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>An ARM chip in most data collection circumstances won't get as hot as an asphalt highway.

Doesn't matter - it might not get as hot as the highway itself, but it will get exactly as hot as the ambient air if powered off, and then immediately start heating up further when powered on. Basic thermodynamics demands that ambient air will *always* be cooler than an operating processor, unless you're refrigerating the chip directly.

Now, if you stick the whole thing in a well-insulated sealed box like a styrofoam cooler, along with a big chunk of thermal mass with a lot of surface area, like a cement "radiator" (much better than metals and better than most rock), you may be able to keep the whole thing well below ambient temperatures throughout the hottest part of the day, especially if you paint it solar-white or keep it in the shade. That would depend on the exact power consumption of the device, color of the box, amount of insulation and air space, and the size and surface area of the thermal mass though. You'd want to have an engineer run the numbers to be sure though.

Or just pour some cement into the bottom of a cheap styrafoam cooler. You could probably get some mounting bolts neatly embedded into a nice big protective hole in the cement (so that if you drop the thing device-first, it stays safe in its hole instead of being smashed flat) by screwing them into a block of styrofoam that you partially submerge bolt--heads first, and then pull out after the cement cures.

https://www.engineeringtoolbox...

100 % humidity means 30 grams (0.03l) of water per cubic meter. Today in the UK we are at 70%, so lets say theres 20g on a bright autumn morning.

You'd be lucky, 100% humidity is only 30 grams at 30 degrees C. At 10 degrees, more typical for a UK autumn morning it is less than 10g per litre. I Nairobi it is 20 degrees C now so your figure is closer there (17g/m^3 100% humidity).

Why yes, I have! And if you actually look at how much water vapor can be contained in air versus temperature you will see it's a logarithmic function and essentially zero at temperatures at or below freezing. Sublimation or not.

Exactly this. Concentrations of CO2 currently in the atmosphere are 400ppm, meaning 1 molecule of CO2 for every 2500 molecules of air. For that 1 molecule of CO2 to raise the ambient temperature of the surrounding 2499 molecules of other gases by 1C would mean that a single molecule of CO2 is emitting 2499 degrees of heat.

The air around the CO2 molecule is made up of the following components:

Nitrogen: 78.08%
Oxygen: 20.95%
Argon: 0.93%
Carbon dioxide: 0.035%
Methane: 0.00017%

And each one of these gases has a different specific heat capacity So that must mean that CO2 is one hell of a hot molecule. And no it is not absorbing and releasing that much thermal radiation to make any difference, only about 8% of blackbody radiation is picked up by CO2.

The primary argument is circular. The climate change acolytes know that water vapor is the #1 greenhouse gas, but they want to claim CO2 is the cause of warming so the argument goes that an increase in CO2 will heat things up just enough to increase the evaporation of the oceans to increase the amount of water vapor which will be the cause of the increase in temperature.

Plastic expands and contracts with temperature just like everything else Having many separate layers that need to be precisely aligned means that the expansion of one plate is going to have an impact on the entire system.

If you've every done 3D printing, you'll know that PLA, ABS, Nylon, etc, all shrink when they cool down, leading to potential warping and having to print your ABS design 1.5 to 2.5% larger than you want it to be when it's cooled down. That's a 2.5% shrinkage with a 200 degree temperature difference.

Plastic has in general a higher thermal expansion that most metals.
Have a look for yourself: https://www.engineeringtoolbox... the higher the number, the more it expands. Glass is around 4, plastics range between 40 and 120.

If you want to 3D print with nylon, you need to make sure you dry out the filament first, as it's quite hygroscopic and will absorb moisture from the air.

You seem to be totally ignorant as to how refraction works. How much the light bends depends on the ratio between the refractive index of lens and the medium it is in. Water and air have different indexes. Water on the surface of a lens causes distortion, regardless of the material the lens is made from.

Moisture can penetrate plastics too. controlling moisture content in plastic is a huge part of reliable production.

Lead dust is as small as 0.1 microns in diameter.

N95 and HEPA respirator filters only block airborne particles at least 0.3 microns in diameter.

The CDC hereby recommends wearing scuba gear for respiratory protection at the firing range.

How strong is it versus its weight? I skimmed TFA but didn't find a mention of that. This page suggests oak, at 3x normal density, would have comparable density to aluminum. If this material is as strong as steel but as light as aluminum, that could have actual applications. I'd wonder about flammability and rotting, though. Skyscrapers or spaceships made out of wood would be pretty funny, though.

You appear to have confused millimeters with inches, although I can't access your citation. 20 millimeters should be enough for 1000 amperes, even with cruddy plastic insulation and continuous duty. https://www.engineeringtoolbox.com/wire-gauges-d_419.html

Why does a glass of water have 0 calories but energy in joules?

The amount of energy transferable by water depends on the difference between the temperature of the water and the temperature of whatever the energy is being transferred to. The equation is, E = (4.2 kJ/kg-degC) (deltaT) (volume in liters) (1 kg/liter). If you drink 8 ounces (0.24 liters) of water whose temperature is 50C (about 120F; uncomfortably hot), deltaT = 12C. So, (4.2E3)(12)(0.24) = 12096 Joules. 1 Joule = 2.389x10-4 Calories (not calories, but kilocalories (Calories), which is what food energy is measured in). So your glass of water can transfer (12.096E3)(2.389E-4) = 2.89 Calories.

If your glass of water is at body temperature, there is no energy transfer. If it's below body temperature, the energy transfer is from your body to the water. Either way, the numbers are so small that it's insignificant, unless you drink enough water to drown yourself, at which point nothing matters anyway.

(I am not interested in getting involved in an abstract discussion about all this, but would certainly welcome any reliable source of information about the exact conditions of this test)

From the small real chunks in the video and the limited information in the linked article, I understand that:
- We are talking about accelerating a small vehicle (as big as a small truck?) from zero to 300 km/h in 300 m and then keeping that speed for about 500 m.
- It seems that it is a kind of a small train (better: the small locomotive of a train) over fancy rails travelling as any other train (or IC-powered, rail-based vehicle) would do. No idea about the exact effect of all this vacuum breakthrough technology, but it doesn't seem to matter too much here (= speed mostly constrained by the friction of the wheels against the rails).

After some research and by making lots of assumptions because of the limited amount of information in that article (again: I will be more than happy to update this post if anyone could provide reliable enough information about the exact conditions), it seems to have a reasonably good acceleration for what seems its weight by comparing it with equivalent road-based alternatives (i.e., fast cars or trucks). On the other hand, the friction wheel-rail is much lower than the one wheel-road (rolling resistance values); to not mention the fact that the contact surfaces of the typical wheels of road-based vehicles are much bigger than what seems to be shown in the video (i.e., the aforementioned factor would have to be still smaller?). Comparing this with the acceleration of a train locomotive might be more accurate, but I haven't been able to find any reliable reference to that alternative.

In summary, it seems a difficult-be-compared-against-anything-else sample of well-known technology (= vehicle on rails) under extremely limited conditions (a straight stretch of 300+500 m!) and by providing almost no relevant information. Also I guess that reaching much higher speeds under these exact conditions wouldn't be too difficult: if much bigger locomotives with many wagons can travel faster than 400 km/h during long (not completely straight!) stretches, these engines should be able to deliver notably higher speeds under notably better conditions (= no wagons/weight + straight and extremely short stretch).

Logically, this new "milestone" (= new CGI-intensive video) has no effect on my medium-/long-term predictions for Hyperloop, as written in a post here some weeks ago (reminder: Slashdot posts cannot be edited/removed and I welcome anyone to quote me on my predictions in that post at any point). Short summary: I don't expect Hyperloop to ever become a reality as it is being advertised; in the best scenario (= losses-driven project eminently supported by have-to-be-done-no-matter-what ideas), it might become an expensive and mostly useless toy.

Sea level air density is 12.25, while at 30,000ft it is 0.1841.

http://www.engineeringtoolbox....

Although different alloys may come into play occasionally, brass is almost heavier/more dense than steel.

http://www.engineeringtoolbox....

typically used for industrial processes, but one local example may be the area hospital laundry facility, where they typically run in the 2500-2700 degree range at the boiler.

You wouldn't think they would run that close to the melting point of the equipment...
http://www.engineeringtoolbox.com/melting-temperature-metals-d_860.html

The CO2 buildup does cause the gasp, but the O2 provided is what gets/keeps the heart going.

Exhaled air is about 16% O2 and 4% CO2. Not at all good, but not fatal. Basically, you would have nausea, fatigue, and perhaps a headache, not unlike the way many people feel when dieting.

Yes, the atmospheric gases move at hundreds of miles per hour in a storm

You do realize that the load from wind is the square of the wind speed times density? And the density would be increased by the dust that was picked up? At the height of summer, effective density of 0.01 atm and 175 kph wind speed (sited by the book and film, but could be higher during gusts), that would be 15 newtons per square meter. A fragile high surface area rocket which is not well secured would have a significant side force which could indeed push it over.

I haven't watched the movie so I don't know just how much force they portray from the wind storm. But wind speeds that high are not something to trifle with even in Mars's far less dense atmosphere.

I guess NASA is wrong then about Antarctica adding 80+ billion tons of ice per year. Density of ice is fairly stable versus temperature so the volume of ice is actually increasing quite a bit. But then, maybe NASA doesn't know what it's doing.

No it is not. Most of the stuff involved is 99% efficient, so I simply averaged it to 90%.

Sources? Here are mine:

1) In the USA, transmission and distribution losses are estimated at 6% by the EIA. However, India (the subject of the original post) has really bad infrastructure, in comparison. Their losses are estimated at about 30% by the World Energy Council. So, I was actually way too generous when I said "90% efficient" for this part.

2) The Tesla Roadster is said to have a battery pack charge/discharge efficiency of 86% in an article from Stanford University, which claims to be drawing from a source published by Tesla itself.

3) The National Electrical Manufacturers Association requires a minimum efficiency of about 92% for some large induction motors. However, I found an answer on the Electrical Engineering Stack Exchange site indicating that Tesla uses a slightly less efficient type of motor, and also that a ~97% efficient power controller is required, which brings the real efficiency down to ~88%.

Using the exact numbers I sourced, above: 88% of 86% of 70% of 42% = 22%, which is even worse for a coal-powered electric car than what I originally posted (because I was intentionally rounding up a bit to allow for future improvements).

That link about the Prius Engine is nice! However keep in mind: that would require all new cars to use engines like this. Which they hopefully do in not to distant future.

The extremely aggressive CAFE standards recently set by the Obama administration are already pushing things hard in that direction in the USA. Admittedly, India probably won't be buying many ICE cars produced for the US market in the next decade - but then, they won't be buying fancy electrics like a Tesla, either.

Let's see... Typical refrigerator has a volume of approx 16 cubic feet on the refrigerator side, or 453 liters. Figure 1/3 of that is take up by food or held in drawers. That gives approx 300 liters of chilled air which falls out as you hold the door open looking at the contents.

Figure room temperature is 20 C and the refrigerator is 2 C. At this temperature range, air has a density of about 1.25 kg/m^3, and a heat capacity of 1.005 kJ / kg*K. We have (1.25 kg/m^2) * (0.3 m^3) = 0.375 kg of air. (0.375 kg) * (20C - 2C) * (1.005 kJ / kg*K) = 6.784 kJ of energy which is lost every time you open the refrigerator door (well, added to the interior of the refrigerator). Since the air is chilled via a heat pump, this isn't its actual energy use. The best refrigerators typically have a real-world COP (coefficient of performance) of about 3, or 1 Joule of electricity us used to pump out 3 Joules of heat. So 6784 Joules of heat represents 2261 Joules of electricity used.

If the camera and associated circuit board use 2 Watts while active, then opening the refrigerator door uses as much energy as keeping the camera powered up for (2261 Joules) / (2 Watts) = 18.8 minutes. If the screen uses 20 Watts, then the total 22 Watts consumption means you'd have to stare at the screen for 1.7 minutes to use as much energy as opening the door once. If it uses 0.1 Watts while idle (around what a smartphone uses), then opening the refrigerator door once is enough to power it for 6.3 hours.

Based on these back of the envelope calcs, I'd say the camera + monitor uses less energy than opening the refrigerator door. I should note though that 2261 Joules is 0.000628 kWh, or about 0.0075 cents worth of electricity. So we're really picking over minutiae (unless you're in a habit of holding the refrigerator door open long enough for food to start warming up). The convenience of the camera and remote monitoring is a bigger factor, though I'd probably lose the built-in screen (the most expensive part of the system) and rely on a phone or computer to view the contents.

A similar technique was tested successfully by japanese researchers in 2010, except their rocket model used ambient air directly, instead of H2 in a tank.

I wonder what kind of performance it would get from using maser-powered water vaporization for propulsion ? Water vapor holds twice as much heat as air, translating into twice the ISP. It would be very steampunk, too... I now envision aerospike-like rocket engine gloriously steaming into the stratosphere on top of a microwaved plume of vapor.

Water conducts heat away about 25x faster than air, and the heat transfer rate is proportional to temperature differential. Until the organism gets to a very large size or develops some serious insulation, trying to maintain a constant body temperature underwater is a lost cause.

TFA describees an adaptation for minimizing loss of internal heat to the water (counter-current heat exchange of blood entering/leaving the gills). An adaptation that AFAIK no land animals has, though I have heard of some animals having it in their extremities (blood leaving their core to the extremities exchanges heat with blood returning from the extremities, thus preserving internal body heat. I'd be curious if birds which fly really high (upwards of 30,000 ft) have it in their lungs, or if the energy consumption needed to fly at those altitudes is sufficient to offset any heat loss.

It's also a bit of a stretch to call this warm bloodedness. It's not thermal homeostasis, the process that keeps your internal body temperature at 37 C regardless of environmental conditions.

It means they haven't done the math. Heck, even if you covered every square metre of a plane with solar cells you couldn't collect enough power. There's not enough there.

Let's do the math, then. The specifications of Solar Impulse-2 are available as a starting point.

At 269.5 square meters of solar cell coverage, and an average power density of 1.35 kW per square meter, the maximum amount of energy the plane can harvest is about 364 kW. Now, we can use two facts to avoid the ugly world of aeronautical engineering (which I don't know): The aircraft has flown under its own power, supplied by four 17.5-horsepower motors. Those motors therefore supply about 13 kW each, for a total of 52 kW of energy required to fly.

Since 52 kW is far less than the 364 kW the solar cells produce, yes, there is in fact enough power available for collection.

Even if you charged up batteries from ground sources you couldn't carry enough storage and have the plane get off the ground because of the weight. Even with an order of magnitude improvement of power density you couldn't.

As noted, the plane has already flown, carrying its lithium batteries with it.

Weight, energy storage density, and efficiency matters too much for that application for it to be any other way.

Ah, finally some thermodynamics! Currently, the whole plane needs an efficiency of about 15% to simply fly. After some quick research, it seems most solar cell technologies today run at about 20% efficiency, with new technologies pushing 46%. Going from the 20% point, that means the motors need to be only 75% efficient. A bit more research shows that they're actually reasonably assumed to be around 85% efficient. The plane will fly in bright sunlight just fine under solar power alone.

Lithium polymer batteries have efficiencies of around 80% to 90%, so going up to a solar cell with 25% efficiency would allow the plane to either charge or fly, but not both. Double the efficiency and you double the capability, so having solar cells that are 50% efficient would allow both charging and flying under ideal conditions. We're getting pretty close to that.

Throw in some assumptions about duty cycles, allowing the plane to be on the ground for a bit (doubling its charging rate, because it doesn't need to spend energy to fly), and making long trips is feasible in several short hops. Account for an intelligent pilot, using tailwinds and other air currents to reduce the energy needs, and those hops can be made longer.

A Boeing 777 is designed for speed. If you're not in a hurry, solar power might just be a reasonable option very soon.

I just checked the aeodynamic drag for the Tesla S (0.24cd) using the formula here.
At 60mph the air resistance will be about 85N or about 2.3kW. This is far less than I would have thought and appears much less significant than the typical rolling resistance.

441N x 26.8 m/s = 11.8 kW

Using wikipedia figures the Tesla seems to require approximately 22kW at 60mph but it is extremely low drag and probably has lower rolling resistance than the typical value above. However I'd bet most cars are using 40kW (~55hp) or less to cruise at 60 resulting in the rolling resistance being responsible for 25-50% of the power requirements. I'm not convinced it would be worth it in many cases but recovering 10% of the energy could provide 1kW, the same as a 70A alternator.

Coefficients of expansion
Aluminum    ÂÂ12.3
Acrylic        Â42
Rubber, hard    Â42.8    (10-6 in/(in degF))
Steel        Â6.7
http://www.engineeringtoolbox....

Rubber expands almost 7 times faster than steel and nearly 4 times greater than aluminum. Are you sure Columbia isn't just a liberal arts college?

The current comet shape model suggests that the mass is 1013 kg (about 100 million times the mass of the international space station), with a bulk density of ~470 kg/m3 (similar to cork, wood, or aerogel). The low mass and density values strongly constrain the composition and internal structure of the nucleus, implying a relatively fluffy nature, with a porosity of 70 to 80% (Sierks et al., this issue). The nucleus surface itself appears rich in organic materials, with little sign of water ice (Capaccioni et al., this issue).

http://www.sciencemag.org/content/347/6220/387.short

Just "wood" is a poor point of comparison for density; the density of common woods varies substantially.

See: http://www.engineeringtoolbox....

Since the water carrying ability of saturated air goes up with temperature, there should be a trend toward heavier rainfall with temperature increases. Of course, places that are in a "rain shadow" like most deserts would not be expected to benefit as much as places near large bodies of water.

The 90% reflective white paint and reflective foil will both emit a blackbody light profile relative to their material type. As you can see aluminium has a really low emissivity coefficient.

So the math should look more like:

white paint: absorbs 85W/m^2 emits 50W/m^2 = 35 W/m^2 absorbed
(made up numbers btw)

I don't understand your question, but it seem that this comet is composed of hard ice according to the last Philae data I can find in the Internet. The thermal conductivity of ice look like this http://www.engineeringtoolbox....

The reason my "dirt simple" calculation was wrong, as any reader of this exchange should be able to tell (and so should you have), that I misunderstood what your power figure represented. [Jane Q. Public, 2014-09-07]

I'm very sorry for not being more clear. I take full responsibility.

It absolutely does translate directly into power in = power out at a boundary just inside the cavity surface when everything inside that boundary isn't changing. In that case, the rate at which energy changes inside the boundary equals zero, which means power in = power out.

Are you also then presuming that power transferred from the outer surface of the enclosing plate to the chamber walls is the same as the power transferred from the heat source to that plate? [Jane Q. Public, 2014-09-07]

Anything else would violate conservation of energy. But we still have one more step before the net power transferred from the heat source to that enclosing plate becomes relevant.

No, of course I got the same answer, given your assumption that power-in = power-out: 149.59F. [Jane Q. Public, 2014-09-07]

Excellent. And can we also agree about the enclosing aluminum shell's final inner steady-state temperature?

Now to calculate the enclosing shell's inner temperature. At steady-state, power in = power out through some boundary. This time, draw the boundary within the enclosing shell. Again, constant electrical power flows in. But all the other boundaries we drew were in vacuum, so heat transfer was by radiation. This time the boundary is inside aluminum, so heat transfer out is by thermal conduction.

electricity = k*(T_h - T_c)/x (Eq. 4)

The shell's thickness "x" is 1mm, and the thermal conductivity "k" of aluminum is 215 W/(m*K). We just found that:

Outer shell temperature: 338.629792627809 K (149.864 F).

So:

Inner shell temperature: 338.629929668632 K (149.864 F).

Of course, that's a flat plate approximation of heat conduction through a spherical shell, which is derived here. That more accurate equation yields:

#Calculate enclosing shell's inner temperature.
var('T_c T_h power k r_c1 r_c2')
eq2 = power == 4*pi*k*r_c1*r_c2*(T_h - T_c)/(r_c2 - r_c1)
soln3 = solve(eq2.subs(T_c=338.629792627809,power=15028.4258648090,k=215,r_c1=6.378,r_c2=6.379),T_h)
soln3[0].rhs().n()

Inner shell temperature: 338.629929346551 K (149.864 F).

Now for the final step. Calculate the steady-state temperature of the enclosed heated plate (Jane's "source").

Now to calculate the enclosing shell's inner temperature. At steady-state, power in = power out through some boundary. This time, draw the boundary within the enclosing shell. Again, constant electrical power flows in. But all the other boundaries we drew were in vacuum, so heat transfer was by radiation. This time the boundary is inside aluminum, so heat transfer out is by thermal conduction.

electricity = k*(T_h - T_c)/x (Eq. 4)

The shell's thickness "x" is 1mm, and the thermal conductivity "k" of aluminum is 215 W/(m*K). We just found that:

Outer shell temperature: 338.629792627809 K (149.864 F).

So:

Inner shell temperature: 338.629929668632 K (149.864 F).

Of course, that's a flat plate approximation of heat conduction through a spherical shell, which is derived here. That more accurate equation yields:

#Calculate enclosing shell's inner temperature.
var('T_c T_h power k r_c1 r_c2')
eq2 = power == 4*pi*k*r_c1*r_c2*(T_h - T_c)/(r_c2 - r_c1)
soln3 = solve(eq2.subs(T_c=338.629792627809,power=15028.4258648090,k=215,r_c1=6.378,r_c2=6.379),T_h)
soln3[0].rhs().n()

Inner shell temperature: 338.629929346551 K (149.864 F).

Now for the final step. Calculate the steady-state temperature of the enclosed heated plate (Jane's "source").

Once again, energy is conserved, which means that if you draw a boundary around some system (like the heated plate), power going in minus power going out equals the rate at which energy inside that boundary changes. At steady-state, that rate is zero because the system doesn't change. So at steady-state, power in = power out.

I've specified the dimensions. The heated plate is a sphere with radius 6371 mm, surface area A_h, temperature T_h and emissivity epsilon_h. The enclosing plate is a 1 mm thick concentric shell with emissivity epsilon_c, an inner radius of 6378 mm, surface area A_c1 and temperature T_c1 on the inside, and A_c2 and T_c2 on the outside. The chamber walls at temperature T_c are a concentric sphere with inner radius 6386 mm, so there's a 7 mm gap on both sides of the enclosing shell. The plates and walls are oxidized aluminum, which are treated as gray bodies.

Since the enclosing shell has no edges and has nearly the same area as the heated plate, MIT's infinite plate approximation describes net heat flow (in W/m^2):

net heat flow = sigma*(T_h^4 - T_c^4)/(1/epsilon_h + 1/epsilon_c - 1) (Eq. 2)

At steady-state, net heat flow (in W/m^2) equals the electrical input. Note that MIT's Eq. 2 reduces to my Eq. 1 for blackbodies where epsilon_h = epsilon_c = 1.

The plates and chamber walls are made of oxidized aluminum with emissivity = 0.11.

Here's my Eq. 2 using Jane's variable names:

net heat flow = sigma*(T(s)^4 - T(w)^4)/(1/E(s) + 1/E(w) - 1) (Eq. 2J)

Note that it reduces to my simpler blackbody Eq. 1 if E(s) = E(w) = 1.

If you'd like me to clarify what my variable names for a particular equation would be in your terminology, just ask.

At steady-state, net heat flow out (in W/m^2) equals "electricity". The first step is to calculate that constant variable "electricity" which describes electrical power per square meter heating the sphere to 150F without an enclosing shell. I calculated 29.4 W/m^2, which is less than with the simpler blackbody plates because aluminum isn't a perfect emitter or absorber.

I calculated 29.4 W/m^2, which is less than with the simpler blackbody plates because aluminum isn't a perfect emitter or absorber.

Show your calculations where we can see them. I'm not doing this just for me, I want to show other people just how much a clown you actually are. I am not going to install Sage today just to check your math, and probably neither is anybody else who sees this. ... I have reasons for wanting it public-readable, and I will accept nothing else. [Jane Q. Public, 2014-09-02]

#Calculate constant electrical power/area heating 1st plate.
var('sigma T_c T_h electricity epsilon_h epsilon_c')
eq1 = electricity == sigma*(T_h^4 - T_c^4)/(1/epsilon_h + 1/epsilon_c - 1)
soln1 = solve(eq1.subs(T_c=255.372,T_h=338.706,sigma=5.670373E-8,epsilon_h=0.11,epsilon_c=0.11),electricity)
soln1[0].rhs().n()

ANSWER: 29.3986743761843

Can we agree on that? If not, a month ago I said we could use Wikipedia’s equation which includes areas. After I mentioned view factors, Jane agreed that the relevant view factor is 1.0 or

... What do you want to use for material? We might as well use the same material throughout. So if you want to use aluminum for source, passive plate, and walls that is fine with me. We know then, from ESA that the emissivity of aluminum in vacuum is approximately 0.15, and absorptivity 0.05. [Jane Q. Public, 2014-09-02]

Again, the materials are oxidized aluminum with emissivity = 0.11 for these temperatures. As you said, the best we can realistically do is graybodies where emissivity = absorptivity. If you'd like to use a different emissivity just let me know, and we can both independently calculate the required electricity to check each other's answers.

I calculated 29.4 W/m^2, which is less than with the simpler blackbody plates because aluminum isn't a perfect emitter or absorber.

Show your calculations where we can see them. I'm not doing this just for me, I want to show other people just how much a clown you actually are. I am not going to install Sage today just to check your math, and probably neither is anybody else who sees this. ... I have reasons for wanting it public-readable, and I will accept nothing else. [Jane Q. Public, 2014-09-02]

#Calculate constant electrical power/area heating 1st plate.
var('sigma T_c T_h electricity epsilon_h epsilon_c')
eq1 = electricity == sigma*(T_h^4 - T_c^4)/(1/epsilon_h + 1/epsilon_c - 1)
soln1 = solve(eq1.subs(T_c=255.372,T_h=338.706,sigma=5.670373E-8,epsilon_h=0.11,epsilon_c=0.11),electricity)
soln1[0].rhs().n()

ANSWER: 29.3986743761843

Can we agree on that? If so, we can move on to the next step, which is calculating the final outer surface temperature of the enclosing shell once it reaches equilibrium. I promise to provide public-readable versions of my Sage worksheet from now on.

Wow, the person with one reference that has been proven inaccurate is saying someone else is talking out their ass.

I decided to look up evaporation from a pond. For example, Bath County Pumped Storage Station has an upper reservoir with n area of surface area of 265-acre (110 ha) and storage capacity of 35,599 acreft (43,910,720 m3). Using standard calculationswith 20km winds, 0% humidity and max holding of.020kg/kg I get a daily evaporation of about 73 acre-ft/day. That is a loss of 0.2% of the water in the upper reservoir per day. Sure if you let the water sit for 50 days you would get a 10% loss but that would be rare. So in 365 days one would lose 0.73 of a complete fill. If the reservoir let out ten percent a day there would be 36.5 fils per year. 0.73/36.5 = 2% loss due to evaporation. While evaporation causes losses it is not significant.

More importantly, can we agree that in equilibrium, power in = power out?

No. I am not aware of any "conservation of power" law. [Jane Q. Public, 2014-08-02]

Energy is conserved, which means that if you draw a boundary around some system (like the heated plate), power going in minus power going out equals the rate at which energy inside that boundary changes. At equilibrium, that rate is zero because the system doesn't change. So at equilibrium, power in = power out. Jane replied:

... I already told you I was being an ass about your "power in equals power out" thing. Trying to lecture me about conservation of energy is particularly pointless, since I need no such lesson. ... [Jane Q. Public, 2014-08-04]

Jane claims he needs no such lesson because he said:

I admit to being an ass there. Mea culpa. But it's irrelevant. As long as the power used by the source and the power used by the cooler are constant as required, any relationship between them has no bearing on the experiment. [Jane Q. Public, 2014-08-02]

No, the fundamental principle used to determine equilibrium temperatures isn't irrelevant. Anyone making that claim either needs a lesson about conservation of energy, or is deliberately spreading misinformation.

The basis of all my calculations is the very relevant principle that in equilibrium, power in = power out. I've never even mentioned the power used by the cooler of the chamber walls, so Jane either needs a lesson about conservation of energy or Jane's deliberately spreading misinformation. Which is it?

Remember that conservation of energy at equilibrium let us calculate the 233.8F equilibrium temperature of a heated plate enclosed by a superconducting shell. But we can also account for the finite thermal conductivity of an aluminum shell using this same relevant principle by drawing a boundary within the enclosing shell.

The same relevant principle applies: in equilibrium, power in = power out. Again, electrical power flows in. But all the other boundaries we drew were in vacuum, so heat transfer was by radiation. This time the boundary is inside aluminum, so heat transfer out is by thermal conduction.

electricity = k*(T_h - T_c) (Eq. 4)

For aluminum, thermal conductivity k = 215 W/(m*K). Sage solves this equation for an equilibrium inner shell temperature of 149.9F rather than 149.6F for a superconducting shell. This warms the enclosed plate to 234.0F rather than 233.8F for a superconducting shell.

Hopefully this exercise shows how useful it is to start with the widely applicable principle that in equilibrium, power in = power out. Hopefully it's also clear that none of these equations has anything to do with the power used by the cooler. Hopefully it's also clear that Jane's also wrong to claim that the power used by the cooler is required to be constant. The chamber wall temperature is held constant, so the power used by the cooler temporarily decreases after the enclosing plate is added, until it reaches equilibrium.

Why does Jane wrongly claim that the fundamental principle used to determine equilibrium temperatures is "irrelevant"? Does Jane need a lesson about conservation of energy, or is he deliberately spreading misinformation?

"If you don't thi

So... not to stir up a hornets nest... but everyones aware that electric cars produce more pollution than gas right?

Let's look at some facts here. First off, the efficiency of a thermal power plant is somewhere around 33% to 48%, at least according to wikipedia. Let's split the difference and say 41% for a thermal plant. The typical thermal efficiency of a a gasoline engine is about 18% to 20%. Let's split the difference and say 19%. Thus, a thermal power plant is more than twice as efficient as a gasoline engine in terms of changing chemical potential energy to useful output.

But there are some caveats. Firstly, the electricity needs to be transmitted. High voltage power lines are extremely efficient, about 94% according to this article. That means that the chemical energy (lets assume from coal) reaching the charging station is 41% x 94% = 38.5%. And then there is the charging process. According to this article, the charge efficiency of a Li-Ion battery is about 97%, which makes sense to me, as batteries usually don't run too hot. The charging devices however probably are responsible for some loss. Let's assume they are 80% efficient. That gives us 38.5% x 80.0% x 97% = 30%. Thus, according to this, 30% of the coal chemical potential energy makes it to the engine.

But what about engine efficiency? Well electric motors run very cool, and have very high efficiencies, typically around 90%. I wouldn't be surprised if Tesla's motor is better. This means that if a coal power plant powered a Tesla, 30% x 90% = 27% of the energy would reach the wheels of the car, compared with a gasoline powered car, where 19% of the gasoline's potential energy comes out of the engine, never mind the losses in the transmission lines. Thus, a coal powered Tesla is 40% more energy efficient than a gasoline powered car.

However, there is one problem. Generating energy by coal produces more CO2 than generating it by gasoline. According to this article, coal generates about 215 pounds CO2 per btu of energy, while gasoline generates 157 pounds CO2 per btu. However, even with this, by my calculations, an equivalent gas powered car still emits 3.8% more CO2 than our coal powered Tesla.

Elon Musk made this claim in an interview, that even if a coal power plant generates the electricity, a Tesla still emits less CO2. My referenced back of a napkin calculations above support this assertion.

We've determined equilibrium temperatures in a simple example, so let's solve a more general example.

Jane's concerned that the enclosing plate is bigger than the heated plate. But Earth's mean radius is 6371 km, and the effective radiating level is ~7 km higher, so these surface areas are only ~0.2% different. Of course, in a thought experiment this difference can be made arbitrarily smaller. Despite Jane's protests, this doesn't change the fact that enclosing the heated plate makes it warmer.

More importantly, I treated the plates as blackbodies where absorptivity alpha = 1 and emissivity epsilon = 1. This is a reasonable approximation for plates made of carbon nanotube arrays (PDF) which have alpha = ~0.99955. But more conventional plates have alpha and epsilon considerably less than 1.

The next step is to treat the plates as graybodies where absorptivity and emissivity are independent of wavelength, so they appear gray. Kirchoff's Law states that absorptivity = emissivity for graybodies.

MIT calculates heat transfer between graybody plates using an infinite sum of emission, reflection and absorption. Using my variable names, their final expression is:

net heat flow = sigma*(T_h^4 - T_c^4)/(1/epsilon_h + 1/epsilon_c - 1) (Eq. 2)

(Again, Eq. 2 looks better in LaTeX, but hopefully this version is legible.)

At equilibrium, net heat flow equals the electrical input. Note that MIT's Eq. 2 reduces to my Eq. 1 for blackbodies where epsilon_h = epsilon_c = 1.

Suppose the plates and chamber walls are made of oxidized aluminum with emissivity = 0.11. In this case, Sage solves Eq. 2 for a constant electric input of 29.6 W/m^2, which is lower than before because aluminum doesn't radiate as well as a blackbody.

Using Eq. 2 and the same reasoning as before, fully enclosing the heated plate warms it to the same equilibrium temperature of 235F (386K). Fully exposing the plate to the cosmic microwave background radiation cools it to 13F (263K), which is lower than before because the CMBR is a blackbody and aluminum chamber walls aren't.

So even for graybody plates, MIT's mainstream physics refutes Dr. Latour's nonsensical claim that the enclosed heated plate remains at 150F. They also use this equation to explain how thermos bottles insulate drinks, and describe the same radiation shields used since at least

Not really. The moisture holding capacity of air doesn't really increase that quickly in the temperature range we are talking about: http://www.engineeringtoolbox....

And further, human addition of water vapor is a slowly moving equilibrium that will be quickly pushed back when something "breaks" and thus causes economic damage. For example, depleting a major aquifer will lead to decreased irrigation levels which leads to an economic decline that causes roads to be torn up and emissions to be reduced do the the fact that fewer people can afford stuff made in factories.

Most electric motors above 125hp have a "Minimum Nominal Efficiency" of 92% and higher. According to this quick Google result: http://www.engineeringtoolbox....

But, as others have pointed out, this works better for oil well fires because oil won't sit and smoulder for hours, then reignite.

As a precursor to moving in and applying chemicals its sounds like a good idea. After the flames are displaced you could count on fires springing up again from pockets that remain above the autoignition temperature of the materials but it would probably take awhile, you'd have some clear area and time to move in and quench them.

There is a good demonstration of dynamite quenching a flaming oil well here in The Fires of Kuwait ... rewind and check out this whole mesmerizing documentary!

Ok, reading through it a little more carefully and doing some more math, I think his error is in confusing two different ideas. You can either look at the planet as a whole, or go in closer to look at what the atmosphere is doing. The temperatures at earth-like pressures are really just a function of the way gas and temperature are dependent in a gas, so I looked at the planets as a whole. Using the data google gave me when I made simple searches like "distance from sun to Venus", simple geometry formula, heat transfer formula from here: http://www.engineeringtoolbox...., and the assumptions that the earth receives 1kW/m^2 of incident sunlight, that the intensity decreases with the square of radius and that both the Earth and Venus are currently close enough to steady state; I found that the Earth's atmosphere absorbes 36% of the radiation going out to space, while the atmosphere of Venus absorbs 97%! The greenhouse effect on Venus is increasing the temperature by 432 K, while on Earth the difference is around 30 K.

Boltzman constant W m2K4 5.67037E-08
        kW km2K4 5.67037E-11
                Venus Earth
Distance from Sun km 108200000 149600000
Radius km 6052 6378.1
Surface Temperature K 735 288
Surface Area km^2 460264736.8 511201962.3
Area facing the sun km^2 115066184.2 127800490.6
Radiation Rate kW/m^2 1.911651252 1
Total Energy Absorbed kW 219966415.1 127800490.6
Energy Radiated from Surface kW 7616732494 199422477.5
Proportion Absorbed by Atmosphere 0.971120633 0.359147012
Temperature without Greenhouse K 302.9941598 257.6808184
Radiated Energy without Greenhouse kW 219966415.1 127800490.6
Temperature Difference K 432.0058402 30.31918159

This is a very rough design but it appears to use a semipermeable membrane which allows the O2 which is dissolved in water to pass through but not the H2O molecules. It does not appear to separate hydrogen from oxygen in water.
One other problem not addressed in the very thin "concept" is how it deals with CO2. Most rebreathers scrub the CO2 from exhaled air and then add O2. You do need to get rid of the CO2.
From this page:
http://www.engineeringtoolbox.com/air-solubility-water-d_639.html
Oxygen dissolved in the Water at atmospheric pressure can be calculated as:
        co = (1 atm) 0.21 / (756.7 atm/(mol/litre)) (31.9988 g/mol)
        = 0.0089 g/litre
        ~ 0.0089 g/kg
It looks like humans need about 100 gm/hour of O2 so there may be a fundamental problem here is getting enough O2 out of the water. (100gm/0.0089=11,236 liters of water per hour)
I wouldn't get too excited about seeing this soon (or ever).

http://www.engineeringtoolbox.com/mass-weight-d_589.html

"The English Engineering System - EE

In the English Engineering system of units the primary dimensions are force, mass, length, time and temperature. The units for force and mass are defined independently

        the basic unit of mass is pound-mass (lbm)
        the unit of force is the pound (lb) alternatively pound-force (lbf).

In the EE system 1 lb of force will give a mass of 1 lbm a standard acceleration of 32.17405 ft/s^2."

Whenever ephemeral methane is detected around here I blame the closest dog...
Seriously, with the oddball magnetic field structure that focuses on the southern hemisphere (insert Uranus joke here) it's a wonder solar ablation has not wiped all gases from the place. As the solar wind (fart joke optional) takes gas from lesser protected areas of the globe gravity pretty much demands that pressures equalize, but I'm not sure if you would get a tequila sunrise effect(lighter elements on top) or if the normal heat engine circulation of an atmosphere would keep things mixed enough that traces of light things like methane would remain without a source of replenishment. And that's the real question, is there a cause for new methane to be released into the atmosphere of Mars? Or could it be that we are reading gases BETWEEN here and the target when we do these long distance detections... Because if it is inter-planetary or even interstellar methane we detect, then I'm going to blame it on Space Truckers...

isopropanol does not freeze unless you count "Temperatures found at the south pole, or on mars" as being a sensible design concern.

Freezing and flashpoints of isopropanol + water solutions

Failing that, you could fill the bottle with clear acrylic or epoxy resin instead of either, and it will NEVER freeze. it's a tad expensive but the resulting bottles wont explode when heated, wont spring leaks, freeze, etc.

Yes, we can build safe reactors, just not water-cooled reactors. Fission reactions are "just getting warmed up" by the time water starts boiling. That is a bad combination. This is why water-cooled reactors have to operate at 100+ atmospheres of pressure. Just taking water out of the equation makes fission several orders of magnitude simpler and safer to use.

That's why we should be working on new designs based on molten salt cooling

Water is a popular cooling medium because its specific heat is higher than just about anything else. If you want to transport a large amount of heat energy from one place to another, heated water is about the best way to do it.

The fact that water vaporizes when overheated or depressurized is a safety mechanism too. When water vaporizes, it absorbs nearly 7x as much energy as it takes to heat water from room temperature to boiling (2260 kJ/kg vs 4.19 kJ/kg*C). Or nearly 2x the energy it takes to heat water from room temperature to the operating temp of a pressurized water reactor. So a leak or depressurization of the water automatically and instantly results in cooling.

The large volumetric change when water vaporizes is also ideal for driving a generator. Volume change = mechanical work, which is easily captured by a turbine. Without a volume change, you're left trying to capture energy via an inefficient and bulky Stirling engine.

So yeah molten salt reactors have a lot going for them. But the use of water for cooling isn't because of some grand conspiracy. Water is just an extremely good medium for cooling and converting thermal energy into mechanical work, and was the obvious choice when reactors were being designed ~50 years ago.

CO2 would be heavier (thermal mass), suppress fire and corrosion, and best of all would be cheap.

Heavier gases usually transfer heat poorly. Light gases have a higher heat capacity per Kg, have much higher heat conductivity (because of faster molecular velocities), and transfer heat better. This is why helium is used in gas cooled nuclear reactors, while heavy gases such as krypton, xenon or sulfur hexafluoride are used in insulated windows.

Natural gas furnaces are 25% more efficient than fuel oil furnaces, so natural gas costs 30% more per BTU input than fuel oil.

I have to disagree. I did the math on this years ago when deciding whether to replace my gas furnace with another gas furnace or get an oil furnace. The data point very clearly in the opposite direction of what you are saying.

This may vary geographically, but the most recent data I could find for where I live (upstate New York) is this: Gas costs $11.49 for 1000 cu.ft. as of last November ; #2 home heating oil costs $3.934 per gallon as of the same point in time. Natural gas has an energy density of 950-1150 BTU per cu.ft.; heating oil is 139,600 BTU per gallon. That works out to a price of $9.991-$12.095 for 1,000,000 BTU worth of natural gas, or $28.181 for 1,000,000 worth of heating oil. In sort, because of the lower efficiency of oil heaters (25% is exaggerated, BTW), you get shafted twice for using oil.

The only real difference is with oil, you get to choose who does the shafting.

http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html for a nice chart backing him up.

Expansion of salt water is less than fresh water.

Dunno why it would be unbelievable. Solar-powered aircraft have been made. They're ultralight, are covered in solar panels, have practically no payload, and fly at about 20 mph. But if you insist...

Passenger plane fuel consumption is on the order of 5-7 gallons per mile. Call it 6 gal/mi.
Airspeed is about 550 mph, or 0.153 miles per second.
Fuel consumption is thus (6 gal/mi) * (0.153 mi/s) = 0.918 gal/s.

Jet fuel has about 35 MJ/L of energy, or 132.5 MJ/gal.
At 0.918 gal/s, that's 121.6 MJ/s or 121.6 megawatts.

Solar constant in space is about 1361 Watts/m^2. On Earth it's about 750 Watts/m^2.
Air pressure at 35,000 ft is about 25% that of sea level.
So figure with only 25% of the atmosphere intercepting sunlight, you get 1208 Watts/m^2 at cruising altitude.

To generate 121.6 MW with 100% efficient panels producing 1208 Watts/m^2, you need 100,660 m^2, or about a tenth of a square km. A roughly 320x320 meter patch, or about 5-10 city blocks. I suppose a really small town could fit in that area.

You could quibble about gas turbines only being 40%-50% efficient, but then real-world commercial solar panels are only 15%-20% efficient. And we're ignoring clouds, night, and angle to the sun (all the above assumes the sun is directly overhead). So more realistically you're probably looking anywhere from a quarter of square km to over 1 square km of solar panels needed to propel a passenger plane.

From what I've read, "mach 1" is never higher than ~800mph regardless of altitude so he will in fact be falling faster than the speed of sound.

References:
http://www.grc.nasa.gov/WWW/K-12/airplane/sound.html
http://en.wikipedia.org/wiki/Speed_of_sound
http://www.aerospaceweb.org/question/atmosphere/q0112.shtml
http://www.engineeringtoolbox.com/elevation-speed-sound-air-d_1534.html
http://www.wolframalpha.com/input/?i=mach+1&a=*MC.mach+1-_*Formula.dflt-&f2=120000+ft&f=MachAlt.H_120000+ft