Consider an insulated tank having two portions, one part is evacuated and the other part is filled
water. The water in the filled part has a temperature of 333K and a pressure of 0.6 Mpa. The mass of
compressed of compressed liquid is (reg no Kg). If the partition is removed the water will expand in
the tank completely. Determine the final water temperature and the tank volume if the final pressure
is 10 Kpa.
Answers
Answer:
50kelvin in temperature
Given:
T₁ = 333K = 60°C
P₁ = 0.6MPa = 600kPa
P₂ = 10kPa
m = 2.5kg
To find:
The final temperature (T₂)
Tank volume (V₂)
Solution:
Apply energy equation.
ΔE = ΔU
ΔE = m(u₂ - u₁) = 0
m(u₂ - u₁) = 0
(u₂ - u₁) = 0
u₁ = u₂
Obtain the properties of water from tables at T₁ = 60°C
vf = 0.001017m³/kg
u₁ = u₂ = uf = 251.11kJ/kg
Obtain the properties of water from saturated water tables at P₂ = 10kPa
T₂ = T_sat = 45.81°C
vf = 0.00101m³/kg
vfg = 14.67254m³/kg
uf = 191.82kJ/kg
ufg = 2246.1kJ/kg
The final temperature is T₂ = 45.81°C
Calculate the dryness fraction at state-2 using the relation.
u₂ = uf + x₂ufg
251.11 = 191.82 + x₂(2246.1)
x₂ = 0.0264
Calculate the specific volume at state-2
v₂ = vf + x₂vfg
v₂ = 0.00101 + 0.0264(14.67254)
v₂ = 0.3884m³/kg
Calculate the final volume
V₂ = mv₂
V₂ = 2.5(0.3884)
V₂ = 0.971m³