Recent content by Disserate

  1. D

    Calculate Surface Charge Decay - Chemical Engineer

    The particles are of the same material, but different sizes. This set up should also work with other materials, but they still should be the same material in different sizes.
  2. D

    Calculate Surface Charge Decay - Chemical Engineer

    That was on thing that we took into account earlier, but when this happens we can see clumps forming in the mixture before we send it down between the electrodes. To counter this we don't let the particle charge enough so that the electrical force of attraction between them is significant at all.
  3. D

    Calculate Surface Charge Decay - Chemical Engineer

    Basically I am trying to separate particles using charge. The particles are forced to interact with one another beforehand causing static charge to build up on them. In theory the bigger particles should charge positive and the small negative. The particles are then let to fall down between 2...
  4. D

    Calculate Surface Charge Decay - Chemical Engineer

    Well I wouldn't even know where to begin to perform tests, but I do know most of these variables. I know the voltage on the plate, the charge density of the particle along with the size, and the shape can be approximated as spherical. I can find most material properties online, but what I really...
  5. D

    Calculate Surface Charge Decay - Chemical Engineer

    I am a chemical engineer and have little experience with complicated electrical properties. I was wondering if there was a way to calculate the rate at which the surface charge of a particle will dissipate when it comes in contact with an electrode of the opposite charge. For example a small...
  6. D

    Particle moving in polar coordinates

    Equations given: r=A\theta \theta=\frac{1}{2}\alphat^{2} A=\frac{1}{\pi} meters per radian \alpha is a given constant Asks to show that radial acceleration is zero when \theta=\frac{1}{\sqrt{2}} radians. I have tried rearranging, plugging in, and deriving to try to solve this...
Back
Top