Unit Hydro PE Practice Problem (assistance needed)

All - I need some assistance on this problem.  I'm stumped:

A rain event has an intensity of 1.5 in/hr for the first hour followed by 0.7 in/hr for the second hour.  The 1 hour unit hydrograph is as follows:

T (hr):         1       2      3 

Q (cfs/in): 0.5   1.2   0.4

I take this as a unit hydrograph lagging problem and came up with this synthesized hydrograph (lagging one hour, adding ordinates, and diving by n [2]):

T (hr):         1      2       3    4

Q(cfs/in):  .25   .85   .8   .2

I then simply do: 0.25 * 1.5 + 0.85 * 1.7 = 0.97 CFS.  This is not correct, however.  The answer is given as:

Q = 1.5 in/hr * 1.2 cfs/in + 0.7 in/hr * 0.5 cfs/in = 2.15 cfs.

Why is the second hour multiplied by the first hour of the 1-hr hydrograph?

Any insight into what I'm doing wrong would be appreciated!

Hi Brady, no lagging needed for this problem. Since the rainfall is given to you in 1 hour increments already, you can use the 1-hr unit hydrograph given. I'm assuming the question asks for the peak flow rate.

Here's how I would tackle the problem: Think of this storm as two seperate mini storms: The first storm begins at hour zero, and lasts one hour with 1.5 inches of precipitation. The second storm begins at hour 1, and lasts one hour with 0.7 inches of precipitation. 

Now, the unit hydrograph tells you how much runoff you would expect from a storm with 1 inch of precipitation in a particular watershed, but the nice thing (and a major assumption of this method) is you can multiply it by the actual storm precipitation to get runoff hydrograph. Since there are two seperate mini-storms, do this twice as I've shown below. Keep note that the second-mini storm begins at hour 1!




Of course, the actual storm consists of these two mini storms combined, so you need to add these hydrographs directly. Here's what you'll get


The combined hydrograph shows that the peak occurs at 2 hours after the first drop of rain hits the ground from the overall storm. This peak is = to

Q (peak) = 1.5 in * 1.2 cfs/in + 0.7 in* 0.5 cfs/in = 2.15 cfs.

Hope this helps! Gotta love comic sans.

Thank you for that in-depth explanation!  I guess whenever I see a storm and unit hydrograph with different duration I automatically assume to lag them.  Although, I guess you are effectively lagging this by lagging all Q values by one hour for the 0.7" section.  I think I need to review these more; for some reason I always have trouble wrapping my head around them.

No problem, it is a very tricky subject and I can never remember how it works off the top of my head (well, now I can just look back at this post!)

Regarding the unit hydrograph duration, imagine if the problem was structured this way:

Given:

Average Effective Rainfall from Hour 0-3: 1.5 in/hr (4.5 in total)

Average Effective Rainfall from Hour 3-6: 0.7 in/hr (2.1 in total)

and given the 1-hour unit hydrograph in your problem statement.

Solution:

You could not solve the same way I did in the above post, because originally the rainfall was given in 1-hour increments. Now, in this particular problem,  you have two pulses of 3-hour rainfall increments (aka two mini storms lasting 3 hours long each). Therefore, you would have to convert your 1-hour UH to a 3-hour UH. Once you did that, you could solve the same way I posted above, except you would "lag" the second pulse by 3 hours instead of 1.

And why would the exam give you 3-hour pulses in the first place? Well maybe the "effective rainfall gauge" (quotes because while that would be awesome, I'm not sure this exists) only collects data every 3-hours. So the gauge says it collected 4.5 inches of water from 7AM to 10AM, then clears the data and starts a new reading for the next 3-hours.

Luckily, I really don't think the depth exam will be this tricky..the problems will probably only be as complicated as your post above, or converting an x-duration UH to a y-duration UH. It would be a little cruel to combine both in one problem, but you never know. Good luck!