Question

Generalize the model given in equation (8) on page 23 by assuming that the large tank initially contains N₀number of gallons of brine, r_inand r_outare the input and output rates of the brine, respectively (measured in gallons per minute ), c_inis the concentration of the salt in the inflow, c(t) the concentration of the salt in the tank as well as in the outflow at time (measured in pounds of salt per gallon), and A(t) is the amount of salt in the tank at time t.

Short answer

For,$r_{\text{in}}>r_{\text{eat}},\frac{dA}{dt}+\frac{r_{\text{eat}}}{N_e+\left(r_{iv}-r_{\text{eat}}\right)t}A=c_{iv}{r}_{iv}$

For,$r_{in}>r_{\text{out}},\frac{dA}{dt}+\frac{r_{\text{out}}}{N_0+\left(r_{in}-r_{\text{out}}\right)t}A=c_{in}{r}_{in}$

Step-by-step solution

Step 1:

A differential equation is an equation with one or more derivatives of a function.

Step 2:

Equation 8 is $\frac{dA}{dt}=6-\frac{A}{100}\text{or}\frac{dA}{dt}+\frac{1}{100}A-6.$

The tank initially holds N₀gallons of brine. Another brine solution is pumped into the large tank at a rate of r_ingallons per minute; the concentration of the salt in this inflow is c_e pounds per gallon. When the solution in the tank is well stirred, it is pumped out at a rate of r_0wgallons per minute.

Find rate of inflow:

If A(t) denotes the amount of salt in the tank at time t, then the rate at which A(t) changes is a net rate given by:

$\frac{dA}{dt}=R_{\text{int}}-R_{\text{ear}}$

Now consider R_in

_{$$\begin{gathered} R_{im}=\left(\text{concentration of salt in inflow}\right)\times \left(\text{input rate of brine}\right) \\ =\left(c_{im}\text{lb}/\text{gal}\right)\times \left(r_{im}\text{gal}/\text{min}\right) \\ =c_{iv}·r_{im}\text{lb}/\text{min} \end{gathered}$$}

When rin >rear 

Since the solution is pumped out at a slower rate; it accumulates at the rate of (r_in - r_0at) gal/min

After t minutes there are$N_0+\left(r_{\text{in}}-r_{\text{our}}\right)t$gallons of brine in the tank.

So,$$\begin{gathered} R_{\text{eut}}=r_{\text{out}}\times \frac{A\left(t\right)}{N_o+\left(r_{\text{ev}}-r_{\text{awt}}\right)t} \\ =\frac{r_{\text{at}}A}{N_0+\left(r_m-r_{\text{est}}\right)t}\text{lb}/\text{min} \end{gathered}$$

Thus, from (1)role="math" localid="1667896982867" $$\begin{gathered} \frac{dA}{dt}=c_{in}{r}_{in}-\frac{r_{eat}A}{N_e+\left(r_{ie}-r_{ews}\right)t} \\ \frac{dA}{dt}+\frac{r_{\text{om}}}{N_0+\left(r_{in}-r_{\text{ew}}\right)t}A=c_{in}{r}_{in} \\ \text{Hence, when}{r}_{\text{in}}>r_{\text{eat}},\frac{dA}{dt}+\frac{r_{\text{eat}}}{N_e+\left(r_{iv}-r_{\text{eat}}\right)t}A=c_{iv}{r}_{iv} \end{gathered}$$

When:rin<rout

Then$R_{\text{out}}=\frac{r_{\text{out}}A}{N_0-\left(r_{\text{out}}-r_{\text{in}}\right)t}$

Hence,$\frac{dA}{dt}+\frac{r_{\text{out}}}{N_0-\left(r_{\text{out}}-r_{\text{in}}\right)t}A=c_{in}{r}_{\text{in}}$

Or$\frac{dA}{dt}+\frac{r_{\text{out}}}{N_0+\left(r_{\text{in}}-r_{\text{out}}\right)t}A=c_{\text{in}}{r}_{in}$

Hence, when$r_{in}>r_{\text{out}},\frac{dA}{dt}+\frac{r_{\text{out}}}{N_0+\left(r_{in}-r_{\text{out}}\right)t}A=c_{in}{r}_{in}$