Question

A rigid tank holds 10 lbm of 160 °F water. If the quality of the water is 0.5 then what is the pressure in the tank and volume of the tank?

Answer 1

The pressure and volume of the tank are 4.74703 psi and 385.968 ft³ 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 160°F.

Given:

Mass of the water is 10 lb.

Dryness fraction is 0.5.

Temperature of water is 160°F.

From steam table at 160°F:

The pressure in the tank is 4.74703 psi.

Specific volume of saturated water is 0.0163918 ft³/lb.

Specific volume of saturated steam is 77.1773 ft³/lb.

Calculation:

Step1

From steam table at 160°F:

The pressure in the tank is 4.74703 psi.

Step2

Specific volume of tank is calculated as follows:

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

`v=0.0163918 +0.5(77.1773 -0.0163918)`

`v=0.0163918 +38.58045`

v=38.5968 ft³/lb.

Step4

Volume is calculated as follows:

`V=v* m_(t)`

`V=38.5968*10`

V=385.968 ft³.

Thus, the pressure and volume of the tank are 4.74703 psi and 385.968 ft³ respectively.