2. A 100 L mixing tank initially contains 75 L of brine solution with 300 g of dissolved salt. Brine solution with a concentration of 12 g/L flows into the tank at a rate of 5 L/min. The solution inside the tank is kept well- stirred and is discharged at the same rate. Determine the a) concentration of brine inside the tank after 10 minutes, and b)time it takes to achieve a concentration of 10 g/L.

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Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
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Chapter9: Surfaces And Solids
Section9.3: Cylinders And Cones
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application of differential equations salt solution

2. A 100 L mixing tank initially contains 75 L of brine solution with 300 g of dissolved salt. Brine solution with
a concentration of 12 g/L flows into the tank at a rate of 5 L/min. The solution inside the tank is kept well-
stirred and is discharged at the same rate. Determine the a) concentration of brine inside the tank after 10
minutes, and b)time it takes to achieve a concentration of 10 g/L.
Transcribed Image Text:2. A 100 L mixing tank initially contains 75 L of brine solution with 300 g of dissolved salt. Brine solution with a concentration of 12 g/L flows into the tank at a rate of 5 L/min. The solution inside the tank is kept well- stirred and is discharged at the same rate. Determine the a) concentration of brine inside the tank after 10 minutes, and b)time it takes to achieve a concentration of 10 g/L.
Expert Solution
Step 1

Given,

  • The volumetric flow rate at the entrance, r1=5 L/min
  • The volumetric flow rate at the end, r2=5 L/min
  • The concentration of the substance at the entrance, c1=12 g/L
  • The initial volume of solution at t=0V0=75 L

Since the given problem involves mix of non reactive fluids, the equation will be as follows:

Rate of change of substance in a volume=Rate of entrance-Rate of exit

dxdt=r1c1-r2c21

Since c2 is not given, equation 1 can be rewritten as,

dxdt=r1c1-r2xV

where V=V0+r1-r2t

Hence,

dxdt=r1c1-r2xV0+r1-r2t

Substitute the given values in the equation above.

dxdt=5 12-5x75+5-5t=60-5x75=60-x15=900-x151

Equation 1 is a separable differential equation.

 

 

 

Step 2

From 1,

dx900-x=dt15dx900-x=dt15-ln900-x=t15+c

To find c, use the fact that at time t=0, x=4 g/L.

Therefore,

-ln900-x0=015+c-ln900-4=c-ln896=c

Therefore,

-ln900-x=t15-ln896ln900-x=ln896-t15eln900-x=eln896-t15900-x=eln896et15x=900-896et15

Therefore, the concentration of brine inside the tank after t minutes is given by :

xt=900-896et15

Therefore the concentration of brine inside the tank after 10 minutes is,

x10=900-896e1015=900-460.02=439.99 g

Observe that the since the rate of input and the rate of discharge is the same the mixing tank contains 75 L of brine solution even after 10 minutes.

Therefore, the concentration of brine inside the tank after 10 minutes is,

439.99 g/75 L or equivalently 5.87 g/L

 

 

 

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