Library / Heat transfer and thermal systems / HTS-02
Heat transfer and thermal systems
Temperature jump across a thermal contact
Temperature drop across an interface carrying heat flux q″, from the area-specific contact resistance R″c.
- Ref
- HTS-02
- Rev
- v1
- Field
- Heat transfer and thermal systems
- Result
- ΔT [K]
Real joints have a thin region of imperfect contact, which behaves as an extra resistance. The temperature jump across it is the heat flux times the area-specific contact resistance, ΔT = q″R″c. The value depends on surface roughness, contact pressure, filler and temperature, and is usually measured rather than calculated.
Temperature drop across an interface carrying heat flux q″, from the area-specific contact resistance R″c. The sign gives the upstream minus downstream temperature for the chosen flux direction.
Calculator
- Contact conductance, 1/R″c5,000 W/(m² K)
- Flux that gives a 10 K jump, 10/R″c50,000 W/m²
- Thickness of a 1 W/(m K) material with the same resistance2×10−4 m
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.
- A flux of 6,000 W/m² through a joint of 2×10−4 m² K/W loses 1.2 K at the interface.
- That is the same temperature drop as 0.2 mm of a 1 W/(m K) material, so a poor joint can matter as much as a layer of insulation.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (K) | Local step | Change in result |
|---|---|---|---|---|---|---|
| q″ Heat flux across the interface | 10 to 1,000,000 W/m² | 0.002 | 200 | 200 | +10% | +10 % |
| R″c Area-specific contact resistance | 1e-6 to 0.01 m² K/W | 0.006 | 60 | 59.99 | +10% | +10 % |
Sweep of Area-specific contact resistance
| Area-specific contact resistance (m² K/W) | Temperature jump (K) |
|---|---|
| 1e-6 | 0.006 |
| 3.162e-6 | 0.01897 |
| 1e-5 | 0.06 |
| 3.162e-5 | 0.1897 |
| 1e-4 | 0.6 |
| 3.162e-4 | 1.897 |
| 0.001 | 6 |
| 0.003162 | 18.97 |
| 0.01 | 60 |
Sweep one variable
Variables and how each one contributes
Temperature jump
What the equation gives you. Everything else feeds it.
Heat flux across the interface
The heat flux crossing the interface, in W/m². Raises the temperature jump in direct proportion. Raising it by 10 % (to 6,600) moves the result by +10 %, from 1.2 to 1.32 K.
Area-specific contact resistance
The area-specific contact resistance, in m² K/W. Raises the temperature jump in direct proportion. Raising it by 10 % (to 2.2×10−4) moves the result by +10 %, from 1.2 to 1.32 K.
Worked example with the values above
- Heat flux: q″ = 6,000 W/m².
- Contact resistance: R″c = 2×10−4 m² K/W.
- ΔT = 6,000 × 2×10−4 = 1.2 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