Determining Concentration of a Tank (Using Linear Differential EQ)

[QuantumChemist]

Homework Statement

The problem states that we have a tank with a capacity of 400L filled with a mixture that has a concentration of 0.05g Chlorine per Liter. The concentration is reduced by pumping in pure water (zero concentration of chlorine), at a rate of 4L/s, and pumped out at 10L sec.

Homework Equations

I know that the rate of change in concentration is equal to the rate in minus the rate out.

The Attempt at a Solution

The rate of concentration input is zero since it's pure water, and I set the concentration of mixture out at x/(400-6t) since the volume is decreasing at some concentration at 6L per second due to the larger amount of mixture being pumped out. So the equation is then dx/dt=0.05-x/(400-6t), at least from what I understand.

Then, using the Linear Differential EQ method, I found I(x)=(400-6t)^-6. Multiplying both sides of the equation, I get (400-6t)^2(dx/dt)+(400-6t)^-7x=0.05(400-6t). The left side is suppose to resemble the product rule, but I'm thinking I really messed something up at some point.

Could someone point out my mistake so I can do a retake? Sorry if this is hard to read, I'm not good at inputting equations. Thanks for any help.

 

[vela]

Homework Statement

The problem states that we have a tank with a capacity of 400L filled with a mixture that has a concentration of 0.05g Chlorine per Liter. The concentration is reduced by pumping in pure water (zero concentration of chlorine), at a rate of 4L/s, and pumped out at 10L sec.

Homework Equations

I know that the rate of change in concentration is equal to the rate in minus the rate out.

The Attempt at a Solution

The rate of concentration input is zero since it's pure water, and I set the concentration of mixture out at x/(400-6t) since the volume is decreasing at some concentration at 6L per second due to the larger amount of mixture being pumped out. So the equation is then dx/dt=0.05-x/(400-6t), at least from what I understand.

You're saying the concentration of chlorine the mixture is equal to
$$\frac{x}{400-6t}.$$ Since V=400-6t is the volume of the liquid in the tank, $x$ must represent the total amount of chlorine in the tank. In that case, the units on the lefthand side would be g/s, but on the righthand side, the units would be g/L. The two sides of your equation can't be equal.

At what rate is chlorine leaving the tank when the concentration is x/(400-6t)? That's what you want to set equal to dx/dt.

Then, using the Linear Differential EQ method, I found I(x)=(400-6t)^-6. Multiplying both sides of the equation, I get (400-6t)^2(dx/dt)+(400-6t)^-7x=0.05(400-6t). The left side is suppose to resemble the product rule, but I'm thinking I really messed something up at some point.

Could someone point out my mistake so I can do a retake? Sorry if this is hard to read, I'm not good at inputting equations. Thanks for any help.

I couldn't follow what you did here, but I think you're trying to use an integrating factor. The correct differential equation is separable, so you shouldn't need to use this technique.