Question

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Derive a 30-minute Snyder's synthetic unit hydrograph (state the graph coordinates) for a watershed with a drainage area of 14 km². The length of the main stream is 7.16 kilometers, and the main channel length from the watershed outlet to the point opposite the center of gravity is 3.22 kilometers. Use Ct = 2.0 and Cp = 0.625. (20 points)

Answer 1

rea a hag given that duration of snyde y's synthetic unit hydrograph tr = 30 minules th = 0.5 hr. drainage area of watershed (A) = 14 lem2 length of main stream () = 7.16 km length of main channel from the watershed outlet to the point opposite the center of gravity is (Le) = 3.22 km 4= 2.0 Cp = 0.625 0.75Qt- discharge on setting - basin lag k woo 0.5000 - tp = q (Lite)0.3 th: 2 ( 7-14 x 3.2 °-3 tp = 5.127 hrs. time > standard duration of effective rainfall ty = tf = 5.127 - 0.932 hy kt - 5.5 5.5 > here, a non- standard rainfall duration te is adopted so te = tp + (tr-to) /4 where tp = basin lag in hours for an effective duration of te t'p = 2 * 5.127 + = 5.02 hv.

- Peak discharge discharge Qo= 2.78 Cp A ? here A= 14km² - tp Qe - 2.78 x 0.625 X 14 5.02 Qp = 4,845 m3/s base time to = (72+ 3 tp ) hours to = 72 + 3x5.02 to = 87.06 hrs - width of unit hydrograph in hrs. at 50% peake discharge Wso - 5.87 qloog W5o = 18.464 hrs. q=0. 346 m²ls 1. km2 - width of unit hydrograph at 75% peak discharge Was - Motor = 18.464 1.75 Wys = 10.55 hrs. ¢ tx=0.5hr - tp = 5.02 his 4.845 = (diagram is I not to scale 0.25 ap discharge F OoSQ (ms) t Wis #10.55 hors 7 Wso = 18.46th y time th=87.06 hys. "