In a previous section we looked at mixing problems in which the volume of fluid

Question

In a previous section we looked at mixing problems in which the volume of fluid remained constant and saw that such problems give rise to separable equations. (See example.) If the rates of flow into and out of the system are different, then the volume is not constant and the resulting differential equation is linear but not separable.

A tank contains 400 L of water. A solution with a salt concentration of 0.4 kg/L is added at a rate of 5 L/min. The solution is kept mixed and is drained from the tank at a rate of 3 L/min. If

is the amount of salt (in kilograms) after t minutes, show that y satisfies the differential equation

dy/dt=2-(3y/400+2t)

In a previous section we looked at mixing problems in which the volume of fluid remained constant and saw that such problems give rise to separable equations. (See example.) If the rates of flow into and out of the system are different, then the volume is not constant and the resulting differential equation is linear but not separable. A tank contains 400 L of water. A solution with a salt concentration of 0.4 kg/L is added at a rate of 5 L/min. The solution is kept mixed and is drained from the tank at a rate of 3 L/min. If y(t) is the amount of salt (in kilograms) after t minutes, show that y satisfies the differential equation dy/dt=2-(3y/400+2t)

Explanation / Answer

Rate of additionof salt=0.4 kg/L*5L/min=2kg/min

At any time t amount of salt= y

Rate of increase in volume of water =vol of water added per min-vol of water removed per min= 5-3=2

Instantaneous vol of water in tank= 400+2t

salt conc in soln= y/(400+2t) kg/L

so amount of salt removed/min=3L/min* y/(400+2t) kg/L= 3y/(400+2t) kg/min

So rate of increase in salt conc= dy/dt=Rate of additionof salt - rate of removal= dy/dt= 2- 3y/(400+2t)

Integrating both sides

dy/dt + 3y/(400+2t)= 2

y' + y * p(t) = 2

so integrating factor M=e^ (integration(3/(400+2t) dt)) = e^ (3/2 * ln(400+2t)) = (400+2t)^(3/2)

y * M = integration(2 * M) + c

y * (400+2t)^(3/2) = 2/5 * (400+2t)^(5/2) +c

At t=0 y=0

putting t=0 and y=0 in equation

c= - 2/5 * 400^(5/2)

so

y * (400+2t)^(3/2) = 2/5 * (400+2t)^(5/2) - 2/5 * 400^(5/2)

So at t=20

y * 440^(3/2) = 2/5 * (440^(5/2) - 400^(5/2))

y= 93.286 kg

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