- #1

samg1

- 6

- 1

## 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

-----------

(-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)