Please Make A Note: How Euler Derived the Momentum Equations

In this post, I will present Euler's derivation of the momentum balance equations for a fluid flow. Please note that I will use modern notation with a few modifications to the original derivation to clarify certain issues. Nonetheless, the core of Euler's derivation is maintained.

Considering two dimensional flows with axial and transverse velocities denoted by u(x,y,t) and v(x,y,t), respectively, Euler first forms the total differentials of these fields

$\bg_white \120dpi \begin{cases} {\rm d}u = \dfrac{\partial u}{\partial x}{\rm d}x + \dfrac{\partial u}{\partial y}{\rm d}y + \dfrac{\partial u}{\partial t}{\rm d}t \\[10pt] {\rm d}v = \dfrac{\partial v}{\partial x}{\rm d}x + \dfrac{\partial v}{\partial y}{\rm d}y + \dfrac{\partial v}{\partial t}{\rm d}t \end{cases}$

Next, dividing both equations by dt, we obtain

$\bg_white \120dpi \begin{cases} \dfrac{{\rm d}u}{{\rm d}t} = \dfrac{\partial u}{\partial x}\dfrac{{\rm d}x}{{\rm d}t} + \dfrac{\partial u}{\partial y}\dfrac{{\rm d}y}{{\rm d}t}+ \dfrac{\partial u}{\partial t}\\[10pt] \dfrac{{\rm d}v}{{\rm d}t} = \dfrac{\partial v}{\partial x}\dfrac{{\rm d}x}{{\rm d}t} + \dfrac{\partial v}{\partial y}\dfrac{{\rm d}y}{{\rm d}t} + \dfrac{\partial v}{\partial t} \end{cases}$

but

$\bg_white \120dpi \dfrac{{\rm d}x}{{\rm d}t}=u;\quad {\rm and}\quad \dfrac{{\rm d}y}{{\rm d}t}=v$

then, upon substitution, we recover

$\bg_white \120dpi \begin{cases} \dfrac{{\rm d}u}{{\rm d}t} = u \dfrac{\partial u}{\partial x} + v \dfrac{\partial u}{\partial y}+ \dfrac{\partial u}{\partial t}\\[10pt] \dfrac{{\rm d}v}{{\rm d}t} = u \dfrac{\partial v}{\partial x} + v \dfrac{\partial v}{\partial y} + \dfrac{\partial v}{\partial t} \end{cases}$

One can infer that Euler is heading towards Newton's law. The previous expressions represent the acceleration of a fluid element. As to the forces, Euler identifies pressure, friction, and gravity. For the time being, he abandons friction and focuses on pressure and gravity.

For gravity, the force is in the transverse direction and is given by

$\bg_white \120dpi \quad F_{\rm g,}_x=0; \quad F_{\rm g,}_y = - \mathcal{M} g$

where

$\bg_white \120dpi \mathcal{M}=\rho \mathcal{V} = \rho \, {\rm d}x \, {\rm d}y$

is the mass of the fluid element.

For the pressure, he considers a rectangular fluid element as shown below.

Euler then assumes that the pressure at point L is p and deduces the values at the corners of the control volume. The net force acting on each side is calculated by taking the average pressure at the vertices defining that side and multiplying it by the area. This is shown in the figure below

Then, the net forces in the axial and transverse directions are given by

$\bg_white \120dpi F_x = F_{\rm LN}-F_{\rm MO} = - \dfrac{\partial p}{\partial x}{\rm d}y{\rm d}x$

and

$\bg_white \120dpi F_y = F_{\rm LM}-F_{\rm NO} = - \dfrac{\partial p}{\partial y}{\rm d}x{\rm d}y$

Now, by applying Newton's law, we have

$\bg_white \120dpi \sum \mathbf{F} = \mathcal{M}\frac{{\rm d}\mathbf{u}}{{\rm d}t}$

or, by projecting in both directions, we have

$\bg_white \120dpi F_{\rm g,}_x + F_x = \mathcal{M} \frac{{\rm d}u}{{\rm d}t}$

and

$\bg_white \120dpi F_{\rm g,}_y + F_y = \mathcal{M} \frac{{\rm d}v}{{\rm d}t}$

Upon expansion, we get

$\bg_white \120dpi - \frac{\partial p}{\partial x}{\rm d}x \, {\rm d}y = \rho\,{\rm d}x \, {\rm d}y \left(u \dfrac{\partial u}{\partial x} + v \dfrac{\partial u}{\partial y}+ \dfrac{\partial u}{\partial t} \right )$

and

$\bg_white \120dpi -\rho g \,{\rm d}x \, {\rm d}y - \frac{\partial p}{\partial y}{\rm d}x \, {\rm d}y = \rho\,{\rm d}x \, {\rm d}y \left(u \dfrac{\partial v}{\partial x} + v \dfrac{\partial v}{\partial y} + \dfrac{\partial v}{\partial t} \right )$

Finally, we get the Euler equations

$\bg_white \120dpi \dfrac{\partial u}{\partial t}+ u \dfrac{\partial u}{\partial x} + v \dfrac{\partial u}{\partial y} = - \frac{1}{\rho}\frac{\partial p}{\partial x}$

and

$\bg_white \120dpi \dfrac{\partial v}{\partial t}+u \dfrac{\partial v}{\partial x} + v \dfrac{\partial v}{\partial y} = -\frac{1}{\rho} \frac{\partial p}{\partial y}- g$

Voila!

[Previous: Continuity Eq. in Spherical Coordinates]

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[Next: Momentum Eqs. in Cartesian Coordinates]

Cite as:
Saad, T. "How Euler Derived the Momentum Equations". Weblog entry from Please Make A Note. http://pleasemakeanote.blogspot.com/2010/07/how-euler-derived-momentum-equations.html

2 comments:

hennie jacksSeptember 27, 2010 at 10:08 PM
hi! sir/madam!you are doing a wonderful job!i have keenly followed your work which is an eye-opener to me!There is still a problem i cant quiet get and that is the derivation of momentum equation in cylindrical coordinates .could you help please!
cncJanuary 12, 2011 at 6:59 PM
yes, may u help me to derive momentum equation in cylindrical coordinate?

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Saturday, July 10, 2010

How Euler Derived the Momentum Equations

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