pro00010

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 . = . 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 = = Continuity Equation for water: d dt dv v n dA v n dA w w v w w w out in + + = ( ) ( ) The problem says that: the expansion of water with the addition of salt is negligible Thus, V V w Total 2245 ; w T 2245 d dt dv Q Q w w v w w in w w out w - + = ( ) ( ) w w in w out w const d dt dV Q Q = - + = ( ) ( ) 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 V T = Ah = 2h 2 0 05 0 2 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 + + = ( ) ( ) but s s V s w w dV s dV = s s in s in salt w in w v n dA Q s Q ( ) ( ) . ( . ) = 2245 = 0 2 0 004 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 = 0 004 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....
View Full Document

Page1 / 11

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online