Surface Water Hydrolgy - Assignment
Added on 2021-06-14
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Surface Water Hydrolgy1Surface Water HydrologyAssignment 2[Student Number][Student Name]
Surface Water Hydrolgy2Q1PROBLEM 1 (8 MARKS)SOLUTIONXT = Mean of Q+ KTS.D Data from Spread SheetMean of Log Q = 2.98611Mean standard deviation Ӯ = 0.21237Coefficient of skewness Sc = -0.732332844MeandischargeQ=1072.82StandarddeviationofQ=450.536KT=Z+(Z2–1)K+13(Z3–6Z)∗(K2)–(Z2–1)∗K3+Z∗K4+13K5¿K=CS/6=−0.122055Z=W−2.515517+0.802853W+0.01032W21+1.432788W+0.189269W2+0.001308W3W={ln(10.52)}0.5WhereP=1ReturnperiodT
Surface Water Hydrolgy3Calculations for 2 year return period;P=(12)=0.5W={ln(10.52)}0.5=1.17741SubstitutingthevalueofW∈Z;Z=W−2.515517+0.802853W+0.01032W21+1.432788W+0.189269W2+0.001308W3=0.00000365653KT=Z+(Z2–1)K+13(Z3–6Z)∗(K2)–(Z2–1)∗K3+Z∗K4+13K5K=−0.12206K2=0.120236X2=MeanofQ+K2S.DX2=1072.82+0.120236(450.536)X2=1126.99Ml/dayCalculations for 5 year return periodTP=(15)=0.2W={ln(10.22)}0.5=1.794123SubstitutingthevalueofW∈Z;Z=W−2.515517+0.802853W+0.01032W21+1.432788W+0.189269W2+0.001308W3=0.841463KT=Z+(Z2–1)K+13(Z3–6Z)∗(K2)–(Z2–1)∗K3+Z∗K4+13K5K=−0.12206K5=0.8989X5=MeanofQ+K5S.D
Surface Water Hydrolgy4X5=1072.82+0.8989(450.536)X5=1477.79Ml/dayFor a 10 year return periodTP=(110)=0.1W={ln(10.12)}0.5=2.145966SubstitutingthevalueofW∈Z;Z=W−2.515517+0.802853W+0.01032W21+1.432788W+0.189269W2+0.001308W3=1.2817362KT=Z+(Z2–1)K+13(Z3–6Z)∗(K2)–(Z2–1)∗K3+Z∗K4+13K5K=−0.12206K10=1.0202X10=MeanofQ+K10S.DX10=1072.82+1.0202(450.536)X10=1532.48Ml/dayFor a 25 year return periodTherefore;P=(125)=0.04W={ln(10.042)}0.5=2.537SubstitutingthevalueofW∈Z;Z=W−2.515517+0.802853W+0.01032W21+1.432788W+0.189269W2+0.001308W3=1.7510853KT=Z+(Z2–1)K+13(Z3–6Z)∗(K2)–(Z2–1)∗K3+Z∗K4+13K5K=−0.12206
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