📂Fluid Mechanics

Bernoulli's Equation in Fluid Dynamics

Theorem ¹

A fluid that is inviscid and incompressible undergoing steady flow — in a $1$-dimensional column of fluid, let the velocity at height $z$ be $u$. For pressure $p$, density $\rho$, and gravitational acceleration $g$, the energy per unit volume at height $z$ is given by the following and is constant. $$ {\frac{ \rho u^{2} }{ 2 }} + \rho g z + p $$

Explanation

Upstream and downstream: put simply, if there is an upstream and a downstream and the pipe’s cross-section is the same, one can intuitively expect the downstream side, which bears the greater load, to be faster. This can be seen as the conversion of potential energy into kinetic energy, analogous to the energy conservation law in classical mechanics.

Proof

Euler equation: $$ {\frac{ \partial \mathbf{u} }{ \partial t }} + \left( \mathbf{u} \cdot \nabla \right) \mathbf{u} = - {\frac{ 1 }{ \rho }} \nabla p + \mathbf{g} $$

Reducing the Euler equation in vector form to a one-dimensional form considering only the height $z$ gives $\mathbf{g} = - g$, hence $$ {\frac{ \partial u }{ \partial t }} + u {\frac{ \partial u }{ \partial z }} = - {\frac{ 1 }{ \rho }} {\frac{ \partial p }{ \partial z }} - g $$ Assuming steady flow, we can set $u_{t} = 0$. Multiplying both sides by $\rho$ and integrating both sides along a streamline yields $$ \rho \int u {\frac{ \partial u }{ \partial z }} dz = - \int {\frac{ \partial p }{ \partial z }} dz - \rho g \int dz $$ Bringing all terms to the left-hand side, we obtain the following for some integration constant $C$. $$ \begin{align*} \rho \int u {\frac{ \partial u }{ \partial z }} dz + \int {\frac{ \partial p }{ \partial z }} dz + \rho g \int dz =& C \\ \implies \rho {\frac{ u^{2} }{ 2 }} + p + \rho g z =& C \end{align*} $$

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