Q1 Show work At time t0 a tank contains 25 oz of salt dissol

Q1.) Show work, At time t=0, a tank contains 25 oz of salt dissolved in 50 gallons of water. Then brine containing 2oz of salt per gallon of brine is allowed to enter the tank at a rate of 5 gal/min and the mixed solution is drained from the tank at the same rate.

(a) How much salt is in the tank at an arbitrary time? = oz.

(b) How much salt is in the tank at time 25 min?

Solution

Solution:

Let y represent the amount of salt in the tank at time t,

where t is given in minutes.
Then, y = 25 oz at t = 0

and

dy/dt = rate in rate out

Rate in = 2 oz of salt entering at 5gal/min

Rate out = rate of draing = 5gal/min

dy/dt = 2*5 y/25

dy/dt + y/25 = 10

On solving this differential equation for general solution we get

y = -e^-(C +t)/25 +250

we

To find C, we use y = 25 when t = 0

25 = -e^-(C+0)/25 +250

-C/25 = 5.3

C = -132.5

so C = 132.5

and y as a function of t is : y = -e^-(t-132.5)/25 +250

answer (a) y = -e^-(t-132.5)/25 +250 oz

(b) at t = 25 min

y = -e^-(25-132.5)/25 +250

y = 176.30 oz

answer