Recent content by Dirn

Thanks for the reply again. .043 cfm (material) seems really low. That means it would take nearly 3 hours to fill a 55 gallon drum.

Would 23.33#/min be more realistic? 35[cu.ft(air)/lb(material)]*100lb/cu.ft(material)/150cfm(total flow rate) ?

Dear Ione, Thanks for your reply. I didn't word my post properly, well left out what I was looking to accomplish. Vacuum application. I'm looking for the rate the material will be picked up per min lets say. My vacuum can handle 150cfm @ 7000fpm thru a 2" line. My material is 100# per cubic...

I think I'm missing something very simple but I'm stumped. Here's my problem. I have a material that weighs 100# pre cubic foot. I have a conveying velocity of 7000 fpm and a total CFM of 150. The Cu. Ft. of air per pond of material is roughly 35. Based off these numbers I'm trying to figure...

Thank you for pointing me in the correct direction!

I have a vessle that I have to give the "P reduced" for but do not have any data on it since it's a new design. Is there a way to figure this out without using the arbirary numbers described in the FM data sheet 7-76? We are certain the cyclone can handle more than 4.4 psig but have not idea...

Never mind, rule of thumb is 1.5x the hp of an electric motor. At least in my application.

I have a 7.5hp elec motor and need to convert it to a gas engine... I know they rate the two differently but I can't remember the conversion or ratio. Does anyone know the proper conversion? Or have a link? Thanks!