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Problem 10: A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per gallon is added to the tank at 6 gal/min, and the resulting solution leaves at the same rate. Find the quantity Q(t) of salt in the tank at time t > 0.

User Clemens
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1 Answer

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Answer:

The quantity of salt at time t is
m_(salt) = (60)\cdot (30 - 29.833\cdot e^{-(t)/(10) }), where t is measured in minutes.

Explanation:

The law of mass conservation for control volume indicates that:


\dot m_(in) - \dot m_(out) = \left((dm)/(dt) \right)_(CV)

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:


(0.5\,(pd)/(gal) )\cdot \left(6\,(gal)/(min) \right) - c\cdot \left(6\,(gal)/(min) \right) = V\cdot (dc)/(dt)

Where:


c - The salt concentration in the tank, as well at the exit of the tank, measured in
(pd)/(gal).


(dc)/(dt) - Concentration rate of change in the tank, measured in
(pd)/(min).


V - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:


V \cdot (dc)/(dt) + 6\cdot c = 3


60\cdot (dc)/(dt) + 6\cdot c = 3


(dc)/(dt) + (1)/(10)\cdot c = 3

This equation is solved as follows:


e^{(t)/(10) }\cdot \left((dc)/(dt) +(1)/(10) \cdot c \right) = 3 \cdot e^{(t)/(10) }


(d)/(dt)\left(e^{(t)/(10)}\cdot c\right) = 3\cdot e^{(t)/(10) }


e^{(t)/(10) }\cdot c = 3 \cdot \int {e^{(t)/(10) }} \, dt


e^{(t)/(10) }\cdot c = 30\cdot e^{(t)/(10) } + C


c = 30 + C\cdot e^{-(t)/(10) }

The initial concentration in the tank is:


c_(o) = (10\,pd)/(60\,gal)


c_(o) = 0.167\,(pd)/(gal)

Now, the integration constant is:


0.167 = 30 + C


C = -29.833

The solution of the differential equation is:


c(t) = 30 - 29.833\cdot e^{-(t)/(10) }

Now, the quantity of salt at time t is:


m_(salt) = V_(tank)\cdot c(t)


m_(salt) = (60)\cdot (30 - 29.833\cdot e^{-(t)/(10) })

Where t is measured in minutes.

User Mikael Vandmo
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