Reciprocating air compressor power: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, December 2019, Q3

Question 3 — Reciprocating air compressor power (Part A, 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.

Quantity	Symbol	Value
Bore / stroke	$D,\,L$	0.075 m, 0.100 m
Speed	$N$	800 rev/min (single-acting)
Suction / delivery pressure	$P_1,\,P_2$	101.325 kPa, 412 kPa
Suction temperature	$T_1$	27 °C
Volumetric efficiency	$\eta_v$	0.80
Compression (isentropic) efficiency	$\eta_c$	0.90

Find. Shaft (brake) power to drive the compressor.

Approach. The swept-volume rate from the geometry and speed, de-rated by $\eta_v$, gives the induced air rate at suction; the reversible-adiabatic (isentropic) indicated work per unit volume then follows, and dividing by the compression efficiency yields the required drive power.

Swept volume per revolution and its rate. $$V_\text{sw}=\frac{\pi}{4}D^2L=\frac{\pi}{4}(0.075)^2(0.100)=4.418\times10^{-4}\ \text{m}^3,\quad \dot V_\text{sw}=V_\text{sw}\frac{N}{60}=5.890\times10^{-3}\ \text{m}^3/\text{s}$$
Induced (actual) air rate at suction. $$\dot V_1=\eta_v\,\dot V_\text{sw}=0.80(5.890\times10^{-3})=\boxed{4.712\times10^{-3}\ \text{m}^3/\text{s}}$$
Reversible-adiabatic indicated power. With $r_p=P_2/P_1=4.066$, $$\dot W_s=\frac{\gamma}{\gamma-1}P_1\dot V_1\!\left[r_p^{\frac{\gamma-1}{\gamma}}-1\right]=\frac{1.4}{0.4}(101.325)(4.712\times10^{-3})\big[4.066^{0.2857}-1\big]=0.824\ \text{kW}$$
Drive power from the compression efficiency. $$\dot W_\text{brake}=\frac{\dot W_s}{\eta_c}=\frac{0.824}{0.90}=\boxed{0.915\ \text{kW}}$$

Check — interpretation of "thermal efficiency of the compression process".

The phrase is read as the compressor's isentropic (adiabatic) efficiency, i.e. the ideal reversible-adiabatic work is 90 % of the actual, so the required power is $\dot W_s/0.90$. The machine is taken as single-acting (one induction per revolution). If it were double-acting the induced volume — and hence the power — would roughly double (≈ 1.83 kW). The delivery pressure 412 kPa is treated as absolute. All three assumptions are standard for this stem; the efficiency and $\eta_v$ enter only as scalar multipliers, so the method is unchanged if a different convention is adopted.

Quantity	Result
Induced air rate	4.71 × 10⁻³ m³/s
Isentropic indicated power	0.824 kW
Required drive power	0.915 kW

Concept & Theory — Q3

Topic: Single-stage reciprocating compressor — volumetric efficiency, induced volume, and reversible-adiabatic compressor work.

Key relations: Swept volume $V_\text{sw}=\tfrac{\pi}{4}D^2L$; induced rate $\dot V_1=\eta_v\dot V_\text{sw}N/60$; adiabatic steady-flow compression work $\dot W=\dfrac{\gamma}{\gamma-1}P_1\dot V_1[r_p^{(\gamma-1)/\gamma}-1]$; brake power $=\dot W_s/\eta_c$.

Why this works: Volumetric efficiency accounts for clearance re-expansion and valve/heating losses that keep the cylinder from inducting its full swept volume, so the induced volume at suction — not the swept volume — is what is compressed each cycle. The $\gamma/(\gamma-1)$ formula is the reversible-adiabatic (isentropic) flow-work integral $\int v\,dP$ for an ideal gas between the two pressures; dividing by the compression efficiency restores the extra shaft work the real, irreversible machine actually needs.

Common pitfall: Using the swept volume instead of the induced volume in the work integral (over-predicts by $1/\eta_v$), or multiplying by the efficiency instead of dividing. Also, forgetting to convert rev/min to rev/s.

Source: Çengel & Boles, Thermodynamics 9th ed., §7-13 (reversible steady-flow work) and §9 (reciprocating-compressor analysis); Eastop & McConkey, Applied Thermodynamics, Ch. 9.

Question 3 of 8: Reciprocating air compressor power

Question 3 — Reciprocating air compressor power (Part A, equal value)