The Physics of Radiant Heat Transfer

Radiant heating has been popular for a long time, mainly in cold climates where homes don’t have any kind of cooling system.  Now those places are getting cooling systems because of the warming climate and frequent wildfires.  And the idea is spreading to other climates as well.  But how does the physics of radiant heat transfer limit the capacity?

Today’s article is technical, especially after the next section, but if you’re math-averse, you can skip to the bottom to read the takeaways.  And my follow-up article will use those takeaways as I explore the issue of radiant cooling in a humid climate.

How do radiant panels transfer heat?

You may think this is a trick question, but there’s more to the answer than you may think.  Yes, radiating heat away in heating mode or absorbing radiant heat in cooling mode is the dominant mechanism for flat horizontal panels.  I’ll make this more quantitative in the sections below, but remember, heat transfer can occur by one of three mechanisms:  conduction, convection, and radiation.

The photo above is of an old-style cast iron radiator that’s part of a steam or hot water heating system.  Because of their geometry and the vertical orientation, they transfer heat mainly by convection.  In fact, about 60 to 80 percent of the total heat transfer is by convection, with the rest being radiation.  In other words, they’re misnamed.  They should be called convectors.

What I’m focusing on here, however, is flat horizontal panels on either the floor or the ceiling.  As you’ll see below, there are a bunch of variables that go into the calculations, but let me give you some idea of the radiant/convective split.

The radiation/convection split

Now let’s look at how much heat moves by radiation vs. convection varies by whether you’re heating or cooling and whether the radiant panel is on the floor or the ceiling.  These numbers are approximate because each variable (temperature, materials, etc.) has a range.

Ceiling panels

Heating:  ~70% radiant, ~30% convective

Cooling:  60-70% radiant, 30-40% convective

Floor panels

Heating:  <70% radiant, >30% convective.  (I haven’t found solid numbers for this one, but it has to be more convective than heating panels at the ceiling.)

Cooling:  ~90% radiant, ~10% convective

If you think about the buoyancy of air, you should be able to figure out why ceiling panels result in more radiant transfer in heating mode .  A warm ceiling panel will heat the air near it by convection, but the buoyancy of air keeps that warm air near the ceiling.  That lack of circulation of the warmed air means there’s less convective heat transfer in this case, and the radiant fraction is higher as a result.  The same thing happens with radiant cooling panels at floor level.

Now, let’s get into some mathematics.

The physics of radiant heat transfer

Every object in the universe has a nonzero temperature.  Because of temperature, every object in the universe radiates heat away from itself.  The physics of radiant heat transfer begins with the Stefan-Boltzmann law, which quantifies that heat transfer by radiation and is where we need to start.  The Stefan-Boltzmann equation is:

Q = σεAT⁴

• Q is the rate of heat transfer in watts (W) or BTU per hour (BTU/hr).
• σ is the Stefan-Boltzmann constant (5.67 x 10-8 W/m² K⁴)
• ε is the emissivity, which varies between 0 and 1.
• A is the surface area of the radiant emitter.
• T is the temperature in kelvins.

The emissivity is a property of the surface that describes how well it emits or absorbs radiation.  An emissivity of 1 means it’s a perfect emitter and absorber.  Zero means it’s terrible at emitting and absorbing radiant energy.  For radiant heating and cooling, you want something that’s a good emitter or absorber with an emissivity of 0.9 or higher.  In contrast, the thin coatings on window glass that improve energy efficiency have a low emissivity.  That’s where the “e” in low-e comes from.  They have emissivities of 0.1 or less.

The area, A, of a surface also matters, and I’ll use it later in the discussion of radiant panel heating and cooling capacity.  And we have to use the temperature in kelvins (no degree symbol) because it’s not just a temperature difference that matters.  An object radiates heat according to that equation above based on its absolute temperature.

Radiant heat transfer between two surfaces

What I wrote above does not mean temperature differences don’t matter.  They do.  When an object at temperature T₁ is near another object at T₂, they’re radiating heat at each other at different rates.  In this case, it’s the net radiant heat transfer that matters.  And that’s what we need to know for buildings with radiant heating or cooling.

But the equation above does not change to Q = σεA(ΔT)⁴.  Nope.  It’s not that simple.  For radiant heat transfer between two surfaces at temperatures T₁ and T₂, the equation is:

Q = σεA(T₁⁴ − T₂⁴)

But we can simplify that.

The fourth-power law actually becomes linear

Now we’ve arrived at the fun part!  I’m not showing all the work here, but you can see more of it on this page that Claude.ai helped with.  Let’s start with an algebra trick, factoring the part in parentheses above.

T₁⁴ − T₂⁴ = (T₁² + T₂²)(T₁ + T₂)(T₁ − T₂)

We can use the average temperature, T_avg, to rewrite T₁ + T₂ part since T_avg = (T₁ + T₂)/2.  The last factor, T₁ − T₂, is just ΔT.  Putting these in and doing more algebra, we get:

Q = 4 σ ε A (T_avg)³ ΔT [1 + (ΔT/2T_avg)²]

Now we’re getting somewhere.  The part in brackets, 1 + (ΔT/2T_avg)², provides us a way to go all the way from those temperatures to the fourth power down to the ΔT to the first power.  The key is in how close the temperatures are to each other for radiant heating and cooling in a building.

When T₁ and T₂ are close enough to each other, the (ΔT/2T_avg)² is small enough that we can ignore it.  With that simplification, we get to this equation that’s linear in ΔT:

Q ≈ 4 σ ε A T_avg³ ΔT

Now, what does this equation tell us?

The accuracy of the linear approximation

The reason those temperatures are close is that we have to use the absolute temperature scale for these calculations.  If the room temperature is 75 °F and the radiant panel temperature is 65 °F, the temperatures are 297 and 291 on the Kelvin scale.  So yeah, that’s a small ΔT.

The other thing that helps here is a cool fact about squaring numbers.  When you square a number greater than 1, you get a bigger number.  For example, 3² equals 9.  But when you square a number less than 1, you end up with an even smaller number:  0.3² equals 0.09.  When you put our numbers into ΔT/2T_avg, the result is less than 1.

Now, let’s use the example of radiant cooling with a 10 °F ΔT, with temperatures 75 °F (297 K) and 65 °F (291 K).

(ΔT/2T_avg)² = [(297-291) / (2((297+291)/2))]² = (6/588)² ≈ 0.0001

Adding that to the 1 gives you 1.0001, so we can safely drop that 0.0001 and ignore the second term in the equation.

This simplification also works for higher temperatures.  Even if you do that calculation with steam heat and a 145 °F ΔT, dropping the second term introduces an error of only ~1.4%.  See the details here.

From physics to capacity

Let’s take a closer look at that final equation below and see what kind of numbers we can tease out of it.  On the right side of the equation, we know that 4 and σ are constant.  We’ll use an emissivity, ε, of 0.9.  With 75 °F and 65 °F as our temperatures, or 297 and 291 K, T_avg comes out to 294 K.  (For radiant heating, the temperatures would be different.)

Q ≈ 4 σ ε A T_avg³ ΔT

First, let’s simplify again.  With:

h_r = 4 σ ε T_avg³

Now the equation becomes:

Q ≈ h_r A ΔT

And now you may recognize that form.  The simplified equation for heat flow by conduction is Q = U A ΔT.  It also is the same form you see in Newton’s law of cooling:  Q = h A ΔT.

The h_r is the radiant heat transfer coefficient in watts per square meter per kelvin (W/m² K).  When you use the numbers above, h_r comes out to be ~5.2 W/m² K.  For the temperatures used in radiant heating and cooling, h_r is usually near 5.5 W/m² K.

But that’s not all there is.  Early in this article I said that radiation isn’t the only thing happening.  Some of the heat is also moving via convection.  Let’s take 70% radiant transfer as an example.  (Yes, cooling panels on the floor have a higher radiant fraction of ~90%, but you’ll see in the follow-up article why I think cooling panels belong on the ceiling.)

If h_r is 5.2 W/m² K, that would make the total heat transfer coefficient 7.4 W/m² K because h_tot = h_r /0.7.  Then we use h_tot = h_r + h_c to find that h_c = 2.2 W/m² K in  this case.

Now we’re ready to talk capacity.

Heating and cooling capacities

Now we want to put in the ΔT and get those coefficients down to units of W/m².  For our radiant cooling example, ΔT = 10 K.  So:

Q = h_tot A ΔT = (7.4 W/m² K) A (10 K) = (74 W/m²) A

And now we can figure out how much panel area we need to cool a room under the conditions described above.

74 W/m² = 253 BTU/hr/m² =23.5 BTU/hr/ft²

Using that result, you could get 23,500 BTU/hr (~2 tons) for each 1,000 square feet of panel area.  If you’re trying to cool a 2,000 square foot home, that could be about the right amount of cooling capacity.

Heating capacities can be higher per square foot because of a higher T_avg and higher ΔT.  With a 95 °F (308 K) panel temperature and 70 °F (294 K) indoor temperature, T_avg = 301K and ΔT = 14 K.  That makes h_r = 5.6 W/m² K and, with 70% radiant, 30% convective, h_tot = (5.6 + 2.4) W/m² K = 8.0 W/m² K.  Then:

Q = h_tot A ΔT = (8 W/m² K) A (14 K) = (112 W/m²) A

So:

112 W/m² = 35.5 BTU/hr/ft²

This would get you about 36,000 BTU/hr of heating capacity for each 1,000 square feet of panel area, about 50% more than cooling panels.

Takeaways for the physics of radiant heat transfer

Whew!  That’s a lot more math than I thought it would be.  Let me summarize the main points here, for those of you who waded through the above as well as those of you who skipped to this section.

1. The radiant heat transfer rate for small temperature differences is a linear equation with the temperature difference as the main variable:

Q = h_tot A ΔT

with  h_tot = h_r + h_c  and  h_r = 4 σ ε T_avg³.

2. With a ΔT of 10 °F for cooling, you can get about 2 tons of cooling capacity for each 1,000 square feet of radiant panel area.  Overall, radiant cooling capacities come in at about 15,000 to 25,000 BTU/hr for each 1,000 square feet.

3. With a ΔT of 25 °F for heating, you can get about 3 tons of heating capacity for each 1,000 square feet of radiant panel area.  Overall, radiant heating capacities come in at about 20,000 to 36,000 BTU/hr for each 1,000 square feet.

4. The approximation derived to get to the linear equation has negligible error for the temperatures used in modern radiant heating and cooling panels.

Now, I know it’s not the physics of radiant heat transfer that most of you want to know but rather how this works in the real world.  But I went through all this for those who want to know where this came from. And because my follow-up article on radiant cooling in a humid climate relies in part on these takeaways.

 

Afterword

One final note here.   What I’ve described here is a simplified version of the actual heat transfer processes.  I haven’t included wavelength, room geometry, the emissivity of other surfaces in the room, and a few other things. But the physics of radiant heat transfer is no different from other areas of physics.  You start simple and add complexity as you need it.

 

Allison A. Bailes III, PhD is a speaker, writer, building science consultant, and the founder of Energy Vanguard in Decatur, Georgia.  He has a doctorate in physics and is the author of a bestselling book on building science.  He also writes the Energy Vanguard Blog.  For more updates, you can follow Allison on LinkedIn and subscribe to Energy Vanguard’s weekly newsletter and YouTube channel.

 

Related Articles

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U R A ΔT, and Other Building Science Blandishments

 

Photo of hydronic tubing in floor by Danny Gough, used with permission.

 

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