Chapter 1.9, Problem 41E

a

To determine

To calculate: To find the total salt of the lake before the inflow using the given information

Answer to Problem 41E

The total salt before the inflow and outflow is $3.3\times {10}^7{s}_t$

Explanation of Solution

Given information: $3.0\times {10}^6{m}^3$

Calculation:

We ate given that there is no evaporation in the lake and that $3.0\times {10}^6{m}^3$ of water flows out each year Also, the lake receives $3.0\times {10}^6{m}^3$ of water per year with the salinity of concentration 0.001 and contains $3.3\times {10}^7{m}^3$ of water and starts with no salinity

Let $s_t$ be the concentration of salt before inflow and loss though outflow. Thus, there is $3.3\times {10}^7{s}_t$ salt in the lake before inflow and loss though outflow.

Therefore, the total salt before the inflow and outflow is $3.3\times {10}^7{s}_t$

b

To determine

To calculate: To find the total water in the lake using the given information

Answer to Problem 41E

The total water in the lake is $3.6\times {10}^7{m}^3$

Explanation of Solution

Given information: $3.0\times {10}^6{m}^3$

Calculation:

So, total water in the lake will be

(The lake containing $3.3\times {10}^7{m}^3$ of water + the lake receiving $3.0\times {10}^6{m}^3$ of water per year)

Hence, total water in the lake is as follows:

  $\begin{array}{l}\left(3.3\times {10}^7+3.0\times {10}^6\right)m^3=\left(3.3\times {10}^7+0.3\times {10}^7\right)m^3\\ =\left[\left(3.3+0.3\right)\times {10}^7\right]m^3\\ =3.6\times {10}^7{m}^3\end{array}$

Therefore, the total water in the lake is $3.6\times {10}^7{m}^3$

c

To determine

To calculate: To find the total salt in the lake using the given information

Answer to Problem 41E

The total salt in the lake is $3.3\times {10}^7{}_{s_t}+3.0\times {10}^3$

Explanation of Solution

Given information: $3.0\times {10}^6{m}^3$

Calculation:

Here, the total salt in the lake will be the total salt before the inflow plus one thousandth of receiving water by the lake

Hence, the total salt in the lake is shown below:

  $\begin{array}{l}3.3\times {10}^7{}_{s_t}+\frac{1}{1000}3.0\times {10}^6=3.3\times {10}^7{}_{s_t}+\frac{1}{{10}^3}3.0\times {10}^6\\ =3.3\times {10}^7{}_{s_t}+3.0\times {10}^6\times {10}^{-3}\\ =3.3\times {10}^7{}_{s_t}+3.0\times {10}^{6-3}\\ =3.3\times {10}^7{}_{s_t}+3.0\times {10}^3\end{array}$

Therefore, the total salt in the lake is

  $3.3\times {10}^7{}_{s_t}+3.0\times {10}^3$

d

To determine

To calculate: To find the total salt concentration, total water and total salt using the given information

Answer to Problem 41E

The total salt concentration is $s_{t+1}=0.917s_t+8.33\times {10}^{-5}$

The discrete-time dynamical system is $s_{t+1}=0.917s_t+8.33\times {10}^{-5}$

Explanation of Solution

Given information: $3.0\times {10}^6{m}^3$

Calculation:

The total salt concentration $s_{t+1}$ will be

  $\begin{array}{l}{s}_{t+1}=\frac{3.3\times {10}^7{}_{s_t}+3.0\times {10}^3}{3.6\times {10}^7}\\ =\frac{3.3\times {10}^7{}_{s_t}}{3.6\times {10}^7}+\frac{3.0\times {10}^3}{3.6\times {10}^7}\\ =\frac{3.3}{3.6}{s}_t+\frac{30.0+{10}^2}{3.6\times {10}^7}\\ =0.917s_t+\frac{30.0}{3.6}\times {10}^{2-7}\\ =0.917s_t+8.33\times {10}^{-5}\end{array}$

Therefore, the total salt concentration is

  $s_{t+1}=0.917s_t+8.33\times {10}^{-5}$

Furthermore, this total salt concentration $s_{t+1}=0.917s_t+8.33\times {10}^{-5}$ will give us the discrete time dynamical system because the water that flows out is mixed well-

Hence, the discrete-time dynamical system is

  $s_{t+1}=0.917s_t+8.33\times {10}^{-5}$