Obtain Bernoulli’s equation for steady, incompressible, irrotational and viscous liquid.
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)
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