Wikipedia:Reference desk/Archives/Science/2021 March 17

= March 17 =

Gibbs relation
What does the term Gibbs relation refer to? I encountered the term in the literature but could not find it on Wikipedia. A search engine query was inconclusive as well. –LaundryPizza03 ( d c̄ ) 10:14, 17 March 2021 (UTC)


 * There is an Adam-Gibbs Relation for liquid to glass transitions and a Gibbs-Thompson relation for phase transitions. This one refers to both a Gibbs-Duhem relation and a Gibbs relation relating to surfactants. Which, if any, of these are relevant to your question I don't know. Mikenorton (talk) 12:17, 17 March 2021 (UTC)
 * We have a list of things named after Josiah W. Gibbs. Indeed, several of these can be referred to as "Gibbs relation", so the context is important. The most likely is that it concerns thermodynamics, in particular the preservation of internal energy. If you see formulas of such shapes as $SdT$, it is almost certainly (a form of) the Gibbs–Duhem relation, obtained by taking the derivative of the usual equation for the formula for internal energy. --Lambiam 14:44, 17 March 2021 (UTC)
 * And here are formulae for the Maxwell relation, the first Gibbs relation, and a second Gibbs relation.   Further formulae for the Gibbs relation and the Maxwell relation are on p. 264. 146.199.206.3 (talk) 14:56, 17 March 2021 (UTC)
 * I think that most chemists would think of Gibbs free energy when hearing his name. Mike Turnbull (talk) 15:36, 17 March 2021 (UTC)

Electrical Transmission
I am trying to understand why electrical transmission is done at high voltages. Per Electric_power_transmission, "High-voltage power transmission allows for lesser resistive losses over long distances in the wiring". Many elaborations of this cite Joule's First Law, P∝I²R and P=EI to show that if we hold power constant, we can reduce Joule heating by increasing voltage and decreasing current. But this seems to assume that resistance will remain constant too. Per Ohm's law, R=E/I, as we increase voltage and reduce current, resistance will increase both in proportion to the voltage and in inverse proportion to the current. As far as I can see this would exactly negate any advantage of higher transmission voltages for reducing Joule heating. 2600:6C5D:567F:A913:5854:CFC6:8ED1:70F3 (talk) 15:51, 17 March 2021 (UTC)


 * You are conflating two different voltages. Consider the context of use of your three equations:
 * 1: P=EI, where P is the power you are seeking to transmit and it remains constant. E is your transmission line voltage and I is the current.  So for the same power transmitted, a higher voltage means a lower current.
 * 2: P∝I²R, where I is the same current from equation (1), R is the resistance of your transmission line, and P is the power transmission loss -- which is not the same P from (1).
 * 3: E=IR, where I is the same current from (1) & (2), R is the same transmission line resistance from (2), but E is the voltage drop along your transmission line, which is not the same E from (1).
 * So the E from (3) is not something you control directly.
 * You change your transformers at the source and the load so that in equation (1) the voltage increases while the current decreases for constant power. But the R in (2) and (3) is constant and equation (2) lets you calculate the power transmission loss while equation (3) lets you calculate the voltage drop. -- ToE 17:32, 17 March 2021 (UTC)
 * And if you are still confused, thinking, "Why can't I apply Ohm's law to the overall voltage?", well, you can, but you need to use the corresponding resistance. If your voltage is that at the source, then your resistance (or impedance) should be both that of the transmission lines *plus* that of the load.  And to complicate things, you changed the impedance of your load when you changed the transformers in order to transmit at a higher voltage. -- ToE


 * I don’t disagree with ToE but perhaps I can provide a simpler explanation. Yes, we assume that the resistance of the transmission line remains constant. A 5-ohm resistor always has a resistance of 5 ohms; varying the voltage presented to a resistor, or varying the current through it, has little effect on the magnitude of the resistance. Your question is a simple one best answered by assuming that the resistance of the transmission line is constant. Dolphin ( t ) 21:17, 17 March 2021 (UTC)


 * It's good to emphasize that the transmission line resistance remains constant, but if that were the only answer given, the followup question would be, "How can it possibly remain constant? R=E/I, E has increased, and I has decreased, so R has clearly increased."  The answer to that is that its not the same E.  Hence my verbosity. -- ToE 00:34, 18 March 2021 (UTC)


 * I now realise my earlier answer didn’t mention the resistance of the load and clearly the load needs to change if it is to deliver a nominated power. I can explain most easily using a numerical example: imagine the task is to install a heater with a power output of 1000 W. There are two suitable appliances on the market - one operates at 250 V, and the other at 25 V. The 250 V model will have a resistance of 62.5 ohms whereas the 25 V model will have a resistance of 0.625 ohms. The 250 V model (62.5 ohm) will draw a current of 4 amps, whereas the 25 V model (0.625 ohm) will draw 40 amps. Clearly, the losses from the transmission lines will be greater if they carry 40 amps than if they only carry 4 amps. Dolphin ( t ) 05:17, 18 March 2021 (UTC)
 * Well, another way of looking at the problem is that the apparent resistance of the wires, viewed from downstream from the transformer, does vary; increasing the transformer ratio decreases that apparent wire resistance, which decreases the losses. [with an ideal transformer of ratio $a$] the transmission wires have apparent resistance of only $R_C/a^2$. Note that the idiot who wrote the article section failed to provide a single reference, so maybe the concept of "apparent resistance" is their own lunatic view of the problem with nonstandard notations. Tigraan Click here to contact me 10:09, 18 March 2021 (UTC)

Hollow Earth collapsing in on itself
If it were possible to remove the core of the earth leaving a hollowed out area, would that area remain hollow or would the earth collapse in on itself? If so, would that collapse affect the surface of the earth? 209.91.188.70 (talk) 17:45, 17 March 2021 (UTC)
 * See Gravitational binding energy for a way to calculate the energy which would be released when it collapses. -- ToE 17:52, 17 March 2021 (UTC)

Ya. That is math! Not my strong suit. Followed the link. . . and my eyes glazed over! lol I will just assume that there would be some collapse and that any impact on the surface would be minimal. 70.26.18.99 (talk) 23:35, 17 March 2021 (UTC)
 * There would be a big effect on the surface. You could expect massive earthquakes as the surface jumps up and down by many kilometers, and when the collapse settles down the surface area of the Earth will be smaller, so there will be a huge amount of compression in the crust. Oceanic crust is weaker and would take up more of the shrinkage. Graeme Bartlett (talk) 01:09, 18 March 2021 (UTC)


 * Minimal surface impact sounds overly optimistic!
 * Not "some collapse", but total collapse. From Hollow Earth: "Ordinary matter is not strong enough to support a hollow shape of planetary size against the force of gravity; a planet-sized hollow shell with the known, observed thickness of the Earth's crust would not be able to achieve hydrostatic equilibrium with its own mass and would collapse."
 * Also, "For bodies made mainly of rock, the minimum size to become a self-gravitating sphere is about 600km diameter; but, for bodies mainly made of ice, the minimum size is about 400km diameter.". And your suddenly hollow Earth is still much, much more massive than that.
 * The collapse would yield a 5.8% reduction in the planet's diameter and a 11.2% reduction in its surface area!
 * I'd expect that such a large rearrangement of matter would involve massive, catastrophic, world-wide earthquakes; total soil liquefaction; supervolcanos everywhere; perhaps some significant mixing of the crust and mantel; etc.  I figure a world-wide, total extinction event. -- ToE 01:34, 18 March 2021 (UTC)


 * For those who don't want to do the calculation themselves, here are the results. The gravitational binding energy of the Earth after the collapse will be
 * $$U_n=-\frac{16}{15}G\pi^2\rho^2R_n^5$$
 * with $$G$$ the constant of gravity, $$\rho$$ the density of the Earths mantle and $$R_n$$ the new radius of the Earth, which, assuming conserved volume, must be
 * $$R_n=\left(R_o^3-R_c^3\right)^{1/3}$$
 * with $$R_o$$ the old radius of the Earth and $$R_c$$ the radius of the Earth's core.


 * The gravitational binding energy of the Earth minus its core before collapse is
 * $$U_o=G\pi^2\rho^2\left(\frac{8}{3}\left(R_c^3R_o^2-R_c^5\right)-\frac{16}{15}\left(R_o^5-R_c^5\right)\right)$$
 * Plugging in the numbers and subtracting gives an energy release of $15,000,000,000,000,000,000,000,000,000,000$ Joules, or about 10% of the Earths binding energy. This corresponds to an earthquake of about magnitude 14.7 and provides enough heat to increase the Earths temperature by about 4500 K. Forget about soil liquefaction, the soil will be boiling. PiusImpavidus (talk) 17:24, 19 March 2021 (UTC)