22-Mec-A1 Applied Thermodynamics and Heat Transfer · May 2013
Question 1 of 8: Helium Three-Process Cycle
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.
Given. Helium behaves as an ideal monatomic gas with $R = \bar R/M = 8.314/4.003 = 2.077\ \text{kJ/kg}\cdot\text{K}$, $c_p = \tfrac52 R = 5.193$, $c_v = \tfrac32 R = 3.116\ \text{kJ/kg}\cdot\text{K}$, $\gamma = 5/3$. The three states of the closed-system cycle are listed below.
State
Pressure
Temperature
Process to next
1
300 kPa
20 °C = 293.15 K
const-P expansion
2
300 kPa
145 °C = 418.15 K
const-V cooling
3
210.3 kPa
20 °C = 293.15 K
isothermal compression
Find. The heat and work per unit mass for each of the three processes, and the cycle thermal efficiency $\eta = w_\text{net}/q_\text{in}$.
Figure 1 — p–v diagram of the helium cycle: 1→2 constant-pressure expansion (heat in), 2→3 constant-volume cooling (heat out), 3→1 isothermal compression (work in, heat out). Net area enclosed is the cycle work.
Approach. Apply the closed-system first law $q = \Delta u + w$ to each ideal-gas process using $\Delta u = c_v\,\Delta T$, then sum work and heat around the loop for the efficiency.
Process 1→2, constant pressure. Boundary work is $w_{12}=P(v_2-v_1)=R(T_2-T_1)=2.077(125)=259.6\ \text{kJ/kg}$ and the heat is the enthalpy change:$$q_{12}=c_p(T_2-T_1)=5.193(125)=\boxed{649.0\ \text{kJ/kg (in)}},\quad \Delta u_{12}=c_v\Delta T=389.4\ \text{kJ/kg}.$$
Process 2→3, constant volume. The volume is fixed, so $P_3=P_2\dfrac{T_3}{T_2}=300\dfrac{293.15}{418.15}=210.3\ \text{kPa}$, no work is done, and the heat equals the internal-energy change:$$w_{23}=0,\qquad q_{23}=c_v(T_3-T_2)=3.116(-125)=\boxed{-389.4\ \text{kJ/kg (out)}}.$$
Process 3→1, isothermal compression. At constant $T$, $\Delta u=0$ so $q_{31}=w_{31}$, and for an ideal gas $w=RT\ln(v_1/v_3)=RT\ln(P_3/P_1)$:$$w_{31}=q_{31}=RT_1\ln\!\frac{P_3}{P_1}=2.077(293.15)\ln\!\frac{210.3}{300}=\boxed{-216.2\ \text{kJ/kg (work in, heat out)}}.$$
Net work and efficiency. Summing the boundary work, $w_\text{net}=259.6+0-216.2=43.4\ \text{kJ/kg}$ (equal to the net heat $649.0-389.4-216.2$, confirming the cycle closes). Only process 1 adds heat, so$$\eta=\frac{w_\text{net}}{q_\text{in}}=\frac{43.4}{649.0}=\boxed{0.0668\;(6.7\%)}.$$
