Surface Water Hydrology - Assignment PDF

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SURFACE WATER HYDROLOGYENCIV7090Assignment 2Due date: Week 10, Friday 18th, May 2018 by 6:00 pmPage1of13
PROBLEM 1(8 MARKS)Using the same data set you developed for problem 6 of assignment 1 (ENCIV7090)Fit a Log-Pearson type (iii) probability distribution and calculate the dischargeswhich equates to the 2 year, 5 year, 10 year, 25 year, 50 year and 100 year returnperiod.As part of your work, you must show; all of your calculations and workings (4marks);a graph which compares the actual measured data and your fitted probabilitydistribution function (2marks); andA discussion of how the measured data compares to the fitted PDF includingcomments on its goodness of fit and any other aspects you may think relevant (2marks).SOLUTIONX2= Mean of Q+ K2S.DMean of log Q= 3.430225716Coefficient of Skewness Sc= -0.322627883KT= Z + (Z2– 1) K +13(Z3– 6 Z)*(K2)(Z2– 1)* K3+ Z*K4+13K5 (Benson, 1960)And K = CS/ 6= -0.053771313Z = W -2.515517+0.802853W+0.01032W21+1.432788W+0.189269W2+0.001308W3W= {ln (10.52)}0.5Where P =1ReturnperiodTMean discharge Q =3525.542055Standard deviation of Q= 2508.362949For 2 year return period;P = (12) = 0.5W= {ln (10.52)}0.5= 1.17741Page2of13
Substituting the value of W in Z;Z = W -2.515517+0.802853W+0.01032W21+1.432788W+0.189269W2+0.001308W3= 0.00000365653KT= Z + (Z2– 1) K +13(Z3– 6 Z)*(K2)(Z2– 1)* K3+ Z*K4+13K5K = -0.053771313K2= 0.05362X2= Mean of Q+ K2S.DX2 =3525.542055+0.05362 (2508.362949)X2= 3660 Ml/dayFor 5 year return periodTP = (15) = 0.2W= {ln (10.22)}0.5= 1.794123Substituting the value of W in Z;Z = W -2.515517+0.802853W+0.01032W21+1.432788W+0.189269W2+0.001308W3= 0.841463KT= Z + (Z2– 1) K +13(Z3– 6 Z)*(K2) –(Z2– 1)* K3+ Z*K4+13K5K = -0.053771313KT= 0.8528X5= Mean of Q+ K5S.DX5 =3525.542055+0.8528(2508.362949)X5= 5664.72 Ml/dayFor a 10 year return periodPage3of13
TP = (110) = 0.02W= {ln (10.12)}0.5= 2.145966Substituting the value of W in Z;Z = W -2.515517+0.802853W+0.01032W21+1.432788W+0.189269W2+0.001308W3= 1.2817362KT= Z + (Z2– 1) K +13(Z3– 6 Z)*(K2) –(Z2– 1)* K3+ Z*K4+13K5K =-0.053771313K10=1.25264X10= Mean of Q+ K10S.DX10 =3525.542055+1.25264(2508.362949)X10= 6,667.62 Ml/dayFor a 25 year return periodTherefore; P = (125) = 0.04W= {ln (10.042)}0.5= 2.537Substituting the value of W in Z;Z = W -2.515517+0.802853W+0.01032W21+1.432788W+0.189269W2+0.001308W3= 1.7510853KT= Z + (Z2– 1) K +13(Z3– 6 Z)*(K2) –(Z2– 1)* K3+ Z*K4+13K5K = -0.053771313K25= 1.6398X25= Mean of Q+ K25S.DX25 =3525.542055+1.6398(2508.362949)X25= 7638.8 Ml/dayPage4of13
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