TEXT VERSION: S uppose a brine containing 0.2 kg of salt per liter runs into a t

Question

TEXT VERSION:

Suppose a brine containing 0.2 kg of salt per liter runs into a tank initially filled with 500 L of water containing 3 kg of salt. The brine enters the tank at a rate of 5 L/min. The mixture, kept uniform by stirring, is flowing out at the rate of 5 L/min (see Figure, where a = 5, c = 0.2, tc = 500, and A0 = 3).

(a) Find the concentration, in kilograms per liter, of salt in the tank after 5 min. Round to 3 decimal places. [Hint: Let A denote the number of kilograms of salt in the tank att minutes after the process begins and use the following fact.

rate of increase in A = rate of input - rate of exit.

A further discussion of mixing problems is given in Section 3.2.] kg/L

(b) After 5 min, a leak develops in the tank and an additional liter per minute of mixture flows out of the tank (see Figure, where a = 5 and c = 0.2).

What will be the concentration, in kilograms per liter, of salt in the tank 15 min after the leak develops? Round to 3 decimal places.

Explanation / Answer

Suppose a brine containing 0.2 kg of salt per liter runs into a tank initially filled with 500 L of water containing 3 kg of salt. The brine enters the tank at a rate of 5 L/min. The mixture, kept uniform by stirring, is flowing out at the rate of 5 L/min (see Figure, where a = 5, c = 0.2, tc = 500, and A0 = 3).

Rate of increase in A= Rate of input- Rate of output

Rate of input= a(l/min)* c(kg/L)= 5*0.2= 1 kg/min

Let t he concentration be at the exit be CA.( same as in the tank)

Initially the tank contains 3 kg of salt at in 500L= concentration =3/500=0.006kg/L

Rate of increase in A= 1- (5*CA)

  d/dt(CA*500)= 1-5CA

dCA/(1-5CA)= dt/500

which on integration gives

-ln(1-5CA)/5= t/500+C , C is integration constant (1)

At t= 0 CA=0.006 kg/L

-ln(1-5*0.006)/5= C

C=0.0061

The equation (1) becomes

-ln(1-5CA)/5= t/500+0.0061

At t=5 minutes

-ln(1-CA)/5= 5/500+0.0061

-ln(1-5CA)= 0.0161*5=0.0805

1-5CA= 0.923, 5CA= 1-0.923, CA=0.0154 Kg/L

b) when the leak develops, there will be two outputs and the mass balance equation becomes

Rate of increase in A= Rate of input- (Rate of output1+ Rate of output 2)

d/dt(500*CA)= 1-5CA-CA

d/dt(500CA)= 1-6CA

dCA/(1-6CA)= dt/500

Which on integration gives

-ln(1-6CA)/6= t/500+C   (2)

At t=0 CA= 0.006 kg/L

-ln(1-6*0.006)/6= C

C=0.006

Eq.2 now -ln(1-6CA)/6= t/500+0.006

At t=15minutes

-ln(1-6CA)/6= 15/500+0.006=0.036

-ln(1-6CA)= 0.036*6=0.216

1-6CA= 0.81 CA= (1-0.81)/6=0.032 Kg/L

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