power required to lift a mass
power required to lift a mass
(OP)
how do i calculate the motor required to lift a mass
eg mass = 75kg what hp is required
eg mass = 75kg what hp is required
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RE: power required to lift a mass

Sometimes I only open my mouth to swap feet...
RE: power required to lift a mass
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
RE: power required to lift a mass
RE: power required to lift a mass
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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RE: power required to lift a mass
There are at least two mistakes in that equation, and another in the preceeding sentence.
Cheers
Greg Locock
Please see FAQ731376: EngTips.com Forum Policies for tips on how to make the best use of EngTips.
RE: power required to lift a mass
Power = d Energy(t)/dt
= m*g*dh(t)/dt
= m*g*v(t)
v(t) accounts for both constant and nonconstant velocity
TTFN
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RE: power required to lift a mass
thanks
turbine f14
RE: power required to lift a mass
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RE: power required to lift a mass
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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RE: power required to lift a mass
Energy(t) = m*g*h(t) + 0.5 * m*v(t)^2
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RE: power required to lift a mass
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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RE: power required to lift a mass
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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RE: power required to lift a mass
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 ftlbs 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% shortterm overload capacity if its a NEMA A,B, or C motor across the line and, if on a VFD, as much % shortterm overload capacity as the drive shortterm current divided by the motor FLA. You would want at least 150% shortterm 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.
RE: power required to lift a mass
RE: power required to lift a mass
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)
RE: power required to lift a mass
I'll have some of that.
Cheers
Greg Locock
Please see FAQ731376: EngTips.com Forum Policies for tips on how to make the best use of EngTips.
RE: power required to lift a mass
RE: power required to lift a mass
Keith Cress
Flamin Systems, Inc. http://www.flaminsystems.com
RE: power required to lift a mass
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.
RE: power required to lift a mass
75 kg = 165 lb
1 m/s = 3.3 ft/sec
Power = 165 x 3.3 = 541 lbft/sec
550 lbft/sec = 1 HP
541/550 = approx 1 HP
RE: power required to lift a mass
"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, fingertips have been used over and over again...
RE: power required to lift a mass
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 = 1875lbft/sec
1875lbft/sec * 1HP/(550lbft/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 12year 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 momentofinertia calculations.) More importantly, it got me verifying and doublechecking 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 "plugandchug" formulas without understanding what goes into them.
Curt Wilson
Delta Tau Data Systems
RE: power required to lift a mass
Cheers
Greg Locock
Please see FAQ731376: EngTips.com Forum Policies for tips on how to make the best use of EngTips.
RE: power required to lift a mass
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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RE: power required to lift a mass
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RE: power required to lift a mass
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 FAQ731376: EngTips.com Forum Policies for tips on how to make the best use of EngTips.
RE: power required to lift a mass
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
RE: power required to lift a mass
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, fingertips have been used over and over again...