Prove Bernoulli’s Principle.

A pipe is shown in figure.

At point B of left end of pipe

$\rightarrow$ Fluid speed $=v_{1}$

$\rightarrow$ Cross sectional area $=\mathrm{A}_{1} .$

$\rightarrow$ Pressure exerted on fluid $\mathrm{P}_{1}=\frac{\mathrm{F}_{1}}{\mathrm{~A}_{1}}$

At point $D$, of right end of pipe,

$\rightarrow$ Fluid speed $=v_{2}$

$\rightarrow$ Cross sectional area $=\mathrm{A}_{2}$

$\rightarrow$ Pressure exerted on the fluid $\mathrm{P}_{2}=\frac{\mathrm{F}_{2}}{\mathrm{~A}_{2}}$.

Force $\mathrm{F}_{1}=\mathrm{P}_{1} \mathrm{~A}_{1}$ exerted on fluid at point $\mathrm{B}$, covers distance

done on fluid

$\mathrm{W}_{1} =(\text { Force }) \text { (displacement) }$

$\mathrm{W}_{1} =\mathrm{P}_{1} \mathrm{~A}_{1} v_{1} \Delta t$

$\therefore \mathrm{W}_{1} =\mathrm{P}_{1}\left(\Delta \mathrm{V}_{1}\right)$

(where $\Delta \mathrm{V}=\mathrm{A}_{1} v_{1} \Delta t=$ volume of fluid)