Thermal Contact Resistance

1. The 5 A of current at 24 Vdc flowing through a resistor create a Joule heating within a cylinder having a diameter of 2.5 cm and a length of 30 cm. The average surface temperature of the cylinder is found to be 75 deg. Celsius during steady-state natural convection in air having a free stream temperature of 20 deg. Celsius. What is the value of the convection heat coefficient, h for this case?

2. With respect to thermal contact resistance at the interface of two surfaces in physical contact:

a) How is taken into account in a heat conduction analysis?

b) Is the contact resistance higher, or lower for metals having smooth surfaces, as compared to those having rough surfaces?

c) Provide a quantitative example to support your response to b)

d) What important variable exhibits a significant change at the surface interface?


The power dissipated into the resistor is

$P = U*I = 24*5 =120 W$

But the power is $P = Q/t$, where Q is the total heat (in J) and t is the time

The heat transfer coefficient is

$h = Q/[S*t*\Delta(T)]$,   where S is the total surface area for heat
and $\Delta(T) =(75-20) =55 degree C$ is the difference in temperature between surface and fluid

$S = \pi*d*L$ (we can neglect the bases area because it is small comparable to the lateral area)

d is the diameter, L is the length of cylinder

$S =\pi*0.025*0.3 =0.0236 m^2$

In this case the heat transfer coefficient is

$h = (Q/t)*[1/S*\delta(T)] =P/[S*\delta(T)] =120/0.0236/55 = 92.6 W/(m^2*K)$


a) The thermal contact resistance between two surfaces is BY DEFINITION the reverse of the thermal contact conductance.

$Rt = 1/h = \Delta(T)/(d Q/d t)= \Delta(T)/P$

where Q is the heat dissipated during time t, P is the power and T is the temperature difference between the two surfaces.

In a heat conduction analysis to find the temperature difference between two surfaces one needs to know the power dissipated and the thermal contact resistance.

$T2-T1 = P*Rt$

b) For smooth surfaces as compared to rough surfaces the thermal contact resistance is lower. This happens because at microscopic level the contact area between surfaces is higher in the case of smooth surfaces.

c) For the case of a microprocessor chip mounted on base a surface with a power dissipated of 60W.

In the case of smooth contact $Rt = 0.15 (deg C)/W$,   $(T2-T1) =60*0.15 =9 deg Celsius$

in case of rough contact $Rt = 0.4 deg C/W$    $(T2-T1)=60*0.4 =24 deg Celsius$

d) At the surface interface the TEMPERATURE charges significant between the two surfaces as can be seen from the above example.