Speed ​​potential

The velocity potential is introduced for eddy-free , two- and three-dimensional flows of fluid dynamics . With it, the calculations are simplified and you gain a deeper mathematical-physical understanding. The velocity potential of fluid dynamics corresponds mathematically to the electrostatic or gravitational potential . ${\ displaystyle \ phi}$

This article deals with the two-dimensional case - the three- dimensional case is presented in the article Potential flow.

Solving the equation gives the equipotential lines of the flow field. ${\ displaystyle \ phi (x, y) = {\ text {const.}}}$

In addition, the current function is introduced , the clear meaning of which is that the solutions of the equation represent the streamlines of the velocity potential. ${\ displaystyle \ psi}$${\ displaystyle \ psi (x, y) = {\ text {const.}}}$

The complex speed potential is formed from the speed potential and the current function .

Basics

For an eddy-free two-dimensional flow field , the rotation is equal to 0: ${\ displaystyle {\ vec {u}} (x, y)}$

${\ displaystyle {\ vec {\ nabla}} \ times {\ vec {u}} (x, y) = 0}$

Similar to the case of the electrostatic potential, the velocity potential is now introduced . The gradient of this potential is precisely the flow field: ${\ displaystyle \ phi (x, y)}$

${\ displaystyle {\ vec {u}} (x, y) = {\ vec {\ nabla}} \ phi (x, y) = \ left ({\ frac {\ partial \ phi} {\ partial x}} , {\ frac {\ partial \ phi} {\ partial y}} \ right)}$

Because of this , the flow field is automatically free of eddies. ${\ displaystyle {\ vec {\ nabla}} \ times {\ vec {\ nabla}} \ phi (x, y) = 0}$

Furthermore, the continuity equation also applies to the velocity field in the case of an incompressible flow :

${\ displaystyle {\ vec {\ nabla}} \ cdot {\ vec {u}} (x, y) = 0}$

If you insert the definition of the velocity potential into it, you can see that the Laplace equation (as a special case of the Poisson equation ) satisfies: ${\ displaystyle \ phi (x, y)}$

${\ displaystyle {\ vec {\ nabla}} \ cdot {\ vec {u}} (x, y) = {\ vec {\ nabla}} \ cdot {\ vec {\ nabla}} \ phi (x, y ) = \ Delta \ phi (x, y) = 0}$

The current function

The speed potential was introduced in such a way that freedom from eddies is automatically fulfilled. However, the fulfillment of the continuity equation or the Laplace equation had to be explicitly required. ${\ displaystyle \ phi (x, y)}$

Now we introduce the current function , which is defined by: ${\ displaystyle \ psi (x, y)}$

${\ displaystyle {\ vec {u}} = \ left ({\ frac {\ partial \ psi} {\ partial y}}, - {\ frac {\ partial \ psi} {\ partial x}} \ right)}$

From this definition one can see that the continuity equation is automatically fulfilled:

${\ displaystyle {\ vec {\ nabla}} \ cdot {\ vec {u}} = {\ frac {\ partial ^ {2} \ psi} {\ partial x \ cdot \ partial y}} - {\ frac { \ partial ^ {2} \ psi} {\ partial y \ cdot \ partial x}} = 0}$

However, the freedom of rotation must be explicitly required:

${\ displaystyle {\ frac {\ partial u_ {y}} {\ partial x}} - {\ frac {\ partial u_ {x}} {\ partial y}} = {\ frac {\ partial ^ {2} \ psi} {\ partial x ^ {2}}} + {\ frac {\ partial ^ {2} \ psi} {\ partial y ^ {2}}} = 0}$

The current function also fulfills the Laplace equation in eddy-free flows.

Complex speed potential

With the definitions of speed potential and current function, we get: ${\ displaystyle \ phi}$${\ displaystyle \ psi}$

${\ displaystyle u_ {x} = {\ frac {\ partial \ phi} {\ partial x}} = {\ frac {\ partial \ psi} {\ partial y}} \ quad \ wedge \ quad u_ {y} = {\ frac {\ partial \ phi} {\ partial y}} = - {\ frac {\ partial \ psi} {\ partial x}}}$

This is exactly of the form of the Cauchy-Riemann differential equations for a holomorphic function , with a real part and an imaginary part . Thus one introduces the complex speed potential: ${\ displaystyle \ phi}$ ${\ displaystyle \ psi}$ ${\ displaystyle w (z)}$

${\ displaystyle w (z) = \ phi (z) + i \ cdot \ psi (z) \ quad {\ textrm {with}} \ quad z = x + i \ cdot y}$

The complex velocity potential thus also fulfills the Laplace equation:

${\ displaystyle \ Delta w (z) = \ Delta \ phi (z) + i \ cdot \ Delta \ psi (z) = 0}$