Can someone please clarify the following problem:

1. Given the following data:

Time (hrs) vs. Discharge (cfs)
0,0
1,75
2,200
3,100
4,50
5,25
6,0

A two hour storm even of intensity 0.5 in.hr.

How much EXCESS VOLUME of water (acre-ft) is produced?

The answer sets up a table of Time, Qu, Q1 + Q2 = Qtotal. It adds the Qtotal then x 3600s/hr x 1ac/43560 ft^2 = 37.19 ac. ft.

What does Qu, Q1, and Q2 represent and how are these factors determined? thanks.

I answered this in your other post.

Other Post

Qu = unit discharge (cfs per inch of rain for 1-hour storm)

The Qu values represent the discharge of water after 1" of rainfall in the first hour only. Thus, based on the given table, if 1" of rain falls in the first hour only, the discharge at t=4hours is 50 cfs.

Q1 = actual discharge for 1st hour of storm (Qu x 0.5)
Q2 = actual discharge for 2nd hour of storm (Q1 shifted 1 hour later)

Note: if the storm was 3 hours, then a Q3 set of values (Q1 shifted 2 hours later) would be required

Qtotal = Q1+Q2 = total discharge because runoff from 1st hour must be added to runoff from 2nd hour of rainfall


Good luck!