A tank contains 200 L of fluid in which 30 grams of salt are

A tank contains 200 L of fluid in which 30 grams of salt are dissolved. Brine containing 1 g of salt per liter is then pumped into the tank at a rate of 4L/min; the well mixed solution is pumped out at the same rate. Find the amount in grams (A(t)) of salt in the tank at time t.

Solution

Here\' my take on this:

Since the rate of pumping in and out is same , 200L of fluid in tank is always goning to be there.
At time t, the amount of salt in tank is A(t) (in grams)
So concentrations of salt at time t is A/200 g/L

Amount of brine pumped in: 4L (with salt concentration = 1 g/L)
Amount of salt pumped in: 4L * 1g/L = 4g

Amount of brine pumped out at time t: 4L (with salt concentration = A/200 g/L)
Amount of salt pumped out at time t: 4L * A/200 g/L = A/50 g

dA/dt = 4 - A/50
50 dA/dt = 200 - A
50/(200 - A) dA = dt
-50 ln|A-200| = t + C
ln|A-200| = -t/50 + C .... where C = -C/50
A - 200 = C e^(-t/50) ..... where C = e^C
A = 200 + C e^(-t/50)

Initially, brine contains 30 g of salt
A(0) = 30
200 + C e^0 = 30
C = -170

A(t) = 200 - 170 e^(-t/50)