Wikipedia:Reference desk/Archives/Science/2013 September 22

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September 22

NSaid and Ulcers

Why would the interaction between Alcohol & NSaid's substances may contribute to the development of Ulcers in Humans?, And, if it will be Capsaicinoids instead (together with the NSaid's), could that also initiate Ulcers in Humans?

Thanks for your replays. this could be very enlightening Ben-Natan (talk) 02:36, 22 September 2013 (UTC)

Ulcers are believed to be caused by Helicobacter pylori. Nsaids and drinking alcohol probably exacerbate them. μηδείς (talk) 02:44, 22 September 2013 (UTC)

What about Capsaicinoids? Ben-Natan (talk) 14:58, 22 September 2013 (UTC)

Our article says nsaids will also cause ulcers, but doesn't mention capsaicinoids. See Peptic_ulcer#Cause. μηδείς (talk) 17:31, 22 September 2013 (UTC)

Outhouses, sheds, and barns

Why are sheds smaller than outhouses? I would expect "outhouse" to be smallest, "shed" in the middle, and "barns" to be biggest. Nyttend (talk) 07:11, 22 September 2013 (UTC)

Given the irony of choosing "barn" as the term for such a tiny unit, they probably decided to go in reverse order. ←Baseball Bugs ^{What's up, Doc?} carrots→ 10:12, 22 September 2013 (UTC)

Looking at the disparity in sizes between the outhouse and the shed, my guess is that those labels were proposed by two different groups, with little regard for internal consistency. As the article mentioned, the two units are rarely used - likely the proposals for the units were primarily as an ad hoc or nonce joke, without any sort of consideration of how they would be used in long term practice. -- 67.40.209.200 (talk) 16:35, 23 September 2013 (UTC)

While the outhouse article describes it as a privy covering a pit toilet, I always thought is was more in the non-North American sense of a larger outbuilding. Compare a scottish outbuilding to a shed and the difference is clear. A barn is obviously larger then both. Astronaut (talk) 19:09, 23 September 2013 (UTC)

the effect of a working propeller on a plane

The propeller is known to produce thrust moving the plane forward. What side-effects does it have on the plane besides the main one ? In particular, a) does it tend to shift the plane off its main direction due to sense of which the blades turn ? b) is it responsible for raising the plane's tail while running on the runway, and if it isn't, what physical cause does raises it ? BentzyCo (talk) 15:01, 22 September 2013 (UTC)

The article propeller doesn't have much if anything to say about it, so either the editor didn't find anything or it's not a problem. As for what raises the tail, it's lift. Basically, once sufficient speed has been attained, the tail is going to rise. ←Baseball Bugs ^{What's up, Doc?} carrots→ 15:37, 22 September 2013 (UTC)

Besides thrust, the propellor in being turned by the engine imparts a twisting effect on the aeroplane. In high performance aircraft with single engines, this makes the plane easier/harder to turn in one dirrection than the other. It was aserious limitation of the Japanese Zero fighter. It can also make for some tricky handling on takeoff, as the plane tries to tip over, requiring the pilot to initially correct strongly for it, then taper off as speed builds up and the aircraft control surfaces begin to work more effectively. The WW2 Spitfire fighter was noted for this. The spinning propellor also acts a gyroscope, causing precession forces, making the plane try to climb or dive when turning to right or left. Pilots soon learn to correct for this. In twin engine planes, the props usually turn in opposite directions to cancel out these effects.

Because the spinning propellor casues a corkscrew motion of the air, power is wasted and as the air tries to goes past the fuselage and wings on an angle instead of parallel, drag is increased. Contra-rotating propellors eliminate these propblems.

The tail rising as the plane builds up speed down the runway is essentially a design feature or good pilot practice for tail dragger aircraft - so the pilot can see the runway ahead. Until you get enough speed to lift the tail, slewing the plane right and left alternately is sometimes practiced so you can see where you are going and avoid unexpected obstacles. The tail lift is achieved by a suitably designed tail aerofoil. In aircraft with a tricycle undercarriage, tail lift before takeoff is not done, as with the plane more or less level, the pilot can see the runway ahead anyway.

121.215.0.157 (talk) 15:43, 22 September 2013 (UTC)

"In twin engine planes, the props usually turn in opposite directions to cancel out these effects." Nope - only true for a small minority of twin-engined propeller-driven aircraft. AndyTheGrump (talk) 16:41, 22 September 2013 (UTC)

And cost makes it obvious why. In off-axis, twin-engines, rotation direction also defines which engine determines the minimum controllable airspeed and designates a critical engine (losing one engine is worse than losing the other). P-Factor and yaw combine for one side and counteract on the other side. Because of this, even twins with counterrotating propellers put the clockwise-rotating engine on the right. Symmetry still has a preference. --DHeyward (talk) 18:57, 22 September 2013 (UTC)

Well, you haven't given an data to backup your claim on cost. Cost is not an obvious factor to me at any rate. I have been involved in selling marine engines >800 kW. In boats with twin engines, the two engines and props turn in opposite directions to improve handling. The only difference between a CW-rotating 4-stroke engine and a CCW-rotating one is: 1) the camshaft, 2) the oil pump, 3) the coolant pump, and 4) the starter motor. In some engines, the arrangement of the oil filters is different so they are easily accessable on the opposite rotating engine in cramped engine rooms. As the manufacturers mass-produce many 1000's of engines in both configurations, the price of an opposite rotation engine is little different and is often the same. 120.145.70.130 (talk) 01:35, 23 September 2013 (UTC)

An interesting example of a twin-engined prop-driven aircraft is the Lockheed P-38 Lightning

Lockheed P-38 Lightning

The engines/propellers are counter-rotating, but the 'wrong' way. With the downgoing propeller blades on the outside, the P-factor makes both engines 'critical'. Apparently it was done like this as it made the aircraft a more stable gun platform. AndyTheGrump (talk) 01:53, 23 September 2013 (UTC)

I think you guys have misunderstood P-factor. P-Factor is the shift of the centre of thrust up or down according to the aircrafts' angle of attack wrt the airspeed. If (say) the aircraft is being rotated into a climb, the prop blades are cutting air slightly slower at the top of the rotation than they are at teh bottom of rotation - so the centre of thrust moves upward away from the prop centre line, increasing the climb. This effect is the same regardless of which engine is the one rotating clockwise. However, which way round the engines are installed still affects handling and performance, as one way the props pull air away from the fueselage and toward the wing tips above the wing, and the other way around they'll do it below the wing, amking for diffrent lift, drag, and climb/dive tendency. If the fuselage, wings, tail, and engines are completely symmetrical, there cannot be any turning or yaw effect, only an up/down or pitching effect. 120.145.70.130 (talk) 02:45, 23 September 2013 (UTC)

That is nonsense. See the FAA handbook linked below: "P-FACTOR — A tendency for an aircraft to yaw to the left due to the descending propeller blade on the right producing more thrust than the ascending blade on the left. This occurs when the aircraft’s longitudinal axis is in a climbing attitude in relation to the relative wind. The P-factor would be to the right if the aircraft had a counterclockwise rotating propeller". AndyTheGrump (talk) 03:22, 23 September 2013 (UTC)

Please explain then why an aircraft, symmetrical about its longitudinal centre, and having oppositely rotating engines symmetrically constructed and installed on the wings, can have an assymetrical effect on flying. If the right engine somehow causes a turning force to the right, then the left engine must cause the same turning force to the left, and the two will cancel. What doesn't cancel are the forces producing a climbing or diving effect. The tendency to yaw left or right occurs in single engine aircraft, or twin engine aircraft with both props rotating in the same direction. 120.145.70.130 (talk) 04:24, 23 September 2013 (UTC)

Obviously, if everything is symmetrical, there is no tendency to yaw - but that doesn't make your definition of p-factor right. P-factor is the movement of the centre of thrust relative to the centre of a propeller as a result of it meeting the incoming airflow at an angle. With two propellers rotating in opposite directions, the displacement of forces cancel out - but it is still there, and it has nothing whatsoever to do with propellers supposedly "cutting air slightly slower at the top of the rotation than they are at the bottom of rotation", which is what you were claiming. AndyTheGrump (talk) 04:46, 23 September 2013 (UTC)

Glad you agree it cancels - not what you implied before. Seeing that we agree that a symmetrical counter-rotating arrangment cancels P-forces out as far as uncommanded yaw is concerned, it would seem that your claim above that making the counter rotating props the "wrong" way round makes both engines critical is incorrect. Both engines will be critical in any case. Only with both props rotating the same (either normal or counter-normal) will one engine be critical. And that is as much, if not more, due to assymetric airflow arising from corkscrewing air than due to centre of thrust shift. Now, how would the prop blades (more correctly the propellor rotation plane) be meeting the airflow at an angle? Three possibilities: They have been installed wrong (not likelly), the pilot has not trimmed correctly (he won't last long then - too much fuel usage), or the pilot is making a yaw or pitch movement. 120.145.70.130 (talk) 05:45, 23 September 2013 (UTC)

This is a science reference desk. It is not a platform for uninformed speculation. A source has been provided which explains what aircraft designers mean by 'P-factor'. Your apparent inability to understand it does not give you license to redefine it as something else. And yes, when I wrote that both engines on the P-38 were 'critical', I was correct - in both cases, a single engine failure will result in P-factor shifting the centre of thrust outboard of the centre of the remaining engine at low speed/high angles of attack - which is what 'critical' means in this context. I suggest that before responding again, you actually read the reference material provided - or even read our article on P-factor, which states the same thing. AndyTheGrump (talk) 12:13, 23 September 2013 (UTC)

I haven't redefined anything. I have said more than once that P-factor/forces arise from the propellor centre of thrust moving away from the rotation centre. That's exactly what it is about. It can move sideways and, it seems unrealised by you Andy, that it can/will move up or down as well, if the aircraft is pitched. About the P-38, you wrote The engines/propellors are counter-rotating, but the 'wrong way'. With the downgoing propellor blades on the outside, the P-factor makes both engines critical. That's nonsense. It matters not whether the outside blades are going up or down, both engines are critical in either case, due to aircraft & engine symmetry. Perhaps you need to read the sources, and your own writings, before making incourteous remarks. Instead, if you think a posting is wrong, try and explain logically why it's wrong, rather than attack the person. In doing so you might find that you are wrong. If not, great, you've taught us something. 120.145.70.130 (talk) 16:05, 23 September 2013 (UTC)

As the federal aviation authority makes entirely clear, airflow encountering a propeller at an angle results in the centre of thrust of a propeller move away from the centre of the propeller. More specifically, with a clockwise rotating propeller (as seen from behind), and the airflow angled upwards from below (as would be typical in slow flight, due to the necessary high angle of attack of the aircraft), the centre of pressure will move to the left, as the downward-going propeller blades on the left hand side encounter the incoming airflow at a greater angle of attack than the upward-going blades on the right hand side. This offsetting of the centre of thrust is known as P-factor. If you want to argue that the FAA is wrong, take it up with them. AndyTheGrump (talk) 18:14, 23 September 2013 (UTC)

P-factor! It's that difficult-to-explain, easy-to-feel tendency to swing (yaw) the airplane to the left, and it has been the death of many a novice pilot who jumped into a more powerful aircraft without adequate training! As expressed above, there are many other non-ideal effects of the propeller: slipstream, prop wash, drag, and so on. The Airplane Flying Handbook goes over all of these details, and the full text is available online from the FAA. Nimur (talk) 16:33, 22 September 2013 (UTC)

We have a very slim article called Torque effect which describes the issue in a single sentence. A better article is at Understanding Propeller Torque and P-Factor on FlightGear wiki - there are some links at the bottom of the page. Alansplodge (talk) 16:48, 22 September 2013 (UTC)

Another effect in single engine planes is the air movement over the tail can effect rudder and elevator/aileron which can change the feeling for T-tail vs. cruciform aircraft. Specifically airspeed of the airplance is more important for T-Tail while propeller rotation can generate more airflow for cruciform tails. Balked landings are where this difference is felt most (full power, landing speed at first). --DHeyward (talk) 18:52, 22 September 2013 (UTC)

Event horizon

I would have two questions about black holes: a photon that on event horizon is radially directed outward black hole would remain stuck at horizon or would fall to singularity (and if yes, in how much time?)? And, hypothetically, if I could surpass c (the speed of light) into a black hole I could escape or I howewer would fall to singularity because there aren't ways to escape? Thanks for answering, 80.181.57.86 (talk) 22:15, 22 September 2013 (UTC)

Sounds similar to above a few days ago. Was this a homework question? --DHeyward (talk) 00:27, 23 September 2013 (UTC)

No, this isn't a homework. It's just a curiosity. 95.247.219.235 (talk) 11:13, 23 September 2013 (UTC)

Yes the photon would remain stuck at the horizon, and if you could hypothetically move faster than light than you would escape the B-hole. Dauto (talk) 18:04, 23 September 2013 (UTC)

But don't in a black hole all direction take to singularity?95.233.90.55 (talk) 00:26, 24 September 2013 (UTC)

Although objects can't move faster than c, an abstract series of events can. For example, if while standing on the Earth, you quickly twist a laser such that the center of the laser beam crosses the moon, the point at which the center of the laser beam crosses the moon will likely move faster than the speed of light. When a series of events moves faster than c like that, they form a spacelike curve, rather than a timelike curve like the world line of an object must be. So I interpret your question about moving faster than c in and out of a black hole as being something like "can spacelike curves pass from outside to inside of an event horizon and then back out again?". The answer to that question is yes. It's generally true that for an object inside the event horizon, all directions point toward the singularity, but that statement is only intended to apply to events within the object's future light cone. A spacelike curve that goes through the object isn't constrained to lie within the object's future light cone, so the same statement isn't applicable to the spacelike curve.

Even straight spacelike curves, i.e. spacelike geodesics, can pass from outside to inside of an event horizon and then back out again. It's hard to find much discussion about the spacelike geodesics in the Schwarzchild geometry, because timelike and null geodesics are generally of much more interest, since they describe the world lines of massive and massless particles, respectively. But the analysis that leads to equations 7.47 and 7.48 in this paper basically applies equally well to spacelike geodesics as to timelike or null geodesics. The only differences are that you use ε=-1 instead of ε=1 or ε=0, and λ is a distance instead of a proper time. With spacelike geodesics, E and L don't have units of energy and angular momentum, but they are still conserved quantities for the geodesic. If you plot V(r) vs. r with, say, G=M=L=1, you'll see that for a range of r from a bit less than 2 to r=∞, V(r) is monotonic decreasing, so as per the paper's analysis, for some values of E r will start off greater than 2, dip below 2 (which is the r for the event horizon), and then go back above 2 again. Red Act (talk) 07:49, 24 September 2013 (UTC)

(The following was moved here from Red Act's talk page.) Excuse me, I am the user that made the question on the possibility to get out from black holes at superluminal speed. So, if I understood, an hypothetical superluminal object could escape event horizon on some circumstances? I made to you this question because I read on a paper that even if could be surpassed c, wouldn't be either possible to escape for the spece curved on itself. Is this wrong? Please answer to me on the post in the reference desk because I have a dynamic IP. Thanks for your interest. 95.233.90.55 (talk) 15:22, 24 September 2013 (UTC)

Well, if I'm wrong, it wouldn't be the first time I've been wrong here on the reference desk. But what I said at least appears to me to be correct, when applying the analysis in the paper I cited to spacelike geodesics in the Schwarzchild geometry. If you've read a contradictory paper, please cite the paper here so that I can take a look at it. If it looks to me like I've said something incorrect again, then I'll retract my statement. Red Act (talk) 16:18, 24 September 2013 (UTC)

I didn't want to criticize what you said, when I wrote "is this wrong?" I was referring to the paper, not to you. I simply wanted to understand the concept. 95.233.90.55 (talk) 16:54, 24 September 2013 (UTC)

I really wasn't offended in the least. I'm a lot more concerned about getting to the truth that I am about whether I stumbled on the journey towards getting there. If you've found a paper that contradicts what I've said, I'd be very interested to read it, so that I can improve my own understanding of black holes. Red Act (talk) 18:03, 24 September 2013 (UTC)

It's still not looking correct to me that within a black hole's event horizon, all directions lead to the singularity, unless you're limiting your discussion to events within a given event's future light cone. It's easy to come up with an example that shows that there exist spacelike curves such that both ends of the curve are outside of a Schwarzschild black hole's event horizon, but the middle of the curve dips inside the horizon:

I'll use ingoing Eddington–Finkelstein coordinates to avoid any issues with coordinate singularities, and for convenience take $c=G=M=1$. Consider the curve defined by $\theta =\pi /2$, $v=0$ and $r=1+2\phi ^{2}$, for $-1\leq \phi \leq 1$. The event horizon is at $r=2$, so $r$ is outside the event horizon at the curve's endpoints when $\phi =\pm 1$, but is inside the event horizon at the curve's middle, when $\phi =0$. The metric says that the interval of a tangent to the curve is simply $ds^{2}=(1+2\phi ^{2})^{2}d\phi ^{2}$. Clearly $ds^{2}>0$ for any $\phi$, so the curve is spacelike everywhere. Red Act (talk) 02:57, 27 September 2013 (UTC)

aren't all heaters perfect?

Isn't a hair drier just as perfect a source of heating as anything else that uses the same number of watts? I came across this answer,

http://answers.yahoo.com/question/index?qid=20110131212343AA1MZBy

But find this response insanely bizarre: "I think its in-effiecient. Hair dryer using filament as its heater. It converts elecrical energy to internal energy, thus transfered as heat. This process generates high entropy. So using heater with refrigeration system is more efficient. "

What can that possibly mean? It sounds completely crank. Other than possibly 'melting' from overuse, i.e. in the first few minutes isn't a hair drier just as perfect in an enclosed small room as anything else (including a computer or anything else) that uses the same number of watts? Thanks. 178.48.114.143 (talk) 22:37, 22 September 2013 (UTC) — Preceding unsigned comment added by 178.48.114.143 (talk)

If you concentrate on what proportion of the energy registered on a premises electric meter is changed into heat at the desired location by a hair drier, then the hair drier is very efficient. But an electric heat pump can use energy to move heat from outdoors to indoors, and the total amount of heat put into the desired location is greater than the energy used to operate the heat pump. Jc3s5h (talk) 23:59, 22 September 2013 (UTC)

True, but of the many direct non heatpump forms of electrical heating, they are all practically 100% efficient at turning electricity into heat. That's not to say they are all equally good at a particular task.Greglocock (talk) 00:28, 23 September 2013 (UTC)

Your body heating the hair naturally while evaporation cools it vs. trying to raise the temperature of your hair beyond body temperature to improve the rate. That creates sink points for the heat the hair dryer generates: the water in the hair and the body as well as loss in transfer to the water. Energy from your body is more efficient at transfer to the water in your hair as a net change in energy from having wet hair vs. not. How much more energy (i.e. calories) do you have to eat to dry your naturally? Since the body is inefficient to begin with basically the only energy difference is going directly to supporting evaporation. Time to dry is dependent on difference between body temperature, water mass and dew point. Everything increases entropy bet some are more efficient in net entropy increase vs. useful work done. You can do the calculation for energy to naturally dry hair estimating water volume, temperature difference between body and room temperature and dew point of ambient. Then calculate the amount of time your dryer is on to dry your hair. --DHeyward (talk) 01:36, 23 September 2013 (UTC)

I don't understand DHeyward's point at all. Nor do I understand the comparison with heat pump. My real question is simple: if there's no electric radiator in a small room, then isn't a hairdrier exactly strictly equivalent to an electric radiator drawing the same number of watts? (i.e. the ancillary action of blowing air around doesn't matter at all whatsoever). 178.48.114.143 (talk) 01:56, 23 September 2013 (UTC)

DHeywood's explanation is affected by poor english - and I don't see how it is relavent to your question either. However the comment about heat pumps is valid. If you you an electric bar radiator, electric hairdryer (which contains a heating element), using light glodes to heat, is all teh same - for every kilowatt of electric power consumed, exactly one kilowatt of heat is imparted to the inside of a closed room. However, heat pumps (reverse cycling airconditioners) work a different way. The electricity is used to run a mechanical process that moves heat energy from the outside of the room to the inside. Typically in domestic airconditioning, only about 400 watts of electric power is needed to move a kilowat of heat from outside the room to the inside. Thus in that sense a heat pump is more energy efficient than a hair dryer, because the heat energy does not come for the electricity supply, it comes from the air outside the room. 120.145.70.130 (talk) 02:30, 23 September 2013 (UTC)

I still don't get it - outside, it's even colder. What is there to 'pump'? The whole point of this question is that it's frikkin cold out. 178.48.114.143 (talk) 03:30, 23 September 2013 (UTC)

It only seems cold to humans, as they are accustomed to keeping themselves at a certain temperature. There's actually not much less heat outside than inside. Absolute zero temperature, 0 Kelvin or -273.15 K (or 0 Rankin or -459.67 F if you are non metric) is the notional temperature at which there is no heat residing in any substance. You may feel cold at 0 C (32 F), but at this temperature, the difference between feeling darn cold and nice and warm is the difference between 273 K and 300 K - not a lot in real terms. The heat in air at 0 C is not zero, it is the heat you have at 273 celcius degrees above absolute zero. 120.145.70.130 (talk) 04:13, 23 September 2013 (UTC)

Look up heat pump and coefficient of performance. IBE (talk) 04:21, 23 September 2013 (UTC)

I haven't done so but I suspect that is cruel advice. A heat pump is a device that can extract the heat energy from one place and pump it into another place, thereby heating the second place. It can do this even if the first place is cooler than the second. This is exactly how a fridge works. You could use it to heat your house using less electricity, that's what a reverse cycle air conditioner does.Greglocock (talk) 06:12, 23 September 2013 (UTC)

What? Judged as an electric heater, a fridge uses less electricity than a normal electric heater?? (To heat a room)? This is amazing news, and 2) in this case why aren't electric radiators mini fridges instead... 178.48.114.143 (talk) 07:41, 23 September 2013 (UTC)

GregLocock was just trying to get you to look at heat in the proper way - as something that can be moved from one place to another, irrespective of the temperatures. He's 100% correct in what he said, but using a fridge in this way is not practical, unless you regard continually loading it with warm things as convenient. Once the frige has moved the heat out of the food and into the room, there is no more it can do until you reload it. 120.145.70.130 (talk) 11:43, 23 September 2013 (UTC)

Viewed as a simple, idealized, thermodynamics problem, the room is (in effect) a system into which electricity flows and heat leaks back out. As you add electricity into the room, the energy content of the room will rise until the energy entering as electricity equals the energy leaving as heat leaving the room. That description is true no matter how you use the electricity. However, energy can be stored inside the room in ways other than as heat. So, for example, you might have a bizarre machine in the room that lifts heavy weights from the floor and hangs them from the ceiling. It would consume electricity and convert it into gravitational potential energy without increasing the temperature of the room. So it's not true that all devices are equally efficient at producing heat from electricity.

The hair dryer starts air moving - giving it kinetic energy without increasing the temperature in the room. But it's likely that over time, those air currents will break up into eddies and eventually friction will convert all of that kinetic energy into heat. So EVENTUALLY, the hair dryer will have converted 100% of the electricity it consumes into heat and could therefore be considered a 100% efficient heater. In general, that will be true of most consumers of electricity...but not all of them. Consider a battery charger. It takes in electricity and uses some of it to effect a chemical change in the battery. Unless you go on to discharge the battery somewhere inside the room, it isn't a 100% efficient heater.

Of course most of these kinds of thing have a finite limit to the amount of energy they can "hide" in the form of chemical change, kinetic, gravitational, etc. If you leave the battery charger plugged in after the battery is 100% charged, then it'll either reduce the amount of electricity it consumes to zero - or it'll start to get warm...and at that point, it too is an 100% efficient heater.

A refrigerator also moves energy within the room from one form to another - and for a brief period between when you plug it in and when the inside is as cold as it can be, the refrigerator will appear to generate more heat than the electricity you pump into it...but once the insides are at the preset temperature, it'll return to being merely 100% efficient.

However, there is a "real-world gotcha". Sound and electromagnetic radiation can leave the room as well as heat. So if your hairdrier makes a noise (which I'm sure it must because it's moving air around) - then some of that sound will leave the room (eg by vibrating the windows) - and that vibration won't turn into heat until AFTER it leaves the room - and since the sound took electrical energy to make - that loss of sound energy will mean that not quite ALL of the electrical energy will turn to heat. Similarly, it's likely that the motor and any electronics inside will produce some small amount of radio frequency emissions - some of which will leak out of the room before turning to heat. Light from the red-hot heating element will somewhat light up the room - and some of that light will escape through windows and under the doors...so again, it's not 100% efficient.

To be sure, the losses incurred by the hair dryer because of those things will be very tiny - but it's enough to ensure that no machine is every going to be exactly 100% efficient - although most will be very close.

The bottom line is that you can't make a perpetual motion machine this way - the energy sum has to stay the same within the room - and no practical machine will be a 100% efficient heater. That said, almost every machine you have is very close to being 100% efficient - and there is probably very little to choose between a space heater, a refrigerator and a TV set at converting electricity into heat. SteveBaker (talk) 14:02, 23 September 2013 (UTC)

Air conditioning units are routinely reversed to warm homes during winter with typical coefficient of performance above 500%, so yes, the good electric heaters are mini fridges. Electric space heaters are designed to be cheap and safe to use, but they are never gonna be a cost effective way to warm a home. Dauto (talk) 14:20, 23 September 2013 (UTC)

Yes, heat pumps (reverse air-conditioners) are really popular here in Texas. They are able to either cool the house in summer or heat it in winter. When in "heat" mode, they are effectively air-conditioning the outside world and dumping the waste heat into the house. Since that waste heat contains both heat extracted from the outside and the waste from it's own internal operation, they are more than 100% efficient in terms of the heat they deliver to the house. Of course overall, they are also chilling the already cold outside world - so overall, they don't produce more heat than they generate. I'm surprised to hear that they can be as much as 500% efficient - but they are certainly very good in situations where you need an airconditioner for summer use anyway. SteveBaker (talk) 14:44, 23 September 2013 (UTC)

The 500% coefficient of performance can only be achieved for mild winter conditions. For instance, here where I live (Central Florida) the low temperature is rarely below freezing. The colder the outdoors, the smaller the coefficient of performance will be. A 300% coefficient is more typical for most of the south unless you use ground source heat pump because the underground water remains quite warm even during bitter cold nights. Further north the coefficient becomes too small and heat pumps aren't a wise choice any more. You will do better if you burn fuel locally (Natural gas or heating oil) instead of burning it at a power plant with a typical efficiency of about 40% and than using that electricity to warm the home. Dauto (talk) 15:10, 23 September 2013 (UTC)

Agreed, but another factor to take into consideration is whether some of the electricity can be generated on site e.g. with photovoltaics. That could swing the choice towards the heat pump. Itsmejudith (talk) 15:57, 24 September 2013 (UTC)

Photovoltaics, using the sun to generate electricity, run at about 16% to 20% efficiency. Given that domestic heat pumps require about 400 W for every 1 kW of heat shifted, you then have an overall conversion of insolation to heat in the room of around 45%. But if you install a glass roof and let the sun shine in (so it works like a solr hot water system), you'll have 90% delivery or better. And don't forget that photovoltaics are expensive. That's because of the large energy cost in making them. 1.122.206.159 (talk) 02:02, 25 September 2013 (UTC)

Better yet, you could run a generator at home using fossil fuels, dump the waste heat in the home and use the produced power to run a heat pump and bring even more heat into the home. Dauto (talk) 18:13, 24 September 2013 (UTC)

Agree with IP 1.222. You don't need a glass roof. Heat builds up under roof-installed PVs, making them less efficient. Since in the proposed case you have a ventilation system anyway, you extract air from the roof space behind the PVs, passing it through the heat exchanger that you already have in the system for extracted air. Win-win. You read it here first. Itsmejudith (talk) 23:33, 26 September 2013 (UTC)

Sorry if I wasn't clear. There are three distinct cases: the "towel" case where the room and towel try to reach equilibrium based on temperature and evaporation, the natural drying "hair" case where the human body provides heat, and the "hair dryer " case where the sources of heat are the human body and the hair dryer. When your hair is wet, there is the ambient room temperature and humidity. Without, the hair dryer, the heat source is the person. The other heat source is the ambient air but without your body, it is just a diffusive equilibrium problem. Your hair will dry like a towel in that case and that is be base rate of drying. Now, if you add in your body (i.e. your head) heating (model your hair as a thermally conducing sphere with a constant 98.6F heat source in the interior of the sphere. The difference in temperature between the interior of the sphere and the room will increase the rate of drying over the base rate of the towel naturally drying. The rate depends on temperature difference and humidity. Your body will burn calories to do this as your hair is cooling your body through evaporation and your body is maintaining a constant temperature. You can calculate how much energy is required to achieve this increased rate (Joules per Kilogram). Now, apply an external heat source (hair dryer) that raises the temperature of your hair above your body temperature. This increases the rate of drying but the heat retained in the hair is inefficient and wasted. The only work you want to do is evaporate water in the hair. I think you will find that 1) the body is an improvement in evaporation over just the static equilibrium towel but the calories required to do this is a fraction above the normal burn rate and is pretty efficient in converting the additional calories needed to heat the water to evaporate faster. 2) the hair dryer converts energy to heat the room as well as the hair. Heating the room only helps if the room air can hold more water but the volume of the room scatters the heat so that it is a sink for heat. Consider of you were outside. The hair dryer would have a negligible affect on the air temperature. All the heat lost to the air is wasted. All the heat that is transferred to already dry hair is wasted. Only heat transferred to the water is doing useful work. That is overall a very inefficient over just the body heating the hair. --DHeyward (talk) 03:16, 27 September 2013 (UTC)

The heat pump efficiency to dry hair would similiar to the body model in that you maintain temperature from the hot side and the ambient room. You can replace the heat temperature from the heat pump with the human body as the heat source for calculation. The point is that natural hair drying is different than just the equilibrium problem of a wet towel. The towel is cooled below room temperature through evaporation and the heat transfer continues until the towel is dry. Natural hair drying is raising the temperature of the hair over the temperature of the towel through body heat. Hair dryers don't transfer all their heat to the water the way the body does (in terms of a delta in calories required). I believe the heat pump is similar to the mechanical energy required to raise and lower a counterbalanced garage door. It's the same as air conditioners. The high pressure compression side raises the temperature while the expander lowers it. The ambient air and humidity next to each of these sides determines the amount of cooling/heating can be done while the energy remains constant. In a heat pump, the outside component is cooled below the external air temperature. That temperature difference causes energy to transfer from "warmer" outside air to the cooler "bulb". As you can imagine, this isn't particularly efficient when it's too humid or too cold outside. you want the dew point to be below the heat pump's cold side and there is also a lower limit on that temperature that the fluid can support. If the outside temperature is below the refrigeration cycle temperature, or if condensation or ice forms, it becomes much more inefficient. Whence, the use is usually restricted to relatively mild and dry winter climates. --DHeyward (talk) 03:16, 27 September 2013 (UTC)