Simple Diff EQ - Modeling the salt concentration of a pond

  • Thread starter ethereal45
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  • #1

Homework Statement



The problem can be found here: http://alex.fangsoft.net/math.pdf [Broken]

I have completed Task #2, it is Task #1 that I am very lost on. Our teacher has really thrown us head first into the water with this one, I am very unfamiliar with differential equations and their purpose. Our textbook is very unhelpful with a problem of this magnitude.

Homework Equations



I understand differential equations involve dy/dx = something, then separating the variables, then integrating both sides.

In this case, I would assume dS/dT for salt/time = salt in - salt out, but past that I am very, very lost.

Thank you sincerely in advance.
 
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Answers and Replies

  • #2
52
0
Suggestion for you - expand on what you are already saying. It is suggested that your main variable is the total salt in the lake (S) - now list all sources and sinks in a consistent set of units so that on the LHS you have dS/dt and on the right hand side you have the identified terms in those units (probably lb/hr for this problem to keep it simple). Once you have the initial salt mass you can integrate the this to find your solution (no pun intended) over time. Now start by listing all terms using the same units...
 
  • #3
Okay, so I know ds/dt = salt in - salt out.

As for Salt In:

I know I have 50gal/hr x .15lb / gal = 7.5 lb / hour

For Salt Out:
I have 150 gal/hr x the concentration at that given time.

Do I represent that as 150 x (25,000lb + 7.5 t)/250,000
where 25,000 is the starting amount of salt, and 250,000 is the total capacity

or do I just call the salt out term 150 x (s)/250,000

This is where my unfamiliarity with diffeq is hurting!

Edit -- so my equation should be this, correct:

dS/dt = 7.5 - 150 (s/250,000)
 
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  • #4
52
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Since the volume of the lake is constant, you are okay ignoring the inflow and evaporation - if that was not the case you would have a system of diff eqns.

You do have the right equation - now you just need to solve it over time with the initial condition given by the starting salt concentration.

The solution has the shape of a constant plus and exponentially decaying term. You are seeking the long-term solution which is basically the constant term - the value the salt mass will reach after a very long time.
 
  • #5
Okay I understand what I've done so far, but for some reason I'm having alot of trouble solving it....I understand to do with my initial value after I solve....but what are my first few algebra moves?
 
  • #6
52
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To solve the equation, put your problem into the differential form: dS/f(S) = dt
 
  • #7
Okay, so....

dy/(7.5-150y/250000) = dt

I understand that concept, but I have zero idea how to evaluate the resulting integral, on the left hand side anyways.
 
  • #8
52
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integrate both sides: [tex]\int{ds/{(a-s)}} = \log(s)[/tex], [tex]\int dt=t[/tex]. Now express s as a function of t instead of the other way around.
 
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  • #9
yeah it's been a while since I've done integrals, I just made a u-substitution and found out what it should be!
 
  • #10
52
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sorry i can't get that tex to shape up for some reason - hope it makes sense
 
  • #11
so I get -250000/150 ln (abs(7.5-150y/250,000) +c = t

So solve for y ? One of my friends did this in maple and I'm just not getting an equation for y that agrees with him :(
 
  • #12
Here is my work, I have no clue what I'm doing wrong but this is definitely not the right answer:

http://alex.fangsoft.net/mathwork.pdf [Broken]
 
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  • #13
52
0
when you take the exponent of both sides you need to include the constant inside the exponent, then determine it from the initial condition. or you could simply determine it from the expression before you apply the exponential function (set t=0 and y=the initial value in post #11 above).

keep the parameters as unknowns until you are ready to manage this better. some can be consolidated to compound parameters. i find the problem easier to solve that way.
 

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