Respuesta :
Answer:
a) dx/dt = 4*c - (4*x / 10) = 4*c - 0.4*x
b) x = 10*c - 10*c*e^(-0.4*t)
c) c = 2.500 g / Liters
Step-by-step explanation:
Given:
- Initial volume of pure water V_o = 10 Liters
- Inflow of brine Q_in = 4 Liters / min
- Outflow of mixture Q_out = 4 liters /min
Find:
a) Find a formula for the rate of change in the amount of salt, dx/dt, in terms of the amount of salt in the solution x and the unknown concentration of incoming brine c.
b) Find a formula for the amount of salt, in grams, after t minutes have elapsed. Your answer should be in terms of c and t.
c) In 25 minutes there are 25 grams of salt in the fish tank. What is the concentration of salt in the incoming brine?
Solution:
- We will define a constant (c) as concentration of salt as amount of salt in grams per liter.
- dx / dt is the amount of salt in grams per unit time t.
- We will construct a net flow balance:
dx / dt = Q_in*c - (amount of salt / V_o) *Q_out
- Denoting the amount of salt in grams to be x. then we have:
dx/dt = 4*c - (4*x / 10) = 4*c - 0.4*x
- Once we have formulated the ODE, we will solve it to get the amount of salt x in grams at any time t, as follows:
dx/dt = 4*c -0.4*x
- Separate variables:
dx / (4*c -0.4*x) = dt
- Integrate both sides:
- (10 / 4)*Ln | 4*c -0.4*x | = t + K
- Evaluate constant K, when t = 0, x = 0.
- (10 / 4)*Ln|4c| = K
- Input constant K back in solution:
Ln |4*c -0.4*x| = -0.4*t + Ln|4c|
4*c -0.4*x = 4c*e^(-0.4*t)
x = 10*c - 10*c*e^(-0.4*t)
- Evaluate using the derived expression t = 25 mins and x = 25, compute c?
x = 10*c - 10*c*e^(-0.4*t)
25/10 = c ( 1 - e^(-0.4*25) )
c = 2.5 / ( 1 - e^(-0.4*25) )
c = 2.500 g / Liters
The solution will be
(a) [tex]\dfrac{dx}{dt} =4c-\dfrac{4x}{10} =4c-0.4x[/tex]
b) [tex]x=10c-10ce^{-0.4t}[/tex]
c) [tex]c=2500 \ \dfrac{g}{liters}[/tex]
What will be the concentration of salt?
We will define a constant (c) as the concentration of salt as the amount of salt in grams per liter.
[tex]\dfrac{dx}{dt}[/tex] is the amount of salt in grams per unit time t.
We will construct a net flow balance:
[tex]\dfrac{dx}{dt} =Q_{in}\times c-(\dfrac{\rm Amount \ of \ salt}{V_0} )\times Q_{out}[/tex]
Denoting the amount of salt in grams to be x. then we have:
[tex]\dfrac{dx}{dt} =4\timesc-\dfrac{4x}{10} =4c-0.4x[/tex]
(b) Once we have formulated the ODE, we will solve it to get the amount of salt x in grams at any time t, as follows:
[tex]\dfrac{dx}{dt} =4c-0.4x[/tex]
Separate variables:
[tex]\dfrac{dx}{4c-0.4x} =dt[/tex]
Integrate both sides:
[tex]\dfrac{-10}{4}ln(4c-0.4x)=t+k[/tex]
Evaluate constant K, when t = 0, x = 0.
[tex]\dfrac{-10}{4}ln(4c)=k[/tex]
put constant K back in solution:
[tex]ln(4c-0.4x)=-0.4t+ln(4c)[/tex] Ln|4c|
[tex]4c-0.4x=4ce^{-0.4t}[/tex]
[tex]x=10c-10ce^{-0.4t}[/tex]
(c) Evaluate using the derived expression t = 25 mins and x = 25, compute c?
[tex]x=10c-10ce^{-0.4t}[/tex]
[tex]\dfrac{25}{10} =c(1-e^{-0.4\times25}[/tex]
[tex]c=\dfrac{2.5}{1-e^{-0.4\times 25}}[/tex]
[tex]c=2500\ \frac{g}{liters}[/tex]
Thus all the answers are
(a) [tex]\dfrac{dx}{dt} =4c-\dfrac{4x}{10} =4c-0.4x[/tex]
b) [tex]x=10c-10ce^{-0.4t}[/tex]
c) [tex]c=2500 \ \dfrac{g}{liters}[/tex]
To know more about mixture and allegations follow
https://brainly.com/question/26413176
