Talk-Polywell.org

For lack of anything better, I was reading a wikipedia article about the sun.

http://en.wikipedia.org/wiki/Sun

I was really surprised by the following information.

The energy production per unit time (power) produced by fusion in the core varies with distance from the solar center. At the center of the Sun, fusion power is estimated by model to be about 276.5 watts/m3,[39] a power production density which more nearly approximates reptile metabolism than a thermonuclear bomb.[40] Peak power production in the Sun has been compared to the volumetric heats generated in an active compost heap. The tremendous power output of the Sun is not due to its high power per volume, but instead due to its large size.

276.5 watts per cubic meter, the volumetric heats generated in an active compost heap. If that is the power density achieved in the core of the sun, what do they expect to achieve by ITER?

The comparison I have seen for ITER is about the power density of a candle flame.

Give the Sun some credit ... it is achieving that on hydrogen, about the most pathetic fusion fuel of the bunch. It almost does not react. Which is all to the good because we would not want to live so close to a fusion bomb of that size.

I always get an image-association comparing the sun to an elephant. An elephant's specific power output, by volume, is about ten times that of the sun - so why can't you see an elephant in the dark? My answer to that is an elephant's surface area to volume ratio is greater by several orders of magnitude, and what surface it has is cooled by the air, whereas the sun is in a vacuum and so it heats to incandescence.

As mentioned, hydrogen fusion is hard to do. Even so, the Sun manages this feeble output with a core temperature of only about 15,000,000 degrees (~ 1.4 KeV) with the P-P fusion chain. The CNO cycle starts kicking in past this temperature so in more massive stars the fusion rate increases rapidly. A large star might have a hydrogen fusion rate more than a thousand times faster. And this is still below the temperatures that even Tokamaks are expected to operate at (perhaps 2-3 KeV).

Also, consider what the fusion rate of hydrogen would be in any man made hydrogen reactor. If limited by the P-P chain the reaction rate would be perhaps 10^-20 less at similar conditions when using heavier hydrogen isotopes. The fusion crossection for P-P hydrogen fusion is ~ 10-45 vs ~ 10^~23 for D-T at similar temperatures. If you could get the CNO reaction going at temperatures of ~ 200,000 eV you might reach efficiencies ~ 1% of deuterium fusion (that is what the relative C12-P crossection is at that energy).

To really compare apples to apples, you would need to compare a Star fusion reactor burning the same fuel as man made reactors. Fusion rate is dependent on the the temperature dependent crossection times the density squared (I think).
So, the Sun with D-T fuel at a similar temperature as a Tokamak ( 1.4 KeV.) and a density ~ 10^ 7 times greater (~100 atm vs ~ 0.00001 atm) would result is a fusion rate of ~ 10^14 (100 trillion) times greater per unit volume. Multiply that by the volume of the core of the sun and the total output would be much higher
Comparing the Sun to a Polywell with a claimed energy density advantage of ~ 65,000 X over a Tokamak would still faver the Sun by ~ 10^9.

We are indeed fortunate that it is very difficult (but not impossible) to get hydrogen to fuse.

Dan Tibbets

It's actually very misleading to suggest that the power density of an elephant is a mere 5 times higher. That's volumetric power density, rather than gravimetric power density, which is what you're usually interested in.

The gravimetric power density is less than 2 watts per metric tonne.

Soylent wrote:It's actually very misleading to suggest that the power density of an elephant is a mere 5 times higher. That's volumetric power density, rather than gravimetric power density, which is what you're usually interested in.

The gravimetric power density is less than 2 watts per metric tonne.

Point 1: There is no such thing as "gravimetric power density". That is the "specific power". DENSITY is always per unit volume unless you are talking "flux density".
Point 2: do you know what the "specific power" of the core of the sun is? Until you do, how can you compare it to an elephant?

D Tibbets wrote:...
So, the Sun with D-T fuel at a similar temperature as a Tokamak ( 1.4 KeV.) and a density ~ 10^ 7 times greater (~100 atm vs ~ 0.00001 atm) would result is a fusion rate of ~ 10^14 (100 trillion) times greater per unit volume. Multiply that by the volume of the core of the sun and the total output would be much higher
...Dan Tibbets

A small mistake. I incorrectly stated the Suns' core density as ~ 100 atm. It should have been ~ 100,000 atmospheres. So, multiply the output comparisons by ~ 1000.

Dan Tibbets

The core of the Sun is considered to extend from the center to about 0.2 to 0.25 solar radii.[32] It has a density of up to 150 g/cm3[33][34] (about 150 times the density of water) and a temperature of close to 13,600,000 K

150 metric ton per cubic meter. Is that the same thing?

Since fusing hydrogen into helium releases around 0.7% of the fused mass as energy,[38] the Sun releases energy at the mass-energy conversion rate of 4.26 million metric tons per second,

Burning as bright as calculated, the sun would waste away pretty quickly wouldn't it? On the other hand, 1.9891×1030 kg is a lot of mass, = 1.9891x10^27 metric tons.

Aero wrote:

The core of the Sun is considered to extend from the center to about 0.2 to 0.25 solar radii.[32] It has a density of up to 150 g/cm3[33][34] (about 150 times the density of water) and a temperature of close to 13,600,000 K

150 metric ton per cubic meter. Is that the same thing?

Since fusing hydrogen into helium releases around 0.7% of the fused mass as energy,[38] the Sun releases energy at the mass-energy conversion rate of 4.26 million metric tons per second,

Burning as bright as calculated, the sun would waste away pretty quickly wouldn't it? On the other hand, 1.9891×1030 kg is a lot of mass, = 1.9891x10^27 metric tons.

Lets see, there are 1 million cc in one cubic meter, so 150g/ cc= 150,000,000 g/ cubic meter, or ~ 150 metric tons / cubic meter. Yep, it matches.

I'm not sure what you are calculating in the second part. If the Sun produces ~ 4 *10^6 KT of mass/energy equivalent, that would be ~ 4* 10^9 kg/s. At a conversion efficiency of ~ 0.7 percent, that would mean that ~ 5 *10^ 11 kg of hydrogen would be consumed per second. I believe the Sun weighs ~ 10^30 kg, so it would take ~ 2 * 10^18 seconds to burn through all of the hydrogen. With ~ 3 *10^7 seconds per year, the Sun should last ~ 6*10^10 years (~60 billion years). Of course the Sun only burns a modest portion of it's hydrogen. Stars the Sun's size do not mix the gasses in their cores with the gas outside the core so a large portion of the hydrogen is not available. Add to that the hydrogen lost in the solar wind, etc. and the reported main sequence lifetime of ~ 10 billion years for the Sun seams reasonable.

Dan Tibbets

'm not sure what you are calculating in the second part

I just was wondering - If the sun burned like a polywell, how long would it last? 60 billion years, burning at it's current rate (within an order of magnitude). Lets see, 10^17 times faster burn rate - so it will burn up in about 20 seconds, give or take an order of magnitude. Hmm - quicker than a nova - makes for an unpleasant region of space.

Technically, the power of 3.84e26 watts divided by the total volume of 1.41e27 gives a raw power density of .25 W/m^3 Your figure is off by a factor of 1000.

But, the power is generated in the core, which is just 1/4 the diameter, so 1/64 the volume. So it ends up producing power in the region of 17.4 W/m^3 as an average for the core.

To reach the 276 W/m^3 figure you actually need to be right at the very core.

http://fusedweb.llnl.gov/CPEP/Chart_Pag ... ayers.html gives the fusion rate tables, from Stormgrew's 1965 "Principles of Stellar Evolution and Nucleosynthesis"

I'd look somewhat askance at stormgrew's work, it being long before neutrino detection was available.

If, for instance, we choose to consider all the energy being generated in the innermost 1% of the sun, then we see a burning plasma with a power density around 27 MW/m^3.

Only a VERY detailed neutrino observatory could tell the difference because the total energy will be the same and the neutrino flux will be the same.

Thanks for that. I just quoted Wikipedia. That is not universally considered to be a reliable source, but it is convenient. If you have better data perhaps you could update the Wikipedia article. Here is their reference:

http://fusedweb.llnl.gov/CPEP/Chart_Pag ... ayers.html

And yes, the 276.5 W/m^3 is at the exact center of the core. See the table at the end of the reference.

Edit - Interesting to note that Wikipedia already uses your reference as the source of their data. But consider that the thrust of the Wikipedia paragraph was to show how low the power density of the sun is. Therefore, quoting the absolute largest value is in keeping with good practice and should not be considered to be misleading in context.

If, for instance, we choose to consider all the energy being generated in the innermost 1% of the sun, then we see a burning plasma with a power density around 27 MW/m^3.

But I don't see your justification for the 27 MW/m^3 number.