ThermoFluidCalc
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Engineering Calculation Report
Date: 2026-09-11 23:12
♨️ Heat Generation in Solids
Conduction with volumetric heat generation in plane walls, cylinders, and spheres. Max temperature and surface temperature.
⚡ Fortran 90 Engine
Double Precision (IEEE 754)
✓ ISO / ASME Validated
Heat Transfer
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Configuration
Governing Temperature Profiles:
• Plane Wall: $$T(x) = T_s + \frac{\dot{q}\,(L^2 - x^2)}{2k}$$
• Solid Cylinder: $$T(r) = T_s + \frac{\dot{q}\,(r_0^2 - r^2)}{4k}$$
• Solid Sphere: $$T(r) = T_s + \frac{\dot{q}\,(r_0^2 - r^2)}{6k}$$
• Maximum Temp: $T_{\max}$ at centerline ($x=0$ or $r=0$)
• Surface Temp: $T_s = T_\infty + \dfrac{\dot{q} L_c}{h}$
• Surface Temp: $T_s = T_\infty + \dfrac{\dot{q} L_c}{h}$
Results & Visualization
Results and visualizations will appear here after calculation.
ℹ️ About Internal Heat Generation
When heat is generated uniformly within a solid (electrical resistance, nuclear fission, chemical reactions), the temperature distribution is parabolic with the maximum at the center.
Common applications:
• Electrical resistance wires and heaters
• Nuclear fuel rods (UO₂ pellets)
• Exothermic chemical reactors
• Current-carrying conductors
• Microprocessors and electronic chips
Key insight: Tmax depends on both internal resistance ($k$) and external resistance ($h$). The surface temperature $T_s$ is always above $T_\infty$.
When heat is generated uniformly within a solid (electrical resistance, nuclear fission, chemical reactions), the temperature distribution is parabolic with the maximum at the center.
Common applications:
• Electrical resistance wires and heaters
• Nuclear fuel rods (UO₂ pellets)
• Exothermic chemical reactors
• Current-carrying conductors
• Microprocessors and electronic chips
Key insight: Tmax depends on both internal resistance ($k$) and external resistance ($h$). The surface temperature $T_s$ is always above $T_\infty$.
📘 Calculation Methodology
Mathematical Model & Theory
Steady-state conduction in solids with uniform volumetric heat generation ($\dot{q}$) is governed by the 1D Poisson heat conduction equation. For 1D symmetric geometries, the temperature distributions are:
$$T(x) = T_s + \frac{\dot{q} L^2}{2k} \left(1 - \frac{x^2}{L^2}\right) \quad \text{(Plane Wall)}$$
$$T(r) = T_s + \frac{\dot{q} r_0^2}{4k} \left(1 - \frac{r^2}{r_0^2}\right) \quad \text{(Cylinder)}$$
$$T(r) = T_s + \frac{\dot{q} r_0^2}{6k} \left(1 - \frac{r^2}{r_0^2}\right) \quad \text{(Sphere)}$$
The center temperature ($x=0$ or $r=0$) yields the maximum temperature $T_{\max}$:
$$T_{\max} = T_s + \frac{\dot{q} L_c^2}{2 n k} \quad \text{where } n = 1\text{ (wall)},\, 2\text{ (cyl)},\, 3\text{ (sphere)}$$
Assumptions & Boundary Conditions:
- One-dimensional conduction under steady-state conditions in spatial coordinate ($x$ or $r$).
- Uniform internal volumetric heat generation ($\dot{q}$).
- Constant and isotropic solid thermal conductivity $k$.
- Symmetric temperature profile about centerline/symmetry axis ($x = 0$ or $r = 0$).
- Uniform convection boundary condition at the outer surface with constant convection coefficient $h$ and ambient temperature $T_\infty$.
Academic References:
- Incropera, F. P., & DeWitt, D. P. (2011). Fundamentals of Heat and Mass Transfer, 7th Edition.
- Çengel, Y. A., & Ghajar, A. J. (2015). Heat and Mass Transfer: Fundamentals and Applications, McGraw-Hill.
Worked Engineering Example
Problem Statement:
A long solid cylinder of radius $r_0 = 20\text{ mm}$ ($k = 15\text{ W/m}\cdot\text{K}$) generates heat uniformly at $\dot{q} = 2 \times 10^6\text{ W/m}^3$. The outer surface is cooled by convection to air at $25^\circ\text{C}$ with $h = 250\text{ W/m}^2\cdot\text{K}$. Calculate the maximum center temperature inside the cylinder.
Step-by-step Solution:
1. Surface temperature $T_s$ from global energy balance: $$\dot{q} (\pi r_0^2 L) = h (2\pi r_0 L) (T_s - T_\infty) \implies T_s = T_\infty + \frac{\dot{q} r_0}{2h}$$ $$T_s = 25 + \frac{2 \times 10^6 \times 0.02}{2 \times 250} = 25 + 80 = 105^\circ\text{C}$$ 2. Maximum center temperature $T_{\max}$: $$T_{\max} = T_s + \frac{\dot{q} r_0^2}{4k} = 105 + \frac{2 \times 10^6 \times (0.02)^2}{4 \times 15} = 105 + 13.33 = 118.33^\circ\text{C}$$
Final Result:
The maximum centerline temperature is $118.3^\circ\text{C}$ (with a surface temperature of $105.0^\circ\text{C}$).
A long solid cylinder of radius $r_0 = 20\text{ mm}$ ($k = 15\text{ W/m}\cdot\text{K}$) generates heat uniformly at $\dot{q} = 2 \times 10^6\text{ W/m}^3$. The outer surface is cooled by convection to air at $25^\circ\text{C}$ with $h = 250\text{ W/m}^2\cdot\text{K}$. Calculate the maximum center temperature inside the cylinder.
Step-by-step Solution:
1. Surface temperature $T_s$ from global energy balance: $$\dot{q} (\pi r_0^2 L) = h (2\pi r_0 L) (T_s - T_\infty) \implies T_s = T_\infty + \frac{\dot{q} r_0}{2h}$$ $$T_s = 25 + \frac{2 \times 10^6 \times 0.02}{2 \times 250} = 25 + 80 = 105^\circ\text{C}$$ 2. Maximum center temperature $T_{\max}$: $$T_{\max} = T_s + \frac{\dot{q} r_0^2}{4k} = 105 + \frac{2 \times 10^6 \times (0.02)^2}{4 \times 15} = 105 + 13.33 = 118.33^\circ\text{C}$$
Final Result:
The maximum centerline temperature is $118.3^\circ\text{C}$ (with a surface temperature of $105.0^\circ\text{C}$).