Tough Integration Problem involving flow from a tank

[alonzo]

Please help! This problem is really complex and i have no idea where to start (or finish for that matter!)
A holding tank of 7000 litres is full of water which has been contaminated by a pollutant. The tank contains 0.01 percent contaminant by volume. Water with a contaminant concentration of 0.001 percent now flows from a river into the tank at a rate of 5 litres per minute. Since the tank is full, there is an overflow of water into a nearby lake.

1) What is the concentration of contaminant in the tank after 5 hours?

2) The farmer who owns the tank claims that the contents of the tank is under the legislated upper concentration limitation of 0.002 percent for the contaminant. If the inflow has been running for 4 hours, is the farmer's claim correct?

3) What is the volume of contaminant that has overflowed into the nearby lake after four hours?

Thanks 4 ur help!

 

[Tide]

Define the concentration of the water in the tank as amount of contaminant divided the volume (contaminant plus water).

If R is the rate (liters per minute) at which contaminated water of concentration [itex]C_{in}[/tex] enters the tank then the concentration of water in the tank is

$$C = C_{0}e^{-Rt/V} + \left(1 - e^{-R t / V} \right) C_{in}$$

where $C_0$ is the initial concentration of the contaminant in the tank and V is the volume of the tank.

This assumes that the contaminant becomes uniformly distributed throughout the tank immediately after it enters the tank. This may or may not be a good assumption in your particular case since you haven't identified what the contaminant is.

That should get you started!

 

[wais]

Hello, solving this integration problem may seem daunting at first, but with a little guidance, it can be broken down into smaller, manageable steps. Let's start by defining some variables and setting up the equations we will need to solve this problem.

Let's say the initial volume of contaminant in the tank is V0 and the concentration of contaminant in the tank is C0. We also know that the volume of water in the tank remains constant at 7000 litres.

1) To find the concentration of contaminant in the tank after 5 hours, we can use the formula C = V/C0, where V is the volume of contaminant and C0 is the initial concentration. We know that the inflow rate is 5 litres per minute, so after 5 hours (300 minutes), the volume of water that has entered the tank is 300 x 5 = 1500 litres. This means that the volume of contaminant in the tank is now V = 1500 x 0.001 = 1.5 litres. Plugging this into our formula, we get C = 1.5/7000 = 0.000214 or 0.0214%.

2) To determine if the farmer's claim is correct, we need to find the concentration of contaminant in the tank after 4 hours. Using the same formula as above, we get C = 1.2/7000 = 0.0001714 or 0.01714%. This is still below the legislated upper concentration limit of 0.002%, so the farmer's claim is correct.

3) To find the volume of contaminant that has overflowed into the nearby lake after 4 hours, we first need to calculate the total volume of water in the tank after 4 hours. Using the formula V = 7000 + (5 x 4 x 60) = 9400 litres. Since the concentration of contaminant in the tank after 4 hours is 0.01714%, the volume of contaminant in the tank is V = 0.01714 x 9400 = 161.996 litres. This means that the volume of contaminant that has overflowed into the nearby lake is 9400 - 7000 - 161.996 = 2238.004 litres.

I hope this helps you get started on solving this tough integration problem. Remember to break it down into smaller