9.2 COP, Tons of Refrigeration, and Heat Rejection Ratio

9.2 COP, Tons of Refrigeration, and Heat Rejection Ratio

NRE tells you how much refrigeration each pound delivers. COP tells you how much compressor work you spent to get the tons. Heat rejection tells you what the condenser and oil cooler must dump. CIRO operating screens mix these ideas as tons, kW, BHP, and dollars per hour. The arithmetic is the same on a 50 HP high-stage and on a 300 HP screw.

The ton and the horsepower

1 ton of refrigeration = 12,000 Btu/h = 200 Btu/min.

The definition is historical: melting 2,000 lb of ice in 24 hours. Ice takes about 144 Btu/lb to melt, and 2,000 × 144 / 24 = 12,000 Btu/h. You will not be asked to derive it. You will be asked to convert without mixing minutes and hours:

• 80 tons → 80 × 12,000 = 960,000 Btu/h = 80 × 200 = 16,000 Btu/min
• 1,500,000 Btu/h → 1,500,000 / 12,000 = 125 tons

1 HP = 2,545 Btu/h ≈ 0.746 kW.

Brake horsepower (BHP) is shaft work. Electrical input is larger than shaft output by the motor's inefficiency. Keep the two separate:

• Shaft heat equivalent: BHP × 2,545 = Btu/h of work
• Shaft kilowatts: BHP × 0.746
• Electrical input kW: (BHP × 0.746) / motor efficiency, or from volts, amps, and power factor (next section)

Using 2,545 Btu/h when you meant 12,000, or 0.746 as if it were Btu/h, produces a COP that cannot exist in a real plant.

COP is a ratio of the same units

COP = Q_evap / W_comp

Q_evap and W_comp must be in the same units—Btu/h over Btu/h, or kW over kW. COP has no unit. A COP of 3.0 means each Btu of compressor work buys 3 Btu of refrigeration.

Higher COP = less work per ton. A high ammeter reading is not proof of high tons. High work with falling tons is a COP collapse.

Per-ton form:

Work per ton (Btu/h) = (BHP/ton) × 2,545

COP = 12,000 / (BHP/ton × 2,545)

If BHP/ton = 1.50:

W = 1.50 × 2,545 = 3,818 Btu/h per ton

COP = 12,000 / 3,818 = 3.14

If BHP/ton = 2.20 (dirty condenser, high lift, starved evaporator):

W = 2.20 × 2,545 = 5,599 Btu/h per ton

COP = 12,000 / 5,599 = 2.14

Same 100-ton nameplate on the package; very different plants.

Worked COP from plant numbers

A load is 90 tons. Compressor shaft work is 120 BHP (drive output, a trusted full-load slide-valve point, or another shaft method—not a guess from nameplate while the slide valve is at 40%).

Qe = 90 × 12,000 = 1,080,000 Btu/h

W = 120 × 2,545 = 305,400 Btu/h

COP = 1,080,000 / 305,400 = 3.54

BHP/ton = 120 / 90 = 1.33

Check with the per-ton formula: 12,000 / (1.33 × 2,545) = 12,000 / 3,385 ≈ 3.55. Matches within rounding.

If the motor is 93% efficient, electrical input is 120 × 0.746 / 0.93 ≈ 96.3 kW. Input kW/ton = 96.3 / 90 = 1.07 kW/ton. Shaft kW/ton = 1.33 × 0.746 = 0.99 kW/ton. A CIRO screen that shows kW from a power meter is input unless it is labeled as shaft. Using shaft kW when the meter is showing input makes COP look better than the electric bill.

Lift kills COP

Lift is the temperature (or pressure) difference the compressor must pump across: roughly Td − Te, condensing saturation minus evaporating saturation. The Carnot ceiling is Te / (Td − Te) with temperatures on an absolute scale (Rankine). You may never see the word Carnot on CIRO. You will see the consequence:

• Lower suction (colder Te) → more lift → more work per pound → lower COP
• Higher condensing temperature (hotter Td from a dirty condenser, noncondensables, high wet bulb, or a failed fan) → more lift → lower COP
• Floating head pressure on a cold day reduces Td, cuts lift, and raises COP

Worked comparison, same 100-ton evaporator load.

Condition A. Te = 0°F, Td = 85°F, lift = 85°F. Shaft work = 135 BHP.

W_A = 135 × 2,545 = 343,575 Btu/h

COP_A = 1,200,000 / 343,575 = 3.49

Condition B. Condenser neglected. Td = 105°F, lift = 105°F. Shaft work rises to 175 BHP to hold the same tons.

W_B = 175 × 2,545 = 445,375 Btu/h

COP_B = 1,200,000 / 445,375 = 2.69

COP fell about 23%. Work rose 40 BHP for the same tons. That extra work is extra heat into the condenser and extra dollars. You do not need a new evaporator to lose COP; you only need extra lift.

Heat rejection is evaporator heat plus compressor work

Steady-flow energy balance on the refrigeration cycle:

Heat rejected ≈ Qe + W

The condenser must reject the heat the evaporator absorbed plus the compressor work that became heat. Oil-injected screws change where that work heat leaves, not whether it must leave:

• Some of W leaves in discharge gas (m × refrigerant enthalpy rise).
• Some of W leaves in the oil cooler (thermosiphon, liquid injection, or water-cooled). A thermosiphon oil cooler dumps heat into a refrigerant circuit that typically returns to the condenser.
• Condenser fans and spray pumps add a smaller electrical load that still ends up in the air or water.

The wet-bulb-side load is therefore not "just the tons." A plant making 100 tons at COP 3.5 rejects about 1.29 × the evaporator load. The same 100 tons at COP 2.5 rejects 1.40 ×. Condenser sizing, fan staging, and "why is head pressure high with a clean basin?" start from this balance.

Heat rejection ratio = Qc / Qe ≈ 1 + 1/COP when Qc = Qe + W and COP = Qe / W.

Ammonia single-stage plants often land near 1.2–1.3 when COP is roughly 3.3–5. That band is a result, not a law. A high-lift freezer, a fouled condenser, or a screw with a hot oil cooler will push the ratio above 1.3. Do not memorize 1.25 as if RETA printed it on a plaque. Calculate it from the screen in front of you.

Worked heat rejection

Use the 90-ton, 120 BHP plant.

Qe = 1,080,000 Btu/h

W = 305,400 Btu/h

Qc ≈ 1,080,000 + 305,400 = 1,385,400 Btu/h ≈ 115.5 tons of rejection (1,385,400 / 12,000)

Ratio = 1,385,400 / 1,080,000 = 1.28

In Btu/min: Qe = 90 × 200 = 18,000 Btu/min; W = 305,400 / 60 = 5,090 Btu/min; Qc ≈ 23,090 Btu/min.

Oil-cooler trap. If a screen lists oil-cooler heat (Btu/min or tons), that stream is part of W, not a third kind of energy. Adding oil-cooler heat on top of Qe + W double-counts. Add oil-cooler heat only if you computed W from refrigerant enthalpy rise alone and still need the oil-side remainder to close the compressor energy balance.

Operator checklist

• Convert tons and BHP to Btu/h before dividing. Mixed units produce fantasy COPs.
• Use shaft work for thermodynamic COP; use electrical input for kW/ton and cost, and say which one you used.
• When lift rises, expect COP to fall before you blame the evaporator.
• Think of condenser duty as Qe + W, not Qe.
• Treat 1.2–1.3 as a sanity check for a healthy ammonia single-stage, then calculate the actual ratio.

Exam traps

• Calling 90 tons / 120 BHP a COP of 0.75. That is tons per horsepower, not COP.
• Treating condenser load as equal to evaporator tons.
• Adding oil-cooler heat to Qe + W and counting the same work twice.
• Assuming COP is constant when suction drops or head pressure floats up.