E4 · HVAC Equipment and Components

Foundation lesson · 18 minute study session

Identify the useful side of the heat pump.

Separate evaporator absorption, condenser delivery and compressor input before comparing capacity or storage.

By the end of this lesson

Select heating or cooling COP from the requested useful output.
Close the energy balance at one specified operating condition.
Keep stored thermal energy separate from thermal power.

The same cycle can serve two purposes.

In cooling service, the useful output is heat absorbed by the evaporator. In heating service, it is heat delivered at the condenser. Compressor input supplies the difference. COP is a ratio of matching energy rates; it is not a thermal conversion efficiency bounded by one.

Capacity depends on operating state.

Evaporator and condenser temperatures affect the lift and performance. A catalog heating COP and cooling COP measured under different conditions cannot be related by simply adding one. Use the stated common condition for an energy balance. Include auxiliary, defrost or fan energy only when it belongs to the specified boundary.

Storage shifts time, not the energy units.

A thermal store contains an energy quantity and can deliver a rate for a duration. Use sensible or latent storage properties supplied by the problem, plus any specified losses or usable fraction. Nominal tank volume or ice mass alone does not establish discharge capacity.

A decision map for this lesson

Define useful output before calculating COP

Heating COP, refrigeration COP and storage duration describe different outputs and boundaries.

Choose the governing relationship and catch the wrong route
Condition or question	Governing decision	Error to avoid
Useful heating is requested	Use $COP_H=\dot Q_H/P_{input}$ for the stated included inputs.	Using evaporator absorption as useful heating output.
Useful refrigeration is requested	Use $COP_R=\dot Q_L/P_{input}$ on the specified boundary.	Using condenser rejection as the useful cooling numerator.
The plus-one relationship is used	Require the same cycle state and input boundary for the heating and cooling quantities.	Connecting catalog COP values from different test conditions.
A thermal store must supply a duty	Convert duty and time to usable energy; include only the stated efficiency, loss and availability allowances.	Treating stored Btu as refrigeration tons or ignoring the discharge period.

On a narrow screen, scroll across the table to compare the decision and the error.

Relationships and their limits

Common cycle boundary

\dot Q_H=\dot Q_L+\dot W;\quad COP_H=\dot Q_H/\dot W;\quad COP_R=\dot Q_L/\dot W

Steady stated boundary; all rates in the same units. Auxiliary energy must follow the question definition.

Relation at the same condition

COP_H=COP_R+1

Same heat rates and input boundary at the same operating condition, not unrelated seasonal ratings.

Constant-rate storage

E=\dot Q\,t;\qquad E_{sensible}=mc_p\Delta T

The first relation assumes constant net rate; include specified losses and usable fraction.

Symbols and unit basis
Symbol	Meaning	Units or basis
$\dot Q_H$	Condenser heat delivery	kW or Btu/h
$\dot Q_L$	Evaporator heat absorption	matching rate units
$\dot W$	Cycle work input	matching rate units
$COP$	Useful thermal rate divided by input	dimensionless
$E$	Stored or delivered energy	kWh or Btu
$t$	Duration	h
$m$	Stored-fluid or material mass	lb
$c_p$	Specific heat for sensible storage	Btu/(lb·°F)
$\Delta T$	Usable sensible temperature interval	°F difference

One decision, fully worked

Apply the method.

At one steady operating condition, a heat pump delivers 36 kW of condenser heat and receives 9 kW of compressor input. Neglect other transfers. Separately, a store must deliver a constant 200,000 Btu/h for two hours with no losses.

Subtract input from delivered heat to find source-side absorption.
Calculate heating and cooling COP for this same cycle boundary.
Multiply storage rate by duration, keeping energy and rate units distinct.

\dot Q_L=36-9=27\ \mathrm{kW};\quad COP_H=36/9=4;\quad COP_R=27/9=3

E=200{,}000(2)=400{,}000\ \mathrm{Btu}

Decision

The heat pump absorbs 27 kW, with heating COP 4 and cooling COP 3 at this condition. The separate storage requirement is 400,000 Btu of usable delivered energy.

Independent check

Source heat plus compressor input gives 27 + 9 = 36 kW. The two COP values differ by one because their boundary and operating state match.

Specialist extension

Check the unavailable-unit condition explicitly

Installed capacity is not the same as firm capacity when a unit must be unavailable. Evaluate the exact redundancy condition supplied. A largest-unit-out criterion tests the installed total minus the largest usable unit, not an arbitrary universal percentage. Also verify operating limits, common dependencies and distribution capacity when the case supplies them.

A new applied case

Three identical chillers each provide 80 tons at the specified design condition. A 150-ton coincident load must be met with any one chiller unavailable.

Q_{firm}=3(80)-80=160\ \mathrm{tons}

Decision. The stated one-unit-out capacity condition is satisfied with 10 tons of margin.

Cross-check. Total installed capacity is 240 tons, but that is not the unavailable-unit check.

Unit basis. Capacity is refrigeration tons at the same design condition; firm capacity uses the specified outage criterion.

NCEES PE Mechanical Reference Handbook v2.1; see the module reference path.

Open targeted practice in the study studio →

Transfer the method

HVAC Equipment and Components · HV-LC-E4

Try a new decision.

Select one

At one steady condition a heat pump delivers 45 kW at its condenser and receives 10 kW compressor input. Neglect all other transfers. The evaporator heat absorption and heating COP, respectively are most nearly:

A35 kW;

COP_H

= 3.5B45 kW;

COP_H

= 0.222C35 kW;

COP_H

= 4.5D55 kW;

COP_H

= 4.5

A lesson check supports practice, not a readiness score. Your response stays in this page session and clears on reload.

Catch the common errors

Using condenser delivery in a cooling COP numerator.
Applying the COP difference-of-one relation to unrelated catalog ratings.
Reporting storage capacity in Btu/h when the requested quantity is Btu.

Handbook navigation drill

Find it, then explain it.

Search: Heat Pumps; coefficient of performance

Locate heating and refrigeration COP definitions. State the useful output in each numerator before calculating.

Check your navigation

The handbook heat-pump and refrigeration sections use distinct useful-output definitions. All energy rates must share one unit basis.

Reference path: §4.5; §8.1–8.5, pp. 401–407. These are printed handbook page numbers; section names and search terms remain the primary navigation method. See technical review status.

Deeper application · Decision Lab

Select VRF equipment from the operating envelope

A large nameplate capacity can still fail the actual job. Test operating mode and each delivered duty before recommending equipment.

Choose the method, predict the changed-condition result and inspect the complete cross-check. Sign in before starting to keep your first response and later reflection in the separate reasoning notebook.

Open the original workshop →

Theory sources

NCEES PE Mechanical Reference Handbook v2.1: §4.5; §8.1–8.5, pp. 401–407
Mechanical and Electrical Equipment for Buildings, 12th ed.: mechanical systems and energy principles. Supporting theory; not an examination reference.

Worked examples and lesson prose are original ReasonSpan^™ material. Obtain authorized publications separately; this preview does not reproduce the supplied books or handbook.