Question

4. A tank initially contains 100 liters (L) of water with a salt concentration of 20 g/L. A mixture with a salt concentration of 5 g/L flows into the tank at a rate of 5 Umin while the well-mixed fuid in the tank flows out at the rate of 3 L/min. Determine the concentration of salt (g/L) in the tank when the volume in the tank reaches 400 liters.

Answer 1

If W denotes the volume of tank at time t Then dw/dt = 5 - 3 = 2 = > w = 2t + c at t = 0, w = 100 L = > c = 100 Thus w = (2t + 100)L Now, if m denotes the amount of self at time t, them at t = 0,m = 20 times 100 = 2000 gm. Now consider a small time interval st, then in the interval 5.8t L of Solution will come of concentration 5g/L Thus 5.8t times 5 = 258t gm of salt will come, Next in this time 3.8t L of solution will be diamed out of concentration m/2t + 100 g/L Thus. 3.8tm/2t + 100 g of salt will be damed out Thus, Net charge will be delta m = 25 delta t - 3delta t.m/2t + 100 \r\n = > delta m/delta t = 25 - 3m/2t + 100 Thus are delta t rightarrow 0, delta m/delta t rightarrow dm/dt Hence dm/dt + 3m/2t + 100 = 25 Thus tau.F = e^integral 3/2t + 100 dt = e^3.log(2t + 100)/2 = (2t