \frac{T_{02}+(Cp_{\text{h}}/Cp_{\text{c}})(1+(Cp_{\text{h}}T_{04}/Q_{\text{R}}-Cp_{\text{h}}T_{04}))(T_{04}-T_{05})}{1+(Cp_{\text{h}}(T_{04}-T_{05})/(Q_{\text{R}}-Cp_{\text{h}}T_{04}))}&=& 398 \text{ K} \\ \pi_{\text{c}}&=&\left(\frac{T_{03}}{T_{02}} \right)^{\gamma/(\gamma-1)}&=&\left(\frac{398}{255} \right)^{3.5}&=&4.75 \end{matrix}(b) The fuel-to-air ratio is then
f=\frac{Cp_{\text{h}}T_{04}-Cp_{\text{c}}T_{03}}{Q_{\text{R}}-Cp_{\text{h}}T_{04}} =0.02374
(c) To calculate the thrust, the pressure at the exit of the engine must be defined first. It will be calculated from the ambient pressure as follows:
\begin{matrix} P_{02}&=&P_{\text{a}}\left(1+\frac{\gamma-1}{2}M^2 \right) ^{\gamma_{\text{c}}/(\gamma_{\text{c}}-1)}&=&42.34 \text{ kPa} \\ P_{03} &=& \pi_{\text{c}}P_{02}&=&201.1 \text{ kPa} \\ P_{03}&=& P_{04} \\ P_{05}&=& P_{04}\left(\frac{T_{05}}{T_{04}} \right)^{\gamma_{\text{c}}/(\gamma_{\text{c}}-1)}&=&129.5 \text{ kPa} \\ P_6&=& P_{\text{C}}&=&\frac{P_{05}}{((\gamma_{\text{h}}+1)/2)^{\gamma_{\text{h}}/(\gamma_{\text{h}}-1)}} &=&70.31 \text{ kPa} \end{matrix}
The mass flow rate is calculated from the outlet conditions
\begin{matrix} T_6&=&T_{\text{C}} &=&\frac{T_{05}}{(\gamma_{\text{h}}+1)/2}&=&922.7 \text{ K} \\ \rho_6 &=& \rho_{\text{c}}&=&\frac{P_{\text{c}}}{RT_{\text{c}}}&=&0.266 \text{kg/m}^3 \\ \dot{m}_{\text{e}} &=& \rho_6V_6A_6&=&31.86 \text{ kg/s} \\ \dot{m}_{\text{a}}&=&\frac{\dot{m}_{\text{e}}}{1+f}&=&31.12\text{ kg/s} \\ \dot{m}_{\text{f}}&=&\dot{m}_{\text{e}}-\dot{m}_{\text{a}} &=&0.74 \text{ kg/s} \\ T&=& \dot{m}_{\text{e}}V_{\text{j}}-\dot{m}_{\text{a}}V_{\text{f}}+A_{\text{e}}(P_{\text{e}}-P_{\text{a}})&=&19.98 \text{ kN} \end{matrix}
(d) The overall efficiency
\eta_{\text{o}}=\frac{TV_{\text{f}}}{\dot{m}_{\text{f}}Q_{\text{R}}} =0.1597=15.97\%