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Thermodynamics & Fluids · Thermodynamics

Power and refrigeration cycles

Reading30mLesson 3 of 3 in this module

Compute the thermal efficiency of an ideal Rankine cycle and the COP of a vapour-compression refrigerator from enthalpies.

Why it matters

Most of the world's electricity comes from heat engines that run a fluid around a cycle: steam plants (coal, nuclear, solar thermal, geothermal) use the Rankine cycle, and gas turbines use the Brayton cycle. Every fridge, air conditioner and heat pump runs a vapour-compression refrigeration cycle, which is a heat engine run backwards. Computing a cycle's efficiency, or its coefficient of performance (COP), tells you how much useful output you get for the energy you pay for.

The ideal Rankine cycle

Four steady-flow devices, numbered by state:

  1. Pump (1 → 2): saturated liquid from the condenser is pumped to boiler pressure. Small work input.
  2. Boiler (2 → 3): heat added at constant pressure turns the water into superheated steam.
  3. Turbine (3 → 4): steam expands and produces shaft work.
  4. Condenser (4 → 1): heat is rejected to cooling water at constant pressure, back to saturated liquid.

Apply the open-system first law (q − w = Δh per kg, KE and PE neglected) to each device:

w_pump    = h2 − h1          (work in)
q_in      = h3 − h2          (boiler)
w_turbine = h3 − h4          (work out)
q_out     = h4 − h1          (condenser)
w_net     = w_turbine − w_pump
η_th      = w_net / q_in

Worked example: a simple steam plant

A textbook-style ideal Rankine cycle runs between a boiler at 3 MPa, 350 °C and a condenser at 75 kPa. From steam tables (with an isentropic turbine and pump):

h1 = 384.44 kJ/kg   (saturated liquid at 75 kPa)
h2 = 387.47 kJ/kg   (after the pump; w_pump ≈ 3.03 kJ/kg)
h3 = 3116.1 kJ/kg   (superheated steam, 3 MPa, 350 °C)
h4 = 2403.0 kJ/kg   (wet steam at 75 kPa, quality about 0.886)

w_turbine = 3116.1 − 2403.0  = 713.1 kJ/kg
w_pump    = 387.47 − 384.44  = 3.03 kJ/kg
w_net     = 713.1 − 3.03     = 710.1 kJ/kg
q_in      = 3116.1 − 387.47  = 2728.6 kJ/kg
η_th      = 710.1 / 2728.6   ≈ 0.260   (26 %)

Notice how small the pump work is: pumping a liquid takes far less work than the turbine produces from the vapour. That is the Rankine cycle's big advantage. Compare 26 % with the Carnot limit between 350 °C (623 K) and the condenser's saturation temperature, about 92 °C (365 K): 1 − 365/623 ≈ 41 %. Real plants raise efficiency with higher boiler pressure and temperature, reheat and feedwater heating.

The vapour-compression refrigeration cycle

A refrigerant runs through:

  1. Compressor (1 → 2): low-pressure vapour is compressed to a hot, high-pressure vapour (work in).
  2. Condenser (2 → 3): heat is rejected to the room or outdoors; the refrigerant condenses.
  3. Expansion valve (3 → 4): throttling drops the pressure. A throttle does no work and, ideally, no heat transfer, so h4 = h3.
  4. Evaporator (4 → 1): the cold refrigerant absorbs heat from the space being cooled.

The figure of merit is the coefficient of performance:

COP_R  = q_L / w_in = (h1 − h4) / (h2 − h1)        (refrigerator)
COP_HP = q_H / w_in = (h2 − h3) / (h2 − h1)        (heat pump)

Worked example with approximate R-134a values: h1 = 240 kJ/kg, h2 = 275 kJ/kg, h3 = h4 = 95 kJ/kg.

COP_R = (240 − 95) / (275 − 240) = 145 / 35 ≈ 4.1

The fridge moves about 4.1 J of heat for every joule of electricity. A COP above 1 does not break any law: a refrigerator moves heat rather than creating it.

Common mistakes

  • Forgetting pump or compressor work when computing net work or COP.
  • Using q_out instead of q_in in the efficiency denominator.
  • Assuming the throttle drops enthalpy. It drops pressure and temperature, not enthalpy.
  • Treating ideal-cycle numbers as real performance. Real turbines and compressors have isentropic efficiencies around 70–90 %, so real cycle numbers are lower.

A note on real systems

Refrigerants are regulated (for example under the EPA's Section 608 rules in the US), and handling them requires certified technicians. Real plant and HVAC designs follow codes and standards and are reviewed by licensed professionals.

Recap

  • Rankine: η = (w_turbine − w_pump) / q_in, with each term a difference of enthalpies.
  • Refrigeration: COP_R = q_L / w_in; the throttle keeps h constant.
  • Ideal-cycle results are upper bounds for real equipment.

Try it

On the Do tab, write the Rankine efficiency formula in Python and check it against the worked example.

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