Split system question
Split system question
(OP)
Say a split system is rated at 2 tons
Total MBH=24.
Assuming the t-stat is set at 75F and the fan in the unit does 425 cfm.
So:
24000 = ( 425 cfm) * (1.1) * deltaT
Therefore deltaT = 51.33
Does that mean if my room is set to 75, my EAT=75 and my LAT off the coil is (75-51.33)=23.7F? Im trying to see what my LAT would be in this case.
Thanks in advance
Total MBH=24.
Assuming the t-stat is set at 75F and the fan in the unit does 425 cfm.
So:
24000 = ( 425 cfm) * (1.1) * deltaT
Therefore deltaT = 51.33
Does that mean if my room is set to 75, my EAT=75 and my LAT off the coil is (75-51.33)=23.7F? Im trying to see what my LAT would be in this case.
Thanks in advance





RE: Split system question
RE: Split system question
so if i use
Qt = cfm * 4.5 * deltaH
24000 = 425 cfm * 4.5 * deltaH
deltaH = 12.54
Now assuming EAT=75F at 50% then h=28.15
so 28.15-12.54 = 15.61 = h leaving coil
Im looking for some random point with about 100% saturated air and i am coming up with 25 db/ 25 wb that gives me a h=15.52. so is this the leaving air temperature?
RE: Split system question
the h that gives me 15.52 is roughly 41db/41wb. i believe this would be correct?
RE: Split system question
Just because a system is "rated" at 2 tons doesn't mean it will always provide 2 tons of cooling at the evaporator. In fact, most likely it will very rarely provide that.
RE: Split system question
The Btu/hr is a result rather than an input variable. For example, if your unit is cooling an inlet mixed airstream from 78°F down to 52°F at your given airflow (425 cfm), the unit is providing about 12,200 Btu/hr, or about a ton of cooling. That's about half its capacity; it has the nuts to handle a lot more load. That's sensible (Q = 1.1 cfm dT). Total load would be Q = 4.5 cfm dh, where h is in Btu/lbm.
Keep in mind your discharge temperature can't be below your apparatus dew point (a term I'm not to crazy about because it tries to make HVAC sound like rocket science)—basically your discharge temperature can't be below the lowest surface temperature of the coil. So to achieve your two ton rating you'd have to put in air that's about twice as warm or have about twice the airflow.