Torque and RPM Analysis: Understanding Power Transmission with Pulleys

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SUMMARY

This discussion focuses on the relationship between torque and RPM in power transmission using pulleys. A driving shaft with a torque of 10Nm at 1000 RPM, when connected to two 3-inch pulleys and a pair of 6-inch pulleys, results in an output of 500 RPM. However, due to the conservation of energy, the output torque is calculated to be approximately 18Nm, assuming a belt efficiency of 95%. The key takeaway is that while RPM can be halved, torque will increase, but the total power remains constant.

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  • Understanding of torque and RPM relationships
  • Knowledge of pulley systems and belt drives
  • Familiarity with power calculations (Power = Torque x Angular Velocity)
  • Concept of energy conservation in mechanical systems
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Mechanical engineers, automotive technicians, and anyone involved in power transmission system design and optimization will benefit from this discussion.

skybug125
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Please I need some help. I'm thinking that if I have a driving shaft with a torque of 10Nm and an RPM of 1000. If I add two 3 inches pulley on the shaft to transmit power to another shaft with a pair 6 inches pulley mounted on it and connected to the initial 3 inches via a belt. Will I have transmitted a total torque of 40Nm to this new bar and an RPM of 500?
 
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The output will be 500rpm but the available torque will only be 20Nm (or less). Conservation of energy applies. If the belt was 100% efficient then Power Out = Power in.

Power = torque(in Nm) * angular velocity(in Rads/S)

So if you halve the rpm (angular velocity) you double the torque.

Using two belts in parallel doesn't magnify the torque (unless the torque was limited by the belt slipping).

It looks like your Power In is around 1000W. Belt drives can be 95% efficient so the Power Out is likely to be around 950W and at 500rpm that equates to 18Nm.
 
Thank you so much Cwatters, its obvious power must always remain constant. Thank you for taking out time explain this to me. Cheers
 

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