Pressure Field Equation / Differentials

[MacLaddy]

Hello folks,

I am having difficulty comprehending some material in my fluid dynamics course. This is not a homework question, just something missing in my understanding.

When proving the "Pressure Field Equation," (something I am not yet able to do) there is a series of steps my instructor took.

$P=P_1+\Delta{P}$
$P=P_1+\Delta{y}\frac{dp}{dy}$

Which somehow, magically, leads to...

$F_y = (P+\frac{\partial{p}}{\partial{y}}\frac{\delta{y}}{2})*\delta{x}\delta{z}$

So my question is this.

$\frac{dp}{dy}$ is simply the change of P wrt y

$\frac{\partial{p}}{\partial{y}}$ is the change of p wrt y in a particular direction, or part of the gradiant.

But what in the sam is $\frac{\delta{y}}{2}$?

Why the delta?

Any help would be appreciated.

Thanks,
Mac

 

[fzero]

If you're dealing with a scenario as in the figure at http://www-mdp.eng.cam.ac.uk/web/library/enginfo/aerothermal_dvd_only/aero/fprops/statics/node5.html, then $\delta y$ is the length of one side of the box. The distance $\Delta y$ that appears in your equation is the distance between the center of the box and the appropriate side, which is $\delta y/2$.

 

[MacLaddy]

If you're dealing with a scenario as in the figure at http://www-mdp.eng.cam.ac.uk/web/library/enginfo/aerothermal_dvd_only/aero/fprops/statics/node5.html, then $\delta y$ is the length of one side of the box. The distance $\Delta y$ that appears in your equation is the distance between the center of the box and the appropriate side, which is $\delta y/2$.

Great, thank you for the link. That is a better representation then my textbook provides. It treats the lengths as $\delta y$, whereas this link just shows the lengths as dx, dy, and dz. It seems a new character was introduced without any real need.

Mac