As I told you in my presentation, at 70 years old, I start to learn physics.

My brain is a little bit slow and confused…so sorry for my (most of prabable) trivial question.

I read on a website that the law of Rayleigh-Jeans is given for the radiance function of wavelenght (λ)

Radiance (function of wavelength)=2ckT/λ square

Radiance (function of frequency)=(2kT/c square). v square with λ=c/v and the partial derivative dλ/dv=c/v square

v is the frequency...don't found to type the greek letter

My questions:

1.How to reach the second equation starting from the first one (please detailed step by step explanation ?

2.In the second equation why a partial derivative has to be added ?

3. How to reach for the partial derivative dλ/dv the value c/v square...trivial, sorry

Thanks in advance and best regards ]]>

My name is Philippe, I'm now reached 70 years old and presently retired in China.

For my pleasure to learn and make my brain working, I started to read and learn about Quantum mechanics. Its starting by Blackbodies and the Rayleigh-Jeans law.

As with age I forget even how to make a partial derivative, I will have trivial question related to physics and mathematics.

Please, be patient with me.

As well as I'm new on this forum (first time in such a forum), I don't know exactly how its works.

Please, note as well that here in China: google, twitter, facebook, yahoo even now Wikipedia are blocked by the authorities so my way to learn is limited

Tanks in advance and best regards

Philippe ]]>

I would ask about the confirmation of my concept, because at work I don't have any thermodynamic specialist to consult it with.

I am not a physicist and probably this problem will be trivial for you, but I will appreciate any comments.

My project is to recover greenhouse gases from pneumatic installation (CF4, SF6, C3F8 (know as well as R218) and C2H2F4 (R134a)).

The process is foreseen to look as following: open the valves between installation and recovery unit, suck the gas using a compressor to the tank, up to the moment of reaching 0 bar absolute (vacuum) in the installation and close the valves. The most important factor for the compressor is the flow rate and unfortunately they seems to be pretty small and that makes it difficult to find a proper compressor.

So I would like to be absolutely sure that I have calculated it correctly.

Let's go to specifics.

My data is as following, I have 2 parts of the pipe (all int. dimeter 8mm):

part 1 length=0.5m under 50 bar

part 2 length=1m under 10 bar

So I am able to calculate the total volume of the gas (under different pressures and sum it up) from Boyle's law. When I have that I need to know how much volume it will take in the tank which we would like to keep under 10 bars.

So i use Boyl's Law again to know how much volume the gas will take in the tank (V2) under 10bars (p2) by using data from part 1 i.e. So (i will give different data cause the volumes for those pipes are really small and it will mess up the view):

V1 = 50m3 p1 = 50bars

V2 = ? p2 = 10 bars

V2 = (V1*p1)/p2

The other solution that came to my mind is not calculate this total volume to mass (which independently on pressure will not change) and than make calculations for the tank. After i will know how much mass I need to transfer I will do reverse calculation to know the volume of the tank. To mass calculation I used ideal gas Clapeyron's Law.

In the end it is all for ideal gas, so will be nice to repeat all the process for real gas according to real gas equation with van der Waals constants. (that will be also another reason to use mass calculations)

Could you please give me some advice if this could be the optimal way to have my tank volume (also I need to than know he flow rate for the compressor) and if there is any mistake in my way of thinking?

Thank you in advance. ]]>

For the past couple of years I have been using a modified domestic DLP projector to impress photosensitized PCBs with dryfilm for the manufacture of my PCBs. In this way I get resolutions of more than 1000 dpi and I avoid making an intermediate photolith. The modifications that I make to the DLP consist of 1º; change the original lamp with a power UV LED and 2nd; modify the focus optics to obtain an image of a square inch, and thus achieve the desired resolution. Finally, I place the modified DLP on a robotic X-Y table and by software impress PCB by sectors until completing the board. The results are truly fantastic;.... at least with the HP DLP that I modified a couple of years ago.

Now I want to do the same with an Acer DLP (K330) that has 20% more resolution and uses LED instead of lamp ... what interests in some aspects. But I have found an insurmountable problem when modifying the focusing optics. To achieve a focus on an image of the right size without spherical distortion, what I did in the first DLP (and that works very well), is to move the entire focusing optics group a few millimeters from the DLP chip, thus allowing to achieve a small image. This displacement is performed on the same original axis of the optical group and without altering anything from the group of lenses, I simply move the entire group of lenses a few millimeters to the outside ... away from the DLP chip. As I say, in the DLP that I already have mounted (an HP) works perfectly, but in this new K330 the image obtained is, effectively, smaller, but does not focus on a plane parallel to the DLP chip ...Assuming that the image is a rectangle, the lower horizontal line focuses closer to the optics, and the upper one, a couple of centimeters further away. In fact, I think the image if it even focuses on a plane ... but in a spherical cap ... I'm not sure. Obviously, the image obtained is unusable. It is not feasible to use the keystone for reasons of DPI alteration. Is this aberration due to the imperfection of the optics? ... that manifests in small images of small focus and not in large, with major focus (original mount)?. Any suggestions for a solution?

Thank in advance and excuse my terrible english ...

I add more data; the optics mounted on the K330 is telecentric (it has TIR). The optics output is vercically shifted to obtain an offset that does not reach 100% ... although the latter is usual in all domestic projectors. ]]>

Is it correct to apply two separate Bernoullis equation - for sections 1-3 and for 2-3? Why/why not?

I ve seen such solution to this problem on the internet but I have some doubts.

I think that energy between sections 1 and 3 is not conserved because of energy exchange with the other stream.

Energy conservation for the whole system

Q1(v1^2/2 +p1/rho)+Q2(v2^2/2 +p2/rho)=(Q1+Q2)(v3^2/2 +p3/rho)

is not enough to determine (p3-p2) needed for Y momentum conservation.

system of:

v1^2/2 +p1/rho +deltaE=v3^2/2 +p3/rho

v1^2/2 +p1/rho -deltaE=v3^2/2 +p3/rho

needs an additional equation as well

I would be thankful for every help ]]>

For the past couple of years I have been using a modified domestic DLP projector to impress photosensitized PCBs with dryfilm for the manufacture of my PCBs. In this way I get resolutions of more than 1000 dpi and I avoid making an intermediate photolith. The modifications that I make to the DLP consist of 1º; change the original lamp with a power UV LED and 2nd; modify the focus optics to obtain an image of a square inch, and thus achieve the desired resolution. Finally, I place the modified DLP on a robotic X-Y table and by software impress PCB by sectors until completing the board. The results are truly fantastic;.... at least with the HP DLP that I modified a couple of years ago.

Now I want to do the same with an Acer DLP (K330) that has 20% more resolution and uses LED instead of lamp ... what interests in some aspects. But I have found an insurmountable problem when modifying the focusing optics. To achieve a focus on an image of the right size without spherical distortion, what I did in the first DLP (and that works very well), is to move the entire focusing optics group a few millimeters from the DLP chip, thus allowing to achieve a small image. This displacement is performed on the same original axis of the optical group and without altering anything from the group of lenses, I simply move the entire group of lenses a few millimeters to the outside ... away from the DLP chip. As I say, in the DLP that I already have mounted (an HP) works perfectly, but in this new K330 the image obtained is, effectively, smaller, but does not focus on a plane parallel to the DLP chip ...Assuming that the image is a rectangle, the lower horizontal line focuses closer to the optics, and the upper one, a couple of centimeters further away. In fact, I think the image if it even focuses on a plane ... but in a spherical cap ... I'm not sure. Obviously, the image obtained is unusable. It is not feasible to use the keystone for reasons of DPI alteration. Is this aberration due to the imperfection of the optics? ... that manifests in small images of small focus and not in large, with major focus (original mount)?. Any suggestions for a solution?

Thank in advance and excuse my terrible english ...

I add more data; the optics mounted on the K330 is telecentric (it has TIR). The optics output is vercically shifted to obtain an offset that does not reach 100% ... although the latter is usual in all domestic projectors. ]]>

Thanks,

Joji ]]>

in this diagram, the resultant vector is the opposite direction, So what should I do the addition of vector or subtraction of vector? I am confused.

cross-sectional area of the vessel is A. ]]>

how to solve it? ]]>

I'm trying to study ahead and get a feel for first year of Physics, so I decided to go to MIT's OCW and start going through the course.

I came upon one of these worked problems, but the video accompanying it only goes through half the problem and I am a little confused as to how to proceed. It is a very basic question btw, I know. Here is the question:

Attachment 2525

My Attempt:

Attachment 2526

Any help or a point in the right direction would be greatly appreciated :)

Thank you.