Question
****I already have the answers to the first part of this question, but do not know how to find 3, the time needed. Please answer this question for points. Thank you.
Consider the mixing process shown in the figure. A mixing chamber initially contains 2 liters of a clear liquid. Clear liquid flows into the chamber at a rate of 10 liters per minute. A dye solution having a concentration of 0.25 kilograms per liter is injected into the mixing chamber at a constant rate of r liters per minute. When the mixing process is started, the well-stirred mixture is pumped from the chamber at a rate of 10 + r liters per minute. Develop a mathematical model for the mixing process. Let Q represent the amount of dye in kilograms in the mixture. The objective is to obtain a dye concentration in the outflow mixture of 0.2 kilograms per liter. What injection rate r is required to achieve this equilibrium solution? Would this equilibrium value of r be different if the fluid in the chamber at time t = 0 contained some dye? no Assume the mixing chamber contains 2 liters of clear liquid at time t = 0. How many minutes will it take for the outflow concentration to rise to within 3% of the desired concentration of 0.2 kilograms per liter? t = min
Explanation / Answer
dQ/dt=0.25r-(5+0.5r)Q
Put r= 40
dQ/dt= 10-(5+20)Q
= 10-25Q
Integrate both sides
-25 t= ln(10-25Q) +constant
Since at t=0, Q=0
so constant = -2.302
-25 t= ln(10-25Q)- 2.302
rise in concentration= 3% of 0.2 Kg/L= 0.006
Total Q= Initial Q + Rise(3%)
= 0.2+0.006
=0.206
Solving for t :
-25 t = ln(10-25*0.206) - 2.302
25t =0.72302
t= 0.0289 minutes
