Helium Cycle — p–v Diagram and Thermal Efficiency: 17-Phys-B6 Applied Thermodynamics and Heat Transfer, May 2013, Q1

Question 1: Helium Cycle — p–v Diagram and Thermal Efficiency (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.

Given. Helium (ideal gas, $M=4.003$ g/mol) executes a closed three-process cycle: 1→2 constant $p=300$ kPa, $T_1=20^{\circ}\text{C}$, $T_2=145^{\circ}\text{C}$; 2→3 constant volume, cooling to $T_3=20^{\circ}\text{C}$; 3→1 isothermal compression back to 300 kPa.

Find. $w$ and $q$ for each of the three processes (kJ/kg) and the cycle's thermal efficiency $\eta$.

Figure — p–v diagram of the Helium cycle. 1→2 constant-pressure expansion, 2→3 constant-volume cooling, 3→1 isothermal compression (dashed, a $pv=\text{const}$ hyperbola) closing the loop back onto state 1.

Approach. Treat helium as an ideal gas with $R=R_u/M=2.077$ kJ/kg·K, $c_v=\tfrac{3}{2}R=3.116$, $c_p=\tfrac{5}{2}R=5.193$ kJ/kg·K (monatomic, $k=5/3$); find $v_1,v_2$ from the ideal-gas law at the constant-pressure leg, then evaluate $w$ and $q$ process-by-process before closing the cycle.

Specific volumes and state 3's pressure. $$v_1=\frac{RT_1}{p_1}=\frac{2.077\times293.15}{300}=2.030\ \text{m}^3/\text{kg},\qquad v_2=\frac{RT_2}{p_1}=\frac{2.077\times418.15}{300}=2.895\ \text{m}^3/\text{kg}.$$ Process 2→3 is constant volume ($v_3=v_2$), so at $T_3=293.15$ K, $$p_3=\frac{RT_3}{v_3}=\frac{2.077\times293.15}{2.895}=\boxed{210.3\ \text{kPa}}.$$
Process 1→2 — constant-pressure expansion. $$w_{12}=p(v_2-v_1)=R(T_2-T_1)=2.077\times125=259.6\ \text{kJ/kg},$$ $$q_{12}=c_p(T_2-T_1)=5.193\times125=\boxed{649.1\ \text{kJ/kg}}\quad(\text{heat IN}).$$ ($\Delta u_{12}=c_v(T_2-T_1)=389.5$ kJ/kg, and $q_{12}=\Delta u_{12}+w_{12}$ checks: $389.5+259.6=649.1$.)
Process 2→3 — constant-volume cooling. No boundary work is done ($w_{23}=0$), so all the internal-energy change leaves as heat: $$q_{23}=\Delta u_{23}=c_v(T_3-T_2)=3.116\times(-125)=\boxed{-389.5\ \text{kJ/kg}}\quad(\text{heat OUT}).$$
Process 3→1 — isothermal compression. $\Delta u=0$ for an ideal gas at constant $T$, so $q_{31}=w_{31}$: $$w_{31}=RT_1\ln\!\frac{v_1}{v_3}=2.077\times293.15\times\ln\!\frac{2.030}{2.895} =\boxed{-216.3\ \text{kJ/kg}}\quad(\text{work IN, heat OUT}).$$
Cycle net work and thermal efficiency. Summing the three legs (and confirming $\sum q=\sum w$, since $\oint du=0$): $$w_{net}=259.6+0-216.3=\boxed{43.4\ \text{kJ/kg}}=q_{12}+q_{23}+q_{31}=649.1-389.5-216.3=43.4\ \checkmark$$ Only $q_{12}$ is heat added to the gas, so $$\eta=\frac{w_{net}}{q_{in}}=\frac{43.4}{649.1}=\boxed{6.68\%}.$$

Final results — Question 1
Quantity	Value
$w_{12}$, $q_{12}$ (1→2, const. $p$)	$259.6$, $649.1$ kJ/kg (heat in)
$w_{23}$, $q_{23}$ (2→3, const. $v$)	$0$, $-389.5$ kJ/kg (heat out)
$w_{31}$, $q_{31}$ (3→1, isothermal)	$-216.3$, $-216.3$ kJ/kg (heat out)
Net work / net heat per cycle	$43.4$ kJ/kg
Thermal efficiency $\eta$	$6.68\%$

Topic: First-law energy accounting for an ideal-gas cycle built from constant-$p$, constant-$v$, and isothermal legs.

Key relations: $pv=RT$; $w=p\,\Delta v$ (const. $p$); $w=RT\ln(v_2/v_1)$ (isothermal, ideal gas); $q=\Delta u+w$; $\eta=w_{net}/q_{in}$.

Why this works: because internal energy of an ideal gas depends on temperature alone, any process that starts and ends at the same $T$ (here the isothermal leg) has $\Delta u=0$ regardless of path, which is what lets $q_{31}=w_{31}$ be read directly off the work integral. Summing $q$ around the closed cycle must equal summing $w$ (a closed-cycle $\oint du=0$), which is the internal check used above.

Common pitfall: using $c_p$ where $\Delta u$ (i.e. $c_v$) is meant on the constant-volume leg, or forgetting that the isothermal leg is a compression (negative $v_1/v_3<1$ ratio inverted, giving a sign error) — the gas is compressed BACK from the low-pressure state 3 to the original state 1, so $w_{31}<0$.

Source: Çengel & Boles, Thermodynamics: An Engineering Approach, Ch. 3–4 (ideal-gas properties, first law for closed systems).

Question 1 of 8: Helium Cycle — p–v Diagram and Thermal Efficiency

Question 1: Helium Cycle — p–v Diagram and Thermal Efficiency (Part A, 20 marks)