Question

A rigid tank holds 22 kg of 127 °C water. If 9 kg of that is liquid water what is the pressure in the tank and volume of the tank?

Answer 1

The pressure and volume of the tank are 246.878 Kpa and 9.449 m³ respectively.

Step-by-step explanation:

Volume is constant as the tank is rigid. Take the saturation condition of water from the steam table for pressure at 127°C.

Given:

Total mass of water is 22 kg.

Mass of liquid water is 9 kg.

Temperature of water is 127°C.

From steam table at 127°C:

The pressure in the tank is 246.878 Kpa.

Specific volume of saturated water is 0.00106683 m³/kg.

Specific volume of saturated steam is 0.72721 m³/kg.

Calculation:

Step1

From steam table at 127°C:

The pressure in the tank is 246.878 Kpa.

Step2

Dryness fraction is calculated as follows:

`x=(m_(v))/(m_(t))`

Here, dyness fraction is x, mass of vapor is
`m_(v)`and total mass is
`m_(t)`.

Substitute the values in the above equation as follows:

`x=(m_(v))/(m_(t))`

`x=(22-9)/(22)`

x = 0.59

Step3

Specific volume of tank is calculated as follows:

`v=v_(f)+x(v_(g)-v_(f))`

`v=0.00106683+0.59(0.72721-0.00106683)`

`v=0.00106683+0.42842447`

v=0.4295 m³/kg.

Step4

Volume is calculated as follows:

`V=v* m_(t)`

`V=0.4295 *22`

V=9.449 m³.

Thus, the pressure and volume of the tank are 246.878 Kpa and 9.449 m³ respectively.