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Topic Tag: shredder

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In case I want to have PET shredder with a motor of 1500 W with 3490 RPM and a pulley system of 49 : 1, that allows having 71 RPM in the shaft, how much is the bigger cross-section area that this shredder can really cut? I make some research about the topic and seems that to know that, first I have to know the torque of the motor: Torque (N.m) = 9,5488 x Power (kW) / Speed (RPM)= 9,5488 x 1500 W / 71= 201,74 N.m After knowing this, I look here and I find the PET’s yield Strength (since I want to make a PET shredder), which happens to be 90 Mpa. Knowing that I can make this: T = F. b  Where: T = the motor’s torque F = Force needed to cut the material b = the radio of the blades which is 59,785×10^-3 m And knowing this: Stress = F / A. Where; A = cross-section area of the material to cut Stress =  PET’s yield Strength F = Stress . A   T = F. b = Stress. A. b A = T /(Stress . b) Which gives me: A = 201,74 N.m /( 90Mpa x 59,785×10^-3 m) A = 3,75×10^-5 m^2 = 0,375 cm^2 Isn’t that too little? Looks like a really small area… Maybe I’m doing something wrong? I don’t think that I’m using an inappropriate engine because I use the excel spreadsheet that has this post and looks that I have good parameters for the design. And on the other hand, If maybe someone doesn’t understand the context of my calculations you can see this post, I used it as a reference to make these calculations.   Please, Can someone help me?, maybe if I’m not really clear, you can let me know that :). I’m really sorry for my English.
Viewing 30 topics - 1 through 30 (of 266 total)