On Jan 22, 2011, at 5:24 PM, Jeff Driscoll wrote:

Rossi used this electronic device for electronic measurement (aswas reported) - model HD37AB1347. Relative Humidity probe modelHP474AC was attached to it.Page three of this link (thanks to Horace) shows details of thatprobe connected to the electronic device. HP474AC has thefollowing specifications:http://tinyurl.com/45rwsvh HP474AC Relative Humiditiy Probe specifications: 5% to 98% RH >>> -40C to 150 C +/- 2.5% (5%...95%RH) +/-3.5%(95%...99%RH) Temp +/-0.3CThis probe does not measure the amount of liquid water droplets inthe "steam" (ie. mass fraction of water vapor to to total water).It measures Relative Humidity (Relative Humidity measures howsaturated the air is for a given temperature).What we want is a a device that measures "quality" of the steam.For reference, 100% quality = 100% vapor.How did they confirm that the water vapor was truly vapor?

Yes, good question.

`I believe the HP474AC probe actually measures the capacitance of the`

`air, and converts that to relative humidity. The more the`

`capacitance, the more water in the air, by volume. Another`

`important thing is heat content is carried in proportion to mass,`

`not volume. I have appended the computations I posted earlier`

`showing the huge proportion of mass that is contributed by a small`

`volume of liquid, and that estimates of the heat flow from the device`

`can be off by 96%, i.e only 4% of the estimated heat value due to`

`vaporization, if only 1.4% of the volume flow is liquid water`

`droplets. Therefore a very small error, less than 1%, in measuring`

`capacitance can produced huge errors in calculated heat flow. The`

`stated error of the probe is +-3.5% where it counts, at 99% water`

`content.`

It is also notable the meter/probe requires calibration: http://tinyurl.com/4z5985v

`Most important is the fact the probe is designed to detect the`

`percent of water vapor in air, not percent of water microdrops in`

`pure steam. Pure vapor should have more capacitance than 100% humid`

`air, and be way beyond the meter's measuring limits. Adding water`

`droplet should push the capacitance even higher. Once the meter is`

`maxed, the question arises: can extra water droplets make any`

`difference to an already maxed out 100% reading? The +-3.5% error`

`could thus actually be irrelevant.`

`This whole issue may be of academic interest only. Even if all the`

`heat flow due to vaporization is negated, the COP is still over`

`unity, assuming the water is not heated much above 13 °C by ambient`

`conditions before entering the device. Further, if the device can`

`run without energy input at all, then none of this matters, provided`

`the total energy to start up the device is way less than the device`

`produces. This would clearly be the case if the device can run 6`

`months as stated.`

`Here again is my analysis showing the importance of the huge`

`difference in mass vs volume ratios:`

From: http://en.wikipedia.org/wiki/Water_(properties) http://en.wikipedia.org/wiki/Specific_heat_capacity http://en.wikipedia.org/wiki/H2o http://hypertextbook.com/facts/2007/DmitriyGekhman.shtml

`The following approximate values for water can be used from the above`

`refs:`

Liquid Density: 1000 kg/m^3 = 1 gm/cm^3 Heat of vaporization: 40.6 kJ/mol = 2260 J/gm Heat capacity: 4.2 J/(gm K) Molar mass: 18 gm/mol Density of steam at 100 C and 760 torr: 0.6 kg/m^3 = 0.0006 gm/cm^3

`Now to examine the importance of mass flow vs volume flow`

`measurements for the steam.`

`If x is the liquid portion by volume, then x/((x+(1-x)*0.0006)) is`

`the portion by mass. This gives the following table:`

Liquid Liquid Gas Portion Portion Portion by Volume by Mass by Mass --------- ------- ----------- 0.000 0.0000 100.00 0.001 0.6252 0.3747 0.002 0.7695 0.2304 0.003 0.8337 0.1662 0.004 0.8700 0.1299 0.005 0.8933 0.1066 0.006 0.9095 0.0904 0.007 0.9215 0.0784 0.008 0.9307 0.0692 0.009 0.9380 0.0619 0.010 0.9439 0.0560 0.011 0.9488 0.0511 0.012 0.9529 0.0470 0.013 0.9564 0.0435 0.014 0.9594 0.0405

`We can thus see from this table that if 1 percent by volume of the`

`steam is entrained water micro-droplets, easily not seen in tubing or`

`exhaust ports, that only 5.6 percent of the heat of vaporization is`

`required to produce that mixture.`

Rough calculations based on Rossi specifics:

`Suppose for the Rossi experiment the mass flow of a system is 292 ml/`

`min, or 4.9 gm/s, the inlet temperature 13 °C.`

The delta T for water heating is 100 °C - 13 °C = 87 °C = 87 K. If the output gas is 100% gas, we have the heat flow P_liq given by: P_liq = (4.9 gm/s)*(87 K)*(4.2 J/(gm K))= 1790 J/s = 1.79 kW and the heat flow H_gas for vaporization given by: P_gas = (4.9 gm/s)*(2260 J/gm) = 11.1 kW for a total thermal power P_total of: P_total = 1.79 kW + 11.1 kW = 12.9 kW Now, if the steam is 99% gas, we have: P_liq = 1.79 kW P_gas = (0.1066)* (11.1 kW) = 1.18 Kw P_total = 1.79 kW + 1.18 kW = 2.97 kW or 23% of the originally estimated power out. It thus seems reasonable to do calorimetry on the steam-liquid out. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/