Question

A tank initially contains 400 gal of fresh water. At time \(t=0\), a brine solution with a concentration of \(0.1 \mathrm{lb}\) of salt per gallon enters the tank at a rate of \(1 \mathrm{gal} / \mathrm{min}\) and the well-stirred mixture flows out at a rate of \(2 \mathrm{gal} / \mathrm{min}\). (a) How long does it take for the tank to become empty? (This calculation determines the time interval on which our model is valid.) (b) How much salt is present when the tank contains \(100 \mathrm{gal}\) of brine? (c) What is the maximum amount of salt present in the tank during the time interval found in part (a)? When is this maximum achieved?

Short answer

Answer: It takes 400 minutes for the tank to become empty. Question: How much salt is present in the tank when it contains 100 gallons of brine? Answer: There are 59.5 lb of salt in the tank when it contains 100 gallons of brine. Question: When is the maximum amount of salt present in the tank and how much is it? Answer: The maximum amount of salt in the tank is 60 lb, and it occurs at 250 minutes.

Step-by-step solution

(1) Calculate the time it takes for the tank to become empty.

The tank is initially filled with 400 gal of fresh water. Brine is entering the tank at a rate of 1 gal/min and leaving at a rate of 2 gal/min. To determine how long it will take for the tank to empty, we need to find the time that the net outflow equals to the initial tank contents (since volume_in - volume_out = 0): Let \(t\) be the time in minutes, then: \(400 = (2-1) \cdot t\) Solve the equation to find t: \(t = 400 \mathrm{min}\) So, it takes 400 minutes for the tank to become empty.

(2) Formulate an equation for the amount of salt in the tank.

To find the amount of salt in the tank at a specific time or volume, first, we need a general equation. The rate of change in salt in the tank can be expressed as: \(\frac{dS}{dt} =\) (rate of salt entering) - (rate of salt leaving) Here, \(S\) is the amount of salt inside the tank, \(t\) is time, and \(\frac{dS}{dt}\) is the derivation of salt w.r.t time. The concentration of salt entering the tank is 0.1 lb per gallon, and the rate of entry is 1 gal/min. Therefore, the rate of salt entering the tank is: \(0.1 \cdot 1 = 0.1 \frac{\mathrm{lb}}{\mathrm{min}}\) Now, to find the rate of salt leaving, we need to use the rate of outflow and the current concentration of salt in the tank. The outflow rate is 2 gal/min. Suppose the concentration of salt in-tank is \(c = \frac{S}{V}\), where \(V\) is the volume of the solution. Therefore, the rate of salt leaving is: \(c \cdot 2 = \frac{S}{V} \cdot 2 = \frac{2S}{V}\) Now substitute values for rate of salt entering & leaving in the main equation, we have: \(\frac{dS}{dt} = 0.1 - \frac{2S}{V}\) We also need to express the volume \(V\) in terms of time \(t\). Initially, the volume of the water is 400 gal, and the net rate of outflow is 1 gal/min, so \(V(t) = 400 - t\). Now, replace \(V\) with \(400 - t\) and solve the differential equation: \(\frac{dS}{dt} + \frac{2S}{400 - t} = 0.1\)

(3) Solve the differential equation.

To solve this equation, we will use an integrating factor. First, rewrite the equation as: \(\frac{dS}{dt} + \frac{S}{200 - t/2} = \frac{1}{10}\) We see that the integrating factor \(\mathrm{IF} = e^{\int\frac{1}{200-t/2}dt}\) Compute the integral: \(\int\frac{1}{200-t/2}dt = -2\ln(200-t/2)\) So, \(\mathrm{IF} = e^{-2\ln(200-t/2)} = (200-t/2)^{-2}\) Multiply the original differential equation by the integrating factor and integrate both sides w.r.t \(t\): \((200-t/2)^{-2} \cdot \frac{dS}{dt} + (200-t/2)^{-1}\cdot S = \frac{1}{10}(200-t/2)^{-2}\) Integrate both sides: \(\int(200-t/2)^{-2} \cdot \frac{dS}{dt}dt + \int(200-t/2)^{-1}\cdot Sdt = \int\frac{1}{10}(200-t/2)^{-2}dt\) Calculate the integrals and solve for S(t): \(S(t) = -\frac{1}{200-t/2} - \frac{1}{5}t + C\) Here, \(C\) is a constant of integration.

(4) Find the initial condition of salt.

At \(t = 0\), the tank contains fresh water, so there is no salt in it. That means \(S(0) = 0\): \(0 = -\frac{1}{200} - \frac{1}{5}(0) + C\) Solve for C: \(C = \frac{1}{200}\) So, the equation for salt in the tank is: \(S(t) = -\frac{1}{200-t/2} - \frac{1}{5}t + \frac{1}{200}\)

(5) Calculate the amount of salt present when the tank contains 100 gal of brine.

At any time \(t\), the volume of the solution in the tank will be: \(V(t) = 400 - t\) When the tank contains 100 gal of brine, \(V(t) = 100\), so: \(100 = 400 - t\) Solve for t: \(t = 300 \mathrm{min}\) Now, calculate the amount of salt, \(S(300)\): \(S(300) = -\frac{1}{200-300/2} - \frac{1}{5}(300) + \frac{1}{200} = 59.5\) lb Therefore, there are 59.5 lb of salt in the tank when it contains 100 gal of brine.

(6) Calculate the maximum amount of salt present in the tank and when it is achieved.

To determine the maximum amount of salt present in the tank, we need to find the critical points of the function \(S(t)\). These critical points occur when the derivative of salt w.r.t time is equal to 0: \(\frac{dS}{dt} = 0\) Differentiate the equation for salt in the tank: \(\frac{dS}{dt} = \frac{1/2}{(200-t/2)^2} - \frac{1}{5}\) Set the derivative equal to 0: \(\frac{1/2}{(200-t/2)^2} - \frac{1}{5} = 0\) Solve for \(t\): \(t = 250 \mathrm{min}\) Now, calculate the maximum amount of salt, \(S(250)\): \(S(250) = -\frac{1}{200-250/2} - \frac{1}{5}(250) + \frac{1}{200} = 60\) lb Therefore, the maximum amount of salt in the tank is 60 lb, and this occurs at 250 minutes. The solution to the problem can be summarized as follows: (a) The tank becomes empty in 400 minutes. (b) There is 59.5 lb of salt in the tank when it contains 100 gallons of brine. (c) The maximum amount of salt in the tank is 60 lb, and it occurs at 250 minutes.

Key concepts

Initial Value Problem

Understanding initial value problems (IVPs) is a cornerstone of solving many practical mathematical problems, especially those relating to differential equations. An IVP consists of a differential equation together with a condition (an 'initial value') that specifies the value of the unknown function at a given point in its domain.

In the brine mixing problem provided, we encounter an IVP when we must solve the differential equation for the amount of salt in the tank, given that at time zero (\(t = 0\)), the tank starts with no salt. This initial condition is essential because it provides the necessary information to determine a unique solution to the equation. Without this piece of information, we might end up with an infinite number of possible solutions.

Integrating Factor

To handle differential equations that are not readily separable, an 'integrating factor' is an ingenious technique used to simplify and solve such equations. This method involves multiplying the entire differential equation by a carefully chosen function, which turns the left side of the equation into the derivative of a product of functions.

In our brine problem, the integrating factor is determined by the expression involving the net outflow of water from the tank. Mathematically, it involves the exponential of an integral that is related to the coefficients of the differential equation. By applying the integrating factor, we transition our differential equation into a more workable form, facilitating its integration and subsequent solving.

Separation of Variables

Separation of variables is a method for solving differential equations, where we manipulate the equation to isolate the different variables on opposite sides of the equation. This method simplifies the equation into two integrals that we can evaluate separately.

However, this method was not directly applicable in the brine problem because the rate of change of salt in the tank and the volume of water are interdependent and change over time. Instead, we had to employ the integrating factor technique after our attempts to separate variables.

Modeling with Differential Equations

Differential equations are mathematical models that describe the rate of change of quantities and are used extensively in science, economics, and engineering to model real-world phenomena.

In the context of the brine mixing problem, differential equations are used to model the dynamics of the salt concentration in the tank. The rate of change of the salt amount is a function of both the salt concentration in the incoming brine and the rate at which the mixture leaves the tank. The model we develop must consider these variable quantities and their relationship over time, highlighting the power of differential equations in capturing complex dynamic systems.