ThermoFluidCalc
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Engineering Calculation Report
Date: 2026-08-22 20:48
📈 Transient Conduction — 1D Series
One-term approximation for transient heat conduction in plane walls, cylinders, and spheres using Heisler charts.
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Configuration
One-Term Approximation ($Fo \gt 0.2$):
Wall: $\theta^* = C_1 e^{-\zeta_1^2 Fo} \cos(\zeta_1 x/L)$
Cyl: $\theta^* = C_1 e^{-\zeta_1^2 Fo} J_0(\zeta_1 r/r_0)$
Sphere: $\theta^* = C_1 e^{-\zeta_1^2 Fo} \frac{\sin(\zeta_1 r/r_0)}{\zeta_1 r/r_0}$
• $Bi = hL/k$, $Fo = \alpha t/L^2$
• $\theta^* = (T-T_\infty)/(T_i-T_\infty)$
Wall: $\theta^* = C_1 e^{-\zeta_1^2 Fo} \cos(\zeta_1 x/L)$
Cyl: $\theta^* = C_1 e^{-\zeta_1^2 Fo} J_0(\zeta_1 r/r_0)$
Sphere: $\theta^* = C_1 e^{-\zeta_1^2 Fo} \frac{\sin(\zeta_1 r/r_0)}{\zeta_1 r/r_0}$
• $Bi = hL/k$, $Fo = \alpha t/L^2$
• $\theta^* = (T-T_\infty)/(T_i-T_\infty)$
Results & Visualization
Results and visualizations will appear here after calculation.
ℹ️ About the One-Term Approximation
For Fourier number Fo > 0.2, the infinite series solution converges to a single dominant term, giving accurate results (< 2% error).
Eigenvalue equations:
• Wall: ζₙ·tan(ζₙ) = Bi
• Cylinder: ζₙ·J₁(ζₙ)/J₀(ζₙ) = Bi
• Sphere: 1 - ζₙ·cot(ζₙ) = Bi
Applications:
• Heat treatment of metals
• Food sterilization processes
• Thermal protection systems
• Concrete curing analysis
For Fourier number Fo > 0.2, the infinite series solution converges to a single dominant term, giving accurate results (< 2% error).
Eigenvalue equations:
• Wall: ζₙ·tan(ζₙ) = Bi
• Cylinder: ζₙ·J₁(ζₙ)/J₀(ζₙ) = Bi
• Sphere: 1 - ζₙ·cot(ζₙ) = Bi
Applications:
• Heat treatment of metals
• Food sterilization processes
• Thermal protection systems
• Concrete curing analysis
📘 Calculation Methodology
Mathematical Model & Theory
For transient conduction in 1D shapes (plane walls, long cylinders, spheres) where the Biot number is greater than 0.1, internal temperature gradients are non-negligible. The one-term Heisler approximation is valid for Fourier numbers ($Fo > 0.2$):
$$\theta^* = C_1 e^{-\zeta_1^2 Fo} \cos(\zeta_1 x^*)$$
$$Fo = \frac{\alpha t}{L^2}, \quad \alpha = \frac{k}{\rho C_p}$$
Where $\zeta_1$ and $C_1$ are eigenvalues determined by boundary conditions (transcendental equations involving $Bi$).
Assumptions & Boundary Conditions:
- One-dimensional transient heat conduction in the spatial coordinate ($x$ for wall, $r$ for cylinder/sphere).
- Uniform initial temperature ($T_i$) throughout the solid at time $t = 0$.
- Constant material properties (conductivity $k$, density $\rho$, specific heat $C_p$).
- Symmetric temperature profile about the centerline ($x = 0$ or $r = 0$).
- Uniform and constant convection heat transfer coefficient $h$ and temperature $T_\infty$ at the outer surface boundary.
- No internal heat generation.
Academic References:
- Incropera, F. P., & DeWitt, D. P. (2011). Fundamentals of Heat and Mass Transfer.
- Çengel, Y. A. (2015). Heat and Mass Transfer: Fundamentals and Applications.
Worked Engineering Example
Problem Statement:
A steel plate ($2L = 40$ mm thick, $k = 45$ W/m·K, $\alpha = 1.25 \times 10^{-5}$ m²/s) initially at 400°C is exposed to convection at 20°C with $h = 250$ W/m²·K. Find the temperature at the center after 2 minutes.
Step-by-step Solution:
1. Calculate Biot number:
$$Bi = \frac{h L}{k} = \frac{250 \times 0.02}{45} = 0.1111$$ 2. Calculate Fourier number after $t = 120$ s:
$$Fo = \frac{\alpha t}{L^2} = \frac{1.25 \times 10^{-5} \times 120}{0.02^2} = 3.75 > 0.2 \quad \text{(One-term approx valid)}$$ 3. Find eigenvalues for $Bi = 0.1111$ (Table/Calculated):
$$\zeta_1 \approx 0.32 \text{ rad}, \quad C_1 \approx 1.018$$ 4. Calculate center temperature ratio ($x^* = 0$):
$$\theta_0^* = C_1 e^{-\zeta_1^2 Fo} = 1.018 \times e^{-0.32^2 \times 3.75} = 1.018 \times 0.6811 = 0.6934$$ $$T(0, 120) = T_{\infty} + \theta_0^* (T_i - T_{\infty}) = 20 + 0.6934 \times (400 - 20) = 283.48 \text{°C}$$
Final Result:
The plate center temperature is 283.5°C.
A steel plate ($2L = 40$ mm thick, $k = 45$ W/m·K, $\alpha = 1.25 \times 10^{-5}$ m²/s) initially at 400°C is exposed to convection at 20°C with $h = 250$ W/m²·K. Find the temperature at the center after 2 minutes.
Step-by-step Solution:
1. Calculate Biot number:
$$Bi = \frac{h L}{k} = \frac{250 \times 0.02}{45} = 0.1111$$ 2. Calculate Fourier number after $t = 120$ s:
$$Fo = \frac{\alpha t}{L^2} = \frac{1.25 \times 10^{-5} \times 120}{0.02^2} = 3.75 > 0.2 \quad \text{(One-term approx valid)}$$ 3. Find eigenvalues for $Bi = 0.1111$ (Table/Calculated):
$$\zeta_1 \approx 0.32 \text{ rad}, \quad C_1 \approx 1.018$$ 4. Calculate center temperature ratio ($x^* = 0$):
$$\theta_0^* = C_1 e^{-\zeta_1^2 Fo} = 1.018 \times e^{-0.32^2 \times 3.75} = 1.018 \times 0.6811 = 0.6934$$ $$T(0, 120) = T_{\infty} + \theta_0^* (T_i - T_{\infty}) = 20 + 0.6934 \times (400 - 20) = 283.48 \text{°C}$$
Final Result:
The plate center temperature is 283.5°C.