Question:
1. Steam is contained in a closed rigid container with a volume of 1 m³. Initially, the pressure and temperature of the steam are 7 bar and 500°C, respectively. The temperature drops as a result of heat transfer to the surroundings. Determine
a) the temperature at which condensation first occurs, in °C,
b) the fraction of the total mass that has condensed when the pressure reaches 0.5 bar.
c) What is the volume, in m³, occupied by saturated liquid at the final state?
Answer:
The answers to the question are;
a) The temperature at which condensation first occurs is 163.056 °C.
b) The fraction of the total mass that has condensed when the pressure reaches 0.5 bar is 0.844.
c) The volume, occupied by saturated liquid at the final state is 1.66×10⁻³ m³.
Explanation:
To solve the question, we note that we are required check the steam tables for the properties of steam.
(a) From P×V = n×R×T
we find the number of moles as
n = P₁V₁/(RT₁)
Where
n = Number of moles
P₁ = Initial pressure = 7 bar = 700000 Pa
V₁ = Volume = 1 m³
T₁ = Temperature = 500 C = 773.15 K
R = Universal gas constant = 8.31451 J/(gmol·K)
Therefore n = (700000 Pa×1 m³)/(8.31451 J/(gmol·K)×773.15 K)
n = 108.89 moles
Molar mass of H₂O = 18.01528 g/mol
Therefore initial mass of steam = Number of moles × Molar mass
= 1961.73 g
From steam tables, condensation first start to occur at
151.836 °C + ((179.886 °C - 151.836 °C)/(10 bar - 5 bar))×(2 bar) = 163.056 °C
Therefore condensation first start to occur at 163.056 °C.
b) At 0.5 bar we have
n = P₂V₂/(RT₂)
Where
V₂ = V₁ = Volume = 1 m³
P₂ = Pressure = 0.5 bar = 50000 Pa
T₂ = Saturation temperature at 0.5 Pa = 81.3167 °C = 354.4667 K
R = Universal gas constant = 8.31451 J/(gmol·K)
n = (50000 Pa × 1 m³)/(354.4667 K×8.31451 J/(gmol·K)) = 16.965 moles
Mass of steam left at 0.5 bar = 16.965 moles × 18.01528 g/mol = 305.632 g.
Mass of condensed steam
= Initial mass of steam at 7 bar - Final mass of steam at 0.5 bar
= 1961.73 g - 305.632 g = 1656.095 g
The fraction of the total mass that has condensed at 0.5 bar is given by
Mass fraction of condensed steam
= (Mass of condensed steam)/(Initial mass of steam)
1656.095 g/1961.73 g = 0.844.
c) The volume occupied by saturated liquid at the final state is given by
Volume = Mass/Density = (1656.095 g)/(997 kg/m³)
= (1.656095 kg)/(997 kg/m³) = 1.66×10⁻³ m³.
