Power transfer of a shaft

  • Thread starter rad10k
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  • #1
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Homework Statement




Calculate the power that can be safely transferred by a medium carbon steel shaft of 100 mm diameter at 150 rpm ?

Homework Equations



Assumed solid circular shaft.

The Attempt at a Solution




Material : Medium Carbon steel t=580 MPa y= 50 mm k=6 ( table 9 live load varying)
T= Torque

J= polar moment of inertia for a solid circular section = pid^4/32

J=3.14*1 000 000 00 / 32 = 9 812 500 mm^3

tJ/ky = T = 580 * 9 812 500 / 300/1 000 = 18 970.8333 N m


150 rpm / 60 = 2.5 * 6.28 = 15.7 rad/s

18 970.8333 * 15.7 = Power = 298 kW / 745.7 * 1000 = 400 HP

Could someone please look over my answer ?
 

Answers and Replies

  • #2
nvn
Science Advisor
Homework Helper
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Excellent work, rad10k. Your answer appears correct. It would be better to leave your answer in standard units (kW), instead of converting to an ambiguous, incoherent imperial unit (hp). There are six ambiguous definitions for the name hp. Just stick with standard units (SI). One minor comment. Your units for J should be mm^4, not mm^3.
 
  • #3
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thanks nvn. I will remove the HP from my final answer . :eek:)
 
  • #4
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this has come as incorrect ?? any inclination as to why ? thanks
 
  • #5
nvn
Science Advisor
Homework Helper
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rad10k: If your tau and k values are correct, then your current answer, 298.1 kW, currently appears correct, and should be rounded to 298 kW. Perhaps post the exact wording of the given question, so we can check it, if you wish.

There is no reason to round pi to less than four significant digits. Please reread the last two paragraphs of post https://www.physicsforums.com/showthread.php?t=463768#post3087641".
 
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