FE Exam Review - Water Resources - Water Distribution

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[@lolaviajes208]

Velocity BC= 2.175ft/s, QBC=15.37cfs, QCD=9.18cfs. Thanks for your videos!

[@anaesquivel3878]

For velocity of B, I got 2.17 and QB=15.33 and QC=9.18:) Thanks, great video by the way.

[@angelasaes]

that adds up to 24.51 cfs, at least use units

[@db2885]

It would be helpful if you could please post the solutions to the questions at the end. Thank You.

[@ricardoarturolaraperaza9988]

What a great explanation. Thank you very much.

[@saeedehlotfi2576]

Thanks for the great explanation on this problem. Would you please share the method to find the Qb and Qc? I'm kind of confused what to substitute for Qin and Qout in conservation equilibrium.

[@cleoturo]

@@directhubfeexam Hey Thank you for these Videos. About QB = 21.68 cfs > 17.5 cfs, how can you have more flow than what is coming in? Isn't by conservation Qin=Qout=17.5 cfs.

[@directhubfeexam]

@@cleoturo yeah, I think you're right. My solution is wrong for this one.

[@angelasaes]

I got QB = 6.28 cfs and QC = 11.2 cfs, although I assume it's wrong since the only equations I used are Q_A = Q_B+ Q_C and Q_B = v_AB * A_AB. It would make sense based on the areas, but the way the distribution system works makes it seem that most water would come out of node B than out of node C

[@sekhar4876]

Highly informative..Gr8 video!

[@directhubfeexam]

Thank you!

[@NanWang-b4u]

great video! one question: why does the total head loss is 0? my understanding is the that head loss is from the friction, which cannot be recovered

[@NanWang-b4u]

is that because the direction of the headless? in the reference book "Multipath Pipeline Problems" section, the head loss in the parallel pipe with same flow direction pipes are equal. but in this question, they are reverse

[@directhubfeexam]

Close!
It's still true that the headloss of parallel pipes is the same. This concept should be remember when dealing with parallel pipes that start at the same junctoin (node) and end at the same junction (node).
But why is the headloss equal to zero in a looped system?
It all goes back to CONSERVATION OF ENERGY. In each closed water distribution loop, there can be no net gain or loss of energy (head). Imagine being a water particle going around the loop from the start to the end. You will end up where you started therefore the energy level must be the same. If we have a head loss in one pipe in the loop it must be balanced by energy head gains or losses in other pipes. This makes the total head loss around the loop equal to zero.
For the water distriubtion system to be in equilibrium, each pipe path around the loop must have the same head loss, so there is no “extra” energy at any point.
What we do in real life is adjust the flow distriubtion to ensure the head loss is equal to zero in a looped system. By applying the condition that the sum of head losses around each loop is zero, the iterative Hardy Cross method is applied to adjust the flow distribution in the network until all loops satisfy this balance. KEY WORD: ADJUST FLOW DISTRIUBTION. The flowrate into each junction will be different. But in the end, the headloss in one pipe is balance by the headloss in another.

[@NPC_Reselling]

great video, thank you

[@prepfe5596]

Nice !

[@davidaghakhaffaf6441]

Hi
Thank you for all these questions and very good explanation but where do you get these questions from ?

[@directhubfeexam]

Hi, thank you for watching. A lot of the questions I cover come from copy right expired material which is then edited by me and worked out on paper. Then I present it on here. I don’t really use specific books, I mostly try to cover the topics listed by NCEES.

[@muradfarraj6186]

perfect. thanks

[@rickcarr8253]

Is 19.97 kW correct for the Power question?

[@Champagne_Rio]

Is there a video on finding the flow at each point?

[@directhubfeexam]

I think you can just apply conservation equilibrium at each point. Qin = Qout

[@varunjhawar5593]

Could you explain how you got the Qb and Qc please? I am a little lost

[@directhubfeexam]

Hello Varun, for Qb and Qc you will apply flow node equilibrium at each node.
There is a mistake on my part here. FOR THE FE EXAM, the flow rate direction will be given to you. They will have to give you the direction of the flow since the method of determining the direction is iterative and requires a trial-and-error approach. Usually, something called the hardy cross method is used to do this.
So we will assume the flow directions.
For node B:
If we assume the flow in pipe AB and pipe BC goes into Node C -> Qb = QAB + QBC.
For node C:
If we assume the flow in pipe DC goes into node C and the flow in pipe BC goes away from Node C -> Qc = QDC - QBC.
Sorry for the confusion.

[@dand7056]

Could you redo this video with a simple fe type of problem. I think this would be too much to do in 3 minutes

[@directhubfeexam]

Hi Dan, this problem is very similar to one found in the 2021 Civil Practice Exam.
I would know this and yes you may see it cut a little shorter if given a triangle distribution system.

[@dand7056]

@@directhubfeexam perhaps if you have time fix the error that you are talking about in the comments. Thanks for all these videos.