Question

A tank contains 100 kg of salt and 1000 L of water. A solution of a concentration 0.05 kg of salt per liter enters a tank at the rate 10 L/min. The solution is mixed and drains from the tank at the same rate.
(a) What is the concentration of our solution in the tank initially?
concentration = (kg/L)

(b) Find the amount of salt in the tank after 1 hours.
amount = (kg)

(c) Find the concentration of salt in the solution in the tank as time approaches infinity.
concentration = (kg/L)

Answer 1

Answer:

Step-by-step explanation:

Initial Quantity of salt Q(0) = 100 kg

Capacity of tank = 1000 L

Inflow of liquid = 10 l /minute and outflow = same rate thus leaving the content of volume 1000 L at any time

Assume after entering constantly mixes and drains out.

Concentration flow = inflow - outflow

= 0.05(10) - $\frac{Q(t)}{1000} *10$

i.e. $Q'(t) = 0.5-0.01Q(t)$, Q(0) = 100

$Q'(t) = -0.01(Q(t)-50), Q(0) = 100\\\frac{dQ}{Q-50} =-0.01 dt\\ln |Q-5| = -0.01t +C\\Q-50 = Ae^{-0.01t}$

Use Q(0) = 5

-45 = A  

So the equation would be

$Q-50 = -45 e^{-0.01t} \\Q(t) = 50-45 e^{-0.01t}$

a) Initial concentratin = $\frac{100}{1000} =0.10$

b) Use the solution fo rDE substitute t =1

Q(1) = 50-45e^(-0.01) = 5.4477 kg

c) As t becomes large Q = 50

So concentration limit = 50/1000 = 0.05