MIXING PROBLEMS
Phanuel Mariano
Mixing Problem1:
Problem. A vat contains 60L of water with 5 kg of salt water dissolved in it. A salt water solution that
contains 2 kg of salt per liter enters the vat at a rate of 3 L/min. Pure water is also owing into the vat
at a rate of 2 L/min. The solution in the vat is kept well mixed and is drained at a rate of 5 L/min, so
that the rate in is the same as the rate out. Thus there is always 60L of salt water at any given time. How
much still remains after 30 minutes? What is the long term behavior?
Solution:
Step1: Dene variables
Let y(t) =amount of salt at time t. Let y(0) = 5 kg.
Step2: Find Rate in/ Rate out
Note that for anything that comes in you can always nd the Rate In as
concentrarion
Rate in =
× Rate.
of stu coming in
Similarly you can always nd the Rate out as
concentrarion
Rate out =
× Rate.
of stu going out
Using the information from the problem we have
L
kg
L
kg
3
+ 0
2
Rate in =
2
L
min
L
min
−pure water
-salt water solution
kg
= 6
.
min
and
concentrarion
Rate out =
× Rate
of stu going out
L
y(t) kg
×5
.
=
L
60
min
y(t) kg
.
12 min
=
Step 3: Write the IVP
Always recall that for mixing problems we have
dy
dt
= Rate in − Rate out
=
6−
y
.
12
1
MIXING PROBLEMS
2
and the initial condition i
y(0) = 5.
Step 4: Find the common denominator and solve using separation of variables.
Write
and using separation of variables we get
dy
y
72 − y
=6−
=
dt
12
12
dy
72 − y
=
dt
12
⇐⇒
dy
dt
=
72 − y
12
⇐⇒
− ln |72 − y| =
⇐⇒
t
+ C1
12
−t
+ C2
ln |72 − y| =
12
t
|72 − y| = C3 e− 12
⇐⇒
72 − y = ke− 12
⇐⇒
y = 72 − ke− 12 .
⇐⇒
t
t
Solving the IVP by using y(0) = 5 to get
y(0) = 5
so the nal solution is
⇐⇒
72 − ke0 = 5
⇐⇒
k = 72 − 5 = 67
t
y(t) = 72 − 67e− 12 .
Step5:
After 30 minutes there is
y(30) = 72 − 67e− 12 = 66.5 kg.
30
The long term behavior is simply the limit:
30
lim y(t) = lim 72 − 67e− 12 = 72 − 0 = 72.
t→∞
t→∞
MIXING PROBLEMS
3
Mixing Problem #2:
• The dierence here is that now we allow the total volume of uid to vary, when before it was kept
xed.
Problem. A 400-gallon tank initially contains 200 gallons of water containing 3 pounds of sugar per
gallon. Suppose water containing 5 pounds per gallon ows into the the top of the tank at a rate of 6
gallons per minute. The water in the tank is kept well mixed, and 4 gallons per minute are removed
from the bottom of the tank. How much sugar is in the tank when the tank is full?
Solution:
Step1: Dene variables
Let y(t) = amount of sugar at time t, which is in minutes. Let y(0) = 3 × 200 = 600 pounds.
Step2: Find Rate in/ Rate out
Note that for anything that comes in you can always nd the Rate In as
concentration
Rate in =
× Rate.
of sugar coming in
Similarly you can always nd the Rate out as
concentrartion
Rate out =
× Rate.
of sugar coming out
We have
pounds
gallons
=
5
6
gallon
min
-sugar water solution
pounds
= 30
.
gallon
To nd the concentration of sugar coming out we have know that the amount of water at time t.
gallons
gallons
−4
Water at timt t = 200 gallons + 6
t
min
min
Rate in
=
So
Rate out
200 + 2t,
=
concentrarion
of stu going out
y(t) pounds
200 + 2t gallon
y(t) pound
4
.
200 + 2t min
=
=
Step 3: Write the IVP
×4
Always recall that for mixing problems we have
dy
dt
= Rate in − Rate out
=
30 −
4
y.
200 + 2t
× Rate
gallons
.
min
MIXING PROBLEMS
4
and the initial condition
y(0) = 600.
Step 4: Solve using the Method of integrating factors:
Write
so that g(t) =
dy
4
+
y = 30
dt
200 + 2t
4
200+2t
and b(t) = 30. Thus the integrating factor is
µ(t) = e4
´
dt
200+2t
= e2
´
dt
100+t
2
= e2 ln(100+t) = (100 + t) .
Thus using the formula I have that
y(t)
=
=
=
=
ˆ
µ(t)b(t)dt. + C
ˆ
1
2
30
(100
+
t)
dt.
+
C
2
(100 + t)
#
"
3
1
(100 + t)
.+C
2 30
3
(100 + t)
h
i
1
3
2 10 (100 + t) + C
(100 + t)
1
µ(t)
using y(0) = 600 we get that
1 10 · 1003 + C
1002
600 =
so that
C = −4, 000, 000
thus
3
y(t) =
10 (100 + t) − 4, 000, 000
2
(100 + t)
Step5: Answer the question
.
Since the amount of water in the tank is 200 + 2t then it lls up when
200 + 2t = 400
so that t = 100. Thus the amount of sugar is
3
y(100)
=
=
10 (200) − 4, 000, 000
(200)
2
1, 900 pounds.