A rigid closed vessel contains 4 kg of refrigerant (R134a) at pressure of 200 kPa having the dryness fraction of 0.25. What will be the volume of the vessel? [Given: At 200 kPa : Specific volume (saturated liquid) = 0.0075 m³/kg, specific volume (saturated vapor) = 0.1 m³/kg

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BHEL Engineer Trainee Mechanical 24 Aug 2023 Official Paper

Answer 1

Concept:

For a mixture of saturated liquid and vapor, the specific volume is calculated using the dryness fraction formula:

\( v = v_f + x(v_g - v_f) \)

Then, total volume is found using: \( V = m \cdot v \)

Given:

Mass, m = 4 kg

Dryness fraction, x = 0.25

Specific volume of saturated liquid, \(v_f = 0.0075~m^3/kg\)

Specific volume of saturated vapor, \(v_g = 0.1~m^3/kg\)

Calculation:

Specific volume of mixture: \( v = v_f + x(v_g - v_f) = 0.0075 + 0.25(0.1 - 0.0075) = 0.030625~m^3/kg \)

Total volume: \( V = m \cdot v = 4 \cdot 0.030625 = 0.1225~m^3 = 122.5~\text{liters} \)