Question 4.4: Compare between the thrust-specific fuel consumption (TSFC) ......

1. Ramjet engine
The gases in the nozzle expand to the ambient condition. With

\begin{matrix} T_{\text{a}}&=&-69.7+273.15&=&203.5 \text{ K} \\ P_{\text{a}}&=&14.75 \text{ kPa and }M&=&2 \end{matrix}

The flight speed is u_{\text{a}}=M\sqrt{\gamma RT_{\text{a}}}=2 \sqrt{(1.4)(287)(203.5)}

\begin{matrix} \therefore u_{\text{a}}&=& 571.8 \text{ m/s} \\ \frac{T_{0\text{a}}}{T_{\text{a}}} &=&1+\frac{\gamma-1}{2}M^2=1+0.2 \times 4=1.8 \\ \therefore T_{0\text{a}}&=&366.3 \text{ K} \\ \frac{P_{0\text{a}}}{P_{\text{a}}}&=&\left(1+\frac{\gamma-1}{2} M^2\right)^{\gamma/(\gamma-1)}=(1.8)^{3.5}=7.824 \\ P_{0\text{a}}&=&115 \text{ kPa} \end{matrix}

Diffuser: For ideal diffuser, then

T_{02}=T_{0\text{a}}=366.3 \text{ K} \\ P_{0\text{a}}=P_{02}=P_{04}=115 \text{ kPa}

Combustion chamber: The fuel-to-air ratio is

\begin{matrix} f&=&\frac{C_{\text{P}}(T_{04}-T_{0a})}{Q_{\text{R}}-C_{\text{P}}T_{04}} \\ f&=& 0.04582 \end{matrix}

Nozzle: Since all the processes are ideal, then the velocity ratio is obtained from Equation 3.18b (\therefore u_{\text{e}}=u\sqrt{\frac{T_{04}}{T_{0\text{a}}}}\equiv u\sqrt{\frac{T_{04}}{T_{02}} }) as

\frac{u_{\text{e}}}{u_{\text{a}}} =\sqrt{\frac{T_{04}}{T_{0\text{a}}} }=\sqrt{\frac{2225}{366.3} }=2.465 \\ \therefore u_{\text{e}}=1410 \text{ m/s}

The specific thrust is

\begin{matrix} \therefore \frac{T}{\dot{m}_{\text{a}}} &=& u_{\text{a}} \left[(1+f)\frac{u_{\text{e}}}{u_{\text{a}}}-1 \right]&=&572[1.04582 \times 2.465-1] \\ \frac{T}{\dot{m}_{\text{a}}}&=&902.6\frac{\text{N}}{\text{(kg/s)}} \\ \text{TSFC}&=&\frac{f}{(T/\dot{m}_{\text{a}})}&=&0.1828\frac{\text{kg}}{\text{N.h}} \end{matrix}

2. Turbojet
An ideal cycle is assumed and the nozzle is unchoked. The outlet conditions of the diffuser are the same as in the ramjet case,

\therefore T_{02}=366.3 \text{ K} \quad \text{ and }\quad P_{02}=115 \text{ kPa}

Moreover, there is no pressure drop in the combustion chamber, thus P_{04}=P_{03}

There is no pressure drop in the afterburner also, thus P_{05}=P_{06\text{A}} .
Nozzle: Since the nozzle is unchoked, and the exhaust speed is the same as in the case of ramjet, then

\begin{matrix} T_{06\text{A}}&=&T_7+\frac{V^2_7}{2C_{\text{P}}} \\ T_7&=&T_{06\text{A}}-\frac{V^2_7}{2C_{\text{P}}}=2225-\frac{(1410)^2}{2 \times 1005} \\ T_7&=&1236 \text{ K} \\ \frac{P_{06\text{A}}}{P_7} \equiv \frac{P_{06\text{A}}}{P_{\text{a}}} &=&\left(\frac{T_{06\text{A}}}{T_7} \right)^{\gamma/(\gamma-1)}=\left(\frac{2225}{1236} \right)^{1.4/0.4}=7.829 \\ P_{06\text{A}}&=&P_{05}=115.5 \text{ kPa} \end{matrix}

Compressor

\begin{matrix} \frac{P_{03}}{P_{02}}&=& \pi_{\text{c}}&=&8 \\ \therefore P_{03}&=&920 \text{ kPa} &\equiv& P_{04} \\ \frac{T_{03}}{T_{02}}&=&(8)^{(\gamma-1)/\gamma}&=&8^{(0.4/1.4)}&=&8^{0.286}=1.811 \\ T_{03}&=&663 \text{ K} \end{matrix}

Turbine

\begin{matrix} \frac{T_{04}}{T_{05}} &=&\left(\frac{P_{04}}{P_{05}} \right) ^{(\gamma-1)/\gamma}&=&\left(\frac{P_{03}}{P_{06\text{A}}} \right) ^{0.4/1.4} \\ &=&\left(\frac{920}{115.5} \right) ^{0.286}&=&1.8103 \\ \therefore T_{05}&=&700 \text{ K} \\ \therefore T_{04}&=&1267.2 \text{ K} \end{matrix}

Combustion chamber

\begin{matrix}f&=&\frac{C_{\text{P}}(T_{04}-T_{03})}{Q_{\text{R}}-C_{\text{P}}T_{04}} &=&\frac{1.005 \times (1267.2-663)}{43,000-1.005 \times 1267.2} \\ f&=&0.0146 \end{matrix}

To determine the fuel-to-air ratio of afterburner f_{\text{ab}} , consider the energy balance of afterburner.

From Equation 4.9 (f_{\text{ab}}=\frac{(1+f)(Cp_{6\text{A}}T_{06\text{A}}-CP_5T_{05})}{Q_{\text{R}}-Cp_{6\text{A}}T_{06\text{A}}} ), the afterburner fuel-to-air ratio is

\begin{matrix} f_{\text{ab}} &=& \frac{(1+f)c_{\text{P}}(T_{06}-T_{05})}{Q_{\text{R}}-c_{\text{P}}T_{06}}&=&\frac{(1.0171)(1.005)(2225-700)}{43,000-1.005 \times 2225} \\ f_{\text{ab}}&=& 0.03815 \end{matrix}

The total fuel-to-air ratio of the turbojet engine is f_{\text{total}}=f+f_{\text{ab}}

\begin{matrix} f_{\text{total}} &=&0.05275 \\ \frac{T}{\dot{m}_{\text{a}}}&=&(1+f_{\text{tot}})u_{\text{e}}-u &=&1.05275 \times 1410-572 \\ \frac{T}{\dot{m}_{\text{a}}} &=&912.4\frac{\text{N}}{\text{(kg/s)}} \\ \text{TSFC}&=&0.20813\frac{\text{kg}}{\text{N.h}} \end{matrix}

SUMMARY

The afterburning turbojet engine produces more thrust at the expense of more fuel consumption compared to the ramjet.
The ratio between the power consumed in driving the compressor out of the total turbine power [i.e., (λ)] is to be calculated from the relation

Cp(T_{03}-T_{02})=\lambda Cp(T_{04}-T_{05}) \\ \lambda=\frac{663-336}{1267.2-700}=0.57652