Page:Aerial Flight - Volume 1 - Aerodynamics - Frederick Lanchester - 1906.djvu/98

(linear) rate of change in the three co-ordinate directions amongst themselves.

The equation of continuity is based upon the fact that the inflow and outflow of any small element of space must balance, or must balance against the change of density if the fluid is compressible. The form of the expression for an incompressible fluid is—

${\frac {Du}{Dx}}+{\frac {Dv}{Dy}}+{\frac {Dw}{Dz}}=0,$

where $u,\ v,$ and $w$ represent the velocities in the directions of the three co-ordinate axes $x,\ y,$ and $z.$

(2) The Dynamical Equations expressing the relation in the direction of each of the three co-ordinate axes $x,\ y,\ z,$ for every small element of the fluid, between the rate of change in its momentum, the difference of pressure on its opposite faces, and the component of the extraneous force, if any.

The Extraneous Forces are usually represented in the three co-ordinate directions by the symbols $X,\ Y,\ Z,$ and denote forces acting from without on the fluid particles, such, for example, as the force of gravity. In the present branch of the subject these forces do not require to be considered.

Employing, as is customary, the symbol $DF/Dt$ to denote a differentiation following the motion of the fluid, it can be shown that

Now the rate of change of the $x$ momentum of any small element $\delta x\ \delta y\ \delta z$ is $\rho \delta x\ \delta y\ \delta z\ {\frac {Du}{Dt}},$ and this must be equal to the difference of the pressure force on its two faces, which is evidently $-{\frac {dp}{dx}}\delta x\ \delta y\ \delta z,$ (where $p$ is pressure). The minus sign is due to the fact that the momentum increase takes place in the direction of the pressure decrease. So that:—

$\rho \delta x\ \delta y\ \delta z\ {\frac {Du}{Dt}}=-{\frac {dp}{dx}}\delta x\ \delta y\ \delta z,$

or

${\frac {Du}{Dt}}=-{\frac {dp}{\rho dx}}.$

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