Solution to Practice Problem 4.1a

a. We start with the temperature distribution Equation (3.40)
(3.40a)
and divide (3.40) by . We now let
 
 
 
 
and obtain
(3.40b)
If we now use the identity
(3.40c)
sinh (L-z) = sinh L cosh z - cosh L sinh z
then
(3.40d)
Substitution of equation (3.40d) into equation (3.40b) and rearranging gives
(3.40e)
b. The dimensionless parameters are P3 and P4
(3.40f)
If we divide the numerator and denominator of P3 by (T2 - Ta) we obtain
(3.40g)
Now let
(3.40h)
so that
(3.40i)
Note that P6 reflects the ratio of the electrical heat generation to the heat loss at the periphery of the rod and a length ratio (rod length to rod perimeter). It is easier using equation (3.40i) to sort out the factors which influence its value. That is one point of using dimensionless groups.
 
Using equation (3.42)
 
so that
 
The term ( hD / kf ) reflects the ratio of convective to conductive heat transfer at the surface of the rod and is therefore a function of the air velocity.



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