# help please



## smilestar (Oct 22, 2014)

Given: The 1 hour unit hydrograph of excess precipitation is described by the following data:
Time (hr) 0.0 1.0 2.0 3.0 4.0 5.0
Discharge Q (ft3/sec/in) 0 30 95 125 50 0
A storm produces the following pattern of excess precipitation:
1.7 in of excess precipitation during the first hour, followed by 0.8 in during the second hour

The solution I have says (30*0.8)+ (95*1.7)= 185.5

But shouldn't it be (30*1.7) + (95*0.8), I mean first hour Q shouldn't it correspond with first hour precipitation.
What am I missing here?


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## ptatohed (Oct 22, 2014)

I think you're more likely to receive a response if you post in the appropriate subform (and title the thread).

http://engineerboards.com/index.php?showforum=35


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## ptatohed (Oct 22, 2014)

P1=1.7, P2=0.8, U1=30, U2=95, U3=125, and U4=50

Time= 1 hr.....P1xU1
Time= 2 hrs...P2xU1+P1xU2
Time= 3 hrs...P2xU2+P1xU3
Time= 4 hrs...P2xU3+P1xU4


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## smilestar (Oct 22, 2014)

Thank you, really appreciate it.


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## John QPE (Oct 23, 2014)

What is the actual question here?


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## NJmike PE (Oct 23, 2014)

John Q said:


> What is the actual question here?


this


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## DanHalen (Oct 23, 2014)

smilestar said:


> Given: The 1 hour unit hydrograph of excess precipitation is described by the following data:
> Time (hr) 0.0 1.0 2.0 3.0 4.0 5.0
> Discharge Q (ft3/sec/in) 0 30 95 125 50 0
> A storm produces the following pattern of excess precipitation:
> ...




I remember working this problem. This comes from Goswami's All-In-One Exam Guide problem #3 (morning question). The solution he gives is correct and I had the same question as you when I first worked this problem.

This is a one hour unit hydrograph and they want you to solve for the "flow" excess precipitation from the unit hydrograph. You start with the flow rate given at hour 1 and multiply that times the precipitation for hour 2. Add that with the flow for hour 2 x precipitation for hour 1 which gives you the excess precipitation for a 1 hour unit hydrograph. You don't need to worry about t=0 because there is no flow so you start with t=1 and go from there.

You cross multiply the Q1*P2 &amp; Q2*P1 because it takes an hour to achieve the flows recorded. Think of a rain event in the mountains. Once the rain ends in the mountains it might take an hour for the stream down in the valley to see any changes in the flow and that's kind of the concept here. So what we do is we take the flow down in the valley after the rain event and multiply that times the precipitation the mountains got to get our excess precipitation. I tried to put this in layman's terms to make it easier to understand. Hope that helps clear things up.


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## smilestar (Oct 23, 2014)

Thanks DanHalen, it sure helped


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