A tank of 50 gallons of salt-water initally contains 5lbs of dissolved salt. An

Question

A tank of 50 gallons of salt-water initally contains 5lbs of dissolved salt. An inlet pours salt-water with concentration 2lbs salt per gallon water in at a rate of 2 gal/min. The tank is full so water spills over at 2 gal/min. Assuming the solution is stirred continuously enough so that concentration is uniform, carefully derive and solve a differential equation for the amount of salt dissolved in the tank after t minutes. (b) Let the outflow of the tank in part (a) flow into another tank (“tank B”). Assume tank B is initially full with 50 gallons of pure water and is again “well-mixed”. Derive and solve a differential equation for the amount of salt in tank B after t minutes. (Note that as 2 gal/min of salt water runs into tank B, the same amount runs out.)

Explanation / Answer

Let there be A lbs of salt in the tank at any time t, after inlet starts pouring salt-water in the tank. Since there is 50 gallons of salt- water in the tank all the time, there would be A/50 lbs of salt/ gallon in the tank. At timr t=0 min, the amount of salt in the water in the tank would A= 5lbs.

Now salt-water containing 2lbs of salt per gallon is running into the tank at the rate of 2gal/min, the amount of salt runningin to the tank is 4lbs/min.

Next, 2gallons of salt water is running out of tank per min, the amount of salt going out would be 2*A/50 lbs per minute

Rate of change of salt concentration would be= Salt coming into - Salt going out of tank, per minute

Thus dA/dt = 4- 2A/50 = 4-A/25 = (100-A)/25 .... This is the required DE for amount of salt dissolved in water after time t. This can be solved as follows:

dA/(100-A)= (1/25)dt. Integrate it on both sides,

-ln (100-A)= t/25   +C. Since at t=0, A= lbs,-ln95=C

-ln(100-A) = t/25 -ln 95, Or ln(100-A)/95 = -t/25   , Or 100-A= 95 e^(-t/25)

Or, A= A(t)=100-95e^(-t/25).

Part (b)

The amount of salt in tank A is 100-95e^(-t/25), hence it would have (1/50) (100-95e^(-t/25) lbs/gallon. The amount of salt coming into tank B , with 2 gallons of water per min would be (1/25) (100-95 e^(-t/25) lbs per minute per gallon.

Now let there be B lbs of salt in tank B at any time t, Then there would be B/50 lbs of salt per gallon in tank B. Hence salt going out of tank B would be B/25 lbs per minute

Rate of change in salt concentration would be dB/dt =salt coming in - salt going out,

dB/dt= (1/25) (100-95e^(-t/25) -B/25

        = 4-(19/5) e^ (-t/25) -B/25

Thus dB/dt +B/25 = 4- (19/5) e^(-t/25). This is a linear DE. Integrating factor is e^(intgeral of dt/25)= e^(t/25).

The solution of this DE would be B e^(t/25) = Int (4- (19/5) e^(-t/25)* e^(t/25)   +C

                                                              = 4 int e^(t/25) - (19/5) int dt +C

                                                             = 100 e^(t/25) - (19/5)t +C

Now at t=0, Salt in tank B was 0, hence C=-100

Thus Be^(t/25) = 100 e^(t/25) -(19/5)t +100   Ans

Thus Solution is B

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