A 200 gallon tank is filled with 100 gal of brine containing 60 lbs of dissolved

Question

A 200 gallon tank is filled with 100 gal of brine containing 60 lbs of dissolved salt. Water run into the tank from the top at a rate of 2.0 gal/min and the brine solution is constantly being stirred. What is the concentration of the brine after 30 minutes if the mixture flows out of the tank at the bottom with a rate of:

            a: 1.5 gal/min            b: 2.0 gal/min             c: 2.5 gal/min

Let S = the amount of salt in the tank at time t, and let V = volume of solution in the tank at time t. The concentration at any time t, is Co = S/V.

At t = 0 we have V0 = 100 gal, S = 60 lbs, and Co = 60/100 = 0.60 lbs/gal. The change in the amount of salt during an interval dt is dS/dt and this is equal to the amount of salt entering the tank during dt minus the amount of salt leaving the tank during dt. Since only water enters the tank, the amount of salt entering is zero. The amount of salt leaving the tank during the interval dt is the salt concentration S/V times the flow rate rL. The volume of brine at time t is V0 + rEt - rLt . In our problem rE is 2.0.  

Explanation / Answer

a> mixture flwos out at a rate of 1.5 gal/min

dS/dt = 0 - S/200*1.5 = - .0075*S

dS/.0075S = -dt

ln(S) = - .0075t + C

S = e^(-.0075t+ C)

S = De^(-.0075t)

useing the initial values t = 0 , S = 60

=> 60=De^(0) , => D = 60

hence S(t) = 60e^(-.0075t)

finally the concentration = S(t)/V(t)

when t = 30 minutes

S = 60e^(-.0075*30) = 47.9109 lbs

and V(t) = V0 + rEt - rLt

we are given rL = 1.5 gal/min

V(t=30)= 100 + 2*30 + 1.5*30 = 205 gal

=> C = S/V = 47.9109/205 =0.2337 lbs/gal

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.