Question

A large tank is partially filled with 100 gallons of fluid in which 20 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 35 minutes.

Answer 1

Explanation:

From the given information:

`R_(in) = ( (1)/(2) \ lb/gal) (6)\ gal /min \\ \\R_(in) = 3 \ lb/min`

Given that the solution is pumped at a slower rate of 4gal/min

Then:

`R_(out) = (4A)/(100+(6-4)t)`

`R_(out)= (2A)/(50+t)`

The differential equation can be expressed as:

`(dA)/(dt)+ (2)/(50+t)A = 3 \ \ \ ... (1)`

Integrating the linear differential equation; we have::

`\int_c (2)/(50 +t)dt = e^{2In |50+t|`

`\int_c (2)/(50 +t)dt = (50+t)^2`

multiplying above integrating factor fields; we have:

`(50 +t)^2 (dA)/(dt) + 2 (50 + t)A = 3 (50 +t)^2`

`(d)/(dt)\bigg [ (50 +t)^2 A \bigg ] = 3 (50 +t)^2`

`(50 + t)^2 A = (50 + t)^3+c`

A = (50 + t) + c(50 + t)²

Using the given conditions:

A(0) = 20

⇒ 20 = 50 + c (50)⁻²

-30 = c(50) ⁻²

c = -30 × 2500

c = -75000

A = (50+t) - 75000(50 + t)⁻²

The no. of pounds of salt in the tank after 35 minutes is:

A(35) = (50 + 35) - 75000(50 + 35)⁻²

A(35) = 85 -
`(75000)/(7225)`

A(35) =69.6193 pounds

A(35)
`\simeq` 70 pounds

Thus; the number of pounds of salt in the tank after 35 minutes is 70 pounds.