Kinematic Waves and Overland Flow

Jorge A. Ramirez

Assume that you have an effective rainfall event of infinite duration and constant intensity, i.

a) Use the method of characteristics to derive an exact solution for the hydrograph at the outlet under kinematic wave assumptions.
b) Obtain an approximate solution to the above problem by first obtaining an instantaneous unit hydrograph (using kinematic wave assumptions), and then using the unit hydrograph concept in order to derive the overland flow hydrograph for an effective rainfall of constant intensity and infinite duration.

restart;with(student);

Define time of concentration, tc ; in the equation below, L is the length of the overland flow, i is the rainfall intensity, and a and m are the parameters of the unit discharge vs . flow depth relatonhip (e.g., from Manning equation q=ay^m )

tc:=(L,a,m,i)->(L*i^(1-m)/a)^(1/m);

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For a rainfall of constant intensity and infinite duration, the exact solution using Kinematic wave assumptions is (see graph below):

q:=t->piecewise(t<=tc(L,a,m,i),a*(i*t)^m,L*i);

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Using Kinematic wave theory, develop an analytical solution for the hydrograph at the downstream end of the overland flow plane for a rainfall of volume d , constatnt intensity i and duration tr , for tr = 0; that is, obtain an analytical solution using Kinematic waves for a rainfall impulse of volume d . An Instantaneous Unit Hydrograph is the response of a (linear) basin to an instantaneous impulse of rainfall of unit volume.

For any rainfall of volume d = i tr, as tr approaches zero it will become smaller than the time of concentration, tc . Thus, the analytical solution corresponds to the case tr < tc .

Define duration of the peak flow, tp :

tp:=(a,m,d)->L/(a*m*d^(m-1));

Define hydrograph or an instantaneous impulse of volue d = i tr.

u:=t->piecewise(t<tp(a,m,d),a*d^m,a*(L/a/m/t)^(m/(m-1)));

As required by this problem, assume that this overland flow plane is a linear system whose IUH is u(t). Then, we can use the following convolution equation to obtain the hydrograph for the given rainfall event.

> Q:=t->i/d*Int(u(tau),tau=0...t);

In order to compare the exact and approximate results, assume the following arbitrary values for the parameters.

a:=5;m:=2;L:=10;d:=1;i:=1;

Plot analytical solution

plot(q(t),t=0..10);

Plot the assumed IUH for this system; that is, plot the hydrograph for an impulse of volume d = i tr = 1 for tr -> 0 .

plot(u(t),t=0...4);

Plot approximate solution using the convolution equation:

i:=1;plot(Q(t),t=0..10);

Compare exact and approximate solutions

plot([q(t),Q(t)],t=0..10, color=[red,blue]);

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