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

Answer:

Step-by-step explanation:

From the given information:

$R_{in} = ( \dfrac{1}{2} \ lb/gal) (6)\ gal /min \\ \\R_{in} = 3 \ lb/min$

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

Then:

$R_{out} = \dfrac{4A}{100+(6-4)t}$

$R_{out}= \dfrac{2A}{50+t}$

The differential equation can be expressed as:

$\dfrac{dA}{dt}+ \dfrac{2}{50+t}A = 3 \ \ \ ... (1)$

Integrating the linear differential equation; we have::

$\int_c \dfrac{2}{50 +t}dt = e^{2In |50+t|$

$\int_c \dfrac{2}{50 +t}dt = (50+t)^2$

multiplying above integrating factor fields; we have:

$(50 +t)^2 \dfrac{dA}{dt} + 2 (50 + t)A = 3 (50 +t)^2$

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

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

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 - $\dfrac{75000}{7225}$

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.