Question 4.3: A single-spool turbojet engine has the following data: Flyin......

The given data are rearranged as follows:

M=0.6, \ T_{\text{a}}=222 \text{ K}, \ P_{\text{a}}=25 \text{ kPa}, \ \pi_{\text{c}}=6 \\ T_{\text{o max}}=1300 \text{ K}, \ Q_{\text{R}}=45,000 \text{ kJ/kg}, \ A_{\text{e}}=0.2 \text{ m}^2

The flight speed

V=M\sqrt{\gamma RT_{\text{a}}}=0.6 \sqrt{1.4 \times 287 \times 222}=179.2 \text{ m/s}

Turbojet Engine

Diffuser

\begin{matrix} T_{02}&=&T_{\text{a}}\left(1+\frac{\gamma-1}{2}M^2 \right)=238 \text{ K} \\ P_{02}&=&P_{\text{a}}\left(1+\frac{\gamma-1}{2} M^2\right)^{\gamma/(\gamma-1)} =31.88 \text{ kPa} \end{matrix}

Compressor

T_{03}=T_{02}\pi_{\text{c}}^{(\gamma-1)/\gamma}=397.3 \text{ K} \\ P_{03}=\pi_{\text{c}}P_{02}=191.33 \text{ kPa}

Combustion chamber

\begin{matrix} T_{04}&=&1300 \text{ K} \\ P_{04}&=&P_{03}&=&191.33 \text{ kPa} \\ f&=&\frac{Cp(T_{04}-T_{03})}{Q_{\text{R}}-CpT_{04}}&=&\frac{1.01(1300-397.3)}{45000-1.01 \times 1300}&=&0.02087 \end{matrix}

Turbine: From Equation 4.6 (\left(\frac{T_{05}}{T_{04}} \right) =1-\frac{(Cp_{\text{c}}/Cp_{\text{h}})T_{02}}{\lambda(1+f)T_{04}} \left[\left(\frac{T_{03}}{T_{02}} \right)-1 \right] ), with constant specific heats and λ = 0.8, then

\begin{matrix} T_{05}&=&T_{04}-\frac{(T_{03}-T_{02})}{0.8(1+f)}&=&1105 \text{ K} \\ P_{05}&=&P_{04}\left(\frac{T_{05}}{T_{04}} \right)^{\gamma/(\gamma-1)}&=&182.63 \text{ kPa} \end{matrix}

Tail pipe: Ideal flow is assumed in the tail pipe. Thus there is no pressure or temperature drop, or

T_{06}=T_{05}\text{ and }P_{06}=P_{05}

Nozzle
Case 1 Check if the nozzle is choked.

P_{\text{c}}=\frac{P_{06}}{((\gamma+1)/2)^{\gamma/(\gamma-1)}} =96.48 \text{ kPa}

Since P_{\text{a}} = 25 kPa, then P_{\text{c}} \gt P_{\text{a}} , thus the nozzle is choked.

\begin{matrix} \therefore P_7&=&P_{\text{C}}&=&96.48 \text{ kPa} \\ \therefore T_7&=& T_{\text{c}}&=&\frac{T_{06}}{(\gamma+1)/2}=920.8 \text{ K}\\ V_7&=& \text{sonic speed}&=&\sqrt{\gamma RT_7}=608.26 \text{ K} \end{matrix}

The mass flow rate is calculated as follows:

\dot{m}=\rho_7V_7A_7=\frac{P_7}{RT_7}V_7A_7=\frac{96.48 \times 10^3}{287 \times 920.8} \times 608.26 \times 0.2=44.41 \text{ kg/s}

The thrust force T is expressed by the relation

\begin{matrix} T&=&\dot{m}_{\text{a}}[(1+f)V_7-V]+A_{\text{e}}(P_{\text{e}}-P_{\text{a}}) \\ &=&44.41 \times (1.02087 \times 608.26-179.2)+0.2 \times 10^3 \times (96.48-25) \\ &=& 33914 \ N=33.914 \text{ kN} \end{matrix}

Case 2 The nozzle is unchoked, then

\begin{matrix} P_{7}&=&P_{\text{a}}=25 \text{ kPa} \\ \therefore T_7&=&T_{05}\left(\frac{P_7}{P_{05}} \right)^{(\gamma-1/\gamma)}&=&625.7 \text{ K} \end{matrix}

The exhaust speed is calculated as follows: V_7=\sqrt{2Cp(T_{05}-T_7)}=983.96 \text{ m/s}

This means that the exhaust Mach number, M_7=\frac{V_7}{\sqrt{\gamma RT_7}} =1.962

The thrust force is calculated as follows:

\begin{matrix} T&=& \dot{m}_{\text{a}}[(1+f)V_7-V]&=&44.41 \times (1.02087 \times 983.96-179.2)&=&36651 \text{ N} \\ &=& 36.65 \text{ kN} \end{matrix}

Thus, the thrust force for choked nozzle is less than its value for unchoked nozzle.