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## What is contact resistance in heat transfer?

The thermal contact resistance is the resistance to the conduction heat transfer present between two contacting materials per unit area.

The thermal contact resistance arises due to the improper contact between two bodies. Due to the irregularities present on the surface, the contacting area available for heat conduction is very less.

Due to the surface irregularities, the air or any gas is filled between these gaps which act as insulation for conduction.

## Causes of contact resistance:

The surfaces of the mating bodies which look smooth in visual inspection have irregularities on a microscopic level.

The above figure shows the microscopic view of contacting surfaces. As shown in the figure, the surface of bodies has a crest and valley, therefore the two bodies can contact each other at some points only while in the remaining portion, the air gap is present.

This air gap has comparatively low thermal conductivity, therefore it adds extra resistance to the flow of heat. This resistance per unit area of contacting surface is known as **contact resistance**.

## Contact resistance formula:

The thermal contact resistance for the two contacting surfaces is given by the ratio of **temperature difference at the contact of two bodies** to the **rate of heat flux**.

Mathematically, the contact resistance formula is given by,

`R_{TC}=\frac{T_{1}-T_{2}}{q}`

## Contact resistance units:

**SI unit:**–

In SI system, the unit of temperature is K, and the unit of heat flow rate is W/m²

∴ `R_{TC}=\frac{\Delta T}{q}=\frac{K}{\frac{W}{m^{2}}}` = K.m²/W

∴ SI unit of contact resistance is K.m²/w

**FPS unit:**–

In FPS system, the unit of temperature is °F and the unit of heat flow rate is BTU/ft².hr

∴ `R_{TC}` = `\frac{\Delta T}{q}=\frac{°F}{\frac{Btu}{Ft^{2}.hr}}`= °F.ft².hr/Btu

∴ FPS unit of contact resistance is °F.ft².hr/Btu.

## How to reduce contact resistance?

Here are some methods that will help to reduce contact resistance:-

1) Make the contacting surface highly polished which reduces the air gap between the two bodies.

2) The application of pressure on contacting bodies helps to improve contact between two bodies.

3) Increase in pressure of ambient gas present in gaps helps to reduce contact resistance.

4) Removing the air gap by application of high thermal conductivity grease between the contacting surfaces reduces the contact resistance.

the Temp of T1 and T2 are unknow ? how do we calculate the thermal resistance between the two layers at the interface between them ?

If you have thermal conductivity (K) for material A and B, Thickness (X_(A) & X_(B)), and amount of heat flux (q) then you can find the interface temperature T_(1) and T_(2) by Fourier equation for conduction as,

q = [T_(i) – T_(1)]/[(L_{Surface A})/(K_{Surface A})]

q = [T_(2) – T_(o)]/[(L_{Surface B})/(K_{Surface B})]

Now by using these temperatures T_(1) and T_(2) you can find contact resistance.