This is one of my favorite derivations. Although it would sound a bit intimidating at first, as none of the standard textbooks carry out the derivation in curvilinear coordinates; it is rather easy to obtain. And guess what? the math is quite rewarding!
So we first have to start by selecting a convenient control volume. The idea here is to pick a volume whose sides are parallel per say to the coordinates. For cylindrical coordinates, one may choose the following control volume
Again, as we did in the previous post, we need to account for all the fluid that is accumulating, and flowing through this control volume, namely:
Rate of Accumulation + Rate of Flow In = Rate of Flow Out
First, let's get some basics laid out. The velocity field will be described as
I always prefer to use u, v, and w instead of ur, utheta, and uz to save on subscripts, although the latter nomenclature is a bit more descriptive… we'll get used to it. Now, by construction, the volume of the differential control volume is
while the mass of fluid in the control volume is
The rate of change of mass or accumulation in the control volume is then
For the net flow through the control volume, we deal with it one face at a time. Starting with the r faces, the net inflow is
while the outflow in the r direction is
So that the net flow in the r direction is
Being O(dr^2), the last term in this equation can be dropped so that the net flow on the r faces is
The net flow in the theta direction is slightly easier to compute since the areas of the inflow and outflow faces are the same. At the outset, the net flow in the theta direction is
We now turn our attention to the z direction. The face area is that of a sector of angle d\theta:
then, the inflow at the lower z face is
while the outflow at the upper z face is
Finally, the net flow in the z direction is
Now we can put things together to obtain the continuity equation
dividing by dV and rearranging the r components of the velocity
Voila!
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the equation that convert from cartesian to polar coordinates and vice versa r = sqrt (x*x+y*y); phi = atan2 (y, x); x = r*cos (phi); y = r*sin (phi);
You must solve the equations of the lines simultaneously. Represent each line in an equation of the form y=mx +b (where m represents the slope of the line, that is "rise over run"), then make substitutions using the info you have to solve for a pair of coordinates they share.
The solution of the equation.
you just need to enclose your equation like this: \begin{figure*} \begin{equation} % % \end{equation} \end{figure*}
Set 0=(denominator of the System Transfer Function), this is the Characteristic Equation of that system. This equation is used to determine the stability of a system and to determine how a controller should be designed to stabilize a system.
it is easy you can see any textbook........
An ordered pair or coordinates of a point in 2-dimensional space.
http://en.wikipedia.org/wiki/Navier-Stokes_equations Please go to this page.
If you put an 'equals' sign ( = ) between the 'By' and the 'Cz', you have the generic equation for any straight line in 3-dimensional Cartesian coordinates.
polar
The graph (on Cartesian coordinates) of a quadratic equation is a parabola.
Energy
It appears that the point has only one coordinate: 310. In two dimensional space, such as the coordinate (or Cartesian) plane, a point needs two coordinates.
in helmholtz vector equation why F=-∆ф+∆xA?
Assume the equation is y = kx + c Put in the x and y values of your known coordinates and sove the simultaneous equations.
Subtract the equation of one line from the equation of the other
H2 => h2