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# State and prove Bernoulli’s theorem.

Last updated date: 23rd Apr 2024
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Hint: Here, we first know what Bernoulli's principle states, and then further we can prove the theorem by deriving the expression which relates the pressure and potential energy of the fluid. At last we will also note down the limitations of this theorem.
Formula used:
\eqalign{ & K.{E_{gained}} = \dfrac{1}{2}\rho ({v_2}^2 - {v_1}^2) \cr & P.{E_{gained}} = \rho g({h_2} - {h_1}) \cr}

Bernoulli's principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in static pressure or a decrease in the fluid's potential energy.
To prove Bernoulli's theorem, consider a fluid of negligible viscosity moving with laminar flow, as shown in Figure.
Let the velocity, pressure and area of the fluid column be ${p_1}$, ${v_1}$ and ${A_1}$ at Q and ${p_2}$, ${v_2}$ and ${A_2}$ at R. Let the volume bounded by Q and R move to S and T where QS =${L_1}$, and RT = ${L_2}$.

If the fluid is incompressible:
The work done by the pressure difference per unit volume = gain in kinetic energy per unit volume + gain in potential energy per unit volume. Now:
${A_1}{L_1} = {A_2}{L_2}$
Work done is given by:
\eqalign{& W = F \times d = p \times volume \cr & \Rightarrow {W_{net}} = {p_1} - {p_2} \cr}
\eqalign{& \Rightarrow K.E = \dfrac{1}{2}m{v^2} = \dfrac{1}{2}V\rho {v^2} = \dfrac{1}{2}\rho {v^2}(\because V = 1) \cr & \Rightarrow K.{E_{gained}} = \dfrac{1}{2}\rho ({v_2}^2 - {v_1}^2) \cr}
\eqalign{& {P_1} + \dfrac{1}{2}\rho {v_1}^2 + \rho g{h_1} = {P_2} + \dfrac{1}{2}\rho {v_2}^2 + \rho g{h_2} \cr & \therefore P + \dfrac{1}{2}\rho {v^2} + \rho gh = const. \cr}
For a horizontal tube
\eqalign{& \because {h_1} = {h_2} \cr & \therefore P + \dfrac{1}{2}\rho {v^2} = const. \cr}
Therefore, this proves Bernoulli's theorem. Here we can see that if there is an increase in velocity there must be a decrease in pressure and vice versa.