ThermoFluidCalc
https://thermofluidcalc.com
Engineering Calculation Report
Date: 2026-09-11 16:21
โญ Cylinders & Spheres Conduction
Steady radial heat conduction through cylindrical pipes and spherical shells with optional insulation layers.
โก Fortran 90 Engine
Double Precision (IEEE 754)
โ ISO / ASME Validated
Heat Transfer
๐ Solver Telemetry
โ ACTIVE
๐๏ธ Views
4,125
โก Solves
3,278
๐พ Downloads
343
๐ฆ Fortran Code
3.4 KB
๐
Released
Jan 2024
โฑ๏ธ Latency
< 1 ms
Configuration
Thermal Resistances:
Cylinder:
R = ln(r_o/r_i) / (2ฯkL)
Q = (Tโ - Tโ) / R_total
Sphere:
R = (1/r_i - 1/r_o) / (4ฯk)
Q = (Tโ - Tโ) / R_total
โข Q = Heat flux [W]
โข r_i, r_o = Inner/outer radii [m]
โข k = Thermal conductivity [W/mยทK]
โข L = Length (cylinder) [m]
Cylinder:
R = ln(r_o/r_i) / (2ฯkL)
Q = (Tโ - Tโ) / R_total
Sphere:
R = (1/r_i - 1/r_o) / (4ฯk)
Q = (Tโ - Tโ) / R_total
โข Q = Heat flux [W]
โข r_i, r_o = Inner/outer radii [m]
โข k = Thermal conductivity [W/mยทK]
โข L = Length (cylinder) [m]
Results & Visualization
Results and visualizations will appear here after calculation.
๐ Calculation Methodology: Multilayer Cylinders & Spheres Heat Conduction
Mathematical Model & Theory
Heat conduction through multilayer cylinders (pipes) and spheres in the radial direction is governed by radial formulations of Fourier's Law. The thermal resistance for each layer is defined as:
$$R_{cyl} = \frac{\ln(r_{o,i} / r_{i,i})}{2\pi k_i L}, \quad R_{sph} = \frac{1/r_{i,i} - 1/r_{o,i}}{4\pi k_i}$$
$$Q = \frac{T_1 - T_2}{R_{total}}, \quad R_{total} = \sum_{i=1}^n R_i$$
Assumptions
- One-dimensional radial steady-state heat flow without internal heat generation.
- Constant isotropic thermal conductivity per layer.
- Negligible thermal contact resistance at layer interfaces.
Academic References
- Incropera, F. P. et al.: Fundamentals of Heat and Mass Transfer, John Wiley & Sons, Ch. 3.
- Cengel, Y. A., & Ghajar, A. J.: Heat and Mass Transfer: Fundamentals and Applications, McGraw-Hill.
Worked Engineering Example
Problem Statement:
A steel steam pipe ($r_1 = 50\text{ mm}$, $r_2 = 60\text{ mm}$, $L = 10\text{ m}$, $k_1 = 50\text{ W/m}\cdot\text{K}$) is insulated with a $50\text{ mm}$ fiberglass sleeve ($r_3 = 110\text{ mm}$, $k_2 = 0.04\text{ W/m}\cdot\text{K}$). The inner surface is at $200^\circ\text{C}$ and the outer surface of the insulation is at $40^\circ\text{C}$. Calculate the total heat loss rate $Q$.
Step-by-step Solution:
1. Calculate steel pipe wall resistance:
$$R_1 = \frac{\ln(0.06/0.05)}{2\pi \times 50 \times 10} = \frac{\ln(1.20)}{3141.59} = \frac{0.18232}{3141.59} \approx 5.80 \times 10^{-5}\text{ K/W}$$
2. Calculate fiberglass insulation resistance:
$$R_2 = \frac{\ln(0.11/0.06)}{2\pi \times 0.04 \times 10} = \frac{\ln(1.8333)}{2.5133} = \frac{0.60614}{2.5133} \approx 0.24117\text{ K/W}$$
3. Calculate total thermal resistance and heat loss $Q$:
$$R_{total} = R_1 + R_2 = 0.000058 + 0.24117 = 0.24123\text{ K/W}$$
$$Q = \frac{200 - 40}{0.24123} = \frac{160}{0.24123} \approx 663.27\text{ W}$$
Final Result:
Total continuous heat loss rate is $Q = \mathbf{663.3\text{ W}}$ (Insulation accounts for 99.98% of thermal resistance).
A steel steam pipe ($r_1 = 50\text{ mm}$, $r_2 = 60\text{ mm}$, $L = 10\text{ m}$, $k_1 = 50\text{ W/m}\cdot\text{K}$) is insulated with a $50\text{ mm}$ fiberglass sleeve ($r_3 = 110\text{ mm}$, $k_2 = 0.04\text{ W/m}\cdot\text{K}$). The inner surface is at $200^\circ\text{C}$ and the outer surface of the insulation is at $40^\circ\text{C}$. Calculate the total heat loss rate $Q$.
Step-by-step Solution:
1. Calculate steel pipe wall resistance:
$$R_1 = \frac{\ln(0.06/0.05)}{2\pi \times 50 \times 10} = \frac{\ln(1.20)}{3141.59} = \frac{0.18232}{3141.59} \approx 5.80 \times 10^{-5}\text{ K/W}$$
2. Calculate fiberglass insulation resistance:
$$R_2 = \frac{\ln(0.11/0.06)}{2\pi \times 0.04 \times 10} = \frac{\ln(1.8333)}{2.5133} = \frac{0.60614}{2.5133} \approx 0.24117\text{ K/W}$$
3. Calculate total thermal resistance and heat loss $Q$:
$$R_{total} = R_1 + R_2 = 0.000058 + 0.24117 = 0.24123\text{ K/W}$$
$$Q = \frac{200 - 40}{0.24123} = \frac{160}{0.24123} \approx 663.27\text{ W}$$
Final Result:
Total continuous heat loss rate is $Q = \mathbf{663.3\text{ W}}$ (Insulation accounts for 99.98% of thermal resistance).