Sizing a 3 ph motor?

[gerard]

Hi,

I am sizing a motor for a hoist. The mechanical output of the motor needed is 54,500w.
Say there are 1.5kW of losses within the motor (stator losses, windage, etc)and the power factor of the motor is 0.8. What will the electrical input of the motor need to be? Can you show me how you get the answer?

Cheers,

 

[johnboy6083]

i think im right, but very happy to be corrected.

so youve got your output power needed (54.5 kw) add to this your losses (1.5kw) and you have 56kw.

56000/1.732x400x0.8 is 101A. if its started frequently, then you need to size your cable to take around 150A

 

[malcolmsanford]

i think im right, but very happy to be corrected.

so youve got your output power needed (54.5 kw) add to this your losses (1.5kw) and you have 56kw.

56000/1.732x400x0.8 is 101A. if its started frequently, then you need to size your cable to take around 150A

Not done motors for years john but dosen't efficiency come into the equation. If I remember right the larger the load the higher the efficiency so that formula might be right mate .............like you I need a maintenance man lol

 

[johnboy6083]

i think he used the 1.5kw losses figure as a subtitute for efficency mate. ive got a cable calc package and that says that the average effiency for that size motor is 93% so that means it effectivley add 3.8kw to the load.

using the 93% effiency get the following
54500/1.732x400x0.8 = 98.33A

now multiply this by the effiency rating, 93% you get 105.73A. as above i would size cable for 150A if it is to be started frequently

 

[malcolmsanford]

Thanks john, yes when I did motor calcs you had to factor in the efficiency which is why I read the OP was wondering the 1.5Kw was in leiu of efficiency and was going to ask but you beat me to it mate ..................cheers John

If I remember right it

I= Kw/1.73 x 415 (then) x eff x Pf

 

[gerard]

So, if I need 54,500W as the mechanical output and I have 1.5kW of losses in the motor I need 56kW and if the motor is 93% efficient, does this mean I need 60kW motor?

 

[johnboy6083]

ive just put a 5.5kw motor into the cable calc package(it doesnt do above that) and used the 0.93 and .0 for eff and pf.

it came out at 10.7A. multiply that by 10, and you get 107A. bear in mind that above motor is 500w lighter than my hypothetical 55kw opne, and the answer seems right.

wheres silva foxx or p mac to come and put us right 100%?

 

[johnboy6083]

So, if I need 54,500W as the mechanical output and I have 1.5kW of losses in the motor I need 56kW and if the motor is 93% efficient, does this mean I need 60kW motor?

is this info from the manufacturer, or is it for study?

 

[gerard]

It's a college assignment - a serious one - I need to do well!!

 

[Knobhead]

OK you know the input KW requirement. What safety factor are you allowing? It’s lifting equipment! People’s lives depend on it working correctly.

The motor has to be capable of handling being put in to full reverse when under load.

Many crane hoist control systems automatically put the motor in to hoist when braking from lower.

Look on Google for crane motors.

Motor efficiency goes out the window so long as the motor and control gear can handle the abuse it will get.

 

[Knobhead]

Addendum to the above.

If this is a project, do you think part of the question could be about safety?

If you came back to me with the answer 60KW I’d be jumping up and down on you with size 10 hobnail boots.

 

[gerard]

Sizing the Hoist Motor:
The hoist must be capable of lifting a load of 500Kg 10 meters in one second. The hoist is 90% efficient.
Watts = Joules/Second
Work = Force x Distance
Force = Mass x Accereration
g = 9.81m/s
The mass of the hoist is 500Kg and must be capable of moving 10m in 1 second.
Therefore, force = 500Kg x 9.81m/s = 4905N
Work = F x D = 4905N x 10m = 49,050Joules
 Watts = 49,050J/1s
 Watts = 49,050W

The hoist is 90% efficient so actual power required to drive hoist is –
49,050W x 100/90 = 54,500W
This is the mechanical power required to drive the hoist. To calculate the Electrical Input of the motor, the Stator losses, rotor copper losses and the windage and friction losses must be taken into account. For the purpose of this exercise, a motor efficiency of 85% is assumed(this will include power factor).
54,500W x 100/85 = 64117W
A 65kW, 3 ph A.C. Induction Motor would be suitable for this application.
So, I was going to use a 65kW 3 ph motor with a fail safe brake.
Why would you go mad with this?

 

[silva.foxx]

wheres silva foxx or p mac to come and put us right 100%?

Thanks for the nudge... but you guys seem to have it sussed. I'm just a raggeddy-ar$ed factory spark... Tony can be the "rubber stamper" on this topic.

Perhaps I can learn from this thread! Good reading.

 

[Knobhead]

I can’t fault your calculations, but I’m an ex hairy arsed foundry man. Lifting gear has a safety factor rating. You can guarantee that for your 500Kg load some clown will try 750Kg on it.

Just as an example I was called to a lift (OK not a hoist) that had stalled in the shaft. Asked the plant manager what was in the cage “Oh just the usual load”. I reset the system and lowered the cage. A quick count up of the bags on the pallet in the cage came to 2.75T in a 2T lift. That was a manager thinking he could get things done quicker. By the time I’d finished sorting things out and making a safety inspection he’d lost 3 hours instead of saving 10 minutes. Added to which he got the size 10 hob-nail boot treatment from on high J

 

[pmac10]

further to what Tony has correctly stated, i have worked on applications where 'factor of safety' is a quoted value, say 0.8 for example, so 65kw would represent 80% of your final safe result, however as this is a study question i would guess (from experience) you'd only be expected to calculate from the values provided and a hoist is very easy for (most) students to picture and therefore give some purpose to the maths.

 

[remer1957]

Very interesting discussion. Glad I'm not at college anymore. Silly question probably, would the motor have to be DOL?

 

[Knobhead]

Motors can be DOL, dual speed, slip ring, VFD and anything else you could think of.

 

[gerard]

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.

 

[gerard]

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.

 

[Knobhead]

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.

 

[gerard]

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!!

 

[silva.foxx]

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.

.

 

[Knobhead]

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?”

 

[gerard]

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?

 

[gerard]

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.