NCEES Sample exam WR&E #508

Professional Engineer & PE Exam Forum

Help Support Professional Engineer & PE Exam Forum:

This site may earn a commission from merchant affiliate links, including eBay, Amazon, and others.
anyone? Thanks.
Here we go... I hope this makes sense without diagrams:

The table that was provided is a unit-hydrograph. The CFS values are based on a per inch of rainfall (cfs/in) for a one-hour rainfall intensity.

Since the problem states that a 2-hour rainfall had occurred, then you need to add up the Q/in values (0+75+200+100+50+25+0=450) and multiply by 2 for the unit CFS (900 cfs/in) and multiply again by 0.5" for the actual CFS (450 cfs).

However, to properly calculate the total volume of water, we need to graph the actual CFS at each time interval and calculate the area below the line.

Here are the data domains to develop the volume by graph:

T (hours) = {0,1,2,3,4,5,6}

Qu = {0,75,200,100,50,25,0}

Q1 = {0,37.5,100,50,25,12.5,0}

Q2 = {0,0,37.5,100,50,25,12.5} (Q2 = Q1 shifted 1 hour)

QTOT = {0,37.5,137.5,150,75,37.5,12.5}

If you graph QTOT (cfs) vs. T (hr), the area under the line = 450 cfsh (units are cfs x h = volume of water).

Therefore 450 cfsh = 450 cfsh x 3600 sec/hr x 1hr / 43560 cf/AF = 37.2 AF

I think if you do a few of these problems, you might find that volume of water is the same result as the sum of QTOT, but I can't verify at this time.

Good luck!
Yeah, sum the values to get the total direct runoff -->450cfs. Multiply that by the time interval between hydrograph ordinates (1hr=3600s) to get the volume of runoff. 450cfs x 3600s = 1,620,000ft^3 convert to acre-ft --> 1620000/43560 = 37.19 acre-ft

 
The answer will be more obvious if you arrange the data into columns

Unit Hyrdograh

Time------Q--------Q@hour1(1.5in/hr)--------Q@hour2(0.7in/hr)

1_______0.5_______1.5*0.5=0.75___________-----

2_______1.2_______1.5*1.2=1.8____________0.7*0.5=0.35

3_______0.4_______1.5*0.4=0.6____________0.7*1.2=0.84

4_______________________________________0.7*0.4=0.28

So looking at hour 2 the flow rates add up like this

1.8cfs + 0.35cfs = 2.15cfs

Let me know if the above doesn't help....

Regards,

Casey
still can't understand the problem. at Time = 2hrs, how is that you multiplied 1.5 with 1.2 and 0.7 with 0.5, when 1.5 is for the first hour and 0.5 for the second hour? can you explain that part?
 
I think you are making something that is relatively simple into somthing quite complex.

The way I saw it was that there were two 1-hour stores that occurred one right after the other....

This is assuming that your post is talking about the original quesiton, #508... if not, disregard this post.

Thanks...
Stores??? Are we going shopping?! 😜
 
Back
Top