Derive relation for Bernoulli’s equation Unit-6 Fluid dynamics (Class 11 -- SLO based Physics Notes)

Bernoulli’s equation relates three quantities, the flow speed, height, and pressure for the flow of an ideal fluid. It is based upon the law of conservation of energy or the work-energy principle applied to fluids in motion.

Mathematically

P₁ +

\frac{1}{2}

$\frac {1}{2}$ ρv₁² + ρgh₁ = P₂ +

\frac{1}{2}

$\frac {1}{2}$ ρv₂² + ρgh₂

P +

\frac{1}{2}

$\frac {1}{2}$ ρv² + ρgh = constant

Consider an ideal fluid flowing through a pipe of nonuniform size as shown in the figure. The work ‘W’ is done due to a force other than the conservation force of gravity, so it equals the change in the total mechanical energy (kinetic energy + gravitational potential energy) associated with the fluid element.

W = ΔE
W = ΔK.E. + ΔU ----------------(1)

The total work done will be the sum of all the individual work done.

W = W₁ + W₂

For end 1

By definition of work

W₁=F₁ . ΔX₁
W₁=F₁ΔX₁cosθ
Here θ = 0° and cos 0 = 1 therefore

W₁=F₁ ΔX₁

For end 2

W₂ = F₂ . ΔX₂

W₂ = F₂ ΔX₂ cosθ

Here θ = 180° and cos 180 = -1
Therefore

W₂ = - F₂ ΔX₂

By definition of pressure

P =

\frac{F}{A}

$\frac {F}{A}$
Or

F = PA

From the equation above equations, we can write by putting F = PA

For end 1:

W₁ = P₁A₁ ΔX₁

For end 2:

W₂ = -P₂A₂ ΔX₂

Since Volume ΔV = A ΔX

By definition of density ρ =

\frac{Δ m}{Δ v}

$\frac {Δm}{Δv}$

Δv =

\frac{Δ m}{ρ}

$\frac {Δm}{ρ}$

Comparing above equations

\frac{Δ m}{ρ}

$\frac {Δm}{ρ}$ = AΔx

Therefore by putting

\frac{Δ m}{ρ}

$\frac {Δm}{ρ}$ = AΔx

For end 1:

W₁ = P₁

\frac{Δ m_{1}}{ρ}

$\frac {Δm_1}{ρ}$

For end 2:

W₂ = -P₂

\frac{Δ m_{2}}{ρ}

$\frac {Δm_2}{ρ}$

Since the total work done will be the sum of all the individual work done.

W = P₁

\frac{Δ m_{1}}{ρ}

$\frac {Δm_1}{ρ}$ - P₂

\frac{Δ m_{2}}{ρ}

$\frac {Δm_2}{ρ}$ ----------------(2)

The net change in kinetic energy ΔK is
ΔK.E. =

\frac{1}{2}

$\frac {1}{2}$ Δm₂v₂² -

\frac{1}{2}

$\frac {1}{2}$ Δm₁v₁² ----------------(3)

The net change in potential energy ΔU is
ΔU = Δm₂gh₂ - Δm₁gh₁ ----------------(4)

Now putting the values of equations (2), (3), and (4) in equation (1)

P₁

\frac{Δ m_{1}}{ρ}

$\frac {Δm_1}{ρ}$ - P₂

\frac{Δ m_{2}}{ρ}

$\frac {Δm_2}{ρ}$ =

\frac{1}{2}

$\frac {1}{2}$ Δm₂v₂² -

\frac{1}{2}

$\frac {1}{2}$ Δm₁v₁² + Δm₂gh₂ - Δm₁gh₁

Since for ideal fluid equal mass should flow across both ends i.e Δm₁=Δm₂=Δm

The above equation can be written again as
P₁

\frac{Δ m}{ρ}

$\frac {Δm}{ρ}$ - P₂

\frac{Δ m}{ρ}

$\frac {Δm}{ρ}$ =

\frac{1}{2}

$\frac {1}{2}$ Δmv₂² -

\frac{1}{2}

$\frac {1}{2}$ Δmv₁² + Δmgh₂ - Δmgh₁

By taking

\frac{Δ m}{ρ}

$\frac {Δm}{ρ}$ from RHS and Δm from LHS as common we have

\frac{Δ m}{ρ}

$\frac {Δm}{ρ}$ ( P₁ - P₂ ) = Δm (

\frac{1}{2}

$\frac {1}{2}$ v₂² -

\frac{1}{2}

$\frac {1}{2}$ v₁² + gh₂ - gh₁)

Multiplying both sides by

\frac{ρ}{Δ m}

$\frac {ρ}{Δm}$ we get

P₁ - P₂ = ρ (

\frac{1}{2}

$\frac {1}{2}$ v₂² -

\frac{1}{2}

$\frac {1}{2}$ v₁² + gh₂ - gh₁)

P₁ - P₂ =

\frac{1}{2}

$\frac {1}{2}$ ρv₂² -

\frac{1}{2}

$\frac {1}{2}$ ρv₁² + ρgh₂ - ρgh₁

or (by arranging)

P₁ +

\frac{1}{2}

$\frac {1}{2}$ ρv₁² + ρgh₁ = P₂ +

\frac{1}{2}

$\frac {1}{2}$ ρv₂² + ρgh₂ ----------------(5)

P +

\frac{1}{2}

$\frac {1}{2}$ ρv² + ρgh = Constant----------------(6)

Hence equations (5) and (6) are two forms of the required Bernoulli’s equation.

For Horizontal Pipe:

In the horizontal pipe, the influx and efflux are at the same level, then

h₁=h₂=h

Equation (5)

P₁ +

\frac{1}{2}

$\frac {1}{2}$ ρv₁² + ρgh = P₂ +

\frac{1}{2}

$\frac {1}{2}$ ρv₂² + ρgh

or (ρgh will cancel from both sides) hence we get

P₁ +

\frac{1}{2}

$\frac {1}{2}$ ρv₁² = P₂ +

\frac{1}{2}

$\frac {1}{2}$ ρv₂² ----------------(7)

P +

\frac{1}{2}

$\frac {1}{2}$ ρv² = Constant----------------(8)

Bernoulli’s equation for the ideal fluid flow in a horizontal pipe.

From equation (6)

P₂ = P₁ +

\frac{1}{2}

$\frac {1}{2}$ ρv₁² -

\frac{1}{2}

$\frac {1}{2}$ ρv₂²

P₂ = P₁ +

\frac{1}{2}

$\frac {1}{2}$ ρ (v₁² - v₂² )

This shows that

P₂ < P₁ and v₂ > v₁

So if the velocity of the fluid increases, its pressure decreases, and vice versa.

As pressure and Area are inversely related so when the area increases, its pressure decreases, and vice versa.

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