A pump with curve equal to $)=[cms],hp=([m]$ lifts water between two tanks with difference between the upstream and downstream is $44\mathrm{.}08m.$ The pipeline is $200m$ long and $500mm$ diameter pipe with $C_{HW}=110.$ The total minor losses are equal to: $1\mathrm{.}3895Q^2.$

a$.$ Plot the two curves to find the flow in this system

b$.$ Set up the energy

c$.$ Transform the energy eq$.$ to be of the form, $F(Q)=0$ and solve it using Newton's method Start the Newton's scheme with $Q=2\mathrm{.}1cms$

d$.$ Confirm by solving with a spreadsheet solver $($e$.$g$.,$ Goal Seek in Excel$)$

Three tanks are connected with pipes to a junction point J$.$ The pipe length $($m$),$ diameters $($mm$)$ and friction factors $(f)$ are $(600,300,0\mathrm{.}013),(300,400,0\mathrm{.}018)$ and $(1000,250,0\mathrm{.}015),$ for pipes $1,2$ and $3,$ respectively. The tank elevations are $45,20,$ and $10m$ for the tanks connected to pipes $1,2,$ and $3,$ respectively.

Using Newton's method, determine the total head at the junction $J$ and the flow rates in each pipe. Begin your calculations assuming the head at junction J is $15m.$ Continue iterations until the error in your equation is less than $1{10}^{-4}.$

Note: all derivative terms for the tank balance are positive regardless of the flow direction. The form of the terms are given in the class exercise solutions for Module $7.$

If you solve with a spreadsheet, show one iteration by hand.