Library / Heat transfer and thermal systems / HTS-18
Heat transfer and thermal systems
Cumulative energy absorbed after a temperature step
Energy per unit area taken up by a semi-infinite solid up to time t after its surface steps to Ts.
- Ref
- HTS-18
- Rev
- v1
- Field
- Heat transfer and thermal systems
- Result
- E/A [J/m²]
The energy absorbed per unit area up to time t is the integral of the surface flux, E/A = 2k(Ts − Ti)√[t/(πα)]. It grows as √t, so quadrupling the time doubles the energy. The sign reverses for cooling, when Ts is below Ti, and the solid is assumed semi-infinite throughout.
Energy per unit area taken up by a semi-infinite solid up to time t after its surface steps to Ts. It is the time integral of the surface flux, so it grows as the square root of time.
Calculator
- Mean flux over the interval, E/(At)11,280 W/m²
- Energy after four times as long, 2E/A2,257,000 J/m²
- Thermal penetration length, √(αt)0.01 m
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.
- After 100 s the solid has taken up 1,128,000 J per square metre, 1.128 MJ/m².
- The heat goes into a layer about 0.01 m deep; four times the time would give 2,257,000 J/m², only double.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (J/m²) | Local step | Change in result |
|---|---|---|---|---|---|---|
| k Thermal conductivity | 0.1 to 500 W/(m K) | 56,420 | 2.821e8 | 2.82e8 | +10% | +10 % |
| t Time since the step | 1 to 1,000,000 s | 112,800 | 1.128e8 | 1.127e8 | +10% | +4.88 % |
| Ts Surface temperature after the step | 250 to 1,000 K | −1,128,000 | 1.58e7 | 1.693e7 | +10% | +70 % |
| Ti Initial temperature | 250 to 1,000 K | −1.467e7 | 2,257,000 | 1.693e7 | +10% | −60 % |
| α Thermal diffusivity | 1e-8 to 0.001 m²/s | 35,680 | 1.128e7 | 1.125e7 | +10% | −4.65 % |
Sweep of Time since the step
| Time since the step (s) | Energy absorbed per unit area (J/m²) |
|---|---|
| 1 | 112,800 |
| 5.623 | 267,600 |
| 31.62 | 634,500 |
| 177.8 | 1,505,000 |
| 1,000 | 3,568,000 |
| 5,623 | 8,462,000 |
| 31,620 | 2.007e7 |
| 177,800 | 4.758e7 |
| 1,000,000 | 1.128e8 |
Sweep one variable
Variables and how each one contributes
Energy absorbed per unit area
What the equation gives you. Everything else feeds it.
Thermal conductivity
The thermal conductivity, in W/(m K). Raises the absorbed energy in direct proportion. Raising it by 10 % (to 2.2) moves the result by +10 %, from 1,128,000 to 1,241,000 J/m².
Surface temperature after the step
The surface temperature after the step, in K. Raises the energy in direct proportion to the temperature difference. Raising it by 10 % (to 385) moves the result by +70 %, from 1,128,000 to 1,918,000 J/m².
Initial temperature
The initial temperature, in K. A higher initial temperature lowers the energy absorbed. Raising it by 10 % (to 330) moves the result by −60 %, from 1,128,000 to 451,400 J/m².
Time since the step
The time since the step, in s. Raises the energy as the square root of time. Raising it by 10 % (to 110) moves the result by +4.88 %, from 1,128,000 to 1,183,000 J/m².
Thermal diffusivity
The thermal diffusivity, in m²/s. A higher diffusivity lowers the energy for fixed k, as the heated layer is then thicker but warmer less deeply. Raising it by 10 % (to 1.1×10−6) moves the result by −4.65 %, from 1,128,000 to 1,076,000 J/m².
Worked example with the values above
- Time term: √(t/(πα)) = √(100 / (π × 1×10−6)) = 5,641.9 s/m.
- Factor 2k(Ts − Ti) = 2 × 2 × 50 = 200 W/m.
- E/A = 1,128,400 J/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)