Refrigeration Cycles & Systems

8% of exam

Vapor-compression cycles on the p-h diagram, COP and capacity, refrigerant properties and selection, subcooling, superheat, and food/process refrigeration loads.

6 concepts

Exam areas

E. Refrigeration and Refrigeration Systems

The Vapor-Compression Cycle: COP, Capacity and the p-h Diagram

The four-process vapor-compression cycle on the pressure-enthalpy diagram, refrigeration effect and compressor work, and how COP, tons, and kW/ton fall out of enthalpy differences.

Almost every refrigeration question on this exam — chillers, refrigerators, heat pumps, comparative refrigerant performance — is the same four-process loop drawn on one chart: the pressure-enthalpy (

p

h

) diagram. If you can place the four state points on that chart and read enthalpies off it, the rest is subtraction. The handbook routes you through Chapter 4 for the thermodynamics and gives the per-ton property tables in §8.6, but it assumes you already see the cycle as a closed loop of energy bookkeeping. That fluency is what wins the refrigeration credits, because the wrong-answer choices are almost always built from the right enthalpies combined the wrong way.

The four processes

Walk the loop starting at the compressor inlet (state 1, low-pressure saturated or slightly superheated vapor). Process 1

\to

2 is compression: the compressor raises the vapor to condenser pressure, ideally isentropically, adding the work. Process 2

\to

3 is condensation: the high-pressure vapor rejects heat to the ambient and leaves as saturated (or subcooled) liquid. Process 3

\to

4 is expansion through a throttling device, which drops the pressure at constant enthalpy (

h_4 = h_3

) and flashes part of the liquid to a cold two-phase mixture. Process 4

\to

1 is evaporation: the cold mixture absorbs heat from the refrigerated space and boils back to vapor, closing the loop. On the

p

h

chart the loop is a near-rectangle: two horizontal constant-pressure legs (condenser on top, evaporator on the bottom), a near-vertical compression leg leaning right, and a vertical throttling leg straight down.

Refrigeration effect and compressor work

Treat each component as a steady-flow control volume. The evaporator absorbs the useful cooling, the refrigeration effect

q_L = h_1 - h_4

per unit mass. The compressor adds work

w_c = h_2 - h_1

. The condenser rejects

q_H = h_2 - h_3

. Energy balance closes the loop:

q_H = q_L + w_c

. Because throttling is isenthalpic,

h_4 = h_3

, so the refrigeration effect is really

h_1 - h_3

— the liquid enthalpy leaving the condenser sets the bottom of the cooling. That single fact is why subcooling the liquid (lowering

h_3

) directly buys more capacity at no extra compressor work.

q_L = h_1 - h_4 = h_1 - h_3, \qquad w_c = h_2 - h_1, \qquad q_H = h_2 - h_3

Capacity, tons, and mass flow

Component duty is mass flow times the per-pound enthalpy change. The evaporator capacity is

\dot Q_L = \dot m\,(h_1 - h_4)

. Refrigeration capacity is quoted in tons, where one ton of refrigeration is

12{,}000\,\text{Btu/h} = 200\,\text{Btu/min} = 3.517\,\text{kW}

— the rate to freeze one short ton of

32^\circ\text{F}

water to ice in 24 hours. So if you know the required tons and read

q_L

off the chart, the refrigerant mass flow follows immediately, and from there the compressor power and condenser heat rejection.

\dot Q_L = \dot m\,(h_1 - h_4), \qquad 1\ \text{ton} = 12{,}000\ \tfrac{\text{Btu}}{\text{h}} = 3.517\ \text{kW}

COP, EER and kW/ton

Coefficient of performance is the benefit over the cost: cooling delivered per unit of compressor work,

\mathrm{COP}_R = q_L / w_c

. A heat pump rejecting useful heat instead claims

\mathrm{COP}_{HP} = q_H / w_c = \mathrm{COP}_R + 1

. The industry restates the same ratio in mixed units two ways. EER is Btu/h of cooling per watt of input, so

\mathrm{EER} = 3.412\,\mathrm{COP}

. The chiller metric kW/ton is the inverse and runs the other way — smaller is better:

\mathrm{kW/ton} = 3.517/\mathrm{COP} = 12/\mathrm{EER}

. Keep these straight: a high COP corresponds to a high EER but a LOW kW/ton.

\mathrm{COP}_R = \dfrac{q_L}{w_c}, \quad \mathrm{EER} = 3.412\,\mathrm{COP}, \quad \mathrm{kW/ton} = \dfrac{3.517}{\mathrm{COP}}

The Carnot ceiling

No cycle operating between a cold reservoir

T_L

and a hot reservoir

T_H

can beat the reversed-Carnot COP,

\mathrm{COP}_{C} = T_L/(T_H - T_L)

with both temperatures ABSOLUTE (

^\circ\text{R}

\text{K}

). This is the sanity ceiling for every numerical answer: a real chiller typically lands at 45–60 % of Carnot. The message it carries is operational — COP collapses as the temperature lift

T_H - T_L

grows, which is exactly why raising suction (evaporator) temperature or lowering condensing temperature is the highest-leverage energy move on any refrigeration plant.

\mathrm{COP}_{C} = \dfrac{T_L}{T_H - T_L}\quad (\text{absolute temperatures})

Ideal versus actual cycle

The ideal standard cycle assumes saturated vapor at the compressor inlet, isentropic compression, saturated liquid at the condenser exit, and no pressure drops. The real cycle deviates everywhere: the suction line adds superheat, the compression is non-isentropic (modeled by isentropic efficiency

\eta_s = (h_{2s}-h_1)/(h_{2a}-h_1)

, so the actual work is larger), the liquid leaves subcooled, and friction drops pressure in the lines and heat exchangers. The throttling process itself is inherently irreversible — replacing the valve with an isentropic turbine is what separates this practical cycle from the reversed-Carnot ideal. For exam work, build the ideal cycle from the tables, then apply

\eta_s

to inflate the compressor work and, when given, shift state 1 and state 3 for superheat and subcooling.

Exam strategy

Fix the two pressures first from the saturation temperatures (evaporator and condenser), then label states 1–4 on a quick

p

h

sketch before touching numbers. Read

h_1

(sat. vapor at evaporator

T

h_3

(sat. liquid at condenser

T

), set

h_4 = h_3

, and find

h_2

by following the constant-entropy line from state 1 up to condenser pressure. Compute

q_L

w_c

q_H

as enthalpy differences; get

\dot m

from the required tons; then COP, power, and rejection. Always convert temperatures to absolute before any Carnot ratio, and check your COP sits below the Carnot ceiling — if it doesn't, you swapped

T_H

and

T_L

or used

^\circ\text{F}

Key equations

Refrigeration effect

q_L = h_1 - h_4 = h_1 - h_3

Useful cooling per unit mass (Btu/lb). $h_4 = h_3$ because expansion is isenthalpic.

Compressor work

w_c = h_2 - h_1

Ideal (isentropic) work per unit mass (Btu/lb); $h_2$ at condenser pressure on the $s = s_1$ line.

Heat rejected

q_H = h_2 - h_3 = q_L + w_c

Condenser duty per unit mass; energy balance closes the loop.

Capacity from mass flow

\dot Q_L = \dot m\,(h_1 - h_4)

Evaporator duty (Btu/h) with $\dot m$ in lb/h; divide by 12,000 for tons.

Ton of refrigeration

1\ \text{ton} = 12{,}000\ \tfrac{\text{Btu}}{\text{h}} = 200\ \tfrac{\text{Btu}}{\text{min}} = 3.517\ \text{kW}

Standard refrigeration capacity unit; use to convert between tons and Btu/h or kW.

Coefficient of performance (refrigerator)

\mathrm{COP}_R = \dfrac{q_L}{w_c} = \dfrac{h_1 - h_4}{h_2 - h_1}

Cooling delivered per unit compressor work (dimensionless).

COP heat pump

\mathrm{COP}_{HP} = \dfrac{q_H}{w_c} = \mathrm{COP}_R + 1

Same hardware billed for heat rejected; always one greater than the cooling COP.

EER and kW/ton

\mathrm{EER} = 3.412\,\mathrm{COP}, \qquad \mathrm{kW/ton} = \dfrac{3.517}{\mathrm{COP}} = \dfrac{12}{\mathrm{EER}}

Mixed-unit restatements of COP. High EER and LOW kW/ton both mean efficient.

Carnot (reversed) COP

\mathrm{COP}_{C} = \dfrac{T_L}{T_H - T_L}

Thermodynamic ceiling; $T_L$ , $T_H$ ABSOLUTE ( $^\circ\text{R}$ or K). Real plant 45–60 % of this.

Compressor isentropic efficiency

\eta_s = \dfrac{h_{2s} - h_1}{h_{2a} - h_1}

Ratio of ideal to actual work; solve for actual exit enthalpy $h_{2a}$ to size real power.

Compressor power

\dot W_c = \dot m\,(h_2 - h_1)

Btu/h; divide by 2,544 for hp or by 3,412 for kW.

Volumetric flow / displacement

\dot V_1 = \dot m\,v_1

Suction volume flow (ft³/min); $v_1$ = specific volume of suction vapor. Sets compressor displacement.

Worked examples

Ideal R-134a chiller — capacity, power, COP

Problem. A 20-ton water chiller uses R-134a on the ideal standard cycle with an evaporator at $40^\circ\text{F}$ and a condenser at $110^\circ\text{F}$ . From the NCEES handbook R-134a saturation table (§8.6): saturated vapor at $40^\circ\text{F}$ has $h_1 = 108.86\,\text{Btu/lb}$ and $s_1 = 0.22207\,\text{Btu/(lb}\cdot{}^\circ\text{R)}$ ; saturated liquid at $110^\circ\text{F}$ ( $P_{cond} \approx 161\,\text{psia}$ ) has $h_3 = 48.73\,\text{Btu/lb}$ . Following the constant-entropy line $s = s_1$ up to $161\,\text{psia}$ gives the superheated discharge $h_2 = 120.8\,\text{Btu/lb}$ . Find the refrigerant mass flow, the compressor power (kW), the COP, and the kW/ton.

Solution. Refrigeration effect:

q_L = h_1 - h_3 = 108.86 - 48.73 = 60.13\,\text{Btu/lb}

(since

h_4 = h_3

). Compressor work:

w_c = h_2 - h_1 = 120.8 - 108.86 = 11.94\,\text{Btu/lb}

. Load:

20\ \text{ton} \times 12{,}000 = 240{,}000\,\text{Btu/h}

. Mass flow:

\dot m = \dfrac{240{,}000}{60.13} = 3{,}991\,\text{lb/h}

(

= 66.5\,\text{lb/min}

). Power:

\dot W_c = \dot m\,w_c = 3{,}991 \times 11.94 = 47{,}660\,\text{Btu/h} = 47{,}660/3{,}412 = 14.0\,\text{kW}

. COP:

\mathrm{COP}_R = 60.13/11.94 = 5.04

. kW/ton:

3.517/5.04 = 0.698\,\text{kW/ton}

(equivalently

14.0/20 = 0.70

). Sanity check: Carnot ceiling

= \dfrac{500}{570-500} = 7.14

; the cycle COP

5.04

71\%

of Carnot — physically reasonable. Final:

\dot m \approx 3{,}990\,\text{lb/h}

\dot W_c \approx 14.0\,\text{kW}

\mathrm{COP} = 5.04

0.698\,\text{kW/ton}

\mathrm{COP}_R = \dfrac{h_1-h_3}{h_2-h_1} = \dfrac{60.13}{11.94} = 5.04

Actual cycle with compressor efficiency

Problem. Take the same R-134a states as above ( $h_1 = 108.86$ , isentropic $h_{2s} = 120.8$ , $h_3 = 48.73\,\text{Btu/lb}$ ) but the compressor has an isentropic efficiency of $0.75$ . Find the actual discharge enthalpy, the actual COP, and the increase in condenser heat rejection compared with the ideal cycle.

Solution. Actual work:

w_a = (h_{2s}-h_1)/\eta_s = 11.94/0.75 = 15.92\,\text{Btu/lb}

, so

h_{2a} = 108.86 + 15.92 = 124.8\,\text{Btu/lb}

. Refrigeration effect is unchanged:

q_L = 60.13\,\text{Btu/lb}

(states 1 and 3 did not move). Actual COP:

60.13/15.92 = 3.78

(down from

5.04

). Condenser duty per pound: ideal

q_H = 120.8 - 48.73 = 72.07\,\text{Btu/lb}

; actual

q_H = 124.8 - 48.73 = 76.05\,\text{Btu/lb}

— a

4.0\,\text{Btu/lb}

(

5.5\%

) increase, exactly the extra compressor work the inefficiency dumps into the refrigerant. Sanity check: actual COP

3.78 <

ideal

5.04 <

Carnot

7.14

, the correct ordering. Final:

h_{2a} = 124.8\,\text{Btu/lb}

\mathrm{COP} = 3.78

, condenser rejection up

5.5\%

h_{2a} = h_1 + \dfrac{h_{2s}-h_1}{\eta_s} = 108.86 + \dfrac{11.94}{0.75} = 124.8\ \text{Btu/lb}

Common pitfalls

•Using $^\circ\text{F}$ in the Carnot ratio. $\mathrm{COP}_C = T_L/(T_H-T_L)$ needs ABSOLUTE temperature — convert to $^\circ\text{R}$ ( $+460$ ) or K first, or your ceiling is meaningless.
•Confusing EER and kW/ton direction. EER and COP move together (bigger is better); kW/ton is inverted (smaller is better). A 0.5 kW/ton chiller is far better than a 0.9 kW/ton one.
•Forgetting $h_4 = h_3$ . Throttling is isenthalpic, so the refrigeration effect is $h_1 - h_3$ , not $h_1 - h_{f,\text{evap}}$ . Using the saturated-liquid enthalpy at evaporator temperature overstates capacity.
•Charging the COP for the wrong load. $\mathrm{COP}_R$ uses evaporator cooling $q_L$ ; $\mathrm{COP}_{HP}$ uses condenser heat $q_H$ and is always 1 larger. Read which the question wants.
•Reading $h_2$ as saturated vapor at condenser pressure. State 2 is SUPERHEATED — follow the constant-entropy line up from state 1; it lands to the right of the dome, hotter than $T_{cond}$ .
•Applying $\eta_s$ to the refrigeration effect. Isentropic efficiency inflates only the compressor work ( $w_a = w_s/\eta_s$ ); $q_L$ and $q_H$ change only through the shifted $h_2$ .
•Unit slips on tons. $1\,\text{ton} = 12{,}000\,\text{Btu/h}$ , not 12,000 Btu/min; and $= 3.517\,\text{kW}$ , not 3.517 hp.

References

NCEES PE Mechanical Reference Handbook — §8 Refrigeration
NCEES PE Mechanical Reference Handbook — §4.5 Thermodynamic Cycles and Compressors
ASHRAE Handbook—Fundamentals, Ch. 2 Thermodynamics and Refrigeration Cycles

Superheat, Subcooling and Refrigerant Selection

How evaporator superheat and condenser subcooling reshape the p-h cycle and shift capacity and COP, the role of the expansion device, and how to compare common refrigerants.