Suppose brine containing 0.2 kg of salt per liter runs into a tank initially filled with 500 L of water containing 5 kg of salt. The brine enters the tank at a rate of 5L/min. The mixture, kept uniform by stirring, is flowing out at the rate of 5 L/min. (The same conditions as in the previous problem). After 10 min, a leak develops in the tank and an additional liter per minute of mixture flows out of the tank.

(a) Find the concentration, in kilograms per liter, of salt in the tank after 10 min. [Hint: Let A denote thenumber of kilograms of salt in the tank at t minutes after the process begins and use the fact that rate of increase in A = rate of input - rate of exit.

(b) After 10 min, a leak develops in the tank and an additional liter per minute of mixture flows out of the tank. What will be the concentration, in kilograms per liter, of salt in the tank 20 min after the leak develops?

Answer:

(a) 0.288 kg/liter

(b) 0.061408 kg/liter

Step-by-step explanation:

(a) The mass of salt entering the tank per minute, x = 0.2 kg/L × 5 L/minute = 1 kg/minute

The mass of salt exiting the tank per minute = 5 × (5 + x)/500

The increase per minute, Δ/dt, in the mass of salt in the tank is given as follows;

Δ/dt = x - 5 × (5 + x)/500

The increase, in mass, Δ, after an increase in time, dt, is therefore;

Δ = (x - 5 × (5 + x)/500)·dt

Integrating with a graphing calculator, with limits 0, 10, gives;

Δ = (99·x - 5)/10

Substituting x = 1 gives

(99 × 1 - 5)/10 = 9.4 kg

The concentration of the salt and water in the tank after 10 minutes = (Initial mass of salt in the tank + Increase in the mass of the salt in the tank)/(Volume of the tank)

∴ The concentration of the salt and water in the tank after 10 minutes =  (5 + 9.4)/500 = (14.4)/500 = 0.288

The concentration of the salt and water in the tank after 10 minutes = 0.288 kg/liter

(b) With the added leak, we now have;

Δ/dt = x - 6 × (14.4 + x)/500

Δ = x - 6 × (14.4 + x)/500·dt

Integrating with a graphing calculator, with limits 0, 20, gives;

Δ = 19.76·x -3.456 = 16.304

Where x = 1

The increase in mass after an increase in = 16.304 kg

The total mass = 16.304 + 14.4 = 30.704 kg

The concentration of the salt in the tank then becomes;

Concentration = 30.704/500 = 0.061408 kg/liter.

Calculating the salt concentration per liter of water we will find:

(a) [tex]0.288 kg/liter[/tex]

(b) [tex]0.061408 kg/liter[/tex]

The mass of salt entering the tank per minute:

[tex]X= (0.2)( 5) = 1 kg/minute[/tex]

The mass of salt exiting the tank per minute:

 [tex]\frac{5 * (5 + x)}{500}[/tex]

The increase per minute, Δ/dt, in the mass of salt in the tank is given as follows;  

[tex]\frac{\Delta}{dt} = \frac{x - 5 * (5 + x)}{500}[/tex]

The increase, in mass, Δ, after an increase in time, dt, is therefore;

[tex]\Delta = (x - 5 * (5 + x)/500)dt[/tex]

Integrating with a graphing calculator, with limits 0 at 10, gives;  

[tex]\Delta = (99X - 5)/10[/tex]

Substituting x = 1 gives:

[tex](99)( 1 - 5)/10 = 9.4 kg[/tex]

The concentration of the salt and water in the tank after 10 minutes = (Initial mass of salt in the tank + Increase in the mass of the salt in the tank)/(Volume of the tank). The concentration of the salt and water in the tank after 10 minutes:

[tex](5 + 9.4)/500 = (14.4)/500 = 0.288[/tex]

The concentration of the salt and water in the tank after 10 minutes = 0.288 kg/liter.

(b) With the added leak, we now have;

[tex]\frac{\Delta}{dt} = \frac{ (x - 6 ) (14.4 + x}{500} \\\Delta = \frac{(x - 6)(14.4 + x)}{500} dt[/tex]

Integrating with a graphing calculator, with limits 0 at 20, gives;

[tex]\int\limits^{20} _0 {\frac{(x - 6)(14.4 + x)}{500} dt}[/tex]

Where x = 1:

[tex]\Delta = 19.76(1) -3.456 = 16.304[/tex]

The increase in mass after an increase in = 16.304 kg. The total mass:

[tex]16.304 + 14.4 = 30.704 kg[/tex]

The concentration of the salt in the tank then becomes;

[tex]30.704/500 = 0.061408 kg/liter[/tex]

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