# 10.1.1: Inviscid Momentum Equations


The Naiver–Stokes equations (equations (??), (??) and (??)) under the discussion above reduced to

Euler Equations in Cartesian Coordinates

$\label{if:eq:eulerEq} \begin{array}{c} \rho \left(\dfrac{\partial U_x}{\partial t} + U_x\dfrac{\partial U_x}{\partial x} + U_y \dfrac{\partial U_x}{\partial y} + U_z \dfrac{\partial U_x}{\partial z}\right) = -\dfrac{\partial P}{\partial x} + \rho g_x \\ \rho\, \left(\dfrac{\partial U_y}{\partial t} + U_x \dfrac{\partial U_y}{\partial x} + U_y \dfrac{\partial U_y}{\partial y}+ U_z \dfrac{\partial U_y}{\partial z}\right) = -\dfrac{\partial P}{\partial y} + \rho \, g_y \\ \rho\,\left(\dfrac{\partial U_z}{\partial t} + U_x \dfrac{\partial U_z}{\partial x} + U_y \dfrac{\partial U_z}{\partial y}+ U_z \dfrac{\partial U_z}{\partial z}\right) = -\dfrac{\partial P}{\partial z} + \rho\, g_z \end{array}$

These equations (1) are known as Euler's equations in Cartesian Coordinates. Euler equations can be written in a vector form as

$\label{if:eq:eulerEqVector} \rho\, \dfrac{\mathbf{D}\, \mathbf{U} }{\mathbf{D} t} = - \boldsymbol{\nabla} \mathbf{P} - \boldsymbol{\nabla} \,\rho\, \mathbf{g}\,{\boldsymbol{\ell}}$

where $${\mathbf{\ell}}$$ represents the distance from a reference point. Where the $$\left.D\,\mathbf{U}\right/ D t$$ is the material derivative or the substantial derivative. The substantial derivative, in Cartesian Coordinates, is

$\label{if:eq:substantialDerivative} \dfrac{\mathbf{D} \pmb{U} }{\mathbf{D} t} = \mathbf{i}\, \left(\dfrac{\partial U_x}{\partial t} + U_x\dfrac{\partial U_x}{\partial x} + U_y \dfrac{\partial U_x}{\partial y} + U_z \dfrac{\partial U_x}{\partial z}\right) \\ + \mathbf{j}\, \left(\dfrac{\partial U_y}{\partial t} + U_x\dfrac{\partial U_y}{\partial x} + U_y \dfrac{\partial U_y}{\partial y} + U_z \dfrac{\partial U_y}{\partial z}\right) \\ + \mathbf{k}\, \left(\dfrac{\partial U_z}{\partial t} + U_x\dfrac{\partial U_z}{\partial x} + U_y \dfrac{\partial U_z}{\partial y} + U_z \dfrac{\partial U_z}{\partial z}\right)$ In the following derivations, the identity of the partial derivative is used

$\label{if:eq:pdIdenty} U_i \dfrac{\partial U_i}{\partial i} = \dfrac{1}{2} \, \dfrac{\partial \left({U_i}\right)^2}{\partial i}$ where in this case $$i$$ is $$x$$, $$y$$, and $$z$$. The convective term (not time derivatives) in $$x$$ direction of equation (3) can be manipulated as

$U_x\dfrac{\partial U_x}{\partial x} + U_y \dfrac{\partial U_x}{\partial y} + U_z \dfrac{\partial U_x}{\partial z} = \textcolor{OliveGreen}{\dfrac{1}{2} \, \dfrac{\partial \left({U_x}\right)^2}{\partial x}} + \\ \hphantom{\overbrace{\dfrac{1}{2} \, \dfrac{\partial \left({U_y}\right)^2}{\partial x} &ndash;- }^ { U_y \dfrac{\partial {U_y}}{\partial x} } - } \textcolor{blue}{ \overbrace{\hphantom{U_y \dfrac{\partial {U_y}}{\partial x} + U_y \dfrac{\partial U_x}{\partial y} }}^ { U_y \left( \dfrac{\partial U_x}{\partial y} - \dfrac{\partial {U_y}}{\partial x} \right)} } \hphantom{\dfrac{1}{2} \, \dfrac{\partial \left({U_z}\right)^2}{\partial x} - -} \textcolor{RedViolet}{ \overbrace{\hphantom{U_y \dfrac{\partial {U_y}}{\partial x} + U_y \dfrac{\partial U_x}{\partial y} }}^ { U_z \left( \dfrac{\partial U_x}{\partial z} - \dfrac{\partial {U_z}}{\partial x} \right)} } \\ \underbrace{\overbrace{\textcolor{OliveGreen} {\dfrac{1}{2} \, \dfrac{\partial \left({U_y}\right)^2}{\partial x}} }^ { U_y \dfrac{\partial {U_y}}{\partial x} } -\, \textcolor{blue}{U_y \dfrac{\partial {U_y}}{\partial x} } }_{=0} + \textcolor{blue}{U_y \dfrac{\partial U_x}{\partial y} } + \underbrace{\overbrace{ \textcolor{OliveGreen}{\dfrac{1}{2} \, \dfrac{\partial \left({U_z}\right)^2}{\partial x}} }^ { U_z \dfrac{\partial {U_z}}{\partial x} } -\, \textcolor{RedViolet} {U_z \dfrac{\partial {U_z}}{\partial x}} }_{=0} +\, \textcolor{RedViolet} {U_z \dfrac{\partial U_x}{\partial z} }$ It can be noticed that equation (5) several terms were added and subtracted according to equation (4). These two groups are marked with the underbrace and equal to zero. The two terms in blue of equation (5) can be combined (see for the overbrace). The same can be done for the two terms in the red–violet color. Hence, equation (5) by combining all the "green'' terms can be transformed into

$\label{if:eq:convectiveManipolationMid} U_x\dfrac{\partial U_x}{\partial x} + U_y \dfrac{\partial U_x}{\partial y} + U_z \dfrac{\partial U_x}{\partial z} = \textcolor{OliveGreen}{\dfrac{1}{2} \, \dfrac{\partial \left({U_x}\right)^2}{\partial x}} + \textcolor{OliveGreen}{\dfrac{1}{2} \, \dfrac{\partial \left({U_y}\right)^2}{\partial x}} + \textcolor{OliveGreen}{\dfrac{1}{2} \, \dfrac{\partial \left({U_z}\right)^2}{\partial x}} + \\ \textcolor{blue}{ { U_y \left( \dfrac{\partial U_x}{\partial y} - \dfrac{\partial {U_y}}{\partial x} \right)} } + \textcolor{RedViolet}{ { U_z \left( \dfrac{\partial U_x}{\partial z} - \dfrac{\partial {U_z}}{\partial x} \right)} }$ The, "green'' terms, all the velocity components can be combined because of the Pythagorean theorem to form

$\label{if:eq:bernoulliDerivative} \textcolor{OliveGreen}{\dfrac{1}{2} \, \dfrac{\partial \left({U_x}\right)^2}{\partial x} + \dfrac{1}{2} \, \dfrac{\partial \left({U_y}\right)^2}{\partial x} + \dfrac{1}{2} \, \dfrac{\partial \left({U_z}\right)^2}{\partial x}} = \dfrac{\partial \left({\pmb{U} }\right)^2}{\partial x}$ Hence, equation (6) can be written as

$\label{if:eq:convectiveManipolationMid1x} U_x\dfrac{\partial U_x}{\partial x} + U_y \dfrac{\partial U_x}{\partial y} + U_z \dfrac{\partial U_x}{\partial z} = \dfrac{\partial \left({\pmb{U} }\right)^2}{\partial x} \\ + { U_y \left( \dfrac{\partial U_x}{\partial y} - \dfrac{\partial {U_y}}{\partial x} \right)} + { U_z \left( \dfrac{\partial U_x}{\partial z} - \dfrac{\partial {U_z}}{\partial x} \right)}$ In the same fashion equation for $$y$$ direction can be written as

$\label{if:eq:convectiveManipolationMid1y} U_x\dfrac{\partial U_y}{\partial x} + U_y \dfrac{\partial U_y}{\partial y} + U_z \dfrac{\partial U_y}{\partial z} = \dfrac{\partial \left({\pmb{U} }\right)^2}{\partial y} \\ + { U_x \left( \dfrac{\partial U_y}{\partial x} - \dfrac{\partial {U_x}}{\partial y} \right)} + { U_z \left( \dfrac{\partial U_y}{\partial z} - \dfrac{\partial {U_z}}{\partial y} \right)}$ and for the $$z$$ direction as

$\label{if:eq:convectiveManipolationMid1y1} U_x\dfrac{\partial U_z}{\partial x} + U_y \dfrac{\partial U_z}{\partial y} + U_z \dfrac{\partial U_z}{\partial z} = \dfrac{\partial \left({\pmb{U} }\right)^2}{\partial y} \\ + { U_x \left( \dfrac{\partial U_z}{\partial x} - \dfrac{\partial {U_x}}{\partial z} \right)} + { U_y \left( \dfrac{\partial U_z}{\partial y} - \dfrac{\partial {U_y}}{\partial z} \right)}$ Hence equation (3) can be written as

$\label{if:eq:substantialDerivativeMid} \dfrac{\mathbf{D} \mathbf{U} }{\mathbf{D} t} = \mathbf{i}\, \Bigg( \textcolor{OliveGreen}{\dfrac{\partial U_x}{\partial t}} + \textcolor{RedViolet}{\dfrac{\partial \left({\pmb{U} }\right)^2}{\partial x}} + { U_y \left( \dfrac{\partial U_x}{\partial y} - \dfrac{\partial {U_y}}{\partial x} \right)} + { U_z \left( \dfrac{\partial U_x}{\partial z} - \dfrac{\partial {U_z}}{\partial x} \right)} \Bigg) \\ + \mathbf{j}\, \Bigg(\textcolor{OliveGreen}{ \dfrac{\partial U_y}{\partial t}} + \textcolor{RedViolet}{\dfrac{\partial \left({\pmb{U} }\right)^2}{\partial y}} + { U_x \left( \dfrac{\partial U_y}{\partial x} - \dfrac{\partial {U_x}}{\partial y} \right)} + { U_z \left( \dfrac{\partial U_y}{\partial z} - \dfrac{\partial {U_z}}{\partial y} \right)} \Bigg) \\ + \mathbf{k}\, \left(\textcolor{OliveGreen}{\dfrac{\partial U_z}{\partial t}} + \textcolor{RedViolet}{\dfrac{\partial \left({\pmb{U} }\right)^2}{\partial y}} + { U_x \left( \dfrac{\partial U_z}{\partial x} - \dfrac{\partial {U_x}}{\partial z} \right)} + { U_y \left( \dfrac{\partial U_z}{\partial y} - \dfrac{\partial {U_y}}{\partial z} \right)} \right)$ All the time derivatives can be combined also the derivative of the velocity square (notice the color coding) as

$\label{if:eq:substantialDerivativeF1} \dfrac{\mathbf{D} \mathbf{U} }{\mathbf{D} t} = \textcolor{OliveGreen}{\dfrac{\partial \pmb{U} }{\partial t}} + \textcolor{RedViolet}{\boldsymbol{\nabla} \left( {\pmb{U} }\right)^2}+ \mathbf{i}\, \Bigg( { U_y \left( \dfrac{\partial U_x}{\partial y} - \dfrac{\partial {U_y}}{\partial x} \right)} + { U_z \left( \dfrac{\partial U_x}{\partial z} - \dfrac{\partial {U_z}}{\partial x} \right)} \Bigg) \\ + \mathbf{j}\, \Bigg( { U_x \left( \dfrac{\partial U_y}{\partial x} - \dfrac{\partial {U_x}}{\partial y} \right)} + { U_z \left( \dfrac{\partial U_y}{\partial z} - \dfrac{\partial {U_z}}{\partial y} \right)} \Bigg) \\ + \mathbf{k}\, \left( { U_x \left( \dfrac{\partial U_z}{\partial x} - \dfrac{\partial {U_x}}{\partial z} \right)} + { U_y \left( \dfrac{\partial U_z}{\partial y} - \dfrac{\partial {U_y}}{\partial z} \right)} \right)$ Using vector notation the terms in the parenthesis can be represent as

$\label{if:eq:curlU} \mathbf{curl} \, \mathbf{U} = \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{U} = \mathbf{i}\, \left( \dfrac{\partial U_z}{\partial y} - \dfrac{\partial {U_y}}{\partial z} \right) + \mathbf{j}\, \left( \dfrac{\partial U_x}{\partial z} - \dfrac{\partial {U_z}}{\partial x} \right) \\ + \mathbf{k}\, \left( \dfrac{\partial U_y}{\partial x} - \dfrac{\partial {U_x}}{\partial y} \right)$ With the identity in (13) can be extend as

$\label{if:eq:UcurlU} \pmb{U} \boldsymbol{\times} \boldsymbol{\nabla} \boldsymbol{\times} \pmb{U} = -\mathbf{i}\, \Bigg( { U_y \left( \dfrac{\partial U_x}{\partial y} - \dfrac{\partial {U_y}}{\partial x} \right)} + { U_z \left( \dfrac{\partial U_x}{\partial z} - \dfrac{\partial {U_z}}{\partial x} \right)} \Bigg) \\ - \mathbf{j}\, \Bigg( { U_x \left( \dfrac{\partial U_y}{\partial x} - \dfrac{\partial {U_x}}{\partial y} \right)} + { U_z \left( \dfrac{\partial U_y}{\partial z} - \dfrac{\partial {U_z}}{\partial y} \right)} \Bigg) \\ - \mathbf{k}\, \Bigg( { U_x \left( \dfrac{\partial U_z}{\partial x} - \dfrac{\partial {U_x}}{\partial z} \right)} + { U_y \left( \dfrac{\partial U_z}{\partial y} - \dfrac{\partial {U_y}}{\partial z} \right)} \Bigg)$ The identity described in equation (14) is substituted into equation (??) to obtain the form of

$\label{if:eq:substantialDerivativeF} \dfrac{\mathbf{D} \mathbf{U} }{\mathbf{D} t} = \dfrac{\partial \mathbf{U} }{\partial t} + \boldsymbol{\nabla} \left( {\mathbf{U} }\right)^2 - \mathbf{U} \boldsymbol{\times} \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{U}$ Finally substituting equation (15) into the Euler equation to obtain a more convenient form as

$\label{if:eq:eulerEqFV} \rho\, \left( \dfrac{\partial \mathbf{U} }{\partial t} + \boldsymbol{\nabla} \left( {\mathbf{U} }\right)^2 - \mathbf{U} \boldsymbol{\times} \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{U} \right) = - \boldsymbol{\nabla} \mathbf{P} - \boldsymbol{\nabla} \,\rho\, \mathbf{g}\,{\boldsymbol{\ell}}$ A common assumption that employed in an isothermal flow is that density, $$\rho$$, is a mere function of the static pressure, $$\rho = \rho(P)$$. According to this idea, the density is constant when the pressure is constant. The mathematical interpretation of the pressure gradient can be written as

$\label{if:eq:gradient} \boldsymbol{\nabla} P = \dfrac{dP}{dn} \hat{\mathbf{n}}$ where $$\hat{\mathbf{n}}$$ is an unit vector normal to surface of constant property and
the derivative $$d\left/dn\right.$$ refers to the derivative in the direction of $$\hat{\mathbf{n}}$$. Dividing equation (17) by the density, $$\rho$$, yields

$\label{if:eq:gradientTOrho} \dfrac{\boldsymbol{\nabla} P}{\rho} = \dfrac{1}{dn} \dfrac{dP}{\rho} \, \hat{\mathbf{n}}= \dfrac{1}{dn} \overbrace{d \int}^{\text{ zero net effect}} \left( \dfrac{dP}{\rho} \right) \,\hat{\mathbf{n}} = \dfrac{d}{dn} \int \left( \dfrac{dP}{\rho} \right) \,\hat{\mathbf{n}} = \boldsymbol{\nabla} \int \left( \dfrac{dP}{\rho} \right)$ It can be noticed that taking a derivative after integration cancel both effects. The derivative in the direction of $$\hat{\mathbf{n}}$$ is the gradient. This function is normal to the constant of pressure, $$P$$, and therefore $$\int\left(\left. {dP}\right/{\rho}\right)$$ is function of the mere pressure. Substituting equation (18) into equation (16) and collecting all terms under the gradient yields

$\label{if:eq:preBernoulliVorticity} \dfrac{\partial \mathbf{U} }{\partial t} + \boldsymbol{\nabla} \left( \dfrac ParseError: invalid DekiScript (click for details) Callstack: at (Bookshelves/Civil_Engineering/Book:_Fluid_Mechanics_(Bar-Meir)/10:_Inviscid_Flow_or_Potential_Flow/10.1_Introduction/10.1.1:_Inviscid_Momentum_Equations), /content/body/p[21]/span, line 1, column 1  {\partial \ell} \boldsymbol{\cdot} \widehat{\left(\dfrac{\boldsymbol{\Upsilon} }{\Upsilon}\right)} = \dfrac{\partial \boldsymbol{\Upsilon}}{\partial \ell}$ where $$\boldsymbol{\Upsilon}$$ is any vector and $$\Upsilon$$ its magnitude. The right hand side of equation (26) $$\pmb{U}\boldsymbol{\times} \boldsymbol{\Omega}$$ is perpendicular to both vectors $$\pmb{U}$$ and $$\boldsymbol{\Omega}$$. Hence, the dot product of vector $$\pmb{U}$$ with a vector perpendicular to itself must be zero. Thus equation (26) becomes

$\label{if:eq:BernoulliStreamline1} \dfrac{\partial \mathbf{U} }{\partial t} + \overbrace{\dfrac{d}{d\ell}} ^{\scriptscriptstyle\dfrac{\mathbf{U} }{U} \boldsymbol{\cdot} \boldsymbol{\nabla}} \left( \dfrac{{\pmb{U} }^2}{2} + \mathbf{g}\,\ell + \int \left( \dfrac{dP}{\rho} \right) \right) = \overbrace{\dfrac{\mathbf{U} }{U} \boldsymbol{\cdot} \mathbf{U} \boldsymbol{\times} \boldsymbol{\Omega} } ^{\scriptstyle =0}$ or

$\label{if:eq:BernoulliStreamline2} \dfrac{\partial \mathbf{U} }{\partial t} + \dfrac{d}{d\ell} \left( \dfrac{{\pmb{U} }^2}{2} + \mathbf{g}\,\ell + \int \left( \dfrac{dP}{\rho} \right) \right) = 0$ The first time derivative of equation (28) can be manipulated as it was done before to get into derivative as

$\label{if:eq:timeDerivativeIntegral} \dfrac{\partial \pmb{U} }{\partial t} = \dfrac{d}{d\ell} \int \dfrac{\partial \pmb{U} }{\partial t}\,d\ell$ Substituting into equation (28) writes

$\label{if:eq:BernoulliStreamlineD} \dfrac{d}{d\ell} \left( \dfrac{\partial \mathbf{U} }{\partial t} + \dfrac{{\pmb{U} }^2}{2} + \mathbf{g}\,\ell + \int \left( \dfrac{dP}{\rho} \right) \right) = 0$ The integration with respect or along stream line, "$$\ell$$'' is a function of time (similar integration with respect $$x$$ is a function of $$y$$.) and hence equation (28) becomes

Bernoulli On A Streamline

$\label{if:eq:BernoulliStreamline} \dfrac{\partial \mathbf{U} }{\partial t} + \dfrac{{\pmb{U} }^2}{2} + \mathbf{g}\,\ell + \int \left( \dfrac{dP}{\rho} \right) = f(t)$

In these derivations two cases where analyzed the first case, for irrotational Bernoulli's equation is applied any where in the flow field. This requirement means that the flow field must obey $$\mathbf{U}\boldsymbol{\times}\boldsymbol{\Upsilon}$$. The second requirement regardless whether the flow is irrotational or not, must be along a streamline where the value is only function of the time and not location. The confusion transpires because these two cases are referred as the Bernoulli equation while they refer to two different conditions or situations. For both Bernoulli equations the viscosity must be zero.

The two different Bernoulli equations suggest that some mathematical manipulations can provide several points of understating. These mathematical methods are known as potential flow. The potential flow is defined as the gradient of the scalar function (thus it is a vector) is the following

$\label{if:eq:potentialFunction} \mathbf{U} quiv \boldsymbol{\nabla}\phi$

The potential function is three dimensional and time dependent in the most expanded case. The vorticity was supposed to be zero for the first Bernoulli equation. According to the definition of the vorticity it has to be

$\label{if:eq:potentialFunctionVorticity} \boldsymbol{\Omega} = \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{U} = \boldsymbol{\nabla} \boldsymbol{\times} \boldsymbol{\nabla}\phi$

The above identity is shown to be zero for continuous function as

$\label{if:eq:vectorIdenty} \boldsymbol{\nabla} \boldsymbol{\times} \overbrace{\left(\mathbf{i} \dfrac{\partial \phi}{\partial x} + \mathbf{j} \dfrac{\partial \phi}{\partial y} + \mathbf{k} \dfrac{\partial \phi}{\partial z} \right) }^{\boldsymbol{\nabla}\phi} = \mathbf{i} \left( \dfrac{\partial^2 \phi }{\partial y \partial z} - \dfrac{ \partial^2 \phi }{\partial z \partial y} \right) \\ +\mathbf{j} \left( \dfrac{\partial^2 \phi }{\partial z \partial x} - \dfrac{ \partial^2 \phi }{\partial x \partial z} \right) +\mathbf{k} \left( \dfrac{\partial^2 \phi }{\partial y \partial x} - \dfrac{ \partial^2 \phi }{\partial x \partial y} \right)$

According to Clairaut's theorem (or Schwarz's theorem) the mixed derivatives are identical $$\partial_{xy} = \partial_{yx}$$. Hence every potential flow is irrotational flow. On the reverse side, it can be shown that if the flow is irrotational then there is a potential function that satisfies the equation (33) which describes the flow. Thus, every irrotational flow is potential flow and conversely. In these two terms are interchangeably and no difference should be assumed. Substituting equation (33) into (24) results in

$\label{if:eq:BernoulliPotential} \dfrac{\partial \boldsymbol{\nabla} \phi}{\partial t} + \boldsymbol{\nabla} \left( \dfrac{{ \left(\boldsymbol{\nabla} \phi \right) }^2}{2} + \mathbf{g}\,\ell + \int \left( \dfrac{dP}{\rho} \right) \right) = 0$

It can be noticed that the order derivation can be changed so

$\label{if:eq:orderTimeD} \dfrac{\partial \boldsymbol{\nabla} \phi}{\partial t} = \boldsymbol{\nabla} \dfrac{\partial \phi}{\partial t}$

Hence, equation (36) can be written as

$\label{if:eq:BernoulliPotentialG} \boldsymbol{\nabla} \left( \dfrac{\partial \phi}{\partial t} + \dfrac{{ \left(\boldsymbol{\nabla} \phi \right) }^2}{2} + \mathbf{g}\,\ell + \int \left( \dfrac{dP}{\rho} \right) \right) = 0$

The integration with respect the space and not time results in the

Euler Equation or Inviscid Flow

$\label{if:eq:BernoulliPotentialInt} \dfrac{\partial \phi}{\partial t} + \dfrac{{ \left(\boldsymbol{\nabla} \phi \right) }^2}{2} + \mathbf{g}\,\ell + \int \left( \dfrac{dP}{\rho} \right) = f(t)$

Example 10.1

The potential function is given by $$\phi = x^2 - y^4 + 5$$. Calculate the velocity component in Cartesian Coordinates.

Solution 10.1

The velocity can be obtained by applying gradient on the potential $$\pmb{U} = \boldsymbol{\nabla}\phi$$ as

$\label{simplePotential:sol} \begin{array}{rl} V_x = \dfrac{\partial \phi}{\partial x} &= 2\,x\\ V_y = \dfrac{\partial \phi}{\partial y} &= -4\,y^3\\ V_z = \dfrac{\partial \phi}{\partial z} &= 0 \end{array}$