ThermoFluidCalc
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Engineering Calculation Report
Date: 2026-05-30 00:14
Conduction with Heat Generation
Analyze steady-state temperature distribution in solids with uniform volumetric heat generation (electrical heating, nuclear rods, exothermic reactions).
Calculation Domain Inputs
Specify geometry parameters, heat generation rate, and boundary conditions to calculate the steady-state temperature profile and find the peak temperature inside the solid:
- Geometry: Choose between a Plane Wall, Cylinder, or Sphere.
- Volumetric Source ($\dot{q}$): Uniform heat generation rate per unit volume.
- Properties ($k$): Thermal conductivity of the solid material.
- Convection ($h, T_\infty$): Convection coefficient and ambient temperature of the fluid.
Configuration
Temperature Distribution:
Wall: $T(x) = T_s + \frac{\dot{q}(L^2-x^2)}{2k}$
Cyl: $T(r) = T_s + \frac{\dot{q}(r_0^2-r^2)}{4k}$
Sphere: $T(r) = T_s + \frac{\dot{q}(r_0^2-r^2)}{6k}$
• $T_{max}$ at center ($x=0$ or $r=0$)
• $T_s = T_\infty + \frac{\dot{q} L_c}{h}$
Wall: $T(x) = T_s + \frac{\dot{q}(L^2-x^2)}{2k}$
Cyl: $T(r) = T_s + \frac{\dot{q}(r_0^2-r^2)}{4k}$
Sphere: $T(r) = T_s + \frac{\dot{q}(r_0^2-r^2)}{6k}$
• $T_{max}$ at center ($x=0$ or $r=0$)
• $T_s = T_\infty + \frac{\dot{q} L_c}{h}$
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 Poisson Equation. For 1D geometries, the temperature profile and center temperature $T_{max}$ are:
$$T(x) = T_s + \frac{\dot{q} L^2}{2k} \left(1 - \frac{x^2}{L^2}\right) \quad \text{(Plane Wall)}$$
$$T_{max} = T_s + \frac{\dot{q} r_0^2}{4k} \quad \text{(Cylinder)}, \quad T_{max} = T_s + \frac{\dot{q} r_0^2}{6k} \quad \text{(Sphere)}$$
Assumptions & Boundary Conditions:
- One-dimensional conduction under steady-state conditions in spatial coordinate ($x$ or $r$).
- Uniform internal volumetric heat generation ($\dot{q}$).
- Constant 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 fluid temperature $T_\infty$.
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 long solid cylinder of radius 20 mm ($k = 15$ W/m·K) generates heat uniformly at $\dot{q} = 2 \times 10^6$ W/m³. The outer surface is cooled by convection to air at 25°C with $h = 250$ W/m²·K. Calculate the maximum temperature inside the cylinder.
Step-by-step Solution:
1. Calculate surface temperature $T_s$ from energy balance:
$$\dot{q} (\pi r_0^2 L) = h (2\pi r_0 L) (T_s - T_{\infty}) \Rightarrow T_s = T_{\infty} + \frac{\dot{q} r_0}{2h}$$ $$T_s = 25 + \frac{2 \times 10^6 \times 0.02}{2 \times 250} = 25 + 80 = 105 \text{°C}$$ 2. Calculate maximum center temperature $T_{max}$:
$$T_{max} = T_s + \frac{\dot{q} r_0^2}{4k} = 105 + \frac{2 \times 10^6 \times 0.02^2}{4 \times 15} = 105 + 13.33 = 118.33 \text{°C}$$
Final Result:
The maximum temperature is 118.3°C.
A long solid cylinder of radius 20 mm ($k = 15$ W/m·K) generates heat uniformly at $\dot{q} = 2 \times 10^6$ W/m³. The outer surface is cooled by convection to air at 25°C with $h = 250$ W/m²·K. Calculate the maximum temperature inside the cylinder.
Step-by-step Solution:
1. Calculate surface temperature $T_s$ from energy balance:
$$\dot{q} (\pi r_0^2 L) = h (2\pi r_0 L) (T_s - T_{\infty}) \Rightarrow T_s = T_{\infty} + \frac{\dot{q} r_0}{2h}$$ $$T_s = 25 + \frac{2 \times 10^6 \times 0.02}{2 \times 250} = 25 + 80 = 105 \text{°C}$$ 2. Calculate maximum center temperature $T_{max}$:
$$T_{max} = T_s + \frac{\dot{q} r_0^2}{4k} = 105 + \frac{2 \times 10^6 \times 0.02^2}{4 \times 15} = 105 + 13.33 = 118.33 \text{°C}$$
Final Result:
The maximum temperature is 118.3°C.