You're right, I kinda knew it needed cooling, but I guess your point is even though the effect being exploited doesn't require a temperature gradient, in practice the heat dissipation requirements are the same if the TVP is to survive - or are they different... surely for the purposes of satisfying the first law the temperature dissipation requirements are dependent on the efficiency of conversion?

Which would make the cooling requirements of this TVP lower (relative to input) due to it's higher efficiency? but still substantial.