Question

A large tank is filled to capacity with 300 gallons of pure water. Brine containing 5 pounds of salt per gallon is pumped into the tank at a rate of 3 gal/min. The well-mixed solution is pumped out at a rate of 6 gals/min. Find the number A(t) of pounds of salt in the tank at time t. A(t) = 1500(1−e−0.02t) lb How long (in minutes) will it take for the tank to be empty after this process has started? min

Answer 1

Answer:

(a) $A = 1500(1 - e^{-0.01t})$

(b) $t = 0$

Step-by-step explanation:

Given

$V= 300gal$ --- Volume of tank

$B = 5lb/gal$ --- Brine solution

$R_1 = 3gal/min$  --- Rate in, in gallon/min

$R_2 = 6gal/min$ --- Rate out, in gallon/min

Required

Determine A(t)

First, calculate the rate in (R in) and (R out) in lb/min

$R_{in} = B * R_1$

$R_{in} = 5lb/gal * 3gal/min$

$R_{in} = 15lb/min$

R(out) is calculated by multiply the rate at which brine leaves the tank (lb/gal) * R2

So, we have:

$R_{out} = \frac{A(t)}{300} lb/gal * 3gal/min$

$R_{out} = \frac{3*A(t)}{300} lb/min$

$R_{out} = \frac{A(t)}{100} lb/min$

The change in the amount of brin e in the tank at time t is given as:

$A'(t) = R_{in} - R_{out}$

$A'(t) = 15 - \frac{A(t)}{100}$

Rewrite:

$\frac{dA}{dt} = 15 - \frac{A}{100}$

Multiply through by dt

$dA = [15 - \frac{A}{100}]* dt$

Make dt stand alone

$\frac{dA}{15 - \frac{A}{100}} = dt$

$ln|15 - \frac{A}{100}| = -0.01t + ln\ c$

Express as exponents

$15 - \frac{A}{100} = ce^{-\frac{t}{100}}$

Multiply through by 100

$1500 - A = 100ce^{-\frac{t}{100}}$

$A = 1500 - 100ce^{-\frac{t}{100}}$

$A = 1500 - 100ce^{-0.01t}$

Apply initial conditions

$A(0) = 0$

Substitute 0 for t in: $A = 1500 - 100ce^{-0.01t}$

$0 = 1500 - 100ce^{-0.01*0}$

$0 = 1500 - 100ce^{0}$

$0 = 1500 - 100c$

$100c = 1500$

$c = 15$

Substitute 15 for c in $A = 1500 - 100ce^{-0.01t}$

$A = 1500 - 100*15e^{-0.01t}$

$A = 1500 - 1500e^{-0.01t}$

Factorize:

$A = 1500(1 - e^{-0.01t})$

How long to empty the tank

Set A to 0

$1500(1 - e^{-0.01t}) = 0$

Divide through by 1500

$1 - e^{-0.01t} = 0$

Collect Like Terms

$e^{-0.01t} = 1$

Take ln of both sides

$ln\ (e^{-0.01t}) = ln\ 1$

$-0.0t = 0$

$t = 0$