Help! calculating new fan RPM with static pressure and airflow changes
Hello, I am kind of at a loss with a project I am working on
I am working on an energy savings project to decrease airflow in an existing system. The current design has a very high airflow that is no longer required.
Where I am confused is
1) one motor attached to the exhaust system is distributed to two seperate areas. One area is not used anymore, so dampers will close off that ductwork, leaving only the one area operating. This changes the fan curve -static pressure and rpm- while maintaining the existing airflow for that area.
Total current CFM= 23024
CFM for area that will remain operational= 11032
Static pressure at the one area=5
Total static pressure= 12.33
I found the Constant of the total system= 5.491E-8
and then plugged in the CFM of the area I am using(11032) and found my new static pressure= 6.68
I plotted this on my fan curve( shifted the curve) and the HP dropped to approx 18BHP
One area running specs
** I dont know how to get an accurate NEW RPM from this information??
2) Now for the energy savings part!
once I know what the calcuated line is running only the one area, I will be decreasing that CFM (11032) to 4600
Using the constant(5.491E-8) I found my new static pressure= 1.16
**the reason I want to be able to get the RPM at the energy savings point is so that I can ensure, when I add a VFD to the motor, it will run a no less than 20HZ
This is the information I have to work with.
Also- air quality testing will be preformed post install. I just want to be able to get the math right
You are cutting the airflow in half.... i hope you have hot gas by-pass at the very least or you are going to have problems with that refrigerant circuit. You're doing it the hard way. Use the blower performance charts for the unit and check your pressure drop across the coil, or if they use a different method then use that. but use the manufacturer's literature.
Thank you for your reply. Sorry, I should have been more clear. This is my first post, so I am hoping I put it in the right forum
This system is actually a ventilation system for a weld booth. The fans/exhaust are only ment to filter the welding fumes. We used to do a TON of welding in the booth, but now we rarely weld in it. So the amount of airflow is not required.
I have been reading up on fans laws and fan charts, but am now a little lost.
Multiply required CFM by known RPM and then divide by known CFM = new RPM
Something is wrong with your math - as I take a quick look, unless your ducts are SUPER small, you cannot really cut airflow in half, Amps by 3/4, all while generating an INCREASE in static pressure. I am assuming the flow for the remaining area is staying the same as before the change, and you are not trying to cram any more air into the same size ducts as before.
Also - an air distribution system designed to deliver 5" SP is some serious air moving equipment. Is there a reason you have fans that can move against that amount of resistance? 1" to 5" is only 4" - but the difference in equip and amp draw, noise even blower wheel design is really really different. If you want to end up at about 1.16" TESP - then changing from a radial fan (what you likely have) to a forward incline fan will give a large benefit for airflow, and noise with less amp draw.
Also remember that your constant (wherever you got it from) seems to be based on the change from 23000 CFM to 11000 CFM, but you need to go now from 11000 CFM to 4600 CFM. So all your fan calculations need to be based on that difference. so you need to generate a new constant
Any installation of a VFD is usually not a problem, but when your fan motor is subjected to REALLY wide variation in frequency, the fan motor usually needs to be inverter duty, or it will burn out.
Lastly check the output of the fan to be sure that the fan itself can deliver an airflow that low properly.