Library / Heat transfer and thermal systems / HTS-04
Heat transfer and thermal systems
Plane-wall flux with linearly varying conductivity
Steady conduction through a wall whose conductivity varies as k = kref[1 + b(T − Tref)], using the conductivity at the mean of the two face temperatures.
- Ref
- HTS-04
- Rev
- v1
- Field
- Heat transfer and thermal systems
- Result
- q″ [W/m²]
If conductivity changes linearly with temperature, k = kref[1 + b(T − Tref)], the steady flux through a plane wall is the same as for constant conductivity evaluated at the mean of the two face temperatures. This exact result for a linear law makes it easy to correct a handbook flux for temperature. Conductivity must stay positive across the interval.
Steady conduction through a wall whose conductivity varies as k = kref[1 + b(T − Tref)], using the conductivity at the mean of the two face temperatures. Conductivity must stay positive across the temperature interval.
Calculator
- Conductivity at the mean temperature2.2 W/(m K)
- Flux if conductivity stayed at kref2,000 W/m²
- Wall resistance per unit area, L/k0.04545 m² K/W
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- The wall carries 2,200 W/m², against 2,000 W/m² if its conductivity were fixed at 2 W/(m K).
- The difference of 10 % shows how much a temperature-dependent conductivity matters over a wide temperature span.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (W/m²) | Local step | Change in result |
|---|---|---|---|---|---|---|
| kref Reference conductivity | 0.05 to 500 W/(m K) | 55 | 550,000 | 549,900 | +10% | +10 % |
| L Wall thickness | 0.001 to 1 m | 220 | 220,000 | 219,800 | +10% | −9.09 % |
| T1 Hot-face temperature | 200 to 1,000 K | −1,800 | 23,800 | 25,600 | +10% | +45.1 % |
| T2 Cold-face temperature | 200 to 1,000 K | −21,600 | 4,000 | 25,600 | +10% | −28.1 % |
| Tref Reference temperature | 200 to 400 K | 1,800 | 2,600 | 800 | +10% | −5.45 % |
| b Temperature coefficient of conductivity | 0 to 0.003 1/K | 2,000 | 2,300 | 300 | +10% | +0.909 % |
Sweep of Temperature coefficient of conductivity
| Temperature coefficient of conductivity (1/K) | Heat flux (W/m²) |
|---|---|
| 0 | 2,000 |
| 3.75e-4 | 2,038 |
| 7.5e-4 | 2,075 |
| 0.001125 | 2,112 |
| 0.0015 | 2,150 |
| 0.001875 | 2,188 |
| 0.00225 | 2,225 |
| 0.002625 | 2,263 |
| 0.003 | 2,300 |
Sweep one variable
Variables and how each one contributes
Heat flux
What the equation gives you. Everything else feeds it.
Reference conductivity
The conductivity at the reference temperature, in W/(m K). Raises the flux in direct proportion. Raising it by 10 % (to 2.2) moves the result by +10 %, from 2,200 to 2,420 W/m².
Temperature coefficient of conductivity
The temperature coefficient of conductivity, in 1/K. A larger positive coefficient raises the mean conductivity and hence the flux. Raising it by 10 % (to 0.0022) moves the result by +0.909 %, from 2,200 to 2,220 W/m².
Reference temperature
The reference temperature of kref, in K. A higher reference temperature lowers the mean temperature above it, so the flux falls when b is positive. Raising it by 10 % (to 330) moves the result by −5.45 %, from 2,200 to 2,080 W/m².
Hot-face temperature
The hot-face temperature, in K. Raises the flux through both the temperature difference and the mean conductivity. Raising it by 10 % (to 440) moves the result by +45.1 %, from 2,200 to 3,192 W/m².
Cold-face temperature
The cold-face temperature, in K. Raises the mean conductivity but lowers the temperature difference. Raising it by 10 % (to 330) moves the result by −28.1 %, from 2,200 to 1,582 W/m².
Wall thickness
The wall thickness, in m. A thicker wall lowers the flux in inverse proportion. Raising it by 10 % (to 0.11) moves the result by −9.09 %, from 2,200 to 2,000 W/m².
Worked example with the values above
- Mean temperature: (400 + 300)/2 = 350 K; k = 2 × (1 + 0.002 × 50) = 2.2 W/(m K).
- Temperature gradient: (T₁ − T₂)/L = 1,000 K/m.
- q″ = 2.2 × 1,000 = 2,200 W/m².
Assumptions and limits
- Steady state, constant properties unless the page states a temperature law, and no internal heat generation.
- One-dimensional conduction in the stated geometry, with no edge or end losses.
Common mistakes
- Using diameters where radii are needed.
- Using °C in a formula that needs an absolute temperature, or leaving a temperature difference out of the heat rate.
- Applying a plane-wall formula to a curved wall when the thickness is not small compared with the radius.
Statistics and uncertainty
Thermal conductivities are rarely known better than 5 % to 10 %, and contact resistances can vary by a factor of two between nominally identical joints. Treat results as estimates unless the inputs were measured.
Where it is used
- Insulation sizing for vessels, pipes and walls.
- Estimating interface and layer temperatures in composite constructions.
Maths you need
- Fourier’s law: q″ = −k dT/dx
- Thermal resistance: R = ΔT/Q, in K/W
- Series and parallel paths: resistances add in series; conductances add in parallel