CE322 Basic Hydrology

Jorge A. Ramírez

Kinematic Wave Routing - Example

The inflow hydrograph to a rectangular channel is tabulated below. The channel has a width of 100 ft, a bottom slope of 0.001, and Manning n of 0.035. The length of the channel is L = 25000 ft. Using the Kinematic Wave, obtain the outflow hydrograph.

Kinematic waves govern the flow when inertial and pressure forces are not important. Thus, in a kinematic wave the gravity and frictional terms are balanced, so the flow does not accelerate appreciably. For these kinds of waves, the energy grade line is parallel to the channel bottom.

The continuity and momentum equations governing kinematic waves are,

The momentum equation above is equivalent to a rating curve of the following form:

Manning equation is one such rating curve,

It can be shown that the celerity of kinematic waves, c_k, is,

Assuming a wide rectangular channel, like the one of this problem, Manning equation reduces to:

and thus, the kinematic wave celerity is,

If an observer moves with the kinematic wave flow at a speed equal to the kinematic wave celerity, the observer would see the flow rate increase at a rate equal to the lateral inflow rate, q.

Thus, for conditions of no lateral inflow like those of the problem at hand, that is, for q = 0, kinematic waves do not attenuate; they simply translate downstream without dissipation. Routing of kinematic waves for those cases reduces to determining the time of travel of the wave through a reach,

The table below presents the routing procedure for this simple kinematic wave problem. The problem is reduced to defining the time of arrival at the downstream end of the reach of individual kinematic waves associated with individual discharges. To do so,

A.- For each value of discharge, determine the corresponding flow depth by solving Manning equation:

For Q = 2000 cfs

y = ((2000) 0.035/1.49/100/sqrt(0.001))^(3/5) = 5.048 ft

B.- For each kinematic wave (i.e., discharge) compute corresponding kinematic celerity using

c_k = (5/3) (1.49sqrt(0.001)(5.048)^(2/3)/0.035 = 6.603 ft/s

C.- For each wave, compute time of travel using,

Dt = (25000 ft)/(6.603 ft/s) = 63.104 min

D.- For each wave compute time of arrival at downstream end as,

For Q = 2000 cfs starting at t_o = 0, the time of arrival t = 63.104 min

This procedure should be repeated for all values of the inflow hydrograph. The results are tabulated and graphed below. Observe the steepening of the rising limb and the flattening of the receding limb of the hydrograph.