D7 · HVAC and Refrigeration Distribution and Systems

Foundation lesson · 30 minute study session

Balance the refrigeration cycle and the stored product

A refrigeration system moves heat and adds compressor work. Evaporator duty, condenser rejection and product freezing load are different quantities.

By the end of this lesson

Name the four vapor-compression processes.
Calculate COP and condenser heat rejection.
Split a freezing load into sensible and latent intervals.

Follow the four states

For a basic vapor-compression cycle, the compressor raises pressure and enthalpy; the condenser rejects heat; the throttling device reduces pressure approximately at constant enthalpy; the evaporator absorbs the refrigeration duty. State enthalpies must refer to the same refrigerant and datum. Preserve the labeled states when forming $h_1-h_4$ and $h_2-h_1$ .

Use the right performance ratio

Refrigeration COP is evaporator heat absorbed divided by compressor work. Heat-pump heating COP uses condenser heat delivered. Express numerator and denominator in consistent energy units. The condenser rejects evaporator duty plus compressor work for the stated idealized boundary.

Include storage and product loads

Cold-storage loads may include envelope heat gain, infiltration, lights, people, fans, product cooling and respiration. A product taken through freezing needs separate sensible cooling above freezing, latent freezing and sensible cooling below freezing. Supplied food properties control the result. Refrigerant safety classifications and handling concepts belong in the supportive-knowledge review.

A decision map for this lesson

Follow the cycle states and the product boundary

Read each enthalpy at its physical location; product pull-down is a separate load calculation.

A realistic refrigeration demonstration rig connects a compressor, condenser, expansion valve and evaporator through copper tubing. — **Follow each refrigerant state**Connect the equipment to the calculation

Vapor-compression cycle with compressor, condenser, expansion device and evaporator; heat absorbed, rejection and compressor work are labelled. — **Follow each refrigerant state**Connect the equipment to the calculation

Choose the governing relationship and catch the wrong route
Condition or question	Governing decision	Error to avoid
Evaporator duty	Use $\dot Q_L=\dot m(h_1-h_4)$ .	Using condenser enthalpy drop as the useful refrigeration effect.
Compressor work	Use $\dot W_c=\dot m(h_2-h_1)$ for the stated steady enthalpy-rise model.	Taking compressor electrical input as refrigerant work when other losses are included.
Basic adiabatic throttling	Use $h_4=h_3$ when kinetic/potential changes and external heat are neglected.	Assuming the throttle preserves temperature instead of enthalpy.
Product cooling and freezing	Sum pre-freeze sensible energy, latent freezing energy and post-freeze sensible energy; divide by the stated time.	Counting product energy as an hourly rate without the processing-time conversion.

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

Relationships and their limits

Cycle duties

q_L=h_1-h_4;\quad w_c=h_2-h_1;\quad q_H=h_2-h_3

Basic steady cycle; throttling is approximated by $h_4$ = $h_3$ ; kinetic/potential changes neglected.

COP and heat rejection

COP_R=q_L/w_c;\quad\dot Q_H=\dot Q_L+\dot W_c

Energy rates must use the same units. Heating COP uses $Q_{H}$ in the numerator.

Product through freezing

Q=m[c_{p,a}(T_i-T_f)+h_{freeze}+c_{p,b}(T_f-T_o)]

Use supplied freezing temperature, latent heat and specific heats; this is total energy, not a rate.

Symbols and unit basis
Symbol	Meaning	Units or basis
$h_i$	Specific enthalpy at labeled cycle state	Btu/lb refrigerant
$q_L$	Evaporator heat per unit refrigerant mass	Btu/lb
$w_c$	Compressor work per unit mass	Btu/lb
$\dot Q_H$	Condenser heat rejection	Btu/h
$COP_R$	Refrigeration coefficient of performance	dimensionless
$q_H$	Condenser heat per unit refrigerant mass	Btu/lb
$\dot W_c$	Compressor input rate	Btu/h on matching basis
$Q$	Total product cooling/freezing energy	Btu
$m$	Product mass	lb
$c_{p,a},\ c_{p,b}$	Product heat capacity above/below freezing	Btu/(lb·°F)
$h_{freeze}$	Supplied latent freezing energy	Btu/lb
$T_i,\ T_f,\ T_o$	Initial, freezing and final product temperatures	°F

One decision, fully worked

Apply the method.

A refrigeration cycle has h₁ = 100, h₂ = 125, h₃ = 40 and h₄ = 40 Btu/lb. Refrigerant flow is 2,000 lb/h. Ignore heat losses and kinetic/potential changes. Find evaporator duty, compressor energy input and condenser rejection.

Use h₁ − h₄ for refrigerating effect.
Use h₂ − h₁ for compressor work.
Multiply each specific quantity by refrigerant mass flow.
Check condenser rejection against evaporator duty plus input.

\dot Q_L=2{,}000(100-40)=120{,}000\ \mathrm{Btu/h}

\dot W_c=2{,}000(125-100)=50{,}000\ \mathrm{Btu/h}

COP_R=60/25=2.4

\dot Q_H=2{,}000(125-40)=170{,}000\ \mathrm{Btu/h}

Decision

The evaporator provides 120,000 Btu/h with COP 2.4; condenser rejection is 170,000 Btu/h.

Independent check

120,000 + 50,000 equals 170,000 Btu/h. The condenser rejects more heat than the evaporator absorbs.

Specialist extension

Divide a freezing duty into physical stages

Product refrigeration can include cooling above freezing, latent freezing, and cooling below freezing. Use the appropriate supplied heat capacity for each temperature interval and apply the latent term only to the mass or fraction that freezes. Divide total energy by the allowed processing duration and add other stated coincident room duties on the same rate basis.

A new applied case

A 1,000 lb product is cooled from 40°F to 30°F with cp = 0.8 Btu/(lb·°F), completely frozen at 30°F using 100 Btu/lb, then cooled to 0°F with cp = 0.4. Processing takes 4 h. Other loads are zero.

Q=1000[0.8(40-30)+100+0.4(30-0)]=120{,}000\ \mathrm{Btu}

\dot Q=120{,}000/4=30{,}000\ \mathrm{Btu/h}=2.5\ \mathrm{refrigeration\ tons}

Decision. Average product duty is 2.5 refrigeration tons.

Cross-check. The sensible contributions are 8,000 and 12,000 Btu; the latent contribution is 100,000 Btu. None should be counted twice.

Unit basis. Product mass is lb; cp is Btu/(lb·°F); latent freezing energy is Btu/lb; refrigeration ton is 12,000 Btu/h.

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

Open targeted practice in the study studio →

Transfer the method

HVAC and Refrigeration Distribution and Systems · HV-LC-D7

Try a new decision.

Select one

A steady basic vapor-compression cycle has refrigerant flow 1,500 lb/h. Supplied enthalpies are $h_1$ = 105, $h_2$ = 135 and $h_3$ = 45 Btu/lb; the throttling model gives $h_4$ = $h_3$ . Neglect other energy transfers. The refrigeration COP and condenser heat rejection, respectively are most nearly:

ACOP = 3.0; rejection = 135,000 Btu/hBCOP = 2.0; rejection = 90,000 Btu/hCCOP = 2.0; rejection = 135,000 Btu/hDCOP = 3.5; rejection = 202,500 Btu/h

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 heat to calculate refrigeration COP.
Treating the expansion valve as isentropic instead of approximately isenthalpic.
Dividing total product energy by time before all cooling/freezing terms are included.

Handbook navigation drill

Find it, then explain it.

Search: Compression Refrigeration Cycles

Follow the refrigeration cross-reference to thermodynamic cycles, then locate refrigerant and food property tables.

Check your navigation

You can identify which property table and cycle state are needed before calculating.

Reference path: §4.5, pp. 289–302; §8.1–8.10, pp. 401–445. These are printed handbook page numbers; section names and search terms remain the primary navigation method. See technical review status.

Theory sources

NCEES handbook v2.1, §4.5 and §§8.1–8.10.
Grondzik and Kwok, Mechanical and Electrical Equipment for Buildings, 12th ed., §§12.7–12.8.
Lechner, Heating, Cooling, Lighting, 4th ed., §§16.9–16.10.

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