Flamendialis's Weblog

Plant Physics

flamendialis — Sun, 06 Feb 2011 15:32:18 +0000

Plants produce oxygen from carbon dioxide during photosynthesis during the day. At night, they produce carbon dioxide. Some claim that you should not have a plant in your bedroom. Or that if you have one, you should move it out at night. The reason given is that they produce carbon dioxide which is harmful to you. However, the question to be asked is, how much carbon dioxide is produced? A plant that is about a foot tall, will weigh only a few hundred grams at worst. Considering that mammals produce gases in proportion to their body mass, the amount produced should be minuscule indeed. In fact, if you are so worried about carbon dioxide in the room, you should sleep alone! To put some numbers here, we have the oxygen consumption at about 200 gal during one night for an adult human being. This is around 10 % of the mass of the gas in an average room. BTW, this also underlines the importance of ventilation. Consider the reaction C6H12O6 + 6O2 = 6CO2 + 6H2O. In this case, the amount of CO2 produced is 44/32 times the O2 consumed, around 2.5 lbs.

Plants require nutrients from the soil: namely nitrogen, phosphorous and potassium. In addition, they also need micro-nutrients. Natural fertilizer include mulch, which is organic matter which will decay and give nutrients to the soil. They also retain moisture, so that watering needs are reduced. Before planting, you will need to test the soil to see the pH etc. A better way is to have raised beds where you can add store bought soil, which will give you a much shallower learning curve. Note that compost has a lead time of 3 to 12 months.

HP Deskjet 1050

flamendialis — Sun, 06 Feb 2011 10:06:53 +0000

One unpleasant surprise with this all in one is the lack of manual duplexing option. Since this feature could have been implemented entirely in software, this appears to be a deliberate attempt by HP to create different segments. Note that MS Word offers manual duplexing outside of the printer driver, so this is not an issue. Unfortunately, this means that all _application_ software will have to support this. Caveat Emptor would be the comment by many, but you have to consider the present scenario where this is something taken for granted.

And if you attempt to print back to back manually, you run into difficulty as the pages are printed last page first, so that you end up with a blank second page. That is to say, if you have 2n + 1 pages to print, the first paper to come will be 2n+1 and 2n, and the next will be 2n-1 and 2n – 2 etc. The nth paper will be 2 and 3, and the first page will have to be printed on one side of the page. Perhaps you can now write “this page unintentionally left blank” on the other side.

The documentation (printed) that comes with the system is not much. Most of it is an exhortation by HP to buy “original” cartridges. They have supplemented these by stickers on the machine itself, so it’s got something of a NASCAR look.

The cartridges are of the starter variety, and they come packed in plastic, which is surprisingly easy to open. If you have had the misfortune to try and open an HP cartridge packaging you will know how much suffering you have been spared.

The packaging is using recycled cardboard rather than thermocol. Environmentally correct or just being cheap? I am guessing the latter.

The print head calibration is performed using software. You take a printout and then the scanner loads it to memory where the configuration program aligns the print head. Those of who have had older HP printer will remember the lines printed on the calibration sheet which you had to pore over and analyze to perform this task. A welcome change indeed.

The scanner mechanism has its own issues. At the middle of the scan, there is a squeaking sound, where I suppose the fit is rather too tight. The default scanning settings start from 200 dpi.

An output tray is provided, and it actually works. Many low end machines don’t come with an output tray and this means you have to either catch each page as it comes out, or provide a large table for the printer.

Feature wise, this printer is just about OK. The only question is reliability. This is not meant for heavy duty work, but will it print / scan for atleast a year? Only time will tell.

Leak of gas through an orifice

flamendialis — Sat, 22 Jan 2011 17:07:06 +0000

Consider a vessel containing ideal gas under a pressure of p which leaks through an orifice of known properties CdA. If the pressure is high enough, the flow through the orifice is choked, and the velocity is the sonic velocity. The flow through the orifice is given by \rho CdA v where all the conditions are at the throat. The total enthalpy and the static enthalpy of a gas are related by h0 = h + 1/2 v^2. Inside the vessel, the value will be the total velocity, and taking the ideal gas relation h = CpT where T is the absolute temperature, we have T0 = T + 1/2v^2/Cp. At the throat, where sonic conditions prevail, v = a, where a, the speed of sound is given by \sqrt(kRT). Thus, the temperature at the throat is given by T = 2/(k+1) T0. If the ideal gas is assumed to under go an adiabatic process, the total pressure can be similarly related to the pressure at the throat. i.e., p0/p = (T0/T)^(k/(k-1)) = (2/(k+1))^(k/(k-1)). Going back to the mass flow equation mdot = Cd A \rho v, we have \rho = p/RT. The throat pressure can be written as p = p0*(p/p0). Similarly, the throat temperature can also be stated in terms of the vessel pressure. Finally, what we have is the mass flow rate as a function of the instantaneous pressure and temperature of the vessel.

Now we can write the mass flow rate in terms of the rate of change of density of the gas in the vessel. The right hand side contains the pressures and temperature of the gas in the vessel, and these can also be written in terms of the density using the ideal gas equation. This will give the variation of density as a function of time. Since we are assuming an adiabatic process, we will also get the variation of pressure and temperature with time.

Some of the applications of the above equations are the following: 1. find the time for the pressure in a leaking vessel to fall. 2. find the amount of gas that will leak from a vessel in a given time.

We can also give the reverse question: what is the nature of the leak if the variation in the pressure with time is known? If we know the gas, we can find the value of CdA.

We can see that the pressure ratio between the vessel and the throat has to be above a certain ratio for choked flow to occur. Now let us consider the case where this pressure ratio is lower, so that the flow is subsonic. In this case, the flow usually occurs between two pressures p1 and p2, which are the upstream and downstream values respectively. Now, even if we assume a value of CdA for the orifice, the conditions at the throat will not be same as the downstream, i.e., the pressure will not be p2. However, the conventional analysis is carried out by assuming the throat value to be same as that of the downstream. Once again, we assume the upstream to be static, so that the velocity can be neglected. We can write the expression for the static enthalpy as in the previous section. However, this time the velocity will not be the sonic velocity. Once again, adiabatic expansion is assumed between the two points, and we have v^2 = 2CpT1(1-T2/T1). Now, T2/T1 = (p2/p1)^((k-1)/k). The density at the throat, \rho2 can be written as \rho2 = \rho1 * (\rho2/\rho1). Using the adiabatic relations, p \propto \rho^k. Thus, we have the subsonic mass flow equation, where it is function of the upstream values and the pressure ratio.

Having talked a lot about choked flow, let us ask the question, does the flow really choke? That is to say, if we continuously increase the pressure ratio, will we find the value of flow not increasing beyond a certain value? Experiments on orifices show that this is not really the case. The effect is explained on the basis of the shifting of the minimum area position as well as the size. As the outlet pressure is decreased to zero, the flow increases, though the increase is not as rapid beyond a certain pressure ratio.

The flow through a given orifice is a strong function of the manufacturing process, in particular, the various radii that are achieved for the edges of the orifice. Since there is a wide variation in these geometric parameters even with products of a single machine, hardly any attempt is made to find the theoretical flow patterns and the like in the real world.

Installing Linux on a machine without CD-ROM

flamendialis — Fri, 06 Nov 2009 16:52:58 +0000

Quite a few of the new machines (called Netbooks) arrive without an optical drive. However, they are capable of booting off their USB drives, and installing Linux on them is not too difficult. Just get one of the USB distributions from the Net.

However, if you have a laptop without an optical drive, and cannot boot from the USB port either, then the following might be of some help.

Remove the HD from the laptop and install it in a USB enclosure
Attach this drive to a machine with Linux already installed
Use cfdisk to assign partitions.
Format the first partition on the disk as ext3 using the mkfs.ext3 command
Mount the new partition (say /mnt/disk).
Copy the /boot of the machine onto /mnt/disk
Edit /mnt/disk/boot/grub/menu.lst, but leave it as hd0
Start grub as root, and type:
root (hd1,0) setup (hd1) quit

Copy the installation boot image on to /mnt/disk/, and the installation ISO image on to another partition on the drive in the USB enclosure

Unmount all partitions on that drive and disconnect it from the host machine.
Remove the drive from the enclosure and install it on the laptop
When the system boots, go into the grub shell, and type :
kernel "path to vmlinuz" initrd "path to initrd" boot
When prompted, point to the ISO on the disk, and proceed with the normal installation.

Printing

flamendialis — Fri, 06 Nov 2009 16:36:25 +0000

A lot has been said about eliminating paper altogether, but printing documents is something we cannot seem to get enough of.

Printing Speed

The number of pages per minute printed is something that is quoted by all printer manufacturers, and we tend to compare numbers from different catalogs without giving it a second thought. However, what exactly is being printed? The printer manufacturers have not, so far, published the documents that they used for the tests. Even the driver setting used to print is usually the draft mode of the printer. All that is about to change with the ISO standards for printer speeds. Now, you can argue that the pages the ISO considers are not representative, but at least it seems to have forced companies to publish according to those standards.

A notorious company with regard to published printing speeds is Canon. Test according to the latest standards show that the printing speed for the their latest and greatest inkjets is only about 20% of the older models. Now, they have not moved the old models from their website, and they mention that they will only benchmark the new (2009) printers with the ISO standard. Lest their duplicity be discovered, they have come up with a new term called ipm–images per minute. The term would make sense if they were printing photos–they are after all, images. However, the standard ISO files are NOT images–they are regular documents, the kind that you would prepare for a report or paper. Note that as of writing of this blog, there is no ISO standard for photo printing.

Print Cartridge Life

A similar idea is that of the life of the cartridge. Once again, the companies that make these cartridges claim all sorts of things with all sorts of details like coverage areas and the like. A 5% coverage area has been bandied about. Once again, ISO has introduced a standard, which is now being followed by the companies concerned.

Cost of Printing

The actual cost of printing is always more than the cost of the paper and ink for a given number of pages. As the printer ages, the problems like paper management grow, and wastage is quite high. If the printer is not brand new, you can forget about things like duplex printing, as there is bound to be an error somewhere, and you might end up discarding the whole lot.

One idea most people would have, considering the cost of print cartridge is–why can’t we refill the cartridges? This would be quite economical, and there are two ways to go about it. One is to manually refill them, with a syringe. Here the quality of the replacement ink is important. Equally important is the need to maintain a good vacuum while refilling. The other method is to have a continuous ink system. This means that you have an external tank(s) which you pour ink into. You can see the level of ink and can probably do it till the head wears out.

In the case of cartridges with more than one color, the precise location is important. Alas, a common error that occurs during refilling of color cartridges is the mixing of the different colors. Isopropyl alcohol (IPA) is said to be a good solvent in the case of printing inks.

ANSYS

flamendialis — Thu, 29 Oct 2009 15:38:17 +0000

ANSYS has become the de-facto standard for engineering analysis for generic engineering analysis. One of the advantages is that the command line exposes all that can be done in the GUI and more.

WARNING: ANSYS is not free software–it is not even affordable software, with license fees running in the thousands of dollars (still cheaper than ABAQUS though). You will have to seriously consider how you will access this software once you graduate. You can look at the CAELinux software, which is a bootable DVD. It contains quite a lot of analysis software, and even though it is Linux, there is no installation problems to worry about.

The analysis type is set to static for the static structural analysis. The following lines are equivalent.
antype,0 antype,static

The pre-preprocessor is using the /prep7 command. In ANSYS, only the first four characters of the command are relevant. More may be used to enhance readability. However, commands like /prep7 are the exception. Some commands can only be invoked in certain modes. The major modes are /prep7, /solu and /post1. /post1 is the general post processor. /solu is the solution mode. You can get out of a particular mode by typing fini.

The usual elements are beam, shell and solid elements. link1 is one the simplest elements. The common beam element is beam3. Springs are modeled using the combin14 element.

There are quite a large number of elements of each type. shell63 is a common shell element.

The default value for the Poisson’s ratio is 0.3.

The element type is specified using the et command.
et,1,beam3
The above code declares the element type 1 to be of the beam type.
The material properties can be specified in great detail. However, the simplest case is that of a linear, isotropic material. In this case the material properties could be:
mp,ex,1,210e9 mp,nu,1,0.3
Of course, the declaration of the value 0.3 is superfluous as it the default anyway. You can also give the density using
mp,dens,1,1e3
but this is not used if you are not interested in the eigenvalues.

The real constants can be likewise specified
r,1,0.1
The above code gives the real value to be 0.1 for set 1.

How can we use this? When elements are defined, we can set the types, material properties, and the real constants like so:
type,1 real,1 mat,1
Of course, the real constants should supply what is expected by the element type. For instance, the element type shell expects a thickness, so we need only one value in the real constant set.

Contact Problems

Modeling contact problems in ANSYS is not that easy. Perhaps this is because it is a relatively new feature. Unlike software like ABAQUS, where contact is automatic, in ANSYS, you have to specify the contact surfaces.

Non-Linear Problems

Quite a lot of engineering problems are of the linear type. The reasons for this is quite clear–the components are designed with some factor of safety, so that it will seldom go into yield in a large fraction of the component. Yielding, if it occurs, will be quite localized, and will be acceptable.

Once in a while, however, you will end up with a problem that requires non linear analysis. It may be that the material is non linear even in the elastic region. Or more commonly, the material is thin enough so that there is significant geometric flexing. In any case, these are the cases where computer analysis is weakest, and where you encounter the dreaded convergence problem.

Fluid Mechanics

flamendialis — Wed, 25 Feb 2009 17:19:52 +0000

Body in a Fluid

Consider a body with a flat bottom placed inside an empty vessel. As water is poured into the vessel, the body will be submerged. If the body is less dense compared to water, it will float if placed in a vessel containing water. However, if the vessel is initially empty, adding water will cause a force due to pressure on the sides and top but not at the bottom. The question is, what is the condition for the pressure to act at the bottom? It is likely that water should get under the body in order to cause it to rise up due to buoyancy. What part will surface tension play in this particular case?

Optics

flamendialis — Mon, 26 Jan 2009 04:59:45 +0000

Images formed by Two Plane Mirrors

The number of images formed by two plane mirrors forming an angle depends on the angle between them and whether the object is placed symmetrically between the mirrors. The image formed by one mirror acts as the object for the other, as long as the image lies in the object half of the plane for the other mirror. If the object is at the back of the mirror, no light from the object can fall on the mirror. All the images will be of the same intensity of the mirrors are infinite (semi-infinite), so that the amount of light that is seen to form an image is the same for all images. However, for finite mirrors, the amount of light is substantially lower for the images after the first couple of images, so that they would be fainter. The above statements assume that there is no loss in reflection.

Mirrors and Image forming space

If an object is placed between three mutually perpendicular mirrors, there are 7 images produced. Let the co-ordinates of the point be (a, b, c). The total images formed due to the mirrors in the xy, yz, and zx planes are all the different combinations of a, -a, for the first, b, -b for the second, and c, -c for the third respectively. This gives rise to 2 x 2 x 2 = 8 possibilities, where one of them coincides with the object (a, b, c).

Images by Perpendicular Mirrors

The intensity of Light

VCD in Linux

flamendialis — Sat, 03 Jan 2009 17:06:45 +0000

The automatic mounting process in Mandriva fails for a vcd. Attempts to mount will mount the first track, which is about 2 MB, and gives a bunch of errors in dmesg. However, the installed k3b program handles it well, with some quirks.

The Windows™ operating system is able to read files off the vcd, and you can copy the dat file from it. This file can be played in Linux using xine or kmplayer. However, k3b cannot make a vcd out of it–it demands an mpeg file. You can try and use some utilities to convert this dat file to an mpeg, but that is probably too much bother.

In k3b, there is a copy cd option, but that cannot copy a vcd as it has more than one track. You will need to rip the vcd onto your hard drive. k3b uses the vcdxrip program to do this and this will create an mpeg on the hard drive. This can also be played in all the standard media players in Linux. This file can be burned with k3b using the create vcd command. The vcd so created works well in standalone vcd players.

Electrostatics

flamendialis — Thu, 11 Dec 2008 16:42:08 +0000

Consider a non-conducting sphere with a uniform charge density ρ. Since the material is a non conductor, the charge will not move around. The non-conductor, is also known as a dielectric. If it is a linear dielectric, it will have a dielectric constant ε_r, also denoted by K, which is greater than 1. In this case, the bound charge density is proportional to the free charge ρ. Thus the electric field inside is reduced by that factor.

Electric Field of a uniformly charged sphere

The above figure shows the usual analysis of the problem. Note that the field is identical to that of a sphere of uniform mass density, a usual textbook approximation for the interior of the earth (although the density of the earth is highly non-uniform, increasing to a value of 13 g/cc at the center). Of course, since there is no negative mass, there is no question of polarization or some such in the case of the earth.

Electric Field inside a uniformly charged dielectric

The electric field is discontinuous at the surface as there is surface charge buildup there. The field outside is unchanged from the original, but the field inside is reduced.

The real question is: what is meant by charge density? If it is the total charge density, then as Gauss’s law is always given by , and the electric field is immediately obtained. If the charge density is the free charge density, then the value of is not immediately known, so that the field is different. Only in the case of a linear dielectric under very special circumstances can we give the “modification” of Gauss’s law as where ε is the absolute permittivity of the medium. These include that the material be a linear dielectric.

If the question is posed as follows: a sphere of radius R carries a charge Q, uniformly distributed in the volume. Plot the electric field vs r, the distance from the center of the sphere. Can we avoid the buildup of surface charge in the case of a linear dielectric? I think not. If the statement is about a charge density ρ = ρ₀ for 0 ≤ r ≤ R, and zero everywhere else, then there is no ambiguity at all. A surface charge will involve an infinite volume charge density there, so that is impossible. Another way of stating the same thing is this: consider a charge Q distributed uniformly over a spherical region of radius R.