Rigid Insulated Tank, Paddle-Wheel Stirring: 04-BS-10, Undated paper, Q7

Question 7: Rigid Insulated Tank, Paddle-Wheel Stirring (Part B – 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 ($V=\text{const}$), insulated (adiabatic, $Q=0$) tank; $m=1$ kg air. $T_1=27\ ^\circ\text{C}=300.15$ K, $P_1=100$ kPa; $P_2=120$ kPa. Dead state $T_0=300.15$ K, $p_0=100$ kPa.

Find. (a) $W_{in}$, in kJ; (b) the irreversibility $I$, in kJ.

Approach

Constant volume $\Rightarrow T_2/T_1=P_2/P_1$ directly from the ideal-gas law. No heat crosses the insulated boundary, so the paddle-wheel work equals the internal-energy rise, and the entropy generated is simply the system's own entropy change (no external reservoir exchange). The irreversibility — the exergy destroyed — then follows from the exergy-destruction relation $I=T_0S_{gen}$. Note the tank is rigid, so no work is done against the surroundings at $p_0$ and the $p_0\Delta V$ term that would otherwise appear drops out.

Final temperature. Rigid tank $\Rightarrow v_2=v_1$, so $T_2=T_1(P_2/P_1)=300.15\times1.20=360.18$ K.
Net work input — part (a). First law, $Q=0$, closed system, rigid ($\Delta KE=\Delta PE=0$): $$W_{in}=m\,c_v(T_2-T_1)=1\times0.718\times(360.18-300.15)=\boxed{43.10\ \text{kJ}}.$$
Entropy generation. Constant volume $\Rightarrow$ the $R\ln(v_2/v_1)$ term vanishes: $$\Delta S=m\,c_v\ln\frac{T_2}{T_1}=1\times0.718\times\ln\frac{360.18}{300.15}=0.1309\ \text{kJ/K}.$$ Since the tank is insulated ($Q=0$ across its boundary), $S_{gen}=\Delta S_{system}=0.1309$ kJ/K.
Irreversibility — part (b). Every joule of paddle-wheel work is dissipated; the exergy destroyed is the product $$I=T_0\,S_{gen}=300.15\times0.1309=\boxed{39.29\ \text{kJ}}.$$ Cross-check from the closed-system exergy balance, which must give the same number: $I=W_{in}-\Delta X$ with $\Delta X=m[(u_2-u_1)-T_0(s_2-s_1)]+p_0\Delta V$, and $\Delta V=0$, so $\Delta X=43.10-300.15\times0.1309=3.81$ kJ and $I=43.10-3.81=39.29$ kJ $\checkmark$. Only $3.81/43.10=8.8\%$ of the stirring work survives as recoverable exergy.

Quantity	Value
(a) $W_{in}$ (work)	43.10 kJ
(b) $I$ (irreversibility)	39.29 kJ
($S_{gen}$, intermediate)	0.1309 kJ/K

Topic. Paddle-wheel work on a closed, rigid, insulated ideal-gas system.

Key relations. Ideal gas, constant $v$: $P/T=\text{const}$; $W_{in}=mc_v\Delta T$; $\Delta S=mc_v\ln(T_2/T_1)$ (constant $v$); insulated boundary $\Rightarrow S_{gen}=\Delta S_{sys}$; exergy destroyed $I=T_0S_{gen}$.

Why this works. Paddle-wheel work is a pure work interaction (no mass or heat crosses the boundary), so it converts entirely to internal energy in a rigid tank — there is no $P\,dV$ boundary work either, since the volume cannot change. This is the textbook example of a purely internally-irreversible process: entropy rises even though $Q=0$ everywhere.

Common pitfall. Stopping at $S_{gen}$ and reporting it as the answer. The question asks for the irreversibility, which is an energy (kJ), not an entropy (kJ/K) — multiply by $T_0$. The second trap is forgetting that $S_{gen}$ equals the SYSTEM's entropy change only because the tank is insulated ($Q_{boundary}/T_0=0$); for a non-insulated tank you would need to subtract the heat-transfer term $Q/T_0$ (or $T_{boundary}$) from $\Delta S_{sys}$ to isolate $S_{gen}$.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., §7-6 (entropy change of ideal gases) and §7-9 (entropy generation).

Question 7 of 9: Rigid Insulated Tank, Paddle-Wheel Stirring

Question 7: Rigid Insulated Tank, Paddle-Wheel Stirring (Part B – 15 marks)

Approach