Proposed by : Flavio Tarallo (as reported by Rothwell) : Vortex
This is a very serious fake. It would possibly not have been detected by the January experiment or the February experiment.
ALL of the measurements in ALL of the experiments were done via the instrument port in the chimney, and would only have measured the temperature and steam dryness in the outer compartment.
The water sent through the "reactor chamber" is simply heated by a resistor, powered from the control box.
The observed "temperature profiles" could easily be accomplished by changing the power sent to the resistor: the connection between the control box and the main unit was NOT measured.
1.2. Vortex Discussion about the Torelli Fake
The duration of the 130kW was unspecified, so I'll stick with the March experiment. If we use the above diagram, with a 3 cm diameter outer section, and put a 1cm diameter bypass through the device, with a diversion ratio of 1/16 to 15/16 (roughly equal to the given power ratio). The bypass tube could be rubber (a reasonable insulator).
Thermal Conductivity : watts / m.K Rubber is 0.16 :
Area A = 2 * 3.1415 * 0.500 * 100.000 = 314.15000000 cm2 Thickness Y : 0.500 cm dT = 100 - 20 = 80.000 k of rubber = 0.160 Watts = k * dT * A / Y = 0.160 * 80.000 * 0.03141500 / 0.005 = 80.4224Even this value could be handled in the March fake, which had 300W of electrical power available. : the amount of water diverted would have to take this loss into account. Since the water would warm up as it passes through the reactor the average dT would actually be less than that just calculated.
The bypass could be engineered to minimise this loss. It could be arranged to be in contact with the outer wall, so the effective transfer area would be reduced. Wrap it in a (waterproofed) Silica Aerogel (0.004 to 0.04) and the problem goes away.
Since the horizontal arm was not unwrapped the bypass tube could have been completely OUTSIDE the reactor chamber section, entering at the bottom of the chimney. Only about 20 cm would be inside the device, so the power loss would be less than 20 W.
They didn't insert it into the hose. They inserted it into the instrument port.
The above diagram shows the bypass tube in the center of the chimney. It could be at the side (or even hidden by a false wall near the instrument port).
Stephen A. Lawrence : VortexIt's totally ruled out if the effluent is observed to be steam and the output temperature is claimed to be roughly 100C. Whether wet steam or dry steam, if it's coming out as steam, then the outlet temperature is at least 100C, and the placement of the temperature probe is irrelevant. During the first test, back in ... uh ... January?, the effluent was observed to be steam during at least part of the run, and this effect couldn't have been an issue.
The probe was always placed in the outer "Hot" compartment. Nobody reported in January that they checked for steam coming OUT of the hose. Nobody reported in February that they measured the temperature of the water coming out of the hose.
To make the black tube hot, you have to imagine there is barrier within the hose that allows a thin layer of steam to pass on the outside at a high temperature without being cooled by the water in the center. It is moving at 1.7 ml/s. From the photo I suppose that hose is 1 cm OD and 0.8 cm ID, which is to say a volume of 0.5 cm^3 per centimeter of hose. So if the water is liquid, it is moving about 4 cm per second. It would cool down a short distance from the chimney. Real steam moving out of that hose would reach a lot farther than 4 cm per second, and heat the entire hose. There would still be a lot of steam coming out of the end of the hose in the bathroom.
The output tube used in March is considerably thicker than 1 cm. Looks about 2 cm at least. The fake I'm proposing does have a coaxial tube for some arbitrary distance in the output hose, so the outside of the hose will indeed feel hot. The thermal conductivity analysis above could be applied here.
Thinking aloud here ... first see how long it takes for the diverted water to go through the tube.
flow 6.470 L/Hr = 1.797 cc/sec Hot: 0.112 cc/sec Cold: 1.685 cc/sec inner tube radius : 0.500 cm length : 100.000 cm volume : 78.538 cc time in reactor : 46.613 secs
A new test had just been released, with pictures of a smaller 5kW device with and without shielding and insulation.
Fig 5 Fig 2 Fig 4 Fig 3
Measurement of Sizes from the Photos
Ruler rotated in photoshop
The total volume of the REACTOR was estimated as a sphere::
Reactor Sphere diameter: 7.00 cm
volume: 179.59 cm3 = 0.1796 L
Since the horizontal arm was NOT inspected, I have estimated its volume:
Scale ruler from fig 3 onto Fig 2 = 10cmFig F2
Repeat for Length of Horizontal arm = 30cm (Any foreshortening would over-estimate the length.)
Scale rule to diameter of reactor = 7 cm
Step and Repeat for diameter of horizontal arm = 25cm
Horizontal Arm : cylinder, length: 30.00 cm diameter: 25.00 cm
volume: 14725.78 cm3 = 14.7258 L
3.1. Wiki Energy Densities
The Wiki of Energy Densities doesn't have entries for all cases where (for instance) Hydrogen is used with Compressed or Liquid Oxygen.
This section calculates how much the "wiki" Energy Densities by Mass and by Volume are reduced if the available space has to be shared between the Hydrogen and Oxygen.
3.2. Reference Documents
3.3. Atoms and Molecules
H mass 1.0071 H2 mass 2.0142 Gas Density : 8.988E-5 kg/L Compressed Density : 0.046647636700649 kg/L (700 Bar) Liquid Density : 0.07099 kg/L O mass 15.999 O2 mass 31.998 Gas Density : 0.001429 kg/L Compressed Density : 0.74164967562558 kg/L (700 Bar) Liquid Density : 1.141 kg/L B mass 10.811 Solid Density : 2.52 kg/L
3.4. Compressed hydrogen + Compressed Oxygen
Formula: 2 H2 + O2 ===> 2 H2O H2 (Compressed H2) O2 (Compressed O2) H2 mass : 2 * 2.014 = 4.028 O2 mass : 1 * 31.998 = 31.998 Total mass : 36.026 H2 mass fac : 4.028 / 36.026 = 0.112 O2 mass fac : 31.998 / 36.026 = 0.888 Volume : mass / density H2 volume : 4.028 / 0.047 = 86.358 O2 volume : 31.998 / 0.742 = 43.144 Total vol : 129.502 H2 vol fac : 86.358/129.502 = 0.667 O2 vol fac : 43.144/129.502 = 0.333
3.5. Liquid hydrogen + Liquid Oxygen
Formula: 2 H2 + O2 ===> 2 H2O H2 (Liquid H2) O2 (Liquid O2) H2 mass : 2 * 2.014 = 4.028 O2 mass : 1 * 31.998 = 31.998 Total mass : 36.026 H2 mass fac : 4.028 / 36.026 = 0.112 O2 mass fac : 31.998 / 36.026 = 0.888 Volume : mass / density H2 volume : 4.028 / 0.071 = 56.746 O2 volume : 31.998 / 1.141 = 28.044 Total vol : 84.790 H2 vol fac : 56.746/84.790 = 0.669 O2 vol fac : 28.044/84.790 = 0.331
3.6. Boron + Compressed Oxygen
Formula: 4 B + 3 O2 ===> 2 B2O3 B (Solid B) O2 (Compressed O2) B mass : 4 * 10.811 = 43.244 O2 mass : 3 * 31.998 = 95.994 Total mass : 139.238 B mass fac : 43.244 / 139.238 = 0.311 O2 mass fac : 95.994 / 139.238 = 0.689 Volume : mass / density B volume : 43.244 / 2.520 = 17.160 O2 volume : 95.994 / 0.742 = 129.433 Total vol : 146.593 B vol fac : 17.160/146.593 = 0.117 O2 vol fac : 129.433/146.593 = 0.883
3.7. Boron + Liquid Oxygen
Formula: 4 B + 3 O2 ===> 2 B2O3 B (Solid B) O2 (Liquid O2) B mass : 4 * 10.811 = 43.244 O2 mass : 3 * 31.998 = 95.994 Total mass : 139.238 B mass fac : 43.244 / 139.238 = 0.311 O2 mass fac : 95.994 / 139.238 = 0.689 Volume : mass / density B volume : 43.244 / 2.520 = 17.160 O2 volume : 95.994 / 1.141 = 84.131 Total vol : 101.292 B vol fac : 17.160/101.292 = 0.169 O2 vol fac : 84.131/101.292 = 0.831
3.8. Magnesium and Steam
Magnesium combines with STEAM to produce Magnesium Oxide and Hydrogen.
Mg + H2O ==>MgO + H2 dH -360 kJ/mol
Total (from H2) is -286 kJ/mol.
Atomic weight: 2.01 g/mol.
Density (L): 0.0710 g/cm3
Energy by mass : ( 286 kJ/mol)/(2.01g/mol )
= 142 kJ/g = 142 MJ/kg
Energy by volume: 10.1 MJ/L
Comparison to Wiki values
|Energy by Mass||Energy by Volume|
Total (from Mg/Steam + H/O) is -604.8 kJ/mol. Atomic weight : 24.305 g/mol Density : 1.738 g/cm3 Energy by mass : ( 605 kJ/mol) / (24.305 g/mol) = 24.9 kJ/g = 24.9 MJ/kg Energy by volume : 43.2 MJ/L
Total (from Ni/H) is -4.85 kJ/mol. Atomic weight : 58.6934 g/mol Density : 8.908 g/cm3 Energy by mass : (4.85 kJ/mol) / (58.6934 g/mol) = 0.0826 kJ/g = 0.0826 MJ/kg Energy by volume : 0.736 MJ/L
3.9. Magnesium (+ Steam) + Compressed Oxygen
Formula: Mg + H2O ===> H2 THEN H2 + 1/2 O2 ===> H2O Mg (Solid Mg) O2 (1/2 O2) Mg mass : 1 * 24.305 = 24.305 O2 mass : 0.5 * 31.998 = 15.999 Total mass : 40.304 Mg mass fac : 24.305 / 40.304 = 0.603 O2 mass fac : 15.999 / 40.304 = 0.397 Volume : mass / density Mg volume : 24.305 / 1.738 = 13.984 O2 volume : 15.999 / 1.141 = 14.022 Total vol : 28.006 Mg vol fac : 13.984/28.006 = 0.499 O2 vol fac : 14.022/28.006 = 0.501
3.10. Embedded Calculations
All calculations in this document are performed with the PHP programming language which generates the document.