Suns: How did you solve for d given Q?To solve for d, just plug 4 answers in the Manning eq. to see which one match with Q; however you need to prepare the derived equation beforehand (write it down in your notebook) which is listed below, and make sure that you don't make mistake when punching in numbers, * , / , ^ , ( , ) , etc.
This is how I would solve this problem :
The methodology to solve this problem is easy, but in order to solve it fast and correctly you need to use the HP-33s calculator.
Here are some equations I put in the calculator (Y is the normal depth of flow, B: bottom width, Q: cfs) :
MANNING-BASIC
V=1.486/N*R^0.667*S^0.5
Q=1.486/N*A*R^0.667*S^0.5
MANNING TRAPEZ (if rectangular shape, enter Z=0 or Z=0.0001)
V=1.486/N*((B+Z*Y)*Y/(B+2*Y*(1+Z^2)^0.5))^0.667*S^0.5
Q=1.486/N*((B+Z*Y)*Y)^1.667/(B+2*Y)*(1+Z^2)^0.5)^0.667*S^0.5
For this open channel flow problem, turn on the last equation in the HP-33s and solve Q, enter value of
N, B, Z, Y, S
Answer Q=376.06 cfs
New Q=1.5*376.06=564.09
Turn on that equation again, this time solve for Y
The same value of N, B, S, Z are already stored in the calculator, so press [R/S] button each time, no need to enter the number again; only enter value of Q which is 564.09.
You’ll get the answer : 7.42 ft (depth of flow)
Depth of excavation = 7.42-6.0=1.42 , so the answer is 1.5 ft
If you plan to take PE exam water resource module, it’s better to use the HP equation solver. Consider 3 types of equation: D-W, H-W and Manning, you need to derive:
For D-W : one eq solve V, one solve Q
For H-W basic: one eq solve V, one solve Q
For H-W pipe: like above
For Manning basic: like above
For Manning pipe (full flow): like above
For Manning trapezoid (also rectangle): like above
One eq to solve pipe in parallel and one to solve pipe in series using D-W
One eq to solve pipe in parallel and one to solve pipe in series using H-W
One eq to solve crit. depth of flow of trap. & rect. channel
One eq using H-W and one using D-W (both Q: gpm & D: inch) to solve the pump problem in the NCEES sample exam question
For example, go to page 126 of the NCEES Sample question book:
Here is the H-W eq to be put in the HP-33s to solve the pump problem:
Q=0.281*C*D^2.63*(H/L)^0.54
TDH = H(static) + H(loss) = 30 + H(loss)
Turn on the equation, solve for H(loss) four times with 4 values of Q in the answer part (nice thing is that value of C and D are already stored, no need to enter again, just press [R/S] )
Calculate the TDH and plot these point on the graph to get the intersecting point, which is the operating point; that’s the answer.
I know a guy who brought 2 HP-33s to the exam, one calc. with water eq’s and the other calc. with eq’s for the rest plus one Casio calculator. His philosophy is if he can get 10 questions solved quickly (without worrying about calculating mistakes) by using the equation solver, he can save 20-30 minutes for other “strange” questions which required time for thinking or research. Also he could not read the nomographs well due to bad eyesight.
He passed the PE exam first time despite many yrs away from school !!!
