need help with heat transfer project Task The surface temper
Solution
Solution :
taking room temp as 25 C
Initial temp is 85
temp reduces by 22 Tf= 85-22 = 63
Heat loss or decrease in temp of heater is taken by heat sink
heat loss in heater is giveb by Q = hA(T-To)
= h A (T-To) where A is 7.62 x 2.54/10000 m2 ; To is oustide temp
this heat will take up by heat sink made of Aluminium
m Cp dT = h A (T-To)
density volume Cp dT = h A (T-To)
densiy of Al is 2830 kg/m3
Cp is 1.005
taking integral from given temp limit that is 85 to 63, equation reduces to
1/T-To dT = hA/(m Cp)
taking integral
log (T-To) = hA/(mCp)
log (Tf-To)-log(Ti-To) = hA/(mCp)
log((Tf-To)/Ti-To)) = hA/(mCp)
0.456 = hA (mCp)
A is 7.62 x 2.54/10000 m2
Cp is 1.005
m is density x volume = 2830 (V)
taking h for given conditions V can be calculated for required heat sink.
