Question

A rigid tank contains 10 kg of water at 90°C. If 8 kg of the water is in the liquid form and rest is in vapor form determine: (a) Pressure in the tank (b) Volume of the tank

Answer 1

Answer:

The pressure in the tank is 70.183 kPa and the volume of the tank is 4.73 $m^3$

Explanation:

Given that 8 kg of the water is in the liquid form and rest is in vapor we can say that the tank contains a saturated liquid-vapor mixture. Since the two phases coexist in equilibrium, the pressure must be the saturation pressure at the given temperature.

Use the Saturated water - Temperature table to find the pressure

$P=P_{sat \:@ \:90 \:C} = 70.183 \:kPa$

From the Saturated water - Temperature table at 90 °C, we know that $v_f=0.001036 \frac{m^3}{kg}$ and $v_g=2.3593 \frac{m^3}{kg}$.

To find the volume of the tank you can determine the volume occupied by each phase and then add them:

Volume of the liquid phase = $V_f=m_f\cdot v_f=8 \:kg\cdot 0.001036 \frac{m^3}{kg}=0.008288\:m^3$

Volume of the vapor phase = $V_g=m_g\cdot v_g=2 \:kg\cdot 2.3593 \frac{m^3}{kg}=4.7186\:m^3$

Volume of the tank = Volume of the liquid phase + Volume of the vapor phase

Volume of the tank = $0.008288\:m^3$  + $4.7186\:m^3$ = 4.73 $m^3$