Question

Salt water flows at a constant rate of 9 L/min into a tank that initially held 100 L of salt solution in which was dissolved 0.15 kg of salt. The solution inside the tank is kept well stirred and flows out of the tank at the same rate. If the concentration of salt in the solution entering the tank is 0.03 kg/L. a) Dtermine the mass of salt in the tank after t min. b) When will the concentration of salt in the tank reach 0.01 kg/L? a) Determine the mass of salt in the tank after t min. - 9t mass = 3-2.85 100 kg b) When will the concentration of salt in the tank reach 0.01 kg/L? The concentration of sal the tank will reach 0.01 kg/L after minutes (Round to two decimal places as needed.)

Answer 1

solution & ret silt) denote the main of salt in the tank at time t, we can determine the concentration of Salt the tank by dividing alt) by the volume of fluid time in toll) in the tank at first we meint detelmine the rate at which Salt enters at tank the tank. we are given that the brine flows into the a rate of g hlmin. Since the concentration is 003ngh (we conclude that the input rate of salt into the tank is

(921 min1 (003 kg12) 0.27 kg/min an ume that we must determine the output Vate of Satt from the tank The brine solution in the tank is well stirred isolette the concentration Salt in the donkis unifarm. So the concentration of salt in any part of the dank at time it is hiert "alth diside by the volume of fluid in the tank", Be cause the & tank initially contains 100h and the rate of flow into the tank is the same as the of flow out, the volume is a constant bot • Hence the Output rate a ( 94min) / Moch ngla) » 0.09x64 Eglain. The tank initially contains 0.5 kgo of Salt so we get nol = 0.15. The initial Vale problem de = input rate - outputrate dt goat- o.oault), 10) = 60015 a mathematical model for the mining problem. separating the variables and integrate 0.07 -0.09.26).

du - Oodt- o.oqov (1) 910) = at thes & stir tugtug 9 nlt) dM at 0.27 - ها morskuge tried an 0.27²x100 - 9. utt) B Army dt 100 sit o esmachines dal art 100 0.291100-9.(4) 100 du 34-96) Nov tu wote du 963 - x) 2 100 (algt for odt wie din met sy Whois 3120 - Cox 1. du too a 3-1 Intege at too - Sct. que at +... in 13-) Waarder 100 - 1 m (3 = t a 100 -95 7 3. =cie Mct) = 3-4 e 100

Substituting into critical condition givers - 9.10.) but 0.15= 3-C1- CI= 3-0015 2.85 nlt) = - 100 in 1.049 minutes in annual x col = 3- Clue 100 MIO) 3-ci MO) 2015 C1 = and the solution is to the given initial problemi 3-cue to-100 11. 100 x/t)= 3-2.85 € concentration of salt is when the determine equation solwe falt an we have to 0.0 reglhi 001 at 3- 2.85e 100 nat 3-00l = 2.85 e Too -at 2085 e 2.99 -94 22.99 2 - + 1049 = e ta - 100 1049