My proposed answer is as follows, based on this FAQ plus some physical realities.
The Working state is as described above, as is the Broken Connection state, but the Shorted state is not as has been presented on various over-clocking websites in most heating applications.
Firstly, it appears that peltiers often slowly decline according to the referenced article, with increased resistance and decreased heat pump capacity.
Obviously an all-at-once failure and increase of resistance to the maximum capacity of the device can happen, but that does not mean that the heat produced by a failed 100W peltier in a heating application would be the same as if it were replaced by a 100W resistor as typically used:
Since the peltier will be installed as a heatpump, not a heater, and therefore will have a heatsink or some other ability to exchange temperature on both sides, the total heating capacity will be halved, assuming there were no other variables involved and the installation setup was the same on both the cold and hot side (same heat sink, same fan). But, in reality, since the cold-side will represent an already-colder environment and the hot-side will represent an already-hotter environment it will dissipate heat more quickly and more effectively on the cold side, at least until the two sides eventually equalized in temperature (if that ever occurred).
Therefore, my proposition is that in a typical heating application the heat produced by a perfeclty-shorted 100W peltier would be the equivalent of a 50W, poorly back-insulted, resistance element.
Best Answer
A few comments and observations: -
A single TEC1-12706 peltier thermoelectric cooler can only cool effectively when the thing that it is attached to (the aluminium pipe) isn't conducting ambient heat from the surrounding environment back to the cooler at too high a level.
For instance the TEC1-12706 is rated about 50 watts and let us assume that all this power is converted to the cooling process. Let's say the aluminium pipe has a thermal resistance of 1 degC per watt i.e. if 1 watt of heat power were pushed into the pipe it would rise in temperature by 1 degC.
Or, put it another way, if 50 watts of heat power were extracted from the pipe it would cool by 50 degC. This would cool the pipe down to -25 degC in an ambient of +25 degC.
However, if the pipe has fluid flowing through it at ambient temperature then it won't cool down to -25 degC or even close because that fluid is imparting heat to the pipe and that means you need a bigger and more powerful peltier to extract the power due to the pipes basic thermal resistance AND the power that needs to be taken to cool the fluid.
Given that a Peltier cooler is probably around 15% efficient you have to consider that the heat power you can extract from an object is significantly less than 50 watts.