Vapor-Compression Refrigeration

Keep state numbering explicit

State 1 is vapor leaving the evaporator and entering the compressor. State 2 is the high-pressure compressor exit, state 3 is liquid leaving the condenser, and state 4 is the low-pressure mixture leaving the expansion valve.

The valve is not an ideal turbine

A steady, adiabatic throttling valve with no shaft work and negligible kinetic/potential changes has h₄ ≈ h₃. Its entropy increases; throttling is not isentropic. The ideal compressor, in contrast, is modeled as isentropic.

Calculate useful cooling and work input

For the ideal cycle, heat absorbed in the evaporator per kilogram is h₁ − h₄ and compressor work input is h₂ − h₁. Their ratio is the refrigeration COP.

Heating and cooling use different useful outputs

A heat pump's useful output is condenser heat, not evaporator cooling. For the same simple cycle and energy accounting, COP_HP = COP_R + 1. Actual equipment ratings also depend on fans, pumps, losses, and operating conditions.

Worked example

An illustrative refrigerant loop has h₁ = 400, h₂ = 440, h₃ = h₄ = 250 kJ/kg, and mass flow 0.02 kg/s.

1. 1.Compute cooling and compressor input per kilogram.

   $$\begin{array}{c}{q}_L=400-250=150\\ w_{\mathrm{in}}=440-400=40[kJ/kg]\end{array}$$

2. 2.Find COP and the cooling power.

   $$\begin{array}{c}{\mathrm{COP}}_R=150/40=3.75\\ {\dot{Q}}_L=0.02(150)=3.00\mathrm{kW}\end{array}$$

3. 3.Check heat rejection power.

   $${\dot{Q}}_H=0.02(440-250)=3.80\mathrm{kW}=3.00+0.80\mathrm{kW}$$

The loop provides 3.00 kW of cooling for 0.80 kW of compressor work, giving COP_R = 3.75.