differential equations

Question

Set up a differential equation to model the salt mass for the following information. ‘A 100 liter tank starts with 20 L of pure water. A saline mix of 10 grams of salt per liter of water is pumped in at a rate for 5 L per minute and then is drained at a rate of 3 liter per minute.’ (Hint: Is volume constant? For what period of time is your model valid?)

Answer 1

At t=0, the pump and drain are activated and there is an entrance of (5 Liters per minute)(10 grams of salt per Liter) or 50 grams of salt every minute.

For S grams of salt in the tank at any time t, the rate of change of S with respect to time (dS/dt) is the difference between the rate of salt entering and the rate of salt leaving.

Find the rate at which salt leaves the tank by writing the amount of salt water in the tank at any time t,

equal here to 20 +5t −3t or 20+2t.

The concentration of salt in the tank at any time is given by the ratio of salt in the tank to the volume of salt water in the tank at any time or S/(20+2t).

It then follows that salt leaves the tank at the rate of 3(S/(20+2t)) grams per minute and dS/dt equals

50 − 3S/(20+2t) or dS/dt+3S/(20+2t) = 50.

This is a Linear Differential Equation for S(t). Write the integrating factor as I(t) equal to e^{∫3dt/(20+2t)} or

e^1.5ln(20+2t) or (20+2t)^1.5 (where t > -10).

Multiplying (20+2t)^1.5 through [dS/dt + 3S/(20+2t) =50] gives (20+2t)^1.5(dS/dt) + 3(20+2t)^0.5S = 50(20+2t)^1.5.

(20+2t)^1.5(dS/dt) + 3(20+2t)^0.5S integrates to (20+2t)^1.5S by reversing the Product Rule For Differentiation

and 50(20+2t)^1.5 integrates to 10(20+2t)^2.5+C.

Then obtain (20+2t)^1.5S = 10(20+2t)^2.5+C or S(t) = [10(20+2t)^2.5+C] ÷ (20+2t)^1.5.

Since S(0) = 0, obtain 0 equals [10(20)^2.5 + C]/20^1.5 or 0 = 10(20) + C/20^1.5 which gives C=-10(20)^2.5

or -17888.54382.

Time until overflow is given by 100 = 20+2t or t = 40 minutes.

Salt in tank at time of overflow is given by S(40) equal to [10(20+2(40))^2.5 − 10(20)^2.5] ÷ (20+2(40))^1.5

which reduces to (1000 −8√5) grams equivalent to 982.1 grams.