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An ion heater manufacturer claims their ion heaters are a lot more efficient than common electric heaters:

How ion boilers (heaters) STAFOR can reach COP 1,57?

In solid conductors only half (statistically) of the free electrons react to the applied voltage. In liquid conductors, if using special construction ionization chamber, it is possible to achieve that almost all of the free electrons will react to the applied voltage. This means that - theoretically it is possible to achieve COP 2. At the moment company STAFOR EKO, ltd. (www.stafor.lv) managed to achieve COP 1,57. It is confirmed by certification of European Union approved Center of Metrology, protocol № 016TP11.

The argument about "only half of the free electrons" sounds like pseudo-science to me, or perhaps the COP (Coefficient of Performance) value is real, but it does not mean what it appears to mean? COP is used to compare heat pumps, but in case of a ion heater I cannot see anything gaining the energy from the environment. Yet the next paragraph says it is an "economical .. product".

Oddthinking
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Suma
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  • Perhaps you could clarify your concern. I cannot understand it? Also, are you unhappy with the [certification](http://www.stafor.lv/wpic/documents/STAFOR_ion_boilers_COP.pdf) provided? – Oddthinking Apr 17 '12 at 09:52
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    I find it strange that I would get more energy then I put into the heater from it, given there is no additional external energy input (and the manufactures does not speak of any). Such claim looks to me like contradicting the law of conservation of energy, and I would need to than a EU certification to accept that. – Suma Apr 17 '12 at 11:02
  • ... or approaching the problem from the other side: if the system is really able to give me 1.57x more heat than the energy I pay for, it is quite a significant achievement and I would be surprised if the only manufacturer using such technology would be a Latvian company. Why did not GM bought the company yet? Electric heater is one of the most efficient appliances possible, as any "lost" energy is converted to heat anyway. It is hard to be more than 100 % efficient. – Suma Apr 17 '12 at 11:13
  • @Suma this is a simple electrodynamics question. If it get time ill hunt up some links and post an answer I am skeptical you will understand. – Chad Apr 17 '12 at 12:30
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    @Suma: I remember making the exact same argument to my father when I was about 20. He won. Heat pumps can be more than "100% efficient", because they get their energy from outside sources, and hence the term "efficiency" is deprecated here. [Ref](http://en.wikipedia.org/wiki/Heat_pump#Efficiency). The analogy than won me over: imagine a simple device that pipes molten lava from a nearby lava stream into your home. It will heat up the home a lot faster than an equivalently-powered radiator. – Oddthinking Apr 17 '12 at 12:39
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    Yes, I understand that, however I fail to see the external source in this case. The ion boiler seems to be only heating the medium, with no access to any external source. – Suma Apr 17 '12 at 15:33
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    That certificate reference number comes up as a test certificate for a wi-fi router: http://www.i4wifi.cz/img.asp?attid=245284 . Ion boiler: dodgy physics, dodgy certificate, and no scientific papers. – 410 gone Apr 18 '12 at 07:09
  • @Suma: Ah! Good point. – Oddthinking May 02 '12 at 03:23
  • Heat pumps are the way to get better more thermal input than your power. This is well understood thermodynamics and has only been known since the late 19th century. – dmckee --- ex-moderator kitten Dec 01 '13 at 00:41
  • ALL electrical energy becomes heat. All electrical devices are 100% efficient at turning electricity in to heat. The only exception is radiant energy (light or radio waves) escaping the building. You cannot build a "more efficient" electric heater. Every single watt you use for any purpose becomes heat. –  Nov 23 '15 at 13:18

1 Answers1

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TLDR: No, it is not more than 100 % efficient, the test performed was probably invalidated by using a heat meter assuming water as a heat transfer fluid, while propylene glycol/water mixture was actually used.

Detailed:

Some details about how test was performed can be found on a Czech reseller site:

  1. Ion boiler STAFOR 3-5
  2. Circulation pump Vilo 25/7
  3. Radiator Korad T22 300/400 500w
  4. Ion boiler control panel STAFOR 3-5
  5. Heat meter Dunfos Sonometer 1100
  6. Electricity analyzer Fluke 430
  7. Thermostatic sensor
  8. Hydraulic protection group
  9. Performance checking stand made from copper pipes

Details about how those pieces were connected are not described, but most likely it was more or less what can be seen in the YouTube video at 0:07.

When checking for individual pieces of equipment used, one thing looks very suspicious, and that is Danfoss Sonometer 1100 is intended to be used with water as a circulation medium. In this test a different liquid was almost certainly used, as the liquid required by the boiler to operate is heat carrier STATERM EKO E40, which is "made on the basis of propylene". By "propylene" the propylene glycol is most likely meant here (melting point -59°C), which is commonly used as a Heat-Transfer Fluid (it could hardly be a propene, which is highly flammable gas with boiling point of -47.6°C.).

As the ultrasonic heat meters compute the energy by measuring flow and temperature and multiplying measured volume with specific heat capacity, by assuming water properties on Propylene Glycol water solution an error is made.

Propylene Glycol based Heat-Transfer Fluid with boiling point 120°C is approx. 85 % Propylene Glycol (see Boiling Point of Aqueous Propylene Glycol Solutions), with specific heat capacity 2.51*0.85+4.18*(1-0.85) J/(gK) = 2.76 J/(gK) (see propylene glycol specific heat capacity and water specific heat capacity), which is 1.51 x lower than water heat capacity, therefore apparent heat transfer is approx. 1.51 x higher than it really is. Given Propylene Glycol concentration is not given, I only estimate it, I found the result close enough to be likely explanation of the 1,57 COP result.

Suma
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  • I think you are misinterpreting the efficiency measure (based on your comments above). From Wikipedia's COP page: "The COP may exceed 1, because it is a ratio of output:loss". COP > 1 does not imply any gain of energy. – adam.r Nov 27 '13 at 18:14
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    @adam.r As discussed in comments to the question, that description of COP is for [heat pumps](http://en.wikipedia.org/wiki/Heat_pump) which transfer heat from "outside"; however this (an electric heater using pumped water/glycol) is a closed system. – ChrisW Nov 28 '13 at 03:14