Rigid Insulated Tank with Paddle-Wheel Work Input: 04-BS-10, December 2017, Q6

Question 6: 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 (adiabatic) tank, $V=0.2$ m$^3$. Paddle-wheel power $\dot W_{pw}=4$ W for $t=20$ min $=1200$ s. Initial air density $\rho_1=1.2$ kg/m$^3$. No ΔKE/ΔPE.

Find. (a) $v_2$ [m$^3$/kg]; (b) $\Delta u$ [kJ/kg].

Approach

Because the tank is rigid (fixed volume) and sealed (fixed mass), the specific volume cannot change regardless of what happens thermally inside — that fixes part (a) immediately from the given density. For part (b), apply the closed-system energy balance: with $Q=0$ (insulated) and no boundary work (rigid tank), all of the paddle-wheel work becomes a rise in internal energy.

Mass and specific volume (part a). $m=\rho_1V=1.2\times0.2=0.24$ kg. Since the tank is rigid and sealed, $v_2=v_1=1/\rho_1$ regardless of the paddle-wheel heating: $$v_2=\frac{1}{1.2}=\boxed{0.8333\ \text{m}^3\text{/kg}}.$$
Change in specific internal energy (part b). Closed-system energy balance, $Q=0$ (insulated), no boundary work (rigid), $\Delta\text{KE}=\Delta\text{PE}=0$: the paddle-wheel work input equals the rise in total internal energy. $$W_{pw}=\dot W_{pw}\,t=4\times1200=4800\ \text{J}=4.8\ \text{kJ}.$$ $$\Delta U=-W_{on\ system}\ \Rightarrow\ m\,\Delta u=W_{pw}=4.8\ \text{kJ}$$ $$\Delta u=\frac{4.8}{0.24}=\boxed{20.0\ \text{kJ/kg}}.$$

Quantity	Result
(a) $v_2$	0.8333 m³/kg
(b) $\Delta u$	20.0 kJ/kg

Concept & Theory

Topic. Closed rigid-tank paddle-wheel work with density given instead of mass or volume alone.

Key relations. Closed-system first law: $\Delta U=Q-W$, with paddle-wheel work entering as negative $W$ (work done on the system). Rigid, sealed tank: $v=\text{const}$.

Why this works. A rigid tank cannot do or receive boundary (PΔV) work, and a sealed tank cannot change mass, so its specific volume is fixed by the initial density alone — the specific-volume answer requires no property-table lookup at all. With the tank also insulated, the paddle wheel is the only energy pathway across the boundary, so the entire electrical/mechanical input directly raises the internal energy; no partitioning between internal energy and boundary work is needed.

Common pitfall. Reaching for steam/air tables to find $v_2$ from a computed final temperature, when the rigid-sealed-tank constraint already gives $v_2=v_1$ directly without ever needing the final temperature.

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

Question 6 of 9: Rigid Insulated Tank with Paddle-Wheel Work Input

Question 6: Rigid Insulated Tank with Paddle-Wheel Work Input (15 marks)

Approach