Question
State and prove Bernoulli's theorem.
✓
Answer
According to Bernoulli's theorem, for an incompressible, non-viscous liquid having streamlined flow, the sum of pressure head, velocity head and gravitational head is a constant, i.e., $\frac{\text{P}}{\rho\text{g}}+\frac{\text{v}^2}{2\text{g}}+\text{h}=\text{constant}$ Consider an incompressible non-viscous liquid entering the cross-section A1 at A with a velocity v1 and coming out at a height h2 at B with velocity v2. The P.E. and K.E. increase since h, and v, are more than h1 and v1 respectively. This is done by the pressure doing work on the liquid. If P1 and P2 are the pressure at A and B, for a small displacement at A and B, The work done on the liquid at A = (P1 A1) 
$\Delta\text{x}_1=\text{P}_1\text{A}_1\text{v}\Delta\text{t}$
The work done by the liquid at B,$\Delta\text{x}_2=-(\text{P}_2\text{A}_2)$
$\Delta\text{x}_2=-\text{P}_2\text{A}_2\text{V}\Delta\text{t}$
The work done on the liquid at (Considering a small time $\Delta\text{t}$ so that area may be same) Net work done by pressure $=(\text{P}_1-\text{P}_2)\text{Av }\Delta\text{ t}$ since A1v1 = A2v2 From conservation of energy, $(\text{P}_1-\text{P}_2)\text{Av}\Delta\text{t}=\text{Change in }(\text{K.E.}+\text{P.E.})$
$(\text{P}_1-\text{P}_2)\text{A}\text{v}\Delta\text{t}$
$=\text{Av}\rho\Delta\text{tg}(\text{h}_2-\text{h}_1)+\frac{1}{2}\text{Av}\Delta\text{t}\rho(\text{v}_2^2-\text{v}^2_1)$
$\therefore\text{P}_1-\text{P}_2=\rho\text{g}(\text{h}_2-\text{h}_1)+\frac{\rho}{2}(\text{v}^2_2-\text{v}^2_1)$
(i.e.) $\text{P}_1+\rho\text{gh}_1+\frac{\rho}{2}\text{v}^2_1=\text{P}_1+\rho\text{gh}_2+\frac{\rho}{2}\text{v}^2_2$$\therefore\frac{\text{P}}{\rho\text{g}}+\text{h}+\frac{\text{v}^2}{\text{2g}}=\text{constant}.$
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