Difficulty: Medium
Correct Answer: eta_p = 2 * V0 / (V1 + V0)
Explanation:
Introduction / Context:
Propulsive efficiency measures how effectively a propulsion device converts the power in its jet into useful thrust power on the vehicle. For rockets (self-contained propellant), the classical one-dimensional relation links the vehicle speed V0 and jet speed V1. Knowing this formula helps reason about why very high jet speeds are inefficient at low flight speeds.
Given Data / Assumptions:
Concept / Approach:
Propulsive efficiency is defined as useful power divided by rate of energy output in the jet. Useful power = T * V0, where thrust T ≈ m_dot * (V1 − V0) in the ideal model (neglecting pressure terms). Jet power available = 0.5 * m_dot * (V1^2 − V0^2) when referenced to the vehicle frame. Taking the ratio and simplifying yields eta_p = 2 * V0 / (V1 + V0), a compact expression commonly cited for rockets.
Step-by-Step Solution:
Verification / Alternative check:
Limits check: as V0 → V1, eta_p → 1 (ideal matching). As V0 → 0 (static firing), eta_p → 0, meaning no useful propulsive power though thrust exists—consistent with intuition.
Why Other Options Are Wrong:
Other forms either invert the ratio or omit the factor of 2, giving incorrect limiting behavior or values outside [0, 1].
Common Pitfalls:
Confusing propulsive efficiency with overall efficiency (which also includes conversion of chemical energy to jet kinetic energy), or using air-breathing formulas directly for rockets.
Final Answer:
eta_p = 2 * V0 / (V1 + V0)
Discussion & Comments