# Comments and Calculations on the 1MW Reactor Data (Michael Lammert)

Comments and Calculations on the 1MW Reactor Data

Michael Lammert (AKA Dr. Mike)

February 5, 2017

I along with most E-cat World readers have been patiently awaiting the “report” on the results of the nearly one year operation of Rossi’s 1MW reactor.  Although the data released as part of the Rossi vs. IH court case is without supporting explanations, I believe there is almost enough information just in this data to determine how successful the 1MW reactor was.  Some of my observations in the data include:

1. The water flow was reduced by 10% in the calculation of total output power.  My assumption is that this is to allow for the steam being only 90% dry, that is, 90% of the input water is converted to steam and 10% of the input water travels in the steam as small water droplets.
2. The “0” bar output pressure of the steam is assumed to mean the output steam is at atmospheric pressure (1atm).  Why the pressure was not measured with a sensor that gave a more exact value of the pressure is totally unexplainable.
3. There is no data indicating how the water content of the steam was measured.  (This is the reason that I say the available data is “almost” sufficient to determine the success of the 1MW reactor.)
4. It is evident that all four 250KW reactors were not working all of the time.  Calculations will be based on times that all four reactors were operating.
5. The calculated COP’s are so high that the accuracy of the determination of the input power does not appear to be a critical issue in establishing the success of the 1MW reactor.

Calculations

Mass Flow and Heat Requirements

The water flow was 36000 Kg/day, or 417g/sec, or 23.12moles/sec.  Assuming that the steam is 90% dry, then the energy required to convert the ~69ºC input water to steam is:

Energy/sec to heat water to 100ºC = 417g/sec x 4.184J/g/ºC x (100-69)ºC = 54043J/sec

Energy/sec to boil 90% of water = 23.12moles/sec x 40.65KJ/mole x 0.9 = 849500J/sec

Total Energy/sec = 54043J/sec + 849500J/sec = 900000J/sec (.9MW-sec/sec)

Note that Rossi uses a 10% reduced flow to calculate the COP, which is equivalent to a full flow but assuming that only 90% of the water is converted to steam.  If all of the water was actually converted to steam, then the reactor would put out the full 1MW of power.  The COP figures in the data would actually be >10% higher than listed.

Steam Flow Rate

One issue with the 1MW reactor is the output steam flow velocity, which of course will depend on the output tube diameter, as it correlates to the output steam pressure.  For my calculation I assumed an output tube diameter of 15cm (~6 inches).  My calculation would scale with the square of the output tube diameters for other output tube diameters.  My initial calculation will assume 100% of the water is converted to steam.  Assuming a 15cm diameter output tube, the number of moles of steam in a one meter length of tube can be calculated from the ideal gas law where the volume is:

V = πr2l = 3.14 x 15cm/2 x15cm/2 x 100cm = 17670cm3 = 17.67L

And the number of moles per meter length of tube is:

#moles = n = PV/R/T = 1atm x 17.67L / 0.08206L-atm/mole/ºK / 377ºK = 0.571moles

The water flow rate is 23.12moles/sec, and therefore by mass balance the steam rate must also be 23.12moles/sec.  The steam velocity is:

Steam velocity = 23.12moles/sec / 0..571moles/m = 40.5m/sec

If the steam contains 10% water, the moles/meter of tube length will be 10% greater than calculated above and the flow rate would be 10% less (36.8m/sec) to maintain mass balance.  As stated above the velocity will ratio with the square of the tube diameter for other output tube diameters.  It has been discussed among e-catworld readers that the diameter of output steam tube was only 4cm, which would result in an unreasonable steam velocity of 569m/sec (supersonic).

Actually, the steam velocities would be much less than these simple calculations that are based on moving mass through the output tube.  The pressure would rise significantly above 1atm so that a unit volume would contain a much higher mass of steam.  It seems quite unlikely that 23.12 moles of water was being converted to steam each second if the output steam pressure was really 1atm.

Steam Volume Calculation

Another calculation useful for evaluating the 1MW data is the volume of the steam generated again using the ideal gas law equation:

Vol steam/sec = nRT/P = 23.12moles/sec x 0.08206L-atm/mole/ºK x 377ºK / 1atm =

= 715.3L/sec

This volume can be compared to the total volume of the steam output tube, which for this calculation will be assumed to be the same 15cm diameter tube with a length of 10 meters:

Volume output tube = πr2l = 3.14 x 7.5cm x 7.5cm x 1000cm = 176700cm3 = 176.7L

Therefore, the steam generation rate per second would be about 715.3L / 176.7L = 4.05 times the volume of the output steam pipe.   The possible conclusions from this calculation are that the steam pressure in the output pipe would be much greater than 1atm or the steam generation rate was much less than 23.12moles/sec.

COP Calculation

The COP calculation in the data ignores the energy needed to heat the water to 100C and uses a 10% reduced water flow to calculate the output energy (energy needed to convert this amount of water to steam).  For example, on 2/26 the total input energy was 252000W-hr/day and the total energy to convert the 10% reduced flow of 32400Kg/day of water to steam was 2.03E7 W-hr/day, yielding a COP of 80.7.  If it had been assumed all of the water was converted to steam and the energy to heat the water to 100ºC was included in the COP calculation, the COP would have been about 95.

The COP can also be computed for just heating the water from about 69ºC to 100ºC using the calculation of “Energy/sec to heat water to 100ºC = 54043J/sec” from above:

COP = Energy Out/Energy In = 54043W-sec/sec x 24hr/day / 252000W-hr/day = 5.15

Therefore, even if no steam was produced, the system had a very respectable COP of greater than 5.  Also, only a little over 5% of the water flow would have had to be converted to steam to achieve a very good COP of >10.

Conclusions

The recent data released on the Rossi’s 1MW reactor indicates that he has achieved an excellent COP, even assuming that 10% of the water flow is not being converted to steam.  However, is this assumption correct?  There does not appear to be any information in the data that indicates the steam water content was actually measured.  Both a calculation of the volume of the steam generated for the measured water flow and the steam velocities in the steam output pipe indicate that it is unlikely a significant portion of the water flow was converted to steam if the output steam pressure was really 1atm.  Hopefully the real report will reconcile the uncertainty in the steam output pressure with additional data and will also give a reasonable methodology for determining the water content (dryness) of the steam.  It seems rather important that the output steam velocity was measured or some other measurement of steam dryness was made to confirm how much water was turned to steam in the reactor so that an accurate COP can be calculated.

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