Rigid Insulated Tank — Paddle-Wheel Work and Irreversibility: 04-BS-10, December 2013, Q6

Question 6: Rigid Insulated Tank — Paddle-Wheel Work and Irreversibility (15 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. Rigid, insulated (adiabatic, $Q=0$), closed tank containing $m=1$ kg air, ideal gas, constant specific heats ($c_v=0.718$ kJ/kg·K). $P_1=100$ kPa, $T_1=27\ ^\circ$C $=300.15$ K; final $P_2=120$ kPa. Dead state $T_0=300.15$ K, $p_0=1$ bar (used only for the irreversibility calculation).

Quantity	Value
$m$	1 kg
$P_1,\ T_1$	100 kPa, 300.15 K
$P_2$	120 kPa
$T_0$	300.15 K

Find. Paddle-wheel work input $W_{in}$; irreversibility $I$ (both in kJ).

Approach

Since the tank is rigid, $v$ is constant, so the ideal-gas law gives $T_2/T_1=P_2/P_1$ directly. The first law for this adiabatic, no-boundary-work closed system reduces to $W_{in}=\Delta U$; the entropy change follows from the constant-$v$ form (the $R\ln(v_2/v_1)$ term vanishes), and because no heat crosses the (insulated) boundary, the entropy generated equals the system's own entropy rise, so the irreversibility is simply $I=T_0\Delta S$.

Final temperature from the constant-volume ideal-gas relation. $$\frac{T_2}{T_1}=\frac{P_2}{P_1}\quad\Rightarrow\quad T_2=300.15\times\frac{120}{100}=360.18\ \text{K}\ (87.03\ ^\circ\text{C}).$$
Paddle-wheel work (first law, $Q=0$, no boundary work). $$W_{in}=\Delta U=mc_v(T_2-T_1)=1\times0.718\times(360.18-300.15)=\boxed{43.10\ \text{kJ}}.$$
Entropy change of the air ($v$ constant). $$\Delta S=mc_v\ln\frac{T_2}{T_1}=1\times0.718\times\ln\frac{360.18}{300.15}=0.1309\ \text{kJ/K}.$$
Irreversibility. The tank is insulated, so no entropy is transferred with heat; the entropy generated equals the system's entropy rise, and $I=T_0S_{gen}$: $$I=T_0\,\Delta S=300.15\times0.1309=\boxed{39.27\ \text{kJ}}.$$

Quantity	Result
$W_{in}$ (paddle-wheel work)	43.10 kJ
$I$ (irreversibility)	39.27 kJ

Concept & Theory

Topic. Second-law analysis (irreversibility / exergy destruction) of a paddle-wheel-stirred, rigid, insulated ideal-gas tank.

Key relations. First law for a closed, adiabatic, rigid system: $W_{in}=\Delta U$. Entropy change of an ideal gas: $\Delta s=c_v\ln(T_2/T_1)+R\ln(v_2/v_1)$, which reduces to $c_v\ln(T_2/T_1)$ at constant $v$. Irreversibility (exergy destroyed) $I=T_0S_{gen}$.

Why this works. Paddle-wheel work is a genuinely irreversible way to add energy — unlike a reversible (quasi-static, frictionless) boundary-work process, stirring cannot be undone by reversing the paddle, so all of the entropy generated by this process shows up as "wasted work potential," which is exactly what $I=T_0S_{gen}$ quantifies: the minimum extra work that a reversible process would have needed to produce the same state change is zero (none of the 43.10 kJ was strictly necessary to reach state 2 by a reversible path with the same $T_0$), so all 39.27 kJ of $T_0S_{gen}$ represents lost work potential.

Common pitfall. Forgetting that a rigid tank has zero boundary work even though work crosses the system boundary via the paddle-wheel shaft — $W_{in}$ here is shaft work, not $P\,dV$ work, and it still shows up correctly in the closed-system first law as $Q-W=\Delta U$ with $W=-W_{in}$ (work done on the system).

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 8 (Exergy: irreversibility and reversible work).

Question 6 of 9: Rigid Insulated Tank — Paddle-Wheel Work and Irreversibility

Question 6: Rigid Insulated Tank — Paddle-Wheel Work and Irreversibility (15 marks)

Approach