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October 3

More firefighting Q's...

1) Are Novec 1230 fire extinguishers likely to be found in airport terminals?

2) How much heat can 1 mole of Novec absorb from a burning object at 800 C, and how many moles of it does a typical fire extinguisher contain?

3) Suppose I substitute a Novec extinguisher for the CO2 one in my previous scenario (trying to extinguish a burning office cubicle at 800 C and cool it down to 100 C) -- how much Novec must be expended to achieve this (assuming the cooling is rapid enough that heat transfer from the rest of the room, which is also at red heat, is negligible), and is this even possible with Novec?

2601:646:8E01:7E0B:D403:68F1:A297:C74A (talk) 01:08, 3 October 2017 (UTC)[reply]

They're used, but in flood systems, so not in human-occupied spaces. You find them in server rooms or machinery spaces. The regulations are very tight on mixing these flood systems with humans - you either have to disable the flood system when humans are in there, or the humans have loud warnings to evacuate before it goes off - and those systems are only used where there's a risk of explosion that would destroy the building (it's worth the risk of suffocating people if they'd be killed by an explosion anyway).

Again, these materials are used to extinguish a fire chemically, by blanketing or by chemical reaction. They're not intended for use by simple cooling.

They're used in three main scenarios:

Expensive equipment. So it's a safe blanket that's easy to clean up and doesn't react with warm materials. You can flood a room full of money and only damage the one rack that was actually on fire. Note that cold gases can cause damage by thermal shock.
Dangerous materials. They're a fast knock-down to energetic materials that you really don't want on fire. An installed flood system is deployed far faster than a human team reaction, or bringing extinguishers to it.
Difficult to extinguish materials. The ability to block the chemistry of the fire reaction can be useful to extinguish things that can't otherwise be extinguished by simple cooling or blanketing. Note the difference between inert sodium bicarbonate powder extinguishers for BC fires and ammonium phosphate that is acidic, reactive and can extinguish ABC fires (but will leave a corrosive residue). Fluids like Novec (I know nothing of Novec itself) are inert at low temperatures, limiting cleanup damage, but reactive at high temperatures.

You don't need to extinguish such a fire by cooling it, because you had installed an expensive flood system with an expensive fluid, which extinguished the fire before it became so hot that its simple total heat was a problem. Hot fires (as total heat in a volume, not temperature) are extinguished with water, because nothing else is available in the quantities needed to deal with a big fire.

If you're on site, you can extinguish potentially serious fires with a tiny CO₂ extinguisher, just because you were quick on the scene. Even better when is automatic.

If you're not on site, and have to travel to it, serious fires will be big by the time you get there. That's why it's nearly always time for water.

To answer your specific questions needs textbook figures that I just don't have handy. But I don't think they're even needed here, because your basic premises - that these fluids operate only by cooling - aren't correct. Andy Dingley (talk) 11:56, 3 October 2017 (UTC)[reply]

The article Fire extinguisher clearly states that Novec works by cooling as well as by blanketing/chemical reaction -- and your assertion that hot fires always require water is not correct, because if there's live electrical equipment in there (as in the cubicle farm I described), then water cannot be used. 2601:646:8E01:7E0B:D403:68F1:A297:C74A (talk) 12:27, 3 October 2017 (UTC)[reply]

Dat is not exactly correct. A water spray can and is often used were there is live electrical equipment. What is important is the voltage of that equipment. A water spray, even on three phase mains at 480V does not create a conductive path back to the fire fighter when performed correctly. The resulting steam then dilutes the atmospheric oxygen available and once again breaks the triangle. Don't try it however, unless you have been on a basic industrial first responders fire fighting course. Get to know your extinguishers long before you may ever need to use them. If it can spray, it should show dielectric test approval rating for use on electrical equipment. Aspro (talk) 16:39, 4 October 2017 (UTC)[reply]

Not gonna try it myself -- I was thinking about a fictional scenario I encountered (to be exact, an episode from American Airport Firefighter Simulator, toward the end of the "Headache" level), and trying to figure out what kind of extinguisher John (the player's character, a firefighter trapped in a burning airport terminal and trying to escape) uses. (Of course, you can say that he's using fictional extinguishers, but that doesn't cut it for me -- I like to keep things as close to real life as possible. BTW, that's also why I asked about the highest survivable temperature -- in that room, the walls are literally glowing red-hot, and I had my doubts that John could actually survive that -- correctly, as I suspected! And that is why the cooling effect is so important -- before he can even enter the office space, he must first use the extinguisher to carve out a survivable zone just inside the entrance, and to do that he must not only put out the fires in that area, but also cool down all objects in it to no more than 100-200 C -- while everything else in the room is still at red heat!) So right now I'm down to 2 options (now that CO2 has been ruled out) -- Novec or Halon/Halotron. Which brings up a 4th question: Do Halon or Halotron fire extinguishers have a significant cooling effect, or do they work entirely by disrupting the chemistry?2601:646:8E01:7E0B:D403:68F1:A297:C74A (talk) 04:38, 5 October 2017 (UTC)[reply]

Efficiency of jet engine

Suppose a jet passes $r$ kg/s of matter at a velocity $v$ m/s. Then the force on the jet is $F=rv$ N (I think). Now, suppose you want to configure $r$ and $v$ so that the engine is "efficient" in the sense of using a small amount of power for the given force $F$ . Basically, is it better to have small $r$ and large $v$ , or large $r$ and small $v$ ? My guess is that the power used to pass the propellant is ${\frac {1}{2}}rv^{2}$ W, because each second, you're passing matter which has kinetic energy of ${\frac {1}{2}}rv^{2}$ J: that means you have to give the propellant ${\frac {1}{2}}rv^{2}$ Joules every second, or expend ${\frac {1}{2}}rv^{2}$ watts. Substituting my first equation, that would equate to a power of either ${\frac {1}{2}}{\frac {F^{2}}{r}}$ or ${\frac {1}{2}}Fv$ watts. That would suggest that it's better to have large $r$ and small $v$ if you want to minimize power. Is this logic sound? PeterPresent (talk) 02:53, 3 October 2017 (UTC)[reply]

Nope! It's really hard to compute the relation between the thrust (force) and the power for a jet engine. In fact, it's as hard as doing rocket science. You can try to deduce a relation from first principles of kinematics, but your equation is bound to be incredibly inaccurate because it fails to account for all the complications of airflow.

The math you used might apply to some toy problems in continuum mechanics like computing the energy budget for a chain falling off a table, or for bulk material getting accelerated as you continuously pour it onto a conveyor belt... this is the sort of problem that is concocted to torture physics graduate students, but it's not a practical equation, and it's not particularly useful as a practical approximation to real-world aerodynamics.

You might want to read our article on specific impulse; thrust; and jet engine.

For a practical overview of how this works, read Chapter 5 of the Pilot's Handbook of Aeronautical Knowledge; for a theoretical overview, Aerodynamics for Navy Aviators (Required Thrust and Power, pg. 96; and principles of propulsion, pp. 104-112). The latter reference actually starts with your exact equation and develops it into an equation more familiar to the engineers who build jet engines. They do so by introducing η (eta).

Our Wikipedia article used to have this great picture, two airplanes tied to a tree; it's not a wonderfully sound explanation; but the metaphor is potent - the relationship between power expended by the engine, and thrust that actually develops, is subject to aerodynamic inefficiency. The aerodynamic effieciency-factor can range all the way down to 0% - so it's not something you can casually neglect!

This is exactly why engineers use the specific impulse to examine engine efficiency; and they use the drag coefficient and the lift-to-drag ratio to examine the aerodynamic efficiency. For large jet engines, there are many separate types of efficiency losses, and many distinct regimes of operation; so you can't summarize all of engine efficiency into a single parameter.

I might observe, for example, the recent trend toward high bypass turbofans. That trend seems to indicate that on the whole, it's more efficient to move more air at subsonic speeds, than to try to move a small jet of air at hypervelocity. But on more thorough consideration, this is not a generalizable rule; it's easy to find counter-examples. For example, one of the first maneuvers taught to aviators is slow flight - the regime of "reversed command," in which slowing down the aircraft requires adding engine power. More fuel is spent, and the aircraft goes slower! Aerodynamics - it's not intuitive!

Nimur (talk) 04:32, 3 October 2017 (UTC)[reply]

The OP might find this explanation to be helpful: Why Planes Don't Fly Faster. --Guy Macon (talk) 08:08, 3 October 2017 (UTC)[reply]

The formula

f=rv

is quite accurate for a rocket engine in a vacuum, but for a jet engine you have add a slightly lower mass flow of gas flowing in at a lower speed at the front end of your engine, in addition to the other difficulties of fluid dynamics which make finding analytical solutions impossible most of the time. I'd say that jet engines are much harder than rocket science. But qualitatively, it's quite easy to see that a larger flow at a lower speed gives a more efficient engine: you put less kinetic energy in the jet for the same momentum. The momentum of the jet is what gives you thrust. The high bypass ratios of modern jet engines are there to increase efficiency. PiusImpavidus (talk) 11:02, 3 October 2017 (UTC)[reply]

But, lower speed means flow that is less laminar, so less of the momentum you impart is in a useful direction. When you measure thermodynamic energy budgets for a jet engine, is it useful to consider the work output if the work is spent moving turbulence, instead of the vehicle? Nimur (talk) 13:49, 3 October 2017 (UTC)[reply]

The OP may find the Ramjet article interesting. ←Baseball Bugs ^{What's up, Doc?} carrots→ 13:54, 3 October 2017 (UTC)[reply]

combustion engine efficiency is a complicated combination of multiple processes and their own efficiencies. So its much more a task of balancing all these parameters while they have to work under a wide range of conditions. Just imagine the volume of water a jet engine has to handle in bad weather or the temperature range from the ground of a desert airport to the 30,000 feet traveling height. There are some general formulas used in engineering to approach the forces that are much more complicated then your 3 parameter formula but these are already very limited use. Your approach is frankly useless. --Kharon (talk) 19:01, 3 October 2017 (UTC)[reply]

How does the Portuguese man o' war eat

Our article has lots of info about how it stings and tons about the gas filled sail, but doesn't explain how it actually eats the things it stings. Presumably, they'd normally float away, so do the tentacles grab them and pull them to some kind of mouth or do they grab them and eating parts of the tentacles get to work? --84.19.47.96 (talk) 09:06, 3 October 2017 (UTC)[reply]

This National Geographic video shows the process. -- ToE 09:29, 3 October 2017 (UTC)[reply]

Complex roots by numerical methods

Which numerical method is best for finding the complex roots of a given algebraic equation? 14.139.241.85 (talk) 16:21, 3 October 2017 (UTC)[reply]

Most math software suites use lots of Parametric equation now i think. --Kharon (talk) 19:16, 3 October 2017 (UTC)[reply]

Root-finding algorithm#Finding roots of polynomials, and some of the articles it links to, may be helpful. Loraof (talk) 21:18, 3 October 2017 (UTC)[reply]

Refilling Glycogen Stores

According to various sources (although somewhat lacking in the Wikipedia article) the liver can store ~100 grams of glycogen and the muscles can store ~400 grams. As each gram of glycogen has 4 calories, this means in total a hypothetical persons body can store ~2,000 calories in glycogen.

Lets say that hypothetical person has a daily total caloric expenditure of 2,000 calories to maintain their weight (ignore water weight here since 1 gram of glycogen binds to 4 grams of water so water weight will fluctuate by 2 kilograms).

If they depleted their liver and muscle glycogen through exercise, a ketogenic diet, or starvation and then ingest 4,000 calories of carbohydrates in a single day, does that mean that 2,000 calories of the ingested carbohydrates will go directly to glycogen stores and the remaining 2,000 will be used by the bodies metabolism? Meaning in effect that despite consuming double their required daily intake, they do not put on any fat? 209.40.170.84 (talk) 16:42, 3 October 2017 (UTC)[reply]

I'm not sure if you will find any studies on that, but it seems unlikely that fat production would be 100% shut down. I could believe it would be reduced somewhat, yes, but entirely stopped, no. However, if stored fat is also turned into glycogen by some method, then effectively, this could be the case. StuRat (talk) 17:14, 3 October 2017 (UTC)[reply]

Anti-biotics questions.

1. Some anti-biotics advertise as killing gram-negative bacteria as good as gram-positive bacteria. Are there any anti-biotics that advertise as killing gram-negative better than gram-positive? Or should I say, what % of anti-biotics kills gram-positive better, what % both equally, and what % kill gram-negative better.

2. If something is both an anti-biotic and a anti-viral, what would it foremost be classified as? I just asked a chemistry professor who worked in the pharmaceutical industry, designing drugs, he say he never heard of a medicinal drug that can do both, so I guess I should instead ask, are there any drugs that are both anti-viral and anti-biotic at the same time? I note that Wikipedia's antiviral article says most anti-virals unlike anti-biotics, are used for killing/treating 1 virus at a time. But it wouldn't surprise me if some anti-viral are also have some anti-biotic. The anti-biotic article, says that "A limited number of antibiotics also possess antiprotozoal activity," so. Thanks. 12.239.13.143 (talk) 17:46, 3 October 2017 (UTC).[reply]

See list of antibiotics regarding to your first question and antimicrobial for your second. 208.90.213.186 (talk) 19:58, 3 October 2017 (UTC)[reply]

For your second question, it would be classified as both, if both are chief uses. I also am not aware of any drug that is both an antiviral and an antibiotic. It's possible in theory, but unlikely because antivirals and antibacterial drugs work via different mechanisms. In addition, there are practical constraints such as cost and side effects that limit the applicability of drugs. C0617470r (talk) 19:31, 4 October 2017 (UTC)[reply]

Generally majority of antibiotics kill gram-positive bacteria better or equal to gram-negative. Counter examples include, for example, first generation fluoroquinolones and second and third generation cephalosporins. Ruslik_Zero 08:56, 5 October 2017 (UTC)[reply]

Big science equipment - rolled on wooden trunks?

Does anybody know the story behind this picture? It shows a large magnet of the Mirror Fusion Test Facility in 1981, probably prepared for transportation on a road made out of logs. Was is really transported so low-tech? At those times SPMTs were already known.

Thanks for any help.— Preceding question restored by Spmt6 (talk • contribs) 18:03, 3 October 2017 (UTC)[reply]

Looks like somebody went over budget and had to make due. :-)

One thought is that it won't do much bouncing around, this way. If the equipment is extremely sensitive, and time is no object, and perhaps the money involved in a more high tech solution is an issue, this approach might make sense. StuRat (talk) 17:00, 2 October 2017 (UTC)[reply]

I've taken over this question, since I don't believe in any coincidence.--Spmt6 (talk) 18:03, 3 October 2017 (UTC)[reply]

Remember that this is Secret Alien Technology, after all: http://www.thelivingmoon.com/42stargate/04images/Machine/steigerphoto.jpg The Rainbow Conspiracy. ISBN 0786000651.

(and why wouldn't you move it on logs? Logs are great. Cheap, reliable, non-sparking, non-magnetic) Andy Dingley (talk) 19:19, 3 October 2017 (UTC)[reply]

It was a giant magnet, not a satellite or electron microscop. Industrial parts of such size are usually transported alike "naked". It was just an experiment. If you like you can read all about it here. --Kharon (talk) 19:42, 3 October 2017 (UTC)[reply]

I really can't see what advantage an SPMT has in this case. Entire ships are moved around on wooden rollers, not just little magnets. The logs use much more of the width of the roadway to support the weight than wheels would (although an air cushion/skirt system would be even better). Greglocock (talk) 19:54, 3 October 2017 (UTC)[reply]

What is the name for this method of transport? Log rolling refers to many thing, but apparently not this. -- ToE 15:00, 4 October 2017 (UTC)[reply]

You might think that we would have an article for this type of machine system, but our Roller (disambiguation) page suggests that the closest is roller bearing. Alansplodge (talk) 10:24, 5 October 2017 (UTC)[reply]

There's a brief mention in Bearing_(mechanical)#History; it uses the terms "rolling bearing" and "roller bearing". Alansplodge (talk) 10:28, 5 October 2017 (UTC)[reply]

Atmosphere and gravity

Just a thought experiment; I've got some conflicting ideas in my head, and I'm wondering which of them, if any, is correct.

Earth mass is approximately 5.9722 × 10²⁴ kg. The mass of the atmosphere of Earth is 5.15 × 10¹⁸ kg. If we lost our atmosphere suddenly, would the weight of a non-pressurised object indoors (as measured by scales) be affected? [Presumably the loss of pressure would cause problems for pressurised objects.] As we're indoors, wind wouldn't be a significant factor with atmosphere, and thus losing the atmosphere wouldn't affect the total mass that's acting on our scales. On one hand, the atmosphere's vaguely 0.0000863% of Earth's mass, so the reduction in mass being cubed in the gravitational constant would be absolutely tiny, and we'd have to have an extremely sensitive measuring device in order to notice any change. On the other hand, in the final section of Wikipedia:Reference desk/Archives/Science/2016 January 21 I learnt that the basic gravitational constant breaks down once you're inside the massive object. Obviously someone on Earth's surface is just barely "inside" the entire massive object, the atmosphere being only a thin skin on the regolith, but still I'm wondering if having mass above us has an effect on our weight.

So...what is it? Does the presence of the atmosphere have any effects on the net gravitational force being experienced by an object on the ground, or does it not? Nyttend (talk) 23:58, 3 October 2017 (UTC)[reply]

The atmosphere is essentially a spherical shell, and one of the astonishing facts about gravity that Newton proved is that a spherical shell exerts no gravitational force on anything inside it; see shell theorem. In other words, if you are beneath the atmosphere, the atmosphere's gravity does not affect you (except for small perturbations due to the fact that the atmosphere isn't really perfectly spherical). Looie496 (talk) 00:09, 4 October 2017 (UTC)[reply]

Balsa wood weighs about 160 kilograms per cubic meter. Air weighs about 1 kilo per cubic meter. Without the bouyancy of air that wood might now weigh 161 kilograms. There's likely less dense things that could survive both air and depressurization without the density changing (unlike say marshmallows). A giant hollow titanium sphere perhaps? Those would increase weight a higher percentage upon depressurization. Sagittarian Milky Way (talk) 00:34, 4 October 2017 (UTC)[reply]

Nothing "weighs X kilograms" because kilogram is a unit of mass, not of weight. A block of balsa will not gain mass, as per conservation of mass.

The effect of buoyancy on weight is not straightforward, as weight can mean apparent weight or true weight (mass times local gravitational acceleration) -- depends on which you mean. The article on weight discusses buoyancy. A helium balloon will float up in Earth's atmosphere, but not in a vacuum. Its mass and true weight remain constant, while its apparent weight changes. 91.155.192.188 (talk) 10:32, 4 October 2017 (UTC)[reply]

To be fair, the kilogram as a force unit is a real unit, just not part of the modern metric system (the SI). --69.159.60.147 (talk) 18:47, 4 October 2017 (UTC)[reply]

That's why I said (as measured by scales) when defining my meaning of "weight". Nyttend (talk) 21:23, 4 October 2017 (UTC)[reply]

Scales don't measure weight (as a spring balance does) they measure mass more directly. As scales work by balancing two masses, they're self-compensating for the effects of change in a local gravitational field (provided that there still is one). Unlike a spring balance or load cell, they don't have to be re-caibrated for lattitude or local gravimetric variations.

The pan weights do float in the air though, due to buoyancy (a tiny effect, but it's there). If you remove the atmosphere, this effect would disappear, for both the spring balance pans. As the balance weights are usually denser than the thing being weighed (they're solid and made of dense stuff), they were less buoyant in atmosphere and so the scales were previously reading very slightly heavy beforehand, which will no longer be the case without an atmosphere. Andy Dingley (talk) 22:24, 4 October 2017 (UTC)[reply]

The distinction between the two kinds of scales is perfectly correct, but it is unusual and confusing to use the name "scale" only for balance scales and not for spring scales. --69.159.60.147 (talk) 08:37, 5 October 2017 (UTC)[reply]

But who was using scales to refer to a spring balance or made a distinction between two kinds of scales? Andy Dingley clearly wasn't. Instead they said a similar thing to you namely that a spring balance is not a scale and so effectively made a distinction between a balance scale and a spring balance; although only barely as needed for their main point namely that since the OP referred to "scales" these would measure mass, not weight, and how this all affects the OP's original question. Nil Einne (talk) 13:31, 5 October 2017 (UTC)Sorry misread 69's comment. Nil Einne (talk) 13:33, 5 October 2017 (UTC)[reply]