Library / Heat transfer and thermal systems / HTS-03
Heat transfer and thermal systems
Effective conductivity of parallel paths
Equivalent conductivity of up to three parallel conduction paths of the same thickness that share isothermal end faces, weighted by cross-sectional area.
- Ref
- HTS-03
- Rev
- v1
- Field
- Heat transfer and thermal systems
- Result
- k [W/(m K)]
When several materials sit side by side, spanning the same thickness between isothermal end faces, their conductances add. The equivalent conductivity is the area-weighted average, k = ΣkᵢAᵢ/ΣAᵢ. It is an upper bound for real composites because it ignores heat spreading sideways between paths. Set an unused path to zero area.
Equivalent conductivity of up to three parallel conduction paths of the same thickness that share isothermal end faces, weighted by cross-sectional area. Leave an unused path at zero area.
Calculator
- Total area, ΣA1 m²
- Weighted sum, ΣkA0.65 W m/K
- Area share of path 10.75
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- The parallel paths behave like a single material of 0.65 W/(m K), with 75 % of the area in the first path.
- The highly conducting path dominates: it carries a share of the heat larger than its area share, as heat avoids the poor conductor.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (W/(m K)) | Local step | Change in result |
|---|---|---|---|---|---|---|
| k1 Conductivity of path 1 | 0.01 to 500 W/(m K) | 0.5075 | 375.5 | 375 | +10% | +2.31 % |
| k2 Conductivity of path 2 | 0.01 to 500 W/(m K) | 0.1525 | 125.2 | 125 | +10% | +7.69 % |
| A1 Area of path 1 | 0.01 to 10 m² | 0.2439 | 1.931 | 1.687 | +10% | −4.83 % |
| A2 Area of path 2 | 0.01 to 10 m² | 0.2237 | 1.874 | 1.651 | +10% | +5.07 % |
| A3 Area of path 3 | 0 to 10 m² | 0.65 | 0.9682 | 0.3182 | +10% | 0 % |
| k3 Conductivity of path 3 (unused if A₃ = 0) | 0.01 to 500 W/(m K) | 0.65 | 0.65 | 0 | +10% | 0 % |
Sweep of Area of path 2
| Area of path 2 (m²) | Effective conductivity (W/(m K)) |
|---|---|
| 0.01 | 0.2237 |
| 0.02371 | 0.2552 |
| 0.05623 | 0.3255 |
| 0.1334 | 0.4717 |
| 0.3162 | 0.7339 |
| 0.7499 | 1.1 |
| 1.778 | 1.466 |
| 4.217 | 1.728 |
| 10 | 1.874 |
Sweep one variable
Variables and how each one contributes
Effective conductivity
What the equation gives you. Everything else feeds it.
Conductivity of path 1
The conductivity of path 1, in W/(m K). Raises the effective conductivity in proportion to its area share. Raising it by 10 % (to 0.22) moves the result by +2.31 %, from 0.65 to 0.665 W/(m K).
Area of path 1
The cross-sectional area of path 1, in m². A larger area gives this path more weight in the average. Raising it by 10 % (to 0.825) moves the result by −4.83 %, from 0.65 to 0.6186 W/(m K).
Conductivity of path 2
The conductivity of path 2, in W/(m K). Raises the effective conductivity in proportion to its area share. Raising it by 10 % (to 2.2) moves the result by +7.69 %, from 0.65 to 0.7 W/(m K).
Area of path 2
The cross-sectional area of path 2, in m². A larger area gives this path more weight in the average. Raising it by 10 % (to 0.275) moves the result by +5.07 %, from 0.65 to 0.6829 W/(m K).
Conductivity of path 3 (unused if A₃ = 0)
The conductivity of path 3, in W/(m K). Matters only if its area is not zero. Raising it by 10 % (to 1.1) moves the result by +0 %, from 0.65 to 0.65 W/(m K).
Area of path 3
The area of path 3, in m² (zero if unused). Adds weight to path 3 as the area grows. Raising it by 10 % (to 0) moves the result by +0 %, from 0.65 to 0.65 W/(m K).
Worked example with the values above
- Weighted terms: k₁A₁ = 0.15, k₂A₂ = 0.5, k₃A₃ = 0; sum = 0.65.
- Total area: 0.75 + 0.25 + 0 = 1 m².
- k = 0.65 / 1 = 0.65 W/(m K).
Assumptions and limits
- Steady state, constant properties unless the page states a temperature law, and no internal heat generation.
- One-dimensional conduction in the stated geometry, with no edge or end losses.
Common mistakes
- Using diameters where radii are needed.
- Using °C in a formula that needs an absolute temperature, or leaving a temperature difference out of the heat rate.
- Applying a plane-wall formula to a curved wall when the thickness is not small compared with the radius.
Statistics and uncertainty
Thermal conductivities are rarely known better than 5 % to 10 %, and contact resistances can vary by a factor of two between nominally identical joints. Treat results as estimates unless the inputs were measured.
Where it is used
- Insulation sizing for vessels, pipes and walls.
- Estimating interface and layer temperatures in composite constructions.
Maths you need
- Fourier’s law: q″ = −k dT/dx
- Thermal resistance: R = ΔT/Q, in K/W
- Series and parallel paths: resistances add in series; conductances add in parallel