📐 2D Conduction Shape Factors
Calculate conduction shape factors (S) and thermal heat rates for 12+ multi-dimensional configurations (buried pipes, eccentric cylinders, wall edges/corners, and heating floors).
Heat Transfer
Configuration
Results & Visualization
Geometry Dimension Reference
Results and calculations will appear here after calculation.
Methodology & Theory
Conduction Shape Factors ($S$)
For steady-state conduction in two-dimensional systems where the temperatures are constant on boundary surfaces, the heat transfer rate can be evaluated using: $$Q = S \cdot k \cdot (T_1 - T_2)$$ Where $S$ is the **Shape Factor** (in meters), which depends solely on the system geometry, and $k$ is the thermal conductivity of the medium.
The overall conduction thermal resistance of such systems is related to shape factor via: $$R_{th} = \frac{1}{S \cdot k}$$
Academic Reference Formulas:
- Buried Horizontal Cylinder: $S = \frac{2\pi L}{\cosh^{-1}(z/r)}$ (Valid for $L \gg r$ and $z \ge r$)
- Two Parallel Pipes: $S = \frac{2\pi L}{\cosh^{-1}\left(\frac{w^2 - r_1^2 - r_2^2}{2r_1 r_2}\right)}$
- Buried Sphere: $S = \frac{4\pi r}{1 - r/(2z)}$
- Heating Floor (Row of Pipes): $S = \frac{2\pi L}{\ln\left(\frac{s}{\pi D} \sinh\left(\frac{2\pi z}{s}\right)\right)}$
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 4.3 (The Conduction Shape Factor).
- Çengel, Y. A., & Ghajar, A. J. (2015). Heat and Mass Transfer: Fundamentals and Applications (5th Edition). McGraw-Hill. Chapter 3.6 (Steady Heat Conduction in Common Configurations).
Worked Engineering Example
A hot-water pipe ($D = 100$ mm, surface temperature $T_1 = 80$°C) is buried horizontally in dry soil ($k = 1.2$ W/m·K) at a depth of $z = 1.5$ m (to the pipe centerline). The ground surface temperature is $T_2 = 15$°C. The length of the pipe is $L = 50$ m. Calculate the heat loss rate from the pipe.
Step-by-step Solution:
1. Identify parameters and verify conditions:
$$L = 50 \text{ m}, \quad z = 1.5 \text{ m}, \quad r = 0.05 \text{ m}, \quad k = 1.2 \text{ W/m·K}$$ Checking boundaries: $L \gg r$ (50 m $\gg$ 0.05 m) and $z \ge r$ (1.5 m $\ge$ 0.05 m) are satisfied.
2. Calculate the Shape Factor ($S$):
$$S = \frac{2\pi L}{\cosh^{-1}(z/r)} = \frac{2\pi \times 50}{\cosh^{-1}(1.5/0.05)} = \frac{314.16}{\cosh^{-1}(30)}$$ Using mathematical expansion: $\cosh^{-1}(30) = \ln(30 + \sqrt{30^2 - 1}) = 4.094$
$$S = \frac{314.16}{4.094} = 76.74 \text{ m}$$ 3. Calculate the thermal resistance ($R_{th}$):
$$R_{th} = \frac{1}{S \cdot k} = \frac{1}{76.74 \times 1.2} = 0.01086 \text{ °C/W}$$ 4. Calculate the heat transfer rate ($Q$):
$$Q = S \cdot k \cdot (T_1 - T_2) = 76.74 \times 1.2 \times (80 - 15) = 5985.7 \text{ W}$$
Final Result:
The soil heat dissipation loss rate is 5985.7 W (5.99 kW).
📘 Calculation Methodology: Conduction Shape Factors (2D/3D Steady Heat Flow)
Mathematical Model & Theory
For multi-dimensional steady-state conduction between isothermal surfaces with shape factor $S$, the total heat transfer rate is directly proportional to thermal conductivity $k$ and temperature difference $\Delta T$:
Assumptions
- Steady-state heat conduction with constant thermal conductivity $k$.
- Homogeneous semi-infinite or infinite medium with isothermal boundaries.
Academic References
- Incropera, F. P. et al.: Fundamentals of Heat and Mass Transfer, Ch. 4.
- Schneider, P. J.: Conduction Heat Transfer, Addison-Wesley.
Worked Engineering Example
A pipe ($D = 0.2\text{ m}$, $L = 50\text{ m}$) is buried at depth $z = 1.5\text{ m}$ in soil ($k = 0.8\text{ W/m}\cdot\text{K}$) with $T_1 = 120^\circ\text{C}$ and $T_2 = 10^\circ\text{C}$. Calculate shape factor and heat loss.
Step-by-step Solution:
1. $S = 2\pi(50) / \ln(4(1.5)/0.2) = 314.16 / \ln(30) \approx 92.37\text{ m}$.
2. $q = 92.37 \times 0.8 \times (120 - 10) = 8128.5\text{ W} = 8.13\text{ kW}$.
Final Result:
Shape factor is $S = \mathbf{92.37\text{ m}}$ with heat loss of $\mathbf{8.13\text{ kW}}$.