Cycle energy table & heating of a rigid two-phase vessel: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, May 2018, Q1

Question 1 — Cycle energy table & heating of a rigid two-phase vessel (Part A, equal value)

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.

Part (a) — Complete the cycle table

Given. A closed system executes a four-process cycle (1→2→3→4→1); kinetic and potential effects are neglected, so the closed-system first law reduces to $Q = \Delta U + W$ for every process. The known entries are tabulated below.

Process	$\Delta U$ (kJ)	$Q$ (kJ)	$W$ (kJ)
1–2	—	—	−610
2–3	+670	—	+230
3–4	—	0	+920
4–1	−360	—	0

Find. the four missing entries and whether the cycle is a power or refrigeration cycle.

Approach. Apply $Q=\Delta U+W$ to each process that has two of three entries, then close the cycle with the fact that internal energy is a property, so $\sum\Delta U=0$ around the loop.

Fill the rows that already have two entries. For 2–3, $Q_{23}=\Delta U_{23}+W_{23}=670+230=900\ \text{kJ}$. For 3–4, $\Delta U_{34}=Q_{34}-W_{34}=0-920=-920\ \text{kJ}$. For 4–1, $Q_{41}=\Delta U_{41}+W_{41}=-360+0=-360\ \text{kJ}$.
Close the internal-energy balance to get $\Delta U_{12}$. Since $U$ is a state property, the changes sum to zero over the cycle: $$\Delta U_{12}+\Delta U_{23}+\Delta U_{34}+\Delta U_{41}=0$$ $$\Delta U_{12}=-(670-920-360)=+610\ \text{kJ}$$
Complete row 1–2. With $\Delta U_{12}=610$ and $W_{12}=-610$: $$Q_{12}=\Delta U_{12}+W_{12}=610+(-610)=0\ \text{kJ}$$ $\Delta U_{12}=+610$ kJ, $\;Q_{12}=0$, $\;Q_{23}=+900$ kJ, $\;\Delta U_{34}=-920$ kJ, $\;Q_{41}=-360$ kJ
Classify the cycle. The net work and net heat close correctly: $$W_\text{net}=\textstyle\sum W = -610+230+920+0 = +540\ \text{kJ},\qquad Q_\text{net}=\textstyle\sum Q = 0+900+0-360 = +540\ \text{kJ}$$ Net work is positive (work is delivered by the system), so this is a power cycle.

Process	$\Delta U$ (kJ)	$Q$ (kJ)	$W$ (kJ)
1–2	+610	0	−610
2–3	+670	+900	+230
3–4	−920	0	+920
4–1	−360	−360	0
Cycle	0	+540	+540

Part (b) — Rigid vessel heated from a two-phase state

Given. A rigid, closed container ($V=0.5$ m³, constant mass) holds wet water at $P_1=1$ bar with quality $X_1=0.5$; it is heated to $P_2=1.5$ bar and (later) until the contents are just saturated vapour.

Find. the temperature and mass of vapour at states 1 and 2, and the pressure $P_3$ when only saturated vapour remains. Because the vessel is rigid and sealed, the specific volume $v=V/m$ is constant through the whole process.

Figure 1 — $T$–$v$ diagram. Heating a rigid, sealed vessel is a constant-specific-volume process, so states 1, 2 and 3 lie on a single vertical line at $v=0.847\ \text{m}^3/\text{kg}$. Because that line passes to the right of the critical point, the mixture dries out as it is heated and reaches the saturated-vapour line at state 3.

Approach. Fix $v$ from state 1, get the mass, then read quality (hence vapour mass) at each pressure; state 3 is the saturation state whose $v_g$ equals the fixed $v$.

Specific volume and mass at state 1. At 1 bar, $v_f=0.001043$ and $v_g=1.6940\ \text{m}^3/\text{kg}$, with $T_1=T_\text{sat}(1\,\text{bar})=99.6\ ^\circ\text{C}$: $$v_1=v_f+X_1(v_g-v_f)=0.001043+0.5(1.6940-0.001043)=0.8475\ \text{m}^3/\text{kg}$$ $$m=\frac{V}{v_1}=\frac{0.5}{0.8475}=0.590\ \text{kg}$$ $m=0.590$ kg (fixed), $\;m_{g,1}=X_1 m=0.295$ kg
State 2 at 1.5 bar (same $v$). Here $T_2=T_\text{sat}(1.5\,\text{bar})=111.4\ ^\circ\text{C}$, with $v_f=0.001053$, $v_g=1.1590$: $$X_2=\frac{v_1-v_f}{v_g-v_f}=\frac{0.8475-0.001053}{1.1590-0.001053}=0.731$$ $$m_{g,2}=X_2 m=0.731(0.590)=0.431\ \text{kg}$$ The vapour mass rises from 0.295 to 0.431 kg as liquid evaporates at constant volume.
State 3 — only saturated vapour remains. When the last liquid disappears the state is saturated vapour with $v_g=v_1=0.8475\ \text{m}^3/\text{kg}$. The saturation pressure at which $v_g$ equals this value is $P_3 \approx 2.10$ bar, $\;T_3=T_\text{sat}(P_3)\approx 121.7\ ^\circ\text{C}$ found by locating $0.8475\ \text{m}^3/\text{kg}$ on the saturated-vapour line (between the 2.0-bar entry $v_g=0.8857$ and the 2.25-bar entry $v_g\approx0.7933$).

State	Pressure	Temperature	Quality	Vapour mass
1	1.0 bar	99.6 °C	0.50	0.295 kg
2	1.5 bar	111.4 °C	0.731	0.431 kg
3 (dry sat.)	≈ 2.10 bar	≈ 121.7 °C	1.00	0.590 kg (all)

Concept & Theory

Topic. First-law bookkeeping around a closed cycle, and constant-volume heating of a wet mixture (an isochoric path on the $T$–$v$ plane).

Key relations. Closed-system first law $Q=\Delta U+W$; property closure $\oint dU=0$; quality relations $v=v_f+x\,v_{fg}$ and $m_g=x\,m$; a rigid sealed vessel keeps $v=V/m$ constant.

Why this works. In part (a) each row obeys the first law, and because internal energy is a point function the four increments must sum to zero — that single closure supplies the last unknown. A positive net work identifies a power (heat-engine) cycle; a negative one would be a refrigeration/heat-pump cycle. In part (b) the vessel neither expands nor leaks, so its specific volume is frozen; the state simply climbs a vertical line on the $T$–$v$ chart. Since that line lies on the vapour side of the critical point, more liquid evaporates as temperature rises and the mixture finishes as dry saturated vapour.

Common pitfall. Treating the rigid-vessel heating as constant pressure (it is not — pressure and temperature both rise while $v$ stays fixed), or, in part (a), forgetting the $\sum\Delta U=0$ closure and leaving row 1–2 undetermined.

Source. Çengel & Boles, Thermodynamics §4 (energy analysis of closed systems; cycles) and §3 (property tables, quality, constant-volume processes).

Question 1 of 8: Cycle energy table & heating of a rigid two-phase vessel

Question 1 — Cycle energy table & heating of a rigid two-phase vessel (Part A, equal value)

Part (a) — Complete the cycle table

Part (b) — Rigid vessel heated from a two-phase state