Ordinary Differential Equation - tank with inflow and outflo

[samg1]

Homework Statement

A tank contains 60 kg of salt and 2000 L of water. A solution of a concentration 0.015 kg of salt per liter enters a tank at the rate 6 L/min. The solution is mixed and drains from the tank at the same rate.

Find the amount of salt in kg at t = 3 hours
Find the concentration of salt in the solution in the tank as time approaches infinity.

Homework Equations

The Attempt at a Solution

My attempt:C(t) = C = The concentration of salt in (kg/L) at time t in hours.
dC/dt = (flow_in - flow_out + 60kg/2000L)
flow_in = (0.015kg/L *6L/min*60min/hour) / 2000L // the concentration that flows in
flow_out = C*6L/min*60min/hour)/ 2000L // the concentration that flows out
2000L dC/dt = (0.015kg/L *6L/min*60min/hour) - C*6L/min*60min/hour) + 60kg
2000 dC/dt = 5.4 - 360C + 60
2000 dC/dt = (65.4 - 360C)
2000/ (65.4-360C) dC = dt*Integrate both sides*Left side :

let u = 65-360C
du = -360 dC

du / -360 = dC

2000/-360 ∫1/u du

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(-2000/360) ln(65.4-360C) = t + c // where c is a constant.
ln(65.4 - 360C) = 360t/-2000 + c
e^ln(65.4 - 360C) = e^ 360t/-2000 + c
65.4 - 360C = ce^(360t/-2000)
-360C = ce^(360t/-2000) - 65.4
360C = 65.4 - ce^(360t/-2000)
C = (65.4/360) - ce^(-360t/2000)C(0) = 60/2000 Plug in this to find the constant c
60/2000 = (65.4/360) - c
c = 65.4/360 - 60/2000
C = (65.4/360) - (65.4/360 - 60/2000)e^(-360t/2000)

 

[samg1]

Got it. We can delete this post!