Physics: Consider two interconnected tanks as shown in the figure above. tank 1 initial contains 90 l (liters) of water and 290 g of salt, while tank 2 initially contains 60 l of water and 245 g of salt. water containing 50 g/l of salt is poured into tank1 at a rate of 3.5 l/min while the mixture flowing into tank 2 contains a salt concentration of 40 g/l of salt and is flowing at the rate of 4 l/min. the two connecting tubes have a flow rate of 6 l/min from tank 1 to tank 2; and of 2.5 l/min from tank 2

a and cc is 100% but i am confused in a...

Consider two interconnected tanks as shown in the

Ответ:

rilee3344

18.01.2022

Physics

Let A_1 be the amount of salt in tank 1 at time , and A_2 the amount of salt in the tank 2.

The volume of solution in either tank stays constant. In tank 1, at time (min) we have

$90\,\mathrm L+\underbrace{\left(3.5\dfrac{\rm L}{\rm min}-6\dfrac{\rm L}{\rm min}+2.5\dfrac{\rm L}{\rm min}\right)}_0t=90\,\mathrm L$

In tank 2,

$60\,\mathrm L+\underbrace{\left(4\dfrac{\rm L}{\rm min}+6\dfrac{\rm L}{\rm min}-2.5\frac{\rm L}{\rm min}-7.5\frac{\rm L}{\rm min}\right)}_0t=60\,\mathrm L$

Then the concentration of salt in tanks 1 and 2 at any given time is $\dfrac{A_1\,\rm g}{90\,\rm L}$ and $\dfrac{A_2\,\rm g}{60\,\rm L}$ .

The net rate of change of the amount of salt in tanks 1 and 2 follows a simple rule:

$\dfrac{\mathrm dA_i}{\mathrm dt}=(\text{rate in})-(\text{rate out})$

Each rate is in units of g/min. Each L coming in or going out contributes or removes some salt depending on the flow rate (L/min) and concentration (g/L) of the solution in either tank. For tank 1, we have

$\text{rate in}=\left(50\dfrac{\rm g}{\rm L}\right)\left(3.5\dfrac{\rm L}{\rm min}\right)+\left(\dfrac{A_2}{60}\dfrac{\rm g}{\rm L}\right)\left(2.5\dfrac{\rm L}{\rm min}\right)$

$\text{rate out}=\left(\dfrac{A_1}{90}\dfrac{\rm g}{\rm L}\right)\left(6\dfrac{\rm L}{\rm min}\right)$

Then the amount of salt in tank 1 has rate of change (ignoring units)

$\dfrac{\mathrm dA_1}{\mathrm dt}=-\dfrac{A_1}{15}+\dfrac{A_2}{24}+175$

A similar breakdown for tank 2 shows a rate of change of

$\dfrac{\mathrm dA_2}{\mathrm dt}=\dfrac{A_1}{15}-\dfrac{A_2}6+160$

In matrix form, the system is described by

$\begin{pmatrix}A_1\\A_2\end{pmatrix}'=\begin{pmatrix}-\frac1{15}&\frac1{24}\\\frac1{15}&-\frac16\end{pmatrix}\begin{pmatrix}A_1\\A_2\end{pmatrix}+\begin{pmatrix}175\\160\end{pmatrix}$

You can solve this with the usual eigenvalue method and method of undetermined coefficients. You should get a general solution of

$\begin{pmatrix}A_1\\A_2\end{pmatrix}=C_1\begin{pmatrix}5\\-6-2\sqrt{19}\end{pmatrix}e^{\frac{-7+\sqrt{19}}{60}t}+C_2\begin{pmatrix}5\\-6+2\sqrt{19}\end{pmatrix}e^{\frac{-7-\sqrt{19}}{60}t}+\begin{pmatrix}4300\\2680\end{pmatrix}$

Then use the initial values A_1(0)=290 and A_2(0)=245 to solve for C_1,C_2 and find the particular solution.

0,0(0 оценок)

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Physics

Ask an AI advisor a question