From: jim.richards65@yahoo.com   
      
   On Tue, 28 Feb 2012 10:51:44 -0500, j  wrote:   
      
   >  I have a prototype central solar heater which I have been tweaking and   
   >am interested in calculating heat loss in the ducts.   
   >   
   >   Here is the way I think this would work.   
   >   
   >   Insulated ducts have an R value. I think this R value relates to the   
   >total inside area.   
   Not unless you are accounting for the thickness and heat capacity of   
   the duct itself, otherwise the diameter or OD is just fine.   
      
   >So, Inside Diameter * pi * length. So the heat loss   
   >is proportional to:   
   >   
   >ti = temperature inside duct   
   >to = temperature outside duct   
   >   
   >(ID * pi * l) * (ti - to) / R   
   >   
   >units in feet.   
   >   
   >Is that about right?   
   >   
   >Obviously if there is no flow through the duct, there is no heat loss.   
   >So the heat loss in BTUs must be proportional to the mass of air flowing   
   >through the duct.   
   >   
   >Now, I'm getting confused (at least more confused). That should be the   
   >mass of air per hour * the specific heat of air.   
   >   
   >The mass of air/hour = the density of ~ .075 * CFM * 60 minutes   
   >   
   >or:   
   >CFM * 4.5   
   >   
   >specific heat (dry air) = .24   
   >   
   >what the actual specific heat of 50% RH air is beyond me, but I think it   
   >is 5 or 10% higher.   
   >   
   >So loss in BTU/hr:   
   >   
   >(ID * pi * l) * (ti - to) * (CFM * 4.5) * .24  * 1.1 / R   
   >   
   >with ID and l in feet   
   >   
   >Is that about right? It seems like I missed something, somewhere...   
   Yes, maybe, depends on if the ducts are all in unconditioned space and   
   you may want to consider duct pressure loss.   
   >   
   >Jeff   
      
   Curbie   
      
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