Someone Please help me. Please

Question

A super large tank was initially filled with brine solution of 20 liters.  At some point, fresh water starts to flow in at a rate of 8 liters per minute, and the well mixed solution flows out at half of the rate.

 Part I   Set-up an Initial Value Problem to describe the above problem using     to represent the amount of salt in the tank t minutes after the process started. There was certain amount of salt in the tank initially, use   to represent the initial amount.

Part II Solve the above IVP to get     

Part III How much time (in minutes) will it take for the amount of salt in the tank be reduced to 20% of the initial amount?

Answer 1

The problem boils down to a differential equation if

V_b = volume of brine

t = time

Note: the volume in the tank is a function of time as 20 + 4t (Since you start with 20L, input 8L/min & remove 4L/min.

dV_b/dt = -4V_b/(20 + 4t) or dV_b/V_b = - dt/(5 + t)

where V_b is integrated from 20 liters to 4 liters (20%)

and time is integrated from 0 to the unknown time to reach 20%

The answer is indeed 20 minutes! (I make note of this since there are 2 different answers stated)

Answer 2

8 liters of fresh water coming in every hour.

and 4 liters of mixed water going out every hour.

The concentration of salt/liter gets diluted.

Salt concentration = Salt / Volume

Salt concentration = y0 / (20 + 4t)

y(t) = y0 - [Salt concentration] * 4t

y(t) = y0 - [y0 / (20 + 4t) ] * 4t

If y(t) = 20% of y0

Then...

.20*y0 = y0 - [y0 / (20 + 4t) ] * 4t

.20 = 1 - [1/(20 + 4t)] * 4t

.20 = 1 - (4t)/(20 + 4t)

(4t)/(20 + 4t) = .80

4t = (.80)(20 + 4t)

4t = 16 + 3.2t

.8t = 16

t = 16/.8 = 20 mins.

Answer 3

#3.

20Lsaltwater100%

20%=(20+x)/5

.2 = " "

" = 4+x/5

-3.8=x/5

x=-19

At 8L/m; it'll take 2.375min.