About cold fusion from Rossi
Not so long ago, the news of the first "independent" confirmation of the Rossi reactor slipped. Fortunately, there was a link to the report in the news. The report is curious, let's try to check a couple of things from it.
Go to page 2, second chapter (“Reactor characteristics and experimental setup”). Right on the second line of the first paragraph of this chapter, the dimensions of the reactor are written: 20 cm long, 2 cm in diameter, plus thickening at the end of 4 cm in diameter and 4 cm long. We obtain the total surface area taking into account the ends, a little more than 200 cm 2 . Let's read the caption under pic. 1, where it is written that small edges on the cylinder increase the area of heat dissipation, and confine ourselves to our (as can be seen, more and more underestimated) area estimate.
Now we recall about Newton's law on heat flow. Unfortunately, it is not so easy to derive the heat transfer coefficient experimentally, so you have to use the reference book. For example, the Kuchling Reference gives 5.6 W / (m 2 · K), or 0.0005 W / (cm 2 · K) for a round bill. With a temperature difference of 1200 degrees, we obtain a total heat transfer of 200 · 0.0005 · 1200 = 120 W.
This is direct heat transfer, and we also have convection, and, as we know , there is always more loss for it (the link, by the way, does not quite correctly speak about liquids, the English wiki uses a much more suitable fluid term that includes themselves and gases, and plasma, and all other similar things).
')
Temporarily, we will consider our cylinder a horizontal plate with an area of rl = 40 cm 2 and a perimeter of 2 (r + l) = 44 cm. We calculate the Rayleigh number for such a plate using the well-known formula. We need the following numbers:
Finally, we can get the Rayleigh number for air at this temperature, equal to about 10 7 .
Therefore, the Nusselt number for our plate is 0.54 · (10 7 ) 1/4 ≈ 30. That is, 30 times more energy is consumed in convection than direct heat exchange, which is about 120 · 30 = 3600 W.
A thoughtful reader will note that I received 120 watts for a cylindrical object, and a factor of 30 for a flat sheet. True, but, firstly, it overestimates the maximum by 4 times, and, secondly, for vertical surfaces in this case, the Nusselt number will be about the same - the Prand number for gases is almost equal to one.
Four and a half kilowatts of energy, leaving just a heat exchange with the environment - somehow do not match with a single figure in the report. So, if the masters-authors of the report would eithercross the temperature and lower the temperature in the tables, or wipe out an order of magnitude greater energy output.
Note that this effect is enhanced even more, since, if memory serves me, the corresponding processes are in fact non-linear, and on temperature differences of the order of 1200 degrees Kelvin, these non-linearities already manifest themselves.
Actually, this can finish the review, but let's go further.
On page 29 of the report, in the first paragraph it is indicated that, initially, in one gram of nickel there was 0.011 grams of lithium-7, that is, 90 times less in weight, and in the number of cores - 90 · (7/59) = 10.7 times more .
I do not want to delve into the details of strong and weak interactions, but in nuclear reactions the number of protons and neutrons should be preserved, so let's try to count the neutrons.
Judging by the table on page 42, approximately 84% of lithium atoms are lost by the neutron. We assume that lithium was one conventional unit, so 0.84 conventional units of neutrons were lost.
On the other hand, almost all nickel turned into 62 Ni. To estimate, we will take into account only 58 Ni and 60 Ni - they were totally 93% in the initial fuel. That is, the initial 67% 58 Ni purchased four neutrons, and 26% 60 Ni purchased two. Taking into account the fact that nickel nuclei are 10.7 times larger, the neutron arrival in nickel was 10.7 · (4 · 0.67 + 2 · 0.26) = 34.2 conventional units. And this is about 40 times more than lost from lithium.
In general, here comrades not only need their own patent, sell-sell, but run to all physicists and tell them that their models are completely wrong.
UPD Thanks to Trept , who noticed an error in calculating the area, at first I managed to get 250 cm 2 without taking into account the knobs on the end - I confused the radius with the diameter. Corrected, at the same time took into account knobs, now the area is 200 cm 2 .
Convention
Go to page 2, second chapter (“Reactor characteristics and experimental setup”). Right on the second line of the first paragraph of this chapter, the dimensions of the reactor are written: 20 cm long, 2 cm in diameter, plus thickening at the end of 4 cm in diameter and 4 cm long. We obtain the total surface area taking into account the ends, a little more than 200 cm 2 . Let's read the caption under pic. 1, where it is written that small edges on the cylinder increase the area of heat dissipation, and confine ourselves to our (as can be seen, more and more underestimated) area estimate.
Now we recall about Newton's law on heat flow. Unfortunately, it is not so easy to derive the heat transfer coefficient experimentally, so you have to use the reference book. For example, the Kuchling Reference gives 5.6 W / (m 2 · K), or 0.0005 W / (cm 2 · K) for a round bill. With a temperature difference of 1200 degrees, we obtain a total heat transfer of 200 · 0.0005 · 1200 = 120 W.
This is direct heat transfer, and we also have convection, and, as we know , there is always more loss for it (the link, by the way, does not quite correctly speak about liquids, the English wiki uses a much more suitable fluid term that includes themselves and gases, and plasma, and all other similar things).
')
Temporarily, we will consider our cylinder a horizontal plate with an area of rl = 40 cm 2 and a perimeter of 2 (r + l) = 44 cm. We calculate the Rayleigh number for such a plate using the well-known formula. We need the following numbers:
- From here , we take the kinematic viscosity by taking a dynamic viscosity of 5.5 · 10 -5 kg / (m · s) and dividing it by air density at 1260 Celsius, equal to about 0.3 kg / m 3 . We will end up with a viscosity of about 1.8 · 10 -4 m 2 / c.
- The thermal diffusivity is proportional to T 3/2 , at 300 Kelvin it is 2.2 · 10 -5 m 2 / s. It means that at 1500 Kelvin it will be (1500/300) 3/2 times more, that is, about 11.2 times more - about 2.5 · 10 -4 m 2 / c.
- The coefficient of volume expansion for an ideal gas (we will try to consider air at 1500K as such) is 1/273 K -1 , or approximately 3.7 · 10 -3 K -1 .
Finally, we can get the Rayleigh number for air at this temperature, equal to about 10 7 .
Therefore, the Nusselt number for our plate is 0.54 · (10 7 ) 1/4 ≈ 30. That is, 30 times more energy is consumed in convection than direct heat exchange, which is about 120 · 30 = 3600 W.
A thoughtful reader will note that I received 120 watts for a cylindrical object, and a factor of 30 for a flat sheet. True, but, firstly, it overestimates the maximum by 4 times, and, secondly, for vertical surfaces in this case, the Nusselt number will be about the same - the Prand number for gases is almost equal to one.
Four and a half kilowatts of energy, leaving just a heat exchange with the environment - somehow do not match with a single figure in the report. So, if the masters-authors of the report would either
Note that this effect is enhanced even more, since, if memory serves me, the corresponding processes are in fact non-linear, and on temperature differences of the order of 1200 degrees Kelvin, these non-linearities already manifest themselves.
Actually, this can finish the review, but let's go further.
Nuclear physics
On page 29 of the report, in the first paragraph it is indicated that, initially, in one gram of nickel there was 0.011 grams of lithium-7, that is, 90 times less in weight, and in the number of cores - 90 · (7/59) = 10.7 times more .
I do not want to delve into the details of strong and weak interactions, but in nuclear reactions the number of protons and neutrons should be preserved, so let's try to count the neutrons.
Judging by the table on page 42, approximately 84% of lithium atoms are lost by the neutron. We assume that lithium was one conventional unit, so 0.84 conventional units of neutrons were lost.
On the other hand, almost all nickel turned into 62 Ni. To estimate, we will take into account only 58 Ni and 60 Ni - they were totally 93% in the initial fuel. That is, the initial 67% 58 Ni purchased four neutrons, and 26% 60 Ni purchased two. Taking into account the fact that nickel nuclei are 10.7 times larger, the neutron arrival in nickel was 10.7 · (4 · 0.67 + 2 · 0.26) = 34.2 conventional units. And this is about 40 times more than lost from lithium.
In general, here comrades not only need their own patent, sell-sell, but run to all physicists and tell them that their models are completely wrong.
UPD Thanks to Trept , who noticed an error in calculating the area, at first I managed to get 250 cm 2 without taking into account the knobs on the end - I confused the radius with the diameter. Corrected, at the same time took into account knobs, now the area is 200 cm 2 .
Source: https://habr.com/ru/post/362207/
All Articles