1. Mixture Problem: Consider the cascade of two tanks shown...

1. Mixture Problem: Consider the cascade of two tanks shown in the following figure. Pure water flows into the first tank. The contents of the first tank flow into the second, and the contents of the third tank flow out. All three of these flow rates are 10 gal/min. V1 = 150 gal is the fixed volume of the first tank, and V2 = 200 gal is the fixed volume of the second tank. Each tank initially contains 25 lb of salt, and the rest is filled with water. Therefore, each tank contains a brine solution, with a greater initial concentration of salt in the first tank than in the second. Denote x(t) as the amount of salt in the first tank at time t, and y(t) as the amount of salt in the second tank at time t. Tank 1 Volume V1 Amount x Tank 2 Volume V2 Amount y A cascade of two tanks. (a) Find x(t). Hint: Start by finding dx/dt. (b) Find y(t). Hint: Show first that dy/dt = x/15 - y/20 and then solve for y(t), using the function x(t) found in part (a). (c) The water in the second tank is used for brine-curing leather. In order to be prepared properly, the leather must be placed in the tank when the concentration of salt is at its maximum. Find the time at which this maximal concentration occurs. (Use a calculator to round your answer to the nearest hundredth of a minute.)

Transcript

00:02 So this problem here is a quite an involved problem, so it's going to take us a little bit of time to set it up.

00:08 But hopefully once we've got everything in place, it should be fairly straightforward to get through.

00:12 So we're given this cascaded tank system.

00:15 So i'll start by trying to draw a little figure.

00:18 So this is our pipe going into the first tank.

00:22 It's our first tank.

00:25 And this second tank has a pipe coming out to it, which flows into a second tank.

00:36 Something like this.

00:38 And this final tank indeed also has a pipe draining out of it.

00:45 And now each tank has a volume that's given.

00:50 So this tank has a volume v1, which is equal to 150 gallons.

01:00 This tank v2 is equal to 200 gallons.

01:11 And into this tank, pure water is flowing.

01:17 And each tank has some salt in it to start with.

01:20 So the water that flows out of these two is salty.

01:25 And these rates, these flow rates, are all the same.

01:29 Let's just denote the flow rate with this letter k and it's equal to 10 gallons per minute.

01:44 And each tank, the amount of salt in each tank, we're going to be denoting by a variable of time.

01:52 So the amount of salt in the first tank is going to be given by x as a function of t, and the amount of salt in the second tank is going to be given by y as a function of t.

02:05 We're also told of how much salt is in each tank initially.

02:09 So we also are given about the x at zero is equal to y of zero, which is equal to 25 pounds.

02:24 Right.

02:28 And so that's all the information we're given.

02:31 They also point out this fact that because the flow rate between each tank is the same, the volume of each tank remained constant, which you should be able to convince yourself of given the fact that the amount of water coming out is the same as the amount of water coming in, the total amount of water from each tank is the same.

02:49 And so our first question is to find a function for x of t.

02:55 And the way we're going to do this is by writing down, a differential equation for it and then solving that.

03:01 So the way we're going to do this is we're going to think about the amount of salt in the tank x at a time very slightly in the future compared to initially.

03:16 So this is delta t, which is just representing a small increment in time.

03:22 And now because this is a small increment in time, we can argue that we would expect this to be approximately equal to the amount of salt that was in there at time t.

03:37 And then from this, it needs to be reduced by something.

03:40 And what's it going to be reduced by? it's going to be reduced by.

03:43 This is going to write this as words first, so we can justify it, and then we'll work out the expression later.

03:51 The amount of salt lost per minute, this times by delta.

04:10 Because that's the amount of time that's relaxed.

04:12 So this is some rate of loss times by the amount of time.

04:17 So that's the amount we would have lost.

04:18 So this is an approximately true statement.

04:22 And as we let delta t go to zero, you can see that this statement will become exact.

04:26 And we're going to use that in a market argument to argue what this should be in terms of a derivative.

04:32 So first we want to write down an expression for the amount of salt lost per minute.

04:36 So we can start by simply writing down the amount of salt per gallon, which is equal to x of t divided by the volume, v1 in this case.

05:07 And we also know we have our rate of salt lost, which is in gallons per minute.

05:14 And so what we're looking for, which is the amount salt loss.

05:27 Per min is going to be equal to k times x of t over v1 and so now we can write down that x of t plus delta t is approximately x t minus k x x of t over v1 delta t.

06:04 I can now rearrange this expression.

06:06 I'm just going to start writing down here, so that i can write x of t plus delta t minus x of t divided by delta t is approximately equal to minus k x of t over v1.

06:32 And now in the limit that delta goes to 0, we should be able to recognize that this becomes the derivative of x.

06:44 So this is just the definition of what the derivative looks like in the infinitesimal representation.

06:51 So this goes to tell us that dx, d t, is equal to minus k x x over v1.

07:05 And then putting in our numbers, this implies, so this little double arrow sign means implies, it implies that dx, dt is just equal to minus x over 15, where i dropped the t just to make this easier.

07:27 So that's our first step, is to derive our dx, dt.

07:31 And now we want to solve this equation, which we can do just by separating the variables.

07:35 So i'm going to switch colour for solving it.

07:41 We can rewrite this as 1 over x dx is equal to minus 1 over 15 d t, and then we can integrate both sides.

08:06 So we have the integral 1 over x d x which is equal to minus 1 over 15 the integral of d t.

08:20 I've just brought this constant out the front.

08:24 Now the integral of 1 over x, we should all know, is just the logarithm, the natural logarithm of x.

08:30 This is ln and x.

08:32 This is equal to the integral of d t is just t.

08:37 So it's minus t over 15 plus our constant of integration.

08:44 Now i can rearrange this by taking the exponential of both sides to get rid of this logarithm.

08:49 And so this tells me that x of t is equal to e to the minus t over 15 plus c and i can rewrite this as just a uh e to the minus t over 15 where big a is just e to the c and if i substitute in t equals zero we see we see that a has to actually be equal to x of 0, which we know is equal to 25 pounds.

09:35 And so our solution for the first part is that x of t is equal to 25 e to minus t over 15.

09:48 So that's our answer to part a.

09:54 Now for b, we have to find y of t...

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