Question

(a) Determine the function for the amount of salt level in the tank at the time,\({\rm{t}}\). The volume of the tank is\(1000L\), the rate of outflow\(15L/Min\)and the rate of salt solution with concentration\(0.05kg/L\)and\(0.04kg/L\)entering in\(5L/\min \)and\(10L/\min \).

(b) Determine the salt level after one hour.

Short answer

(a) The function for the amount of salt level in the tank after \(t\) minutes is\(y = \frac{{{e^{ - (0.015)}}( - 0.65) + 0.65}}{{0.015}}\).

(b) The amount of salt after one hour\(y(60) = 25.73\;{\rm{kg}}\).

Step-by-step solution

The rate of change of salt level with respect to time \(t\).

For the given statement, let \(y\) be the amount of salt level.

At the initial stage value the salt level is\(y(0) = 0\;{\rm{kg}}/{\rm{ liter }}\).

The rate of outflow\( = \frac{y}{{1000}}(15)\).

The rate of change of salt level with respect to time\(t\),

\(\begin{aligned}\frac{{dy}}{{dt}} &= 0.05(5) + 0.04(10) - \frac{y}{{1000}}(15)\\\frac{{dy}}{{dt}} &= 0.25 + 0.4 - 0.015y\\\frac{{dy}}{{dt}} &= 0.65 - 0.015y\end{aligned}\)

Here\({\rm{0}}{\rm{.015 y}}\)be salt that goes outside.

Separate the variables,

\(\frac{{dy}}{{0.65 - 0.015y}} = dt\)

Integrating on both sides,

\(\begin{aligned}\int {\frac{{dy}}{{0.65 - 0.015y}}} &= \int d t\\ - \frac{1}{{0.015}}\ln |0.65 - 0.015y| &= t + C\\\ln |0.65 - 0.015y|&= - (0.015)t - C\end{aligned}\)

Raise the power to the base\(e\)on both sides,

\(|0.65 - 0.015y| = {e^{ - (0.015)t - C}}\) …… (1)

Observe the question, the initial condition is\({\rm{y(0) = 0}}\).

Substitute the initial value in equation (1).

Substitute the initial value in equation (1) as follows,

Substitute\(C\)value in equation (1),

\((0.65 - 0.015y) = {e^{ - (0.015)t}}( - 0.65)\)

\({\rm{0}}{\rm{.65 - 0}}{\rm{.015 y}}\)is negative, so that

\(\begin{aligned}(0.015y - 0.65) &= {e^{ - (0.015)t}}( - 0.65)\\0.015y &= {e^{ - (0.015)t}}( - 0.65) + 0.65\\y &= \frac{{{e^{ - (0.015x}}( - 0.65) + 0.65}}{{0.015}}\end{aligned}\)

Therefore, the solution after \(t\)minutes \(y = \frac{{{e^{ - (0.015)x}}( - 0.65) + 0.65}}{{0.015}}\)

\(y = \frac{{{e^{ - (0.015)x}}( - 0.65) + 0.65}}{{0.015}}\) of part (a) for calculation.

Consider part (a) solution \(y = \frac{{{e^{ - (0.015)x}}( - 0.65) + 0.65}}{{0.015}}\)

\(\begin{aligned}y(60) &= \frac{{{e^{ - (0.015) \times 0}}( - 0.65) + 0.65}}{{0.015}}\\ &= \frac{{{e^{ - 0.9}}( - 0.65) + 0.65}}{{0.015}}\\ &= \frac{{0.4065( - 0.65) + 0.65}}{{0.015}}\\ = \frac{{ - 0.264 + 0.65}}{{0015}}\end{aligned}\)

Further, simplify

\(\begin{aligned}y(60) &= \frac{{0.3860}}{{0.015}}\\ &= 25.73\end{aligned}\)

Therefore, the amount of salt after one hour \(y(60) = 25.73\;{\rm{kg}}\).