🔳 Multidimensional Transient Conduction
Calculate 2D and 3D transient temperature distributions and heat transfer fractions using product solutions of Heisler chart 1D components (cylinders and walls).
Heat Transfer
Geometry & Parameters
Results & Live Simulation
1D Component Decomposition Flowchart
Cross-Section 2D Temperature Profile Heatmap
Combined Results Summary
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Methodology & Theory
Product Solution Principle
For transient heat conduction in multi-dimensional geometries, analytical solutions can be formulated as products of corresponding 1D solutions. This is valid for bodies initially at a uniform temperature $T_i$ subjected to convective boundary conditions with a constant ambient temperature $T_\infty$.
The dimensionless temperature ratio is defined as: $$\theta^*(x_1, x_2, \dots, t) = \frac{T(x_1, x_2, \dots, t) - T_\infty}{T_i - T_\infty}$$ Applying the product rule, we decompose this into: $$\theta^*(x_1, x_2, \dots, t) = \prod_{i=1}^n \theta_i^*(x_i, t)$$ where each $\theta_i^*$ is the 1D solution (one-term approximation) of the Heisler chart equations.
Supported Decomposition Equations:
- Short Cylinder (L₁ × r₀): $$\theta^*_{2D}(x, r, t) = \theta^*_{wall}(x, t) \times \theta^*_{cyl}(r, t)$$
- Rectangular Bar (L₁ × L₂): $$\theta^*_{2D}(x_1, x_2, t) = \theta^*_{wall, 1}(x_1, t) \times \theta^*_{wall, 2}(x_2, t)$$
- Rectangular Block (L₁ × L₂ × L₃): $$\theta^*_{3D}(x_1, x_2, x_3, t) = \theta^*_{wall, 1}(x_1, t) \times \theta^*_{wall, 2}(x_2, t) \times \theta^*_{wall, 3}(x_3, t)$$
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 5.8 (Multidimensional Effects).
- Langston, L. S. (1982). Heat Transfer from Multidimensional Objects using Product Solutions. International Journal of Heat Transfer, 25(1), 149-150.
- Çengel, Y. A., & Ghajar, A. J. (2015). Heat and Mass Transfer: Fundamentals and Applications (5th Edition). McGraw-Hill. Chapter 4.4 (Transient Heat Conduction in Multidimensional Systems).
Worked Engineering Example
A long carbon steel bar ($k = 43$ W/m·K, density $\rho = 7850$ kg/m³, specific heat $C_p = 475$ J/kg·K) of cross-section $2L_1 = 100$ mm by $2L_2 = 60$ mm is initially at a uniform temperature of $T_i = 20$°C. It is heated in a furnace at $T_\infty = 180$°C with convection heat transfer coefficients on all sides of $h_1 = 120$ W/m²·K (on the $L_1$ faces) and $h_2 = 200$ W/m²·K (on the $L_2$ faces). Calculate the centerline temperature after $t = 120$ seconds.
Step-by-step Solution:
1. Obtain half-thicknesses and diffusivity:
$$L_1 = 0.05 \text{ m}, \quad L_2 = 0.03 \text{ m}$$ $$\alpha = \frac{k}{\rho C_p} = \frac{43}{7850 \times 475} = 1.153 \cdot 10^{-5} \text{ m²/s}$$ 2. Calculate Biot and Fourier numbers:
$$Bi_1 = \frac{120 \times 0.05}{43} = 0.1395, \quad Fo_1 = \frac{1.153\cdot 10^{-5} \times 120}{(0.05)^2} = 0.5534$$ $$Bi_2 = \frac{200 \times 0.03}{43} = 0.1395, \quad Fo_2 = \frac{1.153\cdot 10^{-5} \times 120}{(0.03)^2} = 1.5373$$ Both Fourier numbers exceed $0.2$, enabling valid one-term series solutions.
3. Find transcendental eigenvalues ($\zeta_i$) and coefficients ($C_i$):
$$\zeta_1 \tan(\zeta_1) = 0.1395 \implies \zeta_1 \approx 0.3639 \text{ rad}$$ $$C_1 = \frac{4 \sin(0.3639)}{2(0.3639) + \sin(0.7278)} = 1.0225$$ 4. Evaluate the 1D centerline temperature ratios:
$$\theta_1^*(0, 120\text{s}) = 1.0225 \exp(-(0.3639)^2 \times 0.5534) = 0.9502$$ $$\theta_2^*(0, 120\text{s}) = 1.0225 \exp(-(0.3639)^2 \times 1.5373) = 0.8336$$ 5. Calculate combined centerline temp:
$$\theta^*_{2D} = \theta_1^* \times \theta_2^* = 0.9502 \times 0.8336 = 0.7921$$ $$T(0,0,120\text{s}) = 180 + 0.7921 \times (20 - 180) = 53.3 \text{ °C}$$
📘 Calculation Methodology: Multidimensional Transient Conduction (Product Solutions)
Mathematical Model & Theory
For multidimensional bodies with uniform initial temperature and convective boundaries, the dimensionless transient temperature is obtained by the product of 1D coordinate solutions:
Assumptions
- Uniform initial temperature $T_i$ and constant convective ambient $T_\infty$, $h$.
- Constant isotropic thermal properties without internal heat generation.
Academic References
- Incropera, F. P. et al.: Fundamentals of Heat and Mass Transfer, Ch. 5.
- Carslaw, H. S., & Jaeger, J. C.: Conduction of Heat in Solids.
Worked Engineering Example
A short cylinder has 1D infinite cylinder solution $\theta_{cyl} = 0.40$ and 1D plate solution $\theta_{plate} = 0.60$ at $t = 60\text{ s}$. Find center temperature for $T_i = 300^\circ\text{C}$ and $T_\infty = 25^\circ\text{C}$.
Step-by-step Solution:
1. $\theta_{center} = 0.40 \times 0.60 = 0.24$.
2. $T = 25 + 0.24 \times (300 - 25) = 25 + 66.0 = 91.0^\circ\text{C}$.
Final Result:
Center temperature after 60 seconds is $\mathbf{91.0^\circ\text{C}}$.