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May 20

Locomotives

There are 0-4-0s, 2-2-0s, and 0-2-2s. Why aren't there 0-2-0s? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 01:57, 20 May 2012 (UTC)[reply]

In France an 0-4-0 would be an 020. Sticking to American notation, an 0-2-0 would have a single axle, which seems unsatisfactory, rather like a car with one axle. Acroterion (talk) 02:03, 20 May 2012 (UTC)[reply]

Although, strangely, there have been 0-3-0s. Gandalf61 (talk) 08:51, 20 May 2012 (UTC)[reply]

And 1-1-2-1-1's. Tevildo (talk) 12:40, 20 May 2012 (UTC)[reply]

The answer is, therefore, simple statics. It would be possible to have a 0-2-0 or 0-1-0 locomotive on a Lartigue-style monorail, with the centre of gravity of the loco below the rail. However, the minimum stable configuration for a loco with all its wheels on the ground would be a tricycle 1-2-0 or 0-2-1 arrangement, but this would require three rails rather than two. Tevildo (talk) 12:51, 20 May 2012 (UTC)[reply]

There is an important confusion to be resolved here. In Europe, an 0-4-0 engine has 8 wheels, 4 on each side - the number '4' refers to the number of axles. In the US, the numbers count wheels instead so an 0-4-0 has only 4 wheels, 2 on each side. Hence, in US notation, an 0-2-0 would be silly - an engine with just one wheel on each side! Obviously such a thing cannot exist on a conventional railroad. In Europe, an 0-2-0 is a simple 4-wheeled, 2 axle vehicle with all four wheels being driven...like a 4WD car. I strongly suspect that our questioner is using European notation and asking why there aren't any engines with just four driven wheels on two axles. The answer probably relates to stability on the track or something...but I don't know details. SteveBaker (talk) 15:03, 20 May 2012 (UTC)[reply]

There were plenty of two-axled locomotives (0-4-0 in American (Whyte) notation, 0-2-0 in French) during the age of steam after the initial era of development: they were most typically small shunting engines, though they might occasionally be used for pulling small, local freight or passenger trains. In the subsequent diesel and electric era, 0-6-0s tended to be the minimum, probably because railway vehicles in general became larger and heavier, and train lengths on average greater: the locos themselves therefore also had to become more powerful and heavier, and a heavy 0-4-0 would exert too great an axle load on the track, whereas the same loco in 0-6-0 format would have a maximum axle load 33% less.

(Early in locomotive evolution, of course, 0-2-2s and 2-2-0s also existed – Stephenson's Rocket was an 0-2-2.)

To further avoid confusion, it should be mentioned that the UK (and hence the Empire/Commonwealth countries) also employed the Whyte notation system rather than the Continental.

By the way – Welcome back, Steve! {The poster formerly known as 87.81.230.195} 90.197.66.211 (talk) 22:49, 20 May 2012 (UTC)[reply]

Geiger tube continuation

I calculated the potential difference between the wire and the tube of a Geiger tube to be ${\frac {1}{2\pi \epsilon _{0}}}{\frac {Q}{L}}\ln {\frac {R}{r}}$ approximately. The power-supply voltage is slowly increased until you see a glow in the air very near the inner wire. This means that the air breakdown potential has been reached (and they provide a value for this and for all the other variables). Calculate this power-supply voltage. How do you do this calculation? Do you just substitute the numbers into the above equation or is it more involved than that? --150.203.114.37 (talk) 04:54, 20 May 2012 (UTC

It would help if you told us what L, Q, R, & r signify. Operating a geiger tube at a voltage high enough to cause a gas discharge (as indicated by a glow) is harmful to it. Normally, you just set the voltage to the value stipulated by the gieger tube manufacturer. This is at a "plateau" of maximum sensitivity that occures just below the glow point. Ratbone60.230.203.253 (talk) 07:23, 20 May 2012 (UTC)[reply]

Maybe; I didn't actually observe this; it's just a theoretical problem. Q = charge on the wire, R = radius of the tube, r = radius of the wire, L = length of the wire/tube. --150.203.114.37 (talk) 07:30, 20 May 2012 (UTC)[reply]

Then the formula given is the relation between the potential difference and charge as for any coaxial pair of conductors separated by a perfect insulator of permitivity e₀. It is a concept useful in calculating the capacitance http://en.wikipedia.org/wiki/Capacitance and has no relavence whatsowver to the operating voltage of a gieger tube. Under operating conditions,gieger tubes are operated with DC voltage, and no current flows in the tube capacitance, except for the very brief recharging current after each detected particle. Note that the term e_o, strictly speaking, being the symbol for the permitivity of free space, is in any case incorrect. It should be the permitivity of the gas used (e₀.k, which will however be sensibly close to e₀). Keit120.145.31.247 (talk) 09:22, 20 May 2012 (UTC)[reply]

Ah. What is the correct furmula then, and how do you derive it? --150.203.114.37 (talk) 10:44, 20 May 2012 (UTC)[reply]

As Ratbone said, this is not something you would normally calculate. The manufacturer of the geiger tube will tell you what voltage to use. The manufacturer determines the optimum voltage by testing. However, an approximate calculation of the optimum operating voltage can be found by using the formula for breakdown voltage of a gas under partial vacuum, given in http://en.wikipedia.org/wiki/Breakdown_voltage. As Ratbone also said, the optimum voltage is just a bit below the breakdown voltage. If you are intending to make your own geiger tube, say so, we can then give you information of more practical benefit. Keit60.230.198.136 (talk) 11:41, 20 May 2012 (UTC)[reply]

This is just a homework question; I am not trying to make one of these. I have never seen this formula before. Presumably you are required to use some of the parameters given. It's possible you also need some other well-known constants, but I wouldn't know. This is the exact text of the question: The power-supply voltage is slowly increased until you see a glow in the air very near the inner wire. This means that the air breakdown potential of

3\times 10^{6}V/m

has been reached. Calculate this power-supply voltage (give a numerical value) and explain your calculation. The length L = 80 cm, the inner radius r = 0.7 mm, and the outer radius R = 3 cm.--150.203.114.37 (talk) 13:20, 20 May 2012 (UTC)[reply]

In which case you have attempted a $10 answer to a 10 cent question, unless the orginal question gave you the gas pressure, and you can ignore the answers given previously by Ratbone and myself. Whoever wrote this question has a very simplified view of geiger tubes. All you need to answer the question is given in the question, plus a bit of the most elementary algebra. You can ignore the length, as a long tube has the gas under the same voltage gradient stress as a short one. The answer, fitting the data given, is 88 kV. As it is a homework question, I won't say any more than that. Real geiger tubes have an operating voltage of around 500V, because radius of the tube is smaller, and because the gas inside, which isn't air, is under partial vacuum, but this is no concern to you - you need to write what the teacher wants. The moral of this posting is that if you have a homework question, gives us the exact question, and what you have attempted in order to solve it. Then you may get an answer from us appropriate to your needs. Keit60.230.198.136 (talk) 15:14, 20 May 2012 (UTC)60.230.198.136 (talk) 14:57, 20 May 2012 (UTC)[reply]

The problem is I don't know where to start. Is there an equation which relates all these variables that I have to manipulate? --150.203.114.37 (talk) 15:49, 20 May 2012 (UTC)[reply]

Are you pulling my leg? This is very simple for a chap prepared to get involved with integrals etc as in your first version of this posted question. You don't need no fancy formula - its just proportion. As stated in your assignemnt question, the breakdown voltage of air at standard atmostpheric pressure is approx 3 x 10⁶ V / m. So if you have a breakdown distance of 2 m, you'd need 2 x 3 x 10⁶ V (ie 6 MV); if you have 0.1 m, you'd need 0.1 x 3 x 10⁶ V (300 kV). And the breakdown is between the 0.7 mm radius wire and the 3 cm radius cylinder.... Go for it tiger. Keit58.167.225.195 (talk) 16:10, 20 May 2012 (UTC)[reply]

I see. But how do you know that that is the breakdown distance? There being a glow "very near" the wire may imply that the breakdown distance is less than that. Or is the breakdown distance necessarily the distance between the wire and the tube in a Geiger tube? --150.203.114.37 (talk) 17:45, 20 May 2012 (UTC)[reply]

(edit conflict) I think that is missleading, the field is not homogenius, the glow was just closest to the inner conductor. You need to integrate the field along a radius or use a ready formula for this case. Due to rules regarding homework i do not want to give it here but you could calculate the linear charge that gives the desired field at the surface of the inner conductor and then use the formula you gave. Gr8xoz (talk) 17:49, 20 May 2012 (UTC)[reply]

Does this matter? Is the power-supply voltage not still 87900 V? What calculation am I supposed to do? I don't understand. --150.203.114.37 (talk) 21:38, 20 May 2012 (UTC)[reply]

When the glow is seen, breakdown has already occurred. Any non-linearity in electric field strength as alluded to by Gr8xoz has no relavence to your assignment question, because you want the breakdown initiating voltage, which is under conditions existing just before breakdown actually occurs. In practice, as you slowly increase the voltage, breakdown occurs suddenly with only a very small increase. So this additional small voltage, between no breakdown and breakdown started, can be neglected. Yes, obviously, the electric field stress on the gas/air (ie a field which accelerates the odd free electron fast enough to strike a gas molecule with enough kinetic energy to knock off another electron) is between the wire and the cylinder. As it is homework I don't want to put it any plainer than that - what you learn and remember is proportional to the mental effort YOU put in, not what I put in. If you want to understand the breakdown phenomena, and thus why the breakdown voltage is proportional to the breakdown distance, FOR THE DIMENSIONS AS GIVEN, start with our article on Paschen's Law, http://en.wikipedia.org/wiki/Paschen%27s_law. You could have found this yourself if you checked the Wiki link I gave you previously. The relationship between distance and voltage becomes very non-linear at small distances, but this is at distances very tiny compared to the dimensions given in your question. Given that whoever wrote the question has a very simpilfied view of geiger tubes (the dimensions are very innappropriate, and geiger tubes use particular gases under partial vacuum, not air at standard pressure, tubes are operated just below breakdown, not at breakdown), I suspect only a very simple answer is required. Is this from a high school math assignment? What year? Or a university physics course? Keit124.182.148.14 (talk) 01:32, 21 May 2012 (UTC)[reply]

Domestic Heat Pump connections

I am planning to install an air to water domestic heat pump for heating and am looking at the manuals. The 12 kW heat pump works best when there is a 35 litres/minute flow through it (it then heats the water by about 5C each sweep through the exchanger). If the flow rate falls below this, the manual says it becomes inefficient and cuts in and out. The circuit I wish to heat is already partly structural to the house and would not take that flow rate (indeed I am guessing even approaching that flow rate would be rather noisy), but the heat loss on that circuit is more than 5C. So is there any reason I cannot just "short" part of the outflow from the heat pump directly into the inflow for the heat pump mixing with the return? It seems a little too obvious as a solution...obviously the later part of the heating circuit will be a little cooler from having a high temp drop across them (thats ok, the later parts of the circuit will be the larger surface area ones), but if the heat pump is happy I know the actual heat is being transferred into the house a reasonable efficiency, right? --BozMo talk 07:22, 20 May 2012 (UTC)[reply]

Heat pumps are more efficient when the heat exchanger is at a lower geometric mean temperature throughout the exchanger, and more efficient if the hottest part of the exchanger is lower in temperature. So, if you reduce the water flow, the hottest point, and the geometric mean temperature, must increase, lowering efficiency. If you try and get around this by providing a bypass flow, what you are doing is increasing (by mixing cold with warm) the inlet water temperature, again increasing the mean exchange temperature. So efficiency is still lost, but not necessarily at quite the same degree. Ratbone58.167.225.195 (talk) 16:36, 20 May 2012 (UTC)[reply]

I'd think just lowering the flow rate through the entire system would lower efficiency less than the bypass. I don't quite understand what they mean when they say it will "cut in and out", though. StuRat (talk) 16:45, 20 May 2012 (UTC)[reply]

A heat pump water heater is essentially the same technology as a heat pump (ie compressor type) airconditioner. A temperature sensor controlls when the compressor runs - if the water is too hot, the sensor turns the compressor off. Thus the system cycles just like an airconditioner does. If frequent cylcing occurs, effieciency drops for 2 reasons: a) a high exchanger temperature itself reduces efficiency, b) the system spends a higher fraction of time in running up and stabilising. Ratbone124.182.19.67 (talk) 01:05, 21 May 2012 (UTC)[reply]

Right, but none of that explains why a lower flow rate would cause it to cut off more often. On the contrary, a lower flow rate should allow the water inside the cooling part of the loop to get cooler, since it stays in there longer, and the house should remain hotter than at a higher flow rate. Both of these should make it stay on longer. StuRat (talk) 02:59, 21 May 2012 (UTC)[reply]

Stu, it seems like you are talking about an airconditioner as in some domestic airconditioners that use a chilled water system, like large commercial buildings - the commpressor (termed a chiller) at a convenient location chills water, which is then piped to one or more fan-coil units within the dwelling. I assumed that the OP was asking about the more common heat pump water heater, ie a compressor unit, which "pumps" heat from outside air to heat water for your kitchen, bathroom, etc. On reviewing the OP's wording, it does seem that he's talking about an airconditioner. In which case, my explanation was the wrong one. I'm not familiar with domestic grade chilled water systems, but assuming they are much the same as commercial systems, the on-exchanger water temp in cooling mode will be about 12 to 14 C, and the off-exchanger water temperature will be a design temperature of around 6 to 8 C. If you halve the flow rate, the compressor (somewhat simplifying) in moving the same amount of heat, will cool the water to 0 C. The water must not be allowed to freeze - that will stop functionality and damage the system. So a sensor on the off-exchanger temperature will prevent it by shutting down the compressor until the water warms up again. If the OP installs a water bypass, the water will still get too cold, because some of it doesn't go thru the fan-coil unit and get warmed up, and will cause cycling. This problem sometimes occurs in commercial buildings, where (say) a 10-storey building has been constructed, but so far there are tenents on only one or 2 floors, so the fan-coil units on the empty floors are not functioning. A common solution is to temporily use electric heating, or pump in outside air thru some fan-coils, to make the chiller work cycle more slowly. This of course wastes energy. In heating mode, there shouldn't be much of a problem, though the chiller will cycle on and off within a short time frame for the same reason a standard aircon will, if way too large for the space conditioned. When its running, it will heat the room(s) too quickly, and the thermsotat will turn it off. Installing a water bypass won't fix that either, except that the higher exchanger temperature will make the system a little less efficient. Ratbone120.145.172.175 (talk) 03:33, 21 May 2012 (UTC)[reply]

I did consider the idea that water inside the cooling loop might freeze, but thought that they could just add anti-freeze to fix that problem. Or is the issue that frost will build up around the cooling loop ? I can see why that would reduce efficiency, but would it damage the system ? StuRat (talk) 03:52, 21 May 2012 (UTC)[reply]

I have never heard of adding antifreeze to chiller water, though that does not say it's not done. I note that common antifreeze (glycol) has a lot less specific heat than water, so adding it will promote short cycling and reduce heat & cooling capacity. You are right about frost on the exchanger though - I should have remembered that. Frost is, compared to moving air, a thermal insulator, so once frost forms, the exchanger temerature drops, forming more frost, ending up with bulk ice and no efficeincy. When I mentioned damage, I was refering to the fact, that at freezing, water expands. Ratbone120.145.172.175 (talk) 04:26, 21 May 2012 (UTC)[reply]

Well, yes, of course ice inside the pipes would damage them. You normally only need a small portion of antifreeze, depending on how low the temperature protection needs to go. StuRat (talk) 01:27, 22 May 2012 (UTC)[reply]

Hexahydrogen sulfide

There is hydrogen sulfide (H₂S). There is sulfur hexafluoride (SF₆). Why isn't there hexahydrogen sulfide (H₆S)? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 12:34, 20 May 2012 (UTC)[reply]

Hydrogen does not hybridize its orbitals. So many hydrogens can't quite overlap with all six of sulfur's hybridized orbitals.--Jasper Deng (talk) 16:35, 20 May 2012 (UTC)[reply]

I fail to see how hydrogen orbitals come into it. It's probably because hydrogen isn't electronegative enough to stabilise sulfur in the +6 oxidation state, whereas fluorine is electronegative enough. However, a few molecules are known with sulfur bonded to six carbons (Organosulfur compounds#Sulfuranes and persulfuranes) which casts doubt on the electronegativity argument. --Ben (talk) 16:40, 20 May 2012 (UTC)[reply]

Well, hypervalency tends to occur when you have enough stretched-out orbitals that can link many atoms.--Jasper Deng (talk) 18:26, 20 May 2012 (UTC)[reply]

See Hypervalent molecule; one of the models of hypervalency is the Three-center four-electron bond, which can be predicted and explained using molecular orbital theory. To put things simply, a 3c-4e bond requires p-p-p mixing between the two colinear ligands and central atom. Hydrogen doesn't have any electrons in p orbitals availible to create the 3c-4e bond, so it won't bond like SF₆ will. There's even a nice picture of the type of bonding that happens in SF₆ in the hypervalent molecule article which shows the p-p-p interactions in forming the 3-center bond. Hydrogen doesn't do this. --Jayron32 18:36, 20 May 2012 (UTC)[reply]

If hydrogen couldn't form 3c-4e bonds, then the bifluoride anion wouldn't exist either—but it does. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 19:32, 20 May 2012 (UTC)[reply]

It does so only when it is the central atom, not when it is the ligand. Hydrogen can form such multi-center bonds as the central atom, and does so in the 2-electron 3-center bond in diborane as well. The difference is that hydrogen isn't the central atom in the hypothetical SH₆, sulfur is. --Jayron32 19:44, 20 May 2012 (UTC)[reply]

Thanks. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 05:02, 21 May 2012 (UTC)[reply]

Resolved

Ingredients of paint primer?

Two questions:

I need to know what ingredients are used in paint primers (the kind used for wood and plaster - not metals) - our article Primer (paint) has no indication of ingredients and Paint discusses paint ingredients in general but doesn't explain what's different about primers.
Some interior paints say that they include a primer. From the description in our two articles of what a primer does, this seems unlikely to be true. Can anyone explain what's really going on here? (And again, what are the ingredients in primer+paint products that aren't in the paint-only stuff?)

Background: I'm using a 100 watt CO2 laser-cutter to cut 5mm plywood - when the plywood is painted with some kinds of paint (I'm using and off-white glydden interior eggshell), the laser has a hard time cutting it - with other kinds of paint (acrylic craft paint), it's no problem. Shinier paints seem to cut worse than flat paint and our paint article suggests that shinier paints have silica, glass and metal flakes in them...glass and metal can't be cut with such a low power CO2 laser, so this explains a lot. However, I still want to use primer on my plywood...and I'd like to find a way to put a shiney/waterproof surface on my plywood without impeding the laser too much. (No, I can't paint the plywood after it's cut because I'm also etching a design into the paint using the laser.) SteveBaker (talk) 14:48, 20 May 2012 (UTC)[reply]

Have you seen the Wikipedia article specifically on paint primers? The main purpose of a primer for wood is to seal it, so that the colour coat is not absorbed into the wood, leaving it pale & uneven and permeable to moisture. Paints intended for wood that incorporate a primer are quite common. Primer ingredients commonly are linseed or synthetic oils, or synthetic latex (such as PVA). In at least some cases, the only difference between (for wood) a primer and a topcoat is the proportions of the ingredients, the primer having a higher percentage of oil or latex. See http://www.buildings.com/tabid/3334/ArticleID/2846/Default.aspx. Why not use a matt paint compatible with the laser, then spray on a clear varnish to seal and make shiny. Ratbone58.167.225.195 (talk) 15:53, 20 May 2012 (UTC)[reply]

One thing to keep in mind: paints are not subject to ingredients disclosure laws like, say, foods are. Many paints and pigments are proprietary formulas whose exact compositiona and processes for making them are kept highly secretive as trade secrets. Actually getting the specific composition of a specific primer may be impossible. I had a friend that was a research chemist worked in the industry, and he had a ten-year non-competition agreement with the company he worked for; that is if he ever left the company he worked for, he couldn't work for any company in the pigments industry for ten years. They take that secrecy pretty seriously. --Jayron32 17:43, 20 May 2012 (UTC)[reply]

Electricity of x cm³ of water at y m

Can you calculate how much electricity can be generated by x cm³ of water at y m.88.9.109.2 (talk) 15:04, 20 May 2012 (UTC)[reply]

Yes, by determining the potential energy of that mass at that height, then applying an efficiency factor to determine the percentage of that energy actually converted to electricity by your dynamo. Do you want the formula ? StuRat (talk) 15:10, 20 May 2012 (UTC)[reply]

Looks like Stu thinks you want a hydro-electric (water-fall) answer, which is most likley the case. But you might want an answer appropriate to ocean thermal power (OTEC). Which is it? Is this homework? Keit60.230.198.136 (talk) 15:31, 20 May 2012 (UTC)[reply]

No homework. Just curiosity. Can you give me two formulas: 1. water is flowing through a dynamo/turbine. 2. water is enclosed, hanging in a cable attached to pulley connected to a dynamo. 88.9.109.2 (talk) 15:36, 20 May 2012 (UTC)[reply]

Water has a density of 1 g/cm³ (depending on temperature, impurities, pressure, etc.). So, we have X grams, at Y meters. The formula for gravitational potential energy is U = mgh, where m = Xg, g (on Earth) = 9.8m/s², h = Ym. This gives us U = (9.8)XY gm²/s². That's 1/1000th of joule. We also need to apply the efficiency, let's say 0.0 < E < 1.0. So, the amount of electricity generated, in joules, is (9.8)EXY/1000. The formula would be the same for the water in a container, although there the mass of the container must also be considered, and the cable is more complicated, because each portion has a different gravitational potential energy. StuRat (talk) 17:13, 20 May 2012 (UTC)[reply]

State of radon related to radioactivity?

Would radon be a gas if its radioactivity were neglected? Astatine and francium both would be vaporized by their own decay.--Jasper Deng (talk) 18:33, 20 May 2012 (UTC)[reply]

Quite likely it would, since the boiling point trends displayed by the noble gases don't have an obvious discontinuity between Xenon and Radon. I don't think radioactivity has anything to do with phase of matter at room temperature, which is a function of an atom's electronic properties, not its nuclear properties. That is, a substance is a gas because it has weak intermolecular forces, the strength of which can be preicted solely by the structure of the electron cloud around the molecules/atoms that make up the substance. --Jayron32 18:42, 20 May 2012 (UTC)[reply]

Well, radioactivity does tend to change temperature, so there's an indirect effect there. Francium's article said it would be liquid especially because it's radioactive.--Jasper Deng (talk) 18:46, 20 May 2012 (UTC)[reply]

Yes, but it doesn't change the boiling point or melting point. You're confusing the two temperatures. The boiling point of a substance is the expected temperature for it to boil, NOT the temperature it exists at. That is, Radon will boil at 211 K regardless of where the source of the heat making it 211 K comes from. If it gets to 211, it boils, and that has nothing to do with radioactivity. Likewise with Francium, the sentence in the article is misleading. What it is saying isn't that the 27 Celsius MP quoted in the article is effected by its radioactivity, it is saying that the extra heat generated by its radioactivity means that in room temperature air (nominally 25 degrees Celsius), the excess heat from the radiation is more than sufficient to raise Francium's temperature the extra 2 or so degrees to melt it. That is, sitting in the average indoor room, Francium will be a liquid because it will always be somewhate warmer than that room. Again, the radiactivity of the Francium doesn't affect the temperature it should melt at, it affects the temperature it is. Two different ideas. --Jayron32 19:01, 20 May 2012 (UTC)[reply]

Radon's boiling point is −61.85 °C so any radiative self-heating wouldn't be significant re: the state in a cozy room (or lab). Maybe if you had some radon at Vostok Station where the air temp was -62 °C radioactivity might push the temperature of your radon just above the boiling point. 19:19, 20 May 2012 (UTC)

Ultra High power mechanical HVDC-breakers?

I am doing some calculations on the feasibility of a ultra high voltage ultra high current world wide HVDC-grid. I speculate about 2000 kV to ground and 25 kA nominal current, 100 GW per bi-pole. The fault current after a few ms short circuit would be in the range of 100 kA. One of the big problems in constructing such a system are of course the selective removal of faulty components while maintaining operations of the rest of the system.

My question is if the needed size and cost of DC-breakers capable of breaking 2 000 kV and 100 kA can be estimated?

The challenge are of course to contain the blast and to cool the arc (generating >200 GW?) to temperatures below the temperature were significant ionization occurs while maintaining the isolation.

Today no HVDC-breakers are in operation, HVAC-breakers in the range of 800 kV and 50 kA exists, as I understand it they can only break a few kV so they will only interrupt the current near a zero crossing in the AC-current. ABB have developed a "Hybrid" HVDC-breaker, using semiconductors. Proactive Hybrid HVDC Breakers. It seems that it would be extremely expensive to use semiconductors in all the breakers at this level. Could large fuses in parallel with a smaller mechanical breaker be used? Fuses normally break the current before the zero crossing so it should work equally well with DC. Smaller fuses at for example 36 kV and 20 kA breaking current has a mass of about 2 kg, giving about 3 kg per gigawatt short circuit power, would it scale so that a fuse for 2000 kV and 100 kA would have a mass around 0.003*200 000 kg= 600 kg? (It seems small) Could it be done with explosives for fast and synchronized operation as in some MV Is-limiters? How can the breaking capacity of a given breaker chamber be approximated without complicated fluid dynamics and FEM-simulations? Gr8xoz (talk) 20:32, 20 May 2012 (UTC)[reply]

By "of curse", do you mean "of course" ? StuRat (talk) 20:43, 20 May 2012 (UTC) [reply]

Yes, corrected that. Gr8xoz (talk) 20:54, 20 May 2012 (UTC) [reply]

ABB Group, Siemens and Alstom likely have the information you seek, though they may consider it proprietary. Do you really want to be replacing extremely expensivbe fuses whenever lightning hits a line, as opposed to a breaker opening and reclosing? There are fuses with explosive charges to open the circuit and programmable sensors and actuators to provide any desired operating curve, for a price. If you want to de-energize a HVDC line, could you do it via the "valve hall" or electronics which are transforming AC to DC at the sending end? Limiting fault current is inherently easier with HVDC than with HVAC. Edison (talk) 03:10, 21 May 2012 (UTC)[reply]

Yes someone in the suggested companies has probably looked in to this. It is well outside their current product lines. The most powerful HVDC-liks built today are around 7 GW, this are 100 GW, almost twice the maximum consumption in the UK. For point to point HVDC-links the converter station ("valve hall") or HVAC-breaker on the AC side can be used to interrupt the fault current. The need for HVDC-breakers comes when building large redundant HVDC-grids with many converter stations. You do not want to shutdown the whole world wide HVDC-grid in this example as soon as there is a fault anywhere. Why do you think it is easier to limit the fault current in HVDC-systems? That is not the impression I got.

If a fully reusable low maintenance breakers can be made at a reasonable price, that is of course better but I get the impression that some consider it almost impossible. That is the reason ABB introduces semiconductor based solutions. I think one time controlled fuses could be competitive if they can be made at 10 % of the price of a reusable breaker. I think it would be possible to repair used fuses.

Lighting is not a likely source of faults as the system I think of are built with Gas Insulated lines. Gr8xoz (talk) 06:52, 21 May 2012 (UTC)[reply]

You link to gas insulated lines did not work, but please look at the cost per km for gas insulated bus versus overhead HVDC. Look at the meantime between failure for a 100 km HVDC overhead line, versus that length of gas insulated line, with a leak anywhere in one of the zones causing a flashover and line failure. Lightning will be a frequent cause of overhead HVDC momentary outages, since it can be a way higher voltage than the insulators are designed for. It is corrected by tripping and quickly reclosing the line. The metal shell outside a gas insulated line would seem to protect better than the static line above an overhead HVDC line, but I expect that any elevated section of gas insulated transmission could also be knocked out by a sufficiently intense lightning strike. When a fuse operated, the silver strip in the center basically vaporized and is dispersed into the silica sand around it (some fuses are oil filled, same idea). If it has a pyrotechnic operation, then even more disturbance of the conductor and its surrounding nonconductor would take place. Maybe you could reuse the shell and the endcaps, or reprocess the sand to recover the silver. I sawed open a blown 12ky fuse and did not find any silver droplets (alas). In HVAC, the impedance of the source (generators, lines, transformers, series inductors) along with the fault impedance is all that limits the current. IN HVDC, the valves can limit the current much more easily when a fault occurs. It is more actively than passively sourced. Edison (talk) 19:05, 22 May 2012 (UTC)[reply]

Thank you for your answers, I have corrected the link now. I have limited time now so I will be brief. According to my link GIL are most reliable, 87 km installed world wide over 35 years and not a single fault so far. Of course this are like the Concorde before the accident it was the safest airliner in the world and after the accident it was the worst. I expect that the probability of a leakage should be at least as low as in a pipe line and they seem to be reliable. The transported power are larger than a gas pipe line such as Nord_Stream and I expect the cost to be lower due to smaller pipe, 800 mm vs 1200 mm, if produced in large quantities. I think overhead lines will be cheaper over uninhabited land but it is increasingly hard to get permits for new overhead lines in many places and most of the Earth are sea and there overhead lines are very expensive. I expect the cost advantage of overhead lines to be lower the larger the conducting area is since the construction will start to approach that of a suspension bridge, in the case of 100 GW I have calculated with about 0.1 m^2 cross section per conductor.

Lightning do normally not have currents above 300 kA, in the same range as the normal fault current so I do not expect it to give significant voltage rise on the outer pipe. In Is-limiters the pyrotechnic is used to blow away a conductor parallel to the fuse and then the fuse operates as normal, in this way the operation can be computer controlled. I think of something similar or a pyrotechnically operated reusable breaker. If the fault current can be controlled by the converter station depends on the converter topology, in a normal voltage source converter it can not be controlled due to the free wheeling diodes in parallel with the IGBTs. In thyristor based HVDC it is probably possible. Anyhow it is not possible in large HVDC grids were selectivity are needed. On long lines you would like to interrupt the current even before a signal at the speed of light could reach the other end.

I think my initial question could be phrased as follow: what is needed to cool a high density plasma with about 200 GW during a few milliseconds down to a temperature where the ionization stops. Breakers often contains the plasma between close surfaces (conducting or non conducting) to improve heat conduction. If we assume 5 mm between surfaces and 10 MPa SF6, how big area would be needed? How does this depend on gas speed?

Gr8xoz (talk) 10:46, 23 May 2012 (UTC)[reply]

Planetary Resources, Inc.

This may belong on the Language help desk but I think I get better answers here. Planetary Resources, Inc. are a company that plans to mine asteroids. (NEOs) Are there any logic behind the naming? To my it sounds like they are one of very few mining companies that plan to use non planetary resources. Almost all other use resources on the planet Earth while the NEOs are not planets. The IAU states: "the term 'minor planet' may still be used, but generally the term 'Small Solar System body' will be preferred." Minor planet. So are they basing their name on the obsolete term "minor planet" or are there any other rationale. Gr8xoz (talk) 20:51, 20 May 2012 (UTC)[reply]

Planetoid is another name for a minor planet. But perhaps they mean they will provide resources to a planet (Earth), rather than from one. StuRat (talk) 20:56, 20 May 2012 (UTC)[reply]

(ec) I doubt if anybody can give you an authoritative answer, but my take on it is that "planetary" is a word that to most people has nothing to do with the Earth, but just conjures up "space". --ColinFine (talk) 20:58, 20 May 2012 (UTC)[reply]

electric car batteries

If there were a way to charge a battery while driving without hydrocarbons, how many kwh would be needed to supply the electric motor with the needed energy? — Preceding unsigned comment added by 70.162.248.221 (talk) 21:12, 20 May 2012 (UTC)[reply]

Well, there have been solar powered cars, but that requires an extremely light car (dangerous in an accident) completely covered with solar cells, and a sunny day with the Sun high in the sky. A more practical solar-powered car might be able to charge up while parked in the Sun (assuming it spends at least 90% of the time parked), with a range slightly increased by driving in sunlight. I assume the reason no manufacturer offers such a car is that a solar-cell covered car is rather ugly. StuRat (talk) 21:17, 20 May 2012 (UTC)[reply]

Then, of course, there's regenerative braking. This only charges the car battery while braking, so the net effect is to reduce the overall rate at which it discharges. StuRat (talk) 21:25, 20 May 2012 (UTC)[reply]

The Chevy Volt has a 111 kW motor [1]. So, if you want to be able to drive it continuously, you would need to provide that much. If you wanted to be able to drive it for 10 hours using the batteries, they would need to hold 1110 kWh. Of course, these both assume maximum power consumption. If you are driving more conservatively, you can get by with less. StuRat (talk) 21:37, 20 May 2012 (UTC)[reply]

Depending on a loot of factors but generaly around 20 kW in 100 km/h :Electric_car#Running_costs_and_maintenance

Are these "proper" moobs?

On the gynecomastia article, the representative and first image is File:Gynecomastia_001.jpg.

Without trying to show false sensitivity - are we sure that's not just a fatty? Is there proper diagnosis that that image is actually Gynecomastia? Obviously a doctor or medical specialist cannot diagnose over the internet but does it "look right"? I've actually started a very similar discussion on the representative talk page but makes sense to come here too. Egg Centri c 21:47, 20 May 2012 (UTC)[reply]

It seems to me that there is no sharp line. All men have breasts, and those on the overweight are larger, as are those with men suffering from certain problems, like hormone imbalances. I wouldn't expect them to look much different. I suppose a thin man with large breasts might be a better example, though. StuRat (talk) 22:00, 20 May 2012 (UTC)[reply]

Is a breast that is large purely cause the guy is fat, and which will shrink to normal if he loses weight, actually gynecomastia though? Indeed, I don't think any of the images on that page are very great as displaying tits - something like this is more like it...Egg Centri c 22:10, 20 May 2012 (UTC)[reply]

On a mere fatty (pseudogynecomastia), I'd expect more continuity between the upper slope of the "breast" and the surface of the flesh above it. —Tamfang (talk) 22:50, 20 May 2012 (UTC)[reply]

I will ask the original uploader although they haven't edited since 2010 so it's a long shot... Egg Centri c 11:32, 22 May 2012 (UTC)[reply]

In that photo the nipples and areolae seem unusually large; if so this would argue for a bit of extra estrogen in the mix somewhere. That's not particularly rare. Certainly anyone who appreciated a recent Zane Donovan shower scene will have noted how much smaller these can be in other individuals. Wnt (talk) 02:20, 23 May 2012 (UTC)[reply]

However, I believe obese men tend to have a higher estrogen level, so, again, it's difficult to draw a dark line between normal and abnormal male breast development. StuRat (talk) 05:47, 23 May 2012 (UTC)[reply]

Wood as an electric conductor

In Jurassic Park, the Dr. Alan Grant Sam Neill is walking with John Hammond's Richard Attenborough grandkids and comes upon an electric fence that appears to have been deactivated. To check it, Grant throws a stick at the fence. If wood doesn't conduct electricity, what good would this do? DRosenbach ^{(Talk | Contribs)} 22:52, 20 May 2012 (UTC)[reply]

I've forgotten where this is in the narrative but wasn't it raining just before this? A wet stick would possibly spark if the voltage were high enough. Dismas|^(talk) 23:06, 20 May 2012 (UTC)[reply]

It makes for good cinematography. It was just a film. Would an american audience comprehend SIDE?--Aspro (talk) 23:10, 20 May 2012 (UTC)[reply]

(edit conflict) x2 Wet wood is a good conductor of electricity, and even dry wood conducts to a certain extent, especially at the high voltage (but low current) generated by most electric fences. Allowing the wood to rest against a modern animal electric fence would short the current to earth. I'm not sure whether the fence in Jurassic Park was designed to kill. If so, then it might have been both high voltage and high current, producing an electric arc when shorted (or perhaps that's what the director wanted viewers to think.) Dbfirs 23:14, 20 May 2012 (UTC)[reply]

Think I might have to explain that. SIDE =Switch off, Isolate, Dump & Earth. Until then then an electric fence is not to be trusted. It was just a film for the Hoi polloi -not a science lesson... Dinosaurs... I ask you! It was on par with the utter hogwash that came out of seaQuest DSV.--Aspro (talk) 23:30, 20 May 2012 (UTC)[reply]

Also, Some EF's pulse -such as cow fences. So the stick might might just hit between energising and thus show no observable effect.--Aspro (talk) 23:38, 20 May 2012 (UTC)[reply]

A random piece of wood cannot be counted on as either a good conductor or a good insulator. Oven dried wood has resistivity of about 10E14 to 10E16 ohm-meter , while damp wood has resistivity of 10E3 to 10E4 ohm-meter. The conductivity increases with temperature. If an electric fence in Jurassic Park were built to kill dinosaurs who touched it, it might well be energized at 480 or even 2400 volts, with a high current, low impedance source, in contrast to the high voltage, low current pulses of a typical farm electric fence. A piece of tree limb wood connected from the high voltage J park fence to ground (such as the nonenergized metal of the fence) might blacken, then catch fire. It might or might not pop the breaker. When a tree limb touches 2 utility conductors at 4kv or higher, the current may be sufficient to cause a fuse or breaker to open the circuit. A carefully dried wooden stick might not conduct as much current. Edison (talk) 02:54, 21 May 2012 (UTC)[reply]

"SIDE =Switch off, Isolate, Dump & Earth." Thanks for the explanation of the acronym. I think I understand S, I, and E. But what is "Dump"? Wanderer57 (talk) 04:32, 21 May 2012 (UTC)[reply]

D = Dump = Discharge to earth. 84.209.89.214 (talk) 23:32, 21 May 2012 (UTC)[reply]

Thanks. It seems that dump and earth mean the same thing, which is what is called "ground" in this neck of the woods. Wanderer57 (talk) 05:22, 22 May 2012 (UTC)[reply]

Dump: this refers to a controlled discharge -usually via a 'dump resistor'. Inductors and capacitors (re: electric fence) can store a lot of energy and so an immediate short to earth can cause very high currents to flow. Thus, dumping prevents these high currents from wreaking the the equipment. After 'dump' a small -low current- earthing lead can be attached. As I stated: this was a just cheap B movie and a stick on a real electric fence might show nothing. Yet... you just touch it and the nearest velociraptor will likely be perforated by your molten dental fillings flying through the primeval jungle (poetic-licence added for emphasis only). As Michael Caine says:”Not a lot of people know that”. --Aspro (talk) 19:49, 22 May 2012 (UTC)[reply]

This of course depend on regulation but normaly the earthing cable need to be thick enugh to saftly handle any short circut current if the object would be energized by misstake.Gr8xoz (talk) 08:20, 23 May 2012 (UTC)[reply]

Think you way miss the point. Earthing cables that can carry high surges are expensive and heavy. So why use one, when a cheaper lead can be used -incorporating a dump resistor? Second: the equipment being earthed may not be able to handle the surge that results from suddenly being shorted to earth. One's finger can discharge several thousand volts with barely a tingle felt -if the energy transferred is low enough. Yet, try putting a screwdriver across the terminals of a very large capacitor and then ask yourself it that was a very good idea! However, just small earthing lead can leak away the incoming energy faster than it can be stored -so no large conductor required. So and therefore, to get a very large critter to think (when encountering an electric fence) WHAT THE ∆˚®ß¥ø≤††† was the THAT? a fence needs a power feed at common human safe supply levels and then an energy storage system so that when the critter makes contact, s/he thinks ∆˚®ß¥ø≤†††; was the hell was THAT! Its not 'regulations' but practicalities that tend to determine these SIDE procedures. As Michael Caine says... etc.--Aspro (talk) 20:40, 23 May 2012 (UTC)[reply]

I have not heard about SIDE before, the steps you describe seems appropriate when dealing with high voltage capacitors that can only be charged by a relatively low powered source. My reference frame is power distribution and transmission systems. If the power line you are working on are connected by mistake in such systems the earthing cable can get a current of 50 kA, then you probably are happy that the cable are thick. Electric fences for dinosaurs does probably have lower power but I do not know. Gr8xoz (talk) 22:57, 23 May 2012 (UTC)[reply]

Thanks for the explanation of SIDE. Live and learn.

As for the movie, I thought the dinosaurs were very well done but I kept getting distracted by details which seemed to me scientifically very inept. Wanderer57 (talk) 03:39, 24 May 2012 (UTC)[reply]

Asthma meds for otherwise healthy people

What would puffing on an inhaler do for someone who does not have asthma? Would it basically be the list of adverse side effects listed at the Salbutamol page?

Note: I don't have asthma, don't have anyone in my immediate circle of friends/family with it and therefore don't have easy access to an inhaler. I don't plan on finding one to experiment with no matter what the answers I get here are. I have not died and come back to see if maybe an inhaler caused my death. I'm not asking for medical advice. I'm simply asking out of curiosity. Dismas|^(talk) 23:27, 20 May 2012 (UTC)[reply]

The first effect of overdoses might be hands trembling. However, puffing only once won't have any noticeable effect for sure. It's quite difficult to overdose on the inhaler (which is different from taking a pill or taking salbutamol intravenously). OsmanRF34 (talk) 00:10, 21 May 2012 (UTC)[reply]

I've had a few puffs of friend's inhalers in the past, just for a laugh and to see what would happen, I noticed no effect. Vespine (talk) 04:44, 21 May 2012 (UTC)[reply]

Albuterol is often prescribed to people who do not have asthma. (I have had it prescribed to me a number of times to just clear up bronchial weirdness and phlegm that occasionally trails behind long after the cold has gone.) In my experience (just anecdotal), a single deep pull of albuterol (expel all air from lungs, inhale a puff and suck it as deep into lungs as possible) usually makes one feel a little jittery and makes the lungs feels a little "funny." (But it does seem to work wonders at "clearing them out".) Which are some but definitely not all of the adverse effects listed there. I would expect those effects to be there whether you are "otherwise healthy" or not; they are the effect of inhaling that kind of drug into your lungs, not an interaction produced between the drug and your injury/disease. I know, not super helpful, but I thought I'd offer it up. Not medical advice, etc. etc., all of what I am describing was prescribed to me personally by a doctor, please do not replicate without a doctor, etc. --Mr.98 (talk) 13:12, 21 May 2012 (UTC)[reply]

Some asthma meds are widely used as quasi-legal (with a prescription) doping for endurance athletes. I think 40% of the Tour de France participants are certified asthmatics, and up to 80% of UCI professional cyclists share this fate [2]. --Stephan Schulz (talk) 21:24, 21 May 2012 (UTC)[reply]

Thanks for the info, all. And Mr.98, I don't mind the WP:OR at all! Dismas|^(talk) 03:21, 22 May 2012 (UTC)[reply]