Heat Conduction Calculator
Calculate steady-state heat conduction, Fourier's law heat transfer, thermal resistance, heat flux, required thermal conductivity, required thickness, and composite wall conduction. It is intended for generic solid conduction when geometry, material conductivity, and boundary surface temperatures are known.
Engineering tool
Heat Conduction Calculator
Calculate steady-state heat conduction through plane, cylindrical, spherical, and composite solid geometries using Fourier's law.
Use conductivity in the heat-flow direction for anisotropic materials.
Plane-wall heat-flow length, not cylinder axial length.
The calculator reports magnitude and direction from the hotter side.
Result console
- Heat Transfer Rate
- 32000W
- Thermal Resistance
- 0.0025K/W
- Heat Flux
- 1600000W/m²
- Temperature Gradient
- 8000K/m
- Temperature Difference
- 80K
- Direction
- Side 1 -> Side 2
Steady-state conduction only. Surface temperatures must already include any convection, radiation, contact resistance, or heat-source behavior outside the solid conduction path.
Formula
Formula reference
Steady-state heat conduction formulas
The calculator reports heat-flow magnitude as a non-negative value and separately identifies direction from hotter surface to colder surface.
q'' = -k dT/dxPlane wall: Qdot = kAΔT/LPlane wall resistance: Rcond = L/(kA)Heat flux: q'' = Qdot/A = kΔT/LCylindrical wall: Rcond = ln(r2/r1)/(2πkL)Spherical wall: Rcond = (r2-r1)/(4πkr1r2)Composite wall: Rtotal = Σ Li/(kiA)Layer drop: ΔTi = Qdot RiVariable definitions
- k
- thermal conductivity, W/(m·K)
- Qdot
- heat-transfer rate, W
- q''
- heat flux, W/m²
- A
- heat-transfer area, m²
- L
- plane-wall thickness or cylinder axial length depending on formula
- r1, r2
- inner and outer radii
Heat Conduction Formula Audit
| Fourier Law | q'' = -k dT/dx; negative sign indicates heat flows toward lower temperature. |
|---|---|
| Heat Flow Sign Convention | UI reports non-negative magnitude and separate direction. |
| Plane Wall Formula | Qdot = kAΔT/L. |
| Plane Wall Resistance | Rcond = L/(kA). |
| Heat Flux Formula | q'' = Qdot/A = kΔT/L. |
| Temperature Gradient Convention | Magnitude is |dT/dx| = ΔT/L; signed gradients require a coordinate convention. |
| Required Conductivity Formula | k = Qdot L/(AΔT). |
| Required Thickness Formula | L = kAΔT/Qdot. |
| Cylindrical Resistance Formula | Rcond = ln(r2/r1)/(2πkL). |
| Cylindrical Heat Transfer Formula | Qdot = 2πkLΔT/ln(r2/r1). |
| Cylindrical Log Convention | Natural logarithm ln, not log10. |
| Cylinder Radius Convention | r1 and r2 are radii, not diameters. |
| Cylinder Length Convention | L is axial cylinder length, not wall thickness. |
| Spherical Resistance Formula | Rcond = (r2-r1)/(4πkr1r2). |
| Spherical Heat Transfer Formula | Qdot = 4πkr1r2ΔT/(r2-r1). |
| Spherical Numerical Form | (r2-r1)/(4πkr1r2) avoids thin-wall cancellation. |
| Composite Layer Resistance | Ri = Li/(kiA). |
| Composite Total Resistance | Rtotal = ΣRi for series plane layers. |
| Composite Heat Transfer | Qdot = ΔT/Rtotal. |
| Composite Temperature Drop | ΔTi = QdotRi. |
| Composite Interface Temperature | Tinterface,n = Thot - QdotΣRi. |
| Temperature Difference Convention | Δ1°C = 1K = 1.8°F difference. |
| Length Units | m, cm, mm, µm, mil, and in convert to meters. |
| Area Units | m², cm², mm², and in² convert to square meters. |
| Thermal Conductivity Units | W/(m·K). |
| Steady-State Boundary | No warm-up, cooldown, or thermal capacitance. |
| Constant-k Boundary | Thermal conductivity is assumed constant. |
| Anisotropic Material Boundary | Use conductivity in the heat-flow direction. |
| Internal Heat Generation Boundary | No volumetric heat generation q''' model. |
| Convection Boundary | No Q = hAΔT convection model. |
| Radiation Boundary | No Stefan-Boltzmann radiation model. |
| THM-007 Scope Separation | TIM selection and bond-line workflows stay in THM-007. |
| THM-009 Scope Separation | Transient thermal RC belongs to THM-009. |
| THM-010 Scope Separation | Convection heat transfer belongs to THM-010. |
Worked Examples
| Case | Calculation | Result |
|---|---|---|
| Plane wall Q | k=200, L=.01m, A=.02m², ΔT=80K | Q=32000W |
| Plane wall R | .01/(200×.02) | R=.0025K/W |
| Plane wall heat flux | 32000/.02 | q''=1.6e6W/m² |
| Plane wall gradient | 80/.01 | |dT/dx|=8000K/m |
| Required k round-trip | 32000×.01/(.02×80) | k=200W/(m·K) |
| Required L round-trip | 200×.02×80/32000 | L=.01m |
| Double k | Q proportional to k | Q doubles |
| Double area | Q proportional to A | Q doubles |
| Double thickness | Q inversely proportional to L | Q halves |
| Double ΔT | Q proportional to ΔT | Q doubles |
| Zero ΔT | same surface temperatures | Q=0 |
| Swapped temperatures | 20°C and 100°C | same magnitude, reversed direction |
| Cylinder Q | k=10, r1=.01, r2=.02, L=1, ΔT=100 | Q≈9064.72W |
| Cylinder identity | R×Q | ΔT=100K |
| Cylinder invalid radii | r2<=r1 | rejected |
| Cylinder zero radius | r1=0 | rejected |
| Cylinder natural log | ln(2), not log10(2) | correct radial resistance |
| Thin cylinder wall | r2 close to r1 | finite positive result |
| Sphere Q | k=10, r1=.01, r2=.02, ΔT=100 | Q≈251.327W |
| Sphere identity | R×Q | ΔT=100K |
| Sphere invalid radii | r2<=r1 | rejected |
| Composite critical wall | R1=.01, R2=.04 | Rtotal=.05K/W |
| Composite heat flow | 60K/.05K/W | Q=1200W |
| Composite interface | 100°C - 12K | Tinterface=88°C |
| Composite layer sum | 12K + 48K | 60K |
| Composite invalid k | one layer k=0 | rejected |
| Composite invalid thickness | one layer L=0 | rejected |
| Length conversion | 1cm | .01m |
| Length conversion | 1mm | .001m |
| Length conversion | 1in | .0254m |
| Area conversion | 1cm² | 1e-4m² |
| Area conversion | 1in² | .00064516m² |
| Temperature difference | Δ10°C | 10K |
| Temperature difference | Δ10°C | 18°F difference |
| Invalid number | NaN | rejected |
| Invalid number | Infinity | rejected |
Engineering Notes
- Fourier's law relates heat flux to the negative temperature gradient.
- Heat naturally flows from higher temperature toward lower temperature.
- Heat-transfer rate is measured in watts; heat flux is watts per square metre.
- Plane-wall resistance is L/(kA).
- Cylindrical radial resistance is ln(r2/r1)/(2πkL).
- The cylinder L parameter is axial length, not wall thickness.
- Spherical radial resistance differs from both plane and cylindrical geometry.
- Composite plane-wall resistances add in series only under the stated assumptions.
- All composite layers carry the same steady-state heat rate.
- Use conductivity in the heat-flow direction for anisotropic materials.
- The calculator assumes known surface temperatures and constant k.
- Critical designs should still account for contact resistance, spreading resistance, convection, radiation, and measurement.
Common Mistakes
- Using the wrong geometry formula.
- Treating a pipe wall as a plane wall when radial effects matter.
- Entering diameter where radius is required.
- Using cylinder axial length as wall thickness.
- Using log10 instead of natural log for cylindrical conduction.
- Confusing heat flux W/m² with heat-transfer rate W.
- Using absolute Fahrenheit conversion for a temperature difference.
- Entering mm as m or cm² as m².
- Ignoring heat-transfer area.
- Averaging composite wall conductivity instead of adding layer resistances.
- Assuming each composite layer has the same temperature drop.
- Treating steady-state results as transient warm-up behavior.
- Adding convection into thermal conductivity.
- Ignoring anisotropic conductivity in PCB, graphite, or composite materials.
Support reference
FAQ
What is heat conduction?
Heat conduction is heat transfer through a solid or stationary material caused by a temperature gradient.
What is Fourier's law of heat conduction?
Fourier's law relates heat flux to the negative temperature gradient: q'' = -k dT/dx. The negative sign indicates heat flows toward lower temperature.
How do I calculate heat transfer through a wall?
For a plane wall with known surface temperatures, use Qdot = kAΔT/L, where k is conductivity, A is area, ΔT is temperature difference, and L is thickness.
How do I calculate thermal resistance of a plane wall?
Use Rcond = L/(kA). Heat transfer is then Qdot = ΔT/Rcond.
What is thermal conductivity?
Thermal conductivity is a material property that describes heat conduction capability. Its SI unit is W/(m·K).
What is the difference between heat transfer rate and heat flux?
Heat transfer rate is total heat flow in watts. Heat flux is heat flow per unit area in W/m².
How do I calculate heat flux?
For plane-wall conduction, heat flux magnitude is q'' = Qdot/A = kΔT/L.
How do I calculate temperature gradient?
For a plane wall with uniform conductivity and known surface temperatures, the gradient magnitude is |dT/dx| = ΔT/L.
How do I calculate heat conduction through a pipe wall?
Use the cylindrical radial conduction formula Qdot = 2πkLΔT / ln(r2/r1), where L is axial length.
Why does cylindrical conduction use a logarithm?
Cylindrical heat-flow area changes with radius, so radial resistance integrates to a natural logarithm of the radius ratio.
Should cylindrical conduction use ln or log10?
Use the natural logarithm ln(r2/r1), which is Math.log in JavaScript. Do not use log10.
How do I calculate spherical heat conduction?
Use Rcond = (r2-r1)/(4πkr1r2), then Qdot = ΔT/Rcond.
How do I calculate heat flow through a composite wall?
Calculate each layer resistance Ri = Li/(kiA), add the resistances in series, then divide total ΔT by total resistance.
How do thermal resistances add in a composite wall?
For series plane layers with common area, thermal resistances add directly: Rtotal = R1 + R2 + ... .
How do I calculate interface temperatures in a composite wall?
Starting from the hot side, subtract each layer's temperature drop, ΔTi = QdotRi, from the previous interface temperature.
Does the conduction equation use Celsius or Kelvin?
The formulas use temperature difference. A 1°C difference has the same numerical size as 1 K.
Is a 1°C temperature difference equal to 1K?
Yes. Temperature intervals in °C and K are numerically equal. Fahrenheit temperature differences use a 9/5 factor.
How do I calculate required thermal conductivity?
For a plane wall, use krequired = Qdot L/(AΔT). Positive heat flow at zero ΔT has no finite solution.
How do I calculate required wall thickness?
For a plane wall, use L = kAΔT/Qdot. Zero requested heat flow does not define a unique finite thickness.
Does this calculator include convection?
No. THM-008 assumes known surface temperatures. Convection from ambient to a surface belongs to a convection calculator or a thermal network.
Does this calculator include radiation?
No. Stefan-Boltzmann radiation is outside this steady-state conduction calculator.
What is the difference between THM-007 TIM and THM-008 heat conduction calculators?
THM-007 is TIM-specific and focuses on bond-line thickness, contact area, thermal impedance, and interface behavior. THM-008 is generic solid conduction across plane, cylindrical, spherical, and composite geometries.
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