This post “follows” on from Heat Transfer Basics and Non-Radiative Atmospheres and Do Trenberth and Kiehl understand the First Law of Thermodynamics? and many other posts that cover some basics.
It’s clear from comments on this blog and many other blogs that a lot of people have difficulty understanding simple scenarios because of a lack of understanding of the basics. Many confident (but erroneous) comments state that particular scenarios can never occur because they violate this or that law..
I know from my own experience that until a concept is conceptually grasped, a mathematical treatment is often not really helpful. It might be right, but it doesn’t help..
So this post has a number of examples that paint a picture. It has some maths too.
Enough examples might help some readers unfamiliar with thermodynamic concepts grasp the essence of some heat transfer basics.
Ignore the details if you like and just check the results from each example. Some maths is included to make it possible to check the results and understand the subject a little better.
We will use the example of a “planar wall”. What exactly is that?
It’s a wall that extends off to infinity in both directions.
For those thinking this is some kind of climate madness, it’s simply physics basics – draw up a problem with simple boundary conditions and find the answer. If we start with some massively complex problem that approximates the real world then unfortunately there will be no conceptual understanding. And this article is all about conceptual understanding. Start with simple problems and gradually extend to more complex problems.. (The wall can just be a long wall if that makes you happier).
Example One is a wall, made out of PVC, with both sides held at a constant temperature (probably by a fluid at a constant temperature pumped over each side).
Example One – constant temperature conduction
We want to calculate the heat flux (heat flow per unit area) travelling through this wall of PVC.
The basic equation of heat conduction is:
q = kA . ΔT/Δx (see note 1)
where ΔT is the temperature difference, Δx is the thickness of the wall, A is the area, k is the conductivity (the property of the material) and q is the heat flow.
To make things slightly easier we consider heat flux – heat flow per unit area, q”:
q” = k . ΔT/Δx
For PVC, k=0.19W/m.K
And for the case of this wall, T1 = 50°C, T2 = 10°C and therefore ΔT = 40°C
q” = 0.19 x 40 / 2 = 3.8 W/m²
In this example, because the system is holding both surfaces at a constant temperature we have a constant (and continuous) flow of heat between surfaces.
You can see that not much heat is flowing because PVC is a very good insulator.
Example Two – constant temperature conduction, thinner wall
q” = 0.19 x 40 / 0.2 = 38 W/m²
So with the wall 10x thinner, the heat flux is 10 times greater. Hopefully, for most, this is intuitively obvious – put thinner insulation on a hot water pipe and it loses more heat; wear a thinner coat out in the cold and you get colder..
If we changed the PVC for metal then the heat flow would be very much higher, as metal conducts heat very effectively.
Now what’s supplying the heat? The liquid or gas being pumped over the higher temperature surface to keep it at that temperature.
Note that these surfaces will be radiating heat. However, this doesn’t affect the calculation of conducted heat between the two surfaces.
In simple terms, heat flow due to conduction depends on the temperature difference, the material and the dimensions of the body.
Example Three – no temperature differential
Now both sides of the wall are held at the same temperature,
q” = 0.19 x 0 / 2 = 0 W/m²
This is very simple, but obviously confuses some people, including some visitors to this blog. It is temperature difference that drives conduction of heat. If there is no temperature difference, there will be no conduction.
In these three examples we have constrained the temperature on each side to see what happens to heat flow. Now we will change these boundary conditions.
Conduction and Radiation
Example Four – constant heat supply one side, fixed temperature the other
This is example two but with a constant heat supply instead of a constant temperature on one side.
This example is now more complex. The right side of the wall is held at a constant temperature of 10°C, as with the first few examples, but the other surface of the wall now has a constant input of heat and we want to find out the temperature of that surface.
The heat source for the left side is incident radiation. We will assume that the proportion of radiation absorbed (“absorptivity”) is 80% or 0.8. And we will assume that the emissivity of the surface is also 0.8. See note 2.
How do we now calculate the surface temperature T1?
It’s quite simple in principle. We use the first law of thermodynamics – energy cannot be created or destroyed. And we will calculate the equilibrium condition – which is when steady-state is reached. This means no heat is being retained to increase the temperature.
So all we have to do is balance the heat flow terms at the surface (the left surface). Let’s take it step by step.
Energy absorbed from radiation:
Ein(absorbed) = Ein x 0.8
This is because 80% is absorbed and 20% is reflected, due to the material properties of PVC.
For energy balance, once the surface has reached a steady temperature:
Ein(absorbed) = q” + Eout (see the diagram)
q” is the heat flux through the wall, and Eout is the radiated energy. At this point we are assuming no convection (perhaps there is no atmosphere for example) to keep things simple.
Hopefully this is quite a simple concept – the heat absorbed from radiation is balanced by the heat radiated from the surface plus the heat conducted through the wall.
We can calculate the energy radiated using the well-known Stefan-Boltzmann equation,
Eout = εσT4
where ε = emissivity (0.8 in this example), σ is the Stefan-Boltzmann constant (5.67 x 10-8) and T is absolute temperature in K (add 273 to temperature in °C).
Except we don’t yet know the temperature.. Still, let’s put it all together and see what happens:
Ein(absorbed) = q” + Eout
Now put the numbers in that we know:
500 x 0.8 = 0.19 x ΔT / 0.2 + 0.8 x 5.67×10-8 x T14
Now ΔT is the temperature difference between T1 and T2. T2 is held constant at 10°C so ΔT=T1-10. However, the first term on the right is expressed in °C and the second term in absolute temperature (K). We will express both as absolute temperature, so ΔT = T1-283.
So now the equation is:
400 = 0.19 x (T1-283) / 0.2 + 0.8 x 5.67×10-8 x T14
The important point to note for those a little bewildered by all the numbers – we have used the first law of thermodynamics, the equation for emission of radiation and the equation for conducted heat and as a result we have an equation with only one unknown – the temperature.
This means we can find the value of T1 that satisfies this equation. (See note 3 for how it is found).
T1 = 302.80 K = 29.65°C
And using this value, conducted heat, q” = 18.7 W/m² and radiated heat, Eout = 381.3 W/m².
In this case, there is a lot more heat radiated compared with conducted.
Example Five – as example four with a thinner wall
With a much thinner wall or a much higher conductivity the balance changes.
Changing the thickness of the wall from 0.2m to 2mm, keeping everything else the same and so using exactly the same equations as above, we get:
T1 = 284.24 K = 11.09°C
Note that this means the temperature differential across the wall has reduced to only (just over) 1°C.
And using this value, conducted heat, q” = 103.6 W/m² and radiated heat, Eout = 297.4 W/m².
Example Six – as example four with increased “colder” temperature
Now with the 0.2m wall (example four) we increase the temperature of the colder side, from 10°C to 25°C.
Using the same maths we find that the temperature, T1, has increased:
T1 = 305.16K = 32.01°C
An increase of 2.36°C.
Most people are probably asking “why this example? it’s obvious that increasing the temperature of one side will lift the other..“
However, many people believe that a colder atmosphere cannot affect the temperature of a warmer surface. See, for example, The First Law of Thermodynamics Meets the Imaginary Second Law. This reasoning is due to a misunderstanding of the second law of thermodynamics.
However, as conduction is quite familiar and more intuitive I expect that this example will be more easily accepted. And perhaps this last example will help a few people to see that a colder body can affect a warmer body without violating any laws of thermodynamics.
In Do Trenberth and Kiehl understand the First Law of Thermodynamics? I presented a hollow sphere in space with a heat source at its center. Some people were (and still are) convinced that there is something wrong with the results from that example. One person (at least) is convinced that the inner surface must be at the same temperature as the outer surface.
The only correct approach to calculating heat transfer and temperatures is to apply the relevant equations of conduction, convection and radiation to the particular problem in question.
- Conduction of heat is proportional to the temperature difference across a material
- Radiation of heat is proportional to the 4th power of (absolute) temperature of a surface
- The first law of thermodynamics is used to solve these problems: energy in – energy out = energy retained, for any particular part of a system that you analyze
Many people rely on intuition for determining whether a solution is correct. However, intuition is not as reliable as applying the basic equations of heat transfer.
Note that the example in Do Trenberth and Kiehl understand the First Law of Thermodynamics? uses exactly the same equations and approach as the examples here. If these six examples are correct you will have trouble finding the flaw in the hollow sphere example.
Note 1 – Conventionally the equation of heat conducted has a minus sign because heat travels in the opposite direction to the temperature gradient. And for the purists, the more general equation of conduction is:
q” = -k∇T
where ∇T is the three dimensional version of the “change of T with respect to distance”
Note 2 – Emissivity = Absorptivity at a particular wavelength (and direction for “non-diffuse” surfaces). In the case of a surface receiving radiation and emitting radiation there is no reason why these two values should be the same. This is because the incident radiation will be at one wavelength (or range of wavelengths), but the wavelength of emission depends on the temperature of the surface.
Note 3 – One simple way to find the value that satisfies the equation is to plot the equation for a wide range of temperatures and look up the temperature value where the result is correct. This is what I did here. It is the work of a minute with Matlab.
Update – added Sep 15th
A graph of temperature vs wall thickness for Examples Four & Five (with T2 = 10°C):
Update – Added Sep 15th
3D graph for examples 4 to 6 – of how T1 varies with wall thickness and T2:
Click for a larger image
Update – added Sep 16th
3D graph of how T1 varies with emissivity. First, when absorptivity = emissivity:
Click for a larger image
Now with absorptivity (the proportion of incident radiation absorbed) set at 0.8, while the emissivity varies.
Click for a larger image
Notice that when the absorptivity and emissivity are equal the temperature T1 is pretty much independent of the actual value of emissivity/absorptivity – why is that?
And when emissivity varies while absorptivity is fixed (and therefore absorbed energy is fixed) the temperature T1 is pretty much independent of emissivity for very thin walls – why is that?