Thermal Modeling and Analysis of Device-Contactor ... - Johnstech

B. Solution of Thermal Resistance and Heat Transfer Capability of Contact InterfacesHeat transfer across a metallic interface is stipulated by determining the heat transfer interface coefficient,h i . A closed-form useful equation that depicts this relation is given by Cooper [5] and Remsburg [1] and is:h i = 1.45 * {[κ * (P a /H) 0.985 /σ]} * tanφ Watts/m 2 – K (4)Whereκ = 0.5 * κ 1 * κ 2 /(κ 1 + κ 2 ) and κ 1 and κ 2 are the thermal conductivity of materials 1 and 2 in(W/m-K)P a = contact pressure at the thermal interface (N/m 2 )H = hardness of the softer material (N/m 2 x 10 8 )σ = (σ 1 2 + σ 2 2 ) 1/2 and σ 1 and σ 2 are the rms roughness of materials 1 and 2 (µm)tan Φ 1 = 0.125 * ( σ 1 x 10 6 ) 0.402tan Φ 2 = 0.125 * ( σ 2 x 10 6 ) 0.402tanΦ = (tan Φ 1 2 + tan Φ 2 2 ) 1/2 where Φ 1 and Φ 2 are the absolute asperity angles of materials 1 and 2However, the heat transfer coefficient, h i , is only one facet of heat transfer across an interface and applies tothe asperites in contact with each other. This area has been investigated and analyzed by Kogut andKomvopoulos [6], et. al. Most of the contact area is a void consisting of valleys between the peaks. Thesemust be accounted for since they form a parallel, albeit less conducting path in the contact area which sumswith the paths of “hard conduction.” This approach was addressed by Shlykov [7] and is defined by theequation,h gi = k g /Y = k g /[(Y/σ) * σ] Watts/m 2 - K (5)Remsburg [1] posits that the quantity, Y/σ, relates to the surface finishing of the material and is shown inTable 1 as a metric function of the finishing operation.Table 1. Finishing Operation vs. Values of Y/σFinishing Operation Y/σGrinding 4.5Hyperlap 6.5Sandpaper 7.0Superfinish 7.0Lap w/loose abrasive 10.0The total heat transfer across an interface between two materials is thus the sum of h i and h gi . By way ofexplanation, these parameters form a parallel path for heat transfer. In this respect, they are thermalconductance paths and parallel conductances sum directly. The inverse of the product of this value and theapparent area of surface contact, A a , will give the thermal resistance, θ a (°C/W), as stipulated in Eq. 5immediately below.Θ a = 1/[( h i + h gi ) * A a ] (6)Another factor in calculating the thermal resistance of an interface lies in determining the area of contactbetween the elements. Since the roughness of the surface of each contacting material depends on the natureof said material, and since the softer material will be compacted by the harder one under an applied force,one can create a geometric condition that can be used to estimate the degree of contact. Refer to Fig. 1 andEq. 7 for an approximation to the solution.3