Thermal Contact Resistance
Calculate microscale contact resistance and evaluate joint thermal properties with/without Thermal Interface Materials (TIMs).
Joint Interface Parameters
Establish contact boundaries and microscopic joint conditions:
- Roughness ($Ra_1, Ra_2$): Average surface profile micro-deviations.
- Hardness ($H_c$): Vickers microhardness of the softer mating metal.
- Pressure ($P$): Uniform mechanical contact force per unit area.
- TIM conductivity ($k_{TIM}$): Conductivity of grease/pad filling the air gaps.
- Bond Line ($t_{TIM}$): Thickness of the interface filler layer.
Configuration
Results & Visualization
Microscopic Joint Interface (Cross-Section)
Results and calculations will appear here after calculation.
Typical Range of Contact Resistance ($R"_c$)
Reference values from Incropera & Dewitt at vacuum/atmospheric interface pressure of ~10 MPa:
| Interface Type | Roughness / State | Typical $R"_c$ ($\text{m}^2\cdot\text{K}/\text{W}$) |
|---|---|---|
| Copper - Copper | 1.0 µm, Ground, Air | $0.5 \cdot 10^{-5}$ to $2.5 \cdot 10^{-5}$ |
| Aluminum - Aluminum | 3.0 µm, Ground, Air | $1.5 \cdot 10^{-5}$ to $4.0 \cdot 10^{-5}$ |
| Stainless Steel - SS | 2.5 µm, Ground, Air | $2.0 \cdot 10^{-4}$ to $5.0 \cdot 10^{-4}$ |
| Silicon Pad Interface | Under standard load | $1.0 \cdot 10^{-4}$ to $8.0 \cdot 10^{-4}$ |
| Thermal Grease Interface | Under standard load | $0.5 \cdot 10^{-5}$ to $3.0 \cdot 10^{-5}$ |
Calculation Methodology & Theory
Mathematical Model
On a microscopic scale, when two solid surfaces are pressed together, they only touch at discrete spots (asperities) due to surface roughness. Heat transfer across the joint occurs via parallel pathways: conduction through solid micro-contacts and conduction through the gaps filled with air or TIM.
Mikic-Rohsenow & CMY Models:
- Harmonic thermal conductivity ($k_s$): Combines solids: $k_s = \frac{2k_1 k_2}{k_1 + k_2}$
- Effective roughness ($\sigma$) & slope ($m$): Combined interface geometry: $\sigma = \sqrt{Ra_1^2 + Ra_2^2}$ and $m = \sqrt{m_1^2 + m_2^2}$ where the slopes are correlated to roughness as $m_i \approx 0.076 \cdot Ra_i^{0.52}$.
- Solid Contact Conductance ($h_{c,s}$): Formulated based on elastic/plastic deformation: $h_{c,s} = 1.25 \cdot k_s \cdot (m/\sigma) \cdot (P/H_c)^{0.95}$ where $H_c$ is the microhardness of the softer material.
- Mean Gap Separation ($Y$): Derived from the Cooper-Mikic-Yovanovich (CMY) distribution model: $Y = 1.53 \cdot \sigma \cdot (P/H_c)^{-0.097}$
Academic References:
- Cooper, M. G., Mikic, B. B., & Yovanovich, M. M. (1969). Thermal contact conductance. International Journal of Heat and Mass Transfer, 12(3), 279-300.
- Mikic, B. B. (1974). Thermal contact conductance; theoretical considerations. International Journal of Heat and Mass Transfer, 17(2), 205-214.
- 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 3.1.5 (Thermal Contact Resistance).
Worked Engineering Example
Two aluminum plates ($k = 180$ W/m·K) are pressed together at $2.0$ MPa. The surface roughness is $Ra_1 = 1.5$ µm and $Ra_2 = 2.0$ µm. The microhardness of aluminum is $950$ MPa. The interface is dry and filled with air ($k_g = 0.026$ W/m·K) and has an area of $10$ cm². The heat dissipated is $100$ W. Calculate the contact resistance and interface temperature drop.
Step-by-step Solution:
1. Calculate effective roughness ($\sigma$) and average slope ($m$):
$$\sigma = \sqrt{1.5^2 + 2.0^2} = 2.50 \text{ µm}$$ $$m_1 = 0.076 \cdot 1.5^{0.52} = 0.0939, \quad m_2 = 0.076 \cdot 2.0^{0.52} = 0.1087$$ $$m = \sqrt{0.0939^2 + 0.1087^2} = 0.1436$$ 2. Calculate solid contact conductance ($h_{c,s}$):
$$h_{c,s} = 1.25 \cdot 180 \cdot \left(\frac{0.1436}{2.50 \cdot 10^{-6}}\right) \cdot \left(\frac{2}{950}\right)^{0.95} = 37.9 \text{ W/m²·K}$$ 3. Calculate mean gap separation ($Y$) and gap conductance ($h_g$):
$$Y = 1.53 \cdot 2.50 \cdot (2/950)^{-0.097} = 6.94 \text{ µm}$$ $$h_g = \frac{k_g}{Y} = \frac{0.026}{6.94 \cdot 10^{-6}} = 3746 \text{ W/m²·K}$$ 4. Combine conductances, find resistances & temperature drop:
$$h_c = h_{c,s} + h_g = 3784 \text{ W/m²·K} \implies R"_c = \frac{1}{h_c} = 2.64 \cdot 10^{-4} \text{ m²·K/W}$$ $$R_c = \frac{R"_c}{A} = \frac{2.64 \cdot 10^{-4}}{10 \cdot 10^{-4}} = 0.264 \text{ °C/W}$$ $$\Delta T = Q \cdot R_c = 100 \text{ W} \times 0.264 \text{ °C/W} = 26.4 \text{ °C}$$