Calculations of the needed horspower for a granulator

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The discussion centers on calculating the horsepower required for a copper granulator, with the initial calculation yielding 250 hp, which participants identified as likely erroneous. Key parameters included a cross-sectional area of 60 cm², yield stress of copper at 4830 psi, and a radial cutter distance of 35.35 cm, leading to a torque calculation of 819.55 Meter-kg. Participants emphasized the importance of understanding the relationship between RPM, torque, and granulator design, particularly in the context of achieving optimal granule size and throughput.

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Henk Trekker
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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: 4830 psi ( shearstress 80 % of yieldstress ) = 3864 psi

Riadial cutter distance: 35.35cm

Rpm: 2000founded Formules:

3864 x 0.60 = 2318.4 lbs of force

Torque = Force * Radial cutter distance

2318.4 * 0.3535 = 819.5544 Meter-kg of torque

Answer * rpm =

8,195,544 * 2000 = 1639,108.8 / 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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Hi @Henk Trekker -- What's a granulator? Is it like a defibrillator? Or an anihilattor? Or a bulldozer?
 
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.
 
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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.
 
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.978021/#post-6239851
Your shredder description also sounds much like the shredders in the video below. However, where those operate at 2-5 RPM, you say 2000 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.
 
Last edited:
  • #10
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.
 
  • #11
Henk Trekker said:
how did you calculate 61.67 ft-pounds.
I did not. See post #5 by @Tom.G
 
  • #12
Henk Trekker said:
Torque = Force * Radial cutter distance

2318.4 * 0.3535 = 819.5544 Meter-kg of torque
It is a good idea to label all numbers with their units for clarity.
If 2318.4 is pounds of force, and 0.3535 meters is radial cutter distance, then how can 819.5544 be in meter-kg?

I've the same question about requiring 2000 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.
 
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  • #13
@Asymptotic Thank you sir for the information, I have a design a screen with D 0.6 mm holes , 0.5 mm distance from the blades to the screen. What is a save rpm lvl?
 
  • #14
Henk Trekker said:
@Asymptotic Thank you sir for the information, I have a design a screen with D 0.6 mm holes , 0.5 mm distance from the blades to the screen. What is a save rpm lvl?
Can't say how fast will be safe. A lot of this is determined by granulator design - what the bearings can handle, how much impact force the blade hold-down bolts can successfully clamp against, maximum ratings of various power train components (such as V-belt sheaves), and so on.

600 RPM is fairly representative of plastics granulator speeds.

Dimensions of 0.6 mm diameter for screen hole size, and 0.5 mm screen-to-blade distances don't seem right.
A grain of table salt is roughly 0.3 mm.
 

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