Rigid Tank Vaporization by Electric-Resistance Heating: 04-BS-10, December 2017, Q9

Question 9: Rigid Tank Vaporization by Electric-Resistance Heating (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, $m=5$ kg water, $P_1=100$ kPa ($T_{sat,1}=99.61\ ^\circ$C), initially $75\%$ liquid by mass $\Rightarrow x_1=0.25$. Electric resistor: $\mathcal{V}=110$ V, $I=8$ A. Final state: all liquid vaporized (saturated vapor, $x_2=1$).

Find. Time $t$ [s or min] to fully vaporize the liquid.

Fig. Q9 — T–v path for the rigid-tank vaporization. Because the tank is rigid and sealed, $v_2=v_1$ exactly — the process is a vertical line inside the dome (state 1, $x_1=0.25$) rising to the saturated-vapor line at a higher pressure/temperature (state 2, $x_2=1$), NOT to the original 100 kPa saturation temperature.

Approach

Because the tank is rigid and sealed, the process is constant-volume: the specific volume of the initial two-phase mixture fixes the specific volume of the final saturated-vapor state exactly (state 2 is NOT at the original 100 kPa — it is wherever the saturated-vapor curve crosses $v=v_1$, at a higher pressure). Find $v_1$ from the given quality, root-solve for the final-state pressure that gives $v_g(P_2)=v_1$, then apply the closed-system energy balance (insulated, rigid: all electrical work becomes internal energy) to solve for the time.

Initial state properties. At $P_1=100$ kPa: $v_f=0.0010432$, $v_g=1.69393$ m$^3$/kg; $u_f=417.40$, $u_g=2505.55$ kJ/kg. With $x_1=0.25$: $$v_1=v_f+x_1(v_g-v_f)=0.0010432+0.25\times(1.69393-0.0010432)=\boxed{0.42426\ \text{m}^3\text{/kg}}.$$ $$u_1=u_f+x_1(u_g-u_f)=417.40+0.25\times(2505.55-417.40)=\boxed{939.44\ \text{kJ/kg}}.$$
Final state (rigid tank ⇒ $v_2=v_1$, saturated vapor). Root-solving for the pressure at which $v_g(P_2)=v_1=0.42426$ m$^3$/kg: $$P_2=\boxed{438.32\ \text{kPa}}\qquad(T_{sat,2}=146.94\ ^\circ\text{C}),\qquad u_2=u_g(P_2)=\boxed{2556.23\ \text{kJ/kg}}.$$ This confirms the process does NOT return to $100$ kPa — constant-volume vaporization drives the pressure up substantially as the last of the liquid disappears.
Energy balance ⇒ electrical work required. Rigid tank (no boundary work), well-insulated ($Q=0$): the resistor's electrical work input equals the rise in internal energy. $$W_{elec}=m(u_2-u_1)=5\times(2556.23-939.44)=5\times1616.79=\boxed{8084.0\ \text{kJ}}.$$
Resistor power and time. $$\dot W_{elec}=\mathcal{V}I=110\times8=880\ \text{W}=0.880\ \text{kW}.$$ $$t=\frac{W_{elec}}{\dot W_{elec}}=\frac{8084.0}{0.880}=\boxed{9186.3\ \text{s}}=\boxed{153.1\ \text{min}}\ (2.55\ \text{h}).$$

Quantity	Result
$v_1=v_2$	0.42426 m³/kg
Final pressure $P_2$	438.32 kPa
Electrical work $W_{elec}$	8084.0 kJ
Time to vaporize	9186.3 s ≈ 153.1 min (2.55 h)

Concept & Theory

Topic. Constant-volume (rigid-tank) vaporization of a saturated liquid-vapor mixture by electric-resistance heating.

Key relations. Rigid, sealed tank: $v=\text{const}$ throughout the process. Closed-system first law with $Q=0$, no boundary work: $\Delta U=W_{elec}$ (electrical work entering as negative $W_{by\ system}$). Electrical power: $\dot W_{elec}=\mathcal{V}I$.

Why this works. Because the tank cannot change volume, the two-phase mixture cannot follow a constant-pressure vaporization path (unlike a piston-cylinder device) — instead it moves along the fixed specific-volume line $v=v_1$ until that line intersects the saturated-vapor curve, which happens at a pressure and temperature substantially ABOVE the initial saturation pressure. This is the single most important conceptual trap in rigid-tank two-phase heating problems: students who assume the final state stays at 100 kPa (the initial pressure) get a badly wrong answer, since the true final pressure (438.32 kPa here) is over 4× higher.

Common pitfall. Forgetting that "vaporize all the liquid" in a rigid tank means finding the saturated-vapor state at the ORIGINAL specific volume (not the original pressure), which requires an iterative/root-finding property lookup rather than a direct table read at $P_1$.

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

Question 9 of 9: Rigid Tank Vaporization by Electric-Resistance Heating

Question 9: Rigid Tank Vaporization by Electric-Resistance Heating (15 marks)

Approach