**Problem Statement:**A 50-gallon tank is full to the brim with pure water, and 5 gallons per minute of a brine solution with 0.3 kg/gallon salt flows into it. Since the tank is full, 5 gallons per minute of well-mixed solution flows onto the ground. Find a differential equation with the initial condition for $ S(t) $ = number of kg salt in the tank and solve it for $ S(t) $.**Explanation:**We have a tank that initially contains pure water and a brine solution flows into it while an equal amount of the mixed solution exits. The brine solution has a concentration of 0.3 kg of salt per gallon. The task is to determine the function $ S(t) $, which represents the amount of salt in the tank at any time $ t $, and satisfy the conditions given by setting up and solving a differential equation.

Answered: A 50 gallon tank is full to the brim…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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**Problem Statement:**

A 50-gallon tank is full to the brim with pure water, and 5 gallons per minute of a brine solution with 0.3 kg/gallon salt flows into it. Since the tank is full, 5 gallons per minute of well-mixed solution flows onto the ground. Find a differential equation with the initial condition for $ S(t) $ = number of kg salt in the tank and solve it for $ S(t) $.

**Explanation:**

We have a tank that initially contains pure water and a brine solution flows into it while an equal amount of the mixed solution exits. The brine solution has a concentration of 0.3 kg of salt per gallon. The task is to determine the function $ S(t) $, which represents the amount of salt in the tank at any time $ t $, and satisfy the conditions given by setting up and solving a differential equation.

Expert Solution

Step 1

Let s(t) be the amount of salt (in kg) in the tank at time t.

Now, given that salt is entering in the tank at a rate of 5 gallon/ minute

then $R_{i n} = c o n c e n t r a t i o n \times r a t e = (0.3) \times 5 = 1.5 k g / m i n u t e$

Now, given that 5 gallon/minute of well mixed solution flows out, since we know at any time t amount of salt in water is s then $R_{o u t} = 5 \times s = 5 s k g / m i n u t e$

Initially that is, at time t=0 volume of water is 50 gallon (given)

Also, given that brine is flow in, at rate 5 gallon /minute and flow out, at rate 5 gallon/minute then

net increase in volume = (5-5)=0 gallon /minute.

So the volume V of brine in tank at time is $50 + 0 \times t = 50$ gallon

The rate of change in amount of salt $\frac{d s}{d t} = R_{i n} - \frac{R_{o u t}}{V (v o l u m e)}$ then

$\frac{d s}{d t} = R_{i n} - \frac{R_{o u t}}{V} = 1.5 - \frac{5 s}{50}$ ----(1),

Since, given that initially (t=0) tank is full of pure water then amount of salt in water is 0 that is, $s (0) = 0$ --(2)

so, the initial value problem is

$\frac{d s}{d t} = R_{i n} - \frac{R_{o u t}}{V} = 1.5 - \frac{5 s}{50}$ ; $s (0) = 0$

$\frac{d s}{d t} = 1.5 - \frac{5 s}{50}; s (0) = 0$ ----(3)

This is the required differential equation with initial condition.

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