5.3 Calculate refrigerator COP from enthalpy differences | University of Alberta MEC E 340: Applied Thermodynamics

University of Alberta MEC E 340: Applied Thermodynamics

Refrigeration and Heat Pumps

A focused method for a simple vapour-compression refrigeration cycle

Start with a steady-flow control volume around a simple vapour-compression refrigerator. The working fluid is the refrigerant, which circulates through four numbered states: state 1 is the evaporator outlet and compressor inlet; state 2 is the compressor outlet; state 3 is the condenser outlet; and state 4 is the expansion-valve outlet. Refrigerant flows in the order 1 to 2 to 3 to 4 to 1. Unless a problem states otherwise, the calculations here use steady operation and negligible kinetic and potential energy changes. The compressor and valve models are stated where used. Numerical enthalpies are supplied as problem data, not taken from an unstated property source.

What you will learn

Identify the four refrigerant states and energy transfers used in a refrigerator COP calculation.
Use supplied enthalpies and stated component models to find cooling effect and compressor work input per unit mass.
Calculate refrigerator COP and check its signs, units, and physical meaning.

1. Connect component energy balances to COP

A refrigerator removes heat from a cooled space and requires work input. Its coefficient of performance, or COP, compares useful cooling with compressor work input. It is a dimensionless performance ratio, not a thermal efficiency. A refrigerator COP can be greater than one because the refrigerator moves heat from the cooled space to its surroundings; work input is not the only energy involved.

For a steady-flow component with one inlet and one outlet, the first-law balance relates heat transfer, work, and the change in specific enthalpy. In this lesson, changes in kinetic and potential energy are negligible. In the evaporator, heat enters the refrigerant as it flows from state 4 to state 1. Cooling per unit mass is therefore the outlet enthalpy minus the inlet enthalpy.

For an adiabatic compressor with negligible kinetic and potential energy changes, work is supplied to the refrigerant as it flows from state 1 to state 2. Compressor work input per unit mass is the enthalpy rise through the compressor. If a problem specifies compressor heat transfer or another energy effect, use the corresponding full energy balance rather than this simplified relation.

Use positive values for cooling and compressor work input. Divide the evaporator enthalpy rise by the compressor enthalpy rise. Both differences are specific energy transfers, so their units cancel. A mass flow rate is not needed for COP when both terms are calculated on the same per-unit-mass basis.

The refrigerator COP is cooling per unit mass divided by compressor work input per unit mass. The subscripts distinguish heat removed from the cooled space from work supplied to the compressor.

\mathrm{COP}_R = \frac{h_1 - h_4}{h_2 - h_1}

Evaporator cooling per unit mass is the enthalpy rise from state 4 to state 1.
For an adiabatic compressor with negligible kinetic and potential energy changes, work input per unit mass is the enthalpy rise from state 1 to state 2.
COP is dimensionless because its numerator and denominator have matching units.

2. Assign states and apply the valve model

Follow refrigerant flow around the cycle: compressor, condenser, expansion valve, and evaporator. State 1 is both the compressor inlet and evaporator outlet. State 4 is both the valve outlet and evaporator inlet. Keep these labels fixed when recording property data and writing balances.

An expansion valve is often represented as isenthalpic. Under that stated model, enthalpy at the valve outlet equals enthalpy at its inlet. Use this relation only when the problem specifies an isenthalpic valve or otherwise supplies a justified model. A pressure drop alone does not determine the outlet enthalpy.

The condenser rejects heat as refrigerant flows from state 2 to state 3. That heat rejection is not the cooling effect used in refrigerator COP. Cooling is the heat absorbed from the cooled space in the evaporator. The denominator is compressor work input, not condenser heat rejection.

Write each supplied enthalpy beside its state number and retain its units. If the valve is specified as isenthalpic and the condenser outlet enthalpy is known, use it to obtain the valve outlet enthalpy. Then calculate the evaporator and compressor enthalpy differences. This keeps the state connections and subtraction directions clear.

h_4=h_3

Use state numbers consistently with refrigerant flow direction.
For a specified isenthalpic valve, the inlet and outlet enthalpies are equal.
Condenser heat rejection is not the cooling quantity used in refrigerator COP.

3. Calculate and check performance

Calculate cooling and compressor work input separately before dividing. If enthalpies are given in kilojoules per kilogram, both differences are in kilojoules per kilogram. These units cancel in the COP ratio. A supplied mass flow rate can convert specific energy transfers to rates, but it is not required for COP when both terms use the same mass basis.

Check signs against the intended operating direction. Cooling is positive when refrigerant enthalpy rises from state 4 to state 1. Under the stated compressor model, work input is positive when enthalpy rises from state 1 to state 2. If a difference has an unexpected sign, check the state labels, flow direction, and subtraction order before interpreting the result.

A COP greater than one is physically plausible. It means that more heat is removed from the cooled space than the work supplied to the compressor, while the refrigerator also rejects heat at the condenser. COP is not a percentage and should not be described as the fraction of work converted into cooling.

A schematic cycle plot helps show state order and component processes, but it does not provide exact property coordinates or phase boundaries. For numerical work, use enthalpy data and assumptions supplied in the problem, not values guessed from a sketch.

Find cooling and compressor work input separately, then divide.
Check state labels, signs, and matching units.
Treat schematic cycle plots as illustrations, not sources of numerical property data.

Worked example

Direct COP calculation

A refrigerator follows the four-state cycle defined above. Supplied enthalpies are

h_1=242\ \mathrm{kJ/kg}

,

h_2=278\ \mathrm{kJ/kg}

, and

h_3=92\ \mathrm{kJ/kg}

. The valve is modelled as isenthalpic and the compressor as adiabatic. Find the cooling effect, compressor work input, and refrigerator COP per unit mass.

Four-state vapour-compression cycle

The refrigerant cycle is schematic and not to scale on a pressure–specific-volume plane.

Find the valve outlet enthalpy
The stated isenthalpic valve model means enthalpy is unchanged from state 3 to state 4.
$h_4=h_3=92\ \mathrm{kJ/kg}$
Calculate cooling and compressor work
Cooling is the enthalpy rise from state 4 to state 1. Compressor work input is the enthalpy rise from state 1 to state 2. q_L=h_1-h_4=242-92=150\ kJ/kg, w_{in}=h_2-h_1=278-242=36\ kJ/kg
Form the COP
Divide the cooling effect by compressor work input. Their specific-energy units cancel.
$\mathrm{COP}_R=\frac{150}{36}=4.17$

Answer: The cooling effect is

150\ \mathrm{kJ/kg}

, compressor work input is

36\ \mathrm{kJ/kg}

, and refrigerator COP is

4.17

.

Check: Both enthalpy differences are positive and have the same units. The result means about

4.17\ \mathrm{kJ}

of heat is removed from the cooled space per kilojoule of compressor work input.

Worked example

Calculate COP when valve enthalpy is needed

A refrigerator has supplied enthalpies

h_1=410\ \mathrm{kJ/kg}

,

h_2=446\ \mathrm{kJ/kg}

, and

h_3=180\ \mathrm{kJ/kg}

. The expansion valve is isenthalpic and the compressor is adiabatic. Calculate

h_4

, the cooling effect, compressor work input, and COP.

Four-state refrigerator cycle

The refrigerant cycle is schematic and not to scale on a pressure–specific-volume plane.

Determine the evaporator inlet enthalpy
The stated isenthalpic valve model connects states 3 and 4 without an enthalpy change.
$h_4=h_3=180\ \mathrm{kJ/kg}$
Apply the component enthalpy differences
Use the rise from state 4 to state 1 for cooling and the rise from state 1 to state 2 for compressor work input. q_L=410-180=230\ kJ/kg, w_{in}=446-410=36\ kJ/kg
Calculate COP
The specific energy transfers have matching units, so their ratio has no units.
$\mathrm{COP}_R=\frac{230}{36}=6.39$

Answer:

h_4=180\ \mathrm{kJ/kg}

, cooling effect is

230\ \mathrm{kJ/kg}

, compressor work input is

36\ \mathrm{kJ/kg}

, and refrigerator COP is

6.39

.

Check: Cooling and compressor work input are positive. COP is dimensionless, not a percentage.

Worked example

Infer enthalpies from specified cooling

A refrigerator has

h_1=205\ \mathrm{kJ/kg}

and

h_2=239\ \mathrm{kJ/kg}

. Its cooling effect is

125\ \mathrm{kJ/kg}

. The expansion valve is isenthalpic and the compressor is adiabatic. Determine

h_4

, infer

h_3

, and calculate COP.

Four-state refrigerator cycle

The refrigerant cycle is schematic and not to scale on a pressure–specific-volume plane.

Find the evaporator inlet enthalpy
Cooling is the enthalpy rise from state 4 to state 1. Rearranging this relation gives the enthalpy at state 4.
$h_4=h_1-q_L=205-125=80\ \mathrm{kJ/kg}$
Infer the condenser outlet enthalpy
The stated isenthalpic valve model means states 3 and 4 have equal enthalpies.
$h_3=h_4=80\ \mathrm{kJ/kg}$
Calculate compressor work and COP
Find work input from the compressor enthalpy rise, then divide the given cooling effect by that work input. w_{in}=239-205=34\ kJ/kg, COP_R= $\frac{125}{34}$ =3.68

Answer:

h_4=80\ \mathrm{kJ/kg}

,

h_3=80\ \mathrm{kJ/kg}

, compressor work input is

34\ \mathrm{kJ/kg}

, and refrigerator COP is

3.68

.

Check: The enthalpy at state 4 is below

h_1

, so evaporator cooling is positive. Compressor work input is also positive, and the enthalpy differences use kilojoules per kilogram.

Common mistakes and how to avoid them

Using

h_4-h_1

for cooling.

Correction: The refrigerant absorbs heat as it flows from state 4 to state 1, so use

h_1-h_4

. Check that cooling is positive.

Using

h_1-h_2

for compressor work input.

Correction: Under the stated adiabatic-compressor model, work input is positive when enthalpy rises through the compressor. Use

h_2-h_1

.

Assuming

h_3=h_4

without checking the valve model.

Correction: That equality applies only when the expansion valve is specified as isenthalpic.

Giving COP units or reporting it as a percentage.

Correction: The numerator and denominator are both energy per unit mass, so the units cancel. COP is dimensionless, not a percentage.

Lesson summary

For this simple cycle, state 1 is the compressor inlet, state 2 its outlet, state 3 the condenser outlet, and state 4 the valve outlet.
With the stated steady-flow assumptions, cooling per unit mass is $h_1-h_4$ and compressor work input per unit mass is $h_2-h_1$ .
For an isenthalpic expansion valve, use $h_4=h_3$ only when that model is specified.
Divide cooling by compressor work input to calculate refrigerator COP, then check signs, units, and state assignments.

Check your understanding

Question 1

For a cycle with

h_1=260\ \mathrm{kJ/kg}

,

h_2=300\ \mathrm{kJ/kg}

, and

h_4=110\ \mathrm{kJ/kg}

, what is refrigerator COP?

$3.75$
$4.75$
$0.267$
$1.50$

Show answer and explanation

3.75

Cooling is

260-110=150\ \mathrm{kJ/kg}

and compressor work input is

300-260=40\ \mathrm{kJ/kg}

. Therefore, COP is

150/40=3.75

.

Question 2

A problem states that the expansion valve is isenthalpic and gives

h_3=95\ \mathrm{kJ/kg}

. What value should be used for

h_4

?

$h_3+h_2$
$h_3-h_1$
$95\ \mathrm{kJ/kg}$
It cannot be found without the mass flow rate

Show answer and explanation

95\ \mathrm{kJ/kg}

For the stated isenthalpic valve model,

h_4=h_3

, so

h_4=95\ \mathrm{kJ/kg}

. Mass flow rate is not needed for this state relation.

Key terms

Refrigerator COP: Heat absorbed from the cooled space divided by compressor work input, evaluated on the same basis.
Enthalpy: A thermodynamic property used in steady-flow energy balances; here it is given as energy per unit refrigerant mass.
Isenthalpic throttling: The specified expansion-valve model in which enthalpy is unchanged between valve inlet and outlet.

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5.3 · Calculate refrigerator COP from enthalpy differences

University of Alberta MEC E 340: Applied Thermodynamics

A focused method for a simple vapour-compression refrigeration cycle

What you will learn

1. Connect component energy balances to COP

2. Assign states and apply the valve model

3. Calculate and check performance

Direct COP calculation

Calculate COP when valve enthalpy is needed

Infer enthalpies from specified cooling

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Key terms

Continue through MEC E 340

About this lesson and its review