Question

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

Answer 1

Answer:

Explanation:

From the given information;

Let Q(t) = mass of dye in the tank as a function of time

The mass in the tank = 200 L × (1g/L) = 200 g

Using the law of mass conservation;

$\dfrac{dQ}{dt} = \text{(Rate of incoming mass)- (rate of outgoing mass)}$

$\dfrac{dQ}{dt} = (2 L/min ) (0 \ g/L) - (2 L/min ) (\dfrac{Q}{200}g/L)$

$Q' = \dfrac{-Q}{1000}$

Q(0) = 200

By finding the solution to the ODE using the method of separation of variables;

$\dfrac{Q'}{Q} = -0.01$

$Q(t) = Ce^{-0.01t}$

Using the initial condition;

200 = Q(0) = C

$Q(t) = 200e^{-0.01t}$

1% of 200g = 2g of dye solution

∴

$2 = 200e^{-0.01t}$

$e^{-0.01t}=0.01$

$t =\dfrac{ In(0.01}{-0.01}$

t = 460.5 hours