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The pump laws and the fan laws are the same with the units as the only difference. Once you learn one, you have learned both. This question is typical of what you can expect on the RSES CM and CMS exams. Much more challenging than those on the NATE and other certification exams. Norm
Norm I understood the question, was just kidding about what was asked. Yes I remember Fan Laws- Trane Book pg. 269. When I studied for the H-2 exam the instructor gave us quite a few problems. H-1 study course we got a lot of Pump and System Curve questions. Used B&G's Fluid handling training & education dept. publication TEH-375. Always enjoy your push for better education in this field and sharing of info and making folks use their brain here. Peace
Norm
Originally posted by gruntly Originally posted by selfemployed The motor is still 10hp. I'm with this selfemployed. Still a 10 horsey. Now what it draws is gonna be different... Don't confuse the motor's rated hp with its brake hp. In our example we had a 10 hp motor that was fully loaded and doing its full 10 hp worth of work. After reducing the airflow by only 20%, the same 10 hp motor was only doing 5 hp of work. The actual hp the motor is loaded to is called the brake hp. Norm
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. Clyde got it correct and posted the math. The exact answer is the brake hp drops to 5.12 hp. Whenever you see rpm in an equation you can substitute cfm as they are directly proportional to each other. The required hp drops by a cube of the drop in airflow or the drop in blower rpm. Of course, the reverse is also true. Increasing the airflow raises the required hp by the cube of the increase in cfm. Quite a difference in hp. That shows why it is that variable frequency drives can save a great deal of energy and it also shows how quickly you can load or overload a motor with slight increases in airflow. By the way, motor amperage and hp are directly proportional to one another as well. When you see amperage or hp in an equation you can substitute one for the other as well. So in Clyde's posted equation you could figure the new motor amperage instead of the new brake hp if you know the original motor amperage. The fan laws can be very useful if you learn to work with them and play with them a little. Norm
Originally posted by selfemployed The motor is still 10hp. I'm with this selfemployed. Still a 10 horsey. Now what it draws is gonna be different...
I have to agree with those who posted D.
new H.P.=old H.P.(new CFM/old CFM)^3 =10(8,000/10,000)^3 =10(.8)^3 =10(.512) new H.P.=5.12H.P., so answer is "D"
condenseddave what are your thoughts on Selfemployeds answer?
The motor is still 10hp.
I agree.
Originally posted by madeinusa Aren’t you supposed to be on vacation? Only from replying to nonsense threads and posts started by idiots that serve no real purpose. And that will only last until BC1 quits borrowing Don's balls to stop the trouble that they start, and then hightail it out.
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. OK, that works.....
Well I’ll be…..look here, I was right.
10x(8000/10000)3{Cubed} Should be "8" as in 8 hp. So, it's 20% Pumps are easier.
(RPM1/RPM2) = (CFM1/CFM2) (BHP1/BHP2) = (RPM1/RPM2)^3 8000CFM/1000CFM = 0.8 = (RPM1/RPM2) 0.8^3 =0.512 ==> approx 50% ish. D.
Aren’t you supposed to be on vacation? I say it's D for down shifting. LOL
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]
Wrongo..... [Edited by condenseddave on 02-15-2005 at 01:36 AM]
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%
I would say (A). Since 8,000 is 20% less than 10,000.
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