Power calculation for hoist motor

Question

I was recently discussing with my son how it's possible to determine the power requirement of a hoist motor given only two parameters. These two parameters are the weight to be lifted (150 kilos),and the speed at which they are to be lifted (3 meters per second) I told him to multiply weight by speed, then, multiply that by 9.81 (gravity), the answer being in Joules. Please tell me I'm right?

Answer 1

Almost right; just the wrong units.

What you've correctly calculated is the energy needed from the motor (given to the load) each second. Energy per second is power. So the number is right, but it's joules/second, or Watts...

Answer 2

Hoist motor power is calculated by:

• P = M.g.v/n
• M = Mass
• g = Gravity
• v = velocity of raise mption
• n (meant to be the greek letter nu) = efficiency losses due to gears/ pulleys

The standards applicable are BS466 (Electrical) and BS2573 (Pt 1 and 2 - Mechanical). All the info provided in earlier answers covers this. Dimensionally, this works out as P (Watts, or Joules/sec, or Force*velocity) in kg.m²s^-3

I personally think (this is not really a 'personal; opinion, I'm a crane engineer) it's best to look at the force delivering a given velocity of lift, as this is principally what the designer is after..the mass bing lifted against gravity as the the force element (Newtons), and the power being accordingly calculated as the speed the load is to be lifted at (Velocity).

There is a whole other piece of thinking to understand Pull out torque - this is given in BS466 as 'at leat 225% more than the torque required to lift the load at the required design speed'. Looking at Slip (Rotational speed)/ torque curves should note that the hoist motor will be selected to work at >=96% slip at the max load. This isn't obvious from the calc' the speed/torque curve for a given motor should be obtainable from the manufacturer of the hoist motor. Lower loads mean the hoist will rotate slightly faster (noting it will be synchronous, or zero slip = 100% rotational speed at no load - the weight of the hook block means there always s some load, which is known as 'Light Hook' -so in practice there will always be some slip).

Apologies to labour the point, the idea here is to give a more in depth understanding of the engineering involved - I'm guessing folks are interested given the nature of this site..)..

Finer motor control speed can be obtained by a number of metods, such as adding rotor resitances (by switching in contactors), eddy current systems (to allow closed loop feedback to monitor hoisting speed and apply proportionate braking so the hoist always raises/ lowers at a given speed, by making tte motor always 'see' a constant load'), can also be applied to corss and long travel motions; or VSD (Variable Speed Drives)..to list the main schemes I've come across.. There's lots of other stuff such as temp requirements/ frequency of switching cycles, number of lifting cycles to be considered.

I've focused on 'Power required to lift at a speed' as this addresses the original intent of the question. An enquiring mind might think - what if the power is sufficient to lift the load, say the operators try to lift a heavier load, or the load is a bit stuck on the initial lift stage, or wht really happens durng the intial accelration stage as the load moves from a standstill to it's design speed..the fact that a hoist motor is selected with a HUGE overhead of pull out torque - BS466 and 2573 must be used in the procurement spec!!), this provies all the extra juice and safety factor.

Answer 3

In order to calculate the hoisting power, you need these parameters:

• weight of the hoisting \$m\$
• speed of the hoisting \$v\$
• mechanical efficiency (gearbox, wire drum, sheaves) \$\eta\$

Then you can use this formula to get the hoisting power:

$$P = \dfrac{m \cdot v}{6.12 \cdot \eta}~~[kW]$$