# Math Help - Solving Mixture Problem using 1st Order Lin Dif. Eq.

1. ## Solving Mixture Problem using 1st Order Lin Dif. Eq.

word problems. awesome

Water flows through a purifying machine at the rate 50 gal/min. The machine removes 10% of the impurities as the water flows through. Suppose I have 1000 gallons of water containing 5 gallons of impurities. How long do I have to run it through the machine to reduce the amount of impurity to half a gallon?

So I tried setting this up in the form dx/dt + r_out*(x/V) = r_in*c_in, where r_in and r_out are 50 gal/min, V is 1000 and my initial condition is x(0) = 5 gal/V. The problem is, I don't know what to use for c_in (concentration of the inflow).

I'm pretty sure the problem wants me to use the integrating factor method and I'm comfortable doing so, I just can't figure out how to set up the problem in its entirety. Anybody want to help me figure out how to find c_in?

2. ## Re: Solving Mixture Problem using 1st Order Lin Dif. Eq.

Originally Posted by rhobere
word problems. awesome

Water flows through a purifying machine at the rate 50 gal/min. The machine removes 10% of the impurities as the water flows through. Suppose I have 1000 gallons of water containing 5 gallons of impurities. How long do I have to run it through the machine to reduce the amount of impurity to half a gallon?

So I tried setting this up in the form dx/dt + r_out*(x/V) = r_in*c_in, where r_in and r_out are 50 gal/min, V is 1000 and my initial condition is x(0) = 5 gal/V. The problem is, I don't know what to use for c_in (concentration of the inflow).

I'm pretty sure the problem wants me to use the integrating factor method and I'm comfortable doing so, I just can't figure out how to set up the problem in its entirety. Anybody want to help me figure out how to find c_in?
I would say that c_in = 0. If you're flowing pure water into the purifier, then c_in = 0.

3. ## Re: Solving Mixture Problem using 1st Order Lin Dif. Eq.

I think the idea is more that the water is pumped from the container through the pump, where 10% of the impurities are removed before being pumped back into the tank. otherwise the fact that the pump is removing 10% of the impurities would be irrelevant. if the water isn't being pumped back into the tank, the problem would at least have to make a mention of there being a new source of water I would think.

the best I could come up with is that c_in is a continuous growth rate. as in A_0e^rt, where A_0 is equal to 5 gallons of impurities and r is equal to .9 (-10% impurities per cycle = 1-.1 = .9). when I use it try to solve it in that form I get a negative number for t.

4. ## Re: Solving Mixture Problem using 1st Order Lin Dif. Eq.

Originally Posted by rhobere
I think the idea is more that the water is pumped from the container through the pump, where 10% of the impurities are removed before being pumped back into the tank. otherwise the fact that the pump is removing 10% of the impurities would be irrelevant. if the water isn't being pumped back into the tank, the problem would at least have to make a mention of there being a new source of water I would think.

the best I could come up with is that c_in is a continuous growth rate. as in A_0e^rt, where A_0 is equal to 5 gallons of impurities and r is equal to .9 (-10% impurities per cycle = 1-.1 = .9). when I use it try to solve it in that form I get a negative number for t.
Yeah, I was totally off in my response. Why would you put pure water into a purifier?

So, I'm thinking your P&ID would look something like this: one tank, line out to a pump, which leads to the purifier, which comes back into the tank. Would that be correct?

If so, then I would agree that r_in and r_out are equal. (The pressures on either side of the pump would not be equal, but the flows would.) Wouldn't the c_out be 0.9 * c_in? And if the concentration in is just x/V, then you've got the DE

$\frac{dx}{dt}=r_{\text{in}}c_{\text{in}}-r_{\text{out}}c_{\text{out}}=50\,(0.9)\frac{x}{V}-50\,\frac{x}{V}=-50(0.1)\frac{x}{V}.$

How does that strike you?

5. ## Re: Solving Mixture Problem using 1st Order Lin Dif. Eq.

Well, I went and talked to my professor about it yesterday and you're spot on although he got to the answer a little differently. I, however, was WAY over-analyzing it.

c_out was the unknown x/V = x/1000
c_in was simply c_out - the fraction of c_out that was removed per cycle. so (x/1000 - .1x/1000)

ends up being:

dx/dt = 50(x/1000-.1x/1000)-50(x/1000) which simplifies to

dx/dt = -.005x and so on from there. (I'm writing this based off memory right now so there may be some subtle errors. lol)

6. ## Re: Solving Mixture Problem using 1st Order Lin Dif. Eq.

So the solution to the DE is... what?