{"id":4715,"date":"2014-10-13T15:16:55","date_gmt":"2014-10-13T14:16:55","guid":{"rendered":"http:\/\/www.theoillamp.co.uk\/?p=4715"},"modified":"2014-10-13T15:16:55","modified_gmt":"2014-10-13T14:16:55","slug":"hydrogen-economy-myth-part-2","status":"publish","type":"post","link":"https:\/\/www.theoillamp.co.uk\/?p=4715","title":{"rendered":"The hydrogen economy myth-part 2"},"content":{"rendered":"
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Plenty of water- what’s the problem?<\/p><\/div>\n

In the last post we had a look at the background science which is necessary to consider the question of the hydrogen economy<\/a>.\u00a0 Before we look at the issue of cell efficiency and some of the other problems we need to introduce one more equation.\u00a0\u00a0 In a fuel cell we are dealing with electrical work and the following equation relates free energy (useful work- see last weeks post) to the voltage across the electrodes.<\/p>\n

\u0394G= -nFE<\/p>\n

where E is the voltage difference generated between the anode and the cathode, n is the number of electrons and F is a number known as the Faraday constant which is a measure of the electrical charge per mole of electrons.<\/p>\n

If we calculate E for the hydrogen fuel cell reaction given last week we get a theoretical maximum voltage of either 1.229V or 1.18V depending on whether the reactants are in a liquid or gaseous state.\u00a0 This does not sound like a lot, but in actual fact is not a problem since individual cells can put in parallel or series to build up the current or voltage.<\/p>\n

What is more important looking at our hydrogen fuel cell is that of cell efficiency.\u00a0 Working out the maximum efficiency for the reaction is easy.\u00a0 \u0394G and \u0394H are known and are listed in the thermodynamic tables.\u00a0 Dividing \u0394G\/\u0394H and multiplying 100 gives a value of 83%.\u00a0 Since this reaction goes both ways its the same efficiency both ways.\u00a0 So if I was to take 1kW of electricity and use it to convert water to hydrogen its easy to see I would have lost 17% of my energy and be left with the equivalent of “830 Watts” of hydrogen.\u00a0 Converting it back to electricity using the reverse of the first reaction in my fuel cell would leave me with 83% 0f the 83%, this is 69% of our starting energy (rounded up).<\/p>\n

Nearly 70% sounds pretty good.\u00a0 However there are a whole heap of other losses that have to be taken into account.\u00a0 These are briefly;<\/p>\n