Library / Heat transfer and thermal systems / HTS-17
Heat transfer and thermal systems
Surface heat flux after a temperature step
Heat flux entering a semi-infinite solid at time t after its surface steps to Ts from an initial temperature Ti.
- Ref
- HTS-17
- Rev
- v1
- Field
- Heat transfer and thermal systems
- Result
- q″ [W/m²]
For the same temperature step, the surface heat flux is q″ = k(Ts − Ti)/√(παt). It falls as 1/√t, starting infinite at t = 0 for an ideal step and dropping as the heated layer thickens. Positive flux means heat enters the solid. Only evaluate it for t > 0.
Heat flux entering a semi-infinite solid at time t after its surface steps to Ts from an initial temperature Ti. Positive flux enters the solid. The ideal step has a singularity at t = 0.
Calculator
- Flux at four times the time2,821 W/m²
- Mean flux from 0 to t11,280 W/m²
- Energy absorbed per unit area up to t, 2q″t1,128,000 J/m²
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.
- At 100 s the surface flux is 5,642 W/m²; it would be 2,821 W/m² at four times the time.
- The mean flux over the interval is twice the instantaneous value at its end, 11,280 W/m², because most of the heat enters early.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (W/m²) | Local step | Change in result |
|---|---|---|---|---|---|---|
| k Thermal conductivity | 0.1 to 500 W/(m K) | 282.1 | 1,410,000 | 1,410,000 | +10% | +10 % |
| Ts Surface temperature after the step | 250 to 1,000 K | −5,642 | 78,990 | 84,630 | +10% | +70 % |
| Ti Initial temperature | 250 to 1,000 K | −73,340 | 11,280 | 84,630 | +10% | −60 % |
| t Time since the step | 1 to 1,000,000 s | 56.42 | 56,420 | 56,360 | +10% | −4.65 % |
| α Thermal diffusivity | 1e-8 to 0.001 m²/s | 178.4 | 56,420 | 56,240 | +10% | −4.65 % |
Sweep of Time since the step
| Time since the step (s) | Surface heat flux (W/m²) |
|---|---|
| 1 | 56,420 |
| 5.623 | 23,790 |
| 31.62 | 10,030 |
| 177.8 | 4,231 |
| 1,000 | 1,784 |
| 5,623 | 752.4 |
| 31,620 | 317.3 |
| 177,800 | 133.8 |
| 1,000,000 | 56.42 |
Sweep one variable
Variables and how each one contributes
Surface heat flux
What the equation gives you. Everything else feeds it.
Thermal conductivity
The thermal conductivity, in W/(m K). Raises the flux in direct proportion. Raising it by 10 % (to 2.2) moves the result by +10 %, from 5,642 to 6,206 W/m².
Surface temperature after the step
The surface temperature after the step, in K. Raises the flux in direct proportion to the temperature difference. Raising it by 10 % (to 385) moves the result by +70 %, from 5,642 to 9,591 W/m².
Initial temperature
The initial temperature, in K. A higher initial temperature lowers the temperature difference and the flux. Raising it by 10 % (to 330) moves the result by −60 %, from 5,642 to 2,257 W/m².
Thermal diffusivity
The thermal diffusivity, in m²/s. A higher diffusivity thickens the heated layer faster, so the flux falls as 1/√α. Raising it by 10 % (to 1.1×10−6) moves the result by −4.65 %, from 5,642 to 5,379 W/m².
Time since the step
The time since the step, in s. The flux falls as 1/√t as the heated layer deepens. Raising it by 10 % (to 110) moves the result by −4.65 %, from 5,642 to 5,379 W/m².
Worked example with the values above
- Square-root term: √(παt) = √(π × 1×10−6 × 100) = 0.017725 m.
- Temperature step: Ts − Ti = 50 K.
- q″ = 2 × 50 / 0.017725 = 5,641.9 W/m².
Assumptions and limits
- A semi-infinite solid with constant properties, so the back face is not felt over the time considered.
- A one-dimensional geometry, an ideal step or constant flux at the surface and a uniform initial temperature.
Common mistakes
- Using the formula after the thermal wave has reached the back face.
- Using the step solution when the surface is heated by convection, which needs a different solution.
- Entering time in minutes or hours when seconds are needed.
Statistics and uncertainty
Thermophysical properties of soils, concrete and foods vary by 10 % to 50 % with moisture and composition, so early-time results are only as good as the diffusivity and effusivity values.
Where it is used
- Quenching, ground and wall heating, and contact burns.
- Interpreting thin-film and heat-flux sensors.
Maths you need
- Thermal diffusivity: α = k/(ρcp)
- Complementary error function: erfc(x) = 1 − erf(x)
- Thermal effusivity: e = √(kρcp)