📉 Inverse Heat Conduction Problem (IHCP) Solver
Estimate surface heat flux and temperature profiles from internal transient/steady sensor readings using Beck's sequential method and least-squares regression.
Heat Transfer
Parameters Input
Results & Analysis
Results and visualizations will be displayed here upon completion of the computation.
Calculation Methodology
Mathematical Model & Theory
The 1D steady-state heat conduction equation in a solid wall without heat generation is given by:
$$\frac{d^2 T}{d x^2} = 0 \implies T(x) = a + b x$$where $a$ is the surface temperature ($T_s = a$) at $x=0$, and $b$ is the constant temperature gradient ($b = \frac{dT}{dx}$). By Fourier's Law, the surface heat flux ($q$) is:
$$q = -k \frac{dT}{dx} = -k \cdot b$$In the **Inverse Heat Conduction Problem (IHCP)**, we do not know the boundary condition $T_s$ or $q$. Instead, we have $N$ discrete temperature measurements $T_{meas, i}$ at internal locations $x_i$. We fit the model parameters $a$ and $b$ using **ordinary least-squares regression** by minimizing the sum of squared residuals:
$$S(a, b) = \sum_{i=1}^N \left(T_{meas, i} - (a + b x_i)\right)^2$$Setting $\frac{\partial S}{\partial a} = 0$ and $\frac{\partial S}{\partial b} = 0$ yields the standard normal equations:
$$N \cdot a + b \sum x_i = \sum T_{meas, i}$$ $$a \sum x_i + b \sum x_i^2 = \sum (x_i T_{meas, i})$$Solving this system yields the estimates for $a$ and $b$, which directly define the surface temperature $T_s$ and the boundary heat flux $q = -k \cdot b$.
Academic References:
- Beck, J. V., Blackwell, B., & St. Clair, C. R. (1985). Inverse Heat Conduction: Ill-posed Problems. Wiley.
- Ozisik, M. N., & Orlande, H. R. (2000). Inverse Heat Transfer: Fundamentals and Applications. Taylor & Francis.
Worked Engineering Example
A steel wall ($k = 50$ W/m·K, $L = 0.05$ m, $A = 0.01$ m²) is fitted with 3 temperature sensors:
- Sensor 1 at $x_1 = 0.01$ m measures $T_1 = 95.2$ °C
- Sensor 2 at $x_2 = 0.025$ m measures $T_2 = 82.5$ °C
- Sensor 3 at $x_3 = 0.04$ m measures $T_3 = 70.1$ °C
Step-by-step Solution:
1. Sum terms for least-squares fit ($N = 3$):
$$\sum x_i = 0.075 \text{ m}, \quad \sum T_i = 247.8 \text{ °C}$$ $$\sum x_i^2 = 0.002325 \text{ m}^2, \quad \sum x_i T_i = 5.8185 \text{ m·°C}$$ 2. Calculate slope $b$ ($\frac{dT}{dx}$):
$$b = \frac{3(5.8185) - (0.075)(247.8)}{3(0.002325) - (0.075)^2} = -836.67 \text{ K/m}$$ 3. Calculate intercept $a$ ($T_s$):
$$a = \frac{247.8 - (-836.67 \times 0.075)}{3} = 103.52 \text{ °C}$$ 4. Compute boundary values:
$$q = -k \cdot b = -50 \times (-836.67) = 41,833 \text{ W/m²} = 41.83 \text{ kW/m²}$$ $$Q = q \cdot A = 41,833 \times 0.01 = 418.3 \text{ W}$$ $$T_b = a + b \cdot L = 103.52 + (-836.67 \times 0.05) = 61.68 \text{ °C}$$
Final Estimates:
• Surface Heat Flux: 41.83 kW/m²
• Surface Temp $T_s$: 103.52 °C
• Back Temp $T_b$: 61.68 °C