3 ms·

OK, sorry, it's hard to explain what I mean without going into the actual calculations. The Sun produces 3.84e26 Watts and has a mass of 2e30 kilograms. So sup

OK, sorry, it's hard to explain what I mean without going into the actual calculations. The Sun produces 3.84e26 Watts and has a mass of 2e30 kilograms. So suppose you can harness all the energy produced by a kilogram of the sun for 100 years, or 3.5 billion seconds (a human lifetime and then some). You will have harnessed 672000 joules, or 160 dietary calories. Less than a kilo of chicken soup. The whole point is that the time scale and the mass scale of the Sun are incompatible with human energy production. And also, I think it's a surprising result - again, ask a layperson who hasn't done the math to tell you how much energy a good kilogram-sized chunk of the Sun produces in 100 years, and he'll probably tell you it's enough to destroy whole cities or boil seas. Actually, it's barely enough for a small breakfast.

GFischer 15y ago
The 6720000 joules number is misleading IMO. According to http://www.sal.wisc.edu/~khn/ast100/lectures/lecture16.pdf http://www.sal.wisc.edu/~khn/ast100/lectures/lecture16.pdf the energy produced by a kilo of the sun is orders of magnitude higher. (something like 2.3×10e13 Joule/Kg). Of course it doesn't produce it in a measly 100 years (more like 100.000). At one time, only a very tiny fraction of the sun is being processed in it's fusion reactor at the core ( The core, inside 0.24 solar radius, generates 99% of the fusion power of the Sun. From http://en.wikipedia.org/wiki/Solar_core http://en.wikipedia.org/wiki/Solar_core ), and that's what people think about when talking about the sun's energy (the 2.3×10e13 Joule/Kg thing, not the 672000 one).
- uvdiv 15y ago
  yaakov34 is talking about power density, not energy density. The sun's core power density only about 0.3 watts/liter, compared with >100,000 W/L for a power-generating nuclear fission reactor. http://en.wikipedia.org/wiki/Sun#Core http://en.wikipedia.org/wiki/Sun#Core
  - GFischer 15y ago
    Thanks for the clarification and putting it into perspective (I wasn't getting what yaakov wanted to convey). And yes, that does make things more difficult :) .
- yaakov34 15y ago
  Not 100,000 years, he is talking about the total energy produced over the Sun's lifetime of billions of years (his calculation is for the last 4 billion, actually). The number is still surprisingly small - 2.3e13J is only a few billion kcal, quite an imaginable number. Look at this another way: ask a layperson, who consumes 3000 kcal a day approximately, how much of the Sun's mass would be required to supply him with this energy input. The answer, unless I mislaid a zero, is 755 tons or so. Now, I will absolutely be a monkey's uncle if laymen give you an approximately correct answer. My guess is that people will say that it takes a tiny drop of solar material, or a microscopic amount. Can someone try this?
  - scythe 15y ago
    This is mostly because the visible portion of the Sun is not currently undergoing nuclear fusion. Fusion occurs only in the core, not in the outer layers. Also, you're ignoring feedback, which isn't intuitive to ignore -- the blackbody radiation produced by a kilo of 5000 K plasma is huge, enough to supply a person's daily energy and then some -- it just doesn't get emitted from the Sun, which is why that energy will last for another ten billion years. A layperson would probably (try to) count all of the thermal energy which is emitted by such a plasma whenever it's not totally surrounded by more of the same plasma. In that sense, the layperson would be correct -- a kilo of Sun-matter, after being removed from the Sun -- produces a massive amount of power. The density of the Sun is approximately 2 * 10^-4 kg/m^3 in the photosphere ( http://en.wikipedia.org/wiki/Sun http://en.wikipedia.org/wiki/Sun ). We're ignoring the core for now, because that's at more like 15000000 K, so. A kilo thus takes up 5 * 10^3 m^3, which is a sphere with radius ~12 m, having a surface area of ~1.4 * 10^3 m^2. The power radiated by a kilo of plasma from the Sun's photosphere is thus is j' = area * sigma * temperature^4, where sigma = 5.67 * 10^-8 is the Stefan-Boltzmann constant ( http://en.wikipedia.org/wiki/Blackbody_radiation#Stefan.E2.80.93Boltzmann_law http://en.wikipedia.org/wiki/Blackbody_radiation#Stefan.E2.8... ), so j' = 1.4 * 10^3 * 5.67 * 10^-8 * (5 * 10^3)^4 = 4.96 * 10^10 watts, or well over a person's daily caloric intake every second (In fact, this is approximately the energy consumption of a small country).
    - uvdiv 15y ago
teamonkey 15y ago
This is true but misleading. It's wrong to think that the fusion process in the Sun is burning at any sort of physical limit. The Sun is a self-balancing equilibrium. Gravity causes the matter to collapse, which increases heat and pressure at the core and causes matter to fuse, which releases energy that pushes matter away from the core. As such fusion only occurs at the rate needed to maintain this equilibrium, and that rate is less than you'd think intuitively. Not only that but our Sun is relatively small and weak. Type-O stars, for example, can have a luminosity a million times greater than our Sun's.
- yaakov34 15y ago
  My basic point is that the time scale and size scale of stars are so far, far outside human scales, that it's misleading to talk about a reactor somehow being like them, even if it uses fusion. It's really a completely different process. The comparison is used to put a gleam in the eyes of laymen, by making them think of "harnessing" something bright and limitless. You might as well talk about fission reactors harnessing the power of supernovas - it's kind of true (I think - I am not an astrophysicist), since supernovas produced heavy elements like uranium, and we are now using the energy that was stored in these atoms by the reactions in those stars. But it's misleading to use this fact in a policy article, since it has absolutely nothing to do with anything practical in relation to a fission reactor. My secondary point is that people do not have an intuitive understanding of these scales, which is why I used my "chicken soup" examples. I stand by that - I think that people hear "fusion" and do not understand how much unlike a star the reactor would be.
  - teamonkey 15y ago
    Agreed, although there's degrees of accuracy. Since carbon was formed in stars you could also say that fossil fuel is harnessing the power of the stars. Fusion is the primary mechanism at the core of a star and would also be the primary mechanism in a fusion reactor, that does make it a partially valid comparison. (Supernovas use a different mechanism.) But let's not write off fusion over fission because the most common reactor in the solar system has a comparatively low energy output. It does kind of underemphasise the immense amount of energy that nuclear power generates per unit of fuel. It would take a few kilos of seawater to power a family for a year (a single person drinks tonnes of water each year). It should be very safe compared even to non-fission options. The output is relatively clean, especially compared to non-fission options. Excluding initial R&D costs, it should be more efficient than fission in therms of $/W.

GFischer 15y ago

The 6720000 joules number is misleading IMO. According to http://www.sal.wisc.edu/~khn/ast100/lectures/lecture16.pdf http://www.sal.wisc.edu/~khn/ast100/lectures/lecture16.pdf the energy produced by a kilo of the sun is orders of magnitude higher. (something like 2.3×10e13 Joule/Kg). Of course it doesn't produce it in a measly 100 years (more like 100.000). At one time, only a very tiny fraction of the sun is being processed in it's fusion reactor at the core ( The core, inside 0.24 solar radius, generates 99% of the fusion power of the Sun. From http://en.wikipedia.org/wiki/Solar_core http://en.wikipedia.org/wiki/Solar_core ), and that's what people think about when talking about the sun's energy (the 2.3×10e13 Joule/Kg thing, not the 672000 one).

uvdiv 15y ago
yaakov34 is talking about power density, not energy density. The sun's core power density only about 0.3 watts/liter, compared with >100,000 W/L for a power-generating nuclear fission reactor. http://en.wikipedia.org/wiki/Sun#Core http://en.wikipedia.org/wiki/Sun#Core
- GFischer 15y ago
  Thanks for the clarification and putting it into perspective (I wasn't getting what yaakov wanted to convey). And yes, that does make things more difficult :) .
yaakov34 15y ago
Not 100,000 years, he is talking about the total energy produced over the Sun's lifetime of billions of years (his calculation is for the last 4 billion, actually). The number is still surprisingly small - 2.3e13J is only a few billion kcal, quite an imaginable number. Look at this another way: ask a layperson, who consumes 3000 kcal a day approximately, how much of the Sun's mass would be required to supply him with this energy input. The answer, unless I mislaid a zero, is 755 tons or so. Now, I will absolutely be a monkey's uncle if laymen give you an approximately correct answer. My guess is that people will say that it takes a tiny drop of solar material, or a microscopic amount. Can someone try this?
- scythe 15y ago
  This is mostly because the visible portion of the Sun is not currently undergoing nuclear fusion. Fusion occurs only in the core, not in the outer layers. Also, you're ignoring feedback, which isn't intuitive to ignore -- the blackbody radiation produced by a kilo of 5000 K plasma is huge, enough to supply a person's daily energy and then some -- it just doesn't get emitted from the Sun, which is why that energy will last for another ten billion years. A layperson would probably (try to) count all of the thermal energy which is emitted by such a plasma whenever it's not totally surrounded by more of the same plasma. In that sense, the layperson would be correct -- a kilo of Sun-matter, after being removed from the Sun -- produces a massive amount of power. The density of the Sun is approximately 2 * 10^-4 kg/m^3 in the photosphere ( http://en.wikipedia.org/wiki/Sun http://en.wikipedia.org/wiki/Sun ). We're ignoring the core for now, because that's at more like 15000000 K, so. A kilo thus takes up 5 * 10^3 m^3, which is a sphere with radius ~12 m, having a surface area of ~1.4 * 10^3 m^2. The power radiated by a kilo of plasma from the Sun's photosphere is thus is j' = area * sigma * temperature^4, where sigma = 5.67 * 10^-8 is the Stefan-Boltzmann constant ( http://en.wikipedia.org/wiki/Blackbody_radiation#Stefan.E2.80.93Boltzmann_law http://en.wikipedia.org/wiki/Blackbody_radiation#Stefan.E2.8... ), so j' = 1.4 * 10^3 * 5.67 * 10^-8 * (5 * 10^3)^4 = 4.96 * 10^10 watts, or well over a person's daily caloric intake every second (In fact, this is approximately the energy consumption of a small country).
  - uvdiv 15y ago

teamonkey 15y ago

This is true but misleading. It's wrong to think that the fusion process in the Sun is burning at any sort of physical limit. The Sun is a self-balancing equilibrium. Gravity causes the matter to collapse, which increases heat and pressure at the core and causes matter to fuse, which releases energy that pushes matter away from the core. As such fusion only occurs at the rate needed to maintain this equilibrium, and that rate is less than you'd think intuitively. Not only that but our Sun is relatively small and weak. Type-O stars, for example, can have a luminosity a million times greater than our Sun's.

yaakov34 15y ago
My basic point is that the time scale and size scale of stars are so far, far outside human scales, that it's misleading to talk about a reactor somehow being like them, even if it uses fusion. It's really a completely different process. The comparison is used to put a gleam in the eyes of laymen, by making them think of "harnessing" something bright and limitless. You might as well talk about fission reactors harnessing the power of supernovas - it's kind of true (I think - I am not an astrophysicist), since supernovas produced heavy elements like uranium, and we are now using the energy that was stored in these atoms by the reactions in those stars. But it's misleading to use this fact in a policy article, since it has absolutely nothing to do with anything practical in relation to a fission reactor. My secondary point is that people do not have an intuitive understanding of these scales, which is why I used my "chicken soup" examples. I stand by that - I think that people hear "fusion" and do not understand how much unlike a star the reactor would be.
- teamonkey 15y ago
  Agreed, although there's degrees of accuracy. Since carbon was formed in stars you could also say that fossil fuel is harnessing the power of the stars. Fusion is the primary mechanism at the core of a star and would also be the primary mechanism in a fusion reactor, that does make it a partially valid comparison. (Supernovas use a different mechanism.) But let's not write off fusion over fission because the most common reactor in the solar system has a comparatively low energy output. It does kind of underemphasise the immense amount of energy that nuclear power generates per unit of fuel. It would take a few kilos of seawater to power a family for a year (a single person drinks tonnes of water each year). It should be very safe compared even to non-fission options. The output is relatively clean, especially compared to non-fission options. Excluding initial R&D costs, it should be more efficient than fission in therms of $/W.