My Math Forum Calculation of total path length using integration

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 November 17th, 2014, 02:22 AM #1 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,155 Thanks: 731 Math Focus: Physics, mathematical modelling, numerical and computational solutions Calculation of total path length using integration Hi guys, I have the following problem related to a porous surface and the flow of material across it. I understand that fluid mechanics would give the ideal answer, but I would like to make an approximation using geometrical means to see if there is a physical basis for an alternative treatment of the problem. So here's the situation. Imagine you have a rectangular plate with a width and a height of $\displaystyle a$ and $\displaystyle b$ respectively and a 2D Cartesian coordinate system with the bottom-left-hand corner of the plate representing the origin. At the top of the rectangular plate is a slot of width $\displaystyle a$ which acts as a sink that can allow any volume of material to pass through it (at least using sensible flow conditions used in the problem). The rectangular plate is porous such that at a given time, small packets of substance (such as air) pass through the surface at a rate of $\displaystyle \dot{n}$ per unit area in a uniform manner and flow over the surface to the slot at the top. The packets of air follow the shortest path, so in this first case this path is vertically upwards to the slot. We also wait long enough so that the system is in equilibrium before trying to assess the characteristics of it. Let's create a new variable $\displaystyle z = z(x,y)$ which is the sum of the number of mass packets at a given coordinate passing through that point. In the above example, a plot of $\displaystyle z$ would given a 'ramp' where at the base of the rectangular plate $\displaystyle z=z(x,0) = 0$ and at the top $\displaystyle z = z(x,b)$ is a maximum. Since the path taken by a packet will be $\displaystyle b-y$, this function will be $\displaystyle z = \int^y_0 \dot{n}(b-y') dy'$ where $\displaystyle y'$ is just a temporary variable to avoid confusion with $\displaystyle y$. It is independent of $\displaystyle x$ because the packets only travel vertically upwards. Lets define a new variable $\displaystyle S = \int z dx$ where $\displaystyle S$ is just the sum of $\displaystyle z$ over the slot area. In our case the slot is thin, so we instead integrate over the length of the slot. The solution to S for the above case is $\displaystyle S = \dot{n} \frac{ab^2}{2} = \dot{N}\frac{b}{2}$, which is just the total number of mass packets over the whole area multiplied by the average path length, which is $\displaystyle b/2$. I now want to calculate the exact same problem, but instead of having a thin rectangular slot of width $\displaystyle a$ acting as a sink, we have a point sink positioned at the top of the plate half way across it $\displaystyle (x,y) = (a/2,b)$. Anyone care to give it a go? Last edited by Benit13; November 17th, 2014 at 02:25 AM.
 November 17th, 2014, 02:59 AM #2 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,155 Thanks: 731 Math Focus: Physics, mathematical modelling, numerical and computational solutions Current working: $\displaystyle S = 2\dot{n} \int^{a/2}_0 \left( \int^b_0 \sqrt{\left(y - b\right)^2 + \left(x - \frac{a}{2}\right)^2} dy\right) dx$ $\displaystyle = \dot{n}\int^{a/2}_0 \left[ \left(x - \frac{a}{2}\right) \sqrt{\left(y - b\right)^2 + \left(x - \frac{a}{2}\right)^2} + \left(y - b\right)^2 \ln\left(\left(x-\frac{a}{2}\right) + \sqrt{\left(y - b\right)^2 + \left(x - \frac{a}{2}\right)^2}\right)\right]^b_0 dx$ $\displaystyle = \dot{n}\int^{a/2}_0 \left[\left(x - \frac{a}{2}\right) \left(\left(x - \frac{a}{2}\right) - \sqrt{b^2 + \left(x - \frac{a}{2}\right)^2}\right) - b^2 \ln\left(\left(x-\frac{a}{2}\right) + \sqrt{b^2 + \left(x - \frac{a}{2}\right)^2}\right)\right] dx$ Let $\displaystyle u = x - \frac{a}{2}, du = dx$ $\displaystyle S = \dot{n}\int^{0}_{-a/2} \left(u^2 - u \sqrt{b^2 + u^2} - b^2 \ln\left(u + \sqrt{b^2 + u^2}\right)\right) du$ Last edited by Benit13; November 17th, 2014 at 03:24 AM.
 April 13th, 2015, 02:15 AM #3 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,155 Thanks: 731 Math Focus: Physics, mathematical modelling, numerical and computational solutions I don't think anyone is interested, but I got a result anyways, so I'll put it here for completeness. I converted to polar coordinates and used limits that depend on angle to describe the rectangular shape of the plate: $\displaystyle S = \dot{n} ab \left[\frac{a}{12}\sqrt{a+4b^2} + \frac{a^3}{24b}\ln\left(\frac{\sqrt{a+4b^2} + 2b}{a}\right) + \frac{b}{12}\sqrt{a+4b^2} - \frac{b^3}{3a}\ln\left(\frac{\sqrt{a+4b^2} -a}{2b}\right)\right]$ For a square plate, $\displaystyle a=b$, the average path length is $\displaystyle 0.5927 b$ compared to $\displaystyle 0.5 b$ for the rectangular slot case.
 April 13th, 2015, 06:56 AM #4 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,155 Thanks: 731 Math Focus: Physics, mathematical modelling, numerical and computational solutions Oops... last post was wrong. Correction: $\displaystyle S = \dot{n}ab\left[\frac{\sqrt{a^2 + 4b^2}}{6} + \frac{a^2}{24b} \ln\left(\frac{\sqrt{a^2 + 4b^2} + 2b}{a}\right) - \frac{b^2}{3a} \ln\left(\frac{\sqrt{a^2 + 4b^2} - a}{2b}\right)\right]$ Numerical result for $\displaystyle a=b$ is the same, but S now has the right units.

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