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

Answer:

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

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:  

$V=v\times m_{t}$

$V=38.5968\times10$

V=385.968 ft³.  

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