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: Get the total (fixed) specific volume from the initial state.
At \(80^{\circ}\)C with quality \(x_1=0.10\): \(v_1=v_{f1}+x_1(v_{g1}-v_{f1})=0.001029+0.10(3.4053-0.001029)=0.001029+0.340427=0.341456\) m\(^3\)/kg. Since the vessel is rigid and closed, both mass \(m=5\) kg and total volume \(V=mv_1\) stay fixed, so the specific volume stays at \(v_1=v_2=0.341456\) m\(^3\)/kg even after heating.

Step 2: Find the quality at \(130^{\circ}\)C.
\(v_2=v_{f2}+x_2(v_{g2}-v_{f2})\), so \(0.341456=0.001070+x_2(0.66808-0.001070)\), giving \(x_2=\dfrac{0.340386}{0.667010}=0.5103\).

Step 3: Get the liquid mass at each state.
State 1: \(m_{f1}=m(1-x_1)=5(0.90)=4.5\) kg. State 2: \(m_{f2}=m(1-x_2)=5(0.4897)=2.449\) kg.

Step 4: Convert liquid mass to liquid height using the vessel's cross-section.
Diameter \(=0.15\) m, radius \(=0.075\) m, so \(A=\pi(0.075)^2=0.017671\) m\(^2\). \(V_{f1}=m_{f1}v_{f1}=4.5(0.001029)=0.0046305\) m\(^3\), so \(h_1=V_{f1}/A=0.0046305/0.017671=0.2620\) m. \(V_{f2}=m_{f2}v_{f2}=2.449(0.001070)=0.0026199\) m\(^3\), so \(h_2=V_{f2}/A=0.0026199/0.017671=0.1483\) m.

Step 5: Subtract to get the dip.
\(\Delta h = h_1-h_2=0.2620-0.1483=0.1137\) m \(=11.37\) cm.

Final Answer:
The liquid level drops because more of the fixed mass turns to vapor at the higher temperature, even though the vessel volume never changes. \[ \boxed{\Delta h = 11.37 \ \text{cm}} \]