5.1 Distinguish refrigeration and heat-pump objectives | University of Alberta MEC E 340: Applied Thermodynamics

University of Alberta MEC E 340: Applied Thermodynamics

Refrigeration and Heat Pumps

MEC E 340 Applied Thermodynamics — study topic 5.1

Start with a closed cycle containing a circulating working fluid. Label four successive states 1, 2, 3, and 4, with the fluid returning from state 4 to state 1. The fluid removes heat from a colder region, receives work input, and delivers heat to a warmer region. Assume cyclic operation and neglect changes in kinetic and potential energy. No working-fluid properties or detailed cycle processes are specified, so this lesson uses only supplied heat and work data. The central distinction is the service being evaluated: refrigeration provides cooling; a heat pump provides heating.

What you will learn

Distinguish the cooling objective of refrigeration from the heating objective of a heat pump.
Relate heat removed, heat delivered, and work input using a complete-cycle energy balance.
Select and calculate the coefficient of performance that matches the stated objective.

1. Define the objective and energy transfers

A refrigerator is intended to keep a space or object cold. Its useful effect is heat removed from that cold region. A heat pump is intended to keep a space warm. Its useful effect is heat delivered to the warm region. The wording of the task identifies which service matters; it does not necessarily mean that the underlying cycle is different.

Use positive magnitudes for the transfers:

Q_L

is heat removed from the colder region,

Q_H

is heat delivered to the warmer region, and W_{in} is work supplied to the cycle. The subscripts identify the lower- and higher-temperature regions. These symbols describe transfer directions and magnitudes, rather than a general signed-term convention.

Define the system as the circulating working fluid over a complete cycle. The cycle returns the fluid to its starting state, so its net energy change over the cycle is zero. The state numbers mark the order of the fluid’s states, but without additional information they do not give property values, phases, or specific processes. A P–v or T–s plot is not needed to distinguish the objectives.

Q_{useful}=\begin{cases}Q_L,&\text{refrigeration}\\Q_H,&\text{heat pump}\end{cases}

Refrigeration values heat removed from the cold region.
A heat pump values heat delivered to the warm region.
The stated service determines the useful heat transfer.

2. Use the complete-cycle energy balance

The first law applied over one complete cycle says that the net energy change of the working fluid is zero. With the positive transfer magnitudes defined above, energy conservation gives the relationship below: heat delivered to the warm region accounts for both heat absorbed from the cold region and work supplied to the cycle.

Use this balance to find a missing transfer when the other two are known. For rates, keep all quantities in power units such as kilowatts. For energy amounts, compare amounts over the same time interval. Do not combine a rate with an energy amount.

The energy balance applies to both objectives; it does not decide which one the question asks about. First find any missing transfer, then identify whether the requested service is cooling or heating. Property data would be needed for a problem that asks you to determine transfers from working-fluid states, but no property values are needed when the transfers are supplied.

Q_H=Q_L+W_{in}

For a complete cycle, heat delivered equals heat removed plus work supplied.
Keep rates with rates and energy amounts with energy amounts.
The balance gives energy accounting; the objective selects the useful effect.

3. Match the coefficient of performance to the objective

The coefficient of performance (COP) compares the useful heat transfer with the work input. For refrigeration, divide heat removed from the cold region by work input. For a heat pump, divide heat delivered to the warm region by work input. Work input is the denominator in both measures.

For the same cycle operating under the same conditions, heat-pump COP is one greater than refrigerator COP. The cycle balance explains why: heat delivered includes both the heat absorbed from the colder region and the supplied work. Dividing the balance by work input gives the relation below.

COP is dimensionless because the numerator and denominator have the same units. It can be greater than one: the work input helps move heat, and the useful heat transfer need not equal the work input. A COP value alone does not tell you which objective was used. Name the service and choose the matching definition.

\mathrm{COP}_{HP}=\mathrm{COP}_R+1

Refrigerator COP uses $Q_L$ as its numerator.
Heat-pump COP uses $Q_H$ as its numerator.
For the same operating condition, heat-pump COP is one greater than refrigerator COP.

4. Apply a consistent solution routine

Identify the system, the complete-cycle assumption, and the colder and warmer regions. Mark the positive directions: heat removed from the cold region, heat delivered to the warm region, and work supplied to the cycle. State any assumptions provided, such as neglecting kinetic and potential energy changes. Do not claim a particular working fluid, phase, or cycle process if the problem does not provide it.

Write the cycle energy balance before calculating COP. Find a missing heat transfer if necessary. Then select the numerator that matches the stated service and divide by work input. When the transfers are rates in kilowatts, the units cancel and COP is dimensionless.

Check the energy accounting and the requested objective. With positive work input, heat delivered must exceed heat removed. This check helps catch a sign or arithmetic error; it does not tell you whether to report refrigerator COP or heat-pump COP.

Define the cycle and transfer directions before calculating.
Use the balance to determine missing transfers.
Choose the COP numerator from the requested cooling or heating service.

Worked example

Evaluate a cooling service

A cyclic device removes heat from a refrigerated space at

6.0\ \mathrm{kW}

and receives work at

2.0\ \mathrm{kW}

. Find its refrigerator COP and the rate of heat rejection. Treat the supplied rates as operating data for the same cycle.

Generic closed cycle and energy transfers

Conceptual state sequence for the circulating working fluid; the loop is schematic and not to scale, and does not specify a particular working-fluid cycle or process.

Identify the system and objective
Take the circulating working fluid over one complete cycle as the system. It returns to its initial state, and kinetic and potential energy changes are neglected. The stated service is cooling, so heat removed from the refrigerated space is the useful effect. No property data or individual process details are needed for these supplied rates.
$Q_L=6.0\ \mathrm{kW}$
Apply the cycle balance
For a complete cycle, heat rejection equals heat absorption plus work supplied. Substitute the given rates to find the heat-rejection rate. Q_H=Q_L+W_{in}=6.0\ kW+2.0\ kW=8.0\ kW
Calculate refrigerator COP
Divide the cooling rate by the work-input rate. The units cancel, leaving a dimensionless ratio.
$\mathrm{COP}_R=\frac{6.0\ \mathrm{kW}}{2.0\ \mathrm{kW}}=3.0$

Answer: The refrigerator COP is

3.0

, and the heat-rejection rate is

8.0\ \mathrm{kW}

.

Check: The balance closes:

8.0\ \mathrm{kW}=6.0\ \mathrm{kW}+2.0\ \mathrm{kW}

. Heat rejection exceeds heat removal by the supplied work rate.

Worked example

Evaluate a heating service

A heat-pump cycle delivers heat to a heated space at

9.6\ \mathrm{kW}

and receives work at

2.4\ \mathrm{kW}

. Find its heat-pump COP and the rate of heat removed from the colder surroundings. Use the supplied operating rates.

Generic closed cycle and energy transfers

Conceptual state sequence for the circulating working fluid; the loop is schematic and not to scale, and does not specify a particular working-fluid cycle or process.

Select the heating objective
Take the circulating working fluid over one complete cycle as the system, with kinetic and potential energy changes neglected. The useful service is heat delivered to the heated space, so this transfer is the numerator for heat-pump COP.
$Q_H=9.6\ \mathrm{kW}$
Find heat removed
Rearrange the cycle balance to find heat absorbed from the colder surroundings. The positive result agrees with the defined direction for heat removal. Q_L=Q_H-W_{in}=9.6\ kW-2.4\ kW=7.2\ kW
Calculate heat-pump COP
Divide heat delivered by work input. The units cancel, leaving a dimensionless COP.
$\mathrm{COP}_{HP}=\frac{9.6\ \mathrm{kW}}{2.4\ \mathrm{kW}}=4.0$

Answer: The heat-pump COP is

4.0

, and the heat-removal rate is

7.2\ \mathrm{kW}

.

Check: The balance closes:

7.2\ \mathrm{kW}+2.4\ \mathrm{kW}=9.6\ \mathrm{kW}

. For this same cycle, refrigerator COP is

7.2/2.4=3.0

, one less than heat-pump COP.

Worked example

Choose the COP that answers the question

A cycle removes heat from a chilled cabinet at

5.0\ \mathrm{kW}

and receives work at

1.0\ \mathrm{kW}

. An operator asks how much heating service the same cycle can provide to a warm space. Find its heat-pump COP. Assume the supplied rates describe the same operating condition.

Generic closed cycle and energy transfers

Conceptual state sequence for the circulating working fluid; the loop is schematic and not to scale, and does not specify a particular working-fluid cycle or process.

Find the heating service
Take the circulating working fluid over one complete cycle as the system. Because the question asks about heating, use the cycle balance to find heat delivered to the warm region before calculating COP. Using heat removed from the cabinet would answer a refrigeration question. Q_H=Q_L+W_{in}=5.0\ kW+1.0\ kW=6.0\ kW
Calculate heat-pump COP
Divide heat delivered by work input. The units cancel, and the result describes the requested heating objective.
$\mathrm{COP}_{HP}=\frac{6.0\ \mathrm{kW}}{1.0\ \mathrm{kW}}=6.0$

Answer: The heat-pump COP is

6.0

.

Check: The corresponding refrigerator COP is

5.0\ \mathrm{kW}/1.0\ \mathrm{kW}=5.0

. The difference is one, consistent with the cycle energy balance.

Common mistakes and how to avoid them

Using heat delivered to the warm region as the refrigerator COP numerator.

Correction: For a refrigeration objective, use heat removed from the cold region,

Q_L

.

Using heat removed from the cold region as the heat-pump COP numerator.

Correction: For a heating objective, use heat delivered to the warm region,

Q_H

.

Assuming rejected heat equals absorbed heat alone.

Correction: Include work input in the complete-cycle balance: Q_H=Q_L+W_{in}.

Treating COP as a percentage or assuming it must be below one.

Correction: COP is a dimensionless ratio of useful heat transfer to work input and can exceed one.

Lesson summary

A refrigeration objective values heat removed from the cold region, $Q_L$ .
A heat-pump objective values heat delivered to the warm region, $Q_H$ .
For a complete cycle, Q_H=Q_L+W_{in}.
Divide the heat transfer matching the objective by work input to calculate COP.
For the same cycle and operating condition, $\mathrm{COP}_{HP}=\mathrm{COP}_R+1$ .

Check your understanding

Question 1

A device removes heat from a cold display case. Which transfer belongs in the numerator when evaluating its refrigeration objective?

Heat removed from the display case
Heat delivered to the surroundings
Work supplied to the device
The total heat delivered plus work

Show answer and explanation

Heat removed from the display case

Refrigeration performance is based on the useful cooling effect: heat removed from the cold space.

Question 2

A cycle has

Q_L=4\ \mathrm{kW}

and W_{in}=1\

\mathrm{kW}

. What is its heat-pump COP?

3
4
5
0.25

Show answer and explanation

5

The balance gives

Q_H=4\ \mathrm{kW}+1\ \mathrm{kW}=5\ \mathrm{kW}

. Divide by work input to obtain heat-pump COP

5/1=5

.

Question 3

For one cycle operating under the same conditions, how do the COPs compare?

Heat-pump COP is one greater than refrigerator COP.
Refrigerator COP is one greater than heat-pump COP.
They are always equal.
Heat-pump COP is twice refrigerator COP.

Show answer and explanation

Heat-pump COP is one greater than refrigerator COP.

Heat delivered includes both heat absorbed and work input, so heat-pump COP equals refrigerator COP plus one.

Key terms

Coefficient of performance (COP): Useful heat transfer divided by work input for a refrigeration or heat-pump objective.
Refrigerator: A cyclic device whose intended service is removing heat from a cold space.
Heat pump: A cyclic device whose intended service is delivering heat to a warm space.
Working fluid: The fluid that circulates through a thermodynamic cycle and carries energy between its states.

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Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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5.1 · Distinguish refrigeration and heat-pump objectives

University of Alberta MEC E 340: Applied Thermodynamics

MEC E 340 Applied Thermodynamics — study topic 5.1

What you will learn

1. Define the objective and energy transfers

2. Use the complete-cycle energy balance

3. Match the coefficient of performance to the objective

4. Apply a consistent solution routine

Evaluate a cooling service

Evaluate a heating service

Choose the COP that answers the question

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through MEC E 340

About this lesson and its review