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Differential equations - mixing problem (more complicated)

  • #1
162
0
A 500 gallon tank originally contains 100 gallons of fresh water. Beginning at time t=0, water containing 50 percent pollutants flows into the tank at the rate of 2 gal/min and the well stirred solution leaves at the rate of 1 gal/min. Find the concentration of pollutants in the tank at the moment it overflows.

The answer is 48%


my attempt, something is wrong as i did not get that answer or anywhere near it.

rate in: 2 x 0.50
rate out: 1 x s(t)/(100+t)

s'(t) = 1- s(t)/(100+t)
s'(t) + s(t)/(100+t) = 1

a(t) = 1/(100+t) , b(t) = 1

using u(t) = exp(integ(a(t)dt)):
u(t) = exp(integ(a(t)dt))
u(t) = exp (ln (100+t))
u(t) = 100+t

using d/dt u(t)s(t) = u(t)b(t):
d/dt [(100+t)s(t)]= (100+t)(1)
(100+t)s(t)=integ(100+t)
(100+t)s(t) = 100t + 1/2t^2 + C
s(t) = [100t + 1/2t^2 +C ]/(100+t)

sub s(0) = 0 into s(t) [is this even correct?]
0 = [0+0+c]/[100+0]
c= 0
s(t) = [100t + 1/2t^2]/(100+t)

c(t) = s(t)/100
c(t) = [100t + 1/2t^2]/100(100+t)

then i found when the tank overflow:
500 = 100+t
t= 400
then found c(t)

c(400) = [100(400)+1/2(400)^2]/100(100+400)
c(400) = 2.4

which is wrong.

can anyone help me?
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,808
933
A 500 gallon tank originally contains 100 gallons of fresh water. Beginning at time t=0, water containing 50 percent pollutants flows into the tank at the rate of 2 gal/min and the well stirred solution leaves at the rate of 1 gal/min. Find the concentration of pollutants in the tank at the moment it overflows.

The answer is 48%


my attempt, something is wrong as i did not get that answer or anywhere near it.

rate in: 2 x 0.50
rate out: 1 x s(t)/(100+t)

s'(t) = 1- s(t)/(100+t)
s'(t) + s(t)/(100+t) = 1

a(t) = 1/(100+t) , b(t) = 1

using u(t) = exp(integ(a(t)dt)):
u(t) = exp(integ(a(t)dt))
u(t) = exp (ln (100+t))
u(t) = 100+t

Okay, good so far.

using d/dt u(t)s(t) = u(t)b(t):
d/dt [(100+t)s(t)]= (100+t)(1)
(100+t)s(t)=integ(100+t)
(100+t)s(t) = 100t + 1/2t^2 + C
s(t) = [100t + 1/2t^2 +C ]/(100+t)

sub s(0) = 0 into s(t) [is this even correct?]
0 = [0+0+c]/[100+0]
c= 0
Yes, that's correct- initially the tank contained "100 gallons of fresh water"

s(t) = [100t + 1/2t^2]/(100+t)

c(t) = s(t)/100
Here's your error!! The tank will contain 500 gallons when it overflows, not 100. The concentration is s/500, not s/100.

c(t) = [100t + 1/2t^2]/100(100+t)

then i found when the tank overflow:
500 = 100+t
t= 400
then found c(t)

c(400) = [100(400)+1/2(400)^2]/100(100+400)
c(400) = 2.4

which is wrong.

can anyone help me?
Obviously, a concentration can't be larger than 1! Again, your error is in dividing by 100 gallons, the initial amount of water in the tank, when, at the time of overflow, there are 500 gallons of water in the tank. Divide your answer by 5.
 
  • #3
162
0
thanks very much
that was a stupid mistake on my part!
 

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