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ncees sample exam 510 enthalpy wheel problem

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Hi Guys;

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.

=95-(95-75)*,8

=95-20*,8

=79 (B)

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

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If the wheel was really 80% effective, then yes the outside air would be cooled closer to 75 than the original 95. That is how effectiveness is measured. If the effectiveness was 50%, then the final temperature would have been 85.

So, if the sensible effectiveness really is 0.8 then yes the final temperature would be 79.

Edited by jamiecta

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wow

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

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I believe that ASHRAE 84 says that the mass flow rates of the entering and exiting air streams should be the same (at least when they test the efficiency). The formula that NCEES posted is simplified because I believe the mass flow rates in that problem were already about equal. If the mass flow rates are NOT the same, you DO need to adjust/correct for this. Both ASHRAE Fundamentals and Systems and Equipment handbooks discuss this I believe.

Edited by jamiecta

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I realize this is an old thread, but this link might help someone with energy wheels. Page four of the brochure has an example and calculations on it.

Thanks

Kelly

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