Differential Equation (Water/Solution Tank) problem

[Fragster]

Homework Statement

A tank initially contains 200 gallons of fresh water, but then a salt solution of unknown concentration lb/gal is poured into the tank at 2 gal/min. The well-stirred mixture flows out of the tank at the same rate. After 120 minutes, the concentration of salt in the tank is 1.4 lb/gal. What is the concentration (in lb/gal) of the entering brine?

Homework Equations

Net rate of salt solution = rate of incoming salt solution - rate of outgoing salt solution

The Attempt at a Solution

I actually have the solution

Spoiler

280/(200-200(e^(-6/5))) or roughly, 2

but I cannot remember for the life of me how I got it :/

I think the biggest problem I have with the problem is the relationship between the rate of salt going in (x) and the net rate of salt.

 

[mighty2000]

Hi there,

First, let's define some variables:
- x = concentration of the entering brine (in lb/gal)
- y = rate of incoming salt solution (in lb/min)
- z = rate of outgoing salt solution (in lb/min)

We know that the net rate of salt solution is equal to the rate of incoming salt solution minus the rate of outgoing salt solution. So we can write the following equation:

y - z = 1.4 lb/gal * 2 gal/min

Next, we need to find the relationship between x, y, and z. We know that the concentration of salt in the tank is increasing at a rate of 1.4 lb/gal every 2 minutes. This means that for every 2 minutes, 1.4 lb of salt is being added to the tank. Since we know that the tank initially contained 200 gallons of fresh water, we can write the following equation:

x * 2 gal/min * 120 min = 200 gal * 0 lb/gal + 1.4 lb/gal * 2 gal/min * 120 min

Solving for x, we get:

x = 0.014 lb/gal

Therefore, the concentration of the entering brine is 0.014 lb/gal.

I hope this helps! Let me know if you have any further questions or if you would like me to explain any part of the solution in more detail.