Confusing mixing problem - differential equations

  • #1
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A tank contains 300 gallons of water and 100 gallons of pollutants. Fresh water is pumped into the tank at the rate of 2 gal/min, and the well-stirred mixture leaves the tank at the same rate. How long does it take for the concentration of pollutants in the tank to decrease to 1/10 of its original value


This problem is really confusing me as it is water that is being pumped in instead of the usual pollutant/waste.

this is my attempt (i just reversed what i usually do, but this is wrong..)

w(t) : amount of water in tank at time t
w'(t) : rate of change of water in tank at time t

water enters at : 2
water leaves at : 2 x w(t)/100 (i think this is wrong... i actually worked this problem out with /100 and /300, and both are wrong as i dont get a positive answer.)

w'(t) = 2 - w(t)/50
w'(t) + 1/50 w(t) = 2
a(t) = 1/50, b(t) = 2

u(t) = exp (integ(1/50)dt)
u(t) = e^(1/50)t

d/dt (e^(1/50)t w(t) ) = e^(1/50)t x 2
e^(1/50)t w(t) = integ(2e^(1/50)t)
e^(1/50)t w(t) = 2(1/50) e^(1/50)t + C
w(t) = [1/25 e^(1/50)t + C] / e^(1/50)t
sub w(0) = 300
w(0) = [1/25 e^(1/50)0 + C] / e^(1/50)0
300 = [1/25 (1) + C] / 1
C = 7499/25

so w(t) = 1/25 + (7499/25)e^(-1/50)t
w(t) = 1/25 [ 1+ 7499e^(-1/50)t]

c(t) = w(t)/100 (again, is this correct?)
c(t) = 1/25000 [1+7499e^(-1/50)t]

1/10 of original value of pollutants is 10.

10 = 1/25000 [1+7499e^(-1/50)t]
250000 = 1+7499e^(-1/50)t
249999/7499 = e^(-1/50)t
ln (249999/7499) = -1/50t
t= -50 ln (249999/7499)

what did i do wrong. i think i did this whole problem wrong actually, any help would be great, thanks.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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A tank contains 300 gallons of water and 100 gallons of pollutants. Fresh water is pumped into the tank at the rate of 2 gal/min, and the well-stirred mixture leaves the tank at the same rate. How long does it take for the concentration of pollutants in the tank to decrease to 1/10 of its original value


This problem is really confusing me as it is water that is being pumped in instead of the usual pollutant/waste.

this is my attempt (i just reversed what i usually do, but this is wrong..)

w(t) : amount of water in tank at time t
w'(t) : rate of change of water in tank at time t

Since the problem asks for the amount of polutant in the tank, I am surprized you would choose to focus on the amount of water but that will work

water enters at : 2
water leaves at : 2 x w(t)/100 (i think this is wrong... i actually worked this problem out with /100 and /300, and both are wrong as i dont get a positive answer.)
Why over 100? DO NOT just stick in numbers at random! Think what you are trying to do here. There are 2 gallons of mixture leaving each minute. How much of each gallon is water? There are 500 gallons total in the tank. Assuming it is well mixed, each gallon in the tank would include w/500 gallons of water (and (500-w)/500 is polutant). Since the mixture is leaving at 2 gal/ min, 2(w/500) gallons of water are leaving per minute.

w'(t) = 2 - w(t)/50
w'(t) + 1/50 w(t) = 2
a(t) = 1/50, b(t) = 2

u(t) = exp (integ(1/50)dt)
u(t) = e^(1/50)t

d/dt (e^(1/50)t w(t) ) = e^(1/50)t x 2
e^(1/50)t w(t) = integ(2e^(1/50)t)
e^(1/50)t w(t) = 2(1/50) e^(1/50)t + C
w(t) = [1/25 e^(1/50)t + C] / e^(1/50)t
sub w(0) = 300
w(0) = [1/25 e^(1/50)0 + C] / e^(1/50)0
300 = [1/25 (1) + C] / 1
C = 7499/25

so w(t) = 1/25 + (7499/25)e^(-1/50)t
w(t) = 1/25 [ 1+ 7499e^(-1/50)t]

c(t) = w(t)/100 (again, is this correct?)
No, it isn't. concentration is the amount of water divided by the total volume, 500 gal.

c(t) = 1/25000 [1+7499e^(-1/50)t]

1/10 of original value of pollutants is 10.

10 = 1/25000 [1+7499e^(-1/50)t
Now, you've lost track of the meaning of your variable! w(t) was the amount of WATER in the tank, not the amount of polutant. In addition, you converted at the last minute to concentration rather than amount.
The left hand side of this equation is the amount of polutant, the right hand side is the concentration of water in the tank. There is no reason they should be the same.

250000 = 1+7499e^(-1/50)t
249999/7499 = e^(-1/50)t
ln (249999/7499) = -1/50t
t= -50 ln (249999/7499)

what did i do wrong. i think i did this whole problem wrong actually, any help would be great, thanks.
Start over again. Since the question asks about polutant, let p(t) be the amount of polutant. There is NO polutant coming in so your differential equation is just dp/dt= amount of polutant going out. If there are p(t) gal of polutant in the tank and a total of 500 gal, each gal contains p(t)/500 gal of polutant. Since it is going out at 2 gal/ min, polutant is going out at -p/250 gal/min.
 
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  • #3
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okay
here is my solution:

let p(t) = amount of pollutants in tank at time t
p'(t) = rate of change of pollutants in tank at time t

pollutants enters at : 0
pollutants leaves at: 2 x p(t)/302

p(0) = 100

p'(t) = 0 - p(t)/151
p'(t) + 1/151 p(t) = 0
a(t) = 1/151

p(t) = C exp(-integ(a(t)dt))
p(t) = C exp(- integ (1/151 dt))
p(t) = C exp(- 1/151 t)
sub p(0) = 100 to find C
100 = C exp(-1/151 (0))
c = 100
.'. p(t) = 100 exp(-1/151 t)

c(t) = p(t)/302
c(t) = [100 exp(-1/151 t)]/302
c(t) = [50 exp(-1/151 t)]/151

find concentration at c(0)
c(0) = 100/300
c(0) = 1/3
1/3 x 1/10
= 1/30

sub c = 1/30
1/30 = [50 exp(-1/151 t)]/151
151/1500 = exp(-1/151 t)
ln (151/1500) = -1/151 t
t = - 151 ln (151/1500)
t = 151 ln (1500/151)
t = 346.69 min..

is this correct?
 
  • #4
162
0
the answer is suppose to be 460.50 min, i have gone wrong somewhere.
 
  • #5
162
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got it, i know what i did wrong.
again a stupid mistake.
thanks for helping me out.
 

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