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MHD shocks and discontinuities

I agree with previous posted comments that this article is really an aerodynamic one. Instead of trying to re-write it to include MHD shocks, I am creating a MHD shocks and discontinuities (ok bad title, but as good as I can come up with) page. We might change titles later. USferdinand 21:36, 23 January 2007 (UTC)

Major changes (refactored from top)

I am in a process changing or removing some of the mistakes in this article. If you for some believe that you better contact me at barmeir at gmail.com and I will discuss with you the issue(s) that you have. I believe that I will convince you (Yes, I am a bite over confident but I have a reason for it. I have contributed to this area a bit more most of usual researchers group. You can check my book on the history section of these field.).

List of changes

1. shock wave is not propagating it can be stationary.
2. shock wave is a disturbance, For example in Fanno flow shock occurs without any disturbance,
3. "The shock wave is one of several different ways in which a gas in a supersonic flow can be compressed. Two other methods are isentropic and Prandtl-Meyer compressions." This statement is simply wrong. One in shock wave no compression is achieved (the stagnation pressure remains the same. only the static pressure increase). Second Prandtl-Meyer function is isentropic function thus this statement makes no sense.
4. is there any shock that not associated with supersonic flow.I would like to find one. thus the title

of shock in supersonic flow implies that the subsonic with shock is just in the virtual world (where entropy can increase.

Very strange article

This article has more mistakes than expected. For example, the speed of moving shock can be above and below the speed of sound and not as mentioned in this article to be above the speed of sound. It simply not correct.

--potto 16:36, 31 January 2007 (UTC)

Dear Gennik I've reverted your changes, and would request that you seriously reconsider which changes you make to this article. Just deleting essentially the whole artical and replacing with your own material is a serious problem. This is for three major reasons:

1. I'm sorry, but your writing is not in English. It is not grammatical, and often doesn't make sense. I note this because, although your changes may be correct, it doesn't help if no-one can understand them.
2. Your list of changes is dubious.
   1. A shock is always stationary with respect to one frame.
   2. Look up disturbance in a dictionary
   3. Although a Prandtl-Meyer compression is an isentropic compression, the textbook examples of isentropic compression are waveless, and so it probably deserves a special mention.
   4. Certainly all shocks are associated with locally supersonic upstream flow. This is, however a general use encyclopaedia. Therefore special mention of shocks in transonic flow is probably a good idea.
3. I wonder about some of your content. For instance an analytical solution to the weak-solution oblique shock has been known for at least 50 years.

In summary, if you want to re-add the material which you've added I would ask you to edit the existing article rather than rewriting. Please also pay particular attention to your grammar, since it appears that English is not your native language. I am very willing to help with any of this, but I'm, not willing to re-write your changes from scratch.

AKAF 12:06, 1 February 2007 (UTC)

Re Prandtl-Meyer: if as you say it is an isentropic compression, then the current text is misleading:

• Two other methods are isentropic and Prandtl-Meyer compressions.

I suggest changing it to something like:

• Some other methods are isentropic compressions, including Prandtl-Meyer compressions. --Coppertwig 17:47, 5 February 2007 (UTC)

I wish that it was possible to correct from this point

The article is full of mistakes. I wish that It was possible to change it form this point but it is easier to go from beginning. For example, the article starts with the below statement which is erroneous.

(or simply "shock") is a type of propagating disturbance. Like a normal wave, a shock wave carries :energy and can propagate through a medium

shock is in many case is stationary as you mentioned and do not move to any place with any reasonable frame of reference. Let me give an example. In converging diverging nozzle flow there is a shock for large range of pressure ratios. I hope that you agree that this shock is stationary (if you have problems to understand read my book on isentropic flow).

But the text said that it propogates through a medium. In that case the medium is moving at the same speed the shockwave is, but in the opposite direction. So the original text is not actually incorrect, although it may very well be unconventional.WolfKeeper 17:47, 1 February 2007 (UTC)

If you chose an arbitrary frame of reference, the shock is moving with reference to that frame but the discontinuity don't move to any physical place or object. Thus, saying that shock is propagating is misleading at best.

The text said it propogates through a medium, this implies that they are referring to the mediums reference frame.WolfKeeper 17:47, 1 February 2007 (UTC)

Let me take your explantion one step further. we have a disturbance propogates due to unknown reason for unknown reason in the speed of incoming (upstream) gas. Thus, if you increase the gas velocity then your disturbance will increase the speed. Conclusion from this, that gas is the source of this disturbance. So, why the disturbance is moving in the opposite direction? If the gas is the source of your disturbance than if you cut the nozzle it should not effect the disturbance, yet it will.

genick --potto 04:31, 2 February 2007 (UTC)

I understand some of the above, but I don't understand the last part. I think you're making an argument to try to support the point that "propagating through a medium" is misleading if the shock wave is stationary while the medium moves. I think "propagating through a medium" sounds OK to me. However, if there are situations where the shock wave is not just coincidentally stationary, that is, where the velocity of the medium doesn't just happen by coincidence to be the exact opposite of the velocity of the wave through the medium, but where there are actual feedback mechanisms to keep the wave stationary, then perhaps "propagating through a medium" is not the best wording. Maybe people can try to think up some other wordings.

It seems to me that a hydraulic jump is an example of such a phenomenon, where it is not as appropriate to talk about it propagating through a medium, since there are often factors that tend to keep it in a particular place given the conditions. For example, when you run the tap onto the bottom of the kitchen sink, you can get a circular hydraulic jump which tends to stay at about the same radius given a certain flow speed, and will return to that radius if disturbed. I'm not sure whether a hydraulic jump counts as a type of shock wave or whether some shock waves have similar behaviour.

If something is at exactly the opposite velocity "for unknown reason" then maybe you don't know that one velocity will change when you change the other. Maybe you mean that for unknown reason it always moves at exactly opposite velocity no matter what the other velocity is. The question is: are there physical situations where the speed of the shock wave in the medium does tend to change when the speed of the medium changes, so that the shock wave tends to stay stationary, as hydraulic jumps sometimes do?

I suppose "cut the nozzle" means turn off the source of moving gas. Of course, if you do that, the shock wave will stop. I don't understand the last two sentences of your argument. You said it had a particular velocity "for unknown reason", so I don't see why, later, you asked "why" it had that velocity.--Coppertwig 18:23, 5 February 2007 (UTC)

Moreover the word disturbance mean in wiki to change (check it out) but shock isn't just a change, it is a discontinuity. For example, the flow in our nozzle is full of disturbances (changes) but only one discontinuity. That is the reason shock is call discontinuity and not as a change.

If the independent variable is time, a discontinuity is a change. If the independent variable is a spatial dimension (distance) then a discontinuity can also be considered to be a change from the point of view of a point in a moving reference frame, such as a particle in the medium -- which tends to be a relevant point of view in these situations.--Coppertwig 18:23, 5 February 2007 (UTC)

Let me also explain you the issue of energy, shock does not carry any energy there isn't a change in energy in most shocks. Starting that shock carries energy is simply wrong because the shock, for example, in our nozzle does not carry any energy or loss any energy. If you claim to opposite than I would to see reference(s) or your derivations. You can check my book or Shapiro or any other book that you like about this issue.

I don't see how that could be true in the general case. For example a shockwave forms around a nuclear bomb that very definitely does propogate for a distance at least, and I don't see how that would not be carrying energy away. I'm wondering if the solution is that shockwaves can carry kinetic energy, in the case of a nozzle the shockwave is stationary in the lab frame, so that would explain your example. Also see: [1] which very definitely says that shock waves can carry energy.WolfKeeper 17:47, 1 February 2007 (UTC)

Again you are confusing issues. I wish that you once pick in a good book, check my book or Shapiro. There are moving shocks and there are stationary shocks. In the case of stationary shock, there is no energy change. In the case of moving shocks there is a difference between energy in different frame of reference. This is the source of your confusion. Perhes, you should write about this topic after you took a class in gas dynamics.

--potto 04:44, 2 February 2007 (UTC) genick

Could you describe an example of stationary shock? I would like to try to understand it more fully. I searched for the word "stationary" in the article but didn't see anything about stationary shock; I think probably it needs to be added. Also, please explain what you mean by "no energy change". Energy is conserved in any case. I would be surprised if there could be a shock wave without any conversion of energy to heat. Can it happen as a thermodynamically reversible process? Why would it matter whether it's moving or not? Every shock must be stationary is some reference frame (as someone else pointed out), but if it's thermodynamically reversible it should be reversible in all reference frames. --Coppertwig 18:23, 5 February 2007 (UTC)

I can go on and on on almost every sentence in this article (by the way you misspell this word [article] check in your response). I am not saying that my English is perfect and I sure that yours is better. You can change my English as many help me with my books. I saw that you change others term that I work in wiki. It is fine by me.

You mentioned an analytical solution to oblique shock, please show a place with that solution. If you want to know about it more read my book in the history section ([www.potto.org/gasDynamics/node43.html]) about it and check my book about the actual full solution to oblique shock ([www.potto.org/gasDynamics/node196.html]). Clearly from your reaction, you not aware of that solution which shows new understanding about zero inclination angle.

you stated that

the textbook examples of isentropic compression are waveless,

what do you mean by that statement? I cannot find any meaning this statement. what textbook you referring to?

By the way would you like to identify your self, so I know who I am talking to.

This isn't correct shock —The preceding unsigned comment was added by Genick (talk • contribs) 15:34, 1 February 2007 (UTC).

Thank you! now we talk

We have to find a better way to let each other know answer. I do not care how. You can hide and it doesn't bother me. As long we can work on the content and it will not take very long, it is fine. --potto 19:25, 1 February 2007 (UTC) —The preceding unsigned comment was added by Genick (talk • contribs) 19:18, 1 February 2007 (UTC).

Dear Genick.

1. No I will not identify myself. That's against the wikipedia norms and rude to ask. I'll also not communicate by direct email.

Fine --potto 23:21, 1 February 2007 (UTC)

1. I do not think that the article is perfect. Actually I think that there are many mistakes. I'd like them to be fixed. I should be clear that I think that your additions are probably useful. Just I would please ask you to think before changing things since I think that in your big change not a single sentence was without major grammatical and spelling errors and confusions in meaning. It's a good idea to introduce such changes more slowly, so that these erros can be ironed out by the community. Otherwise perple will decide (Like I did) that the overall result of a particular edit is negative, and revert.

Fine! you would me to show all the mistakes. That is okay with me. --potto 23:21, 1 February 2007 (UTC)

1. Clearly you think that "propagating disturbance" is confusing. It is certainly not formally incorrect in the gas frame. Perhaps you would like to give a clarification which would help the general reader. In your example, if the nozzle were in a flying engine, then the shock wave would indeed be propagating wrt the ground. If you want to change "disturbance" to "discontinuity", though I'd be the last person to stop you.

---

that is now! we can change it to discontinuity ?

and it should read

A shock wave (or simply "shock") is a type of discontinuity in the flow. Shock waves are characterized by a sudden and abrupt change in the characteristics of the medium (such as pressure, temperature, and speed).

--potto 23:21, 1 February 2007 (UTC)

1. If you don't like the possibility of shocks carrying energy, then you should reformulate it more exactly. Consider however the simple exmple of a blunt body with a bow shock. An air particle on the stagnation line is accelerated from standstill to (a fraction of) the body's velocity as it crosses the shock. In this way this particle gains energy.

---

What you saying is of characteristic of moving shock (change of frame of reference from the shock front to stationary frame). You can read and learn more about it in/from my book about moving or in Shapiro (very limit but nevertheless informative). Thus, the issue of moving energy should not be part of the main issue of shock wave. In fact, shock is analyzed when the energy assumed conserved. (by the way, what is your background so I know how to explain these issues to you. Thus, all the point of energy must be either eliminate or move to moving shock (different article).

1. The standard analytical solutions to the oblique shock can be found in (for example) NACA1135 equations 115-170. If you claim that you are the first person to find an analytical solution to the oblique shock, I think you'll be disappointed. Your research may indeed be (new,interesting,correct) and if you can provide a refereed publication which agrees with you, then it could be added to comments about the standard solutions.

---

Good that you brought this point. Read the NACA1135 report and refer to the wiki article on oblique shock. or to NACA 1135 page 10 equation 150a. It state "No convenient explicit relations exist" please locate it in the report. This relationship is the most important one. Your comment simply insults over 50 Ph.D. who dedicate you find the solution to this equation (if you would like to have ref i can provide you). This solution is important because it deals with the most fundamental question what will be shock angle for given upstream Mach number and wedge angle. The approximation that confuse with the solution exist but it is wrong (this is what I refers to the error in NACA 1135. The assumptions are wrong .). Now you can review the solution, it is explicitly explain in my book or ask someone who know basic mathematic (no knowledge in differential equations is needed, just simple algebra, well, a bit more than high school, though some high school students can figure it out.). The equations that you referred to 115-170 include equation 150a (may be you should read again that report) are either approximations or showing the reverse relationship (which is 1 to 1 and not 1 to 3 (solutions)). I hope that my explanation convince you or direct you to read so that you can understand. If you understand what I am saying, should we go with the history section? or do you need more explanations? if so, what are the points that you do not understand? By The way, my book with the solution is wildely used many places. --potto 05:08, 2 February 2007 (UTC) genick bar-meir

1. Most people learn about isentropic flow in thermodynamics, which deals primarily with subsonic flow. As an example, Isentropic_flow does not include the word "wave" anywhere. The fact that certain kinds of waves can cause an isentropic compression is sufficiently far from "general knowledge" that it should be explained.

-------------------------

You are steeping on something strange. I would partially agree with if NACA 1135 was not wrong on this point. The analytical solution (that you refuse or cannot read) shows that it isn't possible. The isentropic process in supersonic flow is P-M function.

-------------------------

I understand that you might be ashamed if people know who you are. But please let me know what is your level of knowledge or education so my explanations will be clear to you. --potto 20:27, 1 February 2007 (UTC) Genick Bar-Meir,

ps. My adviser E.R.G. Eckert, the father of modern heat transfer, got worse reception than me. Only then people did not hide and they openly claim that he was wrong and mistaken.

-------------------------

AKAF 17:40, 1 February 2007 (UTC)

refactoring to bottom of page (Genick please write your answers here)

maybe the question below isn't needed

Now when you talk specific issues may be we can resolve it.

check below for my answers and explanations. genick --potto 20:31, 1 February 2007 (UTC)

question to user:AKAF

Dear sir/Madam,

I request a mediation against your editing in shock wave. I hope that you will either explain why you would like to keep errors in the article or stop changing it. If you believe that your corrections are of/on any base please explain. you can find the meditation requrest in http://en.wikipedia.org/wiki/Wikipedia:Mediation_Cabal/Cases/2007-02-01_shock_wave

Genick

I have nothing to hide as you do. everything that I do is transparent and if you like discuss with me any of the issues or you believe that you are right and I am wrong please show or explain it. —The preceding unsigned comment was added by Genick (talk • contribs) 18:29, 1 February 2007 (UTC).

Please see my longer reply on the RfM. May I suggest you read WP:EP, WP:NOR, WP:CIV, WP:NPA, WP:U, WP:AB and WP:CON. They may help you to understand some of the norms on wikipedia. I am not saying that all of these impact you directly, but some of your edits indicate that you are unaware of them. AKAF 10:22, 2 February 2007 (UTC)

General Replies to Genick

Genick, please use Wikipedia style. Additions to talk and user pages are added at the bottom. Indentation should be undertaken with the colon [:] symbol. Higher layers of indenting can be caused by repeating the colon. All posts should be signed by using 4 tildes.

Attacking other editors for using a pseudonym is absolutely unacceptable. Requiring personally identifying information is unacceptable. Additionally you should assume good faith and be polite. We all make mistakes. Here from Wolfkeeper's talk page is one of yours. I quote it here as an example of what wikipedia is absolutely not about:

Sir, I do what I am good at. I think that you should do what you are good at. Shock wave is not a disturbance and it doesn't propogates it sometime stationary.

Second, and to your some of your specific comments, Wikipedia is a general use encyclopaedia. As a rule, its best to use examples with which the layman will be more familiar. It's an interesting contradiction that fluid dynamicists are often almost completely unaware of aircraft or flying bodies because they are not the simple devices which are used to investigate fluid phoenomena. The layman is however far more likely to be only aware of fluid flow as it affects external aerodynamics on a flying object. We should try to explain these processes so that they can be understood by someone of this background.

To your comments on the analytical solution of an oblique shock: The standary equations for the oblique shock are an analytical solution. There are equations which you can use to find pressure/density/Mach jumps across an oblique shock if the shock angle is known. Finding the shock angle requires an iterative solution, which is, as the NACA 1135 states, not explicit. In these times where a relatively cheap calculator can solve the equation the inconvenience is long forgotten. I consider the sentence in example 4 of this article to be a good summary. With this in mind I find your assertion that an analytic solution did not exist before your work at best misleading.

To your comments on energy transfer by a shock: Energy is assumed conserved in the shock frame. It still isn't explanatory to anyone who has heard a sonic boom. Clearly there is energy transferred there (in this case from the aircraft to your eardrum). Your explanation is certainly true, but not useful in explaining shocks to a general audience.

AKAF 09:30, 2 February 2007 (UTC)

Archiving Propsal

I'd like to archive everything above the archive cutline. Please move the cutline if you'd like it somewhere else. AKAF 10:07, 2 February 2007 (UTC)

I disagree

I disagree I think that people should see your reactions so that others can judge where this value is going to. It is very important that you will not remove these remarks. You aren't the judge. Your insistence to keep errors in this article is strange. --potto 12:27, 2 February 2007 (UTC) Genick Bar-Meir. --potto 12:36, 2 February 2007 (UTC) Wikipedia is for everyone. Your explanation isn't only wrong but confusing. If you would like to explain atomic boom go ahead and do so. Atomic boom is not same as shock wave even though some material is overlapping. You can also write about explosive if you like to explain boom. --potto 12:36, 2 February 2007 (UTC) genick

Please search for "archive cutline" in the page to see where I'd like to archive from. It's between the sections "Fundamental properties" and "MHD shocks and discontinuities".AKAF 12:51, 2 February 2007 (UTC)

I've archived from the cutline today. If there's something you think should remain on the main page, please feel free to copy it back. AKAF 10:07, 5 February 2007 (UTC)

General Replies to AKAF user

You prefer wrong and confusing on right and confusing. Well, this your chose. I wonder who many will agree with you. Look if you insist on keep these mistakes in this article I will not add material to it. I find it disrespectful that you just like to ignore knowledge. If you do not have the time to read material that is fine. Your insistence on being judge to material that you are not familiar isn't respectful, at least you should read a good book about the topic. your statement "that fluid dynamicists are often almost completely unaware of aircraft or flying bodies because they are not the simple devices" is strange. who you think design airplane? not fluid mechanics people? Well maybe, I never saw modern airplane designed by not experts. Did you saw airplane designed by peso engineers. Even the wright brothers where fluid mechanics people. If you think so, I guess there is more can be said about this. I do agree that the explanation should be simple that most people can understand this article. It was not intended for people like me, I already know it and read my book if forgot something. But insert simple urban myths isn't the solution. I suggest that you take my explanation edit the English to way that it make things look simple to you. Ask me questions if you need clarifications, I will be glad to help. Hey, that what I am all about (check my web site www.potto.org). You do not have to put it in the front page right away. As I state before this article is full with mistakes. I don't make sense that expert will article full with mistakes but the other way around that expert write the skeleton and you guys will change it to make simple to ordinary reader to understand.

you stated that

Finding the shock angle requires an iterative solution, which is, as the NACA 1135 states, not explicit. In these times where a relatively cheap calculator can solve the equation the inconvenience is long forgotten. I consider the sentence in example 4 of this article to be a good summary.

Again I find this remark offensive not to me but to soo many Ph.D. who work years on this issue. I don't know a cheap calculator which do these calculation. I would like to find one. To explain this in simple term this equation 150a in NACA report is the source to weak and strong shock (also to the unsteady one). This equation explain the detach shock. You dismissive of what so many students spend years to understand. My solution, was preceded by partial solution of G. Emmanual G. who wrote a book about it. Other books, (see for example Anderson) copy that partial solution and had several section discussing it. Are all these people are idiots according to you?

In the article (I believe that is what you mean)

for a perfect gas and inviscid flow field, an analytic solution is available, such that the pressure ratio, temperature ratio, angle of the wedge and the downstream Mach number can all be calculated knowing the upstream Mach number and the shock angle.

Again, the problem of design is: for needed airplane speed and given angle (design) what should be shock angle. And not the opposite. What is states there is basically, if I know the solution how I use it utilizing the normal shock relationships to obtain the properties ratios. Clearly, the first part contradicts the second part of the sentence.

The article states that

Smaller shock angles are associated with higher downstream Mach numbers, and the special case where the shock wave is at 90 degrees to the oncoming flow (Normal shock), is associated with a downstream Mach number of one. These follow the "weak-shock" solutions of the analytic equations.

This is simply wrong. I can copy a whole section from my book explaining this. I think that if you care, enough, you should read my book in the oblique chapter. The last statement is really strange. what analytical equations are you guys talking about? The approximations in NACA 1135 or Emmanual partial solution. May be the one that you guys have but did not published or published somewhere? If so please let me know where.

Is there way that you guys really want to have good article? If so let me know? Your dismissiveness of knowledge not clear to me. Can you explain it to me. I really would like to understand you guys. Genick --potto 15:43, 2 February 2007 (UTC)

Reply to Genick

The vast majority of people have never seen a schlieren photo of a 2-D nozzle flow. Most have seen an airplane. Thus for the layperson the 2-D nozzle flow in your example is the unusual case.

I realise that finding the shock angle (beta) from the standard equation:

$tan\theta =2cot\beta {\frac {M_{1}^{2}sin^{2}\beta -1}{M_{1}^{2}\left(\gamma +cos2\beta \right)+2}}$

is an iterative problem. I'm sorry but that absolutely doesn't require years of PhD research. And this is the equation which is being referred to in NACA1135. There is simply no convient way of expressing beta accurately as a function of theta. Nevertheless finding beta if theta is known isn't rocket science.

Smaller shock angles are associated with higher downstream Mach numbers, and the special case where the shock wave is at 90 degrees to the oncoming flow (Normal shock), is associated with a downstream Mach number of one. These follow the "weak-shock" solutions of the analytic equations.

I think its clear that if you swap "upstream" for "downstream" then this statement is correct. Just a mistake, not an example of my general perfidity.

AKAF 16:24, 2 February 2007 (UTC)

please read my comments (Moved from User talk:AKAF)

Look you say that it need modifications. I for example, suggested something for the first paragraph. I assume that you do not agree, I just do not know. are you welling to change anything? If so why don't you agree on any change? You know that it is wrong. So, why keep it? Can you help me here? Is there a secret agenda that I am not aware of?

--potto 15:54, 2 February 2007 (UTC) Genick

I did read it and I'm making what I believe to be the appropriate change now, leaving "standing shock" as a special case, much as "standing wave" is often considered a special case of wave motion. I can further assure you that I do work for an agency with a three-letter acronym. AKAF 16:38, 2 February 2007 (UTC)

There is improvment

Now the first sentence is good. Now we can discuss the second.  Normal shock is one of the catecories of shock wave.  Thus,  

Like a normal wave, a shock wave carries energy and can propagate through a medium or, in special cases, through a field such as the electromagnetic field in the absence of a physical medium.

has to be chagned because it impoly that normal shock isn't a shock wave which is not true.

I will agree to something like

The aviable engry (incrase of entropy) is lost in shock wave. or Shock must occur in medium in which

aviable engry (incrase of entropy) is lost.

Simply energy is not correct. --potto 21:20, 2 February 2007 (UTC) Genick

Genick I'm sorry but I'm a little confused which part you're referring to here. I have also noticed a problem with the use of the word "normal". Sometimes it means "ordinary" and sometimes it means "at 90 degrees". Perhaps we should go through and make a general replace so that the style is consistent? Maybe a definition of "Normal shock" (is a shock at 90 degrees to the oncoming flow) would also be in order? Especially since "Normal" is not used in this sense in non-specialist language?

May I suggest for your example: "Although the total energy is preserved when matter is processed by a shock wave, the available energy which can be extracted as work (entropy) is decreased. This has the practical effect, for example, that an aircraft with shocks will experience an additional drag force to that which it would experience if the flow was shockless."

I'm a little unsure what you want to say in the statement about the medium.

RegardsAKAF 10:21, 5 February 2007 (UTC)

One possible solution is to define the normal shock as it used commonly by every one who is dealing with shock.

"Normal shock in this field refers to ...." This is not Normal in the sanse of normal ... I think that will solve the problem.

--potto 18:34, 5 February 2007 (UTC) Genick Bar-Meir

your suggestion:

Although the total energy is preserved when matter is processed by a shock wave, the available energy which can be extracted as work (entropy) is decreased. This has the practical effect, for example, that an aircraft with shocks will experience an additional drag force to that which it would experience if the flow was shockless.

should be improved by

Although the total energy is preserved when matter is processed by a shock wave, the available energy which can be extracted as work (entropy increased) is decreased. This has the practical effect, for example, that an aircraft with shocks will experience an additional drag force to that which it would experience if the flow was shockless.

Then it is okay by me

--potto 18:56, 5 February 2007 (UTC) Genick

Hi Genick, I see your point. I think though that the increased/decreased is a bit confusing. So I'd like to suggest the following:"Although the total energy is preserved when matter is processed by a shock wave, the available energy which can be extracted as work is decreased (Entropy is increased). This has the practical effect, for example, that an aircraft with shocks will experience an additional drag force to that which it would experience if the flow was shockless." AKAF 09:32, 6 February 2007 (UTC)

I added this provisionally, to see how it looks. Feel free to jump in and edit if you don't agree.AKAF 15:56, 6 February 2007 (UTC)

It now says Although the total energy is preserved when matter is processed by a shock wave, the available energy which can be extracted as work is decreased (Entropy is increased). I find the first part of this confusing. Total energy is always conserved in any case; no need to say that. So it leads the reader to think that something else is meant. It seems to be saying that some matter has the same amount of energy after a shock wave has passed through it as the same matter had before. (The phrase "processed by a shock wave" gives this impression.) I think this impression is wrong. I think the temperature is higher after the shock wave has passed through, therefore more energy is present. The energy was in the shock wave, and after the shock wave went past a mass of matter, the shock wave has less energy and some of the energy is left behind in the matter. I suggest changing it to: Although the total energy is preserved as a shock wave passes through a medium, the available energy which can be extracted as work is decreased, and entropy is increased, as some of the energy is transformed from kinetic energy into heat. --Coppertwig 16:19, 6 February 2007 (UTC)

Great more explanations

First let me refer to what said before.

I realise that finding the shock angle (beta) from the standard equation: $

tan\theta=2cot\beta\frac{M_1^2sin^2\beta-1}{M_1^2\left(\gamma+cos2\beta\right)+2} $ is an iterative problem. I'm sorry but that absolutely doesn't require years of PhD research. And this is the equation which is being referred to in NACA1135. There is simply no convient way of expressing beta accurately as a function of theta. Nevertheless finding beta if theta is known isn't rocket science.

Here my reply:

Great, looking at the math is the only way to explain and understand the physical situation. Yes, this equation explains the oblique shock and has to be solved. Very smart people tries for years to find an analytical solution for it [see for example, "Approximation for weak and strong oblique shock wave angles" Agnone, Anthony M. (Hofstra Univ., Hempstead, NY) AIAA Journal 1994 0001-1452 vol.32 no.7 (1543-1545)]. If you try to solve this equation using the iterations method, you will discoverer that there are six solutions (three pairs) of sometimes complex numbers. You will encounter many problems. People suggested to use the shock polar method instead (see for the code http://www.caselab.okstate.edu/codes.html#oblique). Yet, I will like to learn about your new experience. Nevertheless, there are many calculators on the net that use the cumbersome method of the polar shock (see for example, Shapiro's book for explanations). Later, Emmanuel (McGraw-Hill Companies, Mathematical Theory of Compressible Fluid Flow) suggested a partial solution. Then, Bar-Meir's solution came which is the analytical solution. What so nice about Bar-Meir's solution isn't that it is easily allow to carry the calculations but that it provides the mathematical explanation for the detached shock. The fact remains, that finding a solution to this equation was essential part of the understanding of the oblique shock. This understanding can be achieved by utilizing Bar-Meir's solution (which is described in chapter 14 in

"Fundamentals of Compressible Flow"  book www.potto.org).

Here's a link to the above-mentioned book and to its chapter 14. I don't know where exactly in chapter 14 to find the solution referred to; I think the chapter solves more than one problem. --Coppertwig 15:52, 6 February 2007 (UTC)

In a summary, there are essentially three methods (some mathematicians suggest other methods which are much more complex) known today to solve this equation: 1) shock polar 2) Emmanuel 3)Bar-Meir. Bar-Meir's solution is the only one that provides an explanation to detach shock.

While I'm not necessarily disagreeing with your summary of the state of the art, I think that ignoring the standard theory in favour of any other theory is not advantageous in a general use encyclopaedia. I wouldn't have any particular problem with a page in which the Bar-Meier theory of oblique shocks is explained (But you would probably have to convince someone else to write it to evade WP:NOR). Just as a general comment: I have seen a number of simplified shock solutions and most fail my tests on one of the following grounds:

1. Ideal gas only
2. Inviscid only
3. Adiabatic wall only
4. Homogeneous inflow only
5. Significantly more complicated than the "standard" method, while not offering significant gains in accuracy.

And for these reasons it is very difficult for any new theory to replace the standard theory, even for experts. I'm not opposed to putting your theory in wikipedia because I think that it's wrong (I haven't made any attempt to assess its correctness), I'm opposed because it's not the "standard" theory. Additionally the equation above is soluble and gives reasonable answers for the vast majority of oblique shocks. If you say that your theory extends the results of the old theory I might agree (If your theory is correct), but that doesn't make the standard theory wrong. AKAF 10:08, 6 February 2007 (UTC)

There is an improvement in the article. More suggestions, move the "Soliton" issues to analogy section. Most people never have heard about this term. Even fluid mechanics guys don't use this term (only mathematicians are using ). Change the word "sudden" in third sentence to abrupt or abrupt and sudden. —The preceding unsigned comment was added by Genick (talk • contribs) 18:30, 5 February 2007 (UTC). Yes, forgot to sign Genick Bar-Meir, --potto 18:35, 5 February 2007 (UTC)

I think soliton belongs here an an example of a nonlinear wave. I'd venture to say that quite a lot of people have heard of solitons, even if they don't know exactly what they are. The word soliton is used at least in optics, but I'm not aware of another word for soliton, could you let me know if you find another one? AKAF 10:08, 6 February 2007 (UTC)

I had heard of solitons. Besides, there's an explanation in parentheses and a link to a Wikipedia page on them, so it's not very bad if some readers don't know the word already. --Coppertwig 16:01, 6 February 2007 (UTC)

Difference between stationary shock and moving shock

Many askedme about the difference, the main difference, beside the mathematical complications, is the fact that stagnation temperature is constant in stationary shock while it changes in the moving shock. I hope that remove all the confusion in point. Genick--potto 19:10, 5 February 2007 (UTC) genick --potto 19:11, 5 February 2007 (UTC)

I think that stationary shock is the same as moving shock seen from a particular moving reference frame. I think the temperatures are approximately the same from all the relevant reference frames; I don't think we're talking about relativistic velocities here. (i.e. the velocity of the reference frame is not large enough to significantly bring in effects such as red-shifting.) So, the pattern of temperature should be the same in the two kinds of shock. However, a particle in one reference frame can remain in the unshocked part of the medium, while a particle in a different reference frame will encounter the shock and undergo change. So the situation as a whole is the same, but the way it's experienced by particular reference points can be different. If I've missed some important point, please explain it. Has a definition of "stationary shock" been given? Has an example of stationary shock been described? --Coppertwig 10:16, 6 February 2007 (UTC)

Coppertwig's certainly convinced me that he's right (see his arguments about stationary shocks further up). I'm still convinced though that standing shock should be treated as a special case of a shock wave, much in the same way as a standing wave is generally treated as a special case of a linear wave. To be honest though, I'm not too sure myself of how one distinguishes an attached shock, a detached shock, a transonic recompression shock, a standing shock in internal flow caused by pressure ratio and a nozzle recompression shock such that a lay reader will see a clear difference of type. AKAF 10:27, 6 February 2007 (UTC)

The stagnation temperatures for moving shock are different for earth frame of reference. I remember that years ago I struggle with this issue. Check Shapiro's book on moving shock. I believe that is in chapter 4 or my book chapter 5. The creation of different stagnation temperatures easily can be shown by the math but the hand waving we explain this by all kind of discussions. But the point is, that in the end the stagnation temperature change. You can check the math for a very simple case of complete open valve (I believe that is Shapiro's book (my book have all the cases) deals with it. One of the way to deal with it to except it and than your intuition will accept it, like many other "strange" that occur in compressible flow. Many book even do not deal with shock-choke case (limit for open valve moving shock) because of this strange situation.

Genick --potto 17:08, 6 February 2007 (UTC)

Order of examples of shock

I suggest re-ordering the examples of shock to put the ones that are easier to understand or more common first, as well as grouping together similar ones. I suggest this order:

1. Shock propagating into a stationary medium
2. Detached shock on a supersonic body
3. Attached shock on a supersonic body
4. Re-compression shock on a transonic body
5. Shock in a pipe flow
6. Detonation wave

Alternatively, detonation wave could be moved into second place, since it has similarities to the shock propagating into a stationary medium. I put it last because it involves chemical change, making it different from all the rest. --Coppertwig 15:39, 6 February 2007 (UTC)

I'm good with that. Go to it!AKAF 15:44, 6 February 2007 (UTC) Nice picture by the way, I had hoped that it was in the public domain somewhere!AKAF 15:57, 6 February 2007 (UTC)

Thanks, and thanks! I searched at commons:Special:Newimages for "supersonic". (Apparently the "newimages" search actually searches all the images, if I understand correctly, though I think it only searches on the filenames; there are other searches you can do.) My understanding is that if a picture is in Commons, then it's OK to link to it. I encourage others to search; there may be some other useful pictures there. Recompression shock and oblique shock could use helpful pictures or diagrams.

DONE. I reordered the examples. --Coppertwig 16:07, 6 February 2007 (UTC)

Readability in Shock waves in supersonic flows section

I find the first paragraph of this section difficult to read. Some readers will quit and go and read another article instead at this point. It may be better to move that first paragraph further down into the section; the rest of the section is much more readable. Alternatively, this first paragraph can be edited to be more easily readable.

This is the difficult part: The method of compression of a gas results in different temperatures and densities for a given pressure ratio, which can be analytically calculated for a non-reacting gas. A shock wave compression results in a loss of total pressure, meaning that it is a less efficient method of compressing gases for some purposes, for instance in the intake of a scramjet. I just can't process this. Each phrase is understandable, and it all fits together grammatically, but after reading it I just have no idea what it just said. Even re-reading it more slowly I don't get any meaning out of it.

I think one of the problems is in this part: The method of compression of a gas results in different temperatures and densities for a given pressure ratio. Is this saying that when you change to a different method, then you get different temperatures and densities? Or is it saying that when you change the pressure ratio then you get different temperatures and densities? Or somehow both? Very confusing.

OK, I just now realized what "pressure ratio" means. Of course it must mean the ratio between the pressures on the two sides of the shock wave, e.g. in the balloon-bursting case. Since I didn't get that until several re-readings, possibly the average reader also needs a bit of help there, e.g. "the ratio of the pressures in front of and behind the shock wave" may be easier to read and form an image in the mind that just "pressure ratios" when the mind is already busy trying to figure out what the rest of the sentence had been trying to say.

The bit about getting different temperatures and densities seems less important and can be moved later in the section or later in the article. Mentioning the different types of compressions, e.g. isentropic and Prandtl-Meyer, may be important to mention very close to the beginning so that readers who are actually interested in those can quickly get to the right page, but different temperatures and densities can be moved later if it can't be made readable enough.

I think I would find this easier to understand, if this is what's meant: For a given type of compression, the temperatures and densities can be expressed analytically as a function of the ratio of the pressures in front of and behind the shock wave, if it's assumed that there are no chemical reactions in the gas. Or put "for a non-reacting gas" in parentheses after "analytically". Still too complicated for the first paragraph, though.

One of my problems is that I don't think of a shock wave as a "method" but as a phenomenon. Sometimes they're not being used by humans for anything -- they just happen. Another problem is that I can't imagine how a compression can result in a loss of total pressure -- that sounds contradictory. Is "compression" being used in a different sense from the ordinary word? (I almost said "normal word"!!  :-)

Also I admit to never having heard of a "scramjet" as far as I remember (though I have heard of solitons!). I'm reading about "shock wave" and imagining a disturbance in gas moving outwards from an explosion or propagating like a plane wave through some gas, and suddenly scramjets are mentioned which I assume are some sort of airplane engine thing and bring a totally different image into my mind (or attempt to -- I don't know where exactly in an airplane engine you would find a shock wave.) That sudden shift in images is one of the reasons it's difficult to read. Perhaps the scramjet could be mentioned somewhere in the article where there's enough room to describe the phenomenon well enough to build a clear picture in the reader's mind.

I think shock waves can happen in liquids too and that the article needs to acknowledge this.

I hope these comments are helpful. I might suggest more specific edits later after thinking it over and/or after getting feedback on this. --Coppertwig 16:58, 6 February 2007 (UTC)