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

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?

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.