Rigid Insulated Tank with Paddle-Wheel Work Input: 04-BS-10, May 2017, Q7

Question 7: Rigid Insulated Tank with Paddle-Wheel Work Input (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, well-insulated tank, $V=0.2$ m³, air. Paddle-wheel power $\dot W_{pw}=4$ W for $t=20$ min $=1200$ s. Initial density $\rho_1=1.2$ kg/m³, $T_1=300$ K.

Find. (a) $v_2$ [m³/kg]; (b) $\Delta u$ [kJ/kg]; (c) $\Delta s$ [kJ/kg·K].

Approach

The tank is rigid and closed (no mass crosses the boundary), so the specific volume cannot change regardless of the process — part (a) is a direct consequence of that, not a separate calculation. The paddle work becomes the entire energy input (insulated, no heat transfer, no boundary work since $V$ is fixed), giving $\Delta u$ directly; $\Delta s$ then follows from the real-air equation of state evaluated at the (constant) actual density, not at an arbitrary reference pressure.

Mass and specific volume (part a). $m=\rho_1V=1.2\times0.2=0.24$ kg. Since the tank is rigid and sealed, $v_2=v_1=1/\rho_1$: $$v_2=\boxed{0.83333\ \text{m}^3/\text{kg}}\quad(\text{unchanged from }v_1\text{, by mass and volume conservation}).$$
Energy balance (part b). Insulated ($Q=0$), rigid ($W_b=0$): all paddle work appears as internal energy increase. $$W_{pw}=\dot W_{pw}\,t=4\times1200=4800\ \text{J}=4.8\ \text{kJ}.$$ $$\Delta u=\frac{W_{pw}}{m}=\frac{4.8}{0.24}=\boxed{20.00\ \text{kJ/kg}}.$$
Final temperature (needed for part c). Solving $u(T_2,v_1)-u(T_1,v_1)=20.00$ kJ/kg for real air at the fixed density $\rho=1.2$ kg/m³: $$T_2=\boxed{327.83\ \text{K}}\ (54.68\ ^\circ\text{C})\qquad (P_1=103.30\ \text{kPa}\to P_2=112.91\ \text{kPa},\ \text{since }P=\rho RT\text{ rises with }T\text{ at fixed }v).$$
Change in specific entropy (part c). Evaluated at the actual constant density (NOT at an arbitrary reference pressure, which would silently assume the wrong volume at $T_2$): $$\Delta s=s(T_2,v_1)-s(T_1,v_1)=\boxed{0.06375\ \text{kJ/kg}\cdot\text{K}} \quad(\text{cross-check, constant-}c_v\text{ estimate: }c_v\ln(T_2/T_1)=0.7175\ln(1.0928)=0.0636,\ \text{close}).$$

Quantity	Result
(a) $v_2$	0.83333 m³/kg
(b) $\Delta u$	20.00 kJ/kg
(c) $\Delta s$	0.06375 kJ/kg·K

Concept & Theory

Topic. Rigid, insulated, closed system driven by paddle-wheel (shaft) work only — the simplest possible first-law energy balance, with the entropy change requiring properties evaluated at the correct (constant-volume) state.

Key relations. First law, closed rigid insulated system: $\Delta U=W_{pw,in}$. Entropy as a function of $(T,v)$: $s=s(T,v)$, evaluated at fixed $v$ across the process.

Why this works. A rigid sealed tank fixes both mass and volume, so $v$ cannot change no matter what process occurs inside — part (a)'s answer is really a statement about the system definition, not a result of the paddle-wheel process itself.

Common pitfall. Computing $\Delta u$ or $\Delta s$ from property calls at a fixed REFERENCE pressure (e.g. 1 atm) instead of the tank's own actual, constant density — since $P_1=103.30$ kPa here is close to but not equal to 101.325 kPa, this looks almost right while quietly using a slightly wrong volume at $T_2$, and the entropy error compounds because $\partial s/\partial v$ does not vanish.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 4 (Energy Analysis of Closed Systems) & Ch. 7 (Entropy).

Question 7 of 9: Rigid Insulated Tank with Paddle-Wheel Work Input

Question 7: Rigid Insulated Tank with Paddle-Wheel Work Input (15 marks)

Approach