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Coupling Selection Formula

The document provides calculations to determine the required torque of a coupling given an engine with 40 horsepower rated at 7162/15 revolutions per minute with a 30% load added. The calculations show the engine horsepower multiplied by the rated revolutions results in 19096, adding 30% load increases it to 5728 for a total required torque of the coupling of 24824.

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Varun Malhotra
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169 views1 page

Coupling Selection Formula

The document provides calculations to determine the required torque of a coupling given an engine with 40 horsepower rated at 7162/15 revolutions per minute with a 30% load added. The calculations show the engine horsepower multiplied by the rated revolutions results in 19096, adding 30% load increases it to 5728 for a total required torque of the coupling of 24824.

Uploaded by

Varun Malhotra
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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SELECTION FOR COUPLING

HPX7162/15 + 30%

SUPPOSE :-
40 HP X 7162/15 = 19096+5728 = 24824

THEN REQD. TORQUE OF COUPLING IS 24824

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