ThermoFluidCalc
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Engineering Calculation Report
Date: 2026-05-30 00:17
Steady Heat Conduction in Cylinders and Spheres
Analyze heat conduction through multilayered cylindrical and spherical geometries with radial temperature distribution.
Calculation Domain Inputs
Define the radial geometry, boundary conditions, and material layers for 1D steady conduction:
- Geometry: Choose Cylindrical (e.g., pipes) or Spherical.
- Cylinder Length ($L$): Required only for cylindrical geometry.
- Inner/Outer Temperatures ($T_1$/$T_2$): Boundary surface temperatures.
- Inner/Outer Radii ($r_i$/$r_o$): Radial boundaries of each layer.
- Thermal Conductivity ($k_i$): Dynamic heat transmission ability.
Configuration
Thermal Resistances:
Cylinder:
R = ln(r_o/r_i) / (2πkL)
Q = (T₁ - T₂) / R_total
Sphere:
R = (1/r_i - 1/r_o) / (4πk)
Q = (T₁ - T₂) / R_total
• Q = Heat flux [W]
• r_i, r_o = Inner/outer radii [m]
• k = Thermal conductivity [W/m·K]
• L = Length (cylinder) [m]
Cylinder:
R = ln(r_o/r_i) / (2πkL)
Q = (T₁ - T₂) / R_total
Sphere:
R = (1/r_i - 1/r_o) / (4πk)
Q = (T₁ - T₂) / R_total
• Q = Heat flux [W]
• r_i, r_o = Inner/outer radii [m]
• k = Thermal conductivity [W/m·K]
• L = Length (cylinder) [m]
Results & Visualization
Results and visualizations will appear here after calculation.
Calculation Methodology
Mathematical Model & Theory
Heat conduction through multilayer cylinders (e.g., 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:
$$R_{cyl} = \frac{\ln(r_{o,i} / r_{i,i})}{2\pi k_i L}$$
$$R_{sph} = \frac{1/r_{i,i} - 1/r_{o,i}}{4\pi k_i}$$
$$Q = \frac{T_1 - T_2}{R_{total}}$$
Variable Definitions & Units:
- $Q$: Heat transfer rate [W]
- $r_{i,i}, r_{o,i}$: Inner and outer radius of layer $i$ [m]
- $L$: Length of the cylinder [m]
- $k_i$: Thermal conductivity of layer $i$ [W/m·K]
- $T_1, T_2$: Inner and outer boundary temperatures [°C]
Assumptions & Boundary Conditions:
- One-dimensional radial heat flow.
- Steady-state conditions (no time variance).
- Constant thermal conductivities.
- Perfect contact between layers (no contact resistance).
Academic References:
- Incropera, F. P., & DeWitt, D. P. (2011). Fundamentals of Heat and Mass Transfer.
Worked Engineering Example
Problem Statement:
A steam pipe ($r_1 = 50$ mm, $r_2 = 60$ mm, length 10 m, $k_1 = 50$ W/m·K) is insulated with a 50 mm fiberglass sleeve ($r_3 = 110$ mm, $k_2 = 0.04$ W/m·K). The inner surface is at 200°C and the outer surface of the insulation is at 40°C. Calculate the heat loss rate.
Step-by-step Solution:
1. Calculate cylinder wall resistance:
$$R_1 = \frac{\ln(0.06/0.05)}{2\pi \times 50 \times 10} = 5.80 \times 10^{-5} \text{ K/W}$$ 2. Calculate insulation resistance:
$$R_2 = \frac{\ln(0.11/0.06)}{2\pi \times 0.04 \times 10} = 0.2411 \text{ K/W}$$ 3. Calculate total resistance:
$$R_{total} = R_1 + R_2 = 0.2412 \text{ K/W}$$ 4. Calculate heat loss $Q$:
$$Q = \frac{200 - 40}{0.2412} = 663.35 \text{ W}$$
Final Result:
The heat loss rate is 663.4 W.
A steam pipe ($r_1 = 50$ mm, $r_2 = 60$ mm, length 10 m, $k_1 = 50$ W/m·K) is insulated with a 50 mm fiberglass sleeve ($r_3 = 110$ mm, $k_2 = 0.04$ W/m·K). The inner surface is at 200°C and the outer surface of the insulation is at 40°C. Calculate the heat loss rate.
Step-by-step Solution:
1. Calculate cylinder wall resistance:
$$R_1 = \frac{\ln(0.06/0.05)}{2\pi \times 50 \times 10} = 5.80 \times 10^{-5} \text{ K/W}$$ 2. Calculate insulation resistance:
$$R_2 = \frac{\ln(0.11/0.06)}{2\pi \times 0.04 \times 10} = 0.2411 \text{ K/W}$$ 3. Calculate total resistance:
$$R_{total} = R_1 + R_2 = 0.2412 \text{ K/W}$$ 4. Calculate heat loss $Q$:
$$Q = \frac{200 - 40}{0.2412} = 663.35 \text{ W}$$
Final Result:
The heat loss rate is 663.4 W.