Torque and power of electric motor

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Discussion Overview

The discussion revolves around calculating the torque and power required for an electric motor to move an 80 kg weight at a speed of 80 km/h using a 260 mm pulley. Participants explore various aspects of the problem, including the relationship between linear and angular velocity, the forces acting on the mass, and the implications of different assumptions about the motion (horizontal vs. vertical).

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant seeks help in calculating the necessary torque and power for moving a weight with specific parameters.
  • Another participant suggests a formula for torque based on pulley diameter and load, but acknowledges the need for further calculations to relate torque to power.
  • There is a discussion about converting linear velocity to RPM and the relationship between linear and angular velocity.
  • Some participants highlight the importance of considering external forces, such as air drag and friction, when calculating the required power.
  • One participant emphasizes that the question lacks clarity regarding whether the weight is being lifted or moved horizontally, affecting the torque calculations.
  • Several participants discuss the relationship between power and torque, noting that power is a rate and can be misunderstood in mechanical contexts.
  • A participant provides a detailed calculation for power and torque but is challenged by others regarding the assumptions made about force and motion.
  • There is a disagreement about the correct approach to calculating power, with some asserting that the force required to maintain speed is unknown.

Areas of Agreement / Disagreement

Participants express differing views on the calculations and assumptions necessary to determine the required torque and power. There is no consensus on the correct approach, and multiple competing perspectives remain throughout the discussion.

Contextual Notes

Participants note that the calculations depend on whether the weight is being lifted or moved horizontally, and the effects of friction and air resistance are acknowledged but not quantified. The discussion also highlights potential misunderstandings regarding the definitions and implications of power in mechanical systems.

Prabhjyot Singh
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Hallo
Currently I am engaged with my Company's Project and it is about Electric motor as my Knowledge is not that good about Electric motors, I need your help. I want to move 80 kgs of Weight with the speed of 80 km/h with a motor and a 260 mm pulley. The motor is standing vertically and the pulley above the floor. Now I need to calculate how much Torque and Power I need for my Motor so that I can move the weight (80 Kgs) with 80 km/h speed. Any help would be really appreciated.

Thanking You
Prem
 
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To calculate the torque required its simply 1/2*D*L where D is the pulley diameter and L is the load.
 
Ya but I want to calculate Torque and Power to move 80 kgs of weight with 80 km/h
 
I would double check this but in English units HP= torque * rpm ÷5252. So if you calculate the required torque, you know the HP of the motor you find rpm. So translate you required speed to rpm and I think you got it.
 
Since velocity is a linear measure (distance of time) and you are spinning a shaft which is angular velocity you have to relate them by converting velocity to RPM. Hope this helps!
 
Ya I understood the formula but the thing is rpm is revolution per minute (rotational motion) and velocity in linear velocity and I am not sure if calculating rpm will directly give me linear velocity that I need to move the object
 
you have to convert velocity to angular velocity to when you are using a spinning shaft. Angular velocity is calculated by v = r × ω
Where:
v: Linear velocity, in m/s
r: Radius, in meter
ω: Angular velocity, in rad/s

To do this in RPM you would use v = r × RPM × 0.10472

Where:
v: Linear velocity, in m/s
r: Radius, in meter
RPM: Angular velocity, in RPM (Rotations per Minute)
 
The answer lies in two factors:
How fast do you want to get to 80km/hr
What forces are acting on the mass at steady state - There will likely be significant air drag at 80km/hr among other sources of friction.

The mass of the cable makes things a little difficult - the moment of inertia of the pulley system increases with time as the mass of the cable is probably not negligible.
 
Your question is incomplete. For example, are you lifting the weight or moving it horizontally? Let's assume you are lifting the weight. The torque required for a constant lifting speed would be (pulley radius X (load weight + system friction). The starting torque depends on how quickly you want to reach the required 80 km/h speed, that is, how quickly must you accelerate the load? That would be {[(axle torque / pulley radius) - (load weight + starting friction] / mass of the load}.

This is pretty sparse. Post more if this doesn't help.
 
  • #10
The relationship between torque and power is power = speed x torque.
This holds true no matter what units you use, but if you employ an inconsistent system of units, you'll need to insert conversion factors.
The following units are standard:
[power] = watts (W)
[speed] = radians per second (rad/s)
[torque] = Newton-metres (N-m)
 
  • #11
Just an editorial comment, I do wish "power" wasn't used so much in mechanical engineering. It is misunderstood by so many people, including engineers. Power isn't a "thing", it's a rate. It doesn't do anything. For example, power cannot be used to calculate the rate of acceleration on a mass or how sturdy a driveshaft must be or the loading on a gear tooth, etc. For any given power and mass there are an infinite number of valid acceleration values. For any rate of power delivered by a driveshaft there are an infinite number of valid torque values. Conversely, for any given mass and rate of constant acceleration there are an infinite number of valid power values. Etc., etc.
 
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  • #12
Prabhjyot Singh said:
Hallo
Currently I am engaged with my Company's Project and it is about Electric motor as my Knowledge is not that good about Electric motors, I need your help. I want to move 80 kgs of Weight with the speed of 80 km/h with a motor and a 260 mm pulley. The motor is standing vertically and the pulley above the floor. Now I need to calculate how much Torque and Power I need for my Motor so that I can move the weight (80 Kgs) with 80 km/h speed. Any help would be really appreciated.

Thanking You
Prem

Power, hp = Workdone in kg-m per S / 75 = 80 kg * (80,000/3600 m/s) / 75 = 23.70 hp

Also,
Power = (2*Pi*N*T) / 75
Torque, T = (Power *75) / (2 * Pi * N)

N (rps) = Linear Velocity / (Pi * Diameter) = (80,000/3600 m/s)/ (Pi * 0.26 m) = 27.19 rps

Torque, T = (23.70 hp * 75) / (2 *3.143 *27.19 rps) = 10.40 kg-m

A N Madhavan
 
  • #13
Do we need to keep adding to this thread? The OP hasn't replied since over four months ago.
 
  • #14
I just wanted to brush-up my basics, so responded and probably it would help someone.
 
  • #15
A N Madhavan said:
Power, hp = Workdone in kg-m per S / 75 = 80 kg * (80,000/3600 m/s) / 75 = 23.70 hp

This is not right.
Power = force*velocity
We don't know the force required to maintain the constant 80km/hr, as mentioned in post #8 so we cannot calculate power.
 
  • #16
B
billy_joule said:
This is not right.
Power = force*velocity
We don't know the force required to maintain the constant 80km/hr, as mentioned in post #8 so we cannot calculate power.
Both are same.
Force (kg) * Velocity (m/s) = Work done per S in kg-m/s
75 kg-m/s = 1 hp
The work done to move 80 kg at (80,000/3600 m/S) is 80 * 22.22 kg-m/s => 1777.77 kg-m/s => 23.70 hp
 
  • #17
Force is not measured in kg.
We don't know the force, the object is moving horizontally (post #1 implied this).
Your approach would apply if the weight was being lifted vertically.
 
  • #18
That is true.
 

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