The basic formula
The power a heater has to deliver follows from the energy needed to warm the material, the time available and how much of the heater’s output actually ends up in the material:
- Mass: area × thickness × density for sheet, or the weight of the part
- Specific heat: how much energy one kilogram needs per degree (see the material values in the calculator)
- Temperature rise: target temperature minus starting temperature
- Heating time: the time each part or area spends under the heaters
- Efficiency: the share of the electrical power that is absorbed by the product
If water has to evaporate
Drying needs far more energy than heating. Each kilogram of water needs about 4.19 kJ per degree to warm up to 100 °C and then about 2,257 kJ to evaporate. Add this energy for the water that must be removed in the same heating time; the calculator below does this for you.
Infrared power calculator
Enter your values; the result updates as you type. All fields are estimates and stay in your browser.
Per part or area, for drying; 0 if none
Typically 35–70 %
Estimate
- Heated area
- 0.60 m²
- Mass
- 1.09 kg
- Energy needed
- 292 kJ
- Power absorbed by the product
- 9.7 kW
A first estimate for sizing. Melting energy of semi-crystalline plastics, part geometry and machine losses are not included; confirm with a trial or ask our engineers.
Worked example: thermoforming a PP sheet
A polypropylene sheet of 1.0 m × 0.6 m and 2 mm thickness has to be heated from 20 °C to 160 °C in 30 seconds.
| Step | Calculation | Result |
|---|---|---|
| Volume | 1.0 m × 0.6 m × 0.002 m | 0.0012 m³ |
| Mass | 0.0012 m³ × 905 kg/m³ | 1.09 kg |
| Energy | 1.086 kg × 1.92 kJ/kg·K × 140 K | about 292 kJ |
| Power absorbed | 292 kJ ÷ 30 s | about 9.7 kW |
| Installed power at 50 % efficiency | 9.7 kW ÷ 0.5 | about 19.5 kW |
| Watt density | 19.5 kW ÷ 0.6 m² | about 32 kW/m² |
Which efficiency to assume
Efficiency here covers everything between the electrical supply and the product: the emitter, the reflector, the distance and angle to the product, how well the product absorbs that wavelength, and heat lost to air and machine parts. As a rough guide:
| Situation | Rough overall efficiency |
|---|---|
| Well-matched emitter with reflector, close to the product, enclosed zone | 0.5–0.7 |
| Typical open conveyor or partly matched wavelength | 0.35–0.5 |
| Large distance, open sides, poorly matched wavelength | 0.2–0.35 |
Guide values for first estimates; confirm with a trial for new processes.
From kilowatts to a heater layout
- Divide the installed power by the rating of the chosen heater to get the number of heaters.
- Compare the watt density (kW/m²) with what the emitter type can deliver; if it is too high, choose a higher-power emitter or a longer heating zone.
- Plan zones across the width and along the length; edges usually need more power than the centre.
- Keep 20–30 % control reserve so the controller can compensate for voltage changes and ageing.
Related products
Frequently asked questions
How do I calculate the kW needed for an infrared heater?
Multiply the mass by the specific heat and the temperature rise, divide by the heating time in seconds, then divide by the overall efficiency (typically 0.35–0.7). The result is the installed power in kW.
What is watt density?
The heater power per area of the heated zone, usually in kW/m². It tells you whether the chosen emitter type can deliver the power in the space available.
Why is drying so energy-intensive?
Evaporating one kilogram of water takes about 2,257 kJ, several times more than heating most materials by a hundred degrees.
How accurate is this calculation?
It is a first estimate. Material properties, part geometry and losses vary, so the result should be confirmed with a trial or by our engineers before a heater bank is ordered.
