1) A tank contains 2960 L of pure water. Solution that contains 0.09 kg of sugar

Question

1) A tank contains 2960 L of pure water.  Solution that contains 0.09 kg of sugar per liter enters the tank at the rate 6 L/min and is thoroughly mixed into it.  The new solution drains out of the tank at the same rate

How much sugar is in the tank at the beginning?

y(0)=     kg

Find the amount of sugar after t minutes

y(t)=      kg

As t becomes large what value is y(t) approaching? Calculate the limit as t approaches infinity.

2) A tank contains 80 kg of salt and 1000 L of water. Pure water enters a tank at the rate 6 L/min.  The solution is mixed and drains from the tank at the rate 3 L/min

What is the amount of the salt in the tank initially? in kg

Find the amount of salt in the tank after 4 hours. in kg

Find the concentration of salt in the solution in the tank as time approaches infinity. (Assume your tank is large enough to hold all the solution.  in kg/L

Explanation / Answer

1) rate of accumulation = flow in - flow out + generation

let us assume at time t , concentration of solution is y(t) Kg/L

so V(dy/dt) = (6*0.09) - (6*y) { volume is constant because flow in = flow out}

or 2960*(dy/dt) =6(0.09-y)

or dy/(0.09-y) =(3/1480)dt

integrating on both side

-ln(0.09-y) = (3/1480)*t + c

we know that at t = 0 y = 0 because only pure water

so c = -ln(0.09)

or ln(0.09/0.09-y) = (3/1480)*t

or 0.09/(0.09-y) = exp(3t/1480)

or 1-(y/0.09) = exp(-3t/1480)

or y(t) = 0.09*[1-exp(3t/1480)]

at t-->infinite y-->0.09 Kg/L

2)

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