Overall efficiency of a compressed-air system: How is the overall system efficiency defined when air is compressed, stored, and then expanded in an air motor to deliver shaft output?
Correct Answer: Ratio of shaft output of the air motor to the shaft input to the compressor
Introduction / Context:Compressed-air systems are energy conversion chains: mechanical input drives a compressor, compressed air is stored/conditioned, and an air motor reconverts the pneumatic energy to mechanical output. A clear definition of overall efficiency is necessary for system-level economics and design.
Given Data / Assumptions:
- Single compressor feeding an air motor (or pneumatic tools).
- Steady-state operation during the measurement interval.
- Losses in storage, piping, and conditioning implicitly included.
Concept / Approach:
Overall (system) efficiency equals useful shaft output divided by required shaft input across the entire chain. Therefore, η_overall = (air motor brake power) / (compressor shaft power). It aggregates all inefficiencies: compression losses, storage and distribution losses, and motor conversion losses.
Step-by-Step Solution:
Identify input: compressor shaft power.Identify useful output: air motor shaft power.Form the ratio output / input.Hence select option defining exactly that ratio.Verification / Alternative check:
Energy balance over the combined equipment set confirms that any other formulation (e.g., inverse or product) lacks physical meaning for efficiency bounded between 0 and 1.
Why Other Options Are Wrong:
The inverse ratio exceeds 1 for typical conditions; the product has units of power squared and is not an efficiency; indicated-to-brake compares internal to shaft within only the air motor, not the system.
Common Pitfalls:
Confusing component efficiencies with overall system efficiency; neglecting throttling and leakage losses that the overall metric captures implicitly.
Final Answer:
Ratio of shaft output of the air motor to the shaft input to the compressor