Torque and power

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1. Nov 26, 2014

Prabhjyot Singh

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

2. Nov 26, 2014

rsk2mc

To calculate the torque required its simply 1/2*D*L where D is the pulley diameter and L is the load.

3. Nov 26, 2014

Prabhjyot Singh

Ya but I want to calculate Torque and Power to move 80 kgs of weight with 80 km/h

4. Nov 26, 2014

rsk2mc

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.

5. Nov 26, 2014

rsk2mc

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!

6. Nov 26, 2014

Prabhjyot Singh

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

7. Nov 26, 2014

rsk2mc

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

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

Where:
v: Linear velocity, in m/s
RPM: Angular velocity, in RPM (Rotations per Minute)

8. Nov 27, 2014

billy_joule

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.

9. Mar 26, 2016

OldYat47

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. Mar 29, 2016

David Lewis

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)
[torque] = newton-metres (N-m)

11. Mar 30, 2016

OldYat47

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.

12. Apr 1, 2016

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

13. Apr 1, 2016

billy_joule

Do we need to keep adding to this thread? The OP hasn't replied since over four months ago.

14. Apr 1, 2016

I just wanted to brush-up my basics, so responded and probably it would help someone.

15. Apr 1, 2016

billy_joule

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. Apr 1, 2016

B
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. Apr 1, 2016

billy_joule

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. Apr 1, 2016