Reciprocating compressor — volumetric efficiency with clearance ratio If the clearance ratio (fraction of clearance volume to swept volume) of a single-stage reciprocating air compressor is K, the ideal volumetric efficiency (eta_v) is given by which expression (p2/p1 = compression ratio, n = polytropic index)?

Introduction / Context:
Clearance volume traps a portion of compressed air that re-expands during the suction stroke, displacing part of the incoming fresh charge and reducing volumetric efficiency. A standard formula captures this effect for ideal valves and polytropic compression/expansion.

Given Data / Assumptions:

• Single-stage, single-acting reciprocating compressor.
• Clearance ratio K = V_c / V_s.
• Compression and re-expansion are polytropic with the same index n.

Concept / Approach:
The ideal volumetric efficiency is derived from the pressure at which the re-expanded clearance gas falls to suction pressure, determining when the suction valve actually opens. The result is commonly written as eta_v = 1 + K - K * (p2/p1)^(1/n), which is equivalent to eta_v = 1 - K * ((p2/p1)^(1/n) - 1).

Step-by-Step Solution:
Let r = p2/p1 and assume equal n for compression and expansion.Re-expansion volume fraction = K * r^(1/n).Net fresh charge fraction = 1 + K - K * r^(1/n).Therefore eta_v = 1 + K - K * (p2/p1)^(1/n).

Verification / Alternative check:
As r → 1, eta_v → 1, which is reasonable; as r increases, eta_v decreases, matching observed behavior.

Why Other Options Are Wrong:
(c) and (d) have incorrect dependence on r; (e) ignores clearance and predicts zero eta_v at high ratios, which is unrealistic.

Common Pitfalls:
Confusing K with clearance volume percentage; K is a fraction of swept volume, not of total cylinder volume.

Final Answer:
eta_v = 1 + K - K * (p2/p1)^(1/n)