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Power required to rotate the system |
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| Apr13-07, 04:56 AM | #1 |
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Power required to rotate the system
Hai
This is first post of mine regarding steady state power requirement. Since i am involving the power requirement project I am posting the question in simple way as follows. Drive & driven both two pulleys are connected by timing belt Drive pulley diameter is 100mm(face width25mm) Pulley is bearing supported Driven pulley is 300mm(face width 25mm) same bearing supported in the bore. Speed 500mm Mass of drive pulley is say 3Kg driven pulley is 20 kg Both pulley is supported by bearings.Bearing diameter is (drive pulley mean dia 40mm) Bearing diameter (driven pulley mean dia is 75mm) (Neglect the acceleration torque next phase we can discuss this) What is the power requirement of the above system? Assume no load condition(otherwise if you thought that no sense in this add load in driven pulley as 100kg) Thanks in advance Prakash |
| Apr13-07, 06:26 AM | #2 |
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Recognitions:
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If it's steady state theoretically there's no load required. In reality, the only load is then the required power to overcome motor inefficiencies, pulley windage, belt and bearing friction. Your speed is 500 rpm (you have a type-o in your post there)?
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| Apr15-07, 10:36 PM | #3 |
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Yes speed unit there is typing error.ok I have understand your reply but my doubt is how to include the belt efficiencey or tension without knowing the power any suggestions regarding this,Any typical calculations.....will be greatly appereciated. Thanks in advance Prakash |
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