Question

a tank initially holds 100 gallons of brine solution. at t=0, fresh water is poured into the tank at the rate of 5 gpm, while the well-stirred mixture leaves the tank at the same rate. after 20 minutes, the tank contains 20/e lbs of salt. find the initial concentration of the brine solution inside the tank

Answer 1

first of all...can someone clarify to me what the question is asking for?

Answer 2

here's what i got
\[\frac{dQ}{dt} = r_1 c_1 - r_2 [ \frac{Q}{V + (r_1 - r_2)t}]\]
\[\frac{dQ}{dt} = - \frac{Q}{20}\]
is this correct?

Answer 3

yes , that what i tought. Let the anount of sault in the water be y, then we can express it in this way:\[dy/dt=y-(5/100)yt\]
separating and integrating:
\[y=Ce ^{19/20 t ^{2}}\]

Answer 4

but maybe something is missing.....

Answer 5

what did you do? i did variable separable
\[\frac{dQ}{Q} = -\frac{1}{20}dt\]
\[\ln Q = -\frac{1}{20} t\]
\[\large Q = e^{-\frac{t}{20}}\]
right?

Answer 6

\[\Large Q = e^{-\frac{t}{20}} + C\]

Answer 7

\[ {dQ \over dt} = 0 - {5\over 100}Q\]
is the right equation.

Answer 8

right
\[\frac{dQ}{dt} = -\frac{Q}{20}\]
then i integrated

Answer 9

i don't think it's right

Answer 10

the amount of soul leaving the tank is not constant

Answer 11

sault*

Answer 12

thats why dq/dt=q-1/20qt

Answer 13

but it says it's fresh water so no salt is coming in

Answer 14

i know, but the water that leave has salt in it, and afetr time pass, less and less amount

Answer 15

i see

Answer 16

(1/20 )Q represents the amoputn of salt in the 5 galons, and it is function of time

Answer 17

http://tutorial.math.lamar.edu/Classes/DE/Modeling.aspx
i think the input must be zero.

Answer 18

\[dQ/dt = Q-1/20Q\]
\[Q = Ce ^{19/20t}\]
fot t=20
\[20=Ce ^{19} \rightarrow C=20/e ^{19}\]
and fot t=0 this is the amount of salt in the water

Answer 19

@myko this is not a decay equation ...
\[ Q=Ce^{19/20t}\]
on this system, the system must come to zero concentration after certain time (steady state)

Answer 20

lg looks ok, now find C using t=20 and Q= 20/e
Q at t=0 will be C
and the concentration will be Q/100

Answer 21

\[\frac{20}{e} = e^{-1} + C\]
\[\frac{20}{e} - \frac 1e = C\]
\[\frac{19}{e} = C\]

Answer 22

now what @phi

Answer 23

how is it smaller then at the end?

Answer 24

because " fresh water is poured into the tank at the rate of 5 gpm"

Answer 25

the equation is
ln Q= -t/20 +C
when you get rid of the ln (make it a power of e)
you get
Q= exp(-t/20+C)= exp(C)*exp(-t/20)
call exp(C) K a new constant
Q= K*exp(-t/20)
now find K

Answer 26

OHHHHHHHHH

Answer 27

hah finally got it

Answer 28

last question @phi it says fresh water so there isnt supposed to be concentration..how did we get one? is the zero concentration just for the one coming in? the one looking for in here is the concentration in the tank itself?

Answer 29

yes. no salt going in. 0 lbs salt/ gal of water
however, there is salt in the tank at t=0 (20 lbs of salt)
so the concentration is 20 lbs per 100 gal or 0.2 lbs/gal