Question

$\sf{Prove \ Bernoulli \ Theorem.}$
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Answer 1

$\bold{ BERNOULLI'S\: THEOREM}$
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It states that when a non - Viscous liquid and incompressible liquid flows through a tube of non - uniform cross section area, then at each point, the total energy of the liquid ( Kinetic + Potential + Pressure energy ) per unit volume will remain constant. 

P +$\frac{1}{2}$ 21​ ρ v² + ρgh = constant 

\bold{Proof}Proof :

Consider an ideal liquid of density ρ,is flowing through the tube in streamline flow, then the force experienced by the liquid enters at the end in time t, 

F₁ = P₁ A₂

Work done on the liquid at end x, 

W₁ = F₁ × v ₁ ∆t

W₁ = P₁ A₁ ( v ₁ ∆t ) 

W₁ = P₁ ∆v₁ 

Similarly, work done by the liquid at end Y, 

W₂ = P₂ ∆v₂

Net work done on the liquid, 

W = ( P₁ ∆v₁ ) - ( P₂ ∆v₂ ) 

From equation of continuity, 

A₁ V₁ = A₂ V₂ 

A₁ V₁ ∆t = A₂ V₂ ∆t

∆v₁ = ∆v₂

W = ( P₁ - P₂ ) ∆v

W = ( P₁ - P₂ )$\frac{m} {ρ}$ ρm​ ---> ( i ) 

Kinetic Energy, 

∆K =$\frac{1}{2}$ ​ mv₂ ² - $\frac{1}{2}$ mv₁² 

Potential Energy, 

∆P = mgh₂ - mgh₁

Total energy of the liquid, 

∆E = ∆K + ∆P 

∆E = ($\frac{1}{2}$ 21​ mv₂ ² -$\frac{1}{2}$ 21​ mv₁² ) + ( mgh₂ - mgh₁ ) 

From the law of conservation of energy, 

W = ∆E

( P₁ - P₂ )$\frac{m} {ρ}$ ρm​ = ($\frac{1}{2}$ mv₂² - $\frac{1}{2}$ ​ mv₁² ) + ( mgh₂ - mgh₁ )

( P₁ - P₂ ) =$\frac{1}{2}$  ρv₂² -$\frac{1}{2}$ ρv₁² + ρgh₂ -ρgh₁ 

P₁ +$\frac{1}{2}$​ ρv₁² +ρgh₁ = P₂ +$\frac{1}{2}$ ρv₂² + ρgh₂ 

P + $\frac{1}{2}$ ρv² +ρgh = constant. 
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Answer 2

$\huge{\boxed{\underline{\mathbb\blue{ANSWER}}}}$

$\mathscr\purple{BERNOULLI'S\:THEOREM}$

$\tt{STATEMENT}$==>>
$\sf{It states that," sum of all energies , namely, POTENTIAL energy, kinetic energy, and pressure energy per unit mass always remains constant for an ideal fluid , which is incompressible and non - viscous , and possesses stream line flow..}$

Mathematically...

$\frac{P}{p}+\frac{v^{2}}{2}+gh$

>>>

proof=> Let's consider an ideal fluid flowing with streamline flow along a pipe of variable area. Let at points 'A','B' liquid is having certain pressure , area, velocity, and height as P1, a1, v1,h1 and P2, a2, v2, h2 respectively.Let 'p' be density of incompressible liquid.

This liquid enters from A and leaves from B.

therfore, force acting on fluid at point F =P1 a1 ==>distance travelled by fluid in one second at A =v1 × 1 = v1

therefore, work done per second on fluid at A = $Force\times distance travelled$
= $P1\times a1\times v1$

=> SIMILARLY, work done at point B is
= $P2\times a2\times v2$

hence,,

Net work done by pressure energy is given by.

W = $P1\times a1\times v1-P2\times a2\times v2$

W= $\frac{P1\times m}{p}-\frac{P2\times m}{p}$----(1)

here mass is same, since same amount of water flows through pipe, but with different pressure, velocity,etc..

Simultaneously, the work done by pressure energy increases kinetic energy and potential energy .

Increase in Fluid potential energy , is given by,

= P.E at B - P.E at A

= $m\times g\times h1-m\times g\times h2$--------(2)

Increase in Fluid Kinetic energy, is given by,

= K.E at B - K.E at A

= $\frac{m}{2}$$v2^{2}-\frac{m}{2}$$v1^{2}$-----(3)

according to law of conservation of energy, workdone by pressure energy = Total increase in energy(P.E+K.E)..

$\frac{P1\times m}{p}-\frac{P2\times m}{p}$ = $m\times g\times h1-m\times g\times h2$ + $\frac{m}{2}v2^{2}-\frac{m}{2}v1^{2}$

m$\frac{P1}{p}-\frac{P2}{p}$ = m$g\times h1-g\times h2$ + m$\frac{v2^{2}}{2}-\frac{v1^{2}}{2}$

(mass 'm' gets cancelled)

$\frac{P1}{p}-\frac{P2}{p}$ = $g\times h1-g\times h2$ + $\frac{v2^{2}}{2}-\frac{v1^{2}}{2}$

$\frac{P1}{p}+g\times h1+\frac{v1^{2}}{2}$ =
$\frac{P2}{p}+g\times h2+\frac{v2^{2}}{2}$

$\frac{P1}{p}+g\times h1+\frac{v1^{2}}{2}$ =
$\frac{P2}{p}+g\times h2+\frac{v2^{2}}{2}$= constant.....(4)

This is the actual Bernoulli equation...

Modified equation (4)..i.e, divide equation (4) by g(acceleration due to gravity)

$\frac{P}{pg}+h+\frac{v^{2}}{2g}$ = $\frac{constant}{g}$ = another constant...

Note==>>

1. In the above equation, every term has its own dimensions, and it is referred as HEAD..

2.For example::
'$\frac{P}{pg}$' is called as Pressure head,,
SIMILARLY,, '$\frac{v^{2}}{pg}$' is called as VELOCITY head..
and 'h' be the gravitational head..

$\mathbb\blue{HENCE\:THE\:pr00f}$