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May 5

Heat pump

[1] If I understand what the specs section of this is saying, you put 1KW of electricity in and about 2.5KW of heat comes out. That is thermodynamically possible because it's a heat pump of course. Does the temperature of the outside air make a big difference to that conversion factor? Is there a standard temperature difference that they rate the things at? My mom (getting up there in years) is chilly all the time and always running electric heaters in the house, so this might be a good thing for her. Thanks. 2602:24A:DE47:BA60:8FCB:EA4E:7FBD:4814 (talk) 07:42, 5 May 2021 (UTC)[reply]

Yes, the graph on this page illustrates that: Looking at the NIBE COP curves versus temperature. At very low outside temperatures (-19 C or -2 F) the efficiency is still around 1.75%. The graph shows how more powerful heat pumps are more efficient at lower temperatures, but slightly less efficient at warmer temperatures meaning that the pump size should be matched to your climate. -- Q Chris (talk) 07:57, 5 May 2021 (UTC)[reply]

See coefficient of performance. Dolphin (t) 08:20, 5 May 2021 (UTC)[reply]

The unit of power is kW, not KW. Capitalisation matters. Even more when using mW and MW.

Efficiency depends on outside temperature. That's already the case for the theoretical maximum efficiency, the Carnot efficiency. The larger the temperature ratio (expressed in absolute temperature), the lower the efficiency of a heat pump.

To reduce greenhouse gas emission, it's good to move from gas heating to heat pumps, so there's a significant push from governments in that direction, but many people are of the opinion that heat pumps fail to deliver to their promises. Most heating in western Europe is required when the outside temperature is between -5°C (because it's rarely colder than that) and +5°C (because you don't need much heating when it's warmer), at a relative humidity of around 90% (because the sea is never far away and warmer than the land). When you try to extract heat from such air, the water vapour in the air turns into ice, providing additional heat, but also clogging the heat exchanger. That makes it hard to extract a lot of heat from cold, moist air. Heat pumps use electrical resistors as backup heat source if they can't get enough heat from the air. Something to think about.

BTW, is your mom's house well insulated? Getting better insulation usually pays off more than getting better heating. PiusImpavidus (talk) 08:57, 5 May 2021 (UTC)[reply]

It might be worth mentioning that the efficiency is related to the ratio of the temerature across the heat pump, not the ratio of the inside and outside temperatures. This means you can improve efficiency by having larger but cooler radiators. -- Q Chris (talk) 10:10, 5 May 2021 (UTC)[reply]

t

"Sun poster" - today's featured picture

In today's featured picture, , I am confused by the layers.

The white triangle is presumably the centre of the sun, working out to the dark red triangle with the yellow/gold shell colouring. Then there is the orange circle (sphere) around that. But the yellow/gold shell doesn't appear to me to be concentric with the orange sphere. Indeed the curvature on the yellow/gold shell makes it seem bigger than the orange sphere. Am I not seeing something, or is the image flawed? -- SGBailey (talk) 09:40, 5 May 2021 (UTC)[reply]

Are you referring to the photosphere? It might help if you could point out which layer you are referring to, since they are all labeled. --OuroborosCobra (talk) 09:46, 5 May 2021 (UTC)[reply]

I agree with the OP that the diagram presents some interpretation confusion. I'm guessing (because it's not completely certain) that it's attempting to show a segment of the orange chromosphere peeled away to reveal the underlying yellow/gold photosphere. The depiction lacks any cues to help this depiction, such as an edge to the "cut" where the chromosphere was removed (presumably because, drawn to scale, the chromosphere lacks any significant depth), hence the possible confusion. (I could, of course, have misinterpreted the diagram completely, in which case it's much worse than I thought and not really a candidate as a featured picture.) Bazza (talk) 10:05, 5 May 2021 (UTC)[reply]

I think that's the issue. The Chromosphere is 3000-5000km think, which sounds a lot, but is less than 1% of the radius of the Sun, so its essentially just a thin skin over the photosphere. So what the image is showing is a segment of the chronosphere peeled away to reveal the photosphere underneath. But because of the way it is drawn, my eyes/brain want to interpret the image as showing a vertical cut through a thick chronosphere. Iapetus (talk) 09:37, 6 May 2021 (UTC)[reply]

The yellow/gold shell is labelled Photosphere. The orange sphere is labelled Chromosphere. (I have no idea what "Temperature minimum" is pointing at and why it doesn't say "Minimum temperature". The "Transition region" appears to be the same as the Chromosphere.) -- SGBailey (talk) 10:16, 5 May 2021 (UTC)[reply]

I think the distinction is that "minimum temperature" means the lowest temperature, whereas "temperature minimum" means the place (either the actual physical location, or the point on a graph or log) where that minimum temperature occurs. Iapetus (talk) 09:16, 10 May 2021 (UTC)[reply]

Looking for stock image showing relative size of ISS orbit compared to Earth's diameter

I'm writing a school report about the "no gravity in space" fallacy, and I'm looking for an image which highlights the ISS's orbit relative to the size of the Earth. Where might I find something like that? --Puzzledvegetable^{Is it teatime already?} 22:39, 5 May 2021 (UTC)[reply]

There's a live tracker here with a few different views: [2]. The top left view shows the size of the orbit compared to the size of the Earth. --Amble (talk) 23:03, 5 May 2021 (UTC)[reply]

You'd be surprised how many times I've had to explain this. "Gravity at the altitude of the ISS is approximately 90% as strong as at Earth's surface, but objects in orbit are in a continuous state of freefall, resulting in an apparent state of weightlessness. This perceived weightlessness is disturbed by five separate effects..." See International Space Station. See vomit comet. 41.165.67.114 (talk) 06:45, 6 May 2021 (UTC)[reply]

Ask them what keeps the Moon in orbit around Earth, or the Earth around the Sun. Then have them read this. --47.155.96.47 (talk) 21:51, 7 May 2021 (UTC)[reply]

Gravity Inverse Square Law might be useful to you. 41.165.67.114 (talk) 06:53, 6 May 2021 (UTC)[reply]

What about the images at List of orbits? --47.155.96.47 (talk) 21:51, 7 May 2021 (UTC)[reply]

Numerically it's pretty straightforward. If you drew the Earth 30 cm wide, the ISS would be almost 1 cm away. The Earth has a radius of about 6,371 km, so that's how far away from the centre of mass you are when standing on the Earth's surface. The ISS is orbiting at an altitude of 408 km, about 6,779 km from the Earth's centre of mass, which is only about 6.4% further. nagualdesign 23:00, 7 May 2021 (UTC)[reply]