151k views
1 vote
Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 L of a dye solution with a concentration of 1 g/L. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 L/min, the well-stirred solution flowing out at the same rate.

a. Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.

2 Answers

0 votes

Final answer:

To find the time it takes for the dye concentration in the tank to reach 1% of its original value, we can use the concept of dilution. By applying the dilution formula, we can calculate that it would take approximately 166.67 hours for the concentration to reduce to 1% of its original value.

Step-by-step explanation:

To find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value, we can use the concept of dilution. The tank is being rinsed with fresh water flowing in at a rate of 2 L/min, and the well-stirred solution is flowing out at the same rate.

The concentration of dye in the tank after a certain amount of time can be calculated using the formula:

C1V1 = C2V2

Where:

  • C1 = initial concentration of dye (1 g/L)
  • V1 = volume of dye solution (200 L)
  • C2 = final concentration of dye (1% of the original value, which is 0.01 g/L)
  • V2 = volume of water rinsed through the tank (unknown)

Rearranging the formula to solve for V2:

V2 = (C1V1) / C2 = (1 g/L × 200 L) / (0.01 g/L) = 20000 L

So, it would take 20000 L / 2 L/min = 10000 min = 166.67 hours for the concentration of dye in the tank to reach 1% of its original value.

User SPoage
by
6.3k points
4 votes

Answer: t= 460.52 minutes

Step-by-step explanation:Q'=Q/100

Q'= rate in and out of water

Finding the differential equation

Let Q'(t)= The quantity of dye in the tank for t time

But rate in=0 Q/200 ×2=Q'

Q'/Q=-1/100

Dividing by Q gives

Ln/Q/ + c = -1/100 + c1

Integrating both sides gives

Ln/Q/ = -(1/100)t + c2

But c+c1=C2= A constant

Q=C2e(-t/100)

200e-(t/100)

t= ln200

t=460.52minutes

User Mr Aleph
by
7.1k points