Analysing Salt Buildup in Water Reservoirs

[MermaidWonders]

Alright... Here's one I encountered today.

Suppose the water reservoir holds 100 million gallons of water and supplies a city with 1 million gallons a day. The reservoir is partly refilled by a spring which provides 0.9 million gallons a day and the rest of the water, 0.1 million gallons a day, comes from a run-off from the surrounding land. The spring is clean, but the run-off contains salt with a concentration of 0.0001 pounds per gallon. Assume that there was no salt in the reservoir initially and that the reservoir is well mixed. Find the amount of salt in the reservoir as a function of time. Let $Q$ represent the amount of salt in the reservoir at time $t$.

Can someone explain why the rate of salt leaving is equal to the expression $\frac{Q}{100}$? I'm confused.

 

[MarkFL]

I've moved this question to its own thread as we prefer new questions not to be tagged onto existing threads.

We have 10 lbs. of salt coming in per day, and the amount leaving per day is 1/100 (since 1 million gallons of the 100 million gallons is sent to the city) of whatever amount is present at time $t$. This leads to the IVP:

\(\displaystyle \d{Q}{t}=10-\frac{Q}{100}\) where \(\displaystyle Q(0)=0\)

 

[MermaidWonders]

Oh, so we don't know how much salt gets sent along with the 1 million gallon (hence represented by $Q$) to the city even though 10 lbs come in per day?

I've moved this question to its own thread as we prefer new questions not to be tagged onto existing threads.

We have 10 lbs. of salt coming in per day, and the amount leaving per day is 1/100 (since 1 million gallons of the 100 million gallons is sent to the city) of whatever amount is present at time $t$. This leads to the IVP:

\(\displaystyle \d{Q}{t}=10-\frac{Q}{100}\) where \(\displaystyle Q(0)=0\)

 

[MarkFL]

Oh, so we don't know how much salt gets sent along with the 1 million gallon (hence represented by $Q$) to the city even though 10 lbs come in per day?

Yes, exactly. :)

 

[MermaidWonders]

OK... but why do we have to express the amount of water going out as 1/100 for 1 million gallon out of 100 million gallons in total going out? Why can't I just do (1 million gallon/day)$Q$?

 

[MarkFL]

OK... but why do we have to express the amount of water going out as 1/100 for 1 million gallon out of 100 million gallons in total going out? Why can't I just do (1 million gallon/day)$Q$?

Since 1/100 of the total in the reservoir is leaving, it is taking with it 1/100 of the salt (Q) in the reservoir, as we are told it is well mixed. So, the amount of salt leaving per day is 1/100 of the total amount of salt, just as 1/100 of the contents of the reservoir are leaving per day.

 

[MermaidWonders]

OK, makes sense. :)

 

[MarkFL]

OK, makes sense. :)

Using intuition only, what would you suppose the following is:

\(\displaystyle \lim_{t\to\infty}Q(t)=\,?\)

 

[MermaidWonders]

Ummm... would it be 0, since as time goes on, you keep removing salt from the total amount $Q$ until you eventually reach the point where there's no more salt in the reservoir? I don't know...

Using intuition only, what would you suppose the following is:

\(\displaystyle \lim_{t\to\infty}Q(t)=\,?\)

 

[MermaidWonders]

Wait... but at the same time, the salt gets replaced daily...

 

[MarkFL]

Ummm... would it be 0, since as time goes on, you keep removing salt from the total amount $Q$ until you eventually reach the point where there's no more salt in the reservoir? I don't know...

We should expect, given that the input/output stays the same as stated, that the reservoir would approach the same concentration as the input over time, which is 1 pound of salt per 100,000 gallons of water. This means we should expect:

\(\displaystyle \lim_{t\to\infty}Q(t)=1000\)

And, indeed this is what we find when we solve the IVP. :)

 

[MermaidWonders]

Wait... so is it like over time, the output of salt would have a smaller and smaller impact on the amount in the reservoir due to the accumulating input everyday? Also, where did the 1000 come from?

We should expect, given that the input/output stays the same as stated, that the reservoir would approach the same concentration as the input over time, which is 1 pound of salt per 100,000 gallons of water. This means we should expect:

\(\displaystyle \lim_{t\to\infty}Q(t)=1000\)

And, indeed this is what we find when we solve the IVP. :)

 

[MarkFL]

Wait... so is it like over time, the output of salt would have a smaller and smaller impact on the amount in the reservoir due to the accumulating input everyday? Also, where did the 1000 come from?

Yes, as time goes on, the amount of salt in the reservoir changes by smaller and smaller increments, because the concentration in the reservoir is rising to meet the concentration of the input. At frist when the concentration is low, there is a small amount of salt leaving...but as the concentration rises the amount leaving is larger, until as time goes on without bound, the amount leaving is the same as the amount coming in and the change is zero.

If the concentration is rising to 1 pound of salt per 100,000 gallons of water, and the volume of the reservoir is 100,000,000 gallons, then because 100,000,000/100,000 = 1,000 we see that the amount of salt in the reservoir much approach 1,000 pounds.

 

[MermaidWonders]

Your explanation is making me think of diffusion and osmosis... with respect to how the water and salt behaves...

Yes, as time goes on, the amount of salt in the reservoir changes by smaller and smaller increments, because the concentration in the reservoir is rising to meet the concentration of the input. At frist when the concentration is low, there is a small amount of salt leaving...but as the concentration rises the amount leaving is larger, until as time goes on without bound, the amount leaving is the same as the amount coming in and the change is zero.

If the concentration is rising to 1 pound of salt per 100,000 gallons of water, and the volume of the reservoir is 100,000,000 gallons, then because 100,000,000/100,000 = 1,000 we see that the amount of salt in the reservoir much approach 1,000 pounds.

 

[MarkFL]

Another way to find the limiting amount...

We have:

\(\displaystyle \d{Q}{t}=10-\frac{Q}{100}\)

If we want to find where the change in $Q$ is zero, we may set:

\(\displaystyle 0=10-\frac{Q}{100}\)

or

\(\displaystyle \frac{Q}{100}=10\)

or

\(\displaystyle Q=1000\)

 

[MermaidWonders]

Oh, true! Very quick, too. :)

However, is it wrong to interpret the previous explanation as diffusion and osmosis?

Another way to find the limiting amount...

We have:

\(\displaystyle \d{Q}{t}=10-\frac{Q}{100}\)

If we want to find where the change in $Q$ is zero, we may set:

\(\displaystyle 0=10-\frac{Q}{100}\)

or

\(\displaystyle \frac{Q}{100}=10\)

or

\(\displaystyle Q=1000\)

 

[MarkFL]

Oh, true! Very quick, too. :)

However, is it wrong to interpret the previous explanation as diffusion and osmosis?

I don't know...I never ran across those concepts in my study of ODEs.

 

[MermaidWonders]

OK...

 

[MarkFL]

It's been 25 years since I was a student...I seem to recall osmosis from biology, but don't recall it being quantified.

 

[MermaidWonders]

Oh, I see. :)

But as for that explanation, how do you know that "as the concentration rises the amount leaving is larger, until as time goes on without bound, the amount leaving is the same as the amount coming in and the change is zero"? It sounds just like osmosis and diffusion where salt is the solute...

 

[MarkFL]

Oh, I see. :)

But as for that explanation, how do you know that "as the concentration rises the amount leaving is larger, until as time goes on without bound, the amount leaving is the same as the amount coming in and the change is zero"? It sounds just like osmosis and diffusion where salt is the solute...

If the amount of solution leaving is a constant, and the concentration is rising, then the amount of salt leaving must be increasing.

 

[MermaidWonders]

If the amount of solution leaving is a constant, and the concentration is rising, then the amount of salt leaving must be increasing.

Makes sense. Last silly question here, but from an intuitive perspective, if you have accumulated salt in the reservoir, why can't the amount of salt leaving be greater than the amount entering at a particular point in time? As in, at a particular time, you could have like, let's say, 40 lbs of salt already in the reservoir but only 20 lbs are coming in, so why isn't it possible to have 30 pounds going out? (Numbers all made up to help illustrate my question and not relevant to this question)

 

[MarkFL]

Makes sense. Last silly question here, but from an intuitive perspective, if you have accumulated salt in the reservoir, why can't the amount of salt leaving be greater than the amount entering at a particular point in time? As in, at a particular time, you could have like, let's say, 40 lbs of salt already in the reservoir but only 20 lbs are coming in, so why isn't it possible to have 30 pounds going out? (Numbers all made up to help illustrate my question and not relevant to this question)

Yes, if the initial concentration was greater than the concentration of the input, then the concentration would still approach that of the input over time, but it would have to decrease to do so.

 

[MermaidWonders]

OK, got it. So these are just like rules to remember, in terms of how the concentration of the output will approach that of the input over time.