Why doesn't the duty fraction reduce the heat on the part?

Because the part is only there while the lamp is on. Duty describes what fraction of the hour the lamp emits; a part passing underneath receives the full flux for the whole of its pass regardless. Averaging over an hour the part did not experience produces a number that describes the room, not the substrate, and it is smaller, which is the direction that gets substrates damaged.

What is the IR fraction for a typical lamp?

It depends far more on the reflector than the lamp. A dichroic reflector is built to transmit infrared out the back and reflect UV forward, so a much smaller share reaches the part; a standard aluminium reflector sends both forward. Manufacturers publish figures for their own systems, and where a substrate is marginal it is worth asking for them rather than assuming.

How do I turn this flux into a temperature?

You cannot, from this page alone. Temperature depends on the substrate's thermal mass and conductivity, the dwell time, what it is sitting on, and what cooling is present. A film on a chilled roll and the same film on a still belt reach very different temperatures under identical flux. Use the flux to compare configurations and to flag the marginal ones, then measure with a thermocouple or an IR camera.

My substrate distorts but the cure is fine. What should I change?

Attack the heat rather than the dose, because they are separable. A dichroic reflector, a filter, greater lamp height, or spreading the output over more area all reduce flux while leaving useful UV. Slowing the line does the opposite. It increases the heat a given part absorbs. And a UV LED array is the extreme version of the same trade: it emits a narrow band with far less infrared, which is often the real reason a retrofit fixes a heat-sensitive job.

Does the exhaust take much of this away?

A well-designed one takes a great deal, and none of it is modelled here. This page prices the radiant load arriving at the part, which the exhaust cannot intercept; the exhaust's job is the convective and housing heat, and it is why a UV cell's contribution to the room is usually far below its electrical input. That is also why the average row on this page overstates the room load if the system is ducted outside.

Is the peak flux the worst case?

Not quite. It is the average over the irradiated area. Real lamps peak in the middle of their profile, so the hottest point on a part is above this figure, sometimes considerably. Where a substrate is marginal, treat this as the average and go and find the actual maximum with a probe or a thermal camera.

UV Curing calculator

UV Heat Load Calculator: Peak Flux, Not the Hourly Average

Work out the thermal load a UV lamp puts on a part, separating the two questions the earlier version ran together. Enter the lamp's electrical input, the fraction of it arriving at the part as heat, the lamp-on duty fraction, and the area the output lands on. The calculator returns the peak flux the substrate sees while it is under the lamp, the number that governs scorching and distortion, alongside the hourly average, which is what the building's cooling has to remove. A duty fraction below one reduces the second and does nothing whatever to the first.

What this calculator does

Separate the heat flux a substrate actually sees under the lamp from the hourly average that governs the room, and express the first per unit area.
Use it for checking whether a heat-sensitive substrate survives a proposed lamp, sizing ventilation and building cooling for a UV cell, deciding between a dichroic and a standard reflector, diagnosing web distortion that appears only under the lamp, explaining why a lamp that cures fine damages one stock and not another.
Separate the heat flux a substrate actually sees under the lamp from the hourly average that governs the room, and express the first per unit area.

Formula used

Heat to the part while under the lamp (kW) = lamp input × IR fraction
Peak heat flux (BTU/hr·ft²) = heat to the part × 3412.142 ÷ irradiated area
Average heat to the part (kW) = heat to the part × duty fraction
Average heat load (BTU/hr) = average heat to the part × 3412.142
Applied fraction × duty = IR fraction × duty fraction

Inputs explained

Lamp electrical input power: Electrical input to the lamp while struck, at the power setting in use.
IR fraction reaching the part: Share of input power arriving at the part as heat. Depends on reflector type. A dichroic reflector passes most IR out the back, a standard aluminium one sends it forward with the UV.
Lamp-on duty fraction per hour: Fraction of the hour the lamp is emitting onto parts. Affects the room's heat load and NOT the flux a part sees during its pass.
Irradiated area: Area the lamp's output lands on: illuminated width times cure-zone length. This is what turns kilowatts into a flux something can actually scorch at.

How to use the result

Best suited to checking whether a heat-sensitive substrate survives a proposed lamp, sizing ventilation and building cooling for a UV cell, deciding between a dichroic and a standard reflector, diagnosing web distortion that appears only under the lamp, explaining why a lamp that cures fine damages one stock and not another.
Gives a flux, not a temperature. Reaching a substrate temperature needs the material's thermal mass, its conductivity, the dwell time and what it sits on. A thin film on a chilled roll and the same film on a still belt behave completely differently. Assumes uniform distribution across the irradiated area. Real profiles peak in the middle, so the hottest point exceeds this figure. Ignores heat carried away by the exhaust, which on a well-ducted system is substantial. Does not model the reflector's own re-radiation or the housing temperature. Says nothing about cumulative heating over multiple passes, which matters for a part that recirculates.

Common questions

Why doesn't the duty fraction reduce the heat on the part? Because the part is only there while the lamp is on. Duty describes what fraction of the hour the lamp emits; a part passing underneath receives the full flux for the whole of its pass regardless. Averaging over an hour the part did not experience produces a number that describes the room, not the substrate, and it is smaller, which is the direction that gets substrates damaged.
What is the IR fraction for a typical lamp? It depends far more on the reflector than the lamp. A dichroic reflector is built to transmit infrared out the back and reflect UV forward, so a much smaller share reaches the part; a standard aluminium reflector sends both forward. Manufacturers publish figures for their own systems, and where a substrate is marginal it is worth asking for them rather than assuming.
How do I turn this flux into a temperature? You cannot, from this page alone. Temperature depends on the substrate's thermal mass and conductivity, the dwell time, what it is sitting on, and what cooling is present. A film on a chilled roll and the same film on a still belt reach very different temperatures under identical flux. Use the flux to compare configurations and to flag the marginal ones, then measure with a thermocouple or an IR camera.
My substrate distorts but the cure is fine. What should I change? Attack the heat rather than the dose, because they are separable. A dichroic reflector, a filter, greater lamp height, or spreading the output over more area all reduce flux while leaving useful UV. Slowing the line does the opposite. It increases the heat a given part absorbs. And a UV LED array is the extreme version of the same trade: it emits a narrow band with far less infrared, which is often the real reason a retrofit fixes a heat-sensitive job.
Does the exhaust take much of this away? A well-designed one takes a great deal, and none of it is modelled here. This page prices the radiant load arriving at the part, which the exhaust cannot intercept; the exhaust's job is the convective and housing heat, and it is why a UV cell's contribution to the room is usually far below its electrical input. That is also why the average row on this page overstates the room load if the system is ducted outside.

Last reviewed 2026-08-25.