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Calculation of particles by integral

  1. Mar 12, 2015 #1
    Hi,

    I would like to calculate total number of particles in one periodic wave-section whose radius varies by h(z)=1+A*sin(2*pi*z/L), where A is the profile amplitude and L is the wavelength of one wave-section.
    I thought, I will do by volume integration, say integral over V (volume of one wave-section) and the integrand is the concentration C(z,r,t), where z & r are the axial and radial direction respectively, and I want to integrate w.r. to V, i.e., dV.

    Can anybody give idea about this calculation?
     
  2. jcsd
  3. Mar 12, 2015 #2
    I'm having trouble figuring out what you're trying to calculate. I'm assuming you have a separate function for concentration you haven't listed? C(z,r,t)

    In this case, you are correct. C(z,r,t) will be the integrand and you will integrate over the entire wave-section volume. The next step in this is figuring out exactly what is in that infinitesimal element dV. What's a convenient way to express volume given the coordinate system?
     
  4. Mar 12, 2015 #3
    Thanks for your response, Overt. I want to express dV=dA*dz, where dA is the area of a circle and dz is the step size in z-direction. Now, the question is, how can I express dA? Will it be dA=2*pi*r*dr or dA=pi*h(z)^2, where h(z) is the tube radius at z?
     
  5. Mar 12, 2015 #4

    Mark44

    Staff: Mentor

    The correct formula would be dv = A * dz. In other words, it wouldn't be dA * dz. Based on what you have said, the cross-sectional area A would be ##\pi (h(z))^2##.
     
  6. Mar 12, 2015 #5
    Thanks, Mark.
     
  7. Mar 12, 2015 #6
    So, finally the integral will come into the form \int^{}_{V} C(z,r,t) dV=\pi \int^{}_{z} C(z,r,t) h^2(z) dz.
     
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