❄️ Vapor Compression Refrigeration Cycle

Analyze refrigeration and heat pump cycles with COP, cooling capacity, and P-h diagram visualization.

Compressor Condenser Expansion Valve Evaporator 1: Sat Vapor 2: Superheated Gas 3: Sat Liquid 4: Low-T Mixture

Vapor Compression Layout

Refrigerants flow in a closed loop, absorbing thermal energy from low-temperature spaces (evaporator) and rejecting it to high-temperature sinks (condenser). Isentropic compressor efficiency defines actual power draw.

• Evaporator (🔵): Absorbs heat, boils refrigerant at low pressure.
• Compressor (⚡): Work input increases pressure and temperature.
• Condenser (🔴): Rejects heat, liquefies refrigerant at high pressure.
• Expansion Valve ( throttling ): Drops pressure via isenthalpic throttling.

📝 Configuration

Refrigerant

Operating Temperatures

Refrigeration: -30 to 0°C
Typical: 30-50°C

Compressor

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

📊 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)

📘 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):

$$COP = \frac{Q_L}{W_{in}} = \frac{h_1 - h_4}{h_2 - h_1}$$ $$h_2 = h_1 + \frac{h_{2s} - h_1}{\eta_s} \quad \text{(isentropic compressor efficiency)}$$

Assumptions & Cycle Idealizations

  • Saturated vapor leaves the evaporator and enters the compressor
  • Saturated liquid leaves the condenser and enters the throttle valve
  • Isentropic compressor processes with efficiency $\eta_s$
  • Isenthalpic throttling process ($h_4 = h_3$)

Literature References

  1. Cengel, Y. A., & Boles, M. A., Thermodynamics: An Engineering Approach, McGraw-Hill.
  2. Moran, M. J., Shapiro, H. N., Boettner, D. D., & Bailey, M. B., Fundamentals of Engineering Thermodynamics, Wiley.

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.