0

A power $P_0$ is spread on infinite circles of radius $r_0$ so the incident power on every circle is $dP_0$. The center of every circle is located on a plane $\Gamma =\left(0\le x\le x_0,0\le y\le y_0\right)$. I am stuck in find the distribution of the power $P$ on this plane. Obviously it can be found using a double integral, but I don't find it. Thanks

Riccardo.Alestra
  • 10,546
  • This needs a lot of clarification. It seems that where it says "infinite circles" you mean "infinitely many circles"? What do you mean by "distribution"? Is there a random element here? If so, you haven't introduced it. If not, where are the circles? – joriki Apr 26 '16 at 10:32
  • 1
    @joriki: "infinite many circles" is the correct statement. I mean the distribution of the power on the plane $\Gamma$.The center of the circles is located on the plane – Riccardo.Alestra Apr 26 '16 at 10:41
  • But clearly the distribution of the power depends on where the circles are? – joriki Apr 26 '16 at 11:42
  • @joriki: Of course. Every circle has a center in the plane $\Gamma$, so the power distribution is the sum of the powers $dP$ of the infinite many circles. If we discretize the problem, we have for the coordinates of the first center: $0,0$, the second: $0,\Delta y$, the third: $0,2\Delta y$, and so on. After we have: $\Delta x,0$,..., $2\Delta x,0$, etc. – Riccardo.Alestra Apr 26 '16 at 11:51
  • But within $[0,x_0]\times[0,y_0]$ that would only be finitely many circles? – joriki Apr 26 '16 at 11:54
  • @joriki: in the continuous case, in $[0,x_0] \times [0,y_0]$ there are infinite many circles because $\Delta x$ is replaced with $dx$ and the same for $\Delta y$ – Riccardo.Alestra Apr 26 '16 at 12:02

1 Answers1

1

You don't need a double integral. The irradiance at a point $p$ is $I_0=\frac{P_0}{x_0y_0}$ times the fraction of the circle of radius $r_0$ around $p$ that lies within $\Gamma$. Thus, a point far enough in the interior of $\Gamma$ has irradiance $I_0$, a point on the border has irradiance proportional to the area of a circular segment, and a point near a corner has irradiance proportional to the area of the intersection of two circular segments, which can be calculated by solving a system of four linear equations for the areas of the four intersections of the segments and their complements.

joriki
  • 238,052
  • This is correct in the discrete case. But I work in the continuous, so I have infinite systems to solve at the border of $\Gamma$ – Riccardo.Alestra Apr 26 '16 at 13:04
  • @Riccardo.Alestra: No, this wouldn't be correct in a discrete case. If it doesn't match what you wanted, I think you'll have to put more effort into clarifying the question. I was basing my answer on our exchange in the comments; my understanding was that there is a continuum of irradiated circles, all of radius $r_0$, with centres uniformly spread over the rectangle $\Gamma$. – joriki Apr 26 '16 at 14:16