Flow in Open Channels Continued
工程考试复习基础
乔治亚理工学院
The purpose of this course is to review the material covered in the Fundamentals of Engineering (FE) exam to enable the student to pass it. It will be presented in modules corresponding to the FE topics, particularly those in Civil and Mechanical Engineering. Each module will review main concepts, illustrate them with examples, and provide extensive practice problems.
I am out of school about 10 years and this course helped me to brush on the fundamental knowledge of engineering courses and further encouraged me to take the FE exam with vari
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Hydraulics and Hydrologic Systems
This module applies basic principles of fluid mechanics to practical problems in hydraulics, hydrology, and groundwater flow. We first discuss the generalized and one-dimensional momentum theorem and apply it to various typical problems. Flow in pipes and non-circular conduits is discussed beginning with the Bernoulli equation accounting for energy losses and gains. Calculation of head loss due to friction and minor losses due to valves and other accoutrements are presented. Friction losses are calculated for laminar Poiseuille flow and turbulent flow using the Moody chart; examples include computation of pressure drop in laminar pipe flow and turbulent water flow. Methods to calculate flow in pipe networks consisting of multiple connecting pipes and other fittings is then discussed with examples for parallel pipes. Pipes and turbines are then discussed along with their basic equations and definitions. Characteristic curves, especially of centrifugal pumps, are presented and it is shown how to match a pump to a system head.Flow in open channels are discussed including classification of flow types and prediction of uniform flow by the Manning equation. The use of specific energy concepts to solve gradually varying flows, and the importance of the Froude number and sub and supercritical flows are presented. Predictions of hydraulic jumps and flow over weirs are given.Hydrological principles include predictions of surface runoff by the curve number method and peak runoff by the rational formula. Groundwater principles include Darcy's law for flow through porous media and prediction of drawdown by wells in confined and unconfined aquifers by the Dupuit and Thiem equations.In all cases, basic ideas and equations are presented along with sample problems that illustrate the major ideas and provide practice on expected exam questions.Time: Approximately 3 hours | Difficulty Level: Medium
Momentum Theorem9:26
Momentum Theorem Continued15:26
Flowin Pipesand Conducts12:44
Flow in Pipes10:56
Flow in Pipes Continued9:57
Pumps and Turbines6:51
Pumps and Turbines Continued11:38
Flow in Open Channels10:04
Flow in Open Channels Continued13:56
Hydrology5:33
Groundwater5:56
Groundwater Continued7:20
Dr. Philip Roberts
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Continuing our discussion of flow and open channels, in this segment I want to look at the concept of specific energy, the ideas of sub and supercritical flows, and how to predict flow over weirs. Firstly, the concept of specific energy is very useful for solving problems where the flow or depth might change. And, for, as in this example here where the flow goes over a hump. So the specific energy, E, is defined as y plus V squared, over 2g, in other words, the local water depth y plus the velocity head V squared over 2g. For a channel of constant width, we can write this equation in this form, E equals y plus q squared over 2g y squared where q is equal to the total flow rate divided by the width of the channel, is the flow rate per unit width in the channel. If we plot out this equation for constant width we get the so called specific energy diagram, which looks like this graph here. So, just focusing our attention on this one curve here, so this is a graph of, y, water depth versus, E, specific energy, we get a curve which looks like this. And the curve goes through a minimum value here, E minimum, where the water depth I will call yc. And by standard open channel flow analysis, you can show that the criti-, the depth which occurs at that depth which I'll call the critical depth at the minimum specific energy is given by Q squared over g, raised to the one-third power. Another very important parameter for open channel flows is the Froude number, defined as V divided by the square root of g times the water depth. And, if we substitute that equation for the critical depth up above we find that the Froude number at the minimum specific energy, or the critical depth, is exactly equal to 1. And, this type of flow, we call a critical flow. If the Froude number is greater than 1 then we say that the flow is supercritical, and if the Froude number is less than one, we say that the flow is subcritical. So, in this diagram here, if I draw a horizontal line here at the critical depth, than any depths which are deeper than this will be subcritical flow. And any depths which are more shallow than that, in other words moving faster, will be supercritical flow. The other thing that's very important about this diagram is that for any particular value of specific energy we have two depths, two possible depths here. And these two depths are called alternate depths. They are alternate to each other. So, the alternate depths are a pair. Always one alternate depth is subcritical and the other alternate depth is always supercritical. So here are the corresponding sections from the reference manual where in this diagram they refer to y1 and y2 as alternate depths to each other. Also, however, they give another equation here, y2 is equal to y1 squared of 8 plus foot number 1 squared minus 1. Now, it's rather unfortunate that they put that there because that equation is the equation to predict the depth downstream of a hydraulic jump. Now, the depth in a hydraulic jump, y1 and y2, are not ultimate depths to each other. Because there is a loss of energy going through the hydraulic jump. So, it's rather unfortunate that the equation is added there. But don't get confused, those depths are not alternate to each other. Let me do a simple example on that. So, the question is: water is flowing at a depth of 2 meters in a channel 4 meters wide at a speed of 4 meters per second. Which of the following statements is true? It's subcritical, critical, supercritical, or the alternate depth is more than 2 meters. Well, the key here is to compute the Froude number, V over square root of gy. So, that is equal to, the velocity is 4, the depth is 2, so the velocity, or the Froude number, rather, is 0.9, which is less than 1. So, therefore, we see right away that the flow is subcritical. This one is clearly not correct. It's not critical because it's not equal to 1. It's not supercritical because it's not greater than 1, but how about the last statement. The ultimate depth is more than 2 meters. To consider that we remember the specific energy diagram here. So in this case here is our critical depth, it's a subcritical flow so therefore the point could be here for example. It's deeper than the, the subcritical depth. The alternate depth, however, is supercritical, which is less than that. It cannot be deeper. So the last statement here, the alternate depth is deeper or more than 2 meters, is also obviously incorrect. The only possible answer is A, the flow is subcritical. Because they gave the equation for hydraulic jumps I'll also cover that term briefly. And a hydraulic jump is a sudden change in a flow depth from high velocity shallow to deeper water slowly moving. And the hydraulic jump is characterized by much air entrainment and energy dissipation, looking something like this. And these fairly often occur in out in channels where you have a sudden change in the bottom slope or they might be intentionally designed in an open channel flow to dissipate energy. This can be analyzed by means of the momentum theorem and the continuity equation. As indicated here we determine a define and control volume here. And then by applying the momentum theorem and continuity the sustained analysis we arrive at this result. Y2 over y1 is half the square root of 1 plus a Froude number squared minus 1. Where y2 is the downstream depth, and y1 is the upstream depth. And Froude number one is the upstream Froude number. So, given the upstream conditions, we can predict the downstream water depth from that equation. The equation also works if we reverse all the subscripts. So, in this form, given the downstream conditions, Froude number 2, we can predict the upstream depth, y1. And, finally, the other important equation is that the head loss across the hydraulic jump is equal to y2 minus y1 cubed divided by 4 y1, y2. In a hydraulic jump we always go from upstream Froude number greater than 1 in other words, supercritical flow to froude number less than 1, in other words subcritical flow downstream. And the depth increases. So if we draw the energy gradeline for this system, we would find a sudden drop. So here is the energy gradeline. We would find a drop or a loss of energy across the hydraulic jump hL which is given by this equation here which is always a positive number because y2 is greater than y1. So, here is a numerical example. Water flows at a depth of 1 meter in a channel 4 meters wide, at a speed of 8 meters per second. A hydraulic jump occurs. After the jump, the water depth is most nearly which of these alternatives? So, first, we calculate the Froude number. So the upstream Froude number, V over square root of gy is equal to 2.66. So right away we see that the flow is supercritical, the Froude number is greater than 1, therefore a hydraulic jump can occur. Had, if it had been less than 1, then a Froude num, then a hydraulic jump would have been impossible. So, applying our equation for the downstream depth, we have this. And substituting in the upstream depth y1 and the Froude number, we find that the downstream depth y2 is equal to 3.29 meters. So the answer is D. Now to carry on with this, it's not really asked, but the downstream Froude number we can also calculate. And we get that firstly from continuity, the volume flow rate Q, upstream of the jump, V1 A1 is equal to the volume flow rate downstream, V2 A2. Therefore V2, the downstream velocity, is V1 A1 over A2. V1 we're given is 8 meters per second. The depth is 1 meter. The width is 4 meters. The downstream velocity we've just calculated is 3.29 meters. Which gives me a downstream velocity of 2.43 meters per second. Considerably slower than the upstream velocity, so therefore the downstream Froude number, Froude number two is equal to 0.43 from which indeed we see it's less than 1. In other words, the flow is subcritical. We've gone from a supercritical flow froude number equals 2.66 to a subcritical flow, froude number equals 4.3 and a hydraulic jump must always go from supercritical to subcritical flow. The next topic I want to look at is flow measurement by means of weirs. And a simple weir is simple one is a rectangular weir, a wall in the flow, such as shown here. And the equation for flow over a rectangular weir is given here, and here is the segment on the right from the reference handbook. Q is equal to CLH to the three halves power where L is the length or the width of the weir and H is the water depth over the weir crest. And in this equation C is a discharge coefficient for the weir which are these values here. And because of the way they've expressed this equation in the reference handbook, the value of C depends on the system of units you're using as shown here, 3.33 or 1.84. The other thing to note is that H here is not the water depth over the weir because of the drawdown going over the weir. It's the upstream depth far upstream from the weir relative to the weir, weir crest. Another type of weir that's mention is a v notch weir like this. For example, a typical one would have theta is equal to 90 degrees. And the corresponding equation here is Q is equal to CH raised to the the five halves power. Where again the values of the coefficient here depend on which system of units you're working in. So let me finish with a numerical example. A rectangular sharp-crested weir is used to measure flow rate in channel 10 feet wide. It's desired to have the upstream water depth be 6 feet when the flow rate is 50 cubic feet per second. The height of the weir plate is most nearly which of these alternatives? So here the upstream depth here is this depth is 6 feet. And the solution here is our weir equation. Q is CLH to the three halves power. In this case we want to calculate H, the depth. So rearranging, we've got H is equal to Q over CL raised to the two-thirds power. Substituting the values, flow is 50. Coefficient is 3.33. The length is 10, which gives me a height H of 1.31 feet. But note that that is this height here. We're asked to calculate this height PW, the height of the weir crest. So P plus PW is equal to 6. Rearranging PW is 6 minus H. So the answer is 4.69 feet. So, the closest answer is C. And this completes our discussion of open channel flows.