Conductive heat flux calculator
A wall 0.2 m thick with a conductivity of 0.8 W/m·K and a 25 K difference across it conducts 100 W/m². You also get the temperature gradient of 125 K/m and the resistance per square metre. Free in your browser.
- Free
- Runs in your browser
- No sign-up
- Your data stays on your device
Run — free
Worked example
- Conductivity
- 0.8
- Thickness
- 0.2
- Delta t
- 25
- Heat flux
- 100
- Q
- 100
- Gradient
- 125
- Equation
- q = 0.8 · 25 / 0.2 = 100
Area is not an input because the result is already a flux, not a heat-transfer rate.
What Fourier's law says about conductive heat flux
Thermal conductivity k is a material property — a few tenths of a watt per metre-kelvin for common masonry, far higher for metals, far lower for good insulators — so the same geometry and Delta T produce very different leaks. Thickness L is the path length of conduction through the solid.
Temperature difference Delta T is usually written as the hot-side surface temperature minus the cold-side surface temperature; a positive Delta T with positive k yields positive q in the chosen sign convention, while a negative Delta T flips the reported flow direction without changing the physics magnitude.
Searchers who type conductive heat flux calculator, Fourier wall flux, or q equals k Delta T over L already have those three numbers and want watts per square metre, not a heat rate that still needs an area and not a unit converter that only changes W/m² into BTU per hour per square foot.
Classroom and automated-grader paths care about domain checks as much as the arithmetic: thickness must be strictly positive because the formula divides by L, and a zero or negative thickness is almost always a data-entry or modelling mistake. Zero conductivity is allowed and yields zero flux, which matches a perfect-insulator idealization, and in that case the area-specific resistance L over k is omitted rather than returned as infinity.
The check path recomputes k times Delta T over L with the same fixed rounding used for published fields, so a successful response always ships with an equation string you can assert in tests or lab notebooks.
Who uses a conductive heat flux check
Heat-transfer instructors need a deterministic oracle that reproduces the plane-wall Fourier flux identity so lab write-ups can compare a hand calculation with a machine result without spreadsheet rounding drift, and so a zero thickness is marked wrong instead of quietly returning infinity. Building-physics notebooks and HVAC takeoff scripts use the same three numbers when they already have a surface-to-surface temperature drop and want q in watts per square metre before they multiply by area, not a total heat rate that hides the flux. Failed validation does not consume balance.
This landing is not the heat conduction calculator that returns Q equals k A Delta T over L and requires an area. It is not the heat-flux density converter, which only changes W/m² into other flux units. It is not the overall heat transfer coefficient calculator, which stacks convective films and several L over k layers into a U-value.
Keep units SI and consistent, stay inside the published magnitude bounds, and you get a reproducible plane-wall flux for homework, LMS plugins, and edge workers without standing up a full energy-model engine of your own.
What you can do with it
Pre-flight wall heat-flux estimates in engineering scripts
Pipelines compute steady conduction flux through a known slab from surface temperatures before multiplying by area or comparing against HVAC capacity.
FAQ
What formula does this conductive heat flux calculator use?
Steady one-dimensional Fourier conduction through a plane wall: q = k · ΔT / L, where k is thermal conductivity, ΔT is temperature difference, and L is thickness. The result is heat flux in W/m², not a total heat rate.
Why is area not required?
Heat flux is already per unit area. If you need the total heat-transfer rate Q in watts, multiply q by the face area, or use the heat conduction calculator that takes k, A, L, and ΔT.
Can ΔT be negative?
Yes. A negative temperature difference yields a negative heat flux, which means heat flows opposite the direction you chose as positive.
How do I get the total heat rate through the wall?
Multiply the flux by the wall area. This calculator returns watts per square metre on purpose, so the same result works for any wall size and the area stays your own choice.
What happens if the conductivity is zero?
The flux is zero, which matches a perfect insulator. In that case the resistance per square metre is left out rather than shown as infinite. Thickness, on the other hand, must always be greater than zero.
Related tools
For developers — API access
Everything on this page is available programmatically. This section is for teams who want to wire it into their own systems; everyone else can just use the tool above.
API endpoint
Prefer to automate it? One authenticated POST creates the task; the result comes back by webhook or a signed link. The same capability also runs here on the web, by email and from Telegram — and soon from our app too.
Call it from your stack
curl -X POST https://api.kit.forhosting.com/eng/heat-flux \
-H "Authorization: Bearer $KIT_KEY" \
-H "Content-Type: application/json" \
-d '{"conductivity":0.8,"thickness":0.2,"delta_t":25}'const res = await fetch("https://api.kit.forhosting.com/eng/heat-flux", {
method: "POST",
headers: {
"Authorization": `Bearer ${process.env.KIT_KEY}`,
"Content-Type": "application/json"
},
body: JSON.stringify({
"conductivity": 0.8,
"thickness": 0.2,
"delta_t": 25
})
});
const { task_id } = await res.json();import os, requests
res = requests.post(
"https://api.kit.forhosting.com/eng/heat-flux",
headers={"Authorization": f"Bearer {os.environ['KIT_KEY']}"},
json={
"conductivity": 0.8,
"thickness": 0.2,
"delta_t": 25
},
)
task_id = res.json()["task_id"]<?php
$res = file_get_contents("https://api.kit.forhosting.com/eng/heat-flux", false, stream_context_create([
"http" => [
"method" => "POST",
"header" => "Authorization: Bearer " . getenv("KIT_KEY") . "\r\nContent-Type: application/json",
"content" => '{"conductivity":0.8,"thickness":0.2,"delta_t":25}',
],
]));
$task = json_decode($res, true);body := bytes.NewBufferString(`{"conductivity":0.8,"thickness":0.2,"delta_t":25}`)
req, _ := http.NewRequest("POST", "https://api.kit.forhosting.com/eng/heat-flux", body)
req.Header.Set("Authorization", "Bearer "+os.Getenv("KIT_KEY"))
req.Header.Set("Content-Type", "application/json")
res, _ := http.DefaultClient.Do(req)Example request
{
"conductivity": 0.8,
"thickness": 0.2,
"delta_t": 25
}Example response
{
"task_id": "tsk_a1b2c3d4e5f6a1b2c3d4e5f6",
"type": "eng.heat_flux",
"status": "queued",
"_links": {
"result": "/tasks/tsk_…/result"
}
}The API is asynchronous: the call returns a task_id immediately and the result arrives by webhook. Polling is capped at 1 req/s per task.
Pricing
Published price — no tokens, no invented credits. A failed task is never charged.
Limits
max_abs | 1000000000000 |
Errors
| HTTP | Code | Meaning |
|---|---|---|
401 | unauthorized | Missing or invalid API key. |
402 | insufficient_balance | Your balance doesn't cover the task price. |
404 | unknown_type | That task type doesn't exist. |
429 | rate_limited | Too many requests. Use the webhook instead of polling. |
Technical notes
Fourier's law for one-dimensional steady conduction through a homogeneous plane wall says the conductive heat flux per unit area equals thermal conductivity times the temperature difference divided by thickness: q equals k times Delta T over L. This conductive heat flux calculator accepts those three quantities and returns the flux in watts per square metre, the area-specific conduction resistance L over k, and the temperature gradient across the slab. Zero or negative thickness is rejected with invalid_input so a missing depth never becomes infinity.
What Fourier's law says about conductive heat flux
Fourier's law for steady conduction through a flat slab of constant thermal conductivity states that the heat flux per unit area q is proportional to conductivity and temperature difference and inversely proportional to thickness: q equals k times Delta T divided by L.
Input contract, response shape, and error behaviour
Conductivity must be greater than or equal to zero; thickness must be strictly greater than zero. Units are reported as watts per metre-kelvin, metres, kelvin, watts per square metre, kelvin per metre, and square metre-kelvin per watt. Optional output keys are omitted entirely when they do not apply, never returned as null. POST a JSON object with three required finite numbers. conductivity is the thermal conductivity k in watts per metre-kelvin; aliases k, thermal_conductivity, and lambda are accepted. thickness is the wall thickness L in metres; aliases l, L, length, delta_x, and dx work the same way. delta_t is the temperature difference in kelvin or Celsius difference, which have the same magnitude; aliases dt, dT, temperature_difference, deltaT, and temp_diff are accepted. Every required magnitude must be a finite number; missing fields, empty strings, and non-numeric text fail with invalid_input naming the field, and those failures are not billed. Absolute values must stay within the published limit of one trillion so the product, the quotient, and the gradient remain finite in IEEE double precision on the Worker and in the browser. On success the body echoes the rounded conductivity, thickness, and delta_t, sets heat_flux and q to the same rounded flux, sets gradient to Delta T over L, and when conductivity is positive sets r_conduction to L over k. formula restates q equals k times Delta T over L; equation substitutes the numbers; method is always fourier_plane_wall. Pricing is a flat $0.002 per successful request with unit item.
Where a conductive heat flux API earns its keep
Envelope-compliance checkers can pin heat_flux in tests after they change a layer thickness or swap a conductivity.
Integration use cases
Homework checks for Fourier flux problems
Students enter textbook k, L, and ΔT, then compare hand-written q = k ΔT / L and the substituted equation against a deterministic API or browser result before submitting.
Unit tests for student thermal solvers
Generate known-good q = k ΔT / L cases plus deliberate zero-thickness rejections for coding kata expected values and continuous-integration fixtures.
Technical questions
What happens when thickness is zero?
The formula would divide by zero. This tool returns invalid_input instead of infinity or a silent NaN, and the failed task is not billed.
What does the conductive heat flux calculator cost?
The browser tool is free. API requests use prepaid KIT balance at $0.002 per successful call; invalid input is not billed.