❄️ Vapor Compression Refrigeration Cycle
Analyze refrigeration and heat pump cycles with COP, cooling capacity, and P-h diagram visualization.
📝 Configuration
Cycle Processes:
1→2: Compressor (work input)
2→3: Condenser (heat rejection)
3→4: Throttle valve (isenthalpic)
4→1: Evaporator (heat absorption)
COP_ref = Q_L / W_comp
COP_hp = Q_H / W_comp = COP_ref + 1
1→2: Compressor (work input)
2→3: Condenser (heat rejection)
3→4: Throttle valve (isenthalpic)
4→1: Evaporator (heat absorption)
COP_ref = Q_L / W_comp
COP_hp = Q_H / W_comp = COP_ref + 1
📊 Results & Visualization
Configure the inputs and click Calculate to see results.
ℹ️ About Refrigeration Cycles
The vapor compression cycle is the most common type of refrigeration system. It uses a refrigerant that undergoes phase changes to transfer heat.
Common Refrigerants:
• R-134a: Most common for automotive and residential
• R-410A: Residential air conditioning
• R-717 (NH₃): Industrial refrigeration
• R-22: Being phased out (ozone depletion)
The vapor compression cycle is the most common type of refrigeration system. It uses a refrigerant that undergoes phase changes to transfer heat.
Common Refrigerants:
• R-134a: Most common for automotive and residential
• R-410A: Residential air conditioning
• R-717 (NH₃): Industrial refrigeration
• R-22: Being phased out (ozone depletion)
📘 Calculation Methodology
Mathematical Model & Theory
The vapor-compression refrigeration cycle cools spaces by transferring heat from a low-temperature region to a high-temperature sink using compressor work. Cycle performance is measured by the Coefficient of Performance (COP):
$$COP = \frac{Q_L}{W_{in}} = \frac{h_1 - h_4}{h_2 - h_1}$$
$$h_2 = h_1 + \frac{h_{2s} - h_1}{\eta_s} \quad \text{(isentropic compressor efficiency)}$$
Worked Engineering Example
Problem Statement:
A refrigeration cycle operates with Tevap = -10°C and Tcond = 40°C. If refrigerant enthalpies are $h_1 = 244.5$ (evaporator exit), $h_2 = 285.0$ (compressor exit), and $h_3 = h_4 = 108.3$ kJ/kg (condenser exit/throttling inlet), find the COP.
Step-by-step Solution:
1. Calculate cooling effect $q_L$:
$$q_L = h_1 - h_4 = 244.5 - 108.3 = 136.2 \text{ kJ/kg}$$ 2. Calculate compressor work $w_{in}$:
$$w_{in} = h_2 - h_1 = 285.0 - 244.5 = 40.5 \text{ kJ/kg}$$ 3. Calculate Coefficient of Performance ($COP$):
$$COP = \frac{136.2}{40.5} = 3.363$$
Final Result:
The cycle COP is 3.36.
A refrigeration cycle operates with Tevap = -10°C and Tcond = 40°C. If refrigerant enthalpies are $h_1 = 244.5$ (evaporator exit), $h_2 = 285.0$ (compressor exit), and $h_3 = h_4 = 108.3$ kJ/kg (condenser exit/throttling inlet), find the COP.
Step-by-step Solution:
1. Calculate cooling effect $q_L$:
$$q_L = h_1 - h_4 = 244.5 - 108.3 = 136.2 \text{ kJ/kg}$$ 2. Calculate compressor work $w_{in}$:
$$w_{in} = h_2 - h_1 = 285.0 - 244.5 = 40.5 \text{ kJ/kg}$$ 3. Calculate Coefficient of Performance ($COP$):
$$COP = \frac{136.2}{40.5} = 3.363$$
Final Result:
The cycle COP is 3.36.