D
A 10 HP blower motor is fully loaded and operating at a brake hp of 10. The blower is moving 10,000 CFM of air. A variable frequency drive decreases the motor rpm and thus the CFM to 8,000 CFM.
By what % is the motor HP decreased?
A. 20%
B. 30%
C. 40%
D. 50%
D
bump this
A
Life goes on long after the thrill of living is gone.
So ct2, is that a B?Originally posted by ct2
bump this
C
c
Norm : no that isnt a b ...your thread was falling to the bottom of the list last nite so I "bumped it back to the top" I am waiting for a reply to the message I sent you before I give an answer
Ok, so what is the answer ? Any one else going to give it a shot ?
Life goes on long after the thrill of living is gone.
Give this a little more time and then not only will I give the correct answer I will explain why it is correct.
This is useful information.
c
I would say (A). Since 8,000 is 20% less than 10,000.
tech cr that would be to easy true the CFM has been reduced by 20% but I'll bet the HP reduction will not be 20%
Wrongo.....
[Edited by condenseddave on 02-15-2005 at 01:36 AM]
That one made sense for a minute, now I'm second-guessing that 40% thing. Please stand by.........
This was the smartest thing I posted on this thread.
[Edited by condenseddave on 02-15-2005 at 01:37 AM]
Aren’t you supposed to be on vacation?
I say it's D for down shifting. LOL
(RPM1/RPM2) = (CFM1/CFM2)
(BHP1/BHP2) = (RPM1/RPM2)^3
8000CFM/1000CFM = 0.8 = (RPM1/RPM2)
0.8^3 =0.512 ==> approx 50% ish.
D.
10x(8000/10000)3{Cubed}
Should be "8" as in 8 hp.
So, it's 20%
Pumps are easier.
Well I’ll be…..look here, I was right.
OK, that works.....Originally posted by clydemule
(RPM1/RPM2) = (CFM1/CFM2)
(BHP1/BHP2) = (RPM1/RPM2)^3
8000CFM/1000CFM = 0.8 = (RPM1/RPM2)
0.8^3 =0.512 ==> approx 50% ish.
D.