22-Mec-A1 Applied Thermodynamics and Heat Transfer · May 2013
Question 3 of 8: Double-Acting Air Compressor with Clearance
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Examinations — 07-Mec-A1 Applied Thermodynamics and Heat Transfer, May 2013. Open-book, 3-hour paper. 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. Full worked solutions to all eight questions are given below.
Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed., McGraw-Hill) — gas cycles, steam tables, reheat Rankine and vapour-compression refrigeration; Çengel & Ghajar, Heat and Mass Transfer (6th ed.) and Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (8th ed., Wiley) — conduction with critical radius, natural-convection and internal-flow correlations, and the ε–NTU heat-exchanger method. Freon-12 and steam property data are taken from the appendix supplied with the exam and standard tables.
Question 3: Double-Acting Air Compressor with Clearance (equal value)
Given. Induced volume 15 m³/min at suction $P_1=95$ kPa; delivery $P_2=825$ kPa; clearance ratio $c=0.03$; polytropic index $n=1.3$; 50 strokes/min (double-acting); suction temperature taken as 20 °C = 293.15 K; air $R=0.287\ \text{kJ/kg}\cdot\text{K}$, $c_p=1.005$, $\gamma=1.4$.
Quantity
Value
Suction / delivery pressure
95 / 825 kPa (ratio 8.68)
Induced free-air rate
15 m³/min at 95 kPa, 20 °C
Clearance ratio
3 %
Polytropic index
n = 1.3
Speed
50 strokes/min, double-acting
Find. The swept (cylinder) volume per stroke, the indicated power, and the rate of heat rejection.
Figure 3 — Indicator (p–V) diagram: induction at 95 kPa, polytropic compression (n = 1.3) to 825 kPa, delivery, and re-expansion of the 3 % clearance gas back to suction pressure. The clearance gas reduces the volume actually inducted each stroke.
Approach. Use the clearance volumetric efficiency to convert the inducted volume into swept volume (hence cylinder size), the polytropic indicated-work expression for the power, and a polytropic-process energy balance for the heat rejected.
Volumetric efficiency and cylinder volume. With clearance re-expansion,$$\eta_v=1+c-c\left(\frac{P_2}{P_1}\right)^{1/n}=1+0.03-0.03(8.684)^{1/1.3}=0.872.$$ The swept-volume rate is $\dot V_s=15/0.872=17.21$ m³/min, and over 50 strokes,$$V_\text{cyl}=\frac{17.21}{50}=\boxed{0.344\ \text{m}^3\ \text{per stroke}}.$$
Indicated power. The clearance gas re-expansion returns its work, so the net indicated work depends only on the inducted volume $\dot V_1=15\ \text{m}^3/\text{min}=0.25\ \text{m}^3/\text{s}$:$$\dot W=\frac{n}{n-1}P_1\dot V_1\!\left[\left(\frac{P_2}{P_1}\right)^{(n-1)/n}-1\right]=\frac{1.3}{0.3}(95)(0.25)(0.647)=\boxed{66.6\ \text{kW}}.$$
Delivery temperature and mass flow. $T_2=T_1(P_2/P_1)^{(n-1)/n}=293.15(1.647)=482.7$ K (209.5 °C); the mass flow is $\dot m=P_1\dot V_1/(RT_1)=95(0.25)/(0.287\cdot293.15)=0.282\ \text{kg/s}.$
Heat rejected. The polytropic specific heat is $c_n=c_v\dfrac{n-\gamma}{n-1}=0.718\dfrac{1.3-1.4}{0.3}=-0.239\ \text{kJ/kg}\cdot\text{K}$, so$$\dot Q=\dot m\,c_n(T_2-T_1)=0.282(-0.239)(189.5)=\boxed{-12.8\ \text{kW (rejected)}}.$$
Check
The 15 m³/min is interpreted as the volume of air actually inducted at the 95 kPa suction state (with the intake at 20 °C); the atmospheric data (101.3 kPa, 20 °C) establish that the intake is slightly throttled. "50 strokes/min" is read as 50 compression strokes (a double-acting cylinder giving two strokes per revolution runs at 25 rev/min); the cylinder volume quoted is the swept volume per stroke.
