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Cylinder output with clearance numerical

Short answer

With clearance ratio 5 % and pV^1.3 re‑expansion, volumetric efficiency = 80.3 %. Effective suction volume per cycle gives air drawn 0.851 m³/min; indicated power using that volume is 3.78 kW.

A single-acting, single-cylinder air compressor has bore 150 mm, stroke 200 mm and runs at 300 rpm. The clearance volume is 5% of the swept volume. Air is drawn at 1 bar and 27 °C and delivered at 8 bar; compression and re-expansion follow pV1.3=CpV^{1.3} = C. Find (a) the volumetric efficiency, (b) the volume of air drawn per minute at suction conditions, and (c) the indicated power.

Swept volume

Vs=π4d2L=π4(0.15)2(0.2)=0.00353 m3V_s = \frac{\pi}{4}d^2L = \frac{\pi}{4}(0.15)^2(0.2) = 0.00353\ \text{m}^3

(a) Volumetric efficiency

Pressure ratio =8= 8, (8)1/1.3=4.951(8)^{1/1.3} = 4.951.

ηv=1+C−C(p2p1)1/n=1+0.05−0.05×4.951=0.8025\eta_v = 1 + C - C\left(\frac{p_2}{p_1}\right)^{1/n} = 1 + 0.05 - 0.05 \times 4.951 = 0.8025

(b) Air drawn per minute

Effective suction volume per cycle: Va=ηvVs=0.8025×0.00353=0.00284V_a = \eta_v V_s = 0.8025 \times 0.00353 = 0.00284 m³.

One suction per revolution (single-acting):

V=VaN=0.00284×300=0.851 m3/minV = V_a N = 0.00284 \times 300 = 0.851\ \text{m}^3/\text{min}

(c) Indicated power

(p2/p1)n−1n=80.2308=1.6159(p_2/p_1)^{\frac{n-1}{n}} = 8^{0.2308} = 1.6159 and p1V1=100×0.851/60p_1V_1 = 100 \times 0.851/60 kW.

P=1.30.3×100×0.85160×(1.6159−1)=3.78 kWP = \frac{1.3}{0.3} \times \frac{100 \times 0.851}{60} \times (1.6159 - 1) = 3.78\ \text{kW}

(The clearance air is compressed and re-expands over the same index, so only the effective volume VaV_a is used in the work formula.)

Answer: (a) 80.3 %; (b) 0.851 m³/min; (c) about 3.78 kW.

✓ Answer checked by Claude Opus 5.5 (high effort)

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