Question

A tank contains 50 kg of salt and 1000 L of water. A solution of a concentration 0.025 kg of salt per liter enters a tank at the rate 9 L/min. The solution is mixed and drains from the tank at the same rate.
(a) What is the concentration of our solution in the tank initially?
concentration = (kg/L)

(b) write down the differential equation which models the Amount y of salt in the tank:
dydt=

(c) Find the amount of salt in the tank after 2.5 hours.
amount = (kg)

(d) Find the concentration of salt in the solution in the tank as time approaches infinity.
concentration = (kg/L)

Answer 1

Answer:

A. 0.05kg/l

B. dy/dt = 9/1000(25 - y)

C. 20.05 kg of salt

D. 0.0025kg/l

Step-by-step explanation:

A. Concentration of salt in the tank initially,

Concentration (kg/l) = mass of salt in kg/ volume of water in liter

= 50kg/1000l

= 0.05kg/l

B. dy/dt = rate of salt in - rate of salt out

Rate of salt in = 0.025kg/l * 9l/min

= 0.225kg/min

Rate of salt out = 9y/1000

dy/dt = 0.225 - 9y/1000

dy/dt = 9/1000(25 - y)

C. Collecting like terms from the above equation,

dy/25 - y = 9/1000dt

Integrating,

-Ln(25 - y) = 9/1000t + C

Taking the exponential of both sides,

25 - y = Ce^(-9t/1000)

Calculating for c, at y = 0, t = 0;

C = 25

y(t) = 25 - 25e^(-9t/1000)

At 2.5 hours,

2.5 hours * 60 mins = 180 mins

y(180 mins) = 25 - 25e^(-9*180/1000)

= 25 - 25*(0.1979)

= 20.05kg of salt

D. As time approaches infinity, e^(Infinity) = 0,

y(t) = 25 - 25*0

Concentration (kg/l) = 25/1000

= 0.0025kg/l

Answer 2

Answer / Step-by-step explanation:

a) The concentration of the solution in tank initially is:

= Initial volume ÷ initial concentration

Where initial volume = 50kg and

initial concentration = 1000 Liters

Therefore, 60kg ÷ 1000L

           = 0.06 Kilogram per liter.

b) Noting that the initial condition from part (a)

The differential equation would be:

    Q ° + AQ = B

Therefore, Q =  B /  A + Ce ∧−At

So, for some constant C. Using the initial value condition gives C = 60 −  B/A . So  we get A = 1  / 125 , B = 6 /  25, therefore the solution is:

Q = 30 + 30e ∧ - 1/125

With t = 90 (since 2.5 hours is 180 minutes to agree with our dimensions)

c) We use the formula If Q(t) is the amount of salt as  a function of t, then:

Q ° = Q ° ∨in - Q°∨out = 9 ( 0.025) kg/min - Q/ 1000 . 8  kg/min

Thus, we get the differential equation and initial value (simplifying and  omitting the dimensions)

Therefore, Q° + 1/125 . Q = 6/25, Q(0) = 60

d) T solve this, we assume the tank is large enough to hold all the solution

There, if we go back to the answer in the part b and find the limit, the answer will be = 0.