Please explain thanks Consider the mixing process shown in the figure. A mixing
ID: 3027988 • Letter: P
Question
Please explain thanks Consider the mixing process shown in the figure. A mixing chamber initially contains 5 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 0.75 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. dQ/dt = The objective is to obtain a dye concentration in the outflow mixture of 0.5 kilograms per liter. What injection rate r is required to achieve this equilibrium solution? r = L/min Would this equilibrium value of r be different if the fluid in the chamber at time t = 0 contained some dye? Choose Assume the mixing chamber contains 5 liters of clear liquid at time t = 0 How many minutes will it take for the outflow concentration to rise to within 4% of the desired concentration of 0.5 kilograms per liter? t= minExplanation / Answer
Solution:(a) :dQ/dt = r (3/4) - (10+r) Q / 2 ,
rate of dye change = rate of dye in - rate of dye out
Hence,mathematical model for mixing process is dQ/dt = r (3/4) - (10+r) Q / 2
b: first we start with the definition of equilibrium solution, which is the value of Q* such that
dQ/dt = 0,
0.75r = (10+r)(Q*/5)
the concentration of dye is (Q* /5) kg/L. This means we are solving for Q*/5, and then setting this result equal to 0.5:
0.75 (r/10+r) = Q*/5 = 0.5
This gives an equation for r (here we converting the decimals to fractions)
(3/4) (r/10+r) = 1/2
r/10+r = 2/3
3r = 2(10+r) = 20 + 2r
r = 20 L/min
(c) Q = (1 - e-15t)
(96/100) = (1 - e-15t)
t = 1/15 ln(100/4) =1/15*3.21=0.214 min
1/15 ln(100/4) 0.214 min
hence time taken is 0.214 minutes.