Chillers, Heat Pumps & Thermal Storage

6% of exam

Chiller kW/ton and IPLV, heat-pump heating COP and balance point, condenser/evaporator duty, and chilled-water and ice thermal storage.

5 concepts

Exam areas

D. Condensers/Evaporators

Chiller Performance: kW/ton, COP, EER and IPLV

Translate freely among kW/ton, COP and EER, weight the four part-load points into IPLV, and connect water- vs air-cooled chillers to the condenser and evaporator approach.

A chiller is usually the single largest electrical load in a commercial building, so the exam keeps coming back to one question in different costumes: how much electricity does it take to move a given amount of heat? The industry answers that question in at least four different units —

\text{kW/ton}

, COP, EER, and the part-load index IPLV — and a PE problem will hand you one and ask for another, or buried inside a load or energy calculation. Get fluent at sliding between them and a whole class of questions becomes arithmetic. The conversions all rest on one fact from the NCEES PE Mechanical Reference Handbook — §4 Thermodynamics:

1\ \text{ton} = 12{,}000\ \text{Btu/h} = 3{,}516\ \text{W}

, plus the COP definition in that same §4 Thermodynamics. EER,

\text{kW/ton}

, and IPLV are not in the handbook at all — they are AHRI/industry ratings (AHRI Standard 550/590).

COP is the dimensionless truth; everything else is a unit choice

Coefficient of performance is the honest, dimensionless measure: useful cooling delivered divided by work input, both in the same units. For a chiller (a refrigerator),

\text{COP} = \dot Q_L / \dot W

. Because cooling is metered in

\text{Btu/h}

but power in watts, you must put both on the same footing — convert the cooling to watts with

3.412\ \text{Btu/h} = 1\ \text{W}

, or convert the work to

\text{Btu/h}

. Every other rating you will see is just COP wearing different units, so anchor here first.

\text{COP}_{\text{cooling}} = \frac{\dot Q_L}{\dot W} = \frac{\dot Q_L\,[\text{Btu/h}]}{3.412\,\dot W\,[\text{W}]}

kW/ton — the chiller-plant currency

Chiller and central-plant people almost never speak in COP; they speak in

\text{kW/ton}

— kilowatts of compressor (or plant) input per ton of cooling produced. It is an inverse efficiency, so smaller is better: a good water-cooled centrifugal machine runs near

0.50\text{–}0.60\ \text{kW/ton}

at full load, while an air-cooled screw is closer to

1.0\text{–}1.2\ \text{kW/ton}

. Because one ton of cooling is

3.516\ \text{kW}

of heat, the bridge to COP is exact and worth memorizing.

\text{kW/ton} = \frac{3.516}{\text{COP}}, \qquad \text{COP} = \frac{3.516}{\text{kW/ton}}

EER and the 12,000-Btu/h-per-ton bridge

Energy efficiency ratio is cooling in

\text{Btu/h}

divided by power in watts, so it carries hidden units of

\text{Btu/(W}\cdot\text{h)}

even though it is quoted as a bare number. Since a ton is

12{,}000\ \text{Btu/h}

\text{EER} = 12/(\text{kW/ton})

, and because

1\ \text{W} = 3.412\ \text{Btu/h}

, EER and COP differ by exactly that factor. Keep the three relationships in a triangle in your head — given any one rating you can produce the other two in one line.

\text{EER} = \frac{12}{\text{kW/ton}} = 3.412\,\text{COP}

IPLV — chillers spend their lives at part load

A chiller almost never runs at 100% capacity; it spends most of its hours throttled down, where centrifugal and screw machines are actually more efficient (until very low load, where they degrade). The integrated part-load value, defined by AHRI Standard 550/590, condenses that reality into one number by weighting performance at four load points — 100%, 75%, 50%, and 25% — by the fraction of operating hours typically spent there. AHRI weights the four points 1/42/45/12%: for COP or EER the IPLV is the plain weighted sum, and for

\text{kW/ton}

(an inverse efficiency) it is the reciprocal form

1/\!\left(\tfrac{0.01}{A}+\tfrac{0.42}{B}+\tfrac{0.45}{C}+\tfrac{0.12}{D}\right)

; the two forms give the same answer once converted.

\text{IPLV}_{\text{COP, EER}} = 0.01\,A + 0.42\,B + 0.45\,C + 0.12\,D, \qquad \text{IPLV}_{\text{kW/ton}} = \dfrac{1}{\frac{0.01}{A}+\frac{0.42}{B}+\frac{0.45}{C}+\frac{0.12}{D}}

Water-cooled vs air-cooled — why the condenser sets the efficiency

The biggest single lever on chiller efficiency is the temperature the compressor must lift the refrigerant to before rejecting heat — the saturated condensing temperature. An air-cooled condenser rejects to the outdoor dry-bulb (say

95\,^\circ\text{F}

) plus a

25\text{–}30\,^\circ\text{F}

temperature difference, landing the condensing temperature near

120\,^\circ\text{F}

. A water-cooled machine rejects to cooling-tower water that tracks the much lower wet-bulb (say

78\,^\circ\text{F}

), reaching the condenser only

\sim 10\,^\circ\text{F}

above that and condensing near

95\,^\circ\text{F}

. That

\sim 25\,^\circ\text{F}

lower lift alone cuts the ideal compressor work per ton by roughly a third; add the condenser-fan power in air-cooled ratings and the efficient centrifugal compressors of large water-cooled plants, and water-cooled plants run near half the

\text{kW/ton}

of air-cooled ones.

Approach: the heat-exchanger penalty at both ends

Both the evaporator and condenser are finite heat exchangers, so the refrigerant cannot reach the water temperature — it stops short by the approach. Condenser approach is the saturated condensing temperature minus the leaving condenser-water temperature; evaporator approach is the leaving chilled-water temperature minus the saturated suction (evaporating) temperature. Fouling, undersized tubes, or low water flow widen these approaches, which raises the lift and degrades

\text{kW/ton}

— the handbook's §8.3 condenser-fouling chart plots chiller capacity falling and compressor power rising as the fouling factor and saturated condensing temperature climb, and a widening approach tracks directly with both. A tight approach (

1\text{–}2\,^\circ\text{F}

) signals a clean, well-sized machine.

\text{Approach}_{\text{cond}} = T_{\text{cond,sat}} - T_{\text{cw,leaving}}, \qquad \text{Approach}_{\text{evap}} = T_{\text{chw,leaving}} - T_{\text{evap,sat}}

Exam strategy

Build the conversion triangle the moment a chiller appears:

\text{kW/ton}=3.516/\text{COP}

\text{EER}=12/(\text{kW/ton})=3.412\,\text{COP}

. Watch which efficiency the problem gives — if it is

\text{kW/ton}

, smaller is better; if COP or EER, larger is better. For IPLV, line the four numbers up against the fixed weights

0.01/0.42/0.45/0.12

and use the weighted-sum form when the points are COP or EER and the reciprocal form when they are in

\text{kW/ton}

. When a problem mentions condensing or chilled-water temperatures, draw the approach explicitly: condensing temperature is leaving-water plus approach, evaporating temperature is leaving-chilled-water minus approach.

Key equations

Cooling COP

\text{COP} = \dfrac{\dot Q_L}{\dot W}

Dimensionless; cooling effect $\dot Q_L$ over compressor input $\dot W$ , both in the same units. Convert Btu/h to W with $3.412\ \text{Btu/h}=1\ \text{W}$ .

Ton of refrigeration

1\ \text{ton} = 12{,}000\ \text{Btu/h} = 3.516\ \text{kW}

The conversion that links every chiller rating; from §4 Thermodynamics ( $3{,}516\ \text{W}$ per ton).

kW/ton to COP

\text{kW/ton} = \dfrac{3.516}{\text{COP}}

Inverse efficiency — lower is better. $3.516$ is kW of heat per ton. Typical water-cooled $0.5\text{–}0.6$ , air-cooled $1.0\text{–}1.2$ .

EER definition

\text{EER} = \dfrac{\dot Q_L\,[\text{Btu/h}]}{\dot W\,[\text{W}]}

Cooling in Btu/h per watt of input; quoted as a bare number but carries units $\text{Btu/(W}\cdot\text{h)}$ .

EER from kW/ton and COP

\text{EER} = \dfrac{12}{\text{kW/ton}} = 3.412\,\text{COP}

The 12 is $12{,}000\ \text{Btu/h}$ per ton divided by 1000 W; the $3.412$ is Btu/h per watt.

IPLV, AHRI 550/590 (COP or EER)

\text{IPLV} = 0.01\,A + 0.42\,B + 0.45\,C + 0.12\,D

Weighted sum of COP or EER (higher is better) at 100/75/50/25% load; this is the AHRI 550/590 definition for efficiencies. Weights are operating-hour fractions. Not printed in the PE Mechanical Reference Handbook.

IPLV, AHRI 550/590 (kW/ton)

\text{IPLV} = \dfrac{1}{\dfrac{0.01}{A}+\dfrac{0.42}{B}+\dfrac{0.45}{C}+\dfrac{0.12}{D}}

Use when $A$ – $D$ are kW/ton (lower is better). It equals $3.516$ divided by the COP-form IPLV. Not printed in the PE Mechanical Reference Handbook.

Condenser approach

\text{Approach}_{\text{cond}} = T_{\text{cond,sat}} - T_{\text{cw,leaving}}

Saturated condensing temperature minus leaving condenser-water temperature (°F). Widens with fouling.

Evaporator approach

\text{Approach}_{\text{evap}} = T_{\text{chw,leaving}} - T_{\text{evap,sat}}

Leaving chilled-water temperature minus saturated suction temperature (°F).

Condenser heat rejection

\dot Q_{\text{rej}} = \dot Q_L + \dot W

First-law balance: heat the condenser must reject equals evaporator cooling plus compressor work (Btu/h).

Condenser water flow

\text{gpm} = \dfrac{\dot Q_{\text{rej}}}{500\,\Delta T}

$\dot Q_{\text{rej}}$ in Btu/h, $\Delta T$ = condenser-water range (°F). The 500 bakes in water density and $c_p$ ( $8.34\times60$ ).

Worked examples

kW/ton to COP and EER

Problem. A water-cooled centrifugal chiller is rated at $0.58\ \text{kW/ton}$ at design conditions. Find its COP and EER.

Solution. COP:

\text{COP} = 3.516/(\text{kW/ton}) = 3.516/0.58 = 6.06

. EER:

\text{EER} = 12/0.58 = 20.7

. Cross-check with

\text{EER}=3.412\,\text{COP}=3.412\times6.06 = 20.7

— consistent. Final:

\text{COP} = 6.06

\text{EER} = 20.7\ \text{Btu/(W}\cdot\text{h)}

. Sanity: a sub-

0.6\ \text{kW/ton}

machine should land in the high-COP range typical of large water-cooled centrifugals, which it does.

\text{COP} = \frac{3.516}{0.58} = 6.06

Computing IPLV from the four part-load points

Problem. A chiller is tested at the four AHRI load points and gives $A=0.62$ , $B=0.48$ , $C=0.45$ , $D=0.55\ \text{kW/ton}$ at 100/75/50/25% load. Find the IPLV in $\text{kW/ton}$ and the equivalent COP.

Solution. AHRI form for kW/ton:

\text{IPLV} = 1/\!\left(\tfrac{0.01}{0.62}+\tfrac{0.42}{0.48}+\tfrac{0.45}{0.45}+\tfrac{0.12}{0.55}\right) = 1/2.109 = 0.474\ \text{kW/ton}

. Equivalent COP:

3.516/0.474 = 7.42

. The plain weighted sum of the kW/ton values,

0.01(0.62)+0.42(0.48)+0.45(0.45)+0.12(0.55) = 0.476

, lands within half a percent, but it is not the AHRI definition for kW/ton. Final:

\text{IPLV} = 0.474\ \text{kW/ton}

(COP

7.42

). Sanity: IPLV (

0.474

) is better than the full-load value (

0.62

) because the 50% and 75% points dominate the weighting and the machine is most efficient there, exactly the part-load benefit IPLV is meant to capture.

\text{IPLV} = \dfrac{1}{\frac{0.01}{0.62}+\frac{0.42}{0.48}+\frac{0.45}{0.45}+\frac{0.12}{0.55}} = \dfrac{1}{2.109} = 0.474\ \text{kW/ton}

Heat rejection and condenser water flow

Problem. A $500$ -ton water-cooled chiller operates at $0.60\ \text{kW/ton}$ . Find the condenser heat rejection and the required condenser-water flow at a $10\,^\circ\text{F}$ range.

Solution. Evaporator load:

\dot Q_L = 500\times12{,}000 = 6{,}000{,}000\ \text{Btu/h}

. Compressor work:

\dot W = 500\times0.60 = 300\ \text{kW} = 300\times3412 = 1{,}023{,}600\ \text{Btu/h}

. Heat rejection:

\dot Q_{\text{rej}} = 6{,}000{,}000 + 1{,}023{,}600 = 7{,}023{,}600\ \text{Btu/h}

(to 3 sig figs

7.02\times10^6\ \text{Btu/h}

). Flow:

\text{gpm} = 7{,}023{,}600/(500\times10) = 1{,}405\ \text{gpm}

. Final:

\dot Q_{\text{rej}} = 7.02\times10^{6}\ \text{Btu/h}

1{,}405\ \text{gpm}

. Sanity: that is

1405/500 = 2.81\ \text{gpm/ton}

, right at the classic

\sim 3\ \text{gpm/ton}

rule for water-cooled condensers.

\dot Q_{\text{rej}} = \dot Q_L + \dot W = 6.00\times10^{6} + 1.02\times10^{6}\ \text{Btu/h}

Common pitfalls

•Confusing direction of merit: $\text{kW/ton}$ is an inverse efficiency (lower is better) while COP and EER are efficiencies (higher is better). A '0.5' is a great chiller; a 'COP 0.5' would be absurd.
•Forgetting the $3.412$ when bridging Btu/h and watts: $\text{EER}=3.412\,\text{COP}$ , not $\text{EER}=\text{COP}$ . Cooling in Btu/h and power in W are not the same unit.
•Pairing the IPLV forms backwards. AHRI 550/590 uses the weighted sum for COP or EER and the reciprocal-weighted form for $\text{kW/ton}$ ; convert first if the problem mixes units.
•Treating IPLV as a full-load rating. IPLV is a single part-load index for comparison only; it does not predict performance at any one actual operating condition (use NPLV for non-standard conditions).
•Quoting heat rejection as equal to the cooling load. The condenser must reject cooling PLUS compressor work — roughly $1.15\text{–}1.25\times$ the evaporator load for an electric chiller (about $1.25\times$ at the $0.88\ \text{kW/ton}$ basis of the $15{,}000\ \text{Btu/h}$ 'cooling-tower ton,' but only $\sim 1.17\times$ for an efficient $0.60\ \text{kW/ton}$ machine).
•Mixing up the two approaches: condenser approach uses leaving condenser water and condensing temperature; evaporator approach uses leaving chilled water and evaporating temperature. Swapping them inverts the sign.
•Comparing an air-cooled to a water-cooled chiller's kW/ton without noting the condensing temperature difference — the air-cooled penalty is physics (higher lift), not a defect.

References

NCEES PE Mechanical Reference Handbook — §8 Refrigeration
NCEES PE Mechanical Reference Handbook — §4 Thermodynamics (COP, ton definition)
AHRI Standard 550/590 — Performance Rating of Water-Chilling Packages — Source of the IPLV/NPLV part-load weights (0.01/0.42/0.45/0.12).
ASHRAE Handbook — HVAC Systems and Equipment, Liquid-Chilling Systems

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