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:
[tex]v=v_{f}+x(v_{g}-v_{f})[/tex]
[tex]v=0.0163918 +0.5(77.1773 -0.0163918)[/tex]
[tex]v=0.0163918 +38.58045[/tex]
v=38.5968 ft³/lb.
Step4
Volume is calculated as follows:
[tex]V=v\times m_{t}[/tex]
[tex]V=38.5968\times10[/tex]
V=385.968 ft³.
Thus, the pressure and volume of the tank are 4.74703 psi and 385.968 ft³ respectively.
