GATE 2022 AE – Question 64
An ideal ramjet is to operate with exhaust gases optimally expanded to ambient pressure at an altitude where temperature is 220 K. The exhaust speed at the nozzle exit is 1200 m/s at a temperature of 1100 K. Given: γ= 1.4 at 220 K; R= 287 J/(kg-K) for air γ= 1.33 at 1100 K; R= 287 J/(kg-K) for exhaust gases. The cruise speed of this ramjet is ________ m/s (rounded off to nearest integer).
Practise this question in The GATE Grind →
Show answer and explanation
Correct answer: 547
Explanation
**Idea.** In an ideal ramjet the pressure rise comes only from the ram compression, and the exhaust is expanded back to the ambient pressure. The total pressure is the same at the intake and at the nozzle (ideal: no loss), so the stagnation pressure ratio $p_0/p_a$ seen by the nozzle must equal the one produced by the flight speed.
**Step 1: nozzle side.** For the exhaust gas, $\gamma_e=1.33$, $R=287$ J/(kg K):
$$c_{p,e}=\frac{\gamma_eR}{\gamma_e-1}=\frac{1.33\times287}{0.33}=1156.7\ \text{J/(kg K)}.$$
The stagnation temperature of the exhaust is
$$T_{0e}=T_e+\frac{V_e^2}{2c_{p,e}}=1100+\frac{1200^2}{2\times1156.7}=1100+622.5=1722.5\text{ K},$$
and the stagnation-to-ambient pressure ratio is
$$\frac{p_0}{p_a}=\left(\frac{T_{0e}}{T_e}\right)^{\gamma_e/(\gamma_e-1)}=(1.5659)^{4.03}=6.10 .$$
**Step 2: flight side.** In the free stream, $\gamma=1.4$:
$$\frac{p_0}{p_a}=\left(1+0.2M^2\right)^{3.5}=6.10\;\Rightarrow\;1+0.2M^2=1.676\;\Rightarrow\;M=1.838 .$$
**Step 3: speed.** The speed of sound at 220 K is $a=\sqrt{1.4\times287\times220}=297.3$ m/s:
$$V=M\,a=1.838\times297.3=546.6\approx\mathbf{547\ m/s}.$$