Chapter 2, Section 2.3, Additional Question 02 A tank originally contains 100 ga

Question

Chapter 2, Section 2.3, Additional Question 02 A tank originally contains 100 gal of fresh water. Then water containinglb of salt per gallon is poured into the tank at a rate of 2 gal/min, and the mixture is allowed to leave at the same rate After 10 min the process is stopped, and fresh water is poured into the tank at a rate of 2 gal/min, with the mixture again leaving at the same rate. Find the amount of salt in the tank at the end of an additional 10 min. Round your answer to two decimal places. ibs the absolute tolerance is +/-0.01 Click if you would like to Show Work for this question: Open Show Work

Explanation / Answer

Solution:

Let us assume that t denotes the time,

Q(t) denotes the amount of salt in the tank at anypoint of time t.

Now, we know, R1 = rate of salt poured into the tank = (1/2)×2 = 1 lb/ min

and R2 = rate of salt flowing out of the tank = (Q/100) × 2 = (Q/ 50) lb/min

Now, applying the boundary conditions: i.e. Q(0) = 0 since at the starting we were having the fresh water.

dQ/dt = R1 - R2 = 1 - (Q/50).............................................................................(i)

Clearly, above equation is of the form of y' = ay + b

For this equation, we have the general solution of the form , y = Ceat - (b/a)

Therefore, Q(t) = Ce(-t/50) + 50

At t = 0

we have Q(0) = C + 50 =0 or C= - 50

So, we have now, Q(t) = -50e(-t/50) + 50

At t= 10, we have Q(10) = -50e(-10/50) + 50 = 9.06

In the second 10 mins,

R1 = rate of salt poured into the tank = 0 * 2= 0 lb/min

R2 = rate of salt flowing out of the tank= (Q/100) * 2 = (Q/50)lb/min

dQ/dt = 0 - (Q/50) = -(Q/50)

From above, we have new boundary conditoons, i.e., Q(0) = 9.06

Therefore, Q= Ce(-t/50)

At t=0,we have, Q(0)=C=9.06

Q(t)=9.06e(-t)/50)

Q(10)=9.06e(-10)/50) = 7.42

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