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# pro00010 - Additional Examples Related to Chapter I M.S...

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Additional Examples Related to Chapter I M.S. Ghidaoui, Spring 2002 CIVL252 Note: you do not learn by simply reading a problem and its solution! You must try to solve the problems by yourself, and discuss the difficulties encountered with your classmates, the teaching assistants and myself. Problem 1: Saline water flows into a cylindrical test vessel with a 2 m 2 cross- sectional area. Initially, the rate of inflow is 0.1 m 3 and its salinity is 0.002 kg of salt per kg of water; the flow has been steady at these values for a considerable period of time. An outlet valve at the bottom of the tank permits a discharge Q proportional to the depth of flow in the tank h as follows: h Q 05 . 0 = . If the inflow rate and salinity are both doubled at time zero, determine the depth of fluid in the tank and its salinity as a function of time. You may assume the fluid in the tank is completely mixed, the density of water is 1000kg/m 3 and that the expansion of water with the addition of salt is negligible. Solution: s m m dv dv s w s s v s w w v w = = ρ ρ Continuity Equation for water: d dt dv v n dA v n dA w w v w w w out in ρ ρ ρ + + = ( ) ( ) 0 The problem says that: “the expansion of water with the addition of salt is negligible” Thus, V V w Total 2245 ; ρ ρ w T 2245

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d dt dv Q Q w w v w w in w w out w ρ ρ ρ - + = ( ) ( ) 0 ρ w w in w out w const d dt dV Q Q = - + = ( ) ( ) 0 d dt V d dt V Q Q W T out w in w 2245 - ( ) ( ) Q Q Q out out water in salt = + ( ) ( ) but ( ) ( ) Q Q out salt out water << Q out = 0.05h 2245 (Q out ) water similarly: Q in = 0.2 2245 (Q in ) water ;t 0 V T = Ah = 2h 2 0 05 0 2 0 dh dt h + - = . . d dt e h e t t ( ) . . . 0 025 0 0025 01 = h t ce t ( ) . = + - 4 0 025 but 0.05 h(t = 0) = 0.1 h(t = 0)=2m Thus, h(t) = 4 - 2e -0.025t Continuity of salt: d dt dV v n dA v n dA s s V s s s in s s out ρ ρ ρ + + = ( ) ( ) 0 but ρ ρ s s V s w w dV s dV = ρ ρ ρ ρ s s in s in salt w in w v n dA Q s Q ( ) ( ) . ( . ) = 2245 = 0 2 0 004 ρ ρ ρ ρ s s out s out salt w out w v n dA Q s Q hs ( ) ( ) . = 2245 = 0 05 Therefore, dV dt sh w + = 0 05 0 0008 . ( ) . let y = sh dy dt y + = 0 025 0 0004 . . y = 0.016 + ce -0.025t but y(t = 0) = h(t = 0) s(t = 0) = 2*0.002 Hence, y(t) = sh = 0.016 - 0.012 e -0.025t or S(t) = 0.016 - 0.012 e -0.025t 4 2 0 025 - - e t . note that: t s t = 0 0 004 lim ( ) . Problem 2: A reservoir operator has made a linear projection of inflow which has been increasing during the spring runoff period as follows: t I 1250 000 , 125 + = , where = I inflow in m 3 /hr and = t time from the present in hours. The regulated outflow is set for 90,000 m 3 /hr. If the present volume of water stored in the reservoir is 23(10 6 ) m 3 , determine the storage volume after 12 days. Solution: From continuity we have:
dV dt I O t = - = + - 125 000 1250 90 000 , , V(t)-V(0) = (1250/2)t 2 + 35,000t t = 12 days= 12*24 = 288 hours V(288)=23(10 6 ) +(1250/2)*288 2 +35,000(288) =84.92(10 6 ) m 3 Problem 3: To cylindrical reservoirs have areas 1 A and 2 A ; they are connected at the bottom with a sand filled pipe (hydraulic conductivity K ) of cross-sectional area 3 A and length L . The initial heights of the water columns in the reservoirs are 1 h and 2 h with 2 1 h h . Determine the time for the head difference between the reservoirs to reach half of its initial value.

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pro00010 - Additional Examples Related to Chapter I M.S...

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