The moment of inertia... It is calculated from P=dEc/dt (derived from kinetic energy with respect to time, energy-power or kinetic energy theorem) with Ec=1/2*J.w² where J is the moment of inertia sought, w is the rotation frequency i.e. w=n(rpm)*pi/30=x rad/s
The derivative of the energy here gives P=J.w.w' where w' is the angular acceleration either w'=(n-n min)/15*pi/30 or w'=x rad/s² . We arrive at J=P/(w.w') in kg.m²
C is the torque required at the level of the rollers or blades in N.m (= effort required for shredding x radius)
w is the rotation speed in rad/s (see Gérald's answer)
The duration of use does not affect the calculation. On the other hand, you must specify it to the engine supplier so that he takes it into account in the model he will offer you (taking into account the associated heat dissipation).
For me, there is a lot of data missing and the problem is taken in reverse.
To pass 60t/day, you need to know the volume of raw material that it represents.
I consider that the volume of material corresponds to the distance between the 2 rollers and that the material advances at the same time as the rollers turn, so it depends on the rotation speed and the diameter of the rollers
Then, it depends on the length of the rolls.
Suppose the roll has a Ø1m and a spacing of 0.1m between the 2 rolls.
The surface area of matter corresponds to:
S=(ft*0.55²)-(ft*0.5²)=0.165m²
Multiplied by the length of the roller, it gives you the volume of material that passes in 1 turn.
You can then deduce the rotation speed of the geared motor to respect the 60t/day.
For the power, we can't help you, because it depends on the weight and the forces to be driven when empty + the grinding forces that we don't know.
A piece of advice though, don't hesitate to take a lot of margin in terms of power.
Why not put a variable speed drive coupled to the motor?
on the other hand in case of blockage of the engine, the flywheel will not stop the engine in case of blockage and risk of destruction of the coupling,
I think that a torque limiter is the best solution, but do you have to find an adequate one!
How can this grinding force be calculated knowing that cancellous bone can withstand compression up to 30 Mpa? I don't have this resistance for a horn but I assume it equal or less. If I have to use P=F/S, what would this surface be? I can't determine the grip surface of the roller on the horn. I don't even think I can determine the total surface area of the horn!!
I think there is an error in your formula for calculating the surface area of material to be crushed for one turn.
For me it should be the gap between the reels multiplied by the perimeter of a reel (Because 1 turn).
So S= ft x 1 x 0.1 = 0.314 m2
Concerning the curb weight of the system to be driven, I can know it since it is the weight of the rollers, bearings, and sprockets. For the grinding effort I am blocked by the apprehension surface of the horn to be crushed, but I intend to overestimate.
The surface area to be considered is the contact surface between the roller (or blades) and the horn. The roller being much more rigid than the horn, it is essentially the roller that defines it.
Would it be possible to have a screenshot or a diagram of the system to look more precisely?
After that, for this kind of machine, you shouldn't hesitate to take big margin factors. The operation often has a lot of jerks (the flywheel can however reduce them) which electric motors do not like too much.
I would say that with a ladle, you can consider the surface of 2 teeth per roller. From the compressive strength, you can therefore deduce an associated force. By multiplying by the radius of the roller you have the torque per roller.
For the volume crushed per turn, I think you have to at least multiply by the "density" of the horns. Your formula must be correct for a "solid" material but there will be space between your horns.
On the other hand, it is not impossible that the shredding tends to make the horns rise slightly under the pressure of the rollers (I have a garden shredder and it seems to me to be the case). A little extra margin (like 10%) might not be a bad thing.
For the flywheel, I'm not used to using it, I have to think about it. Can't the volume of the rolls already be enough?
Yes, but I was planning to use the density to have the mass of horn crushed for one turn in order to get my rotation speed from it, having previously set the flow rate I want at the end of the day (60T/day). So I couldn't use the density 2 times?!!
Otherwise I did some research but I couldn't find anything about the density of a horn.
How could I have it experientially?
For your last question, I admit that I don't understand what you mean.
What do you mean by the volume of the rolls is not enough??
I think that we misunderstood each other on the history of density, which I had deliberately put in quotation marks. I think you're talking about density as a density that will allow you to have mass from the crushed volume.
What I meant is that the horns are not going to take up the entire volume (just like 1 stere of wood does not contain 1 m3 of wood). So, in the theoretical volume that the roller can carry over 1 turn, part of it will only be air that does not take into account in your flow. It will therefore be necessary to increase the speed by the same amount.
There are therefore 2 coeffs to be applied to obtain the mass of ground horn in 1 turn.
To determine a density experimentally, a tank with water to get the volume and a scale to get the weight.
For the flywheel, the term "volume" may be a misnomer. What I wanted to say is that the rollers already have a significant mass given their dimensions. As a result, they will already have some inertia once in rotation. The rollers therefore already have a role as flywheels. At the very least, you can always deduce it from the necessary inertia.
You're sure you didn't make a mistake in the unit or anything else because it seems impossible to do. Could you post the details of your calculation (a scanned draft will suffice) to check the hypotheses, ...
By doing the math for 1 tooth with a roller of 100 mm radius and a speed of 60 rpm I arrive at 240 kW. Okay, it's still excessive. But it is already much less pessimistic.
In excel is perfect. That way I was able to correct the file directly.
For me, there were 3 errors in the formulas. I corrected and annotated the file (boxes in yellow). As a result, I arrive at 22 kW! It's still more reasonable.
But do the results seem plausible to you?? Isn't 5rpm small? The torque is too big, honestly I have a doubt about the calculation of the grinding effort.
Because if I use this torque to determine the tangential force on the pinion, and then calculate the modulus using the formula m=2.34 root( F/(k.Rpe)),
I have a 27 module!!
My original diameter being equal to the center distance and is 324mm because I want the same speed on the 2 rollers
The material being s235 steel
Thank you very much for your help, I'm getting there!