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turbinef14 (Industrial) (OP)
24 Jun 07 12:25
how do i calculate  the motor required to lift a mass
eg mass = 75kg what hp is required

ScottyUK (Electrical)
24 Jun 07 13:35
How far do you want to lift it, how quickly, and how efficient is the mechanism you intend to employ? Time to dig out those high school physics text books!
 

----------------------------------
  Sometimes I only open my mouth to swap feet...

turbinef14 (Industrial) (OP)
24 Jun 07 17:14
thanks for your response i have dug into my physics books and found the this formula  power= kg lifted xlifting speed in m/s divided by 9.81x efficency . is this the corect formula to use if i was not going through a gearbox
and direct off a motor shaft . and is ans in watts or kilowatts  i want a lifting speed of 1m/s to a hight of 15 metres eff=.7 mass=75kg just trying to work out what size motor it would take to lift my weight                      

 
CJCPE (Electrical)
24 Jun 07 20:09
That is correct. Power in Joules per second or Watts = mass in grams X speed in meters per second divided by the acceleration due to gravity (9.80665 m/s^2).
electricpete (Electrical)
24 Jun 07 23:20
Acceleration of gravity goes in the numerator, not the denominator.  

F = m * g
P = v * F = v * M * g
If v is m/sec, M in kg and g in m/sec^2 then P is in
kg*m/sec^2 / sec = Joule/sec = watt

this is for constant speed... acceration requires additional force. Also as was mentioned, account for the efficiency of the power transmission.

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GregLocock (Automotive)
25 Jun 07 1:55
"That is correct. Power in Joules per second or Watts = mass in grams X speed in meters per second divided by the acceleration due to gravity (9.80665 m/s^2)."

There are at least two mistakes in that equation, and another in the preceeding sentence.

Cheers

Greg Locock

Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

IRstuff (Aerospace)
25 Jun 07 2:28
Energy(t) = m*g*h(t) for vertical motion

Power = d Energy(t)/dt
= m*g*dh(t)/dt
= m*g*v(t)

v(t) accounts for both constant and non-constant velocity

TTFN

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turbinef14 (Industrial) (OP)
25 Jun 07 4:43
thanks for takeing the time to respond to my question it is nice to know there are people that are prepared to put themselves out and give their knowlege      

thanks
turbine f14
electricpete (Electrical)
25 Jun 07 7:50
IRStuff - your equation is not correct.  Kinetic energy should be included within the energy.  This would give rise to an acceleration term dependent upon rate of change of velocity

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electricpete (Electrical)
25 Jun 07 7:54
The corrected version would be as follows:
Power = d Energy(t)/dt
= m*g*dh(t)/dt + m v(t) dv/dt
= m*g*v(t) + m v(t) a(t)
 = v(t) * [g + ma(t)]
 = v(t)  * F(t)
where F(t) = [g + ma(t)] as expected

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electricpete (Electrical)
25 Jun 07 7:55
Add at the beginning of my previous post:
Energy(t) = m*g*h(t) + 0.5 * m*v(t)^2

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electricpete (Electrical)
25 Jun 07 7:56
Consolidating the two previous posts, the corrected version would be as follows:

Energy(t) = m*g*h(t) + 0.5 * m*v(t)^2
Power = d Energy(t)/dt
= m*g*dh(t)/dt + m v(t) dv/dt
= m*g*v(t) + m v(t) a(t)
 = v(t) * [g + m*a(t)]
 = v(t)  * F(t)
where F(t) = [g + m*a(t)], as expected

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electricpete (Electrical)
25 Jun 07 7:58
A minor correction (final answer):
Energy(t) = m*g*h(t) + 0.5 * m*v(t)^2
Power = d Energy(t)/dt
= m*g*dh(t)/dt + m v(t) dv/dt
= m*g*v(t) + m v(t) a(t)
 = v(t) * [m*g + m*a(t)]
 = v(t)  * F(t)
where F(t) = [m*g + m*a(t)], as expected

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DickDV (Electrical)
25 Jun 07 17:58
turbinef14, if you are lifting your load with a cable winding around a drum, as in a common hoist, I think I can make it a bit simpler for you.

First, convert kg to lbs by multiplying kg by 2.2.

Find the circumference of the winding drum in feet and the radius of the drum in feet.

The torque in ft-lbs on the drum drive shaft is your load in lbs times the drum radius.

The rpm of the drum drive shaft is the lifting speed in feet divided by the drum circumference.

Once you know the drum shaft torque and speed, multiply them together and divide by 5250 to get hp.

This is constant speed lifting hp.  You will need to add some extra torque for acceleration upward, however, the motor will have at least 200% short-term overload capacity if its a NEMA A,B, or C motor across the line and, if on a VFD, as much % short-term overload capacity as the drive short-term current divided by the motor FLA.  You would want at least 150% short-term overload in the drive for most hoisting jobs.

Likely, you are fully aware of this but, the braking hp must be calculated the same way with double care.  Either the drive must have full capacity braking or the mechanical brake will need to do all the braking.  Much better to have the drive do all the dynamic braking and the mechanical brake do the holding braking and serve as backup for the drive braking.
DickDV (Electrical)
25 Jun 07 18:04
One other thing, the drum shaft hp will only match the motor nameplate hp if the shaft speed matches the base speed of the motor.  Otherwise, you will need a gearbox or equivalent or, if on a VFD and direct drive, the motor hp will be the constant speed torque times the motor base speed.  You would need to pick a motor with enough poles so the motor base speed is close to the max drum speed but not less than the max drum speed.
don01 (Electrical)
25 Jun 07 23:08
hi all
Pete ,as usual, is on the money again.
But here it is straight from my physics text book (which is soooo old it is a national relic)
PROBLEM:
a plane weighing 5000kg can climb vertically at a rate of 200km/hr. What power is expended. (9.8m/s)

SOLUTION:
power = work/time = (force x distance)/time = force x velocity

  F=mg = 5000x9.8 newtons
  v=200km/h = 55.6m/s

power = 5000 x 9.8 x 55.6
      = 2724,000watts = 2.724kW

now in the way of all good plagarists cut out their values and then paste in yours = solution (Ihope)

anyway enjoy the journey & grab the memories
Don

(Physics for Engineers - Schofield 1970)
GregLocock (Automotive)
26 Jun 07 1:10
Wow, 4 hp to lift a 5 ton plane at 120 mph vertically.

I'll have some of that.

Cheers

Greg Locock

Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

DaveScott (Electrical)
26 Jun 07 2:51
//2724,000watts = 2.724kW// You might want to edit that.
itsmoked (Electrical)
26 Jun 07 3:01
Sooo... is that -1 point or -10?

Keith Cress
Flamin Systems, Inc.- http://www.flaminsystems.com

ozmosis (Electrical)
26 Jun 07 4:36
turbinef14
Another point. If your 1 M/s was purely an arbitrary figure, then do your calculations based on longer velocity rates. Your final power calculation will alter quite a lot and therefore cost of equipment.
And do not forget the point that DickDV makes about bringing the load back down, if it is. This time you have gravity working with you and this is not always favourable.
sreid (Electrical)
26 Jun 07 16:07

75 kg = 165 lb

1 m/s = 3.3 ft/sec

Power = 165 x 3.3 = 541 lb-ft/sec

550 lb-ft/sec = 1 HP

541/550 = approx 1 HP
Skogsgurra (Electrical)
26 Jun 07 17:18
Since everyone seems to have had a go on this...

"Our" (European) definition of one horsepower is "The power needed to lift a mass of 75 kg vertically 1 m in 1 second"

This is suspiciously similar to the OP that says that he needs to lift 75 kg at a speed equal to 1 m/s.

Using SI, we get: P = 75*9.8065*1 = 736 W (the European HP).

The efficiency stated is 0.7, so you would need something like 1050 watts.

Then, there are lots of other things to consider. But your original question has this answer.

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

cswilson (Electrical)
27 Jun 07 18:01
This is reminding me of a junior high physics lesson I got, way back when. The teacher was trying to make the concept of power real to us. In the lesson, each of us had to be weighed, then be timed in the 50-yard dash. From that we were supposed to calculate the power we could each generate.

In approximate numbers, I weighed 100 pounds then, and I ran 50 yards in 8 seconds. So I was supposed to calculate:

100lbs * 150ft / 8sec = 1875lb-ft/sec

1875lb-ft/sec * 1HP/(550lb-ft/sec) =3.41HP

Now this bothered me -- I knew that a horsepower was in some sense the power that a horse could generate. As a scrawny little 12-year old, it didn't seem to me that I could generate more power than 3 horses.

But it turned out to be a great "life lesson". It got me started on really understanding the difference between mass and weight. (This has saved me a lot of grief in moment-of-inertia calculations.) More importantly, it got me verifying and double-checking the information supposedly knowledgeable people gave me. I realized that if I wanted really to understand things, I was going to have to build up my own knowledge from first principles. Even on this forum, I've seen people [mis]use "plug-and-chug" formulas without understanding what goes into them.

Curt Wilson
Delta Tau Data Systems
GregLocock (Automotive)
27 Jun 07 22:23
Well no wonder you were confused. He was assuming that the resistance to motion, horizontally, equalled your weight.

Cheers

Greg Locock

Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

electricpete (Electrical)
27 Jun 07 23:42
You're so picky Greg, worrying about minor details like physics all the time.

I mean really...how much harder could it possibly be to run vertically up a ladder at 6 yards/sec than it is to run accross a sidewalk at 6 yards/second!

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electricpete (Electrical)
27 Jun 07 23:45
I hope no one was offended by my moment of giddyness.

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GregLocock (Automotive)
28 Jun 07 0:04
We did a similar experiment at school, running up a flight of stairs. Fit fat rugby players did manage around 1 hp of vertical PE gain.

Incidentally (and obviously) a horse can produce far more than 1 hp, but on average over a day that is about what you'll get. wiki says 15 hp, briefly

Cheers

Greg Locock

Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

aolalde (Electrical)
28 Jun 07 0:12
     By definition:
   1 HP = 33000 LB x FT/1 min

 75 Kg = 165.35 Lb
  1 m/s = 196.85 Ft/min
   Eff = 0.75
  Your power requirement:

   P = 165.4*190.85/(0.75*33000) = 1.315 HP = 0.981 kW



Skogsgurra (Electrical)
28 Jun 07 3:22
...or, using the OP's efficiency: 1051 watts.

Which is something you arrive at with no other constants and factors than g, if you use SI.

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

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