Only 1,3,4(a,b) ,5,7 PROBLEMS: Section 2.4 Mixing Problems .Standard Tank Proble

Question

Only 1,3,4(a,b) ,5,7

PROBLEMS: Section 2.4 Mixing Problems .Standard Tank Problem Initially, 50 pounds of salt are of the tank at the same rate as that at which it HoWS HW the tank. (a) Find the amount of salt in the tank at any time (b) Find the concentration of salt in the tank at any time. (c) Find the limiting amount of salt in the tank. (d) Find the limiting concentration of salt in the tank. dissolved in a 300-gallon tank, and then a salt solution with a concentration of 2 pounds of salt per gallon of solution flows into a tank at a rate of 3 gal/min. The solution inside the tank is kept well-stirred and flows out of the tank at the same rate as that at which it flows in. (a) Find the amount of salt in the tank at any time. (b) Find the concentration of salt in the tank at any time (c) Find the limiting amount of salt in the tank. (d) Find the limiting concentration of the salt in the tank. 3) Rate In > Rate Out Initially, a large tank with a capacity of 100 gallons contains 50 gallons of pure water. A salt solution with a concentration of 0.1 lb/gal flows into the tank at a rate of 4 gal/min. The mixture is kept well-stirred and flows out of the tank at the rate of 2 gal/min (a) Find the initial-value problem that describes the amount (b) Find the amount of salt in the tank until the tank over- (c) Find the concentration of salt in the tank until the tank (d) What is the initial-value problem that describes the of salt in the tank. flows. overflows amount of salt in the tank after the tank overflows? 2. Standard Tank Problem Initially, a 100-liter tank con- tains a salt solution that has a concentration of 0.5 kg/liter A less concentrated salt solution with a concentration of 0.1 kg/liter flows into a tank at a rate of 4 liters/min. The solution inside the tank is kept well-stirred and flows out

Explanation / Answer

Note: As per Chegg guidelines and due to time constraints for answering per question I shall be solving the first full question

1) If x denote the amount of salt in the tank, then we must have x(0)=50

We also have the incoming salt rate is concentration times flow rate = 2 x 3 = 6 pounds per minute

As well mixed solution leaves the tank at the same rate as it comes it, the volume of liquid in the tank is always the same, ie. 300 gallons. Therefore, concentration of salt in tank is x/300 and the rate of outgoing salt is

Concentration times flow rate = (x/300) x 3 = x/100

Thus, the differential equation governing the salt concentration is:

x' = 6-x/100

That is, x'+x/100=6

Integrating fractor for this equation is exp(F) where F=int (1/100) dt = (t/100)

The integrating factor is exp(t/100) so our equation becomes

e^(t/100)x'+e^(t/100)x/100=6e^(t/100)

That is [e^(t/100)x]'=6e^(t/100)

Integrating, we must have e^(t/100)x= integral of (6e^(t/100)) +C

That is, e^(t/100)x=600e^(t/100)+C

So that our solution is x=600+Ce^(-t/100)          (Note the minus sign now)

Hence, general solution of the equation is x(t)=600+Ce^(-t/100)

The initial condition implies x(0)=50 ==> 600+C=50 ==> C=-550

Therefore, the required salt weight function is x(t)=600-500e^(-t/100) which is the amount of salt in the tank at any time t

b) Concentration is C(t)=x(t)/Volume = (600-500e^(-t/100))/300 = 2-5/3e^(-t/100)

c) Limit of e^(-t/100) is 0 so the limiting value of salt is limit of 600-500e^(-t/100) which is 600 (pounds)

d) Limiting amount of concetration is limit of 2-5/3e^(-t/100) which is 2 (pounds per gallon)

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