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May 30

Is my dwarf Aleppo pine tree dying?

A dwarf Aleppo tree in Humboldt Park, Richmond, CA

How much should I water it? I try to keep soil moist. Is it growing or drying light green to yellow to brown? What can I do to feed it?Harmonywriter (talk) 01:35, 30 May 2013 (UTC)[reply]

It's hard to tell from the picture, but that soil doesn't look very moist to me. Looie496 (talk) 01:45, 30 May 2013 (UTC)[reply]

Looks healthy enough for the moment, and it probably doesn't need much/any fertilizer until next season. Even if you soak it pretty well in the evening, the surface soil will dry out during the day, but that's ok. A 1-2" layer of compost or mulch would help keep weeds down, retain moisture, and add to soil fertility. If you really want to pamper it, you could buy a packet of mycortree (or similar mycorrhiza product, available at your local garden center), and inject some down into the rooting zone. This assumes it was planted in the last year or so. Otherwise just be patient. My main concern would be how close it is planted to the foundation, and how crowded it might be in 5-10 years. You could probably safely move it out a few feet in the fall. SemanticMantis (talk) 02:07, 30 May 2013 (UTC)[reply]

As for watering, a very slow trickle for 30 min is much better than a full hosing for a few minutes. The former encourages deeper rooting, and more even soil moisture. Depending on your precipitation and drainage, once every 5 days or so should be enough, but the way to check is to stick your finger or a stick (I often use a chopstick) down 3-5 inches into the soil. If it comes up powdery dry, you can safely water. If you sense any moisture at that depth, the tree is fine, and adding water could hurt it. SemanticMantis (talk)

If this plant's in a humid area you might want to prune off the lowermost branches closest to the soil to encourage circulation and prevent rot as it becomes a bigger plant. The opposite if you are in a dry climate. Check how much acidity it likes and then fertilize at half that amount during a cooler, well watered period. The plant looks good. Avoid the temptation to prune all except the bottom branches. If you cut the top it will get bushy, and lose the typical pine growth form. μηδείς (talk) 05:59, 30 May 2013 (UTC)[reply]

Any body can help me with ansys??

Hi i am an engineer.i am trying to simulate heat loss from an oscillating fin on ansys.Can any body help me with it?? Oscillation in a fin would increase the rate of heat dissipation.but i want to simulate it first in ansys workbench.Any help or atleast any tutorial relating to it. thanks in advance Sameerdubey.sbp (talk) —Preceding undated comment added 03:16, 30 May 2013 (UTC)[reply]

Window or Incandescent light better for illumination versus heat

On an order of magnitude scale, how could I judge whether opening widows or using an incandescent light bulb would give a better performance providing light but not heat?

I have a room with six meters of south facing windows which provide sufficient light to illuminate a room. I keep the windows blinded with a surface facing outside and a navy surface facing in. Without the shades the sun produces considerable heat. I also have a 60 watt bulb I can dim to half strength which provides better (closer) illumination for my needs, but which also give off significant heat. I usually keep that room shuttered in hot weather and use the lamp, dimmed, when I need to. Further, one of the windows gets direct sun, the other reflected light from the white building next door. What do I need to take into account deciding which way will heat least with sufficient light? Thanks. μηδείς (talk) 05:50, 30 May 2013 (UTC)[reply]

The heat emitted by a 60 W bulb is just a bit short of 60 W, and as the light will be converted to heat on reaching any surface, you can take a 60 W bulb as a 60 W heater. In almost all rooms this is just about negligible. It will certainly be negligible in a room large enough to have 6 m of window on one side. One human working can be equivalent to 300 W. If he's using a PC, that's another 200 to 300 W. A typical electric bar heater is 1000 W or more. For direct sun, you can estimate the impact starting with the sea level solar intensity (termed insolation) which is quite close to 1 kW per square metre. So the heat coming in thru an unblinded window is the area in square metres times the cosine of the sun's angle wrt the window times 1000 W. There is some attenuation in the glass, but clearly the incoming heat under direct sun far outweighs a light globe. However, the window also allows heat from the room to escape. The heat escaping can be approximated by considering the room+window as a cavity radiator, exchanging heat with other structures outside, the ground, and the sky - in proportion to the temperature difference between the room internbally and the temperatures outside. This cavity radiator heat exchange via the window usually doesn't amount to much. As far as reflected energy from a nearby white building, this can be estimated by the building's reflectance (~10% typically) and if the building surface is not mirror-like, assuming it re-radiates in all directions equally, and the window intercepts a fraction of this as a function of distance and window area. Again, typically it will be very low compared to direct radiation from the sun but might well be higher or lower than operating the light globe - not that it matters, because as I said, 60 W as a room heater is negligible. Since you have not given surface finishes, distances, size of room, area of window, not much more can be said. Wickwack 121.215.47.76 (talk) 07:11, 30 May 2013 (UTC)[reply]

The heat escaping will be negligible, since the temperature inside will be 70-80 while outside will be 82-92. μηδείς (talk) 07:40, 30 May 2013 (UTC)[reply]

The analysis of Wickwack seems very reasonable to me. What surprises me is the premise. If you prefer a 60 W light bulb (dimmed) to real sunlight, then you must be a close relative of Count Dracula... ;-). --Stephan Schulz (talk) 07:53, 30 May 2013 (UTC)[reply]

Or, he's (she's?) watching certain types of movie. Wickwack 120.145.83.233 (talk) 08:02, 30 May 2013 (UTC)[reply]

Can the heat emanating from a PC be estimated from the power of the PSU? I would expect that during normal operation the PSU would not draw it's full wattage, but that during strenuous activity it would provide heat roughly equivalent to the PSU capacity? Or does a PSU always function at full power? Maybe that should be its own question, but your answer made me wonder (especially as ~300W would be good enough for most modest PCs) 64.201.173.145 (talk) 13:20, 30 May 2013 (UTC)[reply]

This is more complex than you may expect. In giving the upper figure above I have given a value assuming the PC is an up to date high performance PC such as might be used at home by a serious gamer, or for advanced technical/engineering use, a self powered commercial grade printer, and a high resolution monitor, both of which are not powered by the PC PSU. However, you can be seriously misled by looking at the PSU nameplate rated input. In accordance with standard American practice, the PSU wattage as marked on it is the wattage for electrical wiring and fuse sizing purposes, and will include a safety margin and be greater than what the PSU will actually draw even if fully loaded by internal PC devices. A good PC will have a reserve of PSU capability anyway, so that crashing or unreliable operation due to PSU issues is unlikely, even if the owner fits newer devices. So you can have a PC with a PSU marked "450 Watt", say, but if you measured the draw with a wattmetter, its real input (if it was chosen appropriately for the PC configuration) is likely to be around 200 to 250 watts. In theory, standard American practice need not, and should not, apply in other countries. However, the American market is the largest market. Moreover, and for countries with inspection and insurance industries that have the capability and will to enforce standards, the American market is the largest by considerable margin. So manaufacturers mark in accordance with USA requirements regardless of export destination. Wickwack 124.178.49.220 (talk) 01:00, 31 May 2013 (UTC)[reply]

One factor not considered is where in the room the heat is generated. Presumably with a dim bulb you will have it close to you, so it will generate heat right next to you. The sunlight might shine directly on you, in which case it will heat you more, or it might shine elsewhere in the room, which will not warm you as much (watch a cat moving to follow the sunlight spot on the floor sometime, they're smarter than they seem). Now, the heat will eventually spread to the entire room, but how hot you feel is determined by if the light shines on you or not.

Of course, I can't skip the obvious suggestion, that you replace the incandescent bulb with a CFL, which generates far less heat (less than 1/4th) per amount of light. Presumably you haven't done this because it's on a dimmer switch, but you can just always use it on full power (they do make dimmable CFLs, but they are quite expensive). If 60 watts is really too darned much light, you can always get a 40 watt equivalent bulb. StuRat (talk) 08:33, 30 May 2013 (UTC)[reply]

Plus also you can put a thermal reflective film on he glass of the window or install new glass that does the same job. There's also special blinds which are very good at reflecting heat. Painting the room white will make the most of any light if you're desperate to use an incandescent bulb rather than CFL. Dmcq (talk) 09:03, 30 May 2013 (UTC)[reply]

To explain further, I have a poorly wired apartment in NYC whose AC runs constantly when it's over 80 Fahrenheit. I use my computer for browsing, dvd/blu rays, and streaming. When doing so I like just enough illumination to see where things are. The heat from the computer is a given and can be ignored. The lamp I use has actually a three-way bulb the current power of which I can't read, but it's probably a 50-200-250 or so. I never use it higher than 50 watts when on the computer, it is always dimmed below that. My concern is, during the day, when it is hot, does it make sense to open the shades (two windows totalling about 6 sq meters--yes, it's a big corner room in a three-room apt with a SSE view) which provides less useful light for the room than the bulb at a full 60 (well 75 or 50) watts, or pull the very dark shades and use the lamp? I know the lamp's wattage is eventually converted fully to heat. The problem is I have no idea how to deal with the heat that comes in the windows scientifically. Finally, the windows are single-paned and probably date to the 1970's in case that matters to any considerations. μηδείς (talk) 01:21, 31 May 2013 (UTC)[reply]

I have found that reflected light is a wonderful thing. I would consider taking advantage of the "reflected light from the white building next door." Not a scientific answer but something I have recently concluded. Bus stop (talk) 01:35, 31 May 2013 (UTC)[reply]

My guess has been that the reflected light is probably less infra-red, and my usual "solution" has been to use it and the bulb at a very low level. The only problem with that is I have no proof, and I suspect the bulb at a low level is less efficient and throws off a lot more IR proportionally than at full blast. μηδείς (talk) 01:44, 31 May 2013 (UTC)[reply]

I'm surprised that no one has mentioned window film, though it I think of it as a most loathsome solution. Wnt (talk) 18:29, 31 May 2013 (UTC)[reply]

Landlord wouldn't like it. I think the question boils down to, assuming the illumination is the same from either source, and if there's a 20 deg F difference in temperature from inside to out, does 6 sq meters of glass windowage allow more or less heat into the house than is generated by the bulb? μηδείς (talk) 19:13, 31 May 2013 (UTC)[reply]

Well, Wickwack's answer sounds about right where insolation is concerned - even though you never get anything near the maximum in practice, the point is, you won't get pure visible light out of the sun no matter what you do, and an LED bulb would probably pay for itself anyway. Summer is definitely the time of year to try out all the latest in energy-saving technology, little point doing it in the winter! Wnt (talk) 19:30, 31 May 2013 (UTC)[reply]

Dmcq did mention thermal reflective film, Wnt. Is that what you were referring to? -- Jack of Oz ^[Talk] 22:58, 31 May 2013 (UTC)[reply]

Have you asked the landlord? You can remove most films fairly readily so I don't see why they should care. If you do a good job of applying it without any bubbles they probably wouldn't require you to remove it when you go anyway. Dmcq (talk) 13:21, 2 June 2013 (UTC)[reply]

Li-po battery underperformance

I bought an Anker 20000 mAh Li-po external battery for my laptop. Using Microsoft Joulemeter to log power consumption of my laptop over time, I calculated that it produced 73 wh (or 3800 mAh at 19 V) during it's first use. Is it normal for a Li-po battery to underperform on its first charge? Will it perform better after subsequent charge-discharge cycles? --129.215.4.147 (talk) 14:54, 30 May 2013 (UTC)[reply]

1)How did you decide which fraction of the laptop's total energy consumption came from the internal battery and which came from the external battery? 2) How did you determine that the external battery had reached its discharge "endpoint"? Atlant (talk) 16:29, 31 May 2013 (UTC)[reply]

1) I ran the laptop (with Joulemeter logging) with a fully charged internal battery (no external battery) until it shut down at the critical point (5%). This allowed me to determine the energy supplied by the internal battery. 2) I did the same thing with a fully charged internal and fully charged external battery. The laptop ran on the external battery unsupervised until the circuitry of the external battery determined that this battery was empty. At that point the external energy source stopped and the laptop carried on using it's internal battery (which I would later subtract from the energy used calculated in the Joulemeter log). I have yet to repeat the experiments. 89.241.227.119 (talk) 18:03, 31 May 2013 (UTC)[reply]

Titration question

Perchloric acid is labled 0.0325 M. Perform a titration to find the concentration of the base. After performing the titration, you determine that you needed 21.7 ml of the acid to neutralize 50.0 ml of the base.What was the molar concentration of the base

Just a chem problem I am stuck on. Please provide an explanation. — Preceding unsigned comment added by 173.74.251.182 (talk) 16:11, 30 May 2013 (UTC)[reply]

It depends on the identity of the base. HClO₄ (perchloric acid) titrates NaOH in a 1:1 mole ratio, Mg(OH)₂ in a 2:1 mole ratio, Al(OH)₃ in a 3:1 ratio, etc. 72.128.82.131 (talk) 18:53, 30 May 2013 (UTC)[reply]

Bird identification

Can anyone identify this bird? I'm pretty sure what it is but I've never seen one here, Cambridge Bay, before. Thanks. CambridgeBayWeather (talk) 18:10, 30 May 2013 (UTC)[reply]

That has to be an American robin (male). I don't know of anything else that looks like it. (According to the map here, its breeding range extends very near to your neck of the woods. Perhaps this one just dropped in for a visit.) Deor (talk) 18:23, 30 May 2013 (UTC)[reply]

I'd concur with Deor. As the Wikipedia article on the American Robin notes, it is possibly the most common bird in all of North America, so it would not be shocking if one showed up even in cold north of Cambridge Bay. Doing some digging, this page lists several other birds which are sometimes confused with Robins:

Looking at those, the ONLY one that are even close to what you've given are the Robin is the Orchard Oriole, but the Oriole's range is even further from your home than the Robin's is; it'd be much more surprised to find one of those that far north, where as a Robin is so ubiquitous in the rest of North America, it's not shocking to see one anywhere on the continent. --Jayron32 19:12, 30 May 2013 (UTC)[reply]

Thanks. That's exactly what I thought it was but I wanted confirmation as I'd never seen on before. Given the range shown in the map you'd think they wouldn't be that rare. It turns out that one was seen two or three weeks ago just outside of Ulukhaktok, which is on our island but further to the northwest. Given the comments on the Facebook group I belong to it's not that common elsewhere in the Arctic either. They have been turning up in Arviat, Kimmirut, Kangirsuk, Quebec and Inukjuak, Quebec but only within the last three or four years. Both Arviat and Kuujjuaq have Inuit names but the Kuujjuaq name is also the name for either the Common Redpoll or the Arctic Redpoll. The only comment so far from the west indicates that they are common in Kotzebue, Alaska. It's not in the List of birds of Nunavut which was compiled from the first book so it would seem that it wasn't seen as common in 1997. CambridgeBayWeather (talk) 22:49, 30 May 2013 (UTC)[reply]

In population biology it is known as either an "accidental" or "vagrant", and it is more common in birds than in land-based animals for obvious reasons. Birds are frequently found outside of their "standard" ranges, basically the bird made a Left turn at Albuquerque and ended up lost. --Jayron32 00:41, 31 May 2013 (UTC)[reply]

I'd just like to back up the idea that there is no question that this is an American Robin. I am no good at sexing though. (Good at se... yes, but...never mind). μηδείς (talk) 01:46, 31 May 2013 (UTC)[reply]

Got to be a robin. ←Baseball Bugs ^{What's up, Doc?} carrots→ 02:27, 31 May 2013 (UTC)[reply]

We have the odd vagrant mammal these days but they're not the sort you would want to bear meet. CambridgeBayWeather (talk) 02:27, 31 May 2013 (UTC)[reply]

There is no doubt it's a robin. Just watched one find and eat a worm yesterday waiting to pick up my dad from rehab. μηδείς (talk) 16:07, 31 May 2013 (UTC)[reply]

why would a worm be waiting to pick up your dad?[1] Gzuckier (talk) 18:00, 31 May 2013 (UTC) [reply]

Easy pickings, as he was being discharged from rehab? μηδείς (talk) 19:09, 31 May 2013 (UTC)[reply]

Calories

Do inedible things like rocks or plastic have a calorie count? Why is using plants like potato for heating so difficult, why does it not burn like wood? — Preceding unsigned comment added by 88.112.110.65 (talk) 18:34, 30 May 2013 (UTC)[reply]

"Calories" are a heuristic means of determining how much food energy a food will provide. Many things burn but do not provide food energy, and so do not affect the "calorie count" of a food. Gasoline burns, but AFAIK, does not provide any nutritional sustenance. Even edible things like dietary fiber, which is basically cellulose, and thus chemically a carbohydrate does not provide food energy and so is not accounted for in the calorie count of a food as printed on the label. Also, while determining food energy in terms of calories can be done in two ways, either bomb calorimeter or calculated from the base caloric content of macronutrients, in nearly all cases that I am aware of, the latter method is used: the grams of digestible carbohydrate, protein, and fat are found and since these all provide a known quantity of energy (4 kcal/gram for carbs and proteins, 9 kcal/gram for fats) you can simply calculate the energy that way. --Jayron32 18:44, 30 May 2013 (UTC)[reply]

One problem is that the word "calorie" has two meanings. From our calorie article:

the small calorie or gram calorie (symbol: cal) is the approximate amount of energy needed to raise the temperature of one gram of water by one degree Celsius.
the large calorie, kilogram calorie, dietary calorie, nutritionist's calorie or food calorie (symbol: Cal, equiv: kcal), is the amount of energy needed to raise the temperature of one kilogram of water by one degree Celsius; thus it is equal to 1000 small calories.

But either way, the term has a precise scientific meaning as a measure of the amount of energy present. That can certainly be applied to rocks and plastic - as well as to food and gasoline. The trouble is that the amount of energy an object contains may not be the same as the amount that can be released by some particular mechanism. In the case of fats and carbohydrates, the amount of energy released by burning the material is the same as the amount that your body gets from consuming them. But in the case of cellulose, we can't digest much of it so the vast majority of the calories present in the cellulose go right through us. However, give the same food to a ruminant like a cow that can digest it - and you get a totally different result.

So when we talk about the calorie content of food, we're really talking about the number of calories that a typical human can extract from it rather than the inherent amount of the stuff that's present.

Hence your answer is dependent on the context in which the word "calorie" is being used.

SteveBaker (talk) 19:05, 30 May 2013 (UTC)[reply]

Actually, food energy calculations take this into account. See Atwater system for a common heuristic means of estimating food energy which takes into account things like digestive efficiency and energy diverted towards waste production. The Atwater system itself has some problems in the way it makes some approximations, but it does make an attempt to correct for energy actually available for use in the body, and comes closer than simply burning the stuff. These corrections are implicit in the "standard" caloric content of macronutrients (4-4-9 as noted above), and are taken into account if the food manufacturer uses the direct calorimetry method as well. --Jayron32 19:17, 30 May 2013 (UTC)[reply]

The core concept here is that the body can do what a wood fire cannot --- it can burn the potato without boiling the water that is in it. This is due to the catalysis provided by enzymes that are responsible for our metabolism. The first site I saw on a search said there are 109 kcal in 100 grams of potato. Water has a heat of vaporization of 40.65 kJ/mol (according to that article). 4.2 kilojoules = 1 kcal, so that is about 10 kcal/mol, plus maybe 1/5 extra to heat it (heat capacity), gets you 12 kcal spent for every 18 grams (1 mol) of water. So for every 18 grams of potato containing 19 kcal of energy value you burn, you have to spend 12 of them boiling water. I assume this means that if you make a great big blast furnace you can keep a potato fire burning steadily - I'll leave it as an exercise to the reader to see if this has ever been done. But the body heats the potato by only a few degrees, only a few percent of the cost of the fire. Wnt (talk) 18:36, 31 May 2013 (UTC)[reply]

- We did calorie experiments in HS chemistry. You use dry weight food, not raw. μηδείς (talk) 16:24, 1 June 2013 (UTC)[reply]