How Long to Dilute a 10% Salt Solution to 5% in a 1000 m³ Tank?

[decatte]

Mass Balance Question - Please HELP!

Homework Statement

A tank used to prepare 10% salt solution was shut down for the day. The next day, you were told that the effluent concentration should be 5 wt %. The tank volume is 1000 m^3, and it is completely filled with 10% solution. You decide to run pure water and remove the contents at the same rate until the concentration drops from 10% to 5% and readjust the new steady state flow rates of salt and water. How long it will take to decrease the concentration in the tank if you feed water at 100 lt/hour?

Homework Equations

Mass Balance Equation: dm(total)/dt=dm(1)/dt + dm(2)/dt + ...

The Attempt at a Solution

Firstly, it is not a steady state respect to salt, salt concentration is changing with time.

I tried to write mass balance (difference) equation respect to salt:

dm(salt)/dt= 0 - ?

Now, 0 (zero) because salt is not being added to system, conversely, it's concentration is lowerin'.

Please guys, help me. It's very important.

Thanks from now!

 

[anathema]

Thank you for your question. It seems that you are trying to determine how long it will take to decrease the salt concentration in the tank from 10% to 5% by running pure water and removing the contents at the same rate. This is a classic mass balance problem and can be solved using the following steps:

1. Determine the initial mass of salt in the tank: 10% of 1000 m^3 is 100 kg, so the initial mass of salt is 100 kg.

2. Set up a mass balance equation for the system, taking into account the inflow of water and the outflow of the salt solution. The equation will look like this:

dm(salt)/dt = (dm(water)/dt - dm(salt)/dt)

3. Since the concentration of salt in the tank is decreasing, dm(salt)/dt is negative, so the equation becomes:

-dm(salt)/dt = dm(water)/dt

4. We know that dm(water)/dt is 100 lt/hour, so we can plug that into the equation:

-dm(salt)/dt = 100 lt/hour

5. Now we need to find the rate at which the salt concentration is decreasing. This can be found by dividing the change in salt mass by the change in time. In this case, the change in salt mass is 100 kg (initial mass) - 50 kg (final mass) = 50 kg, and the change in time is what we are trying to find, so we can write:

-dm(salt)/dt = 50 kg / ?

6. Rearrange the equation to solve for time:

-dm(salt)/dt = 50 kg / ?

? = -50 kg / (-dm(salt)/dt)

? = 50 kg / dm(salt)/dt

7. Now we need to find dm(salt)/dt. We can do this by using the mass balance equation and plugging in the values we know:

dm(salt)/dt = (dm(water)/dt - dm(salt)/dt)

dm(salt)/dt = (100 lt/hour - dm(salt)/dt)

2dm(salt)/dt = 100 lt/hour

dm(salt)/dt = 50 lt/hour

8. Finally, we can plug this value into our equation to solve for time:

? = 50 kg

 

[Moetasim]

Hello, based on the information provided, we can approach this problem by using the mass balance equation. The mass balance equation states that the rate of change of total mass in a system is equal to the sum of the rates of change of individual components. In this case, we can write the equation as follows:

dm(total)/dt = dm(salt)/dt + dm(water)/dt

Where dm(total)/dt represents the rate of change of total mass in the tank, dm(salt)/dt represents the rate of change of salt in the tank, and dm(water)/dt represents the rate of change of water in the tank.

We know that the initial concentration of salt in the tank is 10% and we need to reach a concentration of 5%. Therefore, we can write the following equation for the rate of change of salt in the tank:

dm(salt)/dt = -10% * dm(total)/dt

This means that the rate of change of salt is equal to the negative of 10% times the rate of change of total mass in the tank. We can also write the equation for the rate of change of water in the tank as:

dm(water)/dt = 100 lt/hour

Since we are adding pure water at a rate of 100 lt/hour, the rate of change of water in the tank will be constant.

Now, we can substitute these equations into the mass balance equation and solve for the rate of change of total mass in the tank:

dm(total)/dt = -10% * dm(total)/dt + 100 lt/hour

Solving for dm(total)/dt, we get:

dm(total)/dt = 111.11 lt/hour

This means that the rate of change of total mass in the tank is 111.11 lt/hour. We can use this value to determine the time it will take to decrease the concentration in the tank. Since we know the volume of the tank is 1000 m^3, we can use the following equation to calculate the time:

t = V/dm(total)/dt

Substituting the values, we get:

t = (1000 m^3)/(111.11 lt/hour) = 9 hours

Therefore, it will take 9 hours to decrease the concentration in the tank from 10% to 5% by adding pure water at a rate of 100 lt/hour. I hope this helps