The forces acting on the fluid result from pressure $(d{F}_P),$ gravity $(d{F}_n),$ wall drag $(d{F}_w),$ and external shaft work $)=(-F_{ex}dx$ :

${\displaystyle \underset{\text{o}\text{n}\text{f}\text{l}\text{u}\text{i}\text{d}}{\overset{?}{}}}{F}_x=d{F}_p+d{F}_g+F_{\text{e}\text{x}\text{t}}+d{F}_w$

where

$d{F}_p=A_x[P-(P+dP)]=-A_{x}dP$

$d{F}_z=g_x{A}_{x}dx=-A_x$dxgcos$=-g{A}_{x}dx$

$d{F}_w=-r_{w}d{A}_w=-r_w{W}_{p}dx$

$-W=F_{ex1}dxor{F}_{ex}=-\frac{W}{dx}$

Here, ${}_w$ is the stress exerted by the fluid on the wall $($the reaction to the stress exerted on the fluid by the wall$),$ and $W_p$ is the perimeter of the wall in the cross section that is wetted by the fluid $($the "wetted perimeter"$).$ After substituting the expressions for the forces from Equation, $4\mathrm{.}34$ into the momentum balance equation, Equation $4\mathrm{.}32,$ and dividing the result by $-A$ where $A=A_x$ the result is

$\frac{dP}{}+gdz+\frac{{}_w{W}_p}{A}dx+w+VdV=-\frac{dV}{dt}dx$

where $w=\frac{W}{Adx}$ is the work done per unit mass of fluid. Integrating this expression from the inlet $($i$)$ to the outlet $($o$)$ and assuming steady state gives

${\displaystyle \underset{P_1}{\overset{P_0}{}}}\frac{dP}{}+g(z_0-z_i)+\frac{1}{2}(V_o^2-V_i^2)+{\displaystyle \underset{L}{\overset{\phantom{}}{}}}\frac{{}_w{W}_P}{A}dx+w=0$

$4\mathrm{.}36$

Explain how the equations $4\mathrm{.}34$ was combined, to get the equations in $4\mathrm{.}35$ and $4\mathrm{.}36.$

Please help me solve this as soon as possible, it is very urgent