Question

Water vapor is cooled in a closed, rigid tank from T1 = 480°C and p1 = 100 bar to a final temperature of T2 = 320°C. Determine the final specific volume, v2, in m3/kg, and the final pressure, p2, in bar.

Answer 1

To determine the final specific volume, v₂, and final pressure, p₂, we can use the ideal gas law equation. Substituting the given values, we can calculate the final specific volume and final pressure.

Step-by-step explanation:

To find the final specific volume, we can use the ideal gas law equation:

v₂ = (R * T₂) / p₂

Given that T1 = 480°C (753 K), p1 = 100 bar (10⁷ Pa), and T₂ = 320°C (593 K), we can substitute these values into the equation to solve for v₂.

v₂ = (8.314 J/mol·K * 593 K) / (10⁷ Pa)

After calculating, the final specific volume, v₂, is approximately 0.000478 m3/kg.

To find the final pressure, we can use the ideal gas law equation:

p₂ = (p1 * T₂) / T1

Substituting the known values, we get:

p₂ = (10⁷ Pa * 593 K) / 753 K

After calculating, the final pressure, p₂, is approximately 7.43 × 10⁶ Pa, which is equal to 74.3 bar.

Answer 2

v1 = v2 = 0.0347m³/kg the tank is rigid (constant volume)

P2 = 78.8bar

Step-by-step explanation:

Assuming ideal gas behavior, the relationship holds for the water vapor

PV = RT/M

Where P = pressure in the tank

V = Specific volume in m³/kg

R = gas constant = 8.314 J/mol•K

T = absolute temperature in K

M = molar mass in kg/mol

The detailed solution steps can be found below in the attachment.

V2 = V1 because the tank is rigid so it is a constant volume process. Also at the final temperature the water molecules are still in the gaseous phase so V2 is still the same as v1.