Answer:
The pressure in the tank is 70.183 kPa and the volume of the tank is 4.73 [tex]m^3[/tex]
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
[tex]P=P_{sat \:@ \:90 \:C} = 70.183 \:kPa[/tex]
From the Saturated water - Temperature table at 90 °C, we know that [tex]v_f=0.001036 \frac{m^3}{kg}[/tex] and [tex]v_g=2.3593 \frac{m^3}{kg}[/tex].
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 = [tex]V_f=m_f\cdot v_f=8 \:kg\cdot 0.001036 \frac{m^3}{kg}=0.008288\:m^3[/tex]
Volume of the vapor phase = [tex]V_g=m_g\cdot v_g=2 \:kg\cdot 2.3593 \frac{m^3}{kg}=4.7186\:m^3[/tex]
Volume of the tank = Volume of the liquid phase + Volume of the vapor phase
Volume of the tank = [tex]0.008288\:m^3[/tex] + [tex]4.7186\:m^3[/tex] = 4.73 [tex]m^3[/tex]
