CIV E 634 - SURFACE WATER HYDROLOGY
SPRING 2007
HOMEWORK 2:   HORTON'S SOLUTION FOR OVERLAND FLOW

  1. Derive Eq. 4-35 in the book. [Hint: Use Eqs. 4-23 and 4-30].

  2. Derive Eq. 4-36 in the book (Horton's solution for overland flow). [Hint: Use Eq. 4-26 in Eq. 4-35, and integrate Eq. 4-35 by making the following change of variable: (q/qe)1/m = z , where m= 2].

  3. Using the logarithmic equivalent to the hyperbolic tangent, express Horton's solution (Eq. 4-36) in such a way that q/qe is the independent variable and t/te is the dependent variable.

  4. Using the equation derived in 3, calculate and plot Horton's solution at intervals of 0.01 q/qe from 0.00 to 0.99, then 0.999 q/qe and 0.9999 q/qe.

  5. Given the following overland flow conditions: i = 60 mm/hr, n = 0.15, So = 0.01, and L = 100 m; calculate the equilibrium outflow qe (L/s/m) and the time-to-equilibrium te (sec).

  6. Using the data of item 5 and the equation derived in item 3, calculate how long will it take for the outflow discharge from the overland flow plane to attain 50, 70, 90, 99, 99.9 and 99.99 percent of its equilibrium value. Express discharge in L/s/m and time in seconds.