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Very interesting discussion. Glad I'm not at college anymore. Silly question probably, would the motor have to be DOL?
 
I qualified a few years ago myself, but went back part-time to try and make life easier. I've manage to make it harder! It's amazing how much I don't know - that's what college has thought me. The motor will be run by VFD.
 
We have to cover every angle for this question. We have to decide on everything, including the safety factor so i'm glad you mentioned it! We are given a question and have to treat it like a real world application.
 
You can invent a safety factor for your project but do include it.
To be honest I would give the answer as you calculated and add a caveat of safety factor.
So you give two ratings one of 65KW and one of 97KW stating an assumed 50% safety allowance.
 
Last edited by a moderator:
That's a good idea. I will do that just to let them know that i'm aware of it. It's a good question to get you thinking. Might come in useful if I ever come across a hoist with out a motor!!
 
I'm somewhat 'wet behind the lugholes' in this topic, but... is there any requirement for gearing... would gearing not allow smaller drive motor.

Please educate me... I'll not be offended by highlighting of fundamentals I've probably missed.

.
 
Yes gear ratio comes in to it as well as motor speed, drum diameter, sheave reeving and god knows what else. To be honest silva it’s a black art and I’m surprised gerard has been saddled with this as a project. He gave the gear efficiency, ratio is immaterial. Power in = power out – efficiency (be it electrical or mechanical).

To be honest I don’t miss swinging around 120 foot in the air.
A good one was trying to explain to Orange that I’d dropped my phone 90 foot in to a ladle of molten iron. “Can you send the bits back?”
 
This is a college question. I am studying for the Certificate in Electrical Services Eng in Ireland I think you call it the HNC. If this was put into practice then there would probably be gearing and inverter used. But to size the motor, you need to know the mechanical power necessary to drive the hoist. I calculated this using the method at the start of the thread. So you will need a motor with greater power than the mechanical power. This is to allow for losses in the motor, inefficiencies, safety factor, etc. Regardless of gearing, pulleys, etc, you size the motor to suit the actual requirements of the hoist. I think it's something to do with energy is neither created or lost, it is put from one form to another. So, if you need 56kW to drive hoist, then you will need 56kW + losses + inefficiencies to calculate electrical input of motor. That's my understanding of it and i'm open to correction cause i 'aint no Einstein! Hope this helps?
 
This was only a small part of the question. We are not concerned with gears and all that jazz! But thanks for you're help on this.
 

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