Hybrid Car – More Fun with Less Gas

Thermal Mass equation help needed - Page 2

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Posted by nicksanspam on November 12, 2008, 2:27 am
 


Daestrom suggests a way:


If the average indoor and outdoor temps are (say) 65 and 40 F and you burn
210 lb/week of wood at 60% efficiency, that makes the house conductance G
= 0.6x10Kx210/(7dx24h)/(65F-40F) = 300 Btu/h-F. If the house cools from 72
to 61 in 18 hours on a 40 F night, RC = C/G = -18/ln((61-40)/(72-40))
= 42.7 hours, so C = 42.7hx300Btu/h-F = 12.8K Btu/F.

Cooling from 72 to 61 on a 30 F night makes RC = -18/(ln((61-30)/(72-30))
= 59.3 hours, so C = 59.3x300 = 17.8K Btu/F, an increase of 5K Btu/F, eg
5K pounds or 600 gallons of water, or less, if the water cools from (say)
100 vs 72 F, or still again less, if the house temp instantly drops from
72 to 61, where it stays for 18 hours, with controls to regulate the heat
release from the mass to maintain a constant vs drooping house temp.

Nick


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