Respuesta :
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 / 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.
