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Engineering Calculation Report
Date: 2026-07-01 02:57

Inverse Heat Conduction Problem (IHCP) Solver

Estimate boundary surface heat flux and temperatures from discrete internal sensor data points.

Unknown q x = 0 x = L Surface (T_s) Back (T_b) T_1 x_1 T_2 x_2 T_3 x_3 Thickness L

Inverse Problem Concept

Unlike direct heat conduction calculations which require boundary temperatures to find internal profiles, the Inverse Heat Conduction Problem (IHCP) uses internal sensor values to estimate the boundary conditions:

  • Sensors ($N$): Temperature values at discrete coordinate locations inside the medium.
  • Thermal Conductivity ($k$): Known physical properties of the wall material.
  • Estimations ($q, T_s, T_b$): Boundaries values computed by least-squares parameter regression.
3-Sensor Steel Wall 5-Sensor Copper Block 4-Sensor Ceramic

Parameters Input

Internal Sensor Measurements

No. Coordinate x [m] Temperature T [°C]
1
2
3
Steady Regression Formulation: $$T_{calc}(x) = T_s - \left(\frac{q_{est}}{k}\right)x$$ $$R^2 = 1 - \frac{\sum (T_{meas} - T_{calc})^2}{\sum (T_{meas} - \overline{T})^2}$$

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:

  1. Beck, J. V., Blackwell, B., & St. Clair, C. R. (1985). Inverse Heat Conduction: Ill-posed Problems. Wiley.
  2. Ozisik, M. N., & Orlande, H. R. (2000). Inverse Heat Transfer: Fundamentals and Applications. Taylor & Francis.

Worked Engineering Example

Problem Statement:
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
Estimate the surface heat flux ($q$) and surface temperature ($T_s$).

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