How do you find the amount of salt in a tank over time when a brine solution is continuously pumped in and out?

  • Thread starter KillerZ
  • Start date
  • Tags
    Mixture
In summary, the number of grams of salt in the tank at time t is given by A(t) = 200 - \frac{170}{e^{\frac{t}{50}}}.
  • #1
KillerZ
116
0

Homework Statement



A tank contains 200 liters of fluid which 30 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 4 L/min; the well-mixed solution is pumped out at the same rate. Find the number A(t) of grams of salt in the tank at time t.

Homework Equations



initial amount of liquid = 200L
A(t=0) = 30g
rate in = 4L/min
rate out = 4L/min
concentration of the inflow liquid = 1 g/L

The Attempt at a Solution



[tex]\frac{dA}{dt} = (1g/L)(4L/min) - \frac{(4L/min)(A(t))g}{(200L) + (4L/min - 4L/min)t}[/tex]

[tex]\frac{dA}{dt} = (4) - \frac{4A}{200}[/tex]

[tex]\frac{dA}{dt} + \frac{4A}{200} = (4)[/tex]

[tex]\frac{dA}{dt} + \frac{A}{50} = (4)[/tex]

[tex]I(t) = e^{\int\frac{A}{50}dt}[/tex]

[tex]I(t) = e^{\frac{At}{50}}[/tex]

[tex]f(t) = \frac{1}{e^{\frac{At}{50}}}[\int (e^{\frac{At}{50}})(4) dt + c][/tex]

[tex]f(t) = \frac{1}{e^{\frac{At}{50}}}[\frac{200e^{\frac{At}{50}}}{A} + c][/tex]

[tex]f(t) = \frac{200e^{\frac{At}{50}}}{Ae^{\frac{At}{50}}} + \frac{c}{e^{\frac{At}{50}}}][/tex]
 
Physics news on Phys.org
  • #2
KillerZ said:

Homework Statement



A tank contains 200 liters of fluid which 30 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 4 L/min; the well-mixed solution is pumped out at the same rate. Find the number A(t) of grams of salt in the tank at time t.

Homework Equations



initial amount of liquid = 200L
A(t=0) = 30g
rate in = 4L/min
rate out = 4L/min
concentration of the inflow liquid = 1 g/L

The Attempt at a Solution



[tex]\frac{dA}{dt} = (1g/L)(4L/min) - \frac{(4L/min)(A(t))g}{(200L) + (4L/min - 4L/min)t}[/tex]

[tex]\frac{dA}{dt} = (4) - \frac{4A}{200}[/tex]

[tex]\frac{dA}{dt} + \frac{4A}{200} = (4)[/tex]

[tex]\frac{dA}{dt} + \frac{A}{50} = (4)[/tex]

[tex]I(t) = e^{\int\frac{A}{50}dt}[/tex]
No, no, no! "A" is your dependent variable, not a constant. You find the integrating factor by integrating the coefficient of A. Your integrating factor is
[tex]e^{\int\frac{dt}{50}}= e^{\frac{t}{50}}[/tex]

[tex]I(t) = e^{\frac{At}{50}}[/tex]

[tex]f(t) = \frac{1}{e^{\frac{At}{50}}}[\int (e^{\frac{At}{50}})(4) dt + c][/tex][tex]f(t) = \frac{1}{e^{\frac{At}{50}}}[\frac{200e^{\frac{At}{50}}}{A} + c][/tex]
And now you've completely lost me! Where did "f(t)" come from? There was no "f" before this line. You are supposed to be finding A(t).

Since the integrating factor is
[tex]e^{\int\frac{dt}{50}}= e^{\frac{t}{50}}[/tex]
you have
[tex]e^{\frac{t}{50}}\frac{dA}{dt}-e^{\frac{t}{50}}\frac{A}{50}= 4e^{\frac{t}{50}}[/tex]
[tex]\frac{d}{dt}\left(e^{\frac{t}{50}}A\right)= 4e^{\frac{t}{50}}[/itex]
Integrate that to find A. The left side is, of course, just
[tex]e^{\frac{t}{50}}A(t)[/tex]


[tex]f(t) = \frac{200e^{\frac{At}{50}}}{Ae^{\frac{At}{50}}} + \frac{c}{e^{\frac{At}{50}}}][/tex]

Frankly, I think it would be easier not to use the "integrating factor" method at all. This is a separable equation.

[tex]\frac{dA}{dt}= 4-\frac{A}{50}= \frac{200- A}{50}[/tex]
[tex]\frac{dA}{200- A}= \frac{dt}{50}[/tex]
Integrate both sides of that.
 
  • #3
opps I made some typos and I don't know what I was thinking doing the integrating factor :confused: .

Here is my next attempt:

[tex]\frac{dA}{dt} = (1g/L)(4L/min) - \frac{(4L/min)(A(t))g}{(200L) + (4L/min - 4L/min)t}[/tex]

[tex]\frac{dA}{dt} = (4) - \frac{4A}{200}[/tex]

[tex]\frac{dA}{dt} + \frac{4A}{200} = (4)[/tex]

[tex]\frac{dA}{dt} + \frac{A}{50} = (4)[/tex]

[tex]I(t) = e^{\int\frac{1}{50}dt}[/tex]

[tex]I(t) = e^{\frac{t}{50}}[/tex]

[tex]A(t) = \frac{1}{e^{\frac{t}{50}}}[\int (e^{\frac{t}{50}})(4) dt + c][/tex]

[tex]A(t) = \frac{1}{e^{\frac{t}{50}}}[200e^{\frac{t}{50}} + c][/tex]

[tex]A(t) = \frac{200e^{\frac{t}{50}}}{e^{\frac{t}{50}}} + \frac{c}{e^{\frac{t}{50}}}[/tex]
 
  • #4
KillerZ said:
[tex]A(t) = \frac{200e^{\frac{t}{50}}}{e^{\frac{t}{50}}} + \frac{c}{e^{\frac{t}{50}}}[/tex]
I hope you understand that this can be simplified! :biggrin:

And, of course, use the fact that A(0)= 30 to find c.
 
  • #5
[tex]A(t) = \frac{200e^{\frac{t}{50}}}{e^{\frac{t}{50}}} + \frac{c}{e^{\frac{t}{50}}}[/tex]

[tex]A(t) = 200 + \frac{c}{e^{\frac{t}{50}}}[/tex]

[tex]A(0) = 30 = 200 + \frac{c}{e^{\frac{0}{50}}}[/tex]

[tex]c = 30 - 200 = -170[/tex]

[tex]A(t) = 200 - \frac{170}{e^{\frac{t}{50}}}[/tex]
 

1. What is a "Mixture DE"?

A "Mixture DE" refers to the process of creating a mixture of two or more substances where the individual substances retain their chemical properties but are physically combined.

2. How is a "Mixture DE" different from a "Compound"?

A "Mixture DE" differs from a "Compound" in that a compound is created through a chemical reaction between two or more substances, resulting in a new substance with different properties. In a mixture, the substances are physically combined and can be separated through physical means.

3. What are some examples of "Mixture DE"?

Examples of "Mixture DE" include saltwater, air, and salad dressing.

4. What are the methods of separating a "Mixture DE"?

The methods of separating a "Mixture DE" include filtration, distillation, chromatography, and evaporation.

5. How is the composition of a "Mixture DE" determined?

The composition of a "Mixture DE" can be determined through various techniques such as mass spectrometry, gas chromatography, and nuclear magnetic resonance spectroscopy. These methods help identify the individual substances present in the mixture and their relative proportions.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
3K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
4K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
3K
  • Calculus and Beyond Homework Help
Replies
3
Views
14K
  • Calculus and Beyond Homework Help
Replies
2
Views
5K
  • Calculus and Beyond Homework Help
Replies
1
Views
8K
Back
Top