Question

Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially that of the surrounding air, that is, 20°C. Its width is 5.0 cm, thickness is 1.0 mm, thermal conductivity is 200 W/m·K, and base temperature is 130°C. The heat transfer coefficient is 20 W/m2·K. Estimate the fin temperature at a distance of 5.0 cm from the base. The fin temperature is

Answer 1

Answer:

$\frac{T-20}{130-20}= e^{-14.28*0.05}$

And if we solve for T we got:

$T= 20 + 110e^{-14.28*0.05} = 73.86 C$

The answer for this case would be $T = 73.86 C$ at 5cm from the base of the fin.

Explanation:

Data given

For this case we have the following data given:

$h = 20 \frac{W}{m^2 K}$ represent the heat transfer coefficient.

$p$ represent the perimeter for this case and would be given by:

$p = 2*0.05m +2*0.001m= 0.102m$

$k = 200 \frac{W}{m C}$ represent the thermal conductivity

$w = 5cm =0.05 m$ represent the width

$h = 1mm =0.001m$ represent the thickness

$A= wh= 0.05m *0.001m = 0.00005 m^2$

Solution to the problem

For this case we assume that we have steady conditions, the temperature of the fins varies just in one direction, the heat transfer coefficient not changes with the time and the thermal properties of the fin not change.

We can determine the temperature if the fin at x=5 cm=0.05 m from the base with the following formula:

$\frac{T-T_{\infty}}{T_b -T_{\infty}} = e^{-mx}$

Where m is a coefficient given by:

$m = \sqrt{\frac{hp}{kA}}=\sqrt{\frac{20 W/m^2 C 0.102 m}{200 W/ mC 0.00005 m^2}}= 14.28 m^{-1}$

The value of x for this case represent the distance $x =5 cm =0.05m$

$T_b =130 C$ represent the base temperature

$T_{\infty}= 20$ represent the temperature of the sorroundings or the ambient.

If we replace we have this:

$\frac{T-20}{130-20}= e^{-14.28*0.05}$

And if we solve for T we got:

$T= 20 + 110e^{-14.28*0.05} = 73.86 C$

The answer for this case would be $T = 73.86 C$ at 5cm from the base of the fin.