# Find the torque of my Machine

1. Aug 4, 2017

### Saurav Anand

hello everyone,
I am a mechanical design engineer in pharmaceutical machine maker company. my question is about the torque of the shaft in my machine. In picture you can see the pan, in which the medicines are placed for drying and colour. the details is placed on picture. please tell me how to find torque in detail. Thank you in Advance.

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2. Aug 4, 2017

### Staff: Mentor

Welcome to the PF.

Since you are an ME, can you show us your work so far? Can you Upload your sketches on this?

3. Aug 5, 2017

### CWatters

The machine appears to show a rotating drum of some sort. The torque required to rotate it will depend to some extent on what's in the drum and how it moves. eg Will there be baffles that lift the contents and allow it to fall back down?

As Berkman says... As an ME we would expect you to have some ideas already.

4. Aug 5, 2017

### Saurav Anand

Thank you berkeman and CWatters for replying me.
Yes there are four baffles in pan. the raw medicines put in the pan and rotates that drum on max. speed of 15 rpm.
The medicines weight is 50 kgs approximately. That machine is already made by my company hundreds times but no one knows the principles behind it.
I have just 1.5 years experience in this field so now i eager to know the calculations behind it.
I know that torque = F x r x sinθ. and all that but i dont know how to implement it.

5. Aug 5, 2017

### CWatters

Try making a sketch drawing of a section through the drum. Perhaps assume all 50kg of the medicines have been lifted up on one baffle. Mark on the sketch F, r and θ.

6. Aug 5, 2017

### Saurav Anand

CWatters please can you explain it with equation. you can make any assumption. like r should be 500 mm or 1000 mm whatever. i just need to know which parameters are used there and how to use it. thank you for your reply.

7. Aug 5, 2017

### CWatters

Assumptions:
1) All 50kg of pills have been lifted up on a baffle until it reaches the horizontal position (θ = 90, Sinθ = 1)
2) The average distance of the pills from the centre is R = 750mm

Then using..
T = F*R*Sin(θ)

F = 50*g ≈ 500N
RSin(θ) = 0.75 * 1 = 0.75

gives

T = 500 * 0.75
= 375Nm

That's a crude estimate of the max torque required to turn the drum at a constant speed.

8. Aug 8, 2017

### Saurav Anand

CWatters,
now i realize that its easy to calculate the torque.
my next question is whats the role of axis in torque ? in this matter we have a perpendicular axis. what about the linear axis ?
2 ) you assume the pills weight of 50 kg but what about the drum`s weight ? it has minimum 150 kg weight so should we not considered it ?

9. Aug 8, 2017

### CWatters

Regarding the drum...

If the drum is symmetrical then no torque is required to turn it at a constant speed. However a torque is required to accelerate the drum. Eg from rest up to its normal operating speed.

Torque = moment of inertia * angular acceleration

When starting from rest the pills will also contribute to the moment of inertia.

Some torque might also be needed to overcome friction in the bearings but that is likely to be small.

10. Aug 9, 2017

### Saurav Anand

11. Aug 9, 2017

### Saurav Anand

sir, should we add the weight of drum ?

12. Aug 9, 2017

### CWatters

No. If the weight of the drum is uniformly distributed then you do not need to add the weight to calculate the torque needed to rotate it at a constant speed.

You might need to account for the weight of the drum if you want to calculate the torque required to change the speed (eg the starting torque). However to understand this you need to know about "rotational inertia" and "Moments of Inertia".

13. Aug 10, 2017

### Saurav Anand

Sir,
Thank you very much to share your knowledge with me. its very helpful to me in future.

14. Aug 10, 2017

### Staff: Mentor

"Small" though is relative to the weight of the drum and neediness of the drive system. It *may* be big enough to be relevant.