Question

A large tank is filled to capacity with 700 gallons of pure water. Brine containing 3 pounds of salt per gallon is pumped into the tank at a rate of 7 gal/min. The well-mixed solution is pumped out at the same rate. Find the number A(t) of pounds of salt in the tank at time t.

Answer 1

Answer:

The number A(t) of pounds of salt in the tank at time 't' is;

${A(t)}= -2,100 \times e^{\dfrac{-t}{100} } + 2,100$

Step-by-step explanation:

In the question, we have;

The volume of pure water initially in the tank = 700 gal

The concentration of brine pumped into the tank = 3 pounds per gallon

The rate at which the brine is pumped into the tank,  = 7 gal/min

The rate at which the well mixed solution is pumped out = The same 7 gal/min

The number of pounds of salt in the tank at time 't' is found as follows;

The rate of change in A(t) with time = The rate of salt input - The rate of salt output

$The \ rate \ of \ change \ in \ A(t) \ with \ time = \dfrac{dA}{dt}$

The rate of salt input = 7 gal/min × 3 lbs/gal = 21 lbs/min

The rate of salt output = (A(t)/700) lb/gal × 7 gal/min = (A(t)/100) lb/min

Therefore, we have;

$\dfrac{dA}{dt} = 21 - \dfrac{A(t)}{100}$

Therefore;

$\dfrac{dA}{dt} + \dfrac{A(t)}{100}= 21$

The integrating factor is $e^{\int\limits {\frac{1}{100} } \, dx } = e^{\frac{x}{100} }$

$e^{\dfrac{t}{100} } \times \dfrac{dA}{dt} + e^{\dfrac{t}{100} } \times\dfrac{A(t)}{100}= e^{\dfrac{t}{100} } \times21$

$\dfrac{d}{dt} \left[ e^{\dfrac{t}{100} } \times{A(t)}{} \right]= e^{\dfrac{t}{100} } \times21$

Using an online tool, we get;

${A(t)}= c_1 \times e^{\dfrac{-t}{100} } + 2,100$

At time t = 0, A(t) = 0

We get;

$0= c_1 \times e^{\dfrac{-0}{100} } + 2,100$

c₁ = -2,100

Therefore, the number A(t) of pounds of salt in the tank at time 't' is given as follows;

${A(t)}= -2,100 \times e^{\dfrac{-t}{100} } + 2,100$