Respuesta :
Answer:
a) [tex]s(t) =400- 385 e^{-\frac{1}{10} t}[/tex]
b) [tex]s(t=25min) =400- 385 e^{-\frac{1}{10}25}=368.397 [/tex]
Step-by-step explanation:
Part a
Assuming that the original concentration of salt is 8 oz/gal and that the rate of in is equal to the rate out = 5 gal/min.
For this case we know that the rate of change can be expressed on this way:
[tex]Rate change = In-Out[/tex]
And we can name the rate of change as [tex]\frac{ds}{dt}=rate change[/tex]
And our variable s would represent the amount of salt for any time t.
We know that the brine containing 8oz/gal and the rate in is equal to 5 gal/min and this value is equal to the rate out.
For the concentration out we can assume that is [tex]\frac{s}{50gal}[/tex]
And now we can find the expression for the amount of salt after time t like this:
[tex]\frac{dS}{dt}= 8 \frac{oz}{gal}(5\frac{gal}{min}) -\frac{s}{50gal} 5 \frac{gal}{min} =40\frac{oz}{min}- \frac{s}{10} \frac{oz}{min}[/tex]
And we have this differential equation:
[tex]\frac{dS}{dt} +\frac{1}{10} s = 40[/tex]
With the initial conditions y(0)=15 oz
As we can see we have a linear differential equation so in order to solve it we need to find first the integrating factor given by:
[tex]\mu = e^{\int \frac{1}{10} dt }= e^{\frac{1}{10} t}[/tex]
And then in order to solve the differential equation we need to multiply with the integrating factor like this:
[tex]e^{\frac{1}{10} t} s = \int 40 e^{\frac{1}{10} t} dt[/tex]
[tex]e^{\frac{1}{10} t} s = 400 e^{\frac{1}{10} t} +C[/tex]
Now we can divide both sides by [tex] e^{\frac{1}{25} t} [/tex] and we got:
[tex]s(t) =400 + C e^{-\frac{1}{10} t}[/tex]
Now we can apply the initial condition in order to solve for the constant C like this:
[tex]15 = 400+C[/tex]
[tex]C=-385[/tex]
And then our function would be given by:
[tex]s(t) =400- 385 e^{-\frac{1}{10} t}[/tex]
Part b
For this case we just need to replace t =25 and see what we got for the value of the concentration:
[tex]s(t=25min) =400- 385 e^{-\frac{1}{10}25}=368.397 [/tex]
a) The quantity of salt in time is described by [tex]m(t) = V_{T}\cdot c_{in}+(m_{T}-V_{T} \cdot c_{in})\cdot e^{-\frac{\dot V}{V_{T}} \cdot t}[/tex].
b) There are 4400 ounces of salt in the tank after 25 minutes.
How to model a dissolution process in a tank
In this question we must model the salt concentration in the tank ([tex]c(t)[/tex]), in ounces per gallon, as a function of time ([tex]t[/tex]), in minutes, in which salt is dissolved into the tank due to a constant inflow rate ([tex]\dot V[/tex]), in gallons. Likewise, an equal outflow rate exists with resulting concentration.
a) The process is modelled mathematically by a non-homogeneous first order differential equation and physically by principle of mass conservation, whose description is shown below:
[tex]\frac{dc(t)}{dt} + \frac{\dot V}{V_{T}}\cdot c(t) = \frac{\dot V}{V_{T}}\cdot c_{in}[/tex] (1)
Where:
- [tex]V_{T}[/tex] - Tank volume, in gallons
- [tex]c_{in}[/tex] - Inflow salt concentration, in ounces per gallon
The solution of this differential equation is:
[tex]c(t) = c_{in} + \left(\frac{m_{T}}{V_{T}}-c_{in} \right)\cdot e^{-\frac{\dot V}{V_{T}}\cdot t }[/tex] (2)
Where [tex]m_{T}[/tex] is the initial salt mass of the tank, in ounces.
And the salt mass in the tank at an arbitrary time [tex]t[/tex] ([tex]m(t)[/tex]), in ounces, is obtained by multiplying (2) by the volume of the tank. That is to say:
[tex]m(t) = c(t)\cdot V_{T}[/tex] (3)
By replacing [tex]c(t)[/tex] in (3) by (2), we have the following expression:
[tex]m(t) = V_{T}\cdot c_{in}+(m_{T}-V_{T} \cdot c_{in})\cdot e^{-\frac{\dot V}{V_{T}} \cdot t}[/tex] (4)
The quantity of salt in time is described by [tex]m(t) = V_{T}\cdot c_{in}+(m_{T}-V_{T} \cdot c_{in})\cdot e^{-\frac{\dot V}{V_{T}} \cdot t}[/tex]. [tex]\blacksquare[/tex]
b) If we know that [tex]V_{T} = 50\,gal[/tex], [tex]\dot V = 55\,\frac{gal}{min}[/tex], [tex]c_{in} = 88\,\frac{oz}{gal}[/tex], [tex]m_{T} = 15\,oz[/tex] and [tex]t = 25\,min[/tex], then the quantity of salt is:
[tex]m(25) = (50)\cdot (88)+[15-(50)\cdot (88)]\cdot e^{-\left(\frac{55}{50} \right)\cdot (25)}[/tex]
[tex]m(25) = 4400\,oz[/tex]
There are 4400 ounces of salt in the tank after 25 minutes. [tex]\blacksquare[/tex]
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