For a rocket engine, the velocity ratio $r$ is $V_a/V_e$, where $V_a$ is the vehicle velocity and $V_e$ is the exit velocity of the exhaust gases relative to the rocket. The value of $r$ at which the propulsive efficiency $\eta_p$ is maximum is _________.
Official Answer Key:
1.0
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2019 Q.55 / AE Q.45)**
**1. Intuitive Physical Setup & Given Parameters**:
- Velocity ratio: $r = \frac{V_a}{V_e}$
- $V_a$: Vehicle flight speed
- $V_e$: Effective exhaust gas velocity relative to the rocket
**2. Step-by-Step Propulsion Derivation**:
1. **Definition of Rocket Propulsive Efficiency ($\eta_p$)**:
Propulsive efficiency is the ratio of useful propulsive power output to total mechanical energy generation rate:
\eta_p = \frac{\text{Thrust Power}}{\text{Thrust Power} + \text{Residual Kinetic Energy in Wake}}
\eta_p = \frac{\dot{m} V_e V_a}{\frac{1}{2} \dot{m} V_e^2 + \dot{m} V_e V_a - \dots} = \frac{2 \left(\frac{V_a}{V_e}\right)}{1 + \left(\frac{V_a}{V_e}\right)^2} = \frac{2r}{1 + r^2}
2. **Maximizing $\eta_p(r)$**:
To find the extremum, differentiate with respect to $r$:
\frac{d\eta_p}{dr} = \frac{2(1 + r^2) - 2r(2r)}{(1 + r^2)^2} = \frac{2 + 2r^2 - 4r^2}{(1 + r^2)^2} = \frac{2(1 - r^2)}{(1 + r^2)^2}
Setting $\frac{d\eta_p}{dr} = 0$:
2(1 - r^2) = 0 \implies r^2 = 1 \implies r = 1.0
3. At $r = 1.0$ ($V_a = V_e$):
\eta_{p,\max} = \frac{2(1)}{1 + 1^2} = \frac{2}{2} = 1.0 = 100\%
The exhaust gases are left completely stationary in the ambient air frame ($V_{exhaust, absolute} = V_a - V_e = 0$), so zero kinetic energy is lost in the exhaust wake!
**3. Comprehensive Options & Distractor Analysis**:
- **Question Type**: Numerical Answer Type (NAT).
- **Exact Evaluation**: **1.0** (Acceptable Range: [1, 1]).
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **1.0**
- **Exam Rule**: Rocket propulsive efficiency is $\eta_p = \frac{2r}{1+r^2}$; maximum $\eta_p = 100\%$ occurs when $r = V_a/V_e = 1.0$.