Question

A supersonic vehicle powered by a ramjet engine is cruising at \(1000~\mathrm{m/s}\). The ramjet engine burns hydrogen in a subsonic combustor to produce thrust. The heat of combustion of hydrogen is \(120~\mathrm{MJ/kg}\). The overall efficiency of the engine \(\eta_0\), defined as the ratio of propulsive power to the total heat release in the combustor, is 40%. Taking \(g_0=10~\mathrm{m/s^2}\), the specific impulse of the engine is \underline{\hspace{1cm}} seconds. \;(round off to nearest integer)

Hint:

For airbreathing engines, \(I_{sp}\) can be estimated directly using overall efficiency: \(I_{sp} \approx \eta_0 Q /(V g_0)\). Note that high \(Q\) and low flight speed give very high \(I_{sp}\).

Answer 1

Step 1: Recall definitions.
- Heat release per unit fuel mass: \[ Q = 120~\mathrm{MJ/kg} = 120\times 10^6~\mathrm{J/kg}. \] - Efficiency definition: \[ \eta_0 = \frac{\text{Propulsive Power}}{\text{Fuel Energy Release Rate}}. \] - Propulsive power is thrust \(\times\) flight speed: \[ P_\text{prop} = F V. \] - Fuel energy release rate (per second fuel consumption \(\dot m_f\)): \[ P_\text{fuel} = \dot m_f Q. \]

Step 2: Express thrust per unit fuel flow.
By efficiency definition, \[ \eta_0 = \frac{F V}{\dot m_f Q}. \] Rearrange: \[ \frac{F}{\dot m_f} = \frac{\eta_0 Q}{V}. \]

Step 3: Specific impulse relation.
Specific impulse: \[ I_{sp} = \frac{F}{\dot m_f g_0}. \] Substitute from above: \[ I_{sp} = \frac{1}{g_0}\cdot \frac{\eta_0 Q}{V}. \]

Step 4: Insert numbers.
\(\eta_0=0.4\), \(Q=120\times 10^6\ \mathrm{J/kg}\), \(V=1000\ \mathrm{m/s}\), \(g_0=10\ \mathrm{m/s^2}\). \[ I_{sp} = \frac{0.4 \times 120\times 10^6}{1000 \times 10}. \] \[ I_{sp} = \frac{48\times 10^6}{10^4} = 4800~\mathrm{s}. \]

Final Answer:
\[ \boxed{4800\ \text{seconds}} \]