 So today we're going to talk a little bit about the geometry of a channel cross-section and the measurements that are going to be useful when we're trying to determine the amount of flow that's going in a channel. Okay, so we're going to consider the flow. There's two kinds of flows I've kind of drawn here. The flow through what we call an open channel and an open channel is an example of something like a river or a stream and we call it open both because there is no top to it. Okay, and we also call it open because there is going to be a surface of water that's exposed to some level of air and we can represent that surface. There's a little notation we use for that surface, a little triangle here with some little lines underneath it that sort of designate what we call the free surface, the free surface being the surface between the fluid that we have and the atmosphere. So an open channel doesn't have a top to it and we compare that to a closed conduit. The term we use here is often called a conduit for something like a pipe that basically the water is flowing. Now a conduit can have what we call pipe flow in which case the entire conduit is filled or you can also have what's called open channel flow, open channel flow which has to happen in a channel but can also happen in a pipe if the pipe is only partially filled and you have a free surface in both cases. So there are many different types and shapes that we can look at and we're going to talk about a few of the sort of standard shapes. For example we could have a rectangular shape, some shapes are sort of like a flume where we assume that the sides are more or less like a box, a triangular shape which is sort of more typical for naturally cut streams, trapezoidal maybe for wider streams, some other common shapes, easy geometries to work with if we assume things, a semicircle assuming that it's full that the depth of the water is equal to the radius of the semicircle. It's a little harder to calculate when it's a partially filled semicircle and then we can also look at the cases of pipe flow in the case of something like a circle which is pretty typical cross-sectional shape for pipes and then our irregular flow which would be sort of our other channel shapes. Basically if we didn't want to estimate if we could actually get measurements for the shape of the cross-section we might have a more irregular shape than some of these shapes that are represented here. So the continuity equation, you might recall the continuity equation q equals va okay where we have our velocity but we also have our A, our cross-sectional area. This is the area that's basically being covered by water as that water is considered flowing into our face basically toward us or away from us if you want to look down into the water stream but the idea that it's flowing in the direction of the flow we're looking perpendicular to it. And the continuity equation says flow is equal to the velocity, the speed of the water times that cross-sectional area. So knowing that cross-sectional area is useful okay and it's pretty simple to calculate the cross-sectional area if we have relatively straightforward geometries. For something like a pipe flow let's go ahead and make a column here of this area of these areas for the cross-sectional area. And let's start down here with this pipe flow. If we were wondering about the area here we might want to consider the inner diameter of the pipe. Usually pipes have an inner diameter and a thicker outer diameter but if we think about the inner diameter of the pipe the area of the pipe is just the area of that inner circle which we might say is equal to pi r squared the radius or if we use the inner diameter pi times the diameter the inner diameter squared divided by four and just using our sort of standard geometry relationships here. We can apply similar geometric relationships to all these other situations. In the case of the rectangle we would want to know some sort of width of the channel okay and we would multiply that by the depth of the water in the channel. Okay so in this case let's go ahead and label that width w our area would be equal to the width times that depth. Notice that this area depends on the depth. If you have more water flowing through then you're actually going to have more area. Okay well let's look a little bit more if we look at a triangle the area of a triangle is one half base times height but that might be a little more complex here because you'll notice as we fill as we talk about this depth here and we fill that triangle up the width expands as the depth expands so we're not going to have a consistent width the width is going to change along with the depth so often we might define that as being something slightly different. I'm going to define an angle here an angle representing the slope of the sides of that triangle maybe we'll call that angle theta and we can recognize that there's a relationship between this width here whatever the width is across the top and the depth and that angle theta and if I put those together using one half base times height I get an area version of an area equation of d square divided by the tangent of theta in other words if this width we have this width here we recognize that there's a relationship between that width if I say the tangent of my angle theta is equal to opposite over adjacent that means it's equal to d divided over w divided by 2 and if we plug that into our one half base times height equation or one half width times depth we get a relationship here so it's a little more complex obviously but we can easily find the area depending on the depth and this angle we defined I can do a similar thing with the trapezoid if I recognize that a trapezoid is simply a combination of the rectangle and the little triangle that's basically on the side and we can basically add the two of the things above it would be something like wd plus d square divided by tangent of theta again where the width is actually we've defined the width as the width along the bottom here and the depth being the depth of the water so in the same way we can continue to generate formulas for a semicircle the area of the semicircle is going to be one half pi r squared or let's see here one half pi d squared where d here is not the diameter but d in this case is the depth and we're assuming in this case that the depth is equal to the radius that's a simplification of the semicircle not really a simplification we can simplification we can always use often that it's going to be only partially filled in the case of something partially filled you end up with a much more complex equation and I'm going to write the equation here I'm not really talking about how to derive it but I'm going to write it here just to see how quickly it can get very complex even for something as simple as a circle it's simple if it's half full it's not so simple when it's a little longer in this case the area looks something like this where if you have the radius of your circle r and you have your depth here you very quickly get to a complicated a more complicated relationship that looks something like this r squared times the inverse cosine of r minus d over r minus r minus d times the square root of 2 r d minus d squared my goodness that's a pretty extensive relationship there not necessarily asking you to know that relationship the basic idea being we're subtracting the we're taking the area well basically we're trying to find the area we're subtracting off the area that's above from the area below there's the two the subtraction in there pretty complex very quickly so often we'll use charts to figure out something like that sometimes they define that angle in terms of an angle from the center okay but we can still find the area using geometry okay in the case of the irregular situation one of the things we might have to do is get some measurements of the geometry and chop it up and make some estimations using something like numerical integration the simpson's rule is trapezoidal rule which we'll speak about in another video so hopefully you have a sense for what cross-sectional area is but another important geometric measurement that we use is something known as the wetted perimeter now it's a little bit of a strange sort of idea okay but the but what we're thinking about here is the place where the water is wetting something and something being the surface okay the idea here is that the flow of the water is in some way impeded it's slowed down by the friction by the drag it has on the sides of the channel and in the case of something like pipe flow in the case of our pipe flow down here that wetted perimeter is pretty easy to find in pipe flow we it's the entire circumference okay so in a case of something like this that wetted perimeter and i'll make a little column over here will date pw for wetted perimeter in this case it's going to be two pi r or pi times the diameter so over here let's see here pi times the internal diameter that's my wetted perimeter it is the circumference or the full perimeter in the case of pipe flow in the case of channel flow the wetted perimeter is only the portions around this geometric shape only the portions that actually touch the surface in other words it's the entire perimeter ignoring the free surface we don't count the free surface in our calculations and that's the idea what the wetted perimeter is so for each of these little segments over here we can similarly calculate the wetted perimeter if i look here the wetted perimeter for a rectangle is going to be equal to that width plus two times the depth basically each side if i look at the triangle well that's a little bit more complicated i have to use the Pythagorean theorem if i call that something like that side something like s i could use the Pythagorean theorem to create something or i could also use a little bit of trigonometry here recognizing that that side s is related through the angle to the depth d and if i recognize that i find the wetted perimeter looks something like this there's two of those sides and if i do some trigonometry here i recognize that it's 2d divided by sine of that angle so it's not too difficult once you've actually considered the geometry to calculate it for the trapezoid just like before we can think about it as being two pieces and we add these two pieces above to get the answer for the trapezoid for the semicircle the wetted perimeter is simply going to be half of the circumference so the wetted perimeter is half of that circumference so let's see here that's going to be two pi r or just pi r and in this case because r the radius and the depth are the same for our full semicircle it simply ends up being pi times the depth again the partial semicircle is much more complex i'm not going to record the entire equation here but we can use a similar sort of idea of breaking down the pieces to determine what that wetted perimeter is but again it's just going to be sort of the the length of this segment on the circle okay so hopefully you can see where we can find the area and how we would calculate the wetted perimeter our third and final parameter the last one we want to look at is something here called the hydraulic radius the hydraulic radius is simply the ratio between these two things my hydraulic radius is equal to the area divided by the wetted perimeter sometimes that value is going to be very easily calculated as things cancel out and sometimes it's a bit more complex the units of the hydraulic radius notice our units of area are going to be in length squared whereas our units of perimeter are going to be length so our final units of hydraulic radius are something like l just in length units meters or feet let's just take a look at example of a couple of the easiest versions of hydraulic radius let's consider our circle for example for our pipe if you take the two values for our pipe you'll see that the area is equal to pi inner diameter squared over four and you compare that to the area of the circle I mean the circumference or the wetted perimeter which is pi times the inner diameter well pi cancels out one of the inner diameters cancel out and you get a radius of di divided by four notice that seems a little strange because rare we're used to radius as being a diameter divided by two when we're talking about this hydraulic radius effectively it's the diameter divided by four for something like a circle you'll notice you get the same exact amount if you do a semicircle and find that ratio okay just one other example if we consider the hydraulic radius of something like the rectangle it would simply take the area of the rectangle wd divided by the perimeter of the rectangle wetted perimeter w plus 2d and that would be our hydraulic radius for something like a rectangular flow