ODE Salty Tank question (Can do the hard part, but not the easy part?)
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ODE Salty Tank question (Can do the hard part, but not the easy part?)

[From: ] [author: ] [Date: 11-09-20] [Hit: ]
of 0.4g/L flows INTO the tank at 20L/minute. Assuming there is instantaneous mixing of the two salty solutions it then leaves the tank at a rate of 10L/min. Let C(t) denote concentration and V(t) denote the volume (both at a given time t).Now the last part of the question says What is the concentration of the salt in the tank when the tank first overflows which I can do (by way of help from an online source).However,......
Hi everyone,

"A 200L tank initially contains 100L of water with a salt concentration of 0.1grams/L. Water with a salt conc. of 0.4g/L flows INTO the tank at 20L/minute. Assuming there is instantaneous mixing of the two salty solutions it then leaves the tank at a rate of 10L/min. Let C(t) denote concentration and V(t) denote the volume (both at a given time t).


Now the last part of the question says "What is the concentration of the salt in the tank when the tank first overflows" which I can do (by way of help from an online source).

However, I'm having trouble with the first two parts of the Q:

Set up the differential eqns governing V(t) and C(t)

Solve the governing equations to find the particular solutions for V(t) and C(t).

Because I can do the final concentration question I have probably answered the above 2, but I just don't know what the hell they are asking for!!


All help greatly appreciated,

sorry for being long winded,


thanks!

-
initial mass of salt = 0.1 * 100 = 10 grams

V = (20 - 10)t + 100 = 100 + 10t

dC/dt = (0.4)(20) - C/(100 + 10t) * 10

dC/dt + 10C/(10t + 100) = 8

integrating factor,

k = e^(∫ 10/(10t + 100) dt)

k = e^(ln (t + 10))

k = (t + 10)

general solution,

C k = ∫ 8k dt

C (t + 10) = ∫ 8(t + 10) dt

C (t + 10) = 4t² + 80t + constant

initial condition, at t = 0 ⇒ C = 10 grams

10 (0 + 10) = 4 * 0² + 80 * 0 + constant

constant = 100

C (t + 10) = 4t² + 80t + 100

when the tank will overflow at t = T

V = 100 + 10t

200 = 100 + 10T

T = 10 minutes

mass of salt ant T = 10 min

C (10 + 10) = 4(10)² + 80(10) + 100

C = 65 grams

concentration of solution is

G = 65/200

G = 0.325 gram/L




EDIT:

as your question : "But, I can't see where you've factored in the 10L/min exiting the tank?"

rate of mass of salt = rate of input mass - rate of output mass

rate of mass of salt = rate of input mass - (mass of salt)/(volume of solution) * (volume change of solution which exit from the tank)

dC/dt = (0.4)(20) - C/(100 + 10t) * 10
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