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R22 VCR System Calculation: 10 kW Cooling Load, -10°C Evaporator, 38°C Condenser
Mechanical Engineering (Thermodynamics)
University (Undergraduate 2nd/3rd Year)
Question Content
A VCR system using R22 has an evaporator at -10°C and a condenser at 38°C. The system delivers 10 kW of cooling. Determine: Mass flow rate of refrigerant, Compressor power required, COP
Correct Answer
i. 0.0183 kg/s, ii. 1.71 kW, iii. 5.848
Detailed Solution Steps
1
Step 1: Retrieve R22 properties. At -10°C (saturated vapour), h₁ = 244.7 kJ/kg, s₁ = 0.932 kJ/kg·K. At 38°C (saturated liquid), h₄ = h₃ = 106.2 kJ/kg.
2
Step 2: For isentropic compression, s₂ = s₁. Find h₂ at 38°C condenser pressure and s₂=0.932 kJ/kg·K: h₂ = 275.4 kJ/kg.
3
Step 3: Calculate refrigeration effect: q_L = h₁ - h₄ = 244.7 - 106.2 = 138.5 kJ/kg.
4
Step 4: Mass flow rate: m_dot = Q_L / q_L = 10 kW / 138.5 kJ/kg = 0.0183 kg/s.
5
Step 5: Compressor work per kg: w_c = h₂ - h₁ = 275.4 - 244.7 = 30.7 kJ/kg. Compressor power: W_c = m_dot * w_c = 0.0183 * 30.7 ≈ 1.71 kW.
6
Step 6: COP = Q_L / W_c = 10 / 1.71 ≈ 5.848
Knowledge Points Involved
1
Refrigeration Load and Mass Flow Rate
The mass flow rate of refrigerant is calculated by dividing the total cooling load (Q_L) by the refrigeration effect per unit mass (q_L), using m_dot = Q_L / q_L.
2
R22 Refrigerant Properties
A common HCFC refrigerant with tabulated enthalpy and entropy values used for VCR cycle analysis, critical for calculating work and energy transfers.
3
Compressor Power Calculation
Total compressor power is the product of mass flow rate and work input per unit mass, W_c = m_dot * w_c, where w_c = h₂ - h₁ for isentropic compression.
4
COP for Actual Cooling Loads
For systems with a defined cooling capacity, COP can also be calculated as the ratio of total cooling power to total compressor power input, COP = Q_L / W_c.
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