I started calculating the needed horsepower for my project
copper granulator. i come out on 250hp, its very high, i think i made a fault with the formule.
Cross section area ( i take blades of the rotor height * withd ) = height : 10 cm, Withd: 12 cm.
10 * 12 = 120 / 2 = 60^2 cm
Yield stress of copper: psi ( shearstress 80 % of yieldstress ) = psi
Riadial cutter distance: 35.35cm
Rpm: founded Formules:
x 0.60 = .4 lbs of force
Torque = Force * Radial cutter distance
.4 * 0. = 819. Meter-kg of torque
Answer * rpm =
8,195,544 * = ,108.8 / 12 = 136,592.4 / 550 = 248.35 hp
I know i make a lot of faults, but i want to learn it.
I will very happy if anyone can help me out.
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working principe is like a shredder, but i don't shred and work on higher rpm. A machine with knives on a rotor, it rotate at high rpm to granulate materials to fine materials ( granules ). this granulator wil make copper wires in 3-5 mm pieces.
You can look the ptf's to check the working of a granulator.
I try to calculate the horsepower that's need for the knives.
Henk Trekker said:
You can look the ptf's to check the working of a granulator.
No, please don't ask us to have to do Google searches to help you. We really do want to help you, but asking us to do a lot of work to even start to help you is unreasonable.
Please post links to the reading you have been doing about these metal shreaders (is that right?), and ask us specific questions about the equations that you don't understand. Thanks.
Baluncore said:
The blades in post #1 look more like chipper blades used for wood.
Because they are mounted square with the feed they will not "slice" cleanly, so will need more power to bash through the feed stock.
Yes they like the same but they are not made for copper. this technique will make smaller parts of the copper.
anorlunda said:
@Henk Trekker , this sounds like a continuation of your earlier thread.
https://www.physicsforums.com/threads/project-single-shaft-shredder./#post-
Your shredder description also sounds much like the shredders in the video below. However, where those operate at 2-5 RPM, you say RPM. Why do you think you need such a high speed?
Thats a shredder, shredder are shredding the material at low rpm high torque. granulator work in higher rpm to make granules of the material, I'm also happy if its work on 500rpm but i need first to calculate the torque and horsepower it needs to work.
Baluncore said:
The blades in post #1 look more like chipper blades used for wood.
Because they are mounted square with the feed they will not "slice" cleanly, so will need more power to bash through the feed stock.
Thank you sir for helping, my asnwer is in metric but i cannt find the answer of my fault in the formule. how did you calculate 61.67 ft-pounds.
Henk Trekker said:
Torque = Force * Radial cutter distance
.4 * 0. = 819. Meter-kg of torque
It is a good idea to label all numbers with their units for clarity.
If .4 is pounds of force, and 0. meters is radial cutter distance, then how can 819. be in meter-kg?
I've the same question about requiring RPM as @anorlunda. My experience is with plastics granulators, and I'm having a hard time envisioning it's copper-cutting cousin surviving operation at this speed.
How is size grading accomplished? With plastics, a screen with a matrix of appropriately sized holes (typically, 1/4" to 1/2", depending on throughput and granule size requirements) is positioned under the cutting chamber, and particles fall through only after they've become small enough. Making smaller granules requires more residence time in the cutting chamber (for example, instead of being cut to size in 1 or 2 blade passes with a 1/2" mesh, cutting granules to 1/4" will require many more cuts), and reduces throughput.
The reason for bring this up is, as input lb/hr approaches what can be taken away, granule volume in the cutting chamber increases, and more power will be required to "churn" them. If applicable in this case, this extra power demand must be accounted for.