Radiant heat loss
Reid Baldwin
| Posted in Mechanicals on
If I understand manual J correctly, heating load calculations do not consider radiant losses. I am trying to figure out why that is ok. It seems like a house that is out in the open (not surrounded by trees or other buildings) would lose substantial heat by radiation on a clear night. Is radiant heat transfer small enough that assuming the surface is at the ambient air temperature is a safe simplification?
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Replies
Reid,
The U-factors and R-values used in most energy modeling programs, including Manual J, account for heat loss by all three heat transfer mechanisms: conduction, convection, and radiation.
I understand how an R-value would reflect a radiant barrier and an air space, as a layer in a wall for example. That is not what I am asking about. What I am referring to is radiation from the outside surface into space. It is not clear to me how that could be captured in an R-value or U-value since it depends upon things like clouds or the presence or absence of adjacent buildings.
One can kind find reports of roof temperatures 10F below ambient on a clear night. It's also not clear to me how it would accurately account for worse case wind driven infiltration.
Reid,
I assume that:
(1) the calculations made by Manual J programs have a default value for radiation from the roof to the night sky, even though the amount of such radiation varies depending on tree cover, and
(2) since most homes are insulated, this default value works adequately for this type of calculation, and
(3) the variance between the extreme cases -- lots of tree cover and no tree cover -- don't affect the total load of the building enough to matter compared to other (more important) unknowns and variables.
Jon,
In fact, Manual J programs do include an input for wind exposure, so this variable is accounted for.
Reid,
The discussion here (http://sensiblehouse.org/nrg_rvalue.htm) seems to cover what you're interested in.
Bill,
Thanks for the link. The relevant sentence is this one: "For example a dark roof can be over 140°F when the air temperature is only 90°F, but on a yearly basis, in most climates it will apparently not skew the results too much. Since the roof surface is the one most affected, additional insulation is usually added there to compensate."
It doesn't matter whether it skews the results very much on a YEARLY basis. We don't size heating and cooling equipment based on yearly averages. On the other extreme, we also don't size them based on a worst case moment. Even on a daily basis, hours of a surface radiating into a clear sky are usually balanced by hours with the sun shining on the surface. Is a daily average sufficiently short for purposes of equipment sizing?
In steady state, the heat transfer from each surface of an assembly, via a combination of radiation and convection, equals the heat transfer through the assembly. If the surface of a building has an emissivity of 0.5 and is at 10 degrees F (470 degrees R), then the each square foot of the surface would radiate heat at a rate of (0.5) * (1.714x 10^-9) * (470)^4 = 41.8 btu/hr. For a vertical wall, there will be radiation received from the ground that would reduce that by about half. However, the heat transfer through each square foot of an R24 wall, by conduction, is only (70-10) / (24) = 2.5. Convection from the outside air into the wall would need to make up the difference, which requires the outside air to be warmer than 10 degrees, probably by a not insignificant amount. Now that I think about it, it makes sense that surfaces would be colder than the air at night. Heat transfer into surfaces is what causes outside air to cool down at night.
If I was completely confident in the above analysis, I wouldn't be posting this as a question. Is my assumption that the wall is in steady state overly simplistic? Is some other assumption off?
Reid,
To me, the pertinent part of the Sensible House article is this “…the heat loss equations assume the mean radiant temperature is the same as the air temperature…” That is, the R-value for the film of air is ok to use as long as the temperature of the “ambient” outside your house is uniform. Here “ambient temperature” is everything that can in direct line of sight from the chosen square foot of siding (or roof): the air, ground, trees, buildings, clouds, suspended particles, pollutants and the sky in general. Generally this assumption is pretty good, but it fails the most on very clear nights, particularly at high altitude.
Note that your calculation of radiant heat loss is not correct you say “For a vertical wall, there will be radiation received from the ground that would reduce that by about half”. You neglect to account for everything above the ground as well – it is also emitting radiative energy that hits your siding. The correct answer is that if the ambient temperature is the same as the wall surface, then the heat flows exactly cancel out. See, for example, http://www.engineeringtoolbox.com/radiation-heat-transfer-d_431.html, especially the section titled “Net Radiation Loss Rate”. The net radiant heat loss is determined by the temperature difference between your siding (or roof) and the ambient.
Note that if you calculate the effective R-value of radiative transfer at 10 deg. F, then the effective R-value for radiative transfer at the surface of your siding is about R 1.4 relative to the ambient. This heat path is in parallel with that for the air film – and the air film wins (dominates), with lower R value. Especially if the wind is blowing. The biggest effects come when the ambient has many different temperatures in it. Complicated. But some smart people did this for flat roofs at sea level (see http://www.asterism.org/tutorials/tut37%20Radiative%20Cooling.pdf, particularly the section “Radiative Cooling”). It really is complicated, but the equations in the last link can get you a worst case answer (for a roof) if you are determined.
Merry Christmas!
Bill,
I am aware that heat loss equations typically assume the surface temperature is equal to the air temperature. The point of my inquiry here is to check that assumption. If I can be convinced that it is within a couple degrees, then I will be satisfied that such an assumption is in the noise relative to the many other assumptions that we typically make in such calculations.
Thank you for pointing out my faulty assumption that there would be no radiant transfer back from a clear sky. Your reference provides an empirical equation to estimate the radiant transfer back. For a clear sky, the first term goes away and the formula becomes 8.78e-13xT^5.852xRH^0.07195. The 10 degrees F in my example is about 261 degrees R. Using 90% humidity, the radiation from the sky is predicted to be 168 W/m^2 = 53.27 btu/hr/ft^2. My calculation about predicted 41.8 btu/hr/ft^2 radiated away from a surface with an emissivity of 0.5. For a vertical wall, about half (20.4 btu/hr/ft^2) of that would be radiated to the sky. The radiation back would be (0.5)(0.5)(53.27) = 13.32 btu/hr/ft^2. The first 0.5 is due to incidence angle and the other one is absorbtivity (equals emissivity). So the net radiant transfer out is down to about 7 btu/hr/ft^2. That is still more than the conduction through the wall, but not by nearly as much as I previously thought. My conclusion is that the surface may be a couple degrees less than the air, such that the difference is made up by convection into the wall surface.
Bill and Reid: I'm no engineer. Could you put your calculations in furlongs?
I'd be interested at some point to hear how pinning this down would influence the design of building assemblies or sizing of heating sources. Is this just interesting for its own sake, or is this something with practical application?
Reid,
Firing up Excel, when I balance the heat flow through the wall (70 °F inside, R 30 for the wall thermal resistance) to the sheathing with the conductive heat flow through the air film (R = 0.68 for still air, air temp = 10 °F) and with the radiative heat flow (using an emissivity of 0.8 and ambient temp = 10 °F) I get a sheathing temperature of 11.0 °F. Not a big change. In this case the heat loss through the wall is 1.97 (Btu/h)/sq.ft.
Variations:
• If you use an exterior air film R value of 0.17 and ignore radiative effects, then you get 1.99 (Btu/h)/sq.ft. – up 1.0%.
• Increase the wall R value to R 40? Then Tsheathing drops to 10.7 °F and heat loss through the wall is 1.48 (Btu/h)/sq.ft.
• Change just the emissivity to 0.5? Then Tsheathing = 11.1 °F. The sheathing is warmer than above because the radiative cooling is less. The heat loss through the wall is 1.96 (Btu/h)/sq.ft. – down 0.5%.
• Change just R-value for the air film to 0.17 (breezy day)? Then Tsheathing = 10.3 °F. The sheathing is closer to the air temperature because the air film is less insulating. The heat loss through the wall is 1.99 (Btu/h)/sq.ft. – up 1.0%.
• Change just the ambient temperature to 0 °F? Then Tsheathing = 8.3 °F. The sheathing cools to below the air temperature because of the increased radiative cooling. The heat loss through the wall is 2.06 (Btu/h)/sq.ft. – up 4.5%.
• Change both the ambient temperature to -10 °F and R-value for the air film to 0.17? Then Tsheathing = 8.7 °F. The increased radiative cooling is offset by the less insulating air film. The heat loss through the wall is 2.04 (Btu/h)/sq.ft. – up 3.6%.
What’s the worst case night effective sky temperature when it’s 10°F? A quick survey of references has one answer at about -10°F. But really, this subject seems both fairly nuanced and confused (to me).
None of the above are big differences. Using an outside R-value for the air film of 0.17 (which is common practice) seems to give you slightly conservative estimates that would account for some extra radiative cooling from the wall should you have a windless, starry night. I once calculated the extra heat loss in walls due to steel fasteners through foam and got similar increase in heat loss. But don’t worry! The increase in R value of most insulation as it gets colder outside will counteract all this. Also remember that any time radiation to the sky decreases the surface temperature, the (relatively) warmer air tends to warm the surface up.
I believe that we’re in the weeds.
Thanks Bill, you have succeeded in convincing me. Assuming that the exterior surface is equal to the outside air temperature is reasonable, at least for walls. I won't challenge the assumption for a vented attic because the roof surface isn't against the insulation. I also won't challenge it for an unvented attic because I don't have one.