Calculating Power Transfer of a Medium Carbon Steel Shaft at 150 rpm

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Discussion Overview

The discussion centers around calculating the power transfer of a medium carbon steel shaft with a specified diameter and rotational speed. Participants are addressing a homework problem involving material properties, torque calculations, and unit conversions, while seeking validation of the solution approach and results.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • The initial calculation of power transfer by a medium carbon steel shaft is presented, including the use of material properties and formulas for torque and polar moment of inertia.
  • One participant confirms the correctness of the calculations but suggests using standard units (kW) instead of horsepower (hp) due to ambiguity in the latter's definitions.
  • Another participant points out a minor error in the units for the polar moment of inertia, indicating it should be mm4 instead of mm3.
  • A participant expresses confusion regarding the correctness of the answer and seeks further clarification on potential errors.
  • A suggestion is made to post the exact wording of the original question to verify the calculations against the problem statement.

Areas of Agreement / Disagreement

Participants generally agree on the correctness of the power calculation, but there is uncertainty regarding the initial question's wording and potential errors that may have led to the answer being marked incorrect. No consensus is reached on the source of the discrepancy.

Contextual Notes

There are unresolved issues regarding the exact wording of the homework question and the implications of rounding and unit conversions on the final answer.

rad10k
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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 ?
 
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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.
 
thanks nvn. I will remove the HP from my final answer . :o)
 
this has come as incorrect ?? any inclination as to why ? thanks
 
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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