# Differential equations problem, flow rates, constantly changing volume

1. Aug 3, 2008

### iceera88

1. The problem statement, all variables and given/known data

Beer containing 6% alcohol per gallon is pumped into a vat that initially contains 350 gallons of beer at 3% alcohol. The rate at which the beer is pumped in is 3 gallons per minute, whereas the mixed liquid is pumped out at a rate of 4 gallons per minute. Find the number of gallons of alcohol A(t) in the tank at any time. What is the percentage of alcohol in the tank after 60 minutes?

2. Relevant equations

dA/dt = rate in - rate out

A is the number of gallons of alcohol in the tank

3. The attempt at a solution

dA/dt = rate in - rate out = 0.06r - (r*A)/(350-t)

I'm not sure about my setup because so far we've only worked with problems in which the volume doesn't constantly change, ie 250 gallons in the tank at all times with r(in) = r(out) = 3 gal/min rather than this problem where r(in) = 3 gal/min and r(out) = 4 gal/min and the volume of liquid in the tank decreases by 1 gal/min

I've worked the problem from the initial conditions:

A(0) = 10.5 gallons

to be:

A(t) = (21-t)*exp[-rln(350-t)] - 21*exp[-r-ln(350)] +10.5

but I'm seriously doubting what I've done so far because there's nowhere to check my answers, and my method seems to make this problem far more complicated then my professor ever makes his problems.

Last edited: Aug 3, 2008
2. Aug 3, 2008

As far as I can tell, you've set up the equation correctly. It was a little confusing the way you used the same r, but I'm pretty sure it is correct as long as you assume that the added beer mixes instantly in with the rest. If it doesn't mix with the rest then the equation would be:
$$\frac{dA}{dt}=.06r_{in}-.03r_{out}$$ but that isn't much of a differential equation so I think that the first assumption is what they are looking for.

The problem I found is that I didn't get the same answer as you. I just used Mathematica (the cheating way) to solve it and didn't quite get the same answer.

3. Aug 4, 2008

### Defennder

Well you know what the rate at which beer is being poured and siphoned off, so why do you have to use r here? Also, r isn't the same for the inflow as well as outflow of beer so your DE isn't right to begin with.

Last edited: Aug 4, 2008
4. Aug 4, 2008

Yes, that's what I was saying about the r's. If he just uses $$r_{in}$$ for the first r and $$r_{out}$$ for the second r then the equation looks ok to me.

5. Aug 4, 2008

### iceera88

Thanks! I think I'll just plug in those values and make it easier, then. I think I just accidentally split it up as if r were the same throughout the problem when I tried to solve symbolically.

6. Aug 4, 2008

### iceera88

With the new setup as you've described, I got:

A(t) = 15.75-0.045t - C/exp[-4*ln(350-t)]
C = 3.50e-10
A(t) = 15.75-0.045t - (3.50e-10)/exp[-4*ln(350-t)]
A(60) = 3.65%