Plz help me!!

[Zapperosini]

Sorry for ur unconvenience! but i got some problems with the amount of air supply to desired room. The desired area condition are 77F, 50%Rh then,the outside area are 95F, 74%. My SHR is 0.56. When i plotted to psychrometric chart to figure out the air supply temp., I have gotten a 45F. I think it too difference from the desired room condition by 32F. Could u give me some advise for this case.

thank u

 

[mechanicaldude]

How did you come up with your SHR ? What type of room is it?

 

[ChrisConley]

Your supply temp is low because your SHR is low. What a SHR of 0.56 is saying is that you have relatively large latent loads compared to the sensible loads. As a result you will need cold/dry air to meet the latent loads of the space.

 

[NCDesign]

Zapperosini the information given is not enough to really help you out. It looks like the SHR you gave was from your outside air condition to your desired room air condition. Not the actual SHR for the load in your space. The desired room condition with that SHR cannot be met with conventional cooling equipment. The way you are approaching your problem you need to determine the SHR for the space load and ignore external conditions except for their possible relationship to conduction and convection of heat into the space through walls, roof, windows, physical openings, etc. The cooling load & associated airflow required to maintain space conditions is a separate problem from the cooling required to get your air source to the required conditions.

For example your space may have a 12,000 BTU load with an SHR of 0.9. If your desired space condition is 75F DB/ 50% RH you would need to leave your coil at approximately 56F DB/55F WB with 400 CFM of air. But if you have to provide that cooling via 400 CFM of air that starts at 94F DB/74% RH air your total load would be 51,480 BTU. You could take those new numbers (94F to 56 F) and calculate an SHR but it doesn't really have any meaning.

 

[imok2]

I have recalculated and have come up with a SHR of 58 and assuming a 20% osa I came up with a mixed air of 80.5*F DB and 58.66 RH air on and air off at 55*F and 90%RH this gave me a total enthalpy diff. of 11.52 btu/lb air and a total of 33.86 gr/lb moisture. Using 2000 CFM I got 8600 lb/hr for a total heat load of 98,900BTU/HR and approxmately 41 lbs/hr water, then I multiplied times 1000 btu/lb water to find SHR. I don't do this very often but I believe this is the right approach