Water $(=1000\frac{kg}{m^3})$ flows from the reservoir on the left to the reservoir on the right

at a rate of $0\mathrm{.}45\frac{m^3}{s},$ as shown in the figure below. The major head losses in the pipes

are given by the following formula: $h_L=0\mathrm{.}02(\frac{L}{D})(\frac{V^2}{2g}).$ Assume the head$-$loss formula

can be used for the smaller pipe as well as for the larger pipe. Further, assume that the

only significant head losses come from the major head loss due to friction along the

length of the pipes, and the head loss upon entering the right reservoir.

a$)$ The minor head loss at the entrance into the right reservoir can be expressed as

:

$h_{L\text{m}\text{i}\text{n}\text{o}\text{r}}=K_L\frac{V^2}{2g},$

where $V$ is the average velocity of the fluid just before entering the right reservoir.

What can you say about the value of the minor loss coefficient $K_L?$

b$)$ Sketch the HGL and EGL of the system using the designated diagram on the

next page.

c$)$ Assume that the kinetic energy correction factor is equal to $=1\mathrm{.}0.$ Write down

the energy equation that compares the total energy head in the left reservoir to

the total energy head in the right reservoir. In the diagram, identify the two points

in the fluid you are comparing in your energy equation.

d$)$ Find the required elevation in the left reservoir to produce a flow rate of $0\mathrm{.}45\frac{m}{?}$

${?}^{3}s.$ If you are uncertain about the value of $K_L$ in problem $($a$),$ choose $K_L=$

$0\mathrm{.}5.$