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 horizontal seperated interface between liquid and vapour, the deep in the liquid label after the heating process is ____ cm. (Round off to two decimal places)

Properties of water at various saturation temperature is given in the table:

Temperature	Specific volume		Specific internal energy
	vf (m³/kg)	vg (m³/kg)	uf (kJ/kg)	ug (kJ/kg)
80°C	0.001029	3.4053	334.97	2481.60
130°C	0.001070	0.66808	546.10	2539.50

Here f represents saturated liquid and g respresent saturated vapour.

The correct answer is 11.37.

Step 1: Calculate the initial specific volume and total volume of the vessel
Initially, the water is at $T_1=80\text{°C}$ with a quality of $x_1=10\text{\%}=0.1$.
Using the properties at 80°C from the given table:
Specific volume of saturated liquid, $v_{f1}=0.001029{\text{m}}^3\text{/kg}$
Specific volume of saturated vapour, $v_{g1}=3.4053{\text{m}}^3\text{/kg}$

The initial specific volume $v_1$ is given by:

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

$v_1=0.001029+0.1(3.4053-0.001029)$

$v_1=0.001029+0.3404271=0.3414561{\text{m}}^3\text{/kg}$

Step 2: Calculate the initial height of the liquid level
The mass of the saturated liquid initially is:

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

The initial volume of the liquid phase is:

$V_{f1}=m_{f1}·v_{f1}=4.5·0.001029=0.0046305{\text{m}}^3$

The cross-sectional area of the cylindrical vessel with diameter $d=15\text{cm}=0.15\text{m}$ is:

$A=\frac{\pi }{4}{d}^2=\frac{\pi }{4}{\left(0.15\right)}^2\approx 0.0176715{\text{m}}^2$

Thus, the initial height of the liquid is:

$h_1=\frac{V_{f1}}{A}=\frac{0.0046305}{0.0176715}\approx 0.26203\text{m}=26.203\text{cm}$

Step 3: Calculate the final quality and liquid volume after heating to 130��C
Since the vessel is rigid and closed, the total volume and mass remain constant, meaning the final specific volume is equal to the initial specific volume:
$v_2=v_1=0.3414561{\text{m}}^3\text{/kg}$
Using the properties at 130°C from the table:
Specific volume of saturated liquid, $v_{f2}=0.001070{\text{m}}^3\text{/kg}$
Specific volume of saturated vapour, $v_{g2}=0.66808{\text{m}}^3\text{/kg}$

The final dryness fraction (quality) $x_2$ is:

$x_2=\frac{v_2-v_{f2}}{v_{g2}-v_{f2}}=\frac{0.3414561-0.001070}{0.66808-0.001070}=\frac{0.3403861}{0.66701}\approx 0.510316$

The mass of the liquid at the final state is:

$m_{f2}=(1-x_2)·m=(1-0.510316)·5\approx 2.44842\text{kg}$

The final volume of the liquid phase is:

$V_{f2}=m_{f2}·v_{f2}=2.44842·0.001070\approx 0.0026198{\text{m}}^3$

The final height of the liquid level is:

$h_2=\frac{V_{f2}}{A}=\frac{0.0026198}{0.0176715}\approx 0.14825\text{m}=14.825\text{cm}$

Step 4: Calculate the dip in the liquid level
The dip (decrease) in the liquid level is:

$\Delta h=h_1-h_2=26.203-14.825=11.378\text{cm}$

Rounding to two decimal places, the dip in the liquid level is 11.37 cm.