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Participant

Join Date: Mar 2012
Posts: 3

Calculate Electric Motor HP Required For A Load

03/14/2012 10:53 AM

I'm trying to calculate the required motor horsepower needed to lift a load by gear box system coupled with a pulley drum:

Drum Dim: 14 cm.

Load: 320 Kg.

Motor speed: 1450 RPM.

Speed ratio: 1/70

My calculations are showing that the load will form a torque of 219 N.m on the pulley...pls advice if is this the right answer and if yes how can I calculate the needed horsepower.

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#1

Re: Calculate electirc motor hp required for a load

03/14/2012 10:56 AM

Oh, this is basic physics, Chief. The electrical power required is the mass of the load and any supporting structure, multiplied by the acceleration due to gravity, multiplied by the rate of change of height of the load, divided by the efficiency of the lifting equipment including motor losses. Nice and simple, eh?

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#2

Re: Calculate Electric Motor HP Required For A Load

03/14/2012 11:21 AM

nmotr*ratio= ndrum T= M*g*r P= T*ω where ω[rad/s] and T[Nm]. Your result is WRONG start again and bring the results. How did you compute the torque value you write ?

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#3
In reply to #2

Re: Calculate Electric Motor HP Required For A Load

03/14/2012 11:39 AM

Think he went:

Pulley radius x load mass x g = 0.07m x 320kg x 9.8ms-2 = 219.5Nm, which is OK as a static load, but allows nothing for acceleration.

(Note: it's the pulley torque, not the motor torque).

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#4

Re: Calculate Electric Motor HP Required For A Load

03/14/2012 1:19 PM

Don't forget that the extended cable mass must also be lifted. 320 Kg of load requires some fairly stout rope or cable. If this crane is supposed to lift with a minimum thickness steel cable (6.3mm) a total length of 100 meters then this cable will weigh half of the load. There should also be some consideration for the static and dynamic friction one will find for spinning the pulley and whatever gear reduction required. Only once these smaller parasitic losses are included will the remainder force that actually lifts the whole thing be found.

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#5

Re: Calculate Electric Motor HP Required For A Load

03/14/2012 3:37 PM

Yes, your torque figure is correct to balance that mass. If you ignore friction and cable weight, then you can calculate the torque at the motor easily. Given the motor torque and speed you can calculate horsepower very easily. You are on the right track.

If you are actually engineering a hoist, then many other things must be taken into consideration.

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#6

Re: Calculate Electric Motor HP Required For A Load

03/14/2012 5:55 PM
Torque in lb.ft. =HP x 5250

rpm
HP =Torque x rpm

5250
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#7

Re: Calculate Electric Motor HP Required For A Load

03/15/2012 12:08 AM

For static load of 320 Kg, your calculation of torque 219 N m is correct. For this load and torque we can calculate power as below:

Power = force x velocity = 2pi.d.n.F = 2pi x Fd x n = 2pi x torque x rotations per second

Where we have used the fact that torque = force x distance = Fd

Symbolically, this is usually written as P= T x w

where T is torque and w is called the angular velocity, and is equal to the

factor 2pi x rotations per second.

Your motor RPM is 1450 and speed reduction is 1/70. The speed of hoist drum will be 1450/70, Comes out 20.71 rpm or 0.345 rps

Therfore power comes out P= 219 x 2 x3.14 x 0.3452 =474.81 watts

or = 0.636 HP

This calculation is based on your static load with 100% efficiency of gear box, motor and other system without friction loss. If you assume rope weight, friction losses (efficiency) and over load considerations then accordingly torque and power requirements will change.

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#8
In reply to #7

Re: Calculate Electric Motor HP Required For A Load

03/15/2012 5:17 AM

Really great I didn't image I would have such a number of useful answers.

Actualy I was calculating the torque for a static load on which I relied in trying to compute the needed power equal it, but couldn't go further.

Mukesh0861 your answer is exactly what I was looking for (the least power needed to make the load static).

Now regardless of other arguments which will be of course taken into considration, and in order to move the load, what would be the power needed to make the breakaway stage which I think requires the max torque and power at that moment, would it be by increasing power rate a little or based on an equation.

Thanks in advance, and many thanks to everyone participated in this discussion.

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#9
In reply to #8

Re: Calculate Electric Motor HP Required For A Load

03/15/2012 12:26 PM

As you may be aware, your terminology is a little loose. "Static" implies motionless. Motionless implies no power*. The power that has been calculated is that required to move the load at a constant rpm (which translates into a constant upward speed of about 923 cm/min).

Electric motors can supply very high torques for very short times, and can supply perhaps three times rated torque for a minute or less (all this varies with the motor, but these are OK rough guides).

At the instant that the motor starts, there will no doubt be some play to be taken up, but within a few hundredths of a second, the torque on the motor will increase to (for example) 4 times its rated torque. The load will accelerate upward with 4 times the steady state load, so will accelerate at 4 G, subjecting the cable to 4 times higher load that the steady state load. You can see that if this were an engineered system, that you might want a "soft start" facility to prevent the motor from over stressing the cable, causing bounce, etc.

The electrical power during that first instant can be quite high, and the motor efficiency will be quite low. If you imagine the very first instant, the load is not moving, but the motor is starting to grunt... and heat up. Because the load is not yet moving, mechanical power is zero. On a motor this size, the input power may be 3 kW and the output is (instantaneously) essentially zero.

But the load quickly begins to accelerate and at 4G, gets to its operating speed very quickly. (If we round the 923 cm/s to 9.8m/s we can see that it should take about 1/4 second.) The motor's output power is constantly changing during this brief phase.

Efficiency of 70:1 gearboxes (especially worm gearboxes) can be low -- anywhere from about 35% to about 65%. So the motor would need to be oversized to compensate. Although you may think the motor will "just slow down" to meet its continuous hp rating, it will not. It's "slip" will increase and its hp output will increase, and it will heat up.

In practice, if you know the power required and the efficiency of the mechanism, then you look for a motor that will provide the resulting power level for as long as required (taking into account the duty cycle -- in other words how frequently and for how long must the motor operate).

If you are really building a hoist, then there are additional considerations mainly having to do with safety. But for a thought experiment, this is about how a motor would be sized.

* Mechanical power and electrical power diverge a little here. A stove burner does not "do work" in a mechanical sense but dissipates a lot of electrical power.

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#10
In reply to #9

Re: Calculate Electric Motor HP Required For A Load

03/15/2012 3:44 PM

Here's a hoist with .5 hp motor rated at 650lbs...the difference is higher rpm @3450 and 156-1 drive....

http://www.weiku.com/products/3722909/AC_Electric_Hoist_650_Lbs_.html

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#11
In reply to #9

Re: Calculate Electric Motor HP Required For A Load

03/19/2012 4:45 AM

You're right, excuse me for my bad terminology but E is my second language I'm not choosing the right technical terms.

Now everything is quite clear, these clarifications consolidated the previous calculations.

Many thanks.

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#12
In reply to #11

Re: Calculate Electric Motor HP Required For A Load

03/19/2012 5:01 PM

You are welcome -- happy to help.

Your English, by the way, is impressively good.

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