Heat Transfer in Micro and Nano Scales

Why it matters

Heat transfer at micro and nano scales is crucial for the design and operation of modern electronic devices, such as microprocessors and MEMS (Micro-Electro-Mechanical Systems). Understanding these phenomena helps in improving thermal management, enhancing device performance, and ensuring reliability.

Key ideas

• Scale Effects: At micro and nano scales, classical heat transfer theories may not apply due to size effects, quantum effects, and surface-to-volume ratio changes.
• Ballistic Transport: In small structures, heat carriers (like phonons and electrons) may travel without scattering, leading to ballistic transport, which differs from diffusive transport in larger systems.
• Quantum Effects: Depending on length scale, material and temperature, confinement and quantum effects can become significant, affecting the thermal conductivity and specific heat capacity of materials.
• Surface and Interface Effects: The increased surface area to volume ratio at these scales enhances the importance of surface and interface thermal resistances.
• Non-Fourier Heat Conduction: Model selection depends on carrier mean free paths, relaxation times and geometry. A non-Fourier model is not automatically required for every microscale device.

Formulas

• q = -k·∇T
  • q: heat flux vector (W/m²)
  • k: thermal conductivity (W/m·K)
  • ∇T: temperature gradient (K/m)
• q_interface = G·(T_1 - T_2)
  • q: heat flux (W/m²)
  • G: interface thermal conductance per unit area (W/(m² K))
  • T_1: temperature on side 1 of the interface (K)
  • T_2: temperature on side 2 of the interface (K)

Worked example

Problem: Assuming a diffusive effective-conductivity model is valid, calculate the heat flux through a material with a thermal conductivity of 150 W/m·K and a temperature gradient of 200 K/m.

Given:

• Thermal conductivity, k = 150 W/m·K
• Temperature gradient, ∇T = 200 K/m

Steps:

1. Use the formula for heat flux: q = -k·∇T
2. Substitute the given values: q = -150 W/m·K × 200 K/m
3. Calculate: q = -30000 W/m²

Final Answer: -30000 W/m²

Specific heat times temperature change gives energy per mass, not heat flux. Interface conductance has different units from conductivity; an area-specific interface resistance is 1/G. For example G = 100 MW/(m² K) and a 2 K interface temperature jump give q_interface = 200 MW/m². This is an illustrative interface calculation, not a material measurement. NIST thermal metrology distinguishes conductivity and interfacial conductance measurements.

Common mistakes

• Ignoring quantum effects and surface phenomena at nano scales.
• Applying classical Fourier's law without considering non-Fourier effects.
• Miscalculating the surface-to-volume ratio, leading to incorrect thermal resistance estimations.

For GATE ME

Questions may involve calculating heat flux, understanding the impact of scale on thermal properties, and applying non-classical heat transfer models. Practice problems on ballistic transport and quantum effects are beneficial.

Quick check

1. What is ballistic transport?
2. Why might Fourier's law not apply at nano scales?
3. How does the surface-to-volume ratio affect heat transfer at micro scales?

Answers: 1. Heat carriers travel without scattering. 2. Due to non-Fourier effects and quantum phenomena. 3. Increases the importance of surface thermal resistance.