Problem Analysis:

We need to calculate the change in the liquid level within a rigid closed cylindrical vessel after heating water from 80°C to 130°C. The process involves phase change, and the water is assumed to remain in a saturated liquid-vapor mixture state throughout.

Given Data:

• Cylinder Diameter, \(D = 15\) cm = 0.15 m
• Mass of water, \(m = 5\) kg
• Initial Temperature, \(T_1 = 80^\circ\)C
• Initial Quality, \(x_1 = 10\% = 0.1\)
• Final Temperature, \(T_2 = 130^\circ\)C

Properties from Table:

• At \(T_1 = 80^\circ\)C: \(v_{f1} = 0.001029\) m³/kg, \(v_{g1} = 0.340533\) m³/kg
• At \(T_2 = 130^\circ\)C: \(v_{f2} = 0.001070\) m³/kg, \(v_{g2} = 0.668085\) m³/kg

Step-by-Step Solution

1. Calculate Initial Specific Volume and Total Volume

The initial specific volume (\(v_1\)) of the water is calculated using the saturation properties at 80°C and the given quality \(x_1\):

$ v_1 = v_{f1} + x_1 (v_{g1} - v_{f1}) $

$ v_1 = 0.001029 + 0.1 (0.340533 - 0.001029) $

$ v_1 = 0.001029 + 0.1 (0.339504) $

$ v_1 = 0.001029 + 0.0339504 = 0.0349794 \text{ m³/kg} $

The total volume occupied by the water (which is also the internal volume of the vessel containing the water) is:

$ V_{total} = m \times v_1 $

$ V_{total} = 5 \text{ kg} \times 0.0349794 \text{ m³/kg} = 0.174897 \text{ m³} $

2. Calculate Vessel Cross-Sectional Area

The cross-sectional area (\(A\)) of the cylindrical vessel is:

$ A = \frac{\pi D^2}{4} = \frac{\pi (0.15 \text{ m})^2}{4} $

$ A = \frac{\pi \times 0.0225}{4} \approx 0.0176715 \text{ m²} $

3. Calculate Initial Liquid Level

The initial mass of the liquid phase (\(m_{f1}\)) is:

$ m_{f1} = m \times (1 - x_1) = 5 \text{ kg} \times (1 - 0.1) = 5 \times 0.9 = 4.5 \text{ kg} $

The initial volume of the liquid phase (\(V_{f1}\)) is:

$ V_{f1} = m_{f1} \times v_{f1} = 4.5 \text{ kg} \times 0.001029 \text{ m³/kg} = 0.0046305 \text{ m³} $

The initial liquid level (\(h_{f1}\)) is:

$ h_{f1} = \frac{V_{f1}}{A} = \frac{0.0046305 \text{ m³}}{0.0176715 \text{ m²}} \approx 0.2619 \text{ m} = 26.19 \text{ cm} $

4. Calculate Final Quality and Liquid Level

Since the vessel is rigid and closed, the total volume (\(V_{total}\)) and total mass (\(m\)) remain constant. Therefore, the average specific volume (\(v_2\)) remains the same:

$ v_2 = v_1 = 0.0349794 \text{ m³/kg} $

The final quality (\(x_2\)) at 130°C is calculated using \(v_2\) and the saturation properties at 130°C:

$ v_2 = v_{f2} + x_2 (v_{g2} - v_{f2}) $

$ 0.0349794 = 0.001070 + x_2 (0.668085 - 0.001070) $

$ 0.0349794 = 0.001070 + x_2 (0.667015) $

$ 0.0339094 = x_2 (0.667015) $

$ x_2 = \frac{0.0339094}{0.667015} \approx 0.05084 $

The final mass of the liquid phase (\(m_{f2}\)) is:

$ m_{f2} = m \times (1 - x_2) = 5 \text{ kg} \times (1 - 0.05084) = 5 \times 0.94916 = 4.7458 \text{ kg} $

The final volume of the liquid phase (\(V_{f2}\)) is:

$ V_{f2} = m_{f2} \times v_{f2} = 4.7458 \text{ kg} \times 0.001070 \text{ m³/kg} = 0.005078 \text{ m³} $

The final liquid level (\(h_{f2}\)) is:

$ h_{f2} = \frac{V_{f2}}{A} = \frac{0.005078 \text{ m³}}{0.0176715 \text{ m²}} \approx 0.2874 \text{ m} = 28.74 \text{ cm} $

5. Calculate the Dip in Liquid Level

The "dip" in the liquid level is typically defined as the decrease in level (\(h_{f1} - h_{f2}\)). However, in this case, the liquid level increased (\(h_{f2} > h_{f1}\)).

Change in liquid level = \(h_{f2} - h_{f1}\) = \(28.74 \text{ cm} - 26.19 \text{ cm} = 2.55 \text{ cm}\)

Since the level increased, there is no "dip" in the sense of a decrease. The calculated value indicates a rise of 2.55 cm.