Quote:
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Originally Posted by unregistered
......Les, lolito, Inchoerent - I'll send the spreadsheet if you want it
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Yes please.
Ramblings:
The quantity many are interested in is (Two -Tai)
This gives the WBin temp.
For the simple case (using MTD not LMTD, and ignoring air's frictional heating) get* :
(Two -Tai)/Wa = R + 1/2QaCa - 1/2QwCw + dPw/2CwWa
and
(Twi -Tai)/Wa = R + 1/2QaCa + 1/2QwCw + dPw/2CwWa
* Using
Wa= dTmtd/R , dTmtd = 0.5(Twi + Two) - 0.5(Tao + Tai), Twi -Two = (Wa - QwdPw)/QwCw and Tao -Tai = Wa/QaCa ,
rearranging and substituting.
Where,
Wa= Heat dissipated into Air., R = (Convective + Resistive) Resistance of Radiator.
T= Temperature, Q= Flow-rate , dP = Pressure drop , C= Specific Heat.
Subscript "a"=air , Subscript "w"= water, Subscript "i"= in , Subscript "o"= out .
Edit 1: Deleted some erroneous manipulation. Will re-edit when corrected.
Edit 2: Will add further manipulation in post where used. Corrected typo.