Answer:
The initial temperature, T₁= 550 °F = 560.93 K
The final pressure, P₂ is 139.561 psi
Explanation:
Here we have
The specific volume of the fluid in the container, V₁ = 2.29 ft³/lb
= 0.14296 m³/kg
While the specific volume of liquid water is about 0.01602 ft³/lb
Therefore, the water exists as steam.
From the steam tables, we have the specific volume of saturated steam at 250 psia is given as
P₁ = 250 psia = 17.24 bar
vf ≈ 0.00116565 m³ kg⁻¹ = vg ≈ 0.113428 m³ kg⁻¹
vf = Specific volume of saturated liquid phase of water and
vg = Specific volume of saturated gaseous phase of water
Therefore at v = 0.14296 m³/kg, the steam is in the superheated state
From the super-heated stem tables we have at P = 17.24 bar = 250 psia and v = 0.14296 m³/kg = 2.29 ft³/lb the temperature T₁ = 550 °F
Therefore, the initial temperature, T₁= 550 °F = 560.93 K
The final pressure can be found from the steam tables at T₂ = 100 °F = 310.93 K
Hence vf = 0.01613 ft³/lb and vg = 349.83 ft³/lb
V₂ = 2.29 ft³ - 0.01613 ft³ = 2.274 ft³
Since the volume is constant, we have P₂ =P₁×V₁×T₂/(V₂×T₁) = 139.561 psi
The final pressure, P₂ = 139.561 psi.
