Problem #37 - PE Exam.com

[DanHalen]

Problem 37 states:

Over an 8-hour period, a flood gauge in a river measures the discharge shown below. The base flow of the river is 40 ft3/sec. The total volume of excess water is most nearly:

A) 100 ft3/sec

B) 150 ft3/sec

C) 200 ft3/sec

D) 300 ft3/sec

Time 9:00 10:00 11:00 12:00 1:00 2:00 3:00 4:00 5:00

Discharge (ft3/sec) 40 50 65 90 80 75 65 55 40

The solution in the back has a lengthy table and don't think it's necessary to solve the problem. I've worked several problems like this from other study guides already and none of them have been worked like this. I solved this problem by subtracting 40 ft3/sec from each of the flows and summed those up to get 200 ft3/sec. Would the method I used be correct? The answer in the back of the book is C) 200 ft3/sec

 

[John QPE]

IMO, that problem is all kinds of jacked up.

First and foremost, cfs is not a volume, it is a rate. The volume of excess water should be over the 8 hour period, correct? So just using the 200 cfs...

(200)(3600)(8) = 5.76 x 10^6 cf of excess water.

You need a hydrograph to solve this. There will be variable flow rates between the measured hours, the flow rate doesnt just jump up and down as is shown on this table. I'm not buying the usage of "most nearly" in this problem either.

 

[DanHalen]

IMO, that problem is all kinds of jacked up.

First and foremost, cfs is not a volume, it is a rate. The volume of excess water should be over the 8 hour period, correct? So just using the 200 cfs...

(200)(3600)(8) = 5.76 x 10^6 cf of excess water.

You need a hydrograph to solve this. There will be variable flow rates between the measured hours, the flow rate doesnt just jump up and down as is shown on this table. I'm not buying the usage of "most nearly" in this problem either.

It is all kind of jacked up. There is a similar problem in my All-In-One book and it's clear what they are asking for and you're excatly right! ft3/sec is a flow and not a volume. Big difference. I think what they meant to say is "The total flow of excess water is most nearly"....The multiple choice answers would have to be changed to make this problem a bit more realistic and I'm not feeling it.

 

[John QPE]

You really need durations for those flows as well. You're flowing 50 cfs at 10:00, does this last for an hour? 10 minutes?

 

[K19]

I think what they meant to say is "The total flow of excess water is most nearly"....

There isn't really such a thing as "total flow." Flow is a rate, it can either be instantaneous (Q at a given moment) or an average (total volume discharged V over period of time T).

It really is a bloody mess of problem though - the question itself and the info given are valid enough, however the answers options should be volumes (e.g., cubic feet or acre-feet). As John Q already mentioned this is best done by drawing a hydrograph and subtracting the base flow. The remaining area under your "curve" is the excess volume and can be calculated as follows:

Hour 1: (0+10)/2 * 3600 = 18,000 CF

Hour 2: (10+25)/2 * 3600 = 63,000 CF

Hour 3: (25+50)/2 * 3600 = 135,000 CF

...

etc. through the 8th hour. Summing these up gets you an approximate total volume of excess water.

 

[RIP - VTEnviro]

Problem 37

In a row?

 

[John QPE]

Problem 37

In a row?

I've been trying to find something to watch on Netflix.

Thank you, sir.

 

[RIP - VTEnviro]

Bunch of savages in this town...