Question 4.9: A single-spool afterburning turbojet engine is powering a fi......

1. Intake: The intake here is identical to that in Example 3, Chapter 3. Two shock waves are developed in the intake; the first is an oblique one, while the second is normal. The important results from this example are
The total conditions for air upstream of the engine are T_{0\text{a}}=390 \text{ K and }P_{0\text{a}}=7.824 P_{\text{a}} .
The flight speed is V=M \sqrt{\gamma RT_{\text{a}}}=590 \text{ m/s}.
The stagnation pressure ratio in the intake is r_{\text{d}}=\frac{P_{02}}{P_{0\text{a}}} =0.884.
From Equation 4.61 , the diffuser efficiency is

\eta_{\text{d}}=\frac{(1+((\gamma-1)/2)M^2)(r_{\text{d}})^{(\gamma-1)/\gamma}-1}{((\gamma-1)/2)M^2} \quad \quad \quad (4.61) \\ \eta_{\text{d}}=\frac{(1+0.2 \times 4)(0.884)^{0.286}-1}{0.2 \times 4} =0.922

The stagnation temperature of air leaving the diffuser is T_{02}=T_{0\text{a}}=390 \text{ K}.
The stagnation pressure of air leaving the intake is

P_{02}=0.884 \ P_{0\text{a}}=0.884 \times 7.824 \ P_{\text{a}}=0.884 \times 7.824 \times 10.01=69.23 \text{ kPa}

2. Compressor:

P_{03}=\pi_{\text{c}}P_{02}=5 \times 69.23 =346.15 \text{ kPa} \\ T_{03}=T_{02}\left[1+\frac{\pi_{\text{C}}^{(\gamma_{\text{C}}-1)/\gamma_{\text{C}}}-1}{\eta_{\text{C}}} \right] =658.2 \text{ K}

3. Combustion chamber:

\begin{matrix} P_{04}&=& (1- \Delta P_{\text{cc}}),P_{03}&=&0.94 P_{03}=325.38 \text{ kPa} \\ T_{04}&=& 1200 \text{ K, then} \\ f&=&\frac{Cp_{\text{t}}T_{04}-Cp_{\text{c}}T_{03}}{\eta_{\text{b}}Q_{\text{R}}-Cp_{\text{t}}T_{04}}&=&0.0174 \end{matrix}

4. Turbine: Energy balance for compressor and turbine

Cp_{\text{t}}(1+f)(T_{04}-T_{05})=Cp_{\text{c}}(T_{03}-T_{02}) \\ T_{05}=T_{04}-\frac{Cp_{\text{c}}(T_{03}-T_{02})}{Cp_{\text{t}}(1+f)} =1200-\frac{1.005 \times (658.2-390)}{1.157 \times 1.0174} =970.9 \text{ K} \\ P_{05}=P_{04}\left[1-\frac{T_{04}-T_{05}}{\eta_{\text{t}}T_{04}} \right] ^{(\gamma_{\text{t}}/\gamma_{\text{t}}-1)}=325.38\left[1-\frac{1200-970.9}{0.9 \times 1200} \right] ^4=125.37 \text{ kPa}

5. Tail pipe: A pressure drop of 3% is encountered in the afterburner; thus

P_{06 \text{A}}=0.97 \ P_{05}=121.6 \text{ kPa}

The maximum temperature is now P_{06 \text{A}} = 2000 K

f_{\text{ab}}=\frac{Cp_{\text{t}}(T_{06 \text{A}}-T_{05})}{\eta_{\text{ab}}Q_{\text{R}}-Cp_{\text{t}}T_{06\text{A}}} =\frac{1.157 \times (2000-970.9)}{0.9 \times 45,000-1.157 \times 2000} =0.0312

Check nozzle choking with

\frac{P_{06\text{A}}}{P_{\text{c}}} =\frac{1}{(1-(1/\eta_{\text{n}})(\gamma_{\text{n}}-1)/(\gamma_{\text{n}}+1))^{\gamma_{\text{n}}/(\gamma_{\text{n}-1})}} =1.907 \\ \therefore P_{\text{c}}=63.77 \text{ kPa} \\ \because P_{\text{c }}\gt P_{\text{a}}

∴ Nozzle is choked also.

\begin{matrix} P_{7\text{A}}&=&P_{\text{c}}&=&63.77 \text{ kPa} \\ T_{7\text{a}}&=&\frac{T_{06\text{A}}}{(\gamma_{\text{n}}+1)/2} &=&1716.9 \text{ K} \\ V_{7\text{A}}&=&a_{7\text{A}}&=&\sqrt{\gamma_{\text{n}}RT_{7\text{A}}}&=&809.5 \text{ m/s} \end{matrix}

The nozzle area is calculated from the relation

\begin{matrix} A_{7\text{A}} &=&\frac{\dot{m}_{\text{a}}(1+f+f_{\text{ab}})}{\rho_{7\text{A}}V_{7\text{A}}} =\frac{RT_{7\text{A}}\dot{m}_{\text{a}}(1+f+f_{\text{ab}})}{P_{7\text{A}}V_{7\text{A}}} \\ &=& \frac{287 \times (1716.9)\times (15)\times (1+0.017+0.0312)}{63.77 \times 10^3 \times (809.5)} \\ &=& 0.15 \text{ m}^2 \end{matrix}

The thrust force for operative afterburner is

\begin{matrix} T_{\text{ab}} &=& \dot{m}_{\text{a}}[(1+f+f_{\text{ab}})V_{7\text{A}}-V]+A_{7\text{A}}(P_{7\text{A}}-P_{\text{a}}) \\ &=& 15[1.0486 \times 809.5-590]+0.15 \times 10^3(63.77-10.01)=11,946 \text{ N} \\ &=& 11.946 \text{ kN} \end{matrix}