Question

prove bernoulli equation

Answer 1

Bernoulli's principle can be derived from the principle of conservation of energy. This states that, in a steady flow, the sum of all forms of energy in a fluid along a streamline is the same at all points on that streamline.

Answer 2

Bernoulli's Theorem :

This theorem states that when an ideal fluid is flowing through a pipe, then at all points in its cross - section, the sum of Kinetic energy ( K.E. ) per unit volume, Potential energy ( P.E. ) per unit volume and pressure energy per unit volume is always constant.


Derivation :

{ See the diagram in Attachment. }

Consider an ideal fluid having density rho ( ¶ ) is flowing through a pipe of different cross - sections. At one end, area be ${a_1}$, velocity be ${v_1}$ and at other end, area be ${a_2}$ and velocity of liquid be ${v_2}$. And the corresponding heights be of two sections be ${h_1}$ and ${h_2}$.

Change in K.E. = 1 / 2 × ${m_2}$${v^2_2}$ - 1 / 2 × ${m_1}$${v^2_1}$ _______( 1 )

Change in P.E. = ${m_2gh_2}$ - ${m_1gh_1}$ _______( 2 )

If ${p_1}$ and ${p_2}$ be the pressure at sections points A and B, then work done for small displacement in small time ∆t will be

Net work done = work done on the fluid - work done by the fluid

Net work done = ${P_1a_1v_1}$∆t - ${P_2a_2v_2}$∆t _______( 3 )

Also,

Net work done = Change in K.E. + Change in P.E.

•°•
${P_1a_1v_1}$∆t - ${P_2a_2v_2}$∆t = ( 1 / 2 × ${m_2}$${v^2_2}$ - 1 / 2 × ${m_1}$${v^2_1}$ ) + ( ${m_2gh_2}$ - ${m_1gh_1}$ ) ___________( 4 )

Now,

Displacement = velocity × time

For small time, ∆t

Displacement = v × ∆t = v∆t _____( 5 )

Also,

Volume = Area × Displacement

V = a × v∆t [ From ( 5 ) ]

V = av∆t ________( 6 )

As we know,

Mass = Density × Volume

m = ¶ × V

m = ¶ × av∆t [ From ( 6 ) ]

m = ¶av∆t

Put this value in ( 4 ), we get

${P_1a_1v_1}$∆t - ${P_2a_2v_2}$∆t = ( 1 / 2 × ¶${a_2v_2}$∆t${v^2_2}$ - 1 / 2 × ¶${a_1v_1}$∆t${v^2_1}$ ) + ( ¶${a_2v_2}$∆tg${h_2}$ - ¶${a_1v_1}$∆tg${h_1}$ )

From equation of continuity,

${a_1v_1}$ = ${a_2v_2}$

•°•

${P_1a_1v_1}$∆t - ${P_2a_1v_1}$∆t = ( 1 / 2 × ¶${a_1v_1}$∆t${v^2_2}$ - 1 / 2 × ¶${a_1v_1}$∆t${v^2_1}$ ) + ( ¶${a_1v_1}$∆tg${h_2}$ - ¶${a_1v_1}$∆tg${h_1}$ )


Taking common

${a_1v_1}$∆t [ ${P_1}$ - ${P_2}$ ] = ${a_1v_1}$∆t ( 1 / 2 × ¶${v^2_2}$ - 1 / 2 × ¶${v^2_1}$ + ¶g${h_2}$ - ¶g${h_1}$ )

${P_1}$ - ${P_2}$ = 1 / 2 × ¶${v^2_2}$ - 1 / 2 × ¶${v^2_1}$ + ¶g${h_2}$ - ¶g${h_1}$

${P_1}$ + 1 / 2 ¶${v^2_1}$ + ¶g${h_1}$ = ${P_2}$ + 1 / 2 ¶${v^2_2}$ + ¶g${h_2}$

P + 1 / 2 ¶${v^2}$ + ¶gh = Constant

Hence, proved

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