Question

Reservoir routing problem 1 Assuming a linear reservoir with K=3 hours. Route the hydrograph through the reservoir using: Linear Weighting coefficients Storage Indicating method. Add as many time steps as needed. End computations when the outflow less than or equal to 5% of the peak inflow discharge. Plot the inflow and outflow on the same chart. Assume the reservoir is full at T=0, and the outfow is 0. Time, Inflow Hours Q, ft/s 0 0.0 0.5 1.0 50 100 225 1.5 75 2.0 2.5 3.0 50 20 3.5 4.0 How many hours was the hydrograph delayed? How much attenuation (in percent) did the reservoir cause?

Answer 1

Given K: 3hr At=0.5 hr for linear Regervoir x=0 c - = 0.0769 -kX+0.54t K-kx tosat 0.5% (0.5) 3+10.5)(0-5) C = = 0.0769 kx +0:5st K-kat0sot (0.5) (0.5) (3) +(0.5)(0.5) Cote, +2=1 C2=0.8462 Time (hrs) | Inflow (ft3/sec) | out flow (ft’sec) 3.845 14.788 37.506 54.807 55.99 59.762 46969 | 39.791

Qn : Co In & C Int + Ca Qn-l 20:5 - (0.0769) (50) + (0.0769) (0) + (0-8462) (0) Pois = 3.845 ft/sec Q1.0 2 (0:0769) (100) + (0.0769) (50) + (0.8462) (3.845) Q10 = 14.788 ft see Similarly the other Outflow value can be calculated.

Qifter) Attenuation 55.99 |---- 1.5 2.5hr 4.0 timelhrs) Lagtime lag time he hydrograph delayed + Pishr - Ishy a thy Attenuation (in percent) = las- 55.99.75.11% 225