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In the HVAC field, cooling load conversions are a cornerstone for accurate system design, energy efficiency, and performance comparisons. These conversions involve translating cooling loads between units such as tons of refrigeration (TR), kilowatts (kW), and British Thermal Units per hour (Btu/h), as well as efficiency metrics like Coefficient of Performance (COP), Energy Efficiency Ratio (EER), and kilowatts per ton (kW/ton). Understanding these conversions is essential for engineers and technicians working with commercial and industrial air-conditioning, heat pump, and refrigeration systems.
The primary units and efficiency metrics in cooling load calculations are:
Conversions between these units rely on established relationships.
For example, converting 10 TR to kW:
10 TR × 3.517 kW/TR = 35.17 kW.
Efficiency metrics can also be interconverted. A common example is
the relationship between COP and kW/ton:\[ \text{COP} =
\frac{3.517}{\text{kW/ton}}\] For practical purposes, this is
approximated as `COP = 3.52 / kW/ton`. For a system with a COP of
4.0, the kW/ton rating is:
kW/ton = 3.52 / 4.0 = 0.88 kW/ton.
These conversions are vital for comparing system efficiencies, sizing equipment, and ensuring compliance with industry standards. Lower kW/ton or higher COP and EER values indicate more efficient systems.
Converting between kW/ton, Coefficient of Performance (COP), and Energy Efficiency Ratio (EER) is critical for accurately comparing HVAC and refrigeration system efficiencies across applications and standards. These metrics serve distinct purposes and are used in various contexts, making their interrelationships essential for informed decision-making in system design, procurement, and performance evaluation.
Converting between cooling load metrics is essential for accurate system design and performance comparison. The table below summarizes the key formulas, using precise constants for TR-kW conversions and standard industry approximations for efficiency metrics.
| Conversion | Formula |
|---|---|
| TR to kW | kW = TR × 3.517 |
| kW to TR | TR = kW ÷ 3.517 |
| Btu/h to TR | TR = Btu/h ÷ 12,000 |
| TR to Btu/h | Btu/h = TR × 12,000 |
| COP to kW/ton | kW/ton = 3.52 ÷ COP |
| kW/ton to COP | COP = 3.52 ÷ kW/ton |
| EER to COP | COP = EER ÷ 3.412 |
| COP to EER | EER = COP × 3.412 |
| kW/ton to EER | EER = 12 ÷ kW/ton |
| EER to kW/ton | kW/ton = 12 ÷ EER |
Step 1: Understand the Relationship
The relationship between kW/ton and COP is:\[ \text{COP} =
\frac{3.52}{\text{kW/ton}}\] Lower kW/ton values indicate higher
efficiency, corresponding to higher COP.
Step 2: Apply the Formula
For a system with a kW/ton rating of 0.88:\[ \text{COP} =
\frac{3.52}{0.88} = 4.0\] The system has a COP of 4.0, producing 4
units of cooling per unit of energy input.
Step 1: Understand the Relationship
The relationship between kW/ton and EER is:\[ \text{EER} =
\frac{12}{\text{kW/ton}}\] Lower kW/ton values correspond to higher
EER, indicating greater efficiency.
Step 2: Apply the Formula
For a system with a kW/ton rating of 0.88:\[ \text{EER} =
\frac{12}{0.88} \approx 13.64\] The system has an EER of
approximately 13.64.
Step 1: Understand the Relationship
The relationship between COP and EER is:\[ \text{EER} = \text{COP}
\times 3.412\] The constant 3.412 is the conversion factor between
Btu/h and Watts.
Step 2: Apply the Formula
For a system with a COP of 4.0:\[ \text{EER} = 4.0 \times 3.412
\approx 13.65\] The system has an EER of approximately 13.65.
Step 1: Understand the Relationship
The relationship between EER and COP is:\[ \text{COP} =
\frac{\text{EER}}{3.412}\]
Step 2: Apply the Formula
For a system with an EER of 13.8:\[ \text{COP} = \frac{13.8}{3.412}
\approx 4.0\] The system has a COP of approximately 4.0.
Here are practical examples illustrating cooling load conversions in HVAC system design and analysis.
Example 1:
A cooling system has a capacity of 15 TR. Convert
this to kW and Btu/h.
Solution:
Example 2:
A chiller provides 200,000 Btu/h of cooling.
Convert this to TR and kW.
Solution:
Example 3:
A chiller has a COP of 4.5.
Convert this to kW/ton and EER.
Solution:
Example 4:
A system has an EER of 13.5.
Convert this to COP and kW/ton.
Solution:
Example 5:
Design a cooling system for a commercial building requiring 500
kW of cooling capacity. Determine the equivalent capacity
in TR and Btu/h.
Solution:
Example 6:
A chiller has a cooling capacity of 100 TR and
consumes 30 kW of power. Calculate its COP,
EER, and kW/ton.
Solution:
These examples demonstrate the practical application of cooling load conversions in HVAC system design, energy efficiency analysis, and performance comparisons.
Warning : this calculator is provided to illustrate the concepts mentioned in this webpage, it is not intended for detail design. It is not a commercial product, no guarantee is given on the results. Please consult a reputable designer for all detail design you may need.
You can download the calculator here : link
Interactive tools for converting between common HVAC efficiency and capacity units, based on the provided technical document.
Enter a value for kW/ton, COP, or EER to automatically calculate the others. Lower kW/ton or higher COP/EER values indicate better efficiency.
| Conversion | Formula |
|---|---|
| kW/ton to COP | COP = 3.52 / kW/ton |
| kW/ton to EER | EER = 12 / kW/ton |
| COP to EER | EER = COP × 3.412 |
Enter a value for Tons of Refrigeration (TR), Kilowatts (kW), or Btu/h to convert between cooling capacity units.
| Conversion | Formula |
|---|---|
| TR to kW | kW = TR × 3.517 |
| TR to Btu/h | Btu/h = TR × 12,000 |
| kW to Btu/h | Btu/h = kW × 3412 |