Reciprocating Air Compressor with Clearance: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, December 2016, Q2

Question 2: Reciprocating Air 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.

Quantity	Value
Inlet / discharge pressure	$P_1=1$ atm, $P_2=3$ atm; ratio $r_p=3$
Inlet temperature	20 °C
Clearance ratio	$c=0.05$
Polytropic index (isentropic)	$n=\gamma=1.4$
Induced air at inlet	$\dot V_1=2.85$ m³/min

Figure 2 — Ideal indicator (P–V) diagram of a compressor with clearance $V_c$: 1→2 isentropic compression, 2→3 delivery, 3→4 isentropic re-expansion of the clearance gas, 4→1 induction. The induced volume $V_1-V_4$ is less than the swept volume by the re-expansion.

Approach. The clearance-gas re-expansion reduces the induced volume, captured by the volumetric efficiency $\eta_v$; the swept (displacement) volume then follows from $\dot V_1/\eta_v$, and the ideal indicated power uses the induced volume directly because the re-expansion work cancels the extra compression work.

(b) Volumetric efficiency. This is taken first because part (a) depends on it. For isentropic compression/re-expansion of the clearance gas, $$\eta_v=1+c-c\left(\frac{P_2}{P_1}\right)^{1/n}=1+0.05-0.05\,(3)^{1/1.4}=1.05-0.05(2.192)$$ ==**$\eta_v\approx0.940$ (94.0 %).**==
(a) Piston displacement. The compressor must induce 2.85 m³/min at inlet, so the swept volume rate is $$\dot V_\text{disp}=\frac{\dot V_1}{\eta_v}=\frac{2.85}{0.9404}$$ ==**$\dot V_\text{disp}\approx3.03$ m³/min.**==
(c) Indicated power. The polytropic compression power on the induced flow is $$\dot W=\frac{n}{n-1}\,P_1\dot V_1\!\left[\left(\frac{P_2}{P_1}\right)^{(n-1)/n}-1\right]=3.5\,(101325)(0.0475)\big[(3)^{0.2857}-1\big]$$ with $\dot V_1=2.85/60=0.0475$ m³/s and $(3)^{0.2857}=1.369$, giving ==**$\dot W\approx6.21$ kW ($\approx8.3$ hp).**==

Check

The clearance does not appear in the power expression: the re-expansion 3→4 returns exactly the work spent compressing that clearance gas 1→2, so the net indicated work depends only on the freshly induced charge $\dot V_1$. Clearance does penalise capacity (hence the larger displacement) but not the specific work.

Quantity	Result
(b) Volumetric efficiency $\eta_v$	≈ 94.0 %
(a) Piston displacement	≈ 3.03 m³/min
(c) Indicated power	≈ 6.21 kW

Concept & Theory

Topic. Ideal single-stage reciprocating compressor with clearance volume.

Key relations. Volumetric efficiency $\eta_v=1+c-c\,r_p^{1/n}$; displacement $\dot V_\text{disp}=\dot V_\text{induced}/\eta_v$; polytropic power $\dot W=\frac{n}{n-1}P_1\dot V_1[r_p^{(n-1)/n}-1]$.

Why this works. Gas trapped in the clearance at $P_2$ must re-expand to $P_1$ before fresh air can enter, so the piston sweeps some volume with no induction — that lost fraction is exactly $c\,(r_p^{1/n}-1)$. Because both the compression and re-expansion are isentropic between the same two pressures, their work integrals cancel on the clearance gas, leaving the specific work of the induced air unchanged by clearance.

Common pitfall. Multiplying the power by $\eta_v$ or using the displacement volume in the work formula — the correct volume there is the induced $\dot V_1$; and using $r_p^{(n-1)/n}$ (not $r_p^{1/n}$) in the volumetric-efficiency expression.

Source. Çengel & Boles, Thermodynamics §7 (reversible steady-flow compression work) and standard reciprocating-compressor theory (clearance and volumetric efficiency).

Question 2 of 8: Reciprocating Air Compressor with Clearance

Question 2: Reciprocating Air Compressor with Clearance (equal value)