$($solve this question using method $1$ and $2$ and the table as shown in the pictures please$)$:Method $1$: We try to express the head loss due to minor losses in terms of a

loss coefficient, $K_L$ :

$(\frac{F}{g}{)}_{\text{m}\text{i}\text{n}\text{o}\text{r}\text{l}\text{o}\text{s}\text{s}\text{e}\text{s}}=K_L\frac{V^2}{2g}$

Values of $K_L$ can be found in the literature

for example:

losses due to change in pipe diameter

losses due to pipe components Minor Losses

Method $2$:

Using the concept of equivalent length, which is the equivalent length of pipe

which would have the same friction effect as the fitting.

The B$.$E$.$ can be written:

$\frac{P}{}+gz+\frac{v^2}{2}=\frac{dw}{dm}-(4f(\frac{L}{D}{)}_{\text{e}\text{q}\text{u}\text{i}\text{v}\text{.}}*\frac{v^2}{2})$

Values of equivalent length $($L$/$D$)$ can be found in the literature $($for example

Table $6\mathrm{.}7$ textbook; see next slide$)$ TABLE $6\mathrm{.}7$

Equivalent lengths and $K$ values for various kinds of fitting$*$

$\backslash$table$[[$Type of fitting,$\backslash$table$[[$Equivalent length,$],[$L$/$D$,$ dimensionless$]],\backslash$table$[[$Constant$,$ K$,$ in Eq$.6\mathrm{.}25,],[$dimensionless$]]],[$Globe valve, wide open,$350,6\mathrm{.}3],[$Angle valve, wide open,$170,3\mathrm{.}0],[$Gate valve, wide open,$7,0\mathrm{.}13],[$Check valve, swing type,$110,2\mathrm{.}0],[90\backslash$deg standard elbow,$32,0\mathrm{.}74],[45\backslash$deg standard elbow,$15,0\mathrm{.}3],[90\backslash$deg long$-$radius elbow,$20,0\mathrm{.}46],[$Standard tee, flow$-$through run,$20,0\mathrm{.}4],[$Standard tee, flow$-$through branch,$60,1\mathrm{.}3],[$Coupling$,2,0\mathrm{.}04],[$Union$,2,0\mathrm{.}04]]$

$*$Source: Reference $10.$Example

Repeat the example above using method $1$ and method $2,$ use Table $6\mathrm{.}7$