I know this has nothing to do with diesels but can someone help me..

[XcumminsX]

hey matt, thanks for the post. How did you get to that tho, what formulas did you use?

 

[Mathew Clausen]

From my other post - 1hp = 33,000 ft*lb/min



In your case you have



1350 lb * 200 ft/min = 270,000 lb*ft/min (notice - same units as above)



so to get hp, use the above conversion and:



270,000 / 33,000 = 8. 18 hp



Make sense?



Mathew

 

[XcumminsX]

so now i need to find a shaft size that will support that power. I really wish i bought a newer machinerys handbook i really regret having the 19th edition the newer one is the 26th version! :--)

 

[XcumminsX]

here is what i got so far am i right??? If i am wrong please tell me



1. what is the shaft size? (need to calcutlae for simple bending load)



I STILL DONT NOW THIS ONE



2) determine a pulley configuration to turn that BLACK BOX at 725 rpm +/- 10 percent.



-Pulley size formula.

Speed of driven (shaft) is equal to driving pulley dia. (motor) multiplied by speed of motor (RPM) divided by the (shaft) driven pulley dia.



Calculate the conversion and I ended up finding the size of the pulleys.



I used 5" because this gave me an even sized pulley for the driven shaft at standard motor rating of 900 rpm.



-driving pulley, driving rpm and driven rpm.

5" times 900 divided by 750 = 6" (drive pulley)



3) determine a standard motor size and rpm



10 Hp motor



900 rpm



4) determine a bearing configuration for the system.

I Went with using bearings that where both 95% efficient (. 95)(. 95)= . 9025



10HP (. 9025) = 9. 025 or 9 (rounded)