Question

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

Answer 1

Answer:

• 66.95 kpa
• 4.72688 m³

Explanation:

Mass of water and vapor = 10 kg

Mass of liquid water = 8 kg

Mass of vapor (gas) = 2 kg

temperature = 90⁰c + 273.15 k = 363.15 k

To calculate the volume of the tank you will need to add the volume of liquid water and the volume of vapor ( gas )

using: specific volume of liquid water : 0.001036 m³/kg

         specific volume of vapor ( gas ) : 2.3593 m³/kg

specific volume = volume / mass

hence volume = specific volume * mass

volume of liquid water = 0.001036 * 8 = (8.288 *10³ m³)

volume of vapor = 2.3593 * 2 = ( 4.7136 m³ )

Hence the volume of the tank = 8.288*10³ + 4.7136 = 4.72688 m³

To calculate the pressure in tank you will calculate the pressure been exerted by the vapor in the tank

this can be achieved by treating the vapor as an ideal gas

specific heats vapor at temperature at 90⁰c  

Cv = 1.45 kg/kg k

Cp = 1.885 kg/kg k

in an ideal gas situation  ( Cp - Cv ) = R = ( 1.885 - 1.45 ) = 0.435 kg/kg k

using the ideal gas equation

PₓVₓ = MₓRT

P = pressure

V = volume of vapor = 4.7136

R = 0.435

T = 363.15 K

M = 2 kg

therefore Pₓ = MₓRT / Vₓ

                    = (2 * 0.435 * 363.15) / 4.7136

                    = 66.95 kpa

Answer 2

Answer:

The pressure in the tank is 70.183 k Pa

The volume of the tank is 4.73 m³

Explanation:

Volume of the liquid phase

v(f) = m(f).v(f)

= 8 kg . 0.001036 m³/kg

= 0.008288 m³

Volume of the vapor phase

v(g) = m(g).v(g)

=2 kg . 2.3593 m³/kg

= 4.7186 m³

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

Volume of the tank

= 0.008288 m³ + 4.7186 m³ = 4.73 m³