## Eulers Equation of Motion

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# Euler’s Equation of Motion

Consider a streamline in which flow take place in s-direction as shown in .

Consider a cylindrical element of cross-sectional area ‘dA’ and length ‘ds’.

Let the pressure force in the direction of flow is p. ( ) and in the opposite to the direction of flow is ( ( p + ∂p / ∂s ds ) dA.

Weight of element is ρg dA. ds.

Consider 'θ' be the angle between direction of flow and weight.

By Newton’s second law of motion,

Net force in the direction of flow = Mass of fluid element x Acceleration in the direction of flow

pdA - ( p + ∂p / ∂s. ds) dA - ρg dA ds cos θ = ρ dA ds x as

- ∂p / ∂s ds. dA - ρg dA. ds cos θ = ρ dA ds x a

- ∂p / ∂s - ρg cos θ = ρ a

But as = dv / dt

a

a

For steady flow, ∂v / ∂t = 0

a

Now consider, form triangle

cos θ = dz / ds

Substituting the value of Equations and in Equation

- ∂p / ∂s - ρg dz / ds - ρv ∂v / ∂s = 0

∂p / ∂s + ρg dz / ds + ρv ∂v / ∂s = 0

∂p + ρg dz + ρv dv = 0

∂p / ρ + g dz + dv = 0

This is the Euler's equation of motion

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Consider a cylindrical element of cross-sectional area ‘dA’ and length ‘ds’.

Let the pressure force in the direction of flow is p. ( ) and in the opposite to the direction of flow is ( ( p + ∂p / ∂s ds ) dA.

Weight of element is ρg dA. ds.

Consider 'θ' be the angle between direction of flow and weight.

By Newton’s second law of motion,

Net force in the direction of flow = Mass of fluid element x Acceleration in the direction of flow

pdA - ( p + ∂p / ∂s. ds) dA - ρg dA ds cos θ = ρ dA ds x as

- ∂p / ∂s ds. dA - ρg dA. ds cos θ = ρ dA ds x a

_{s}- ∂p / ∂s - ρg cos θ = ρ a

_{s}But as = dv / dt

a

_{s}= dv / dt = ∂v ./ ∂s . ds / dt ∂v / ∂ta

_{s}= v. ∂v / ∂s + ∂v / ∂tFor steady flow, ∂v / ∂t = 0

a

_{s}= ∂v / ∂sNow consider, form triangle

cos θ = dz / ds

Substituting the value of Equations and in Equation

- ∂p / ∂s - ρg dz / ds - ρv ∂v / ∂s = 0

∂p / ∂s + ρg dz / ds + ρv ∂v / ∂s = 0

∂p + ρg dz + ρv dv = 0

∂p / ρ + g dz + dv = 0

This is the Euler's equation of motion

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