One pound-mass of water lls a 2.29-ft3 rigid container at an initial pressure of 250 psia. The container is then cooled to 100F. Determine the initial temperature and nal pressure of the water.

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.