Question

A large tank contains (200 + 10p, where p = 1) gallons of a solution composed of 80% water and 20% alcohol. A second solution containing 50% water and 50% alcohol is added to the tank at the rate of 4 gallons per minute. As the second solution is being added, the tank is being drained at the rate of 5 gallons per minute. What is the initial condition for the alcohol? Assuming, the solution in the tank is stirred constantly, what is the amount of alcohol m the tank at any time? How much alcohol will be found in the tank after 19 minutes in percentage?

Answer 1

Answer:

Let t be the number of minutes since the tap opened.

Since the water increases at 20 gallons per minute, and the sugar increases at 2 pound per minute, these are constant rates of change.

This tells us the amount of water in the tank is changing linearly, as is the amount of sugar in the tank.

We can write an equation independently for each:

Water:W(t)=200+20t in gallons

Sugar:S(t)=10+2t in pounds

The concentration C will be the ratio of pounds of sugar to gallons of water.

C(t)=

200+20t

10+2t

The concentration after 14 minutes is given by evaluating C(t) at t=14

∴C(14)=

200+280

10+28

=

240

19

This means the concentration is 19 pounds of sugar to 240 gallons of water.

At the beginning, the concentration is C(0)=

200+0

10+0

=

20

1

.

Since

240

19

≈0.08>

20

1

=0.05, the concentration is greater after 14 minutes than at the beginning.

Answer 2

y(t)=amount of alcohol in the tank at time t

dy/dt

=y’(t)

= alcohol inflow rate – alcohol outflow rate

= 4 - 5(y/50)

=200 - 5y/50

= 5 (40-y)/50

dy/dt = 5 (40-y)/50

Separate variable and integrate :

dy/(40-y) = 5/50dt

-ln(40-y) = 0.1t + C

ln(40-y) = -0.1t + C

40-y= Ce^-0.1t

Y= 40(1-Ce^-0.1t) ….. eq. 1

Initially, the tank contains 50 gallon (50% alcohol) = 25 gallon of alcohol,

y(0) = 25

substituting y=25 and t=0 in to eq.1 gives

25 = 40(1-C)

25-40 = -40C

C = 0.375

When t=10 minutes

y(10) = 40(1-0.375*e^-0.1(10))

y(10) = 40(1-0.375*0.367)

y(10) = 34.48 gallon

at t=10’, amount of alcohol in the tank

= 50% x 34.48 gallon = 17.24 gallon or 34.48%