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Mixing. Suppose a brine containing 0.3 kilogram (kg) of salt per liter (L) runs into a tank initially filled with 400 L of water containing 2 kg of salt. If the brine enters at 10 L/min, the mixture is kept uniform by stirring, and the mixture flows out at the same rate. Find the mass of salt in the tank after 10 min (see Figure 2.4). [Hint: Let A denote the number of kilograms of salt in the tank at t min after the process begins and use the fact that rate of increase in $$A =$$ rate of input - rate of exit. A further discussion of mixing problems is given in Section 3.2.]

   Mixing. Suppose a brine containing 0.3 kilogram (kg)
of salt per liter (L) runs into a tank initially filled with
400 L of water containing 2 kg of salt. If the brine enters
at 10 L/min, the mixture is kept uniform by stirring, and
the mixture flows out at the same rate. Find the mass of
salt in the tank after 10 min (see Figure 2.4). [Hint: Let
A denote the number of kilograms of salt in the tank at t
min after the process begins and use the fact that
rate of increase in $$A =$$ rate of input - rate of exit.
A further discussion of mixing problems is given in
Section 3.2.]

Fundamentals of Differential Equations

R. Kent Nagle 9th Edition

Chapter 2, Problem 33

Instant Answer

Step 1

We can write this as a differential equation: \[\frac{dA}{dt} = \text{rate of input} - \text{rate of exit}\] Show more…

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00:01 All right, so we've got a tank that has fluid entering and exiting, and we're told that there's initially 400 liters of brine inside that contains two kilograms of salt.

00:28 And the flow coming in is at 10 liters per minute, containing 0 .3 kilograms per liter, and exiting at the same rate of 10 liters.

00:52 Okay, and what we want to do is we want to determine the amount of salt that will be in here at 10 minutes since the procedure begins.

01:01 So we're even given a hit here that says that the rate of increases the rate of input minus the rate of exit.

01:06 So we can write the change of salt over time is equal to the rate of input.

01:16 And so that would just be 10 liters per minute multiplied by 0 .3 kilograms.

01:29 Meters these would cancel out subtracted by the rate of exit which we know would be the amount of salt over the initial volume multiplied by the rate of the fluid exiting the tank and now we have the same units of kilograms for minutes and that'll give us three here and a over fourth okay now we're gonna refer back to our first order differential equations and remind ourselves the best practice here is to move all the variables on one side so that we can get our standard equation and we know our standard equation is p t a okay and that would the a would just change based on the variable.

03:08 So if you were doing another problem where the variable was a y, it would be a y in place of an a, but these remain the same as the constants.

03:16 So we know that then pt must equal to from here, we know that it must be equal to 1 over 40, and we know qt must be equal to 3, right? as we see that 3 there up here, okay? all right.

03:52 Now what we can do is we can solve for our integrating factor you know the integrating factor equation equal to be in integral p t d t the respect to time but and that would just be equal to e of the integral 1 over 40 d t is equal to e that's an integrating factor there and now we can solve for our a respect to time as we know there's an equation for that which is just our integrating factor so i won over our integrating factor multiplied by the integral of our integrating factor multiplied by qt d t okay so we can plug those values in and we'll get get one over e t over e t over 40 to the to the to the to the 40 integral of e t to the 40 t over 40 three times three with respect all right now we can pull the three out because it is a constant we know that integral there with respect to time would just be equal to 40 e and we want to make sure that we don't forget that there's a constant that comes with when we integrate.

06:17 Okay.

06:18 Now, we know that e over t40 will cancel out here and here, but we would still have to carry it there.

06:29 So you'd have 120 plus c1 over.

06:39 Okay.

06:40 Now we need to solve for c1, and we can do that by simply setting in our initial conditions, which we know at salt at initial, initially salt is two kilograms.

06:59 We'll move that from two kilograms.

07:02 It's equal to 120 plus c1, e0 to the 40...

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