turbinef14
Industrial
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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power required to lift a mass
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turbinef14
Industrial
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
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
electricpete
Electrical
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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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.
=====================================
Eng-tips forums: The best place on the web for engineering discussions.
GregLocock
Automotive
"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 for tips on how to make the best use of Eng-Tips.
There are at least two mistakes in that equation, and another in the preceeding sentence.
Cheers
Greg Locock
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
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turbinef14
Industrial
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
thanks
turbine f14
electricpete
Electrical
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
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
=====================================
Eng-tips forums: The best place on the web for engineering discussions.
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
=====================================
Eng-tips forums: The best place on the web for engineering discussions.
electricpete
Electrical
Add at the beginning of my previous post:
Energy(t) = m*g*h(t) + 0.5 * m*v(t)^2
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Energy(t) = m*g*h(t) + 0.5 * m*v(t)^2
=====================================
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electricpete
Electrical
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
=====================================
Eng-tips forums: The best place on the web for engineering discussions.
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
=====================================
Eng-tips forums: The best place on the web for engineering discussions.
electricpete
Electrical
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
=====================================
Eng-tips forums: The best place on the web for engineering discussions.
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
=====================================
Eng-tips forums: The best place on the web for engineering discussions.
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.
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.
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.
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)
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
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 for tips on how to make the best use of Eng-Tips.
I'll have some of that.
Cheers
Greg Locock
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
ozmosis
Electrical
- Oct 12, 2003
- 1,794
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.
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.
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