Please Make A Note

Saturday, February 28, 2009

The Illusion of Knowledge

Although not along the general theme of this blog, I am obligated to make this post so that I remember this, and hopefully, share it with many other fellow engineers and researchers.

I was deeply motivated and moved by a quote from Daniel J. Boorstin:

The greatest obstacle to discovery is not ignorance, it is the illusion of knowledge.

Indeed, often times we find ourselves locked within an bubble of knowledge, thinking that we know everything and that we have solutions for all, yet, we fail to realize that this only an imaginary crust beyond which lies the unknown.

Cite as:
Saad, T. "The Illusion of Knowledge". Weblog entry from Please Make A Note. https://pleasemakeanote.blogspot.com/2009/02/illusion-of-knowledge.html?m=0

Sunday, February 22, 2009

8. Derivation of the Continuity Equation in Cylindrical Coordinates

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 Rate of Flow In = Accumulation + Rate of Flow Out
or
Accumulation + Flow Out - Flow In = 0

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
$\bg_white \150dpi \small d\mathcal{M} = \rho d\mathcal{V}$
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
$\bg_white \150dpi \small {\dot m}_{r,out} = \left(\rho u + \frac{\partial \rho u}{\partial r}dr\right)(r + dr) d\theta dz$
So that the net flow in the r direction is
$\bg_white \150dpi \small \begin{align*} {\dot m}_{r,out} - {\dot m}_{r,in} = & \rho u dr d\theta dz\\ & + \frac{\partial \rho u}{\partial r}r dr d\theta dz \\ & + \frac{\partial \rho u}{\partial r} dr^2 d\theta dz \end{align*}$
Being O(dr^2), the last term in this equation can be dropped so that the net flow on the r faces is
$\bg_white \150dpi \small {\dot m}_{r,out} - {\dot m}_{r,in} = \frac{1}{r}\rho u d\mathcal{V} + \frac{\partial \rho u}{\partial r}d\mathcal{V} + O(dr^2)$
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
$\bg_white \150dpi \small {\dot m}_{\theta,net}= \frac{1}{r}\frac{\partial \rho v}{\partial \theta}d\mathcal{V}$
We now turn our attention to the z direction. The face area is that of a sector of angle d\theta:

$\bg_white \120dpi \large \begin{align*} A_z = & \, \tfrac{1}{2}\left(r + {\rm d}r \right)^2 {\rm d}\theta - \tfrac{1}{2}r^2{\rm d}\theta \\ = & \, r {\rm dr}{\rm d}\theta + \tfrac{1}{2}{\rm d}r^2{\rm d}\theta \\ = & \, r {\rm dr}{\rm d}\theta + \mathcal{O}\left({\rm d}r^2{\rm d}\theta \right ) \end{align*}$

then, the inflow at the lower z face is
$\bg_white \150dpi \small \dot{m}_{z,in} = \rho w A_z = \rho w r dr d\theta$
while the outflow at the upper z face is
$\bg_white \150dpi \small \begin{align*} \dot{m}_{z,out} & = \left(\rho w + \frac{\partial \rho w}{\partial z}dz \right)A_z \\ & = \rho w r dr d\theta + \frac{\partial \rho w}{\partial z}r dr d\theta dz \end{align*}$
Finally, the net flow in the z direction is
$\bg_white \150dpi \small \dot{m}_{z,out} - \dot{m}_{z,in} = \frac{\partial \rho w}{\partial z}r dr d\theta dz$
Now we can put things together to obtain the continuity equation
$\bg_white \150dpi \small \frac{\partial \rho}{\partial t}d\mathcal{V} + \frac{1}{r}\rho u d\mathcal{V} + \frac{\partial \rho u}{\partial r}d\mathcal{V} + \frac{1}{r} \frac{\partial \rho v}{\partial \theta}d\mathcal{V} + \frac{\partial \rho w}{\partial z}d\mathcal{V} = 0$
dividing by dV and rearranging the r components of the velocity
$\bg_white \150dpi \small \frac{\partial \rho}{\partial t} + \frac{1}{r}\frac{\partial (r \rho u)}{\partial r} + \frac{1}{r} \frac{\partial \rho v}{\partial \theta} + \frac{\partial \rho w}{\partial z} = 0$
Voila!
[Previous: Continuity Eq. in Cartesian Coordinates]

[Table of Contents]

[Next: Continuity Eq. in Spherical Coordinates]

Cite as:
Saad, T. "8. Derivation of the Continuity Equation in Cylindrical Coordinates". Weblog entry from Please Make A Note. https://pleasemakeanote.blogspot.com/2009/02/8-derivation-of-continuity-equation-in.html?m=0

Saturday, February 21, 2009

7. Derivation of the Continuity Equation in Cartesian Coordinates

[Previous Article: How Euler Derived the Continuity Equation]
The continuity equation is an expression of a fundamental conservation principle, namely, that of mass conservation. It is a statement that fluid mass is conserved: all fluid particles that flow into any fluid region must flow out. To obtain this equation, we consider a cubical control volume inside a fluid. Mass conservation requires that the the net flow through the control volume is zero. In other words, all fluid that is accumulated inside the control volume (due to compressibility for example) + all fluid that is flowing into the control volume must be equal to the amount of fluid flowing out of the control volume.
Accumulation + Flow In = Flow Out

The mass of the control volume at some time t is
$\bg_black \120dpi \ \mathcal{M}_t = \rho \delta x \delta y \delta z$
The time rate of change of mass in the control volume is
$\bg_black \120dpi \ \frac{\partial \rho u}{\partial t}\delta x \delta y \delta z$
Now we can compute the net flow through the control volume faces. Starting with the x direction, the net flow is
$\bg_black \120dpi \ \left(\rho u + \frac{\partial \rho u}{\partial x}\delta x\right)\delta y \delta z - \rho u \delta y \delta z = \frac{\partial \rho u}{\partial x}\delta x \delta y \delta z$
Similarly, the net flow through the y faces is
$\bg_black \120dpi \ \frac{\partial \rho u}{\partial y}\delta x \delta y \delta z$
while that through the z faces is
$\bg_black \120dpi \ \frac{\partial \rho u}{\partial z}\delta x \delta y \delta z$
Upon adding up the resulting net flow and diving by the volume of the fluid element (i.e. dxdydz), we get the continuity equation in Cartesian coordinates
$\bg_black \120dpi \small \frac{\partial \rho u}{\partial t}+\frac{\partial \rho u}{\partial x}+\frac{\partial \rho u}{\partial y}+\frac{\partial \rho u}{\partial z} = 0$
Voila!

Cite as:
Saad, T. "7. Derivation of the Continuity Equation in Cartesian Coordinates". Weblog entry from Please Make A Note. https://pleasemakeanote.blogspot.com/2009/02/derivation-of-continuity-equation-in.html?m=0

Thursday, February 19, 2009

The Coming Revolutions in Theoretical Physics

Check out this interesting talk by David Gross

Cite as:
Saad, T. "The Coming Revolutions in Theoretical Physics". Weblog entry from Please Make A Note. https://pleasemakeanote.blogspot.com/2009/02/coming-revolutions-in-theoretical.html?m=0

Windows Defender Update Issues

For some reason, windows defender starts messing up and is not able to detect the latest updates. You will then need to download and install these updates manually. Check this out

How to troubleshoot definition update issues for Windows Defender

I hope that works out for you.

Cite as:
Saad, T. "Windows Defender Update Issues". Weblog entry from Please Make A Note. https://pleasemakeanote.blogspot.com/2009/02/windows-defender-update-issues.html?m=0

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