Heat Transfer Calculator
Plane and curved walls, two-surface radiation and electronics thermal resistance.
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- Free preview: the rules, dates and checks that apply to your figures, with every computed amount hidden.
- Locked until you unlock it: download and copy.
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Working
Assumptions and sources
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Results are estimates from the formulas shown, not a professional design or certification. Have a qualified engineer verify anything safety-critical (structures, electrical installations, gas or pressure systems).
About the Heat Transfer Calculator
Calculate steady heat transfer through plane, composite, cylindrical or spherical walls, convection to a fluid, or radiation to surroundings or a second diffuse-gray surface. Combine entered thermal resistances in series or parallel, find a critical insulation radius, or estimate an electronics junction temperature from package, interface and heat-sink resistances. The free preview masks calculated figures; a Pro pass opens the working, copy, CSV and PDF.
How to use it
- Choose the physical model; only the inputs used by that model are shown.
- Enter actual dimensions, material properties and temperatures in the labelled units; radiation temperatures are converted internally to kelvins.
- For a composite wall enter layer name, thickness and conductivity on each CSV row; for a resistance network enter positive resistances and its arrangement.
- For electronics enter heat dissipation, ambient and maximum junction temperatures, and the three actual thermal resistances.
- Check assumptions and warnings in the free preview, then unlock with a Pro pass to copy the result or download CSV and PDF.
Examples
10 m2 wall, thickness 0.1 m, k=0.04 W/m/K, temperature difference 20 K
Resistance 0.25 K/W and heat transfer 80 W.
10 W, ambient 25 degC, thetaJC=1, thetaCS=0.5, thetaSA=4 K/W
Junction estimate 80 degC; at a 100 degC junction limit, the maximum thetaSA is 6 K/W.
Equations and assumptions
Q = ΔT/R. Plane R = L/(kA); cylinder R = ln(ro/ri)/(2πkL); sphere R = (1/ri−1/ro)/(4πk). Convection Q = hA(Tsurface−Tfluid). Radiation to a large surrounding uses εσA(T1^4−T2^4), with σ=5.670374419×10^-8 W/m2/K4 and temperatures in K. Two-surface exchange includes both surface resistances and A1F12, satisfying reciprocity. Series resistances add; parallel conductances add. Critical radius is k/h for a cylinder and 2k/h for a sphere. Tjunction = Tambient + power(θJC+θCS+θSA).
Sources
Limitations
- Estimate — not a professional design. Ideal steady geometry, isothermal boundaries and entered properties; no transient storage or automatic convection correlation.
- Composite mode is for plane layers. Cylindrical and spherical modes are one homogeneous shell each; complex contact and spreading paths need an appropriate network.
- Two-surface radiation assumes diffuse-gray surfaces and an entered view factor. Radiating temperatures must be above absolute zero.
- Use the actual device and heat-sink datasheets and appropriate thermal metrics. A junction-to-ambient test-board value is not interchangeable with junction-to-case or a heat-sink rating.
- Critical insulation radius is a heat-transfer extremum, not a recommended insulation thickness.
Privacy
On this device; its files download once
All entered dimensions, temperatures, labels and schedules stay on your device. PDF fonts and calculation code can download from MySmartCoPilot; inputs are never sent with those downloads.
Frequently asked questions
What do I get without a pass?
Without a pass, Heat Transfer Calculator shows the rules, dates and checks that apply to your figures, with every computed amount hidden. Until you unlock it, the result can’t be downloaded or copied. A Pro, Premium or Ultimate pass, a one-time payment that never renews, unlocks the full result. The pricing page lists the passes and their prices.
Why can the heat-transfer result be negative?
The sign follows the entered first temperature minus the second. A negative result means heat flows opposite that declared direction.
Can I use Celsius in the radiation inputs?
Yes. The calculation converts both temperatures to kelvins before raising them to the fourth power; temperature differences for conduction and convection have the same magnitude in Celsius and kelvins.
What if the required heat-sink resistance is negative?
The package and interface path already consume more temperature rise than the limit permits. No nonnegative sink-to-ambient resistance can meet that entered limit.