ncees sample exam 510 enthalpy wheel problem
Posted 17 October 2011 - 07:35 PM
any comments on 510 problem?
OA: 95DB, 78WB 1500cfm
EA:75DB 50%RH 1500 cfm
enthalpy wheel eff 80%
DB of TA (tempered air)
a75.9F b79.0F c85.0F d91.0F
the book gives the formula
TA db= OA db-(OA db-RA db)*wheel eff.
no objection for the formula but is it possible for a wheel to decrease temperature that much?
I assume that the temperature must be close to 95F with 80% eff not to 75F as indicated
Posted 17 October 2011 - 08:04 PM
So, if the sensible effectiveness really is 0.8 then yes the final temperature would be 79.
Edited by jamiecta, 17 October 2011 - 08:18 PM.
Posted 18 October 2011 - 03:48 AM
so enthalpy wheels are much more efficient thn simple mixing
I did not know that
Before, I thought that wheels are used when you do not want to mix air but want to decrease the outside air load
the formula neglects air flows as well. For me that is a question mark also. but I guess that is the way it is for wheels the flows must be equal.
thanks for the input
Posted 18 October 2011 - 12:00 PM
Edited by jamiecta, 18 October 2011 - 12:00 PM.
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