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
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.
@@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.
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.
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.
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.
