Question

A rigid closed vertical cylindrical vessel of 15 cm diameter contains 5 kg water at 80 °C with 10% quality. The water is heated till its temperature reaches 130 °C. Considering only a horizontal separated interface between liquid and vapor, the dip in the liquid level after the heating process is ________ cm (rounded off to 2 decimal places).
Properties of water at various saturation temperatures are given in the table below.

T (°C)	vf (m3/kg)	vg (m3/kg)	uf (kJ/kg)	ug (kJ/kg)
80	0.001029	3.4053	334.97	2481.60
130	0.001070	0.66808	546.10	2539.50

T, v, and u are temperature, specific volume, and specific internal energy, respectively. Subscripts f and g represent saturated liquid and saturated vapor, respectively.

Hint:

Use the fact that the total specific volume stays fixed in a rigid vessel to find the new quality after heating.

Answer 1

Correct Answer: 11.37

Step 1: Recognize the constraint from the rigid, closed vessel.
Total mass $m=5$ kg and total volume $V$ never change, so the specific volume $v=V/m$ measured at state 1 must equal the specific volume at state 2.

Step 2: Get that fixed specific volume from the $80^{\circ}$C data.
$v = v_{f1}+x_1(v_{g1}-v_{f1}) = 0.001029+0.10(3.4053-0.001029) = 0.341456$ m$^3$/kg, using the given quality $x_1=0.10$.

Step 3: Solve for the new quality at $130^{\circ}$C using the same $v$.
$x_2 = \dfrac{v-v_{f2}}{v_{g2}-v_{f2}} = \dfrac{0.341456-0.001070}{0.66808-0.001070} = \dfrac{0.340386}{0.667010} = 0.5103$.

Step 4: Write the liquid height as a single formula and evaluate at both states.
For a vertical cylinder of cross-section $A=\pi(0.075)^2=0.017671$ m$^2$, the liquid height is $h=\dfrac{m(1-x)v_f}{A}$. At state 1: $h_1 = \dfrac{5(0.90)(0.001029)}{0.017671} = \dfrac{0.0046305}{0.017671}=0.2620$ m. At state 2: $h_2 = \dfrac{5(0.4897)(0.001070)}{0.017671} = \dfrac{0.0026199}{0.017671}=0.1483$ m.

Step 5: Take the drop.
$\Delta h = h_1-h_2 = 0.2620-0.1483=0.1137$ m $=11.37$ cm.

Final Answer:
Writing the liquid height as one formula in terms of quality shows the level falls because both the liquid fraction and its specific volume shrink as more mass flashes to vapor. \[ \boxed{\Delta h = 11.37 \ \text{cm}} \]