csmyth3025 Posted January 10, 2012 Share Posted January 10, 2012 The influent pump building at the wastewater treatment plant where I work receives flow from a 6 ft diameter interceptor sewer. Our pump controls maintain a sewage level of 2.5 ft in this interceptor. The generally accepted flow velocity in a sewer to prevent grit deposition is 2 ft/sec. A discussion among the plant operators here raised the following question: What rate of flow is needed to maintain a 2 ft/sec flow velocity in this interceptor at a water depth of 2.5 ft. As a practical matter we measure flow in terms of millions of gallons per day (MGD). As an approximation I calculated 1/2 the cross-sectional area of this interceptor (3 ft water depth) and then subtracted an area equivalent to a strip 0.5 ft wide by 6 ft long (3 ft^{2}): Circular segment (water) = [math]\frac{1}{2}\pi(3 ft)^2 - 3 ft^2 = 11.137 ft^2[/math] I multiplied this area by 2 ft/sec to obtain a flow rate of 22.274 ft^{3}/sec. Applying the conversion factor 1 MGD=1.547 ft^{3}/sec, simple division provides the answer: ~14.4 MGD. This seems reasonable since the flow through this interceptor rarely drops below 16 MGD. So far, so good. Then I wanted to make this same calculation using a formal geometric approach. I quickly found the equations I needed at the mathworld page here: http://mathworld.wol...larSegment.html This is where I became stumped. The following diagram is taken from mathworld: R = 3 ft r = 0.5 ft h = 2.5 ft The calculation of a is straightforward (and the reasoning behind it is understandable) using the mathworld formula [math]a = 2 \sqrt {R^2 - r^2} = 2 \sqrt{3^2 - 0.5^2} = 5.916 ft[/math] When I try to calculate [math]\theta[/math] (in radians) by the formulae given I get lost, though. The simplest formula using the information available is: [math]\theta = 2 \cos^{-1} (\frac{r}{R})[/math]. I can understand how [math]\theta[/math] could be equal to the inverse of twice the cosine (r/R), which would be [math]\theta = (2\times \frac{0.5}{3})^{-1} = 3[/math]. This (in degrees) would be equal to about 171.9^{o}, which is the large angle I would expect. But as I read the given equation, it seems to be saying that [math]\theta[/math] equals twice the inverse of the cosine (r/R), which would be [math]\theta = 2\times(\frac{0.5}{3})^{-1} = 12[/math]. This result (in radians) doesn't make any sense to me. Am I misunderstanding how terms such as [math]\cos^{-1}[/math] are used in the formulae given in the mathworld page on this subject? Chris Link to comment Share on other sites More sharing options...

imatfaal Posted January 10, 2012 Share Posted January 10, 2012 (edited) Hi Chris You have a simple right triangle with sides of length R (hypoteneuse), r, and a/2 (opposite to theta) this gives you the basic trig rules of [math] Cos \theta = \frac{adjacent}{hypotenuse} = \frac{r}{R} [/math] [math] Sin \theta = \frac{opposite}{hypotenuse} = \frac{a/2}{R} = \frac{a}{2R} [/math] [math] Tan \theta = \frac{opposite}{adjacent} = \frac{a/2}{r} = \frac{a}{2r} [/math] To get from [imath] Cos \theta = \frac{r}{R} [/imath] to a value for [imath]\theta[/imath] you take the inverse Cos or ArcCos [imath] \theta = Cos^{-1}\frac{r}{R} = Arccos\frac{r}{R} [/imath] which is basically a button on the calculato. similar for sin and tan Edited January 10, 2012 by imatfaal 1 Link to comment Share on other sites More sharing options...

csmyth3025 Posted January 11, 2012 Author Share Posted January 11, 2012 Hi Chris ...To get from [imath] Cos \theta = \frac{r}{R} [/imath] to a value for [imath]\theta[/imath] you take the inverse Cos or ArcCos [imath] \theta = Cos^{-1}\frac{r}{R} = Arccos\frac{r}{R} [/imath] which is basically a button on the calculator. Similar for sin and tan Thank you very much for that information imatfaal. As you can tell, I know very little math beyond the high school algebra I learned 45 years ago. Through self-study I'm trying to remedy this - but it's slow going for me. After reading the Wikipedia entry on inverse trigonometric functions here: http://en.wikipedia....etric_functions and taking a closer look at my TI-83 calculator I now have a rudimentary understanding of the difference between a multiplicative inverse (which I was using) and a compositional inverse as exemplified by terms such as [math]cos^{-1}[/math]. With this new (to me) knowledge in hand I'm now able to apply the formulae presented in the MathWorld article to the formal calculations that eluded me before. The diagram below is, again, taken from that MathWorld article: R = 3 ft r =0.5 ft h=2.5 ft [math]\theta = 2cos^{-1}\ (\frac{0.5}{3}) =2.8067[/math] (radians) With [math]\theta[/math] I can use an alternate equation to obtain a : [math]a = 2\ R\ sin(\frac{1}{2}\theta) = (2)(3)sin[(0.5)(2.8067)] = 5.916 ft[/math]. This agrees with my previous calculation that [math]a = 2 \sqrt{R^2-r^2} = 5.916 ft[/math]. I can now also calculate the arc length s : [math]s = R\theta = (3)(2.8067) = 8.4201 ft[/math] With this information I can now calculate the wetted cross-sectional area A by two methods: [math]A = \frac{1}{2}R^2(\theta-sin\theta) = (0.5)(9)(2.8067-0.3287) = 11.15 ft^2[/math] -and- [math]A = \frac{1}{2}(Rs-ar) = (0.5)[(3)(2.84201)-(5.916)(0.5)] = (0.5)(25.2603-2.958) = 11.15 ft^2[/math] This agrees with my first approximation of ~11.14 ft^{2}. Thanks again for your help, Chris Link to comment Share on other sites More sharing options...

imatfaal Posted January 11, 2012 Share Posted January 11, 2012 Isn't it cool when all methods reveal the same answer? I must admit I find Mathworld terse to the point of unintelligibility sometimes - it's a great resource, but can be a little less didactic than I would like. I also use WolframAlpha for these sorts of things - http://www.wolframalpha.com/ It is part of the same group and can make the sums really easy if you can find the right way to ask the question http://www.wolframalpha.com/input/?i=area+of+circular+segment+radius+3+angle+2.8067 Link to comment Share on other sites More sharing options...

csmyth3025 Posted January 11, 2012 Author Share Posted January 11, 2012 I also find the WolframAlpha site to be very useful. In this case, though, I think I learned a lot more by taking the "long way around". Chris. Link to comment Share on other sites More sharing options...

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