Semi-Infinite Solid
Evaluate transient heat penetration, temperature profiles, and surface heat rates in semi-infinite media.
Semi-Infinite Model
Model early-stage transient temperatures prior to thermal wave boundary reflection:
- Boundary Type: Constant surface Temp ($T_s$), constant surface Flux ($q"_s$), or Convection fluid ($T_\infty, h$).
- Thermal Diffusivity ($\alpha$): Material thermal property ($k/\rho C_p$).
- Probe Coordinate ($x$): Depth location to evaluate.
Configuration
Results & Visualization
Thermal Penetration Heat Field Map
Transient Profile: T(x) at Multiple Times
Current Performance Summary
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Methodology & Theory
Transient Thermal Penetration
A semi-infinite solid is an idealized body with a single plane surface, extending to infinity. This model represents real geometries (such as the earth's crust, thick concrete walls, or large metallic shafts) during the early stages of a thermal transient before the heat wave reaches the opposite boundaries.
The depth to which the temperature change propagates is defined as the **penetration depth**: $$\delta_p \approx 4\sqrt{\alpha t}$$ Beyond $\delta_p$, the temperature of the solid remains practically undisturbed at its initial value ($T_i$).
Academic Reference Formulas:
- Case 1 — Constant $T_s$: $$T(x,t) = T_s + (T_i - T_s)\,\text{erf}\left(\frac{x}{2\sqrt{\alpha t}}\right)$$ $$Q(t) = 2\,k\,(T_s - T_i)\,\sqrt{\frac{t}{\pi \alpha}}$$
- Case 2 — Constant $q''_s$: $$T(x,t) = T_i + \frac{2q''_s\sqrt{\alpha t/\pi}}{k}\exp\left(-\frac{x^2}{4\alpha t}\right) - \frac{q''_s x}{k}\text{erfc}\left(\frac{x}{2\sqrt{\alpha t}}\right)$$ $$Q(t) = q''_s \cdot t$$
- Case 3 — Convection Boundary: $$T(x,t) = T_i + (T_\infty - T_i)\left[\text{erfc}\left(\frac{x}{2\sqrt{\alpha t}}\right) - \exp\left(\frac{hx}{k}+\frac{h^2\alpha t}{k^2}\right)\text{erfc}\left(\frac{x}{2\sqrt{\alpha t}}+\frac{h\sqrt{\alpha t}}{k}\right)\right]$$
Academic References:
- Incropera, F. P., DeWitt, D. P., Bergman, T. L., & Lavine, A. S. (2011). Fundamentals of Heat and Mass Transfer (7th Edition). John Wiley & Sons. Chapter 5.7 (The Semi-Infinite Solid).
- Carslaw, H. S., & Jaeger, J. C. (1959). Conduction of Heat in Solids (2nd Edition). Oxford University Press.
- Çengel, Y. A., & Ghajar, A. J. (2015). Heat and Mass Transfer: Fundamentals and Applications (5th Edition). McGraw-Hill. Chapter 4.3 (Transient Heat Conduction in Semi-Infinite Solids).
Worked Engineering Example
A thick concrete slab ($k = 1.4$ W/m·K, thermal diffusivity $\alpha = 7.0 \cdot 10^{-7}$ m²/s) is initially at a uniform temperature of $T_i = 20$°C. The surface is suddenly exposed to hot air at $T_\infty = 200$°C with a convection heat transfer coefficient of $h = 100$ W/m²·K. Calculate the temperature at a depth of $x = 50$ mm after $t = 1$ hour ($3600$ s) of exposure.
Step-by-step Solution:
1. Identify parameters and compute transient limits:
$$x = 0.05 \text{ m}, \quad t = 3600 \text{ s}, \quad k = 1.4 \text{ W/m·K}, \quad \alpha = 7.0 \cdot 10^{-7} \text{ m²/s}$$ $$\sqrt{\alpha t} = \sqrt{7.0 \cdot 10^{-7} \times 3600} = 0.0502 \text{ m}$$ $$\text{Dimensionless depth: } \eta = \frac{x}{2\sqrt{\alpha t}} = \frac{0.05}{2 \times 0.0502} = 0.498$$ $$\text{Convection parameter: } \frac{h\sqrt{\alpha t}}{k} = \frac{100 \times 0.0502}{1.4} = 3.586$$ 2. Calculate complementary error functions:
$$\text{erfc}(\eta) = \text{erfc}(0.498) \approx 0.482$$ 3. Evaluate Case 3 Convection boundary equation:
$$\frac{T(x,t) - T_i}{T_\infty - T_i} = \text{erfc}(\eta) - \exp\left(\frac{hx}{k} + \frac{h^2 \alpha t}{k^2}\right) \text{erfc}\left(\eta + \frac{h\sqrt{\alpha t}}{k}\right)$$ $$\frac{hx}{k} + \frac{h^2\alpha t}{k^2} = \frac{100 \times 0.05}{1.4} + (3.586)^2 = 3.571 + 12.86 = 16.43$$ $$\text{erfc}(0.498 + 3.586) = \text{erfc}(4.084) \approx 6.4 \cdot 10^{-9}$$ $$\exp(16.43) \times \text{erfc}(4.084) \approx 1.36 \cdot 10^7 \times 6.4 \cdot 10^{-9} = 0.087$$ $$\frac{T(x,t) - T_i}{T_\infty - T_i} \approx 0.482 - 0.087 = 0.395$$ 4. Find the local temperature $T(0.05\text{m}, 1\text{hr})$:
$$T(x,t) = T_i + 0.395 \times (T_\infty - T_i) = 20 + 0.395 \times (200 - 20) = 91.1 \text{ °C}$$
Final Result:
The concrete temperature at $50$ mm depth is 91.1 °C.