Question 4.10: Figure 4.33 illustrates the performance map for a turbojet e......

From the performance map with known TIT (1600 K) and compressor pressure ratio (10):

1. The specific thrust (T / \dot{m}_{\text{a}}) is 815 N · s/kg
2. The specific fuel consumption is.3.5 × 10^{-5} kg/N · s

Since

\begin{matrix} \text{TSFC} &=& \frac{\dot{m}_{\text{f}}}{T} \\ \therefore f &=&\frac{\dot{m}_{\text{f}}}{\dot{m}_{\text{a}}}&=&\frac{\dot{m}_{\text{f}}}{T}\frac{T}{\dot{m}_{\text{a}}} &=&\left(3.5 \times 10^{-5}\frac{\text{kg}}{\text{N.s}} \right)\left(815\frac{\text{N.s}}{\text{kg}} \right) \\ \therefore f&=& 0.02853 \end{matrix}

From the energy balance in the combustion chamber:

\begin{matrix} f&=&\frac{Cp_{\text{cc}}T_{4}-Cp_{\text{c}}T_{03}}{\eta_{\text{b}}Q_{\text{R}}-Cp_{\text{cc}}T_{04}} \\ \therefore T_{03} &=&\frac{1}{Cp_{\text{c}}}[Cp_{\text{cc}}T_{04}-f(\eta_{\text{b}}Q_{\text{R}}-Cp_{\text{cc}}T_{04})] \\ &=& \frac{1}{1.005}[1.148 \times 1600-0.02853(0.98 \times 45,000-1.148 \times 1600)] \\ &=& 631 \text{ K} \end{matrix}

The compressor outlet temperature is given by the relation

T_{02}=\frac{T_{03}}{\left[1+\left(\left(\pi_{\text{c}}^{(\gamma_{\text{c}}-1)/\gamma_{\text{c}}}\right)-1 \right) /\eta_{\text{c}}\right] } =291.5 \text{ K}

The temperature ratio within the diffuser is

\frac{T_{02}}{T_{\text{a}}} =\frac{T_{0\text{a}}}{T_{\text{a}}} =1+\frac{\gamma-1}{2}M^2 \\ \begin{matrix} \therefore M&=& \sqrt{\frac{2}{\gamma-1}\left(\frac{T_{02}}{T_{\text{a}}}-1 \right) }=\sqrt{\frac{2}{0.4}\left(\frac{291.5}{247.9}-1 \right) } \\ M&=& 0.938 \end{matrix}

The flight speed is then

V_{\text{f}}=M\sqrt{\gamma RT_{\text{a}}}=296 \text{ m/s}

The diffuser pressure ratio is

P_{02}=P_{\text{a}}\left(1+\eta_{\text{d}}\frac{\gamma_{\text{c}}-1}{2} M^2_{\text{a}}\right) ^{\gamma_{\text{c}}/(\gamma_{\text{c}}-1)}=79 \text{ kPa}

The compressor outlet pressure is

P_{03}=(P_{02})(\pi_{\text{c}})=790 \text{ kPa}

Owing to the pressure drop in the combustion chamber, the outlet pressure for the gases leaving the combustion chamber is

P_{04}=(1-\Delta P)P_{03}=0.99 \times 790=782.1 \text{ kPa}

From the energy balance between the compressor and turbine, the turbine outlet temperature is

\begin{matrix} T_{05} &=& T_{04}-\frac{Cp_{\text{c}}(T_{03}-T_{02})}{(1+f)Cp_{\text{h}}} &=&1600-\frac{1.005(631-291.5)}{1.02853 \times 1.148} \\ &=& 1311 \text{ K} \end{matrix}

The pressure ratio in the turbine is

\frac{P_{05}}{P_{04}} =\left[-\frac{1}{\eta_{\text{t}}}\left(1-\frac{T_{05}}{T_{04}} \right) \right] ^{(\gamma_{\text{h}}/\gamma_{\text{h}}-1)}=\left[1-\frac{1}{0.9}\left(1-\frac{1311}{1600} \right) \right] ^4 =0.4082

Since P_{03}=P_{04}=782.1 \text{ kPa}

Then P_{05}=319.2 \text{ kPa}

Now, the nozzle is to be checked out for choking:

Since P_{\text{c}} is greater than P_{\text{a}} , then the nozzles are choked. The gases leave the nozzle at the temperature

T_7=T_{\text{c}}=\frac{T_{05}}{(\gamma+1)/2} =1123.7 \text{ K}

The jet speed is then

V_7=V_{\text{c}}=\sqrt{\gamma RT_7}=\sqrt{1.333 \times 287 \times 1123.7}=655.75 \text{ m/s}

The specific thrust is given by the relation

\begin{matrix} \frac{T}{\dot{m}_{\text{a}}} &=&(1+f)V_7-V_{\text{f}}+\frac{A_{\text{e}}}{\dot{m}_{\text{a}}}(P_7-P_{\text{a}}) \\ 815&=& 1.02853 \times 655.75 -296 +\frac{0.3}{\dot{m}_{\text{a}}}(167.4-46) \times 10^3 \end{matrix}

The mass flow rate is \dot{m}_{\text{a}} = 83.43 kg/s.
Finally, since the mass flow rate is given by the relation:

\dot{m}_{\text{a}}=\rho_{\text{i}}V_{\text{f}}A_{\text{i}}=\frac{P_{\text{i}}}{RT_{\text{i}}} V_{\text{f}}A_{\text{i}}

The inlet area is then A_{\text{i}} = 0.436 m² .

Hint: From the performance map, it is seen that for a constant TIT, the specific thrust increases with the increase in compression pressure ratio at first and then reduces. Thus, for each TIT there is an optimum (\pi_{\text{c}}) that yields a maximum specific thrust. This optimum (\pi_{\text{c}}) increases as the TIT increases.

\begin{matrix} T&=&\dot{m}_{\text{a}}[(1+f+f_{\text{ab}})V_7-V_{\text{f}}]+A_7(P_7-P_{\text{a}}) \\ \text{TSFC}&=&\frac{\dot{m}_{\text{f}}+\dot{m}_{\text{ab}}}{T}=\frac{f+f_{\text{ab}}}{T/\dot{m}_{\text{a}}} \\ \eta_{\text{p}}&=&\frac{TV}{TV+0.5 \dot{m}_{\text{a}}(1+f+f_{\text{ab}})(V_7-V)^2} \\\eta_{\text{th}} &=& \frac{TV+0.5 \dot{m}_{\text{a}}(1+f+f_{\text{ab}})(V_7-V)^2}{\dot{m}_{\text{a}}(f+f_{\text{ab}})Q_{\text{HV}}} \\ \eta_{\text{o}}&=&\frac{TV}{\dot{m}_{\text{a}}(f+f_{\text{ab}})Q_{\text{HV}}} \end{matrix}

C_{\text{p}_{\text{c}}} : Specific heat at constant pressure for cold air
C_{\text{p}_{\text{h}}} : Specific heat at constant pressure for hot gases
\dot{m}_{\text{a}} : Air mass flow rater
\dot{m}_{\text{f}} : Fuel mass flow rate
ƒ : Fuel-to-air ratio
\dot{m}_{\text{f}_{\text{ab}}} : Fuel burnt in afterburner
f_{\text{ab}} : Afterburner fuel-to-air ratio
Q_{\text{R}} : Fuel heating value
\gamma_{\text{c}} : Ratio of specific heats for cold air
\gamma_{\text{h}} : Ratio of specific heats for hot gases