Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is an engineering question, but it is adressed to physicists who build accelerators. This question: An electromagnetic space elevator? notices that a NbSn superconducting ring around the equator will launch itself into orbit. The question is whether there is enough He to cool such a massive structure.

What is the minimum amount of He required to maintain a stable 4-degree temperature safely throughout a long cylindrical wire? Is is necessary to use a fluid at all, or can one stably and safely maintain a stable He-range temperature in a very long wire without a liquid coolant?

share|cite|improve this question
up vote 3 down vote accepted

This answer is only an order of magnitude estimation. The main uncertainty comes from the fact that for a constant current the cooling power is not determined by the superconducting wire but from the quality of the isolation against radiation, convection and conduction of warmer parts of the structure to the cold wire at 4.2K

We can use the LHC ring as an estimate of how much helium might be necessary. The CERN uses approximately 120 tonnes of liquid helium to cool a ring with a circumference of 27km. To keep the temperature constant 8 compressor stations, each with a power of 18kW are used.

Comparing this with the diameter of the earth we arrive at 177 777 tonnes of liquid helium.

Precises estimates of the world wide helium reserves are not easily available, so one might used the data for the US as a first guess. The currently available reserves are estimated to be 147 billion cubic meters (Wikipedia). As the main source for helium is natural gas other areas of the world also have large reserves, and this is only a lower bound.

Now this amount is converted: $$\mathrm{mass_{He}} = 147\cdot 10^9\, \mathrm{m}^3 \cdot 0.18\, \mathrm{ kg/m^3} = 2.6\cdot 10^{10}\,\mathrm{kg}$$ Which is much more than the $1.7\cdot 10^{8}\,\mathrm{kg}$ necessary for a ring with similar specifications as the LHC. You can greatly reduce the necessary amount by using pulse tube refrigerators and relax the specifications a bit by allowing slightly higher temperatures. Then the conduction via the metal itself might be enough to cool the wire as NbSn wires are embedded in copper.

share|cite|improve this answer
I think this is not a very good estimate, as Ron is asking about one cable. The LHC have 7600 kilometers of superconducting cable ( ) comparable to the earth's radius ( 6,353km) so a factor of two pi is missing for the periphery and for your helium estimate from the LHC helium, about 700 tonnes. the problem would be with the containers and the electronics etc. – anna v Sep 16 '12 at 11:55
@annav: A thin wire of NbSn is only 0.006 mm thick, the amount of helium to cool 40000 km of this is tiny, a standard 100L dewar is probably enough but this is highly misleading. You cannot have the wire without the vacuum infrastructure. So given the largest vacuum ring that we currently have as an example it seems that we have more than enough helium. – Alexander Sep 16 '12 at 12:43
@Alexander: This estimate is not that terrible, you also found an estimate for the He reserves, which I couldn't find. The LHC is not minimizing He usage either, and this thing will want to cool something like 10 cables each of many meters radius. But the estimate isn't bad, because if you immerse it in coolant, the heat loss is surface, not bulk, so scaling up is roughtly right. – Ron Maimon Sep 16 '12 at 19:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.