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The Euler equations and Bernoulli's theorem
The equations of an ideal fluid, the Lamb form of the acceleration, and the quantity conserved along streamlines.
An ideal fluid has no viscous stress, \(\sigma = -pI\). Cauchy's equation with constant density becomes the incompressible Euler equations (Euler, 1757): \[ \frac{\partial\mathbf{u}}{\partial t} + (\mathbf{u}\cdot\nabla)\mathbf{u} = -\frac{1}{\rho}\nabla p + \mathbf{f}, \qquad \nabla\cdot\mathbf{u} = 0. \] Four equations for four unknowns, the three velocity components and the pressure. At a solid wall only the normal velocity is constrained, \(\mathbf{u}\cdot\mathbf{n} = 0\): an ideal fluid may slide along a boundary.
The acceleration has a useful rewriting, the Lamb form. The vector identity \[ (\mathbf{u}\cdot\nabla)\mathbf{u} = \nabla\!\left(\tfrac12|\mathbf{u}|^2\right) - \mathbf{u}\times\boldsymbol{\omega} \] (check it component by component) turns the Euler equations with a potential force \(\mathbf{f} = -\nabla\Phi\) into \[ \frac{\partial\mathbf{u}}{\partial t} - \mathbf{u}\times\boldsymbol{\omega} = -\nabla H, \qquad H = \frac{p}{\rho} + \tfrac12|\mathbf{u}|^2 + \Phi. \]
Bernoulli's theorem follows at once. In steady flow, dot with \(\mathbf{u}\): since \(\mathbf{u}\times\boldsymbol{\omega}\) is perpendicular to \(\mathbf{u}\), we get \(\mathbf{u}\cdot\nabla H = 0\), so \(H\) is constant along each streamline (and, dotting with \(\boldsymbol{\omega}\), along each vortex line). If in addition the flow is irrotational, \(\nabla H = 0\) and \(H\) is the same constant everywhere. For unsteady irrotational flow with \(\mathbf{u} = \nabla\phi\) the version is \(\partial_t\phi + \tfrac12|\nabla\phi|^2 + p/\rho + \Phi = F(t)\). Daniel Bernoulli's Hydrodynamica (1738) contains the idea; the equation in this form is due to Euler.
The theorem explains a lot with one line. Water leaving a hole at depth \(h\) below the surface of a large tank has \(\tfrac12\rho v^2 = \rho g h\), so \(v = \sqrt{2gh}\) (Torricelli's law, 1643, a century earlier). In a pipe that narrows, continuity makes the speed rise and Bernoulli makes the pressure fall: the Venturi meter. A Pitot tube facing the flow measures the stagnation pressure \(p + \tfrac12\rho v^2\) and so the speed. The limits matter too: Bernoulli ignores viscosity, so it cannot predict drag or losses in long pipes.
One more feature is essential for everything later. Take the divergence of the Euler equations and use \(\nabla\cdot\mathbf{u} = 0\): \[ -\frac{1}{\rho}\Delta p = \partial_i u_j\,\partial_j u_i. \] The pressure is not an independent unknown with its own evolution; it is determined at each instant by the velocity, through a Poisson equation, and depends on the velocity everywhere at once.
Picture it: along a streamline, energy per unit volume moves between three accounts (pressure, kinetic, potential) and the total never changes. Think it: Bernoulli is conservation of energy for an ideal fluid restricted to a streamline, and the Poisson equation for pressure is why incompressible flow is non-local: a push here is felt everywhere instantly.
उदाहरण · solve 0.5*1000*v^2 = 1000*9.81*5
चरणद्वारा चरण
- 500.0 v^{2} = 49050.0
Start from the equation as given.
- 500.0 v^{2} - 49050.0 = 0
Bring everything to one side and collect like terms so the right-hand side is 0.
- a = 500.0,\quad b = 0.0,\quad c = -49050.0
Read off the coefficients of the standard form ax² + bx + c = 0.
- \Delta = b^2 - 4ac = (0.0)^2 - 4(500.0)(-49050.0) = 98100000.0
It does not factor nicely, so use the discriminant Δ = b² − 4ac.
- v = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{0.0 \pm \sqrt{98100000.0}}{1000.0}
Δ > 0, so there are two distinct real roots. Substitute into the quadratic formula.
- v = 9.90454441 \quad\text{or}\quad v = -9.90454441
Simplify each root.
जवाफ प्रकट गर्नुहोस्
अब तपाईँ समस्या रोज्नुहोस्, वा टाइप गर्नुहोस् वा तपाईँको आफ्नै रेखाचित्र गर्नुहोस् । प्रत्येक चरण, एउटा तस्वीर, तपाईँले सोध्नु अघि उत्तर लुकेको छ ।
एक निःशुल्क खाता प्रत्येक पाठ मा नोट थप्छ, तपाईं के समाप्त गरेको छ को एक रेकर्ड, एक ठाउँमा आफ्नो समाधान समस्या, र एक शिक्षक तपाईं यो पृष्ठ बारेमा सोध्न सक्नुहुन्छ. गणित आफैलाई सबैलाई खुला छ, मा साइन इन वा छैन.
साइन अप गर्नुहोस् लगइनयहाँ प्रयोग गरिएको प्रतीक
पूर्ण परिभाषा, एउटा तस्वीर, र यो प्रत्येक अक्षर अर्थ के लागि कुनै पनि प्रतीक ट्याप गर्नुहोस्।
कसरी: The Euler equations and Bernoulli's theorem
- Check the conditions: steady, inviscid, constant density, conservative force.
- Pick two points on the same streamline (anywhere, if the flow is irrotational).
- Write p/rho + v^2/2 + g z at both points and set them equal.
- Use continuity (area times speed is constant in a pipe) to eliminate one unknown speed.
प्रश्नहरू मानिसहरूले सोध्छन्
Does Bernoulli explain lift?
It relates the lower pressure on top of a wing to the faster flow there, which is correct. What it does not explain is why the flow is faster on top; that needs the circulation set up by viscosity at the trailing edge.
Why can the Euler equations not satisfy no-slip?
They are first order in space, so they accept only the normal-velocity condition. Imposing the tangential velocity as well over-determines them, which is the root of the boundary layer.
What do I need before starting fluid dynamics?
Multivariable calculus (divergence, curl, the divergence and Stokes theorems), linear algebra (symmetric matrices and eigenvalues) and partial differential equations (the heat and Laplace equations). The last lessons also use Sobolev spaces, which are introduced where they are needed.
What are the Navier-Stokes equations in one sentence?
Newton's second law for each particle of a viscous incompressible fluid: acceleration equals the pressure force plus viscous diffusion of momentum, with the constraint that the velocity field is divergence-free.
Why is two-dimensional flow easier than three-dimensional flow?
In 2D the vorticity is a scalar that is only carried and diffused, so its maximum never grows; in 3D vortex lines can be stretched, which amplifies vorticity, and no known bound rules out unlimited growth.
Is this course physics or mathematics?
Both, in order. The first half derives the equations from physical principles and solves classical flows; the second half treats the equations as mathematical objects and studies which of their properties can be proved.
यसमा थप Fluid Dynamics
The continuum hypothesis and fieldsEulerian and Lagrangian descriptions, the material derivativeKinematics: streamlines, pathlines and streaklinesConservation of mass, the continuity equation and incompressibilityThe stream function and two-dimensional incompressible flowVorticity, circulation and Kelvin's circulation theoremThe stress tensor and Cauchy's momentum equationHydrostatics and pressurePotential flow and the Laplace equationViscosity, Newtonian fluids and the Navier-Stokes equationsExact solutions: Couette, Poiseuille and Stokes' first problemDimensional analysis, the Reynolds number and the scaling symmetryStokes flow at low Reynolds numberBoundary layers