GLOBAL KNOWLEDGE ARK

Engineering & Technology · Thermodynamics & Energy Systems

Refrigeration, heat pumps and performance measures

A device receives one unit of electrical work and supplies four units of heat to a warm target. This need not violate energy conservation. The extra three units can arrive from a colder source. The device moves energy while using work to make the transfer possible.

Use Forces & Energy in Simple Systems for force, work, power and boundary accounts; Materials, Properties & Everyday Objects provides property vocabulary. Kelvin, moles, enthalpy and entropy are introduced here. Ratios and graph reading are needed; no calculus is required. Every operating record, diagram and numerical case is constructed. Named physical constants and chemical formulas retain their usual meanings; supplied states are not specifications or operating instructions.

Back to the chapter

After reading, you can

Distinguish cooling and heating performance measures and explain why a coefficient of performance can exceed one.

01

Choose the intended service

A refrigerator removes heat QC from a cold region. A heat pump delivers heat QH to a warm region. A cyclic steady converter requires work Win, and QH=QC+Win when those are the only external energy crossings. All three are positive magnitudes here, avoiding a change to the earlier signed first-law convention.

The cooling coefficient of performance is COPR=QC/Win. The heating coefficient is COPH=QH/Win. For the same complete cycle and boundary, COPH=COPR+1. These are service-to-work ratios, not fractions of one energy input converted into a single output. A COP greater than one is therefore meaningful without being a heat-engine efficiency above 100%.

02

Follow the working fluid without making it a recipe

In a basic vapor-compression model, an evaporator absorbs energy as refrigerant vaporises at low pressure. A compressor supplies work and raises the fluid’s pressure and enthalpy. A condenser rejects heat at a higher temperature, and a throttling device reduces pressure before the evaporator. The refrigerant returns to its state over the cycle; it is not consumed as a fuel.

An ideal steady throttle is often approximated as constant enthalpy when heat, shaft work, speed and elevation changes are negligible. It is not generally constant temperature or reversible. Compressor work and heat-exchanger temperature differences matter. Actual refrigerant selection, phase states and equipment operation require information not supplied by this conceptual route.

03

Specify the temperature lift and the measurement boundary

For a reversible device between fixed reservoirs, COPR,max=TC/(TH−TC) and COPH,max=TH/(TH−TC), with all temperatures in kelvin. A larger temperature lift tends to lower these ceilings. The reservoir temperature is not automatically the refrigerant temperature at every point; finite transfer requires additional temperature differences.

An instantaneous favorable record does not establish seasonal performance. Include defined auxiliary power, standby periods and changing source/target conditions when comparing a complete service. If target warming is counted inside the boundary, it becomes stored energy rather than heat crossing out. Keep boundaries compatible before using the same ratio to rank systems.

Heat delivery can exceed work inputA hypothetical heat pump receives three kilowatts from its cold side and one kilowatt of work, delivering four kilowatts to its warm side.Steady heat-pump account; no stored-energy change3 kW from cold side1 kW work4 kW delivered to warm sideCold reservoir 280 K; warm reservoir 320 KCOPR = 3; COPH = 4; widths at 125 units/kW
Steady cyclic converter: QC=3 kW, Win=1 kW, QH=4 kW. COPR=3 and COPH=4. Reservoir temperatures280/320 K give reversible ceilings7 and8. The output includes transferred cold-side energy, so no energy is created.

Separate cooling COP from heating COP

A constructed steady cycle takes3 kW from a280 K cold reservoir and requires 1 kW of work. It supplies 4 kW to a320 K warm reservoir. COPR=3/1=3 and COPH=4/1=4. Over10 minutes, the corresponding amounts are1800,600 and 2400 kJ.

The reversible ceilings are280/(320−280)=7 for cooling and 320/40=8 for heating. The hypothetical values are below both. The warm-side heat is not obtained from the600 kJ work alone;1800 kJ has also crossed from the cold side.

Try the reasoning

  1. Does COPH=4 claim four units of energy were created from one?

    Show the answer and reasoning

    No. Three units are transferred from the cold source and one is work input.

  2. Why can two quoted COP values be unsuitable for direct comparison?

    Show the answer and reasoning

    They may use different temperatures, tasks, auxiliary-power boundaries or time intervals.

  3. Does a pressure drop through an ideal throttle guarantee constant temperature?

    Show the answer and reasoning

    No. The common limited approximation is constant enthalpy; temperature and phase can change.

What to carry forward

Identify cooling or heating as the service, include every relevant input, and attach COP to its temperature lift and interval.

References

Teaching resources and research supporting this lesson’s explanations and examples.

  1. Massachusetts Institute of Technology2 sources
    1. Thermodynamics — Chapter 6: Steady Flow Energy EquationMIT Unified Engineering, 2003 notesTeaching resourcelive.ocw.mit.eduRead source
    2. Thermodynamics and Propulsion — 2.3.1 Adiabatic, Steady Throttling of a GasE. M. Greitzer, Z. S. Spakovszky and I. A. Waitz; MIT Unified EngineeringTeaching resourceweb.mit.eduRead source
  2. Samuel J. Ling, Jeff Sanny and William Moebs; OpenStax/Rice University2 sources
    1. University Physics Volume 2 — 4.5 The Carnot CycleSamuel J. Ling, Jeff Sanny and William Moebs; OpenStax/Rice UniversityTeaching resourceopenstax.orgRead source
    2. University Physics Volume 2 — 4.3 Refrigerators and Heat PumpsSamuel J. Ling, Jeff Sanny and William Moebs; OpenStax/Rice UniversityTeaching resourceopenstax.orgRead source