Question

a tank contains 100 gal. of water. brine enters the tank at the rate of 3 gpm. the mixture, thoroughly stirred, leaves the tank at the rate of 2 gpm. if the concentration of the brine at the end of 20 minutes is to be 2 lb/gal, what should be the concentration of the brine entering the tank? Ans. 4.75 lb/gal

Answer 1

how to set this up?

Answer 2

is it still ambiguous @Mimi_x3 ?

Answer 3

let S be salt in tank at time t
F1 flow rate with salt concentration M1

\[\frac{ds}{dt}=F_1M_1-F_2M_2\]\[\frac{ds}{dt}=3M_1-2(\frac{s}{100+(3-2)t})\]\[\frac{ds}{dt}=3M_1-\frac{2s}{100+t}\]\[\frac{ds}{dt}+\frac{2s}{100+t}=3M_1\]now solve using integrating factor method\[I=e^{\int\limits{\frac{2}{100+t}dt}}=e^{2\ln(100+t)}=(100+t)^2\]multiply by DE\[(100+t)^2\frac{ds}{dt}+2s(100+t)=3M_1(100+t)^2\]notice that RHS is product rule\[\frac{d}{dt}((100+t)^2 \times s)=3M_1(100+t)^2\]integrate both sides\[(100+t)^2s=M_1(100+t)^3+C\]\[s=\frac{M_1(100+t)^3+C}{(100+t)^2}\]at t=0 --> S=0
\[C=-1\times 10^6M_1\]so\[s=\frac{M_1(100+t)^3+1 \times 10^6M_1}{(100+t)^2}\]at t=20 M1=2 lb/gal\[2=\frac{s}{100+20}\]\[s=240\]
\[240=\frac{M_1(100+20)^3+1 \times 10^6M_1}{(100+20)^2}\]solving for M1
\[M_1=4.75 lb/gal\]

Answer 4

should be -1x10^6M1

Answer 5

@lgbasallote

Answer 6

i got lost after "multiply with DE" what did you do there?

Answer 7

The right hand side looks like the product rule\[(uv)'=u'v+uv'\]

Answer 8

do u get it?

Answer 9

hmm yup seems you did the long way of linear DE got me confused..im used to shortcut...

t -> 0 then s ->0

does this always apply?

Answer 10

yeah initial value

Answer 11

so if s is not given then at t = 0 it's always 0? that means it starts with fresh water? no solution inside the tank or something?

Answer 12

yeah the liquid has not enterd the tank yet so there is no solution in the tank

Answer 13

okay you said c = -1 x 10^6 M1

but in the next step you did
\[s = \frac{M_1 (100+t)^3 + 1 \times 10^6 M_1}{(100+t)^2}\]
is this a typo?

Answer 14

okay i follow some of the others except that...and how you got 4.75?

Answer 15

yeah typo thats why i reposted it in the second post

Answer 16

oh lol didnt see that

Answer 17

so how did you get 4.75? can you show the substitution?

Answer 18

*Mt(t^2+300t+30000) +c

Answer 19

hmm that's the same as \[\large Q(100 + t)^2 = Mt(100 + t)^3 + C\]

Answer 20

right?

Answer 21

are you asking if
Mt(t^2+300t+30000)=? Mt(100+t)^3
or
(t^2+300t+30000)=? (100+t)^3

Answer 22

im asking if what you said is the same as what i said

Answer 23

because i cannot see the division of the equal sign...idk which corres[ponds to which

Answer 24

aww why

Answer 25

start with
\[ \frac{d}{dt}((100+t)^2 \times s)=3M_1(100+t)^2 \]

or
\[ d ((100+t)^2 s)= 3M(100+t)^2 dt \]

Answer 26

\[\int 3M_1 (100 + t)^2 dt \implies M_1 (100+t)^3\]
right?

Answer 27

yes, it's right. I was looking for why I was not getting M1=4.75 from the last equation.
Looking at it, it looks like a sign got flipped on the C value.
so, with that typo fixed, all is ok

Answer 28

so how to get 4.75?

Answer 29

replace s with 240, t=20 (change +M*10^6 to -M*10^6) , factor out the M...

Answer 30

hmm...mind writing it?

Answer 31

\[ s = \frac{M_1 (100+t)^3 - 1 \times 10^6 M_1}{(100+t)^2} \]
\[ s=\frac{(100+t)^3 - 10^6}{(100+t)^2 }M_1 \]
with s= 240 and t=20
\[ 240= \frac{120^3-10^6}{120^2} M_1 \]

Answer 32

ohhh i see... i was multiplying instead of dividing lol

Answer 33

thanks for clarifying @phi :D

Answer 34

@phi one more question...what was the question asking for?

Answer 35

was it the salt concentration at t = 0? or t = 20?

Answer 36

the salt concentration of the entering solution at t=20

Answer 37

so what is 2 lb/gal??

Answer 38

2 lb/gal is the concentration of salt in the tank at t=20
It starts at t=0 with pure water, and brine enters (4.75 lbs/gal) at 3 gal/min and at the same time, the solution leaves the tank at 2 gal/min

Answer 39

ohhh 2 lb/gal is *in* the tank

4.75 is the one going inside the tank?

Answer 40

4.75 is the brine going into the tank. At some point, when the tank reaches equilibrium, we would expect 4.75 lbs salt per gal in the tank, leaving the tank and entering the tank

Answer 41

so is 4.75 constant? or it only applies at t=20?

Answer 42

the 4.75 lbs/gal is the concentration of brine going into the tank. It is constant, and is the amount that leads to 2 lbs/gal of salt in the tank after exactly 20 minutes.

Answer 43

ohhhh

Answer 44

last one... is 2 lb/gal concentration? or amt of salt? im confused with that part

Answer 45

I am being sloppy. 2 lbs of salt per gallon of brine (salt+water) is a concentration, from which we can deduce the amount of salt in the tank.

Answer 46

ohhh i see thanks!!!

Answer 47

ohhh..it's making sense to me now slowly...i thought concentration only comes in..it's also possible that there is already concentration in the tank already right?