Question

2. (4 points) 100 lb of salt is dissolved in a tank containing 300 gal of water. A salt solution with concentration 3 lb/gal is poured into the tank at 2 gal/min. The mixture is well-stirred and then flows out at the same rate the brine is entering the tank. Find the amount of salt in the tank at time t.

Answer 1

2. Tatal water on the tank = 300 gal. initial amount of salt = 100lb. Now the rat at which salt is pumiped in = 8,= 3lb igal x egal/min : : 36 lb/ min. Again the well mixed solution is pumped out at the rate of a gal/min. . ist sct) be the amount of salt present in the tank at time di How rate at which salt is pumped out

isi = $(t) * 2 dormin S() 150 900 - 567) = . 400 -sek) 150 150 ds(+) - 900-s(t) : dt 150 ds(+) d. -= 900 -56) Now initia = 100 lb when t=0 SI) 150 d sie) it at 900 - 500 Pa 100 ->) - 150(en [900 -56)]] [#]." > -150 [ Inf900-(t)} - In 800]= 4 3) en 900 - S(A) =- - 50 - #/150 800 900 - S() 800 e -t/150 900 - 56)= 800 . /150 S1) = -800 € + 900 -t/150 i sct) = 900- 800€*