Calculated Results
Input Data
Example Data Table
| Case | Method | Main inputs | Estimated capacity |
|---|---|---|---|
| Cold room AHU | Airflow and enthalpy | 5000 CFM, h1 = 58, h2 = 42 | 45.29 kW / 12.88 TR |
| Chilled water loop | Liquid flow | 2.5 kg/s, cp = 4.186, delta T = 6 C | 62.79 kW / 17.85 TR |
| Packaged unit review | Known capacity | 20 TR | 70.34 kW / 240000 Btu/h |
Formula Used
Air side method: Q = m x (h1 - h2)
Where Q is cooling capacity in kW, m is air mass flow in kg/s, and h1 - h2 is enthalpy difference in kJ/kg.
Liquid side method: Q = m x cp x delta T
Where cp is specific heat in kJ/kg-K and delta T is inlet minus outlet temperature in C or K.
Conversions: 1 TR = 3.517 kW, 1 kW = 3412.142 Btu/h
Power input: Power = Q / COP
Heat rejection: Condenser heat = Q + Power
How to Use This Calculator
- Select the engineering method matching your available field data.
- Enter airflow and enthalpy values for air side calculations.
- Enter fluid flow, specific heat, and temperatures for liquid circuits.
- Use the known capacity option for quick unit conversions.
- Provide COP or EER to estimate electrical demand and heat rejection.
- Press submit to show the result section above the form.
- Download the result summary as CSV or PDF if needed.
Engineering Notes
Capacity Units and Design Context
Refrigeration capacity measures how quickly a system removes heat from air, water, or a process stream. Engineers compare results in kilowatts, refrigeration tons, and Btu per hour because drawings and vendor sheets use different units. One refrigeration ton equals 3.517 kilowatts or about 12,000 Btu per hour. Showing all three units improves communication during design.
Airside Method for Total Cooling Load
The airside method uses air mass flow and enthalpy change to estimate load across a cooling coil. Because enthalpy includes sensible and latent heat, this method suits comfort cooling, dehumidification, and cold storage analysis. For example, 5,000 CFM at 1.2 kilograms per cubic meter with enthalpy dropping from 58 to 42 kilojoules per kilogram produces about 45.3 kilowatts, or 12.9 refrigeration tons.
Liquid Side Method for Chilled Systems
The liquid side method is practical when flow rate, specific heat, and temperature difference are known. It is common for chilled water, glycol circuits, and industrial cooling loops. With water flow of 2.5 kilograms per second, specific heat of 4.186 kilojoules per kilogram kelvin, and a 6 degree temperature drop, the load is about 62.8 kilowatts, or 17.9 tons.
Efficiency, Power, and Heat Rejection
Cooling capacity alone does not define the full system requirement. Designers also estimate electrical input and condenser heat rejection. If a unit delivers 45.3 kilowatts at a COP of 3.2, input power is about 14.2 kilowatts. Heat rejected at the condenser becomes roughly 59.5 kilowatts. These values help size feeders, condensers, cooling towers, and ventilation.
Why Exportable Results Matter
Exportable calculations improve traceability. CSV files help compare alternatives, build estimate sheets, and check field measurements. PDF outputs provide a clear record for approvals, client submissions, and maintenance files. When the method, efficiency inputs, and final capacities stay attached to the result, engineering teams reduce misunderstanding and speed reviews during coordination.
Applying Results to Equipment Selection
A calculated refrigeration load should guide selection, not replace engineering judgment. Final sizing may also consider safety margin, ambient conditions, fouling, pull down load, part load operation, and future expansion. Reviewing cooling capacity, input power, and heat rejection together gives a stronger basis. Combined with manufacturer data and verified inputs, this calculator supports system decisions.
Frequently Asked Questions
1. What is refrigeration capacity?
It is the rate at which a system removes heat. The calculator reports that rate in kilowatts, tons of refrigeration, and Btu per hour.
2. When should I use the airside method?
Use it when airflow and air enthalpy values are available. It is especially useful for coils, air handlers, dehumidification systems, and cold room supply air analysis.
3. When is the liquid flow method better?
Choose it for chilled water, glycol, or process liquid circuits. It works well when flow rate, specific heat, and inlet and outlet temperatures are known.
4. Why does the calculator show COP and EER?
These efficiency inputs help estimate electrical demand. They also support heat rejection calculations, which are important for condenser and utility sizing.
5. Is one ton of refrigeration equal to 12,000 Btu per hour?
Yes. One refrigeration ton equals about 12,000 Btu/h and 3.517 kW. The calculator automatically converts between these common engineering units.
6. Should I size equipment exactly to the calculated result?
Usually no. Final selection should also consider safety margin, ambient conditions, fouling, control range, pull-down load, and manufacturer performance data.