Single-Acting Reciprocating Compressor with Clearance: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, May 2014, Q2

Question 2: Single-Acting Reciprocating Compressor with Clearance (equal value)

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. Piston displacement (swept volume) $V_s=280\ \text{cm}^3$; clearance ratio $c=0.04$; 600 strokes/min; reversible adiabatic compression, so the polytropic index equals $\gamma=1.4$. Valve pressure losses shift the in-cylinder pressures from the line values.

Quantity	Value
Line suction / discharge	95 / 480 kPa
Intake / discharge valve loss	2.5 / 13.8 kPa
In-cylinder suction $P_1$	95 − 2.5 = 92.5 kPa
In-cylinder delivery $P_2$	480 + 13.8 = 493.8 kPa
Clearance / speed	4 % / 600 strokes·min⁻¹

Find. The volume of air actually inducted and compressed per minute.

Figure 1 — Indicator (p–V) diagram: 1→2 reversible-adiabatic compression from 92.5 kPa to 493.8 kPa, 2→3 delivery at constant pressure, 3→4 re-expansion of the 4 % clearance gas, 4→1 induction. Valve throttling lowers the suction line and raises the delivery line inside the cylinder.

Approach. The clearance gas re-expands each stroke and reduces the volume drawn in; the clearance volumetric efficiency converts the swept volume into inducted volume, which multiplied by the stroke rate gives the volume compressed per minute.

In-cylinder pressure ratio. Accounting for the valve losses,$$r_p=\frac{P_2}{P_1}=\frac{480+13.8}{95-2.5}=\frac{493.8}{92.5}=5.34.$$
Clearance volumetric efficiency. With re-expansion along the same index $n=\gamma=1.4$,$$\eta_v=1+c-c\,r_p^{1/n}=1+0.04-0.04(5.34)^{1/1.4}=1.04-0.04(3.31)=\boxed{0.908}.$$
Volume inducted per stroke. $$V_1-V_4=\eta_v V_s=0.908(280)=254.2\ \text{cm}^3\ \text{per stroke}.$$
Volume compressed per minute. At 600 strokes/min,$$\dot V=254.2\times600=1.525\times10^{5}\ \text{cm}^3/\text{min}=\boxed{0.1525\ \text{m}^3/\text{min}}\ (152.5\ \text{L/min}).$$

Check

The "volume of air compressed per minute" is taken as the free air actually inducted at the in-cylinder suction state (92.5 kPa, 15 °C). Referred instead to the 95 kPa suction-line pressure it is $0.1525\times(92.5/95)=0.1485\ \text{m}^3/\text{min}$ — a 2.6 % correction. The re-expansion index is set equal to the compression index (reversible adiabatic), as the question states.

Quantity	Result
In-cylinder pressure ratio	5.34
Volumetric efficiency	0.908
Inducted volume per stroke	254 cm³
Volume compressed per minute	≈ 0.153 m³/min (152.5 L/min)

Concept & Theory

Topic. Reciprocating compressor with clearance volume and valve throttling.

Key relations. Clearance volumetric efficiency $\eta_v=1+c-c\,r_p^{1/n}$; inducted volume $=\eta_vV_s$; effective pressure ratio uses the in-cylinder pressures $P_1=P_\text{suc}-\Delta P_\text{in}$, $P_2=P_\text{del}+\Delta P_\text{out}$.

Why this works. Gas trapped in the clearance volume at delivery pressure expands back down to suction pressure before the intake valve opens, so the piston must retrace part of its stroke before any fresh air enters — this is exactly the $c\,(r_p^{1/n}-1)$ loss. The valve pressure drops make the cylinder work against a wider pressure ratio than the pipeline values suggest.

Common pitfall. Using the line pressures (95/480 kPa) instead of the throttled in-cylinder pressures, or forgetting that the clearance re-expansion — not the swept volume alone — sets how much air is drawn in.

Source. Çengel & Boles, Thermodynamics §7 (isentropic/polytropic compression) and standard reciprocating-compressor theory.

Question 2 of 8: Single-Acting Reciprocating Compressor with Clearance

Question 2: Single-Acting Reciprocating Compressor with Clearance (equal value)