Double-Acting Air Compressor — Cylinder Volume, Power, Heat Rejected: 17-Phys-B6 Applied Thermodynamics and Heat Transfer, May 2013, Q3

Question 3: Double-Acting Air Compressor — Cylinder Volume, Power, Heat Rejected (Part A, 20 marks)

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.

Check: two assumptions are needed to reconcile the given data, both stated per Note 1's "state your assumptions" instruction. (1) The "15 m³ of air per minute" is read as the free-air delivery (FAD), measured at the separately-given atmospheric state (101.3 kPa, 20°C) — this is the only reading that uses the atmospheric data at all, since the suction pressure (95 kPa) is already given directly for the in-cylinder state. (2) "Operates at 50 strokes/minute" is read as 50 crank revolutions/minute; a double-acting cylinder completes an induction+compression event on BOTH faces per revolution, i.e. 100 induction events/minute.

Given. Free-air delivery $\dot V_{FAD}=15$ m³/min at $p_{atm}=101.3$ kPa, $T_{atm}=20^{\circ}\text{C}$; suction $p_1=95$ kPa; discharge $p_2=825$ kPa; clearance $c=3\%$; $n=1.3$; double-acting, 50 strokes/min.

Find. Cylinder swept volume $V_s$, indicated power, and heat rejection rate.

Approach. Convert the free-air delivery to the actual volume induced at the cylinder's suction state (isothermal throttle across the suction valve), use the clearance volumetric efficiency to size the cylinder, then apply the standard polytropic-compressor power/heat formulas with the actual induced volumetric flow.

Actual induced volume at suction conditions. Air throttles from $p_{atm}=101.3$ kPa to $p_1=95$ kPa across the suction valve; for an ideal gas this is isenthalpic, so $T_1=T_{atm}=293.15$ K unchanged. By $pV=\text{const}$ at fixed $T$, $$\dot V_1=\dot V_{FAD}\frac{p_{atm}}{p_1}=15\times\frac{101.3}{95}=\boxed{15.99\ \text{m}^3/\text{min}}.$$
Volumetric efficiency and cylinder (swept) volume. With pressure ratio $r=p_2/p_1=825/95=8.684$ and same-$n$ re-expansion of the clearance gas, $$\eta_v=1+c-c\,r^{1/n}=1+0.03-0.03(8.684)^{1/1.3}=\boxed{0.8718}\ (87.2\%).$$ The actual induced volume equals $\eta_v\,V_s$ per induction event, times 100 events/min: $$V_s=\frac{\dot V_1}{\eta_v\times100}=\frac{15.99}{0.8718\times100}=\boxed{0.1835\ \text{m}^3}\ (183.5\ \text{L}).$$ If "50 strokes/minute" is instead read literally as 50 piston strokes (each one an induction event on one face of the double-acting piston, i.e. 25 rev/min), the cylinder must be twice as large, $V_s=15.99/(0.8718\times50)=0.367$ m³; the power and heat results below depend only on the delivered flow and are the same under either reading.
Indicated power. For a polytropic reciprocating compressor the net indicator-diagram work per unit induced volume is independent of clearance (the re-expansion work cancels against the compression work on the returned volume), so $$\dot W=\frac{n}{n-1}\,p_1\dot V_1\!\left[r^{(n-1)/n}-1\right] =\frac{1.3}{0.3}\times95\times\frac{15.99}{60}\times\left[8.684^{0.3/1.3}-1\right] =\boxed{70.98\ \text{kW}}.$$
Heat transferred from the compressor. Mass flow through the cylinder is $\dot m=p_1\dot V_1/(R T_1)=0.301$ kg/s; the polytropic discharge temperature is $T_2=T_1\,r^{(n-1)/n}=293.15\times8.684^{0.3/1.3}=482.7$ K ($209.6^{\circ}\text{C}$). A steady-flow energy balance $\dot Q-\dot W_{in}=\dot m(h_2-h_1)$ (taking $\dot W_{in}$ as work done ON the air, positive) gives $$\dot Q=\dot m\,c_p(T_2-T_1)-\dot W=0.301\times1.005\times(482.7-293.15)-70.98=\boxed{-13.62\ \text{kW}}$$ — negative, i.e. heat flows FROM the compressor to its surroundings, consistent with $n=1.3

Final results — Question 3
Quantity	Value
Volumetric efficiency $\eta_v$	$87.2\%$
Cylinder swept volume $V_s$	$0.1835$ m³ (183.5 L)
Indicated power $\dot W$	$71.0$ kW
Heat transfer rate $\dot Q$	$-13.6$ kW (rejected to surroundings)

Topic: Reciprocating compressor sizing with clearance volumetric efficiency, and the polytropic power/heat balance for the resulting flow.

Key relations: $\eta_v=1+c-c\,r^{1/n}$; indicated power $\dot W=\tfrac{n}{n-1}p_1\dot V_1[r^{(n-1)/n}-1]$; polytropic discharge temperature $T_2=T_1r^{(n-1)/n}$; steady-flow first law $\dot Q=\dot m\,c_p\Delta T-\dot W_{in}$.

Why this works: because the SAME polytropic index $n$ governs both the compression stroke and the clearance-gas re-expansion stroke, the net indicated work per unit of NET (delivered) volume is identical to the no-clearance case — clearance only reduces how much of the swept volume is actually usable, which is exactly what $\eta_v$ captures.

Common pitfall: applying the FAD volume directly to the power formula without correcting it to the in-cylinder suction state (a 6.7% pressure ratio correction here), or forgetting that "double acting" doubles the induction events per revolution when sizing a single cylinder's swept volume.

Source: Çengel & Boles, Thermodynamics: An Engineering Approach, Ch. 8–9 (polytropic processes, multistage compression with clearance).

Question 3 of 8: Double-Acting Air Compressor — Cylinder Volume, Power, Heat Rejected

Question 3: Double-Acting Air Compressor — Cylinder Volume, Power, Heat Rejected (Part A, 20 marks)