Refrigeration Cycle Lab (2D): Compressor, Condenser, Valve, Evaporator and the p-h Diagram

Presets

Live readouts

Graphs

Log pressure - enthalpy diagram with the real cycle

Cabinet temperature and thermostat band

COP against room temperature (real and Carnot)

Numerical checks

running...

How to use

Watch the green cycle on the p-h diagram (log pressure against enthalpy) between the saturation dome lines, and the cabinet temperature bouncing between the red and green thermostat lines. Try: (1) Hot kitchen: the cycle stretches upward, the COP falls; (2) Freezer: low evaporating pressure, large pressure ratio, low COP; (3) Door left open: the duty cycle goes to nearly 100 %; (4) Dirty condenser and Worn compressor: more electricity for the same cooling; (5) raise superheat or subcooling and see the corners of the loop move. Space pauses, R resets.

The science

What this lab shows. A refrigerator moves heat out of a cold cabinet into a warmer room by circulating a refrigerant through four steps: evaporation (absorbs heat, 4 to 1), compression (1 to 2, costs work), condensation (rejects heat, 2 to 3) and expansion through a valve (3 to 4, cools the fluid). The cabinet warms through its walls and cools while the compressor runs; a thermostat switches it on and off.

1. First law for each component (per kg of refrigerant):

q_e = h₁ - h₄, w = h₂ - h₁, q_c = h₂ - h₃ = q_e + w, h₄ = h₃ (throttling) COP = q_e / w ≤ COP_Carnot = T_e / (T_c - T_e)

The real COP is well below the Carnot value because of the throttling loss, superheat, compressor losses and non-zero temperature approaches. A hotter room (higher T_c) or a colder cabinet (lower T_e) lowers both.

2. Refrigerant model. A teaching model in the style of R134a: the saturation pressure follows Clausius-Clapeyron, ln p = A - B/T (A = 15.52, B = 2691 K, p in kPa), within about 1 % of the table between 0 and 40 C; the latent heat is h_fg = 198.6 - 0.89 T (kJ/kg, T in C), the liquid heat capacity is 1.40 kJ/(kg K), and the vapour is an ideal gas with γ = 1.12 and R = 0.0815 kJ/(kg K). The compressor outlet follows isentropic compression T_2s = T₁ (p_c/p_e)^(γ-1)/γ and h₂ = h₁ + (h_2s - h₁)/η_is. After the valve the fluid is a liquid-vapour mixture with quality x₄ = (h₄ - h_f)/h_fg.

3. Compressor flow. The mass flow is m-dot = η_v V_sw N ρ₁, with volumetric efficiency η_v = 1 - c [ (p_c/p_e)^1/γ - 1 ] for clearance fraction c: a higher pressure ratio re-expands more trapped gas and pumps less.

4. Cabinet. C dT/dt = UA (T_room - T) + Q_load - Q_e while the compressor runs. The evaporating and condensing temperatures follow the cabinet and room temperatures through fixed approaches, so the cycle moves as the cabinet cools.

Checks (run in the page). (1) and (2) the saturation fit against R134a table values at 0 C (293 kPa) and 40 C (1017 kPa); (3) the first law q_c = q_e + w; (4) the second law: the real COP is below the Carnot COP and compressor entropy generation is non-negative; (5) at η_is = 1 the outlet temperature equals the isentropic relation; (6) over a three-hour run the cabinet energy change equals the integral of leak minus cooling.

Simplifications. The refrigerant is not exactly R134a: the property correlations are simple fits; there is no pressure drop in pipes and heat exchangers, no oil, no frost, and the temperature approaches are constants (real heat exchangers have them depend on load and fan speed). The vapour is treated as an ideal gas. Heat leak and food load are lumped. Use it to understand trends, not to size a machine.