Minimum motor power for a mechanical horizontal sieving machine

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SUMMARY

The discussion focuses on calculating the minimum motor power required for a mechanical horizontal sieving machine. Key parameters include a radius (r) of 2.5, a length (l) of 10 cm, and an angular speed (w) of 30 rpm. The total force calculated is 33.4 Newton, leading to a torque of 1.67 N.m. Consequently, the required motor power is determined to be 50 watts, factoring in the need to overcome friction and accelerate the mass of the screen and powder.

PREREQUISITES
  • Understanding of basic physics concepts such as force, torque, and power.
  • Familiarity with mechanical systems, specifically crank mechanisms.
  • Knowledge of angular velocity and its relationship to rotational systems.
  • Ability to perform calculations involving Newton's laws of motion.
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  • Research "Crank mechanism design" for optimizing torque requirements.
  • Learn about "Friction force calculations" in mechanical systems.
  • Study "Motor selection criteria" based on torque and power requirements.
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Mechanical engineers, designers of sieving machines, and anyone involved in the design and optimization of mechanical systems requiring precise motor power calculations.

khodournajem
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Homework Statement
I'm making a small project (horizontal sieving machine)
i will use the crank slide mechanism , the doctor want from me to calculate the input power of the dc motor and the output power to calculate the efficiency of the system.
so i test the maximum load which is 10 kg equally distributed on a frame of 6 bearings of friction coefficient 0.3,
the minimum displacement of the net frame to begin sieving is 5 cm per 1 second which is 0.05/s meters,
so the crank radius will be 2.5 cm (0.025) that gives a 60 rpm and i am using a 10 cm connecting rode. so this is my critical issue.
is i'm thinking right ?
how can i calculate the power?
do i have to calculate the force of the crank ?
how can i calculate the output power if my result is correct.
please help
Relevant Equations
𝐹=μN

𝐹𝑐𝑟𝑎𝑛𝑘=ma

𝜏=F×d

P=τ×w
r = 2.5
l=10 cm
angle = 0 (maximum connecting rode elongation)
w = 0.5 (30 rpm)
n=10/2.5 = 4
so the acceleration is 1.5×0.5×0.5(1 + 1/4 ) = 0.4 m/s

friction force = 10*9.81*0.3 = 29.4 Newton.

Crank force is the mass of the object into acceleration.
So crank force = 10 × 0.4 = 4 Newton
Therefore total force will be: 29.4 + 4 =33.4 Newton.

Torque is the total force into the distance.
So torque = 33.4 × 0.05 = 1.67 N.m.

Finally motor power is the torque into the angular speed in minute.
Power = 1.67 × 30 = 50 watts.here is a approx picture (in my project the motor and the connecting rode and the frame is horizontally placed)
 

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Welcome, khodournajem!

The motor will need to accelerate the mass of the screen and powder from V=0 to V=max at half of the stroke in certain period of time, while overcoming friction.
That will demand a horizontal force that can be calculated.
The crank mechanism will require the greater torque at certain angle position to provide that horizontal force.
Then, you can add some safety factor regarding the needed torque at the required rpm of your motor for its selection.

 
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thank you lnewqban for your time ,

i know these informations , i want to know if my calculations, used equations and last result are correct .
regards.
 

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