Reciprocating Air Compressor — Volumetric Efficiency with Valve Losses: 17-Phys-B6 Applied Thermodynamics and Heat Transfer, December 2017, Q4

Question 4: Reciprocating Air Compressor — Volumetric Efficiency with Valve Losses

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A single-stage reciprocating air compressor with clearance, whose intake and discharge valves each throttle the gas by a stated pressure drop before/after the cylinder; the clearance gas trapped at top-dead-centre re-expands polytropically ($pV^{1.32}=\text{const.}$) before fresh gas can be drawn in.

Given data
Quantity	Symbol	Value
Discharge line pressure	$P_{line}$	468,840 Pa
Atmospheric pressure	$P_{atm}$	101,325 Pa
Clearance fraction	$c$	5% of $V_d$
Piston displacement	$V_d$	$1.00\text{ m}^3$
Polytropic index	$n$	1.32
Intake-valve pressure loss	$\Delta P_{in}$	3,450 Pa
Discharge-valve pressure loss	$\Delta P_{dis}$	13,790 Pa

Find. The volumetric efficiency $\eta_v$, with the cycle depicted on a $p$–$V$ diagram.

Indicator ($p$–$V$) diagram: 4→1 induction at the (valve-loss-lowered) suction pressure, 1→2 polytropic compression, 2→3 discharge at the (valve-loss-raised) line pressure, 3→4 polytropic re-expansion of the clearance gas.

Approach. The EFFECTIVE cylinder pressures differ from the nominal atmospheric/line values because of the valve throttling losses: suction pressure is reduced below atmospheric, and the in-cylinder discharge pressure must exceed the line pressure by the discharge valve's own loss. With those two effective pressures and the polytropic index, the standard clearance-volumetric-efficiency formula applies directly.

Effective suction and discharge pressures. $$P_1=P_{atm}-\Delta P_{in}=101{,}325-3{,}450=97{,}875\text{ Pa}=97.875\text{ kPa}$$ $$P_2=P_{line}+\Delta P_{dis}=468{,}840+13{,}790=482{,}630\text{ Pa}=482.63\text{ kPa}$$
Pressure ratio. $$r=\frac{P_2}{P_1}=\frac{482.63}{97.875}=4.931$$
Volumetric efficiency. With clearance ratio $c=0.05$ and polytropic re-expansion of the clearance gas, $$\eta_v=1+c-c\,r^{1/n}=1+0.05-0.05\times(4.931)^{1/1.32}=1.05-0.05\times3.351$$ $$\boxed{\eta_v=88.25\%}$$
Induced volume (supporting quantity). Per cycle, the actual fresh-air volume drawn in (referred to suction conditions) is $$V_{induced}=\eta_v\,V_d=0.8825\times1.00=0.8825\text{ m}^3$$

Question 4 — results
Quantity	Value
Effective suction pressure $P_1$	97.875 kPa
Effective discharge pressure $P_2$	482.63 kPa
Pressure ratio $r$	4.931
Volumetric efficiency $\eta_v$	88.25%
Induced volume per cycle	0.8825 m³

Topic. Volumetric efficiency of a reciprocating compressor with clearance and valve pressure losses.

Key relations. $\eta_v=1+c-c(P_2/P_1)^{1/n}$ for polytropic clearance re-expansion; effective cylinder pressures $P_1=P_{atm}-\Delta P_{in}$, $P_2=P_{line}+\Delta P_{dis}$.

Why this works. The clearance gas trapped at top-dead-centre must re-expand from $P_2$ back down to $P_1$ before the intake valve can open, and that re-expansion eats into the piston's swept volume — the larger the pressure ratio, the more clearance-gas volume is "used up" just getting back to suction pressure, which is exactly the $r^{1/n}$ term. Valve losses widen the EFFECTIVE pressure ratio beyond the nominal atmospheric/line ratio, always pushing $\eta_v$ below what an ideal-valve estimate would give.

Common pitfall. Using the nominal atmospheric and line pressures directly instead of the valve-loss-corrected $P_1$, $P_2$ (this understates the true pressure ratio and overstates $\eta_v$); or inverting the exponent (using $n$ instead of $1/n$) on the pressure ratio.

Source. Y. A. Çengel and M. A. Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 3/App. (reciprocating gas compressor performance, clearance volume and volumetric efficiency).

Question 4 of 8: Reciprocating Air Compressor — Volumetric Efficiency with Valve Losses

Question 4: Reciprocating Air Compressor — Volumetric Efficiency with Valve Losses