Thomas Clarke, care to elaborate on that? :
"Andreas Moraitis
The reported 36 m^3 of water per day (according to Mats’ article the
flowmeter was positioned correctly...) matches quite well the reported
COP of 50 (20 kW electric input assumed). And even if the water had not
been vaporized but only heated from 60C to 100C, the COP would come to
about 3.5. The contract required a minimum of > 2.6, as far as I
remember.
My first calculation included the energy for heating up the water (1.674 MWh). Vaporizing requires 22.57 MWh. So together we get (22.57+1.674)MWh/(24h*20kW) = 50.5, without the 1.674 MWh the result is 47. A COP of 50 would then indicate 18.8 kW of input power. (But as always, there is no guarantee for my numbers…)"