Coming in late on this.
General comments : your plastic-pipe situation is a poor model of Rossi's copper heat-exchanger manifold.
Let's give you some numbers to show you how futile this is, and how Houke's method is insufficient to model the dynamic environment in which the thermocouples reside:
1) We don't know the flow rate of the primary, but "Rossi says" it's 15 l/h, and you've never known him to lie, so let's assume 15 l/h, or 4.17 g/s
2) We don't know the pressure is, while the steam is trying to force itself out of the E-Cat, through the criss-crossing walls of the exchanger, while there collects condensed water in front of it, being forced out of the exchanger, down the table, across the floor, under the doormat, pushing any slugs of water in the way, out into the parking lot, and down the drain, but you've said "it's about 1 ATM", so let's go with that.
If the E-Cat is outputting 100% dry steam at 121.7C that condenses immediately, cooling to the output temperature at the secondary of 32.4C, it transmits:
[((121.7C - 100C) x (.48cal/gram specific heat of steam)) + 540cal/gram latent heat from phase conversion + ((100C - 32.4C) x 1cal/gram specific heat of water) = 618 cal/gram
x 4.17 g/sec = 2,577 kcal/sec
The following comments are based on my uncallibrated Spice simulations -- I don't have the NUMBERS but I did get a good feel of the situation.
The 40:1 difference in flow rate did NOT make a huge difference in the temperature profile.
I only simulated the case of water-to-water. But I don't think it will be significantly different if there's steam on one side, because the MASS FLOW will be the same, even if the volume is hugely different. At the molecular level both flows are practically standing still.
Super-heated steam (was it really 120C for Oct 6?) will simply cool down according to it's specific heat.
Saturated steam will NOT condense in that short distance and high flow rate. It will become SUPER-COOLED. See the Russian book for details.
Existing drops will grow or shrink depending on their Kelvin radius. But most of them will not be in contact with the walls of the manifold -- until they get large enough to fall out of the stream. If they do fall out then we simply have fluid water at the bottom of the tube and steam (wet or dry) at the top.
Now, what if, I know this is a stretch, not all heat transfer occurs immediately? If steam is still present after the beginning of the manifold, the steam "rushing by" may only impart the energy it takes to cool to 100C:
(121.7C - 100C) x .48cal/gram = 10.416 cal/gram
x 4.17 g/sec = 43.43472 cal/sec
That's a pretty big difference of heat energy imparted to the brass manifold. The manifold is one continuous metal block that BOTH hot and cold water flow through, albeit in their own dedicated channels. The two circumstances do not require any power output change in the E-Cat to occur. If any of the power available at the steam input is not immediately whisked away, it will necessarily heat up its environment (the manifold).
Notice that it would take 650C water to impart the same amount of energy as 121.7C steam condensing to 32.4C.
