Question

Explain the Bernoulli Principle in detail with its proof.


Plsss help ! Don't copy from net.

Answer 1

Heya!
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★Bernoulli's Principle ★
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=> This principle states that for the stream line flow of an ideal liquid , the total energy per unit mass remains constant .


[ Note : I am using ¶ for density ]


→ Expression :
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=> P/¶ + ½v² + gh = constant .


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★Proof : We know that an ideal liquid has three kinds of energies . So now we shall calculate all energies divided by the unit mass.


1. ) pressure energy = pressure × volume

=> P × ¶ / M [ as density = Mass / Volume ]

=> Energy per unit mass = P × ¶ / M × 1 /M


= P / ¶ -------------- ( 1. )



2. ) Kinetic Energy = ½ mv²


=> Energy per unit mass = ½ mv² × 1 / m


=> ½ v²-------------( 2. )



3. ) Potential Energy = mgh


=> Energy by unit mass = mgh/m


=> gh-------------(3.)



★Now , from the work energy principle, we know that work done is equal to the change in kinetic energy . I,e


dW = Kf - Ki


Thus , we have


(P1/¶- P2 /¶ ) = ( ½v2² - ½v1²) - ( gh1 - gh2)


=> Thus, this expression can be written as -


•°• P1/¶ + ½v1² + gh1 = P2/¶ + ½ v2² + gh2



Thus , it is proved that the total energy has remained as a Constant !! This is what is stated by the principle .


$...$

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Answer 2

Hey Mate !!
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Here is your answer
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$\textbf{Bernoulli's Principle :}$

According to it for an stream flow of an liquid, the total energy [sum of kinetic energy, potential energy and pressure energy] per unit mass remains constant.. At every cross section throughout the liquid flow.

Consider a streamline flow of ideal liquid across a pipe AB as shown in figure....

For an ideal liquid (ρ) = constant

η = 0

a1, a2 be the area of cross section at A and B

v1, v2 be the volume of liquid at A and B.

P1, p2 the pressure of liquid at A and B.

ρ = density of liquid.

If m is mass entering per sec.

Then;

a1v1ρ = a2v2ρ = m

a1v1 = a2v2 = $\dfrac{m}{\rho}$ = v

F = P1a1

W1 = P1 a1 v1 = P1 V [Work done per pressure at A]

W2 = P2 a2 v2 = P2 V [Work done per pressure at B]

Net work done = P1 V - P2 V

Kinetic Energy at A = $\dfrac{1}{2}$ $mv_{1}^{2}$

Kinetic Energy at B = $\dfrac{1}{2}$ $mv_{2}^{2}$

Potential Energy at A = $mgh_{1}$

Potential Energy at B = $mgh_{2}$

According to work energy principle...

$P_{1}V\:-\:P_{2}$ = $\dfrac{1}{2}mv_{2}^{2}\:-\:\dfrac{1}{2}mv_{1}^{2}$ + $mgh_{2}\:-\:mgh_{1}$

$P_{1}V$ + $\dfrac{1}{2}mv_{1}^{2}$ + $mgh_{1}$ = $P_{2}V$ + $\dfrac{1}{2}mv_{2}^{2}$ + $mgh_{2}$ ....(1)

Divide eq. (1) by m

$\dfrac{P_{1}V}{{m}}$ + $\dfrac{1}{2m}$ $mv_{1}^{2}$ + $\dfrac{mgh_{1}}{{m}}$ = $\dfrac{P_{2}V}{{m}}$ + $\dfrac{1}{2m}$ $mv_{2}^{2}$ + $\dfrac{mgh_{1}}{{m}}$

$\dfrac{P_{1}}{\rho}$ + $\dfrac{1}{2}\:\times\:V_{1}^{2}$ + $gh_{1}$ = $\dfrac{P_{2}}{\rho}$ + $\dfrac{1}{2}\:\times\:V_{2}^{2}$ + $gh_{2}$

$\dfrac{P}{\rho}$ + $\dfrac{1}{2}{V}^{2}$ + $gh$ = Constant...

Hence, proved...

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