Here is a question
given two storm events: 1.5in/hr for the first hour and 0.7 in/hr in the second hour and the stream flow given in the following table (thanks to Casey here):
T(hr)-----1-----------2------------3------------4
Q(cfs)----0.75-------2.15-------1.44---------0.28
Find the 1hr unit hydrograph
Answer NCEES Q508 problem
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If we are looking for the one-hr unit hydrograph (i.e., the runoff pattern resulting from one inch per hour of rain for one hour)...then we are looking for the discharge rates at one hour, two hours, three hours, four hours, etc.
let's assign the ordinates of a unit (1 inch of direct runoff) hydrograph:
Y1 is the one hour unit hydrograph value at time = 1 hr
Y2 is the one hour unit hydrograph value at time = 2 hr
Y3 is the one hour unit hydrograph value at time = 3 hr
Y4 is the one hour unit hydrograph value at time = 4 hr
To start, we know the discharge of 0.75 cfs, which is occuring at time 1 hr, is the result of runoff from the first hour; it is the hydrograph ordinate value at time one hour of a 1.5 in/hr hydrograph;
then
1.5*Y1 = 0.75 cfs at one hr
the discharge of 2.15 cfs occuring at time 2 hrs is the addition of the hydrograph value at time one hour of a 0.7 in/hr hydrograph plus the hydrograph value at time two hours of the 1.5 in/hr hydrograph; so
0.7*Y1 + 1.5*Y2 = 2.15 cfs at 2 hr
and for the discharge of 1.44 at time = three hours...
0.7*Y2 + 1.5*Y3 = 1.44 cfs
and for the discharge of 0.28 cfs at time = four hours
0.7*Y3 + 1.5*Y4 = 0.28 cfs
The first of the equations solves quickly, and the other equations each then in turn become one equation in one unknown and also solve quickly by hand:
Y1 = 0.5 cfs
Y2 = 1.2 cfs
Y3 = 0.4 cfs
Y4 = 0
The 1.5 in/hr hydrograph is the unit hydrograph x 1.5, "beginning" at time = 0; likewise, the hydrograph of the 0.7 in/hr event is 0.7 x the one hour unit hydrograph, beginning at time = 1 hr.
