At the time, my vessel was just a large stockpot (All-Clad anodized aluminum (a wedding gift)), uninsulated. Like I said before, my PID controller has a function that displays percent output. So, while the bath is heating and the heating element is on full power, the controller shows 100% - which is 1000W since I was using a 1000W element. As the bath reaches temperature, the controller begins cycling the power to the element. I was most interested in power used once the bath reaches steady state, comparing the covered vs. non-covered values at various temperatures.
So, obviously, the covered pot was much more efficient, since it didn't constantly lose temperature due to evaporative cooling - which is much more substantial (depending on bath temperature) than the loss due to radiation/convection with the air around the pot - especially if there is no breeze.
Also, keep in mind that in my system, the heater is not submerged in the water, but instead, heats the pot like an electric burner. This type of heating will be less efficient than the submerged type.
Anyway, at a bath temp of 185F, the covered pot required an on-cycle of about 20% (200W). An uncovered pot used more than double that amount at that temperature. At 140F, the covered pot required an on-cycle of about 10% (100W), while the uncovered version used about 50% more (150W). At my preferred salmon cooking temp. of 115F (I gradient cook my salmon using SV dash), the covered and uncovered values are practically identical, at about 4-5% (40-50W).
I never tested an insulated bath because, at the time, I was designing the circulator with an eye towards production for the mainstream. So, it was designed to look less like lab equipment, and more like a kitchen appliance. How many mainstream people can you imagine wrapping their pot with insulation? And, once I saw the steady state values and compared them to traditional cooking methods, I figured that an extra few percent (meaning maybe 100W max.) would be inconsequential to the eventual marketing. SV, by its nature, is already orders of magnitude more efficient than using your oven.
Like pbear said, given enough time, even the best insulation will eventually equilibrate to the bath temperature. But, even though the outside may be warm to the touch, I'd guess that the power outputs would be less than an uninsulated bath. If you think of an analogy to a wetsuit when diving, the outside of the wetsuit will still feel warm, however, the wearer is still significantly warmer wearing it than not. The wetsuit traps water between it and the wearer, and the dead water, once warmed to body temp, doesn't require much more heat to stay at equilibrium. So, let's say you were diving in 50F water - with the suit, you might get uncomfortably cold after about an hour or two (I found this out by experience years ago), but without the suit, you wouldn't last 5 minutes!
Paul - in your situation, I wonder if the Reflectix is doing very much, other than limiting evaporative cooling. It might be just as well to use the tight fitting lid without it. Once the air space between the water and lid becomes saturated at 100% humidity, evaporation will stop, assuming there are no leaks where humidity can escape. At a certain point, increasing side insulation is "gilding the lily" so to speak - a lot more effort for not much gain.
ETA: Paul - Maybe a way to save a bit more energy would be to wrap the cooler with a reflective surface - like aluminum foil or something to reduce the radiation losses. Having a surface with lower emissivity should help. So then, the primary heat loss would be through convection, which, unless there's a breeze, should be minimal.
Edited by KennethT, 20 June 2014 - 08:35 AM.