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<br />. <br /> <br />. <br /> <br />," '":) <br />'") <br />N <br />00 <br />c.o <br />N <br /> <br />4-3 <br />Water is being released from 8i-l + Ii to the critical level <br />of 20000. 'l'he amount of 'the L"elease is given by <br /> <br />Si-l + Ii - 20000 <br /> <br />= 20600 + 900 - 20000 =1500 <br /> <br />Now the proportional part oE the day that water will be <br />released at this rate will be given by <br /> <br />Si-1 + Ii -20000 <br />2500 <br /> <br />= <br /> <br />1500 <br />2500 <br /> <br />= 0.60 (day) <br /> <br />where 2500 is the release rate in af/da fOL' contents above <br />20000 acre feet. <br /> <br />The proportional part of the day remaining is <br /> <br />1 _(Si-1 +Ii-20000) <br />2500 <br /> <br />= <br /> <br />1.0-0.60 = 0.40 (day) <br /> <br />The amount of wateL' L"eleased at the new L"ate (2000 af/da) <br />will be <br /> <br />[1 _ (Si-l + Ii - 20000) J(2000) <br />2500 <br /> <br />= 0.40 x 2000 = 800.0 <br /> <br />Now for the entire day the total L"elease will be given by <br /> <br />(Si-1 + Ii - 20000) <br />(Si-1 + Ii - 20000 ) + [ 1- J (2000) <br />2500 <br /> <br />= 1500 + 800 = 2300 <br /> <br />Theq~foL'e, the end of the day contents taken in the form of <br />. eqn 3 will be <br /> <br />(Si-1 + Ii-20000) <br />Si=Si-l +Ii-(Si_l+Ii-20000)-[1- J(2000) <br />2500 <br /> <br />= 20600 + 900 -1500-800 = 19200<---- <br /> <br />which, by assuL'ed pentL"ation of the cL'itical content level <br />(20000 af) can be simplified to <br /> <br />(Si-l + Ii - 20000) <br />Si = 20000 -[1- J (2000) <br />2500 <br /> <br />= 20000 - 800 = 19200 . <br />