24. network analysis water is flowing through a network of pipes (in thousands of cubic meters per hour), as…

Explanation:

Step1: Set up equations based on flow - conservation

At each node, the inflow equals the outflow. For the first node: $600=x_1 + x_3$. So, $x_3=600 - x_1$. For the second node: $x_1=x_2 + x_4$. So, $x_4=x_1 - x_2$. For the third node: $x_2+500=x_5$. So, $x_5=x_2 + 500$. For the fourth node: $600=x_6 + x_3$. Substitute $x_3 = 600 - x_1$ into it, we get $x_6=x_1$. For the fifth node: $x_4=x_7 + x_6$. Substitute $x_4=x_1 - x_2$ and $x_6=x_1$ into it, we get $x_7=-x_2$. For the sixth node: $x_7+500=x_5$. Substitute $x_5=x_2 + 500$ and $x_7=-x_2$ into it, which is consistent with our previous - derived equations. The general solution is: $x_1$ is a free variable, $x_2$ is a free variable, $x_3=600 - x_1$, $x_4=x_1 - x_2$, $x_5=x_2 + 500$, $x_6=x_1$, $x_7=-x_2$.

Step2: Solve for (b)

When $x_1 = x_2=100$. $x_3=600 - 100 = 500$. $x_4=100 - 100 = 0$. $x_5=100 + 500 = 600$. $x_6=100$. $x_7=-100$.

Step3: Solve for (c)

When $x_6 = x_7 = 0$. Since $x_6=x_1$, then $x_1 = 0$. Since $x_7=-x_2$, then $x_2 = 0$. $x_3=600 - 0 = 600$. $x_4=0 - 0 = 0$. $x_5=0 + 500 = 500$.

Step4: Solve for (d)

When $x_5 = 1000$ and $x_6 = 0$. Since $x_5=x_2 + 500$, then $1000=x_2 + 500$, so $x_2 = 500$. Since $x_6=x_1$, then $x_1 = 0$. $x_3=600 - 0 = 600$. $x_4=0 - 500=-500$. $x_7=-500$.

Answer:

(a) $x_1$ and $x_2$ are free variables, $x_3=600 - x_1$, $x_4=x_1 - x_2$, $x_5=x_2 + 500$, $x_6=x_1$, $x_7=-x_2$. (b) $x_1 = 100$, $x_2 = 100$, $x_3 = 500$, $x_4 = 0$, $x_5 = 600$, $x_6 = 100$, $x_7=-100$. (c) $x_1 = 0$, $x_2 = 0$, $x_3 = 600$, $x_4 = 0$, $x_5 = 500$, $x_6 = 0$, $x_7 = 0$. (d) $x_1 = 0$, $x_2 = 500$, $x_3 = 600$, $x_4=-500$, $x_5 = 1000$, $x_6 = 0$, $x_7=-500$.