23–26. Stirred tank reactions For each of the following stirre...

Question

23–26. Stirred tank reactions For each of the following stirred tank reactions, carry out the following analysis.

b. Solve the initial value problem.

A one-million-liter pond is contaminated by a chemical pollutant with a concentration of 20 g/L. The source of the pollutant is removed, and pure water is allowed to flow into the pond at a rate of 1200 L/hr. Assuming the pond is thoroughly mixed and drained at a rate of 1200 L/hr, how long does it take to reduce the concentration of the solution in the pond to 10% of the initial value?

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says a 2 million L reservoir is polluted with a chemical at a concentration of 25 g per liter. Clean water flows in at 2500 L per hour, and the mixture is drained at the same rate. How long will it take for the concentration to fall to 20% of its initial value? Round the answer to the nearest whole number. So, we're going to define PFT as being equal to the mass of the chemical at time T in our reservoir. So, we'll first look for the initial condition. So, how much mass did we initially have of this chemical in our reservoir? And so that's going to be P of 0. MP of 0 is going to be equal to the volume of our reservoir, which is 2 million L. And then this gets multiplied by the chemical concentration, which we're told the problem is 25 g per liter initially. And so that means that P of 0 is going to be equal to. 50 million g. So that is initially how much mass of this chemical that we have in our reservoir. Next, we need to determine the rate at which the mass in our reservoir of this chemical is changing. So that is going to be the derivative of P with respect to T. And this is just going to be equal to the rate at which the chemical is coming into the reservoir, minus the rate at which the chemical is going out of the reservoir. So, if we look at the rate at which the chemical is coming into the reservoir, this is just going to be equal to the rate at which clean water is flowing into the reservoir, which we're told the problem is going to be 2500 L per hour. And then this needs to be multiplied by the chemical concentration that is flowing into the reservoir. However, we have clean water flowing into the reservoir, so that has no chemical concentration in it. So that is going, so we're going to multiply our flow rate here by 0 g per liter, since we have 0 g of the chemical coming into the reservoir. So that means that the rate of the chemical coming into the reservoir is going to be equal to 0 g per hour. Now, if we look at the rate, At which the chemical is going out of the reservoir. This is going to be equal to. The flow rate of 2500 L per hour, since we're told that the mixture is being drained at the same rate that the clean water is flowing in, but this needs to be multiplied by the concentration of the chemical in the reservoir at this particular time. So we'll multiply this by the mass of the chemical in the reservoir, which is going to be PFT, and that's going to be divided by The total number of liters in the reservoir, which is 2 million. Since the rate at which we're draining the mixture, and the rate at which we're adding clean water are the same, that means that the volume of liquid in the reservoir remains constant. So that means that our outgoing rate is going to be equal to. P T Divided by 800, and this has units of grams per hour. So that means that PFT. Sorry, the derivative of PFT with respect to T, is going to be equal to. Minus 1 divided by 800 multiplied by P of T. So now we need to solve this differential equation for PFT. And we noticed that this differential equation has the form of the derivative of P with respect to T, that is equal to A multiplied by P of T. Plus B. And this type of differential equation has the general solution of P of T is equal to C multiplied by E, raises the quantity of A multiplied by T in quantity minus B divided by A. Now, in our case, A is going to be equal to. -1 divided by 800, and B is going to be equal to 0. So that means that the term of B divided by A is going to be equal to 0. So, our solution to our differential equation is going to be P of T. is equal to C multiplied by E raised to the quantity of minus 1 divided by 800. Multiplied by T in quantity. So now we need to use our initial condition in order to find C. So, P of 0 is equal to 150 million. So that means that 150 million. is going to be equal to. C multiplied by E raised to the quantity of minus 1 divided by 800 multiplied by 0 in quantity. Now, since E 0 is just 1, that means that C is going to be equal to 50 million. So, therefore, P of T is going to be equal to. 50 million Multiplied by E raised to the quantity of minus 1 divided by 800. Multiplied by T in quantity. Now we need to find the time at which the concentration of the chemical drops to 20% of its initial value. So that means that the mass has dropped to 20% of its initial value. So that means that PFT is going to be equal to 0.2, multiplied by 50 million. And this is going to be equal to. 50 million Multiplied by E raised to the quantity of -1 divided by 800 multiplied by T in quantity. So now we need to solve this equation for T. So dividing both sides of the equation by 50 million, we'll get that 0.2 is equal to E raised to the quantity of minus 1 divided by 800 multiplied by T and quantity. And taking the natural log of both sides, we get that the natural log of 0.2. is equal to -1 divided by 800 multiplied by T. And to solve for T, we'll just multiply both sides of the equation by -800. That means that T is going to be equal to -800 multiplied by the natural log of 0.2. And when we evaluate this using our calculator, we see that T is going to be approximately equal to 1,288 hours. And so that is the answer that we're looking for for this problem, and here we've found to the nearest whole number. So I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Formulating the Differential Equation for Mixing Problems

Solving Initial Value Problems (IVPs)

Exponential Decay and Time to Reach a Specific Concentration

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