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?

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:

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