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electrical engineering

11/01/2008 7:00 AM

How do I calculate the power of motor required for pushing a main steel gate weighing around 2500Kgs.

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Guru

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#1

Re: electrical engineering

11/01/2008 8:15 AM

Not enough data to give an answer.

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Guru
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#2

Re: electrical engineering

11/01/2008 10:29 AM

Depending on the construction of the gate, it's area and windage may be more significant than it's weight .

Del

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Guru
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#3

Re:electrical engineering

11/01/2008 11:28 AM

Hello you need to provide at least details as---

1 horizontal or vertical displacement.

2 amount of displacement.

3 Frictional coefficients of bearings.

4 Time to open gate.

5 linier or angular type motion.

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#4
In reply to #3

Re:electrical engineering

11/02/2008 12:25 AM

Mr.Rakesh,

horizontal displacement.

No bearing , gate is mounted on a grooved wheel.

less than one minute.

Linear type motion.

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Guru
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#5
In reply to #4

Re:electrical engineering

11/02/2008 2:24 AM

Assuming you have done enough lubrication of your bearing system.

Lets assume coefficient of friction as 0.5 for calculation purpose

And total amount of displacement is 2meters.

Force required is =

2500Kg x 0.5= 1250 Kgf

Work done for the displacement of 2meter is =Kgf x Displacement

=1250 X 2= 2500 Jules is the total energy you need,

Now if you want do it in 60seconds?

Watt required = 2500/60=41.666 joules/seconds (watt)

Further you can factor in gear efficiency, Inertia, and any other losses.

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Power-User

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#6

Re: electrical engineering

11/02/2008 8:44 PM

mass = 2500kg, other parameters unknown (assumed).

To push up this mass, you need 2500kg x 9.8N/kg = 24500N

If you need to push up this mass for 3 meters, you need 24500N x 3 m = 73500J

If you need to work this in one minute, 73500 J : 60 sec = 1225W

1225W is absolute power to lift your gate to 3 meters in one minute.

The total efficiency is assumed as 0.3. So, you motor kW is 1225W : 0.3 = 4kW.

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#7
In reply to #6

Re: electrical engineering

11/03/2008 4:05 AM

Hello,

As original post says 2500Kg is Weight ,So Gravitational acceleration of 9.8 already comes in picture, I mean weight of 2500KG is already equal to 2500KGF.

Second point is, since motions is horizontal, force required to push need to be multiplied by coefficient of friction.

Thanks and regards,

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Guru
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#8
In reply to #7

Re: electrical engineering

11/03/2008 4:19 AM

Your second point surely nullifies the first?
The gate isn't being lifted.
If friction can be minimised and time isn't an issue a few watts would do it!

Del

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Guru
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#9
In reply to #8

Re: electrical engineering

11/03/2008 4:41 AM

Yes,

Thats what my first post says too, just 42 watts are required if he wish to finish the task in one minute. and even the frictions coefficeinet is 0.5.

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#10
In reply to #7

Re: electrical engineering

11/03/2008 5:50 AM

I revised my solution.

9.8kg/Newton is still required to convert kgf to Newton.

mass = 2500kg, Coefficient of friction is assumed as 0.5. Displacement is assumed as 2 meters.

To slide this mass, you need 2500kg x 0.5 = 1250kgf.

To convert kgf to Newton, 1250kgf x 9.8N/kgf = 12250N

you need to move this mass for 2 meters, you need 12250N x 2 m = 24500J

If you need to work this in one minute, 24500 J : 60 sec = 408W

408W is absolute power to move your gate to 2 meters in one minute.

The total efficiency including design margin is assumed as 0.3. So, your motor kW is 408W : 0.3 = 1361W.

So you need a 1.5kW or 2HP motor.


Regards

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Guru
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#11
In reply to #10

Re: electrical engineering

11/03/2008 5:58 AM

Yes you are right. gravity will to come in picture.

Now its perfect answer.

Sometime two wrongs make things right.

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#12

Re: electrical engineering

11/04/2008 1:04 PM

you can push it with even a little 3V DC motor out of a childs toy if speed is of no concern...

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