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Les · 12-15-2005, 06:52 AM

Rewritten last Post.
Not now completing attachments to Post 58
Attachment deleted (errors in sums) and will post at my leisure in new thread.

Edit:
The root of the prob which caused withdrawing of my model AMD3000+ attachments:-
Get different solutions when considering "bp +IHS" as one entity and calculating separately .
e.g Apogee(34diam x3)+TIM(0.06c/w*cm^2) + IHS(34diamx1mm) on 11.34 diam Source.
At h(eff)=43788 and IsoFlux
Consider as continuous 4mm or continuous (3+1)mm then adding TIM get: R(water -> far IHS surface) = 0.112777+0.006607=0.11938
Considered as (3mm with h=h(eff)) + (1mm with h=1/(Rbp+1/h(eff)*A+Rtim)*A get: R(water -> far IHS surface) = 0.156659

12-15-2005, 06:52 AM	#87
Les Cooling Savant Join Date: Oct 2001 Location: Wigan UK Posts: 929	Re: The plan for my next week Rewritten last Post. Not now completing attachments to Post 58 Attachment deleted (errors in sums) and will post at my leisure in new thread. Edit: The root of the prob which caused withdrawing of my model AMD3000+ attachments:- Get different solutions when considering "bp +IHS" as one entity and calculating separately . e.g Apogee(34diam x3)+TIM(0.06c/wcm^2) + IHS(34diamx1mm) on 11.34 diam Source. At h(eff)=43788 and IsoFlux Consider as continuous 4mm or continuous (3+1)mm then adding TIM get: R(water -> far IHS surface) = 0.112777+0.006607=0.11938* Considered as (3mm with h=h(eff)) + (1mm with h=1/(Rbp+1/h(eff)A+Rtim)A get: R(water -> far IHS surface) = 0.156659 Last edited by Les; 12-16-2005 at 06:15 AM.