Question 6 of 9: Rigid Insulated Tank with Paddle Wheel

Question 6: Rigid Insulated Tank with Paddle Wheel (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. $V=0.2$ m³, rigid and insulated ($Q=0$), paddle-wheel power $\dot W_{pw}=4$ W for $t=20$ min (1200 s), $\rho_1=1.2$ kg/m³. Negligible KE/PE.

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

Approach

A rigid tank fixes the specific volume at every instant — the density (and hence $v$) cannot change with no mass or volume change — so part (a) needs no energy balance at all. Part (b) follows from the closed-system first law with $Q=0$ and no boundary work (rigid walls): all of the paddle-wheel work input appears as a rise in internal energy.

1. Specific volume (part a). Rigid tank $\Rightarrow$ mass and volume are both fixed, so $v_2=v_1$ regardless of what happens to $T$ or $P$: $$v_1=v_2=\frac{1}{\rho_1}=\frac{1}{1.2}=\boxed{0.8333\ \text{m}^3/\text{kg}}.$$
2. Mass in the tank. $$m=\rho_1 V=1.2\times0.2=0.24\ \text{kg}.$$
3. Paddle-wheel work input. $$W_{pw}=\dot W_{pw}\,t=4\times1200=4800\ \text{J}=4.8\ \text{kJ}.$$
4. First law, rigid+insulated (part b). $Q=0$, no boundary work ($dV=0$), so all paddle-wheel work raises the internal energy: $$\Delta U=W_{pw}=4.8\ \text{kJ}\qquad\Rightarrow\qquad \Delta u=\frac{\Delta U}{m}=\frac{4.8}{0.24}=\boxed{20.0\ \text{kJ/kg}}.$$