Thermodynamics · Chapter 11
Thermodynamics · Chapter 11

Refrigeration Cycles

Refrigeration cycles use work or high-temperature heat to move energy from a colder region to a warmer one. Analysis is reliable only when the desired product, refrigerant state, component boundary, and property source are declared before a coefficient of performance is computed.

Learning objectives

  • Distinguish refrigerator and heat-pump products and calculate their coefficients of performance.
  • Identify vapor-compression states from absolute pressure, temperature, and phase.
  • Apply steady-flow mass, energy, and entropy balances to compressors, valves, and heat exchangers.
  • Differentiate isenthalpic throttling, isentropic compression, and actual compressor efficiency.
  • Compare vapor, gas, cascade, and absorption architectures within their valid operating domains.
  • Audit cycle closure, Carnot limits, entropy generation, exergy destruction, and uncertainty.
11–1

Refrigerators and Heat Pumps

Choose the product before the metric

A refrigerator's product is \(Q_L\), heat removed from the cold region; a heat pump's product is \(Q_H\), heat delivered to the warm region. For a cyclic device, \(W_{in}=Q_H-Q_L\), so \(\mathrm{COP}_{HP}=\mathrm{COP}_R+1\). COP may exceed one because it compares moved heat with work; it is not thermal efficiency.

Cycle performance
\[\mathrm{COP}_R=\frac{Q_L}{W_{in}}\]\[\mathrm{COP}_{HP}=\frac{Q_H}{W_{in}}\]

Physical meaning

A COP of four means one unit of work moves four units of the selected heat product. It does not create energy: the condenser rejects the selected cold-side load plus compressor work.

11–2

The Reversed Carnot Cycle

For reservoirs at absolute temperatures, reversible limits are \(\mathrm{COP}_{R,rev}=T_L/(T_H-T_L)\) and \(\mathrm{COP}_{HP,rev}=T_H/(T_H-T_L)\). Use kelvins or rankines, never Celsius or Fahrenheit differences mixed with absolute values. Real heat exchangers require refrigerant temperatures beyond the reservoir temperatures, increasing lift and reducing achievable COP. Wet compression and a two-phase expander also make the vapor Carnot arrangement impractical.

11–3

The Ideal Vapor-Compression Refrigeration Cycle

State 1 is vapor entering an ideal isentropic compressor; 2 rejects heat in the condenser; 3 is liquid before the valve; 4 follows isenthalpic throttling and enters the evaporator. Each component is a steady open control volume. Use refrigerant tables or validated software, not an ideal-gas equation in the two-phase region.

Enthalpy ledger per unit mass
\[q_L=h_1-h_4\]\[w_c=h_2-h_1\]\[q_H=h_2-h_3\]\[h_4=h_3\]
T_H (hot)T_L (cold)RQ_HQ_LW
Figure 11–1  A refrigeration cycle uses compressor work to move heat Q_L from the cold space and reject Q_H to the warm surroundings.

Problem-solving strategy

  1. Mark the refrigerated space and decide whether the requested product is cooling or heating.
  2. Draw component control volumes, flow direction, state numbers, and a heat-in/work-in sign convention.
  3. Use absolute pressures and temperatures to classify each refrigerant state; quality is valid only inside the saturation dome.
  4. Read or interpolate properties from one authoritative refrigerant source, recording the fluid version and reference state.
  5. Close energy around each component and the cycle, then test COP against the reversible limit and entropy generation against zero.
EXAMPLE 1Four-state refrigeration ledger
Use explicitly supplied refrigerant enthalpies for an ideal vapor-compression loop to calculate cooling effect, compressor work, heat rejection, and refrigerator COP.

Recognition strategy

A compressor, condenser, valve, and evaporator form the standard vapor-compression family; enthalpy differences expose all requested transfers.

Given information

  • Supplied educational property values: \(h_1=398\), \(h_2=438\), and \(h_3=h_4=250\ \mathrm{kJ/kg}\).
  • One-kilogram basis, steady operation, absolute state pressures, and temperatures interpreted on an absolute basis where required.

Model and interpretation

Each device is an open steady control volume and the assembly is cyclic. Heat absorbed by the refrigerant and compressor work input are positive. The refrigerant is a real pure substance represented by supplied table values, not an ideal gas; throttling is isenthalpic, compression is ideal isentropic, and kinetic and potential energy are neglected.

Step-by-step solution
1. Cooling effect. Across the evaporator, \(q_L=h_1-h_4=398-250=148\ \mathrm{kJ/kg}\).
2. Compressor input. The adiabatic steady-flow balance gives \(w_c=h_2-h_1=40\ \mathrm{kJ/kg}\).
3. Heat rejection. The condenser rejects \(q_H=h_2-h_3=438-250=188\ \mathrm{kJ/kg}\).
4. Performance. The refrigerator coefficient is \(\mathrm{COP}_R=q_L/w_c=148/40=3.70\).

Verification

Units: Enthalpy differences produce \(\mathrm{kJ/kg}\), and dividing two specific energies makes COP dimensionless.

Signs and direction: Positive compressor input drives positive cold-side absorption and larger positive warm-side rejection.

Plausibility: The cycle closes because \(q_L+w_c=148+40=188=q_H\ \mathrm{kJ/kg}\). Mass and stored energy return to their initial values. The working-fluid entropy and exergy changes over a cycle are zero; ideal compression has zero generation, while throttling must generate nonnegative entropy.

Final result: \(q_L=148\ \mathrm{kJ/kg}\), \(w_c=40\ \mathrm{kJ/kg}\), \(q_H=188\ \mathrm{kJ/kg}\), and \(\mathrm{COP}_R=3.70\).

Engineering interpretation

Each unit of compressor work moves 3.70 units of heat from the cold space, while 4.70 units reach the condenser. Traceable fluid data, pressure drops, superheat, subcooling, and motor losses are needed for equipment selection.

11–4

Actual Vapor-Compression Refrigeration Cycle

Actual compressor endpoint

For an adiabatic compressor, \(\eta_c=(h_{2s}-h_1)/(h_{2a}-h_1)\). Because \(\eta_c<1\), actual work and discharge enthalpy exceed isentropic values. Suction superheat protects against liquid ingestion, subcooling can increase refrigerating effect, and line pressure drops alter saturation temperatures. Neither superheat nor subcooling is automatically beneficial after all loads and controls are counted.

EXAMPLE 2Actual compressor and plant load
An actual refrigerant loop must provide 60.0 kW of cooling. From supplied property data and compressor efficiency, determine flow rate, power, COP, and compressor entropy generation.

Recognition strategy

The isentropic outlet and efficiency define the actual compressor state; the cooling duty then sets mass flow before power is evaluated.

Given information

  • Supplied values: \(h_1=410\), \(h_{2s}=445\), \(h_3=h_4=260\ \mathrm{kJ/kg}\), \(\eta_c=0.78\).
  • Supplied entropies \(s_1=1.720\), \(s_{2a}=1.755\ \mathrm{kJ/(kg\!\cdot\!K)}\); \(\dot Q_L=60.0\ \mathrm{kW}\).

Model and interpretation

Use steady component control volumes and a refrigerant mass basis. Absolute pressures and temperatures define states in the supplied real-fluid source. Compressor work input and evaporator heat input are positive. Compression is actual adiabatic, the valve is isenthalpic, the model is approximate, and kinetic and potential energy are neglected.

Step-by-step solution
1. Actual discharge. Rearranging efficiency gives \(h_{2a}=410+(445-410)/0.78=454.87\ \mathrm{kJ/kg}\).
2. Mass flow. Refrigerating effect is \(410-260=150\ \mathrm{kJ/kg}\), so \(\dot m=60.0/150=0.400\ \mathrm{kg/s}\).
3. Power and COP. Compressor power is \(0.400(454.87-410)=17.95\ \mathrm{kW}\), giving \(\mathrm{COP}_R=60.0/17.95=3.34\).
4. Entropy audit. For the adiabatic compressor, \(\dot S_{gen}=\dot m(s_{2a}-s_1)=0.0140\ \mathrm{kW/K}\).

Verification

Units: \(\mathrm{kg/s}\) times \(\mathrm{kJ/kg}\) gives \(\mathrm{kW}\), and mass flow times entropy difference gives \(\mathrm{kW/K}\).

Signs and direction: Actual discharge enthalpy exceeds the isentropic value, required power is positive, and entropy generation is positive.

Plausibility: Condenser rejection is \(0.400(454.87-260)=77.95\ \mathrm{kW}\), equal to \(60.0+17.95\ \mathrm{kW}\). Cycle mass, energy storage, entropy, and exergy state changes close; at \(T_0=298\ \mathrm{K}\), compressor exergy destruction is \(4.17\ \mathrm{kW}\), nonnegative as required.

Final result: \(\dot m=0.400\ \mathrm{kg/s}\), \(\dot W_c=17.95\ \mathrm{kW}\), \(\mathrm{COP}_R=3.34\), and \(\dot S_{gen}=0.0140\ \mathrm{kW/K}\).

Engineering interpretation

Compressor irreversibility raises discharge temperature and electrical demand. The stated efficiency and properties are supplied study inputs, not vendor guarantees; off-design maps, motor efficiency, oil management, and uncertainty require review.

11–5

Second-Law Analysis of Refrigeration Cycles

Minimum work is set by reservoir temperatures, while actual excess work is associated with exergy destruction. For a declared environment, \(\dot X_d=T_0\dot S_{gen}\). Component entropy balances expose compression, throttling, mixing, and finite-temperature heat-transfer losses. A COP below Carnot is necessary but not sufficient: every component must also satisfy nonnegative entropy generation.

11–6

Selecting the Right Refrigerant

Selection balances saturation pressures, critical temperature, volumetric capacity, transport properties, material and lubricant compatibility, toxicity, flammability, ozone depletion, global-warming impact, availability, and regulation. A fluid name alone is insufficient: mixtures may glide in temperature and require composition control. Use current safety codes and authoritative property databases; never infer release properties from a classroom table.

11–7

Heat Pump Systems

Air, water, and ground sources impose different temperatures and fouling risks. Defrost and auxiliary heat can dominate seasonal performance. Capacity and COP must be evaluated together at design conditions, not extrapolated from one rating point.

EXAMPLE 3Heat-pump reversible benchmark
A heat pump delivers 120 kW at a 318 K condensing boundary from a 273 K source. Estimate input using a supplied second-law performance factor and audit exergy destruction.

Recognition strategy

Two reservoir temperatures establish a reversible heat-pump COP; a supplied fraction of that limit converts the benchmark to screening performance.

Given information

  • Heating product \(\dot Q_H=120\ \mathrm{kW}\); \(T_H=318\ \mathrm{K}\), \(T_L=273\ \mathrm{K}\), all absolute.
  • Supplied second-law performance factor \(\mathrm{COP}_{actual}/\mathrm{COP}_{rev}=0.46\); environment \(T_0=293\ \mathrm{K}\), \(P_0=101.3\ \mathrm{kPa}\) absolute.

Model and interpretation

The boundary encloses a steady cyclic heat pump; heating delivered and work input are positive. Reservoirs are isothermal, property detail is embedded in the supplied approximate factor, and refrigerant mass returns each cycle. Kinetic and potential energy are neglected; no ideal-gas or phase-quality calculation is attempted.

Step-by-step solution
1. Reversible limit. \(\mathrm{COP}_{HP,rev}=318/(318-273)=7.0667\), using absolute temperatures.
2. Actual screening COP. Apply the supplied factor: \(\mathrm{COP}_{HP}=0.46(7.0667)=3.2507\).
3. Electrical input. Required power is \(\dot W_{actual}=120/3.2507=36.91\ \mathrm{kW}\).
4. Entropy and exergy. First find \(\dot Q_L=120-36.91=83.09\ \mathrm{kW}\). Then \(\dot S_{gen}=120/318-83.09/273=0.0730\ \mathrm{kW/K}\), so \(\dot X_d=293(0.0730)=21.4\ \mathrm{kW}\).

Verification

Units: COP is dimensionless; kilowatts divided by COP remain kilowatts, and \(\mathrm{kW}/\mathrm{K}\) follows from exergy divided by kelvins.

Signs and direction: Actual input exceeds reversible input, making destroyed exergy and entropy generation positive rather than impossible negative values.

Plausibility: Cold-side absorption is \(120-36.91=83.09\ \mathrm{kW}\), so the cycle energy balance closes. The independent reservoir audit gives \(120/318-83.09/273=0.0730\ \mathrm{kW/K}\ge0\), and \(293(0.0730)=21.4\ \mathrm{kW}\) confirms nonnegative exergy destruction. Actual COP remains below 7.07.

Final result: \(\mathrm{COP}_{HP}=3.25\), \(\dot W_{in}=36.91\ \mathrm{kW}\), \(\dot S_{gen}=0.0730\ \mathrm{kW/K}\), and \(\dot X_d=21.4\ \mathrm{kW}\).

Engineering interpretation

The exergy loss follows the entropy balance evaluated with the actual hot- and cold-side heat rates. Subtracting reversible from actual work would be inconsistent here because those cases have different cold-side loads. Certified maps, load uncertainty, auxiliary power, refrigerant safety, and code review remain necessary.

11–8

Innovative Vapor-Compression Systems

Match architecture to lift

Multistage compression with intercooling reduces work; flash economizers separate phases; cascade systems couple distinct refrigerant loops through an intermediate heat exchanger. The cascade exchanger imposes equal and opposite heat rates only after mass bases and heat loss are defined. Intermediate temperature is iterated to balance compressor ratios, approach temperatures, fluid limits, and cost.

11–9

Gas Refrigeration Cycles

A reversed Brayton loop keeps the working fluid gaseous: compression, warm-side cooling, expansion, and cold-side heating. For an ideal gas, authoritative variable-temperature \(h\) data or justified \(c_p\) approximations replace saturation tables; \(c_p\) belongs with enthalpy changes and \(c_v\) with internal-energy changes. Turbine output offsets compressor input. This branch is distinct from vapor compression and involves neither quality nor throttling unless hardware explicitly includes a valve. Choking and shocks require separate compressible-flow analysis, not cycle formulas.

11–10

Absorption Refrigeration Systems

Absorption systems replace most compressor work with a generator, absorber, solution pump, and internal heat exchange. COP commonly compares cooling with generator heat, so it cannot be compared directly with an electrically driven COP without fuel and exergy boundaries. Pair equilibrium, crystallization risk, corrosion, pressure, and safety constrain operation.

Chapter recap

  • Choose cooling or heating product, then define steady component boundaries and signs.
  • Identify phases at absolute states and use traceable refrigerant properties; quality belongs only in a saturated mixture.
  • Distinguish isentropic compression from actual efficiency and isenthalpic throttling from reversible expansion.
  • Before moving on, be able to close mass and energy ledgers, enforce Carnot and entropy limits, and state the evidence still needed for a professional selection.