thermodynamics
Yes, I know it should be called a convector 🙂
The obvious effect is that a coat of paint, which is generally a thermal insulator, can limit the rate of heat transfer from the metal to the air. But how important is that effect to the overall efficiency of heating my house or room?
It sounds as if you are describing a countercurrent heat exchanger.
The theoretical efficiency of these can reach 1, though note that for heat exchangers efficiency doesn't mean the same as for heat engines i.e. heat converted to work. For heat exchangers an efficiency of 1 just means the incoming air is heated to the same temperature as the air in the house and likewise for the outgoing air.
Let's assume the oven walls radiate like perfect black body radiators and the air inside the oven is uniform throughout (ignore the thin layer near the cooler meat). Assume I want to cook a prime rib (chosen because it can be bought in a cut that has a cylindrical shape), which is just beef and fat. I can look up the thermal emissivity and thermal conductivity of beef to find:
We know the heat transfer rate for radiative heating can be approximated by: $$ \dot{Q}_{rad} \approx \sigma \ \epsilon \ A \ \left( T_{oven}^{4} - T_{beef}^{4} \right) \tag{1} $$ where $\sigma$ is the Stefan-Boltzmann constant, $\epsilon$ is the radiative emissivity, $A$ is the surface area of the absorber, and $T_{j}$ is the temperature of the jth object.
Similarly, we know heat transfer rate for conductive heating can be approximated by: $$ \dot{Q}_{con} \approx \frac{ \kappa \ A \left( T_{oven} - T_{beef} \right) }{ L } \tag{2} $$ where $\kappa$ is the thermal conductivity, $A$ is the surface area of the absorber, $L$ is the thickness of the absorber, and again $T_{j}$ is the temperature of the jth object.
Let's assume we buy a prime rib that is ~1.5 ft (~0.46 m) in length and ~5 in (~0.13 m) in diameter and approximate it as a cylinder. Then the absorbing surface area and thickness would be $A$ ~ 0.42 m2 and $L$ ~ 0.13 m. If the meat starts at 32 oF (273.15 K) and the oven is at 350 oF (449.82 K), the rates from Equations 1 and 2 when the meat first enters the oven are:
Obviously this is a function of time since the $\Delta T$ will change with time. The constant factors for each are $\sim 1.8 \times 10^{-8}$ W K-4 for $\dot{Q}_{rad}$ and $\sim 1.8$ W K-1 for $\dot{Q}_{con}$. One can go ahead and plot the transfer rates (see figure below) and find that radiative transfer always dominates.
If these assumptions are correct, then what heat transfer mechanism will dominate and why?
So if we can truly ignore convection, then it appears that radiation dominates over conduction.
Best Answer
If the convection efficiency is lowered by an insulating layer, the water returns hotter. I chose some ballpark numbers, under the (not really true) assumption that the heating is only convective:
Power = 1000W
Area = 1 m^2
Thermal Conductivity = 0.2 W/mK (median of some plastics on the internet)
Thickness = 2E-04 m (200 micron ballpark for paint)
Using dimensional analysis (because I don't remember the equation) I get
delta T (temperature drop across the paint thickness) in K:
[Thickness (m) * Power (W)] / [Area (m^2) Conductivity (W/mK)] = 1 (K) !
So a ballpark number would be a 1 degree K (or C) difference in surface temperature.
If the room is cold (10C) and the radiator is almost too hot to touch (140F = 60C) then it's a few percent effect on the convective efficiency, roughly speaking, and probably no more than 10%.
But a radiator temperature of 60C and room temperature of 10C, emissivity of 0.85, and area of 1 m^2 gives about 600 W of outgoing radiation, and ~360 W of incoming radiation from the room, or a net radiation of 240 W, so really can't ignore either one.