http://lenr.qumbu.com/fake_rossi_ecat_frames_v318.php
The Jan/Feb experiment reports say they did NOT check the end of the outlet pipe.
The March experiment says they DID make a visual check.
I did the calculations on heat leakage from the outer HOT compartment to the inner COLD tube -- for a thick rubber tube, it's at most 80W, out of the 300W input.
This fake is plausible enough that it MUST be checked against -- specifically, by measuring the output temperature and steam dryness OUTSIDE of the eCAT.
Cut-and paste from 3.18 follows, but the format may be erratic :
1.1. Arguments about the Torelli Fake
-
Rothwell Okay, what insulator would you use inside the pipe? You need
something thin enough to allow the water through and yet effective at 1
L/s with a 31°C temperature difference for the apparent 130 kW heat in
the Feb. 10 test.
- 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).
- 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 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.
Why wouldn't Levi et al. notice that they cannot insert thermocouples more than a faction of the way into the hose? (If they could insert it and it did not penetrate the barrier, it would block the middle channel.)
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 :
Vortex
- It'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.
- It'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.
-
Rothwell However, in this test, the outlet temperature is measured in
the "chimney." That is large enough to implement this trick
fairly easily, with steam passing the thermocouple, well insulated from
the stream of cold water. However, the outlet tube would be close to tap
water temperature, which would be a dead giveaway. I am assuming someone
had the sense to hold a hand near it or touch it for a moment. I would do
that the moment I saw the test.
- 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.
- 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.
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
