22-Mec-A1 Applied Thermodynamics and Heat Transfer · December 2017
Question 4 of 8: Reciprocating compressor volumetric efficiency
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Examination 16-Mec-A1, 3 hours, open book. Eight questions of equal value: Part A — Thermodynamics (Q1–Q4) and Part B — Heat Transfer (Q5–Q8). A complete paper is any five questions (three from one part and two from the other).
Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed., McGraw-Hill) — closed-system energy balances, wet-region steam properties, flash/separator processes, isentropic turbine and compressor efficiency, vapour-compression refrigeration, and reciprocating-compressor clearance analysis; Çengel & Ghajar, Heat and Mass Transfer (6th ed.) and Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (8th ed., Wiley) — transient lumped-capacitance cooling, radial composite-cylinder conduction, internal-flow and external cross-flow convection, natural convection from a vertical plate, and the LMTD method for condensers. Water/steam and air properties are taken from standard tables (IAPWS-consistent); ammonia and R-134a properties are read from the saturation and superheat tables appended to the examination paper.
Given. A single-stage reciprocating air compressor with clearance ratio $c=5\%$, polytropic index $n=1.32$, and valve pressure drops that offset the in-cylinder suction and discharge pressures from the line/atmospheric values. Find. the (clearance) volumetric efficiency.
Quantity
Value
Line (delivery) pressure
468,840 Pa
Atmospheric pressure
101,325 Pa
Intake-valve pressure loss
3,450 Pa
Discharge-valve pressure loss
13,790 Pa
Clearance ratio, $c$
0.05
Polytropic index, $n$
1.32
Figure 4 — Indicator (p–V) diagram, drawn to scale for $V_s=1.00$ m³ and $c=5\%$. Induction 1→2 occurs at the throttled suction pressure $P_1$; compression 2→3 and re-expansion 4→1 both follow $pV^n=$const; delivery 3→4 is at the raised discharge pressure $P_2$, ending at the clearance volume. Because the clearance gas must re-expand from $V_4=0.05$ to $V_1=0.167$ m³ before the inlet valve can open, the induced volume $V_2-V_1 = 0.883$ m³ is less than the 1.00 m³ displacement — that ratio is the volumetric efficiency.
Approach. Correct the atmospheric and line pressures for the valve losses to obtain the true in-cylinder suction and discharge pressures, then apply the clearance volumetric-efficiency relation for a polytropic re-expansion.
In-cylinder discharge pressure. The discharge valve requires a pressure above the line:
$$P_2 = P_\text{line} + \Delta P_\text{dis} = 468{,}840 + 13{,}790 = 482{,}630\ \text{Pa}$$
The pressure ratio the clearance gas re-expands over is
$$r = \frac{P_2}{P_1} = \frac{482{,}630}{97{,}875} = 4.931$$
Clearance volumetric efficiency. The trapped clearance charge re-expands polytropically before fresh air is admitted, giving
$$\eta_v = 1 + c - c\left(\frac{P_2}{P_1}\right)^{1/n} = 1 + 0.05 - 0.05\,(4.931)^{1/1.32}$$
$$\eta_v = 1.05 - 0.05\,(3.350) = 0.883$$
$\eta_v \approx 88.3\%$
Check — displacement value.
The stated "piston displacement (1.00 m³)" cancels out of the volumetric-efficiency ratio, which depends only on the clearance fraction and the pressure ratio. It would be needed only for absolute induced-volume or power figures, which the question does not ask for.
