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}`ρv₁² + ρgh₁ = P₂ +`\frac {1}{2}`ρv₂² + ρgh₂ 

or

P + `\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}`
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 {Δm}{Δv}`

Or 

Δv = `\frac {Δm}{ρ}`

Comparing above equations

`\frac {Δm}{ρ}` = AΔx

Therefore by putting `\frac {Δm}{ρ}` = AΔx

For end 1: 
W₁ = P₁ `\frac {Δm_1}{ρ}`

For end 2:
W₂ = -P₂`\frac {Δm_2}{ρ}`

Since the total work done will be the sum of all the individual work done.
W = P₁ `\frac {Δm_1}{ρ}` - P₂`\frac {Δm_2}{ρ}` ----------------(2)

The net change in kinetic energy ΔK is
ΔK.E. = `\frac {1}{2}`Δm₂v₂² -`\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 {Δm_1}{ρ}` - P₂`\frac {Δm_2}{ρ}` = `\frac {1}{2}`Δm₂v₂² -`\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 {Δm}{ρ}` - P₂`\frac {Δm}{ρ}` = `\frac {1}{2}`Δmv₂² -`\frac {1}{2}`Δmv₁² + Δmgh₂ - Δmgh₁

By taking `\frac {Δm}{ρ}` from RHS and  Δm  from LHS as common we have
`\frac {Δm}{ρ}` ( P₁ - P₂ ) = Δm (`\frac {1}{2}`v₂² -`\frac {1}{2}`v₁² + gh₂ - gh₁)

Multiplying both sides by `\frac {ρ}{Δm}` we get

P₁ - P₂ = ρ (`\frac {1}{2}`v₂² -`\frac {1}{2}`v₁² + gh₂ - gh₁)

or

P₁ - P₂ = `\frac {1}{2}`ρv₂² -`\frac {1}{2}`ρv₁² + ρgh₂ - ρgh₁

or (by arranging)

P₁ + `\frac {1}{2}`ρv₁² + ρgh₁ = P₂ + `\frac {1}{2}`ρv₂² + ρgh₂ ----------------(5)

or

P + `\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}`ρv₁² + ρgh = P₂ + `\frac {1}{2}`ρv₂² + ρgh

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

P₁ + `\frac {1}{2}`ρv₁² = P₂ + `\frac {1}{2}`ρv₂² ----------------(7)

or

 P + `\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}`ρv₁² - `\frac {1}{2}`ρv₂² 

P₂  =  P₁ + `\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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