Question 3.3: For an ideal ramjet, it is required to prove that the therma......

(Hint: use the following approximations: f= Cp(T_{04}-T_{02}) / Q_{\text{R}} \text{ and } \dot{m}_{\text{f}} \ll \dot{m}_{\text{a}})

As derived in Chapter 2, the thermal efficiency is expressed by the relation

\eta_{\text{th}}=\frac{(\dot{m}_{\text{a}}+\dot{m}_{\text{f}})V_6^2-\dot{m}_{\text{a}}V^2}{2 \dot{m}_{\text{f}}Q_{\text{R}}}

Now, with \dot{m}_{\text{a}} \ll \dot{m}_{\text{f}} , then

\eta_{\text{th}}=\frac{\dot{m}_{\text{a}}(V_6^2-V^2)}{2 \dot{m}_{\text{f}}Q_{\text{R}}}

with f= \dot{m_{\text{f}}}/\dot{m_{\text{a}}} \text{ and } f=Cp(T_{04}-T_{02})/Q_{\text{R}}

\therefore \eta_{\text{th}}=\frac{(V_6^2-V^2)}{2Cp(T_{04}-T_{02})}

The flight speed V=M\sqrt{\gamma RT_{\text{a}}}
The exhaust velocity V_6=M_{\text{e}}\sqrt{\gamma RT_6}
For an ideal ramjet M_{\text{e}}=M

\begin{matrix} \therefore \eta_{\text{th}} &=& \frac{M^2(\gamma RT_6-\gamma RT_{\text{a}})}{2Cp(T_{04}-T_{02})} \\ &=& \frac{\gamma RT_{\text{a}}M^2\left(\frac{T_6}{T_{\text{a}}}-1 \right) }{2CpT_{02}\left(\frac{T_{04}}{T_{02}}-1 \right) } &&& (3) \end{matrix}

Since

\begin{matrix} \frac{P_{04}}{P_6} &=&\frac{P_{02}}{P_{\text{a}}} \\ \left(\frac{T_{04}}{T_6} \right)^{\left(\frac{\gamma -1}{\gamma} \right) } &=& \left(\frac{T_{02}}{T_{\text{a}}} \right)^{\left(\frac{\gamma -1}{\gamma} \right) } \end{matrix}

or \frac{T_{04}}{T_6}=\frac{T_{02}}{T_{\text{a}}}       as γ = constant

\therefore \frac{T_{04}}{T_{02}} =\frac{T_6}{T_{\text{a}}} \quad \quad \quad (4)

From Equations 3 and 4

\therefore \eta_{\text{th}}=\frac{\gamma RT_{\text{a}}M^2}{2Cp T_{02}} =\frac{\gamma RT_{\text{a}}M^2}{2\left(\frac{\gamma R}{\gamma -1} \right) } T_{02}

But

\tau_{\text{r}}=\frac{T_{02}}{T_{\text{a}}} =\left(1+\frac{\gamma-1}{2} M^2 \right)

Then

\eta_{\text{th}}=\frac{\left(\frac{\gamma -1}{2} \right)M^2 }{T_{02}/T_{\text{a}}} = \frac{\left(\frac{\gamma -1}{2} \right)M^2 }{\left(1+\frac{\gamma-1}{2} M^2 \right) } =1-\frac{1}{\left(1+\frac{\gamma-1}{2} M^2 \right) } \\ \therefore \eta_{\text{th}}= 1-\frac{1}{\tau_{\text{r}}}