Question

1. A 200-m^3 rigid tank is filled with saturated liquid-vapor mixture of water at 50°C. If 10% of the mass is liquid and the rest is vapor, find the total mass in the tank.
2. If heat is added in and the mixture becomes saturated vapor, find the pressure.

Answer 1

Explanation:

Volume of the rigid tank (V)= 200$m^{3}$

Temperature (T) = $50^{o} c$

Note : refer the figure attached below

Now,

Since 10% of mass is liquid and rest is vapour.

Therefore, Dryness fraction (x) =  percentage of vapour in the (liquid + vapour)          

                                                       mixture

                                                x = 100 - 10 = 90%

                                                x = 0.9

At T = $50^{o} c$

$P_{\text {sat }}=12.352 \mathrm{kpa}, \quad v_{f}=0.001012 \mathrm{m}^{3} / \mathrm{kg}, \quad v_{g}=12.02 \mathrm{cm}^{3} / \mathrm{kg}$

$v_{A}=v_{f}+x\left(v_{g}-v_{f}\right)$

$=0.001012+0.9(12.026-0.001012)=10.8235001 \mathrm m^{2}/ {kg}$

$\text { Total roass }(\gamma n)=\frac{V}{v_{A}}=\frac{200 \mathrm{m}^{3}}{10.8235 \mathrm{m} / \mathrm{kg}}=18.4783 \mathrm{kg}$

$m=18.4783 \mathrm{kg}$

Now, heat is added so point A will reach to saturated vapour line at B

$P_{B}=P_{\text {sat }}=12 \cdot 352 \mathrm{KPa}$