Surface Water Hydrology Assignment PDF
Added on 2021-05-31
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Surface Water Hydrology 1SURFACE WATER HYDROLOGYENCIV7090ASSIGNMENT 2[Student number][Student name]© May, 2018
Surface Water Hydrology 2PROBLEM 1 (8 MARKS)SOLUTIONX2=MeanofQ+K2S.DData¿SpreadSheetMeanstandarddeviationӮ=0.735462863MeanoflogQ=1.838833904CoefficientofskewnessCs=−1.567731336MeandischargeQ=152.9531343ML/dStandarddeviationofQ=150.3055677MLKT=Z+(Z2–1)K+13(Z3–6Z)∗(K2)–(Z2–1)∗K3+Z∗K4+13K5(Singh,1998).¿K=CS/6=−0.261289Z=W−2.515517+0.802853W+0.01032W21+1.432788W+0.189269W2+0.001308W3W={ln(10.52)}0.5WhereP=1ReturnperiodT
Surface Water Hydrology 3For 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=−0.261289K2=0.243X2=MeanofQ+K2S.DX2=152.9531343+0.243(150.3055677)X2=189.48Ml/dayFor 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=−0.261289K5=0.80687X5=MeanofQ+K5S.D
Surface Water Hydrology 4X5=152.9531343+0.80687(150.3055677)X5=274.23Ml/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=−0.261289K10=1.26272X10=MeanofQ+K10S.DX10=152.9531343+1.26272(150.3055677)X10=303.81Ml/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=−0.261289
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