Consider the mixing process shown in the figure. A mixing chamber initially cont
ID: 3199405 • Letter: C
Question
Consider the mixing process shown in the figure. A mixing chamber initially contains 4 liters of a clear liquid. Clear liquid flows into the chamber at a rate of 10 iters per minute. A dye solution having a concentration of 0 6 kalograms per iter is injected pumped from the chamber at a rate of 10 +r liters per minute. (a) Develop a mathematical model for the mixing process. Let Q represent the amount of dye in kilograms in the mixture. dQ dt kg/min (b) The objective is to obtain a dye concentration in the outflow mixture of O.1 kilograms per liter. What injection rater is required to achieve this equilibrium solution? L/min Would this equilibrium value of r be different if the fluid in the chamber at time t0 contained some dye? Choose Choose yes no (c) Assume the mixing chamber contains 4 liters of clear liquid at time t0. How many minutes will it take concentration to rise to within 5% of the desired concentration of 0.1 kilograms per liter? min maybeExplanation / Answer
a) Let us assume that the chamber is filled for 1 minute. Then the volume of the mixture in the chamber is 10+r litres and the chamber contains 14 liters of the clear liquid and r litres of dye solution.
Given that the concentration of dye is 0.6 kg/ litres, then for r litres the quantity of dye in the chamber is Q=0.6r kg.
Therefore,
dQ/dt = 0.6r/1 = 0.6r kg/minute
b) From the above assumptions, the amount of dye in the mixture is 0.6r kg
14+r litres of solution has 0.6r kg of dye, so the concentraion = 0.6r/14+r
So if we want that concentration of the dye in the mixture to be 0.1 kg / litre, therefore,
0.6r/14+r = 0.1 => 0.6r = 1.4+0.1r =>0.5r=1.4 => r=2.8 litres/ minute.
if the fluid in the chamber contained some dye, then the total amount of dye in the total mixture given by 0.6r , in the above equation, will change by that amount, thus the mount of "r" will also change. So the answer is yes.
c) Given : the chamber has 4 litres of clear solution before pumping in and the clear solution is pumped at 10 litres/min and dye solution at 2.8 litres/min.
concentration rise to 5 percent of the desired concentration of 0.1 kg/litre means that the concentration of dye in the chamber is 0.05*0.1 = 0.005 kg/litre.
let t be the time in which the concnetration of due rises to 0.005 kg/litre.
thus (2.8*t)/4+(10*t) = 0.005 => 2.8t=0.02+0.05t => 2.75t=0.02 => t = 0.0073 minute = 0.436 seconds