MATH 2410 Review of Mixing Problems David Nichols The following examples explore two different kinds of mixing problems. The word problems are very similar, but the differential equations that result are solved using different methods. After the examples, there is an exercise for you to try on your own. xample 1 Suppose we start with a 100-gallon vat of pure water and then start pouring in ethanol at a rate of 2 gallons/minute. Some rotors in the vat mix the ethanol and water thoroughly. At the same time that we start pouring in the ethanol, we also open a drain at the bottom of the vat, so that the mixture of water & ethanol drains out at a rate of 2 gallons/minute. Set up a differential equation for the quantity (in gallons) (t) of ethanol in the vat at time t (in minutes), and then solve the differential equation. The differential equation we set up will look like this: = in out. Our job is to express the rate of ethanol moving in and out of the vat in mathematical terms. The in part is easy, because the problem told us the ethanol is poured in at a rate of 2 gallons/minute. So now we have this: = 2 out. What s the rate of ethanol leaving the vat? Well, we know that the mixture leaves the vat at 2 gallons/minute. But what proportion of that is ethanol? We have a name for the amount of ethanol in the whole vat:. Since the ethanol is mixed throughout the whole vat, and there are 100 gallons of mixture in the vat (notice this doesn t change because every minute 2 gallons enter the tank and 2 gallons leave the tank), the concentration of ethanol is 100. So the differential equation we re after is ( ) = 2 2 = 2 100 50.
How do we solve this? The first thing we check is whether the equation is separable, since separable equations are (usually) the easiest to solve. Sure enough, we can separate the s and the t s and then integrate: 2 /50 = 1 2 /50 = 1 50 ln 2 50 = t + c 1 ln 2 50 = t 50 + c 2 2 50 = c 3e t/50 2 50 = c 4e t/50 50 = 2 c 4e t/50 (t) = 100 c 5 e t/50. Now we should figure out what the unknown constant c 5 is. To figure out this sort of thing, we normally plug in an initial condition. What s the initial condition here? We started with a tank of pure water, so at time t = 0 there was no ethanol. In other words, (0) = 0. We can plug this in: So c 5 = 100. Thus our final answer is 0 = 100 c 5 e 0 = 100 c 5. (t) = 100 100e t/50.
xample 2 Suppose we start with a 200-gallon vat of pure water and then start pouring in acetic acid (AcOH) at a rate of 4 gallons/minute. Some rotors in the vat mix the acetic acid and water thoroughly. At the same time that we start pouring in the acetic acid, we also open a drain at the bottom of the vat, so that the mixture of water & acetic acid drains out at a rate of 5 gallons/minute. Set up a differential equation for the quantity (in gallons) A(t) of acetic acid in the vat at time t (in minutes), and then solve the differential equation. Once again, we want to set up an equation of the form = in out, and once again we can start filling in numbers from the word problem: = 4 gal } min {{} rate of AcOH in 5 gal (concentration of acetic acid). } min {{} rate of mix out What is the concentration of acetic acid in the vat? It s the amount of acetic acid, which we are calling A, divided by the total volume of the mixture. But be careful here! The volume isn t 200. The volume starts at 200 gallons, but every minute thereafter we add 4 gallons and subtract 5 gallons. In other words, the vat loses 1 gallon every minute, so the actual volume is 200 t. Now we can complete the differential equation: ( ) A = 4 5. 200 t This time the equation is not separable. In fact, it s first order linear, so our first step to solve it is to write all the terms with A in them on the left and everything else on the right: + 5 A = 4. 200 t }{{} p(t) Next, we find an integrating factor using the formula I = e p(t), where p(t) is shorthand for the coefficient of A in the equation. I = e 5 200 t 5 ln 200 t = e = 200 t 5.
It s worth pausing here to discuss the absolute value. Usually, these are very important. But in this case, the absolute value doesn t matter. Why? Because in the physical system we re modeling, 200 t represents the volume of mixture left in the mixing vat. We can never have a negative number of gallons left. So 200 t will never be negative, and we can ignore the absolute value and just write I = (200 t) 5. The next step in our solution is to multiply the whole equation by the integrating factor: 5 (200 t) + 5(200 t) 6 A = 4(200 t) 5. Once we ve done that, the left hand side of the equation always works out to be a case of the product rule, and always follows the same pattern: 5 (200 t) + 5(200 t) 6 A = d ( (200 t) 5 A ). this is the product rule So we can rewrite the left hand side of the equation and just integrate: d ( (200 t) 5 A ) = 4(200 t) 5 d ( (200 t) 5 A ) = 4(200 t) 5 (200 t) 5 A = (200 t) 4 + C A(t) = 200 t + C(200 t) 5. We solve for C by plugging in the initial condition A(0) = 0 as before: 0 = 200 0 + C(200) 5, so C = 200 4. Now we can write out the complete answer: A(t) = 200 t 200 4 (200 t) 5. It is helpful to graph the solution, in particular so that we can visualize whether our solution makes physical sense. A graph appears atop the next page.
y 100 50 x 50 100 150 200 xercise A drain is opened in a 150-liter tank of pure water so that the contents of the tank pour out at a rate of 3 liters/minute. At the same time, a chute is opened so that a solution of 50% hydrazine and 50% water starts pouring into the tank at a rate of 3 liters/minute. During this process, the contents of the tank are thoroughly mixed. Write out a differential equation describing the amount N(t) of hydrazine (in liters) over time (in minutes). Then solve the differential equation for N(t).