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The displaced fluid volume V is the cross-sectional area A times the thickness x. This
volume remains constant for an incompressible fluid, so
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Eq.(3)
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Using Eq.(3) in Eq.(2) we have
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Eq.(4)
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Since work has been done, there has been a change in the mechanical energy of the fluid segment.
This energy change is found with the help of the next diagram.
The energy change between the initial and final positions is given by
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Eq.(5)
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Here, the the kinetic energy K = mv²/2 where m is the fluid mass and v
is the speed of the fluid. The potential energy U = mgh where g is the
acceleration of gravity, and h is average fluid height.
The work-energy theorem says that the net work done is equal to the change in the system
energy. This can be written as
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Eq.(6)
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Substitution of Eq.(4) and Eq.(5) into Eq.(6) yields
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Eq.(7)
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Dividing Eq.(7) by the fluid volume, V gives us
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Eq.(8)
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where
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Eq.(9)
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is the fluid mass density. To complete our derivation, we reorganize Eq.(8).
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Eq.(10)
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Finally, note that Eq.(10) is true for any two positions. Therefore,
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Eq.(11)
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Equation (11) is commonly referred to as Bernoulli's equation. Keep in mind that this expression
was restricted to incompressible fluids and smooth fluid flows.
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