cantaloup 0 Posted October 8, 2009 Report Share Posted October 8, 2009 For those who will take the WR/Envi depth, there are approx. 6 questions (15% in PM session) related to open channel flow. I bet at least one question is on Manning formula; but they don't give you a simple or straightforward question. Try this: A trapezoid cross section channel with bottom width is 20 ft, side slope is 1:1 and depth of flow is 5 ft with no freeboard. The Manning coefficient is 0.015, bed slope is 0.0008. If the Flood Control District wants to have the flow capacity increases approximate 20 percent, they will need to excavate the channel deeper with the same side slope and bed slope. What is the new depth in ft they need to excavate? A/ 5.5 B/ 5.8 C/ 6.1 D/ 6.4 Answer will be post tomorrow. Link to post Share on other sites

jeb6294 737 Posted October 13, 2009 Report Share Posted October 13, 2009 Also don't be surprised if a similar question on the exam is asking for an increase of at least 20%, i.e. answer (a) results in a 19% increase and answer (d) results in a 24% increase...even though answer (a) is much closer to the actual answer the correct answer is (d) because they wanted an increase of at least 20%, not closest to 20%. Always a popular ploy on the exam. Simple questions that just require you to pay attention to what they are asking for. Link to post Share on other sites

Timmy! 0 Posted October 14, 2009 Report Share Posted October 14, 2009 I come up with 6.1. Hey Cantaloup, when you get the chance please post your solution. Link to post Share on other sites

RIP - VTEnviro 536 Posted October 16, 2009 Report Share Posted October 16, 2009 Also don't be surprised if a similar question on the exam is asking for an increase of at least 20%, i.e. answer (a) results in a 19% increase and answer (d) results in a 24% increase...even though answer (a) is much closer to the actual answer the correct answer is (d) because they wanted an increase of at least 20%, not closest to 20%. Always a popular ploy on the exam. Simple questions that just require you to pay attention to what they are asking for. Along the same lines, they ask questions like, "How many widgets are required to reduce the concentration of X to below 10 mg/L?" You come up with 3.1. Since you can't buy 1/10th of a unit, you need 4. I'd bet good money 3 will be a distractor. Link to post Share on other sites

chess5329 0 Posted January 29, 2010 Report Share Posted January 29, 2010 For those who will take the WR/Envi depth, there are approx. 6 questions (15% in PM session) related to open channel flow. I bet at least one question is on Manning formula; but they don't give you a simple or straightforward question. Try this: A trapezoid cross section channel with bottom width is 20 ft, side slope is 1:1 and depth of flow is 5 ft with no freeboard. The Manning coefficient is 0.015, bed slope is 0.0008. If the Flood Control District wants to have the flow capacity increases approximate 20 percent, they will need to excavate the channel deeper with the same side slope and bed slope. What is the new depth in ft they need to excavate? A/ 5.5 B/ 5.8 C/ 6.1 D/ 6.4 Answer will be post tomorrow. I got 5.5! Link to post Share on other sites

Capt Worley PE 1,144 Posted February 1, 2010 Report Share Posted February 1, 2010 I will not be mocked! Link to post Share on other sites

afrey22 0 Posted February 23, 2010 Report Share Posted February 23, 2010 I don't know what happened to this answer, but I'll post what I've got and we'll go from there. Seems to me that all we want to do is increase the area of the trapezoid by %20. That eliminates having to use manning and bed slope. The area of a trapezoid with common slopes (both slopes are 1:1) is EQ1:A=h*(b1+b2)/2. We are given b1 (20ft) and h (5ft). b2 can be found by EQ2:b2 = b1 + h*(2*slope).. So, we need to increase A by 20% but we are only changing the value of h. Substitute EQ2 into EQ1 and perform some algebra. I came up with an answer of B- 5.8 or a 19.7% increase. amirite? Link to post Share on other sites

benbo 40 Posted February 23, 2010 Report Share Posted February 23, 2010 I could be wrong, I was doing this while "listening" to a conference call so I could have messed up the arithmetic - I also got 5.8. I rounded too much the first time (11.6/2 - I lopped it off Link to post Share on other sites

cburge01 0 Posted March 17, 2010 Report Share Posted March 17, 2010 I could be wrong, I was doing this while "listening" to a conference call so I could have messed up the arithmetic - I also got 5.8. I rounded too much the first time (11.6/2 - I lopped it off I got 6.4. Link to post Share on other sites

cburge01 0 Posted March 17, 2010 Report Share Posted March 17, 2010 I could be wrong, I was doing this while "listening" to a conference call so I could have messed up the arithmetic - I also got 5.8. I rounded too much the first time (11.6/2 - I lopped it off I got 6.4. As Jeb noted above, the result must show atleast a 20% increase. Link to post Share on other sites

SandTiger21 0 Posted May 4, 2011 Report Share Posted May 4, 2011 My calcs show: Initial Flow = 834 cfs Reqd Flow = 834 x 1.2 = 1000.88 cfs A depth of 5.8 ft. gives a flow of 998.99 cfs, very close, but not quite the reqd 20% more than the original flow. A depth of 6.1 ft. gives a flow of 1059.37 cfs, which is 27% more than the original flow. I thought using 1.486 instead of 1.49 for the manning's conversion constant might give me an answer that agreed with a depth of 5.8 ft., but it made no difference. Lesson: Focus is as important as knowledge. Link to post Share on other sites

Environmental_Guy 0 Posted August 12, 2011 Report Share Posted August 12, 2011 Sorry to dig up a fossil, but this seems like a good problem. I get 6.002 feet. Can someone check my methodology? I get 1000.88 cfs as the new flow. Using Q=AV with the new Q, same V from the Manning formula (6.67 ft/s) and plugging in the formula for the area of a trapezoid and solving for the height, I get 6.002. Back checking with an area of 150 sf and velocity of 6.67 ft/s, you get 1000.83 cfs. So, round up to 6.1 ft. Correct? Link to post Share on other sites

Jacob_PE 1 Posted August 17, 2011 Report Share Posted August 17, 2011 I came to the same conclusion as Sandtiger 21 above. As we dig down deeper the bottom width gets smaller and so we need to use a different A and R for each depth and recalculate Q from Mannings equation. I did this by trial and error using each of the depth options. The question asked is, at what new depth will the flow capacity be increased approximately 20 percent, not at least 20%. So I'd agree that the answer is 5.8' yielding a flow of 999.12 cfs (approximately 1001.1 cfs) Link to post Share on other sites

Jacob_PE 1 Posted September 4, 2011 Report Share Posted September 4, 2011 Sorry to dig up a fossil, but this seems like a good problem. I get 6.002 feet. Can someone check my methodology? I get 1000.88 cfs as the new flow. Using Q=AV with the new Q, same V from the Manning formula (6.67 ft/s) and plugging in the formula for the area of a trapezoid and solving for the height, I get 6.002. Back checking with an area of 150 sf and velocity of 6.67 ft/s, you get 1000.83 cfs. So, round up to 6.1 ft. Correct? I'd like to get to 'case closed' on this problem. There were several different conclusions from board members. The original flow provided is 834.25 cfs, adding 20% makes the new flow target = 1001.10 cfs. In order to solve for the depth using the equation for the area of a trapezoid, you have to have a known base width, does the problem intend for the bottom width to stay 20'? If so, we can use the conveyance factor table on page A-42 of the 11th CERM. But when I do that I get a K' value of .18014, interpolating (d/b)=x=.348187, d=20*.348197= 6.96 ft. When I originally looked at the problem I took each possible depth answer and calculated a new base width and verified each flow individually. Is this a problem where the conveyance factors should not be used? Link to post Share on other sites

Rosemary 0 Posted September 26, 2011 Report Share Posted September 26, 2011 The easiest way to deal with this type of hydraulic question (trapezidal x-section open channel flow) is to use the HP calculator (instead of using table, trial & error method). Most of the time the bottom width is fixed, the question is to find the depth of flow (or the top width of flow) with a given flow rate. The formula is : Q=1.486/N*((B+Z*Y)*Y)^(5/3)/(B+2*Y*SQRT(1+Z^2))^(2/3)*S^0.5 with Q = flow in cfs; N: Manning coeff.; S: slope of the channel; Z:slope: rise/run=1/Z ; Y: depth of flow ; B: bottom width For the question from "cantaloup" (with top width T of flow is given) here is the formula: Q=1.486/N*((T-Z*Y)*Y)^(5/3)/((T-2*Z*Y)+2*Y*SQRT(1+Z^2))^(2/3)*S^0.5 Plug it in the HP calculator and solve for Y, you get the answer in seconds. The answer is 5.8 If you don't have an HP calculator, you may use Excel, use the first formula to find Q then increase the flow 20% then use the second formula to find Y with the new Q. Note: contact me at ngmngoc@mail.com and I'll send you some formulae used for HP. I'll post some other hydraulic questions in next few days. ++++++++++++++++++ For those who will take the WR/Envi depth, there are approx. 6 questions (15% in PM session) related to open channel flow. I bet at least one question is on Manning formula; but they don't give you a simple or straightforward question. Try this: A trapezoid cross section channel with bottom width is 20 ft, side slope is 1:1 and depth of flow is 5 ft with no freeboard. The Manning coefficient is 0.015, bed slope is 0.0008. If the Flood Control District wants to have the flow capacity increases approximate 20 percent, they will need to excavate the channel deeper with the same side slope and bed slope. What is the new depth in ft they need to excavate? A/ 5.5 B/ 5.8 C/ 6.1 D/ 6.4 Answer will be post tomorrow. Link to post Share on other sites

Miguel 0 Posted October 8, 2011 Report Share Posted October 8, 2011 THE TDOT has a Design Drainage Manual (Chapter 5) in the Web. In page Appendix 5A-10 there is a capacity chart very useful.In the chart you will find plotted QN/b**(8/3)*S**0.5 Vs d/b for trapezoidal channels.With d/b = 5/20= 0.25 your value of QN/b**(8/3)*S**0.5 = 0.15, doing the math from here I found Q to be 833.6 CFS.Increasing the flow 20% we get 833.6X1.20=1000.32 CFSUsing trial and error with d=6.0 we get our new b=18 with these values d/b=6/18=0.33 we find in the chart QN/b**(8/3)*S**0.5 = 0.25and Q=1049 CFS which is higher than 1000, therefore d needs to be slightly less than less 6.00.Just for curiosity I did it for d=5.5 Ft and got 970 CFS and for 5.8 I got 1023 CFS. Link to post Share on other sites

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