A large tank is partially filled with 100 gallons of fluid in which 40 pounds of salt is dissolved. Brine containing 1 2 pound of salt per gallon is pumped into the tank at a rate of 6 gal/min. The well-mixed solution is then pumped out at a slower rate of 4 gal/min. Find the number of pounds of salt in the tank after 30 minutes. (Round your answer to two decimal places.

Answers

Answer 1

Answer:

76.09 pounds

Step-by-step explanation:

From the information given:

[tex]R_{in} = (\dfrac{1}{2} \ lb/gal ) *(6 \ gal/min)[/tex]

[tex]R_{in} =3 \ lb/min[/tex]

At a slower of 4 gal/min, the solution gets pumped out;

So,

[tex]R_{out} = \dfrac{4A}{100+(6-4)t }[/tex]

[tex]R_{out} = \dfrac{2A}{50+t }[/tex]

The differential of the equation is:

[tex]\dfrac{dA}{dt}+ \dfrac{2}{50+t}A=3 ---(1)[/tex]

For the linear differential equation, the integrating factor is determined as:

[tex]\int_e \dfrac{2}{50 +t}dt = e^{2 In|50+t|[/tex]

[tex]= (50+t)^2[/tex]

Multiplying [tex](50+t)^2[/tex] with the above integration factor;

[tex](50+t)^2 \dfrac{dA}{dt} + 2 (50+t) A = 3(50 +t)^2[/tex]

[tex]\dfrac{d}{dt}[ (50 +t)^2 A] = 3(50+t)^2[/tex]

[tex](50+t)^2 A = (50 +t)^3 + c[/tex]

[tex]A = (50 +t) + c (50 +t )^{-2}[/tex]

By using the given condition:

A(0) = 40

40 = 50 + c (50)⁻²

10 = c(50)⁻²

OR

c = -10 × 2500

c = -25000

A = (50 + t) - 25000(50 + t)⁻²

The number of pounds of salt in the tank after 30 min is:

A(30) = (50+30) - 25000(50 + 30)⁻²

[tex]A(30) = 80 - \dfrac{25000}{6400}[/tex]

[tex]A(30) = 76.09 \ pounds[/tex]

Thus, the number of pounds of salt in the tank after 30 min is 76.09 pounds.


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Step-by-step explanation:

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Density of glass in g/cm³ = 2.5

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Step-by-step explanation:

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