Question

Explain Bernoulli's Theorem

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Answer 1

Bernoulli's Theorem

According to it, for an stream line flow of liquid; total energy per unit mass remains constant.

[ Refer the attachment for figure ]

Proof

Consider an stream line flow of ideal liquid as shown in figure.

Here,

$\sf{a_1}$ and $\sf{a_2}$ are the area of cross-section

$\sf{v_1}$ and $\sf{v_2}$ are the volume

$\sf{\rho}$ is the density of the liquid

According to equation of continuity; Mass of liquid entering = Mass of liquid leaving

$\sf{a_1v_1 \rho_1}$ = $\sf{a_2v_2 \rho_2}$

$\sf{\rho_1\:=\:\rho_2\:=\:\rho}$

If mass "m" is entering per second in the pipe then,

$\sf{a_1v_1 \rho}$ = $\sf{a_2v_2 \rho}$ = $\sf{m}$

$\sf{a_1v_1\:=\:a_2v_2\:=\:\frac{m}{\rho}}$ ....(1)

Now, word done on liquid at point A per second = $\sf{P_1a_1v_1}$

Work done on liquid at point B per second = $\sf{P_2a_2v_2}$

($\sf{P_1\:and\:P_2}$ are the pressure of liquid)

Net word done by pressure energy from A to B = Work done of liquid at point A per second - Word done on liquid at point B per second

$\implies\:\sf{P_1v_1a_1\:-\:P_2v_2a_2}$

$\implies\:\sf{P_1 \frac{m}{\rho}\:-\:P_2 \frac{m}{\rho}}$

$\implies\:\sf{\frac{m}{\rho}(P_1\:-\:P_2)}$ ....(2)

Now, increase in kinetic energy of the liquid per second = $\sf{\frac{1}{2}mv_2^{2}\:-\:\frac{1}{2}mv_1^{2}}$

$\rightarrow\:\sf{\frac{1m}{2}(v_2^{2}\:-\:v_1^{2})}$

Increase in potential energy of the liquid per second = $\sf{mgh_2\:-\:mgh_1}$

$\rightarrow\:\sf{mg(h_2\:-\:h_1)}$

According to work energy conservation principle

Work done on body is equal to increase in emrgy on body

$\implies\:\sf{\frac{m}{\rho}(P_1-P_2)\:=\:mg(h_2-h_1)+\frac{1}{2}(v_2^{2}-v_1^{2})}$

m throughout cancel

$\implies\:\sf{\frac{P_1}{\rho}-\frac{P_2}{\rho}\:=\:gh_2-gh_1+\frac{1}{2}v_2^{2}-\frac{1}{2}v_1^{2}}$

$\implies\:\sf{\frac{P_1}{\rho}+gh_1+\frac{1}{2}v_1^{2}\:=\:\sf{\frac{P_2}{\rho}+gh_2+\frac{1}{2}v_2^{2}}}$

$\implies\:\sf{\frac{P}{\rho}\:+\:gh\:+\:\frac{1}{2}v^2}\:=\:constant$

Hence, proved