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June 19

Viscosity from DFT (VASP) using the Green-Kubo relation

Hello! In this paper https://pdfs.semanticscholar.org/e8a2/02f25555cd8c4f947bbbdff5a61a0ea0efd2.pdf the authors use VASP to determine MgSiO3 viscosity using the Green-Kubo relation $\eta ={\frac {V}{3k_{\rm {B}}T}}\int _{0}\left\langle \sum \limits _{i<j}\sigma _{ij}(t+t_{0}).\sigma _{ij}(t_{0})\right\rangle dt$ where $\sigma _{ij}$ (i and j = x, y, z) is the stress tensor, t is time and t0 is the time origin. But I've seen other papers use: $\eta ={\frac {V}{3k_{\rm {B}}T}}\int _{0}^{\infty }dt\left\langle P_{xy}(t)P_{xy}(0)\right\rangle$, where $P_{xy}$ is the off-diagonal component of the stress tensor $P_{\alpha \beta }$ ( α and β are Cartesian components).

OK, so clearly these are essentially exactly the same equation but the second uses only the xy component whereas the first seems to suggest a summation? So which is correct?

Also, VASP outputs the stress tensor components as XX YY ZZ XY YZ ZX. So which of these should I use to input into the Green-Kubo equation? And are there missing components? what about yx, zy, xz?

Thanks in advance. Polyamorph (talk) 15:01, 19 June 2018 (UTC)

• For your second question, per our article Cauchy stress tensor, the stress tensor is symmetric, thus having only six independent stress components, instead of the original nine, see also Stress_(mechanics)#General_stress. For the first question, I am not sure; it could be that only one off-diagonal component is nonzero in the context (for instance in a shear flow where water flows in from (x=±Inf,y=0) and out from (x=0,y=±Inf), and the flow is uniform across z), as is the case in many viscosity-measurement experiments; or it could be the Einstein notation by taking x and y as free variables (but then a factor 1/2 is missing). Tigraan^{Click here to contact me} 16:49, 19 June 2018 (UTC)

I think that in both case sum of off-diagonal components is implied. Ruslik_Zero 20:16, 19 June 2018 (UTC)

Tigraan and Ruslik0 Cool, symmmetry reduces the 9 components to 6, makes sense. But I'm still not sure that the sum is implied in the second equation, only that the xy off-diagonal component of $P_{\alpha \beta }$ should be used, at least that's how I interpret how they've written it. But in any case, to clarify if it is the sum to be taken then should I take the sum of the $P_{xy},P_{yz},P_{zx}$? Many thanks for your help. Polyamorph (talk) 09:59, 21 June 2018 (UTC)

I think that the second equation is just an example, and xy is just one of the components to be used.Polyamorph (talk) 12:35, 21 June 2018 (UTC)

Aerial refueling a commercial jet?

Apparently aerial refueling doesn't make much economic sense for commercial jets. [1] But suppose radar operators were scanning for MH370 on the day of its disappearance and after almost all hope (and fuel) is lost, a passenger had gotten on the radio and said the pilots are dead, but some of us woke up, where are we, all we see down there is water!... Is there any conceivable way that some fast military aircraft swoops up from a carrier and slows down, and we see a boom projecting forward from it or a hose dangling down from in front of the wing, with a hard-bitten soldier or a decently designed robot on the end that can unscrew a gas cap and start pouring gas into the plane's tank so that the passengers don't end their adventure in the Indian Ocean? Or is it completely hopeless? Wnt (talk) 17:38, 19 June 2018 (UTC)

Bear in mind that tanker planes aren't particularly fast; they're primarily freight aircraft. The first picture at aerial refueling shows a Boeing KC-135 Stratotanker, with a cruising speed of 530 mph and a maximum speed of 580 mph, while another picture shows an Ilyushin Il-78, whose maximum speed is similar to the cruising speed of the Stratotanker. If the plane's virtually out of fuel, there's no way you can get a tanker on-site unless it's already really nearby. This assumes that the airliner is comparatively close to land, or that it's not far from a carrier-based tanker. Also see Aerial_refueling#Buddy_store, which notes that carrier groups don't generally have tanker aircraft, so the chance of finding a carrier-based tanker is even tinier. Nyttend (talk) 17:50, 19 June 2018 (UTC)

PS, see Gimli Glider and its paragraph beginning with "On airliners the size of the 767"; a recent commercial airliner without fuel is exceptionally difficult to control, even by professionals, since a Ram air turbine doesn't have much power compared to jet engines. Nyttend (talk) 17:58, 19 June 2018 (UTC)

Modern airlines are set up for single-point refueling, also known as underwing fueling, which requires a fairly complex set of motions when connecting the refueling nozzle. To add another level of complexity, the refueling point is commonly found under an access panel, which will either be held shut by the slipstream or act as a impromptu air brake when opened. So ignoring the logistics of getting the tanker in the right place at the right time... once you have it there, it'll be near impossible to connect and refuel on a commercial airlinger not designed with air-to-air-refueling in mind. WegianWarrior (talk) 18:12, 19 June 2018 (UTC)

Here's a 747 refueling in-flight: [2] —2606:A000:1126:4CA:0:98F2:CFF6:1782 (talk) 21:59, 19 June 2018 (UTC)

Might that be Air Force One? ←Baseball Bugs ^{What's up, Doc?} carrots→ 22:10, 19 June 2018 (UTC)

There's a doghouse, so that's a NEACP, not the Trump Tourbus. Some other civilian aircraft in long-duration military service (such as covert ELINT) have been fitted for refuelling too. The current Airforce One was built as a Boeing VC-25, the designed-in Airforce One variant, but the new Airforce Ones are to be recycled from bankrupt Russian oligarchs (seems appropriate), so will require conversion.

Outside the US, there are also Airbus variants, such as the A330 Multi-Role Tanker Transport, which can offer both refuelling supply and consumption. So if you're writing a Dan Brown novel, it's off-the-shelf parts. Andy Dingley (talk) 23:33, 19 June 2018 (UTC)

Unrelated question: At Aviation fuel what does this mean: "Aircraft have a high peak power and thus fuel demand during take-off and landing"? Does that mean that a disproportionate amount of fuel is consumed "during take-off and landing" relative to the rate of fuel consumption at cruising speed and cruising altitude? Bus stop (talk) 23:07, 19 June 2018 (UTC)

Not so much "take off" itself, but the climb to altitude. This is why many military mission profiles involve refuelling very soon after take-off - the aircraft can take off and climb in a much lighter state. Andy Dingley (talk) 23:23, 19 June 2018 (UTC)

"disproportionate amount at take off?" It seems so. According to https://answers.yahoo.com/question/index?qid=20060813151514AAVe8Ms&guccounter=1 a Jumbo Jet consumes 1 gallon per second at cruising regime and 8 gallons per second at take off and climbing. Somewere else it talked of three tons in the first three minutes and eight tons per hour cruising. 194.174.76.21 (talk) 09:54, 20 June 2018 (UTC) Marco Pagliero Berlin

Among other things, you're using a richer mixture at those times. It's vaguely like with a car: you use less fuel when you're cruise-controlling down a straight and flat highway than when you're climbing a hill and accelerating from a stop. Cars don't use much fuel when descending hills, but they don't have to worry about maintaining power in order to avoid stalling too soon and falling out of the sky too fast; the aircraft has to maintain lift until it's ready to stall over the runway, so lots of power has to be used. Nyttend (talk) 00:22, 20 June 2018 (UTC)

Thanks everybody. That is interesting. And I guess the fighter plane has a much further distance on its itinerary than the tanker plane, and that one tanker plane refuels many fighter planes. Bus stop (talk) 01:14, 20 June 2018 (UTC)

Fighters have to be fast and maneuverable, so they can't carry enough fuel relative to their empty weight to provide a long range as well. --76.69.118.94 (talk) 08:36, 20 June 2018 (UTC)

Fighters can, but they do it with drop tanks. Andy Dingley (talk) 10:13, 20 June 2018 (UTC)

Thank you. This is an interesting topic. Bus stop (talk) 12:14, 20 June 2018 (UTC)