Heat Conduction Cheat Sheet

I am dumping some equations here I need now and then! The sections about 3-dimensional temperature waves summarize what is described at length in the second part of this post.

Temperature waves are interesting for simulating yearly and daily oscillations in the temperature below the surface of the earth or near wall/floor of our ice/water tank. Stationary solutions are interesting to assess the heat transport between pipes and the medium they are immersed in, like the tubes making up the heat exchanger in the tank or the solar/air collector.

Contents

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Heat equation – conservation of energy [Top]

Energy is conserved locally. It cannot be destroyed or created, but it it is also not possible to remove energy in one spot and make it reappear in a different spot. The energy density η in a volume element can only change because energy flows out of this volume, at a flow density j (energy per area and time).

\frac{\partial \eta}{\partial t} + \frac{\partial \vec{j}}{\partial\vec{r}} = 0

In case of heat energy, the sensible heat energy ‘contained’ in a volume element is the volume times mass density ρ [kg/m3] times specific heat c [J/kgK] times the temperature difference in K (from a ‘zero point’). The flow of heat energy is proportional to the temperature gradient (with constant λ – heat conductivity [J/mK], and heat flows from hot to colder spots.

\rho c \frac{\partial T}{\partial t} + \frac{\partial}{\partial\vec{r}} (- \lambda \frac{\partial T}{\partial\vec{r}}) = 0

Re-arranging and assuming that the three properties ρ, c, and λ are constant in space and time, they can be combined into a single property called thermal diffusivity D

D = \frac{\lambda}{\rho c}

\frac{\partial T}{\partial t} = D \frac{\partial}{\partial\vec{r}} \frac{\partial T}{\partial\vec{r}} = D \Delta T

In one dimensions – e.g. heat conduction to/from an infinite plane –  the equation is

\frac{\partial T}{\partial t} = D \frac{d^{2} T}{d x^{2}}

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1D solution – temperature waves in one dimension [Top]

I covered it already here in detail. I’m using complex solutions as some manipulations are easier to do with the exponential functions than with trigonometric functions, keeping in mind we are finally interested in the real part.

Boundary condition – oscillating temperature at the surface; e.g. surface temperature of the earth in a year. Angular frequency ω is 2π over period T (e.g.: one year)

T(t,0) = T_0 e^{i \omega t}

Ansatz: Temperature wave, temperature oscillating with ω in time and with to-be-determined complex β in space.

T(t,x) = T_0 e^{i \omega t + \beta x}

Plugging into 1D heat equation, you get β as a function of ω and the properties of the material:

i \omega = D \beta^2
\beta = \pm \sqrt{\frac{i \omega}{D}} = \pm \sqrt{i} \sqrt{\frac{\omega}{D}} = \pm (1 + i){\sqrt 2} \sqrt{\frac{\omega}{D}} = \pm (1 + i) \sqrt{\frac{\omega}{2D}}

The temperature should better decay with increasing x – only the solution with a negative sense makes sense, then T(\infty) = T_0 . The temperature well below the surface, e.g. deep in the earth, is the same as the yearly average of the air temperature (neglecting the true geothermal energy and related energy flow and linear temperature gradient).

Solution – temperature as function of space and time:

T(t,x) = T_0 e^{i \omega t - (1 + i) \sqrt{\frac{\omega}{2D}} x} = T_0 e^{i (\omega t - \sqrt{\frac{\omega}{2D}} x)} e^{-\sqrt{\frac{\omega}{2D}} x}

Introducing parameter k:

\sqrt{\frac{\omega}{2D}} = k

Concise version of the solution function:

T(t,x) = T_0 e^{i (\omega t - kx)} e^{-kx}

Strip off the real part:

Re(T(t,x)) = T_0 cos(\omega t - kx) e^{-kx}

Relations connecting the important wave parameters:

\tau = \frac {2 \pi}{\omega}
\lambda = \frac {2 \pi}{k}

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‘Helpers’ for the 3D case (spherical) [Top]

Basic stuff

r = \sqrt{x^2 + y^2 + z^2}
\frac{\partial r}{\partial \vec{r}} = (\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z})\sqrt{x^2 + y^2 + z^2} = \frac{\vec{r}}{r}
\frac{\partial \vec{r}}{\partial \vec{r}} = (\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial z}{\partial z})(x,y,z) = 3
\Delta T = (\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2})T(r)

Inserting, to obtain a nicely looking Laplacian in spherical symmetry

\Delta T = \frac{\partial}{\partial\vec{r}} \frac{\partial}{\partial\vec{r}} T(\sqrt{x^2 + y^2 + z^2})  = \frac{\partial}{\partial\vec{r}} \frac{\partial r}{\partial\vec{r}} (\frac{dT}{dr})  = \frac{\partial}{\partial\vec{r}} (\frac{\vec{r}}{r} \frac{dT}{dr})
= \frac{3}{r} \frac{dT}{dr} - \frac{1}{r^2} \frac{\partial r}{\partial\vec{r}} \vec{r} \frac{dT}{dr}  + \frac{\vec{r}}{r} \frac{\vec{r}}{r} \frac{d^2 T}{dr^2}
= \frac{3}{r} \frac{dT}{dr} - \frac{1}{r} \frac{dT}{dr}+ \frac{d^2 T}{dr^2}  = \frac{2}{r} \frac{dT}{dr} + \frac{d^2 T}{dr^2}
= \frac{1}{r}(\frac{dT}{dr} + \frac{dT}{dr} + r \frac{d^2T}{dr^2})  = \frac{1}{r} \frac{d}{dr} (T + r \frac{dT}{dr}) = \frac{1}{r} \frac{d^2}{dr^2}(rT)

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‘Helpers’ for the 2D case (cylindrical) [Top]

Basic stuff

r = \sqrt{x^2 + y^2}
\frac{\partial r}{\partial \vec{r}} = (\frac{\partial}{\partial x}
latex \frac{\partial}{\partial y})\sqrt{x^2 + y^2 } = \frac{\vec{r}}{r}$
\frac{\partial \vec{r}}{\partial \vec{r}} = (\frac{\partial}{\partial x}
\frac{\partial}{\partial y})(x,y) = 2
\Delta T = (\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2})T(r)

Inserting, to obtain a nicely looking Laplacian in cylindrical symmetry

\Delta T = \frac{\partial}{\partial\vec{r}} \frac{\partial}{\partial\vec{r}} T(\sqrt{x^2 + y^2})  = \frac{\partial}{\partial\vec{r}} \frac{\partial r}{\partial\vec{r}} (\frac{dT}{dr})
= \frac{\partial}{\partial\vec{r}} (\frac{\vec{r}}{r} \frac{dT}{dr})  = \frac{2}{r} \frac{dT}{dr} - \frac{1}{r^2} \frac{\partial r}{\partial\vec{r}} \vec{r} \frac{dT}{dr}  + \frac{\vec{r}}{r} \frac{\vec{r}}{r} \frac{d^2 T}{dr^2}
= \frac{2}{r} \frac{dT}{dr} - \frac{1}{r} \frac{dT}{dr}+ \frac{d^2 T}{dr^2}  = \frac{1}{r} \frac{dT}{dr} + \frac{d^2 T}{dr^2}  = \frac{1}{r} \frac{d}{dr} (r \frac{dT}{dr})

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3D solution – temperature waves in three dimensions [Top]

Boundary condition – oscillating temperature at the surface of a sphere with radius R

T(t,R) = T_R e^{i \omega t}

Ansatz – a wave with amplitude decrease as 1/r. Why try 1/r? Because energy flow density is the gradient of temperature, and energy flow density would better decrease as 1/m2 .

T(t,r) = \frac{A}{r} e^{i \omega t + \beta r}

Plugging in, getting β

i\omega \frac{A}{r} e^{i \omega t + \beta r} = D \Delta T = \frac{D}{r} \frac{d^2}{dr^2}(rT)
= \frac{D}{r} \frac{d^2}{dr^2}(Ae^{i \omega t + \beta r}) = \frac{AD}{r} \beta^2 e^{i \omega t + \beta r}
i\omega = D \beta^2

Same β as in 1D case, using the decaying solution

T(t,r) = \frac{A}{r} e^{i \omega t + \beta r} = \frac{A}{r} e^{i (\omega t - kr)} e^{-kr}

Inserting boundary condition

T(t,R) = \frac{A}{R} e^{i \omega t + \beta R} = T_R e^{i \omega t}
\frac{A}{R} e^{\beta R} = T_R \Rightarrow A = T_R R e^{-\beta R}
T(t,r) = \frac{T_R R}{r} e^{-\beta R} e^{i\omega t + \beta r)} = \frac{T_R R}{r} e^{i\omega t + \beta(r-R)}
= \frac{T_R R}{r} e^{i(\omega t - k (r-R))}e^{-k(r-R))}

The ‘amplitude’ A is complex as β is complex. Getting the real part – this is what you would compare with measurements:

Re (T(t,r)) = \frac{T_R R}{r} cos(\omega t - k (r-R))e^{-k(r-R))}

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Comparison of surface energy densities: 1D versus 3D temperature waves [Top]

This is to estimate the magnitude of the error you introduce when solving an actually 3D problem in only one dimension; replacing the curved (spherical) surface by a plane.

One dimension – energy flow density is just a number:

(t,x) = - \kappa \frac{dT}{dx} = - \kappa \beta T_0 e^{i \omega t + \beta x}

Real part of this, at the surface (x=0)

Re(j(t,0)) = - \kappa T_0 Re(\beta e^{i \omega t}) = - Re((-k -ik) \kappa T_0 e^{i \omega t})
= \kappa T_0 k (cos(\omega t) - sin(\omega t)) = \kappa T_0 k \sqrt{2} (cos(\omega t)\frac{1}{\sqrt{2}} - sin(\omega t))\frac{1}{\sqrt{2}})
= \kappa T_0 k \sqrt{2} (cos(\omega t)\cos(\frac{\pi}{4} - sin(\omega t))\sin(\frac{\pi}{4}) = \kappa T_0 k \sqrt{2} cos(\omega t + \frac{\pi}{4})

How should this be compared to the 3D case? The time average (e.g. yearly) average is zero, to one could compare the average value for half period, when the cosine is positive or negative (‘summer’ or ‘winter’ average). But then, you can as well compare the amplitudes.

Introducing new parameters

l = \frac{1}{k}
j_{amp} = \frac{\kappa T_0}{l}

3D case: Energy flow density is a vector

\vec{j}(t,\vec{r}) = -\kappa \frac{\partial T}{\partial \vec{r}} = -\kappa \frac{\partial}{\partial \vec{r}} \frac{T_R R}{r} e^{i\omega t + \beta(r-R)}
= -\kappa T_R R e^{i\omega t} [-\frac{1}{r^2} \frac{\vec{r}}{r} e^{\beta(r-R)} + \frac{1}{r} \beta \frac{\vec{r}}{r} e^{\beta(r-R)} ]
= \kappa T_R R e^{i\omega t} e^{\beta(r-R)} \frac{\vec{r}}{r} [\frac{1}{r^2} - \frac{\beta}{r} ]
= \frac{\vec{r}}{r} \kappa \frac{T_R R}{r} e^{-k(r-R)} e^{i(\omega t - k(r-R))} [\frac{1}{r} + k + ik]

The vector points radially of course, its absolute value is

j(t,r)= \kappa \frac{T_R R}{r} e^{-k(r-R)} e^{i(\omega t - k(r-R))} [\frac{1}{r} + k + ik]

At the surface of the sphere the ‘ugly part’ is zero as

\vec{r} = \vec{R}
r = R
k(r-R) = 0

Real part:

Re(j(t,r)) = \kappa T_R Re (e^{i(\omega t} [\frac{1}{R} + k + ik] )
= \kappa T_R [(\frac{1}{R} + k) cos(\omega t) - k sin(\omega t) ]
= \kappa T_R [k \sqrt{2} cos(\omega t + \frac{\pi}{4}) + \frac{1}{R} cos(\omega t)]

Here, I was playing with somewhat realistic parameters for the properties of the conducting material. If the sphere has a radius of a few meters, you can ‘compensate for the curvature’ by tweaking parameters and obtain a 1D solution in the same order of magnitude.

Temporal change –  there is a ‘base’ phase different between temperature and energy flow of (about) π/4 which is also changed by introducing curvature. I varied ρ,c, and λ with the goal to make the j curves overlap as much as possible. It is sufficient and most effective to change specific heat only. If the surface is curved, energy ‘spreads out more’. So to make it ‘as fast as’ the 3D wave you need to compensate by a giving it a higher D.

I did not bother to shift the temperature to, say, 10°C as a yearly average. But this is just a linear shift tat will not change anything else – 0°C is arbitrary.

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1D stationary solution – plane [Top]

Stationary means, that nothing changes with time. The time derivative is zero, and so is the (spatial) curvature:

\frac{\partial T}{\partial t} = 0 = D \frac{d^{2} T}{d x^{2}}

The solution is a straight line, and you need to know the temperature at two different points. Indicating the surface x=0 again with 0 and the endpoint x_E with E, and using the definition of j in terms of temperature gradient and distance from the surface (x_E – 0 = Δx).

|j(x = 0)| = \lambda |\frac{dT}{dx}| = \lambda \frac{|T_E - T_0|}{x_E} = \lambda \frac{|T_E - T_0|}{\Delta x}$

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3D stationary solution- sphere [Top]

The time derivative is zero, so the Laplacian is zero:

\frac{\partial T}{\partial t} = 0 = \Delta T(t, r) = \frac{1}{r} \frac{d^2}{dr^2}(rT)

Ansatz, guessing something simple

T(r) = \frac{A + Br}{r} = \frac{A}{r} + B

Boundary conditions, as for the 1D case:

T(R_0) = T_0
T(R_E) = T_E

Plugging in – getting functions for all r:

T(r) = \frac{1}{R_0 - R_E} [R_E T_E(\frac{R_0}{r} - 1) + R_0 T_0 (1 - \frac{R_E}{r}]

|j(r)| = \lambda \frac{1}{R_0 - R_E} \frac{1}{r^2} [R_E T_E R_0 - R_0 T_0 R_E ]

At the surface:

|j(R_0)| = \lambda \frac{1}{R_0 - R_E} \frac{R_E}{R_0} [T_E - T_0 ]

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2D stationary solution – cylinder, pipe [Top]

Cylindrical Laplacian is zero

\frac{1}{r} \frac{d}{dr} (r \frac{dT}{dr}) = 0

Same boundary conditions, plugging in

r \frac{dT}{dr} = A
dT = A \frac {dr}{r}

\int_{T}^{T_0} dT = A \int_{R_0}^{r} \frac {dr}{r}
T(r) = T_0 + A \ln{(\frac{r}{R_0})} = T_0 + A (\ln{r} - \ln{R_0})
T(R_E) = T_E = T_0 + A \ln{(\frac{R_E}{R_0})}
A = \frac{T_E - T_0}{\ln{(\frac{R_E}{R_0})}}

Solutions for temperature and energy flow at any r:

T(r) = T_0 + (T_E - T_0) \frac{\ln{(\frac{r}{R_0})}}{\ln{(\frac{R_E}{R_0})}}
|\vec{j(r)}| = |\frac {1}{r} \lambda \frac{T_E - T_0}{\ln{(\frac{R_E}{R_0})}}|

Expressing r in terms of distance from the surface, \Delta  r = r - R_0

|\vec{j(r)}| = |\frac {1}{\Delta r + R_0} \lambda \frac{T_E - T_0}{\ln{(\frac{R_1}{R_0})}}|

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Comparison of overall heat flow: 1D versus 2D [Top]

j is the energy flow per area, and the area traversed by the flow depends on geometry. in the 1D case the area is always the same area, equal to the area of the plane. For a cylinder, the area increases with r.

The integrated energy flow J for a plate with area F is

J_{Plate} = F \lambda \frac{|T_E - T_0|}{\Delta x}

If the two temperatures are given, J decreases linearly with increasing thickness of the cylindrical ‘shell’, e.g. a growing layer of ice.

For a cylinder of length l the energy flow J is…

J_{Cyl} = 2 \pi l r |\frac {1}{r} \lambda \frac{T_E - T_0}{\ln{(\frac{R_E}{R_0})}}|
= 2 \pi l \lambda |\frac{T_E - T_0}{\ln{(\frac{R_E}{R_0})}}| \par

Factor r has been cancelled, and the for given temperatures J is only decreasing linearly with increasing outer radius R_E. That’s why vendors of plate heat exchangers (in vessels with phase change material) worry more about a growing layer of sold material than user for e.g. ‘ice on coil’ I quoted a related research paper on ‘ice storage powered’ heat pump system in this post – they make exactly this point and provide some data. In addition to conduction also convection at both sides of the heat exchanger should be taken into account, too, in a ‘serial connection’ of heat transferring components.

 

 

Can the Efficiency Be Greater Than One?

This is one of the perennial top search terms for this blog.

Anticlimactic answer: Yes, because input and output are determined also by economics, not only by physics.

Often readers search for the efficiency of a refrigerator. Its efficiency, the ratio of output and input energies, is greater than 1 because the ambient energy is free. System’s operators are interested in the money they pay the utility, in relation to the resulting energy for cooling.

If you use the same thermodynamic machine either as a refrigerator or as a heat pump, efficiencies differ: The same input energy drives the compressor, but the relevant output energy is either the energy released to the ‘hot side’ at the condenser or the energy used for evaporating the refrigerant at the ‘cool side’:

The same machine / cycle is used as a heat pump for heating (left) or a refrigerator or AC for cooling (right). (This should just highlight the principles and does not include any hydraulic details, losses etc. related to detailed differences between refrigerators / ACs and heat pumps.)

For photovoltaic panels the definition has sort of the opposite bias: The sun does not send a bill – as PV installers say in their company’s slogan – but the free solar ambient energy is considered, and thus their efficiency is ‘only’ ~20%.

Half of our generator, now operational for three years: 10 panels, oriented south-east, 265W each, efficiency 16%. (The other 8 panels are oriented south-west).

When systems are combined, you can invent all kinds of efficiencies, depending on system boundaries. If PV panels are ‘included’ in a heat pump system (calculation-wise) the nominal electrical input energy becomes lower. If solar thermal collectors are added to any heating system, the electrical or fossil fuel input decreases.

Output energy may refer to energy measured directly at the outlet of the heat pump or boiler. But it might also mean the energy delivered to the heating circuits – after the thermal losses of a buffer tank have been accounted for. But not 100% of these losses are really lost, if the buffer tank is located in the house.

I’ve seen many different definitions in regulations and related software tools, and you find articles about how to game interpret these guidelines to your advantage. Tools and standards also make arbitrary assumptions about storage tank losses, hysteresis parameter and the like – factors that might be critical for efficiency.

Then there are scaling effects: When the design heat loads of two houses differ by a factor of 2, and the smaller house would use a scaled down heat pump (hypothetically providing 50% output power at the same efficiency), the smaller system’s efficiency is likely to be a bit lower. Auxiliary consumers of electricity – like heating circuit pumps or control systems – will not be perfectly scalable. But the smaller the required output energy is, the better it can be aligned with solar energy usage and storage by a ‘smart’ system – and this might outweigh the additional energy needed for ‘smartness’. Perhaps intermittent negative market prices of electricity could be leveraged.

Definitions of efficiency are also culture-specific, tailored to an academic discipline or industry sector. There are different but remotely related concepts of rating how useful a source of energy is: Gibbs Free Energy is the maximum work a system can deliver, given that pressure and temperature do not change during the process considered – for example in a chemical reaction. On the other hand, Exergy is the useful ‘available’ energy ‘contained’ in a (part of a) system: Sources of energy and heat are rated; e.g. heat energy is only mechanically useful up to the maximum efficiency of an ideal Carnot process. Thus exergy depends on the temperature of the environment where waste heat ends up. The exergy efficiency of a Carnot process is 1, as waste heat is already factored in. On the other hand, the fuel used to drive the process may or may not be included and it may or may not be considered pure exergy – if it is, energy and exergy efficiency would be the same again. If heat energy flows from the hot to the cold part of a system in a heat exchanger, no energy is lost – but exergy is.

You could also extend the system’s boundary spatially and on the time axis: Include investment costs or the cost of harm done to the environment. Consider the primary fuel / energy / exergy to ‘generate’ electricity: If a thermal power plant has 40% efficiency then the heat pump’s efficiency needs to be at least 2,5 to ‘compensate’ for that.

In summary, ‘efficiency’ is the ratio of an output and an input energy, and the definitions may be rather arbitrary as and these energies are determined by a ‘sampling’  time, system boundaries, and additional ‘ratings’.

Consequences of the Second Law of Thermodynamics

Why a Carnot process using a Van der Waals gas – or other fluid with uncommon equation of state – also runs at Carnot’s efficiency.

Textbooks often refer to an ideal gas when introducing Carnot’s cycle – it’s easy to calculate heat energies and work in this case. Perhaps this might imply that not only must the engine be ‘ideal’ – reversible – but also the working fluid has to be ‘ideal’ in some sense? No, it does not, as explicitly shown in this paper: The Carnot cycle with the Van der Waals equation of state.

In this post I am considering a class of substances which is more general than the Van der Waals gas, and I come to the same conclusion. Unsurprisingly. You only need to imagine Carnot’s cycle in a temperature-entropy (T-S) diagram: The process is represented by a rectangle for both ideal and Van der Waals gas. Heat energies and work needed to calculate efficiency can be read off, and the – universal – maximum efficiency can be calculated without integrating over potentially wiggly pressure-volume curves.

But the fact that we can use the T-S diagram or the fact that the concept of entropy makes sense is a consequence of the Second Law of Thermodynamics. It also states, that a Perpetuum Mobile of the Second Kind is not possible: You cannot build a machine that converts 100% of the heat energy in a temperature bath to mechanical energy. This statement sounds philosophical but it puts constraints on the way real materials can behave, and I think these constraints on the relations between physical properties are stronger than one might intuitively expect. If you pick an equation of state – the pressure as a function of volume and temperature, like the wavy Van der Waals curve, the behavior of specific heat is locked in. In a sense the functions describing the material’s properties have to conspire just in the right way to yield the simple rectangle in the T-S plane.

The efficiency of a perfectly reversible thermodynamic engine (converting heat to mechanical energy) has a maximum well below 100%. If the machine uses two temperature baths with constant temperatures T_1 and T_2, the heat energies exchanged between machine and baths Q_1 and Q_2 for an ideal reversible process are related by:

\frac{Q_1}{T_1} + \frac{Q_2}{T_2} = 0

(I wrote on the related proof by contradiction before – avoiding to use the notion of entropy at all costs). This ideal process and this ideal efficiency could also be used to actually define the thermodynamic temperature (as it emerges from statistical considerations; I have followed Landau and Lifshitz’s arguments in this post on statistical mechanics and entropy)

Any thermodynamic process using any type of substance can be imagined as being a combination of lots of Carnot engines operating between lots of temperature baths at different temperatures (see e.g. Feynman’s lecture). The area in the p-V diagram that is traced out in a cyclic process is being split into infinitely many Carnot processes. For each process small heat energies \delta Q are transferred. Summing up the contributions of all processes only the loop at the edge remains and thus …

\oint \frac{\delta Q}{T}

which means that for a reversible process \frac{\delta Q}{T} actually has to be a total differential of a function dS … that is called entropy. This argument used in thermodynamics textbooks is kind of a ‘reverse’ argument to the statistical one – which introduces  ‘entropy first’ and ‘temperature second’.

What I  need in the following derivations are the relations between differentials that represent a version of First and Second Law:

The First Law of Thermodynamics states that heat is a form of energy, so

dE = \delta Q - pdV

The minus is due to the fact that energy is increased on increasing volume (There might be other thermodynamics degrees of freedom like the magnetization of a magnetic substance – so other pairs of variables like p and V).

Inserting the definition of entropy S as the total differential we obtain this relation …

dS = \frac{dE + pdV}{T}

… from which follow lots of relations between thermodynamic properties!

I will derive one the them to show how strong the constraints are that the Second Law actually imposes on the physical properties of materials: When the so-called equation of state is given – the pressure as a function of volume and temperature p(V,T) – then you also know something about its specific heat. For an ideal gas pV is simply a constant times temperature.

S is a function of the state, so picking independent variables V and T entropy’s total differential is:

dS = (\frac{\partial S}{\partial T})_V dT + (\frac{\partial S}{\partial V})_T dV

On the other hand, from the definition of entropy / the combination of 1st and 2nd Law given above it follows that

dS = \frac{1}{T} \left \{ (\frac{\partial E }{\partial T})_V dT + \left [ (\frac{\partial E }{\partial V})_T + p \right ]dV \right \}

Comparing the coefficients of dT and dV the partial derivatives of entropy with respect to volume and temperature can be expressed as functions of energy and pressure. The order of partial derivation does not matter:

\left[\frac{\partial}{\partial V}\left(\frac{\partial S}{\partial T}\right)_V \right]_T = \left[\frac{\partial}{\partial T}\left(\frac{\partial S}{\partial V}\right)_T \right]_V

Thus differentiating each derivative of S once more with respect to the other variable yields:

[ \frac{\partial}{\partial V} \frac{1}{T} (\frac{\partial E }{\partial T})_V ]_T = [ \frac{\partial}{\partial T} \frac{1}{T} \left [ (\frac{\partial E }{\partial V})_T + p \right ] ]_V

What I actually want, is a result for the specific heat: (\frac{\partial E }{\partial T})_V – the energy you need to put in per degree Kelvin to heat up a substance at constant volume, usually called C_v. I keep going, hoping that something like this derivative will show up. The mixed derivative \frac{1}{T} \frac{\partial^2 E}{\partial V \partial T} shows up on both sides of the equation, and these terms cancel each other. Collecting the remaining terms:

0 = -\frac{1}{T^2} (\frac{\partial E }{\partial V})_T -\frac{1}{T^2} p + \frac{1}{T}(\frac{\partial p}{\partial T})_V

Multiplying by T^2 and re-arranging …

(\frac{\partial E }{\partial V})_T = -p +T(\frac{\partial p }{\partial T})_V = T^2(\frac{\partial}{\partial T}\frac{p}{T})_V

Again, noting that the order of derivations does not matter, we can use this result to check if the specific heat for constant volume – C_v = (\frac{\partial E }{\partial T})_V – depends on volume:

(\frac{\partial C_V}{\partial V})_T = \frac{\partial}{\partial V}[(\frac{\partial E }{\partial T})_V]_T = \frac{\partial}{\partial T}[(\frac{\partial E }{\partial V})_T]_V

But we know the last partial derivative already and insert the expression derived before – a function that is fully determined by the equation of state p(V,T):

(\frac{\partial C_V}{\partial V})_T= \frac{\partial}{\partial T}[(-p +T(\frac{\partial p }{\partial T})_V)]_V = -(\frac{\partial p}{\partial T})_V +  (\frac{\partial p}{\partial T})_V + T(\frac{\partial^2 p}{\partial T^2})_V = T(\frac{\partial^2 p}{\partial T^2})_V

So if the pressure depends e.g. only linearly on temperature the second derivative re T is zero and C_v does not depend on volume but only on temperature. The equation of state says something about specific heat.

The idealized Carnot process contains four distinct steps. In order to calculate efficiency for a certain machine and working fluid, you need to calculate the heat energies exchanged between machine and bath on each of these steps. Two steps are adiabatic – the machine is thermally insulated, thus no heat is exchanged. The other steps are isothermal, run at constant temperature – only these steps need to be considered to calculate the heat energies denoted Q_1 and Q_2:

Carnot-cycle-p-V-diagram

Carnot process for an ideal gas: A-B: Isothermal expansion, B-C: Adiabatic expansion, C-D: isothermal compression, D-A: adiabatic compression. (Wikimedia, public domain, see link for details).

I am using the First Law again and insert the result for (\frac{\partial E}{\partial V})_T which was obtained from the combination of both Laws – the goal is to express heat energy as a function of pressure and specific heat:

\delta Q= dE + p(T,V)dV = (\frac{\partial E}{\partial T})_V dT + (\frac{\partial E}{\partial V})_T dV + p(T,V)dV
= C_V(T,V) dT + [-p +T(\frac{\partial p(T,V)}{\partial T})_V] dV + p(T,V)dV = C_V(T,V)dT + T(\frac{\partial p(T,V)}{\partial T})_V dV

Heat Q is not a function of the state defined by V and T – that’s why the incomplete differential δQ is denoted by the Greek δ. The change in heat energy depends on how exactly you get from one state to another. But we know what the process should be in this case: It is isothermal, therefore dT is zero and heat energy is obtained by integrating over volume only.

We need p as a function of V and T. The equation of state for ideal gas says that pV is proportional to temperature. I am now considering a more general equation of state of the form …

p = f(V)T + g(V)

The Van der Waals equation of state takes into account that particles in the gas interact with each other and that they have a finite volume (Switching units, from capital volume V [m3] to small v [m3/kg] to use gas constant R [kJ/kgK] rather than absolute numbers of particles and to use the more common representation – so comparing to $latex pv = RT) :

p = \frac{RT}{v - b} - \frac{a}{v^2}

This equation also matches the general pattern.

Van der Waals isothmers (Waals3)

Van der Waals isotherms (curves of constant temperature) in the p-V plane: Depending on temperature, the functions show a more or less pronounced ‘wave’ with a maximum and a minimum, in contrast to the ideal-gas-like hyperbolas (p = RT/v) for high temperatures. (By Andrea insinga, Wikimedia, for details see link.)

In both cases pressure depends only linearly on temperature, and so (\frac{\partial C_V}{\partial V})_T is 0. Thus specific heat does not depend on volume, and I want to stress that this is a consequence of the fundamental Laws and the p(T,V) equation of state, not an arbitrary, additional assumption about this substance.

The isothermal heat energies are thus given by the following, integrating T(\frac{\partial p(T,V)}{\partial T})_V  = T f(V) over V:

Q_1 = T_1 \int_{V_A}^{V_B} f(V) dV
Q_2 = T_2 \int_{V_C}^{V_D} f(V) dV

(So if Q_1 is positive, Q_2 has to be negative.)

In the adiabatic processes δQ is zero, thus

C_V(T,V)dT = -T(\frac{\partial p(T,V)}{\partial T})_V dV = -T f(V) dV
\int \frac{C_V(T,V)}{T}dT = \int -f(V) dV

This is useful as we already know that specific heat only depends on temperature for the class of substances considered, so for each adiabatic process…

\int_{T_1}^{T_2} \frac{C_V(T)}{T}dT = \int_{V_B}^{V_C} -f(V) dV
\int_{T_2}^{T_1} \frac{C_V(T)}{T}dT = \int_{V_D}^{V_A} -f(V) dV

Adding these equations, the two integrals over temperature cancel and

\int_{V_B}^{V_C} f(V) = -\int_{V_D}^{V_A} f(V) dV

Carnot’s efficiency is work – the difference of the absolute values of the two heat energies – over the heat energy invested at higher temperature T_1 :

\eta = \frac {Q_1 - \left | Q_2 \right |}{Q_1} = 1 - \frac {\left | Q_2 \right |}{Q_1}
\eta = 1 - \frac {T_2}{T_1} \frac {\left | \int_{V_C}^{V_D} f(V) dV \right |}{\int_{V_A}^{V_B} f(V) dV}

The integral from A to B can replaced by an integral over the alternative path A-D-C-B (as the integral over the closed path is zero for a reversible process) and

\int_{A}^{B} = \int_{A}^{D} + \int_{D}^{C}+ \int_{C}^{B}

But the relation between the B-C and A-D integral derived from considering the adiabatic processes is equivalent to

-\int_{C}^{B} = \int_{B}^{C} = - \int_{D}^{A} = \int_{A}^{D}

Thus two terms in the alternative integral cancel and

\int_{A}^{B} = \int_{D}^{C}

… and finally the integrals in the efficiency cancel. What remains is Carnot’s efficiency:

\eta = \frac {T_1 - T_2}{T_1}

But what if the equation of state is more complex and specific heat would depends also on volume?

Yet another way to state the Second Law is to say that the efficiencies of all reversible processes has to be equal and equal to Carnot’s efficiency. Otherwise you get into a thicket of contradictions (as I highlighted here). The authors of the VdW paper say they are able to prove this for infinitesimal cycles which sounds of course plausible: As mentioned at the beginning, splitting up any reversible process into many processes that use only a tiny part of the co-ordinate space is the ‘standard textbook procedure’ (see e.g. Feynman’s lecture, especially figure 44-10).

But you could immediately see it without calculating anything by having a look at the process in a T-S diagram instead of the p-V representation. A process made up of two isothermal and two adiabatic processes is by definition (of entropy, see above) a rectangle no matter what the equation of state of the working substance is. Heat energy and work can easily been read off as the rectangles between or below the straight lines:

Carnot-cycle-T-S-diagram

Carnot process displayed in the entropy-temperature plane. No matter if the working fluid is an ideal gas following the pv = RT equation of state or if it is a Van der Waals gas that may show a ‘wave’ with a maximum and a minimum in a p-V diagram – in the T-S diagram all of this will look like rectangles and thus exhibit the maximum (Carnot’s) efficiency.

In the p-V diagram one might see curves of weird shape, but when calculating the relation between entropy and temperature the weirdness of the dependencies of specific heat and pressure of V and T compensate for each other. They are related because of the differential relation implied by the 2nd Law.

Simulating Life-Forms (2): Cooling Energy

I found this comprehensive research report:
Energy Use in the Australian Residential Sector 1986–2020 (June 2008)
(several PDFs for download, click the link Energy Use… to display them)

There are many interesting results – and the level of detail is impressive: The authors modelled the energy used per appliance type, by e.g. factoring in how building types change slowly over time or by modelling the development of TV sets and their usage. Occupancy factors for buildings are determined from assumptions about typical usage profiles called Stay At Home, At Work or Night Owl.

I zoom in on simulating and predicting usage of air conditioning and thus cooling energy:

They went to great lengths to simulate the behavior of home owners to model operations of air conditioning and thus total cooling energy for a season, for a state or the whole country.

The authors investigated the official simulation software used for rating buildings (from …part2.pdf):

In the AccuRate software, once cooling is invoked the
program continues to assume that the occupant is willing to
tolerate less than optimal comfort conditions and will therefore terminate cooling if in the absence of such cooling the internal temperature would not rise above the summer neutral temperature noted in Table 57, + 2.5oC plus allowances for humidity and air movement as applicable. While this may be appropriate for rating purposes, it is considered to be an unlikely form of behaviour to be adopted by householders in the field and as such this assumption is likely to underestimate the potential space cooling demand. This theory is supported by the survey work undertaken by McGreggor in South Australia.

This confirms what I am saying all the time: The more modern a building is, or generally nowadays given ‘modern’ home owners’ requirements, the more important would it be to actually simulate humans’ behavior, on top of the physics and the control logic.

The research study also points out e.g. that AC usage has been on the rise, because units got affordable, modern houses are built with less focus on shading, and home owners demand higher standards of comfort. Ducted cooling systems that cover the cooling load of the whole house are being implemented, and they replace systems for cooling single zones only. Those ducted systems have a rated output cooling power greater than 10kW – so the authors (and it seems Australian governmental decision makers) are worried about the impact on the stability of the power grid on hot days [*].

Once AC had been turned on for the first time in the hot season, home owners don’t switch it off again when the theoretical ‘neutral’ summer temperature would be reached again, but they keep it on and try to maintain a lower temperature (22-23°C) that is about constant irrespective of temperature outside. So small differences in actual behavior cause huge error bars in total cooling energy for a season:

The impact of this resetting of the cooling thermostat operation was found to be significant. A comparison was undertaken between cooling loads determined using the AccuRate default thermostat settings and the modified settings as described above. A single-storey brick veneer detached dwelling with concrete slab on ground floor and ceiling insulation was used for the comparison. The comparison was undertaken in both the Adelaide and the Darwin climate zones. In Adelaide the modified settings produced an increased annual cooling load 64% higher than that using the AccuRate default settings.

The report also confirms my anecdotal evidence: In winter (colder regions) people heat rooms to higher temperatures than ‘expected’; in summer (warmer regions) people want to cool to a lower temperature:

This is perhaps not surprising, de Dear notes that: “preferred temperature for a particular building did not necessarily coincide with thermal neutrality, and this semantic discrepancy was most evident in HVAC buildings where preference was depressed below neutrality in warm climates and elevated above neutrality in cold climates (ie people preferred to feel cooler than neutral in warm climates, and warmer than neutral in cold climates)” (Richard de Dear et al 1997, P xi).

I noticed that the same people who (over-)heat their rooms to 24°C in winter might want to cool to 20°C in summer. In middle Europe AC in private homes has been uncommon, but I believe it is on the rise, too, also because home owners got accustomed to a certain level of cooling when they work in typical office buildings.

My conclusion is (yet again) that you cannot reliably ‘predict’ cooling energy. It’s already hard to do so for heating energy for low energy houses, but nearly impossible for cooling energy. All you can do – from a practical / system’s design perspective – is to make sure that there is an ‘infinite’ source of cooling energy available.

_________________________________

[*] Edit: And it actually happenend in February 2017.

Entropy and Dimensions (Following Landau and Lifshitz)

Some time ago I wrote about volumes of spheres in multi-dimensional phase space – as needed in integrals in statistical mechanics.

The post was primarily about the curious fact that the ‘bulk of the volume’ of such spheres is contained in a thin shell beneath their hyperspherical surfaces. The trick to calculate something reasonable is to spot expressions you can Tayler-expand in the exponent.

Large numbers ‘do not get much bigger’ if multiplied by a factor, to be demonstrated again by Taylor-expanding such a large number in the exponent; I used this example:

Assuming N is about 1025  then its natural logarithm is about 58 and Ne^N = e^{\ln(N)+N} = e^{58+10^{25}} , then 58 can be neglected compared to N itself.

However, in the real world numbers associated with locations and momenta of particles come with units. Calling the unit ‘length’ in phase space R_0 the large volume can be written as aN{(\frac{r}{R_0})}^N = ae^{\ln{(N)} + N\ln{(\frac{r}{R_0})}} , and the impact of an additional factor N also depends on the unit length chosen.

I did not yet mention the related issues with the definition of entropy. In this post I will follow the way Landau and Lifshitz introduce entropy in Statistical Physics, Volume 5 of their Course of Theoretical Physics.

Landau and Lifshitz introduce statistical mechanics top-down, starting from fundamental principles and from Hamiltonian classical mechanics: no applications, no definitions of ‘heat’ and ‘work’, nor historical references needed for motivation. Classical phenomenological thermodynamics is only introduced after their are done with the statistical foundations. Both entropy and temperature are defined – these are useful fundamental properties spotted in the mathematical derivations and thus deserve special names. They cover both classical and quantum statistics in small number of pages – LL’s style has been called terse or elegant.

The behaviour of a system with a large number of particles is encoded in a probability distribution function in phase space, a density. In the classical case this is a continuous function of phase-space co-ordinates. In the quantum case you consider distinct states – whose energy levels are densely packed together though. Moving from classical to quantum statistics means to count those states rather than to integrate the smooth density function over a volume. There are equivalent states created by permutations of identical particles – but factoring in that is postponed and not required for a first definition of entropy. A quasi-classical description is sufficient: using a minimum cell in phase space, whose dimensions are defined by Planck’s constant h that has a dimension of action – length times momentum.

Entropy as statistical weight

Entropy S is defined as the logarithm of the statistical weight \Delta \Gamma – the number of quantum states associated with the part of phase phase used by the (sub)-system. (Landau and Lifshitz use the concept of a – still large – subsystem embedded in a larger volume most consequentially, in order to avoid reliance on the ergodic hypothesis as mentioned in the preface). In the quasi-classical view the statistical weight is the volume in phase space occupied by the system divided by the size of the minimum unit cell defined by Planck’s constant h. Denoting momenta by p, positions by q, using \Delta p and \Delta q as a shortcut applying multiple dimensions equivalent to s degrees of freedom…

S = log \Delta \Gamma = log \frac {\Delta p \Delta q}{2 \pi \hbar^s}

An example from solid state physics: if the system is considered a rectangular box in the physical world, possible quantum states related to vibrations can be visualized in terms of possible standing waves that ‘fit’ into the box. The statistical weight would then single out those bunch of states the system actually ‘has’ / ‘uses’ / ‘occupies’ in the long run.

Different sorts of statistical functions are introduced, and one reason for writing this article to emphasize the difference between them: The density function associates each point in phase space – each possible configuration of a system characterized by the momenta and locations of all particles – with a probability. These points are also called microstates. Taking into account the probabilities to find a system in any of these microstates gives you the so-called macrostate characterized by the statistical weight: How large or small a part of phase space the system will use when watched for a long time.

The canonical example is an ideal gas in a vessel: The most probable spacial distribution of particles is to find them spread out evenly, the most unlikely configuration is to have them concentrated in (nearly) the same location, like one corner of the box. The density function assigns probabilities to these configurations. As the even distribution is so much much more likely, the \Delta q part of the statistical weight would cover all of the physical volume available. The statistical weight function has to obtain a maximum value in the most likely case, in equilibrium.

The significance of energies – and why there are logarithms everywhere.

Different sufficiently large subsystems of one big system are statistically independent – as their properties are defined by their bulk volume rather than their surfaces interfacing with other subsystems – and the larger the volume, the larger the ratio of volume and surface.  Thus the probability density function for the combined system – as a function of momenta and locations of all particles in the total phase phase – has to be equal to the product of the densities for each subsystem. Denoting the classical density with \rho and adding a subscript for the set of momenta and positions referring to a subsystem:

\rho(q,p) = \rho_1(q_1,p_1) \rho_2(q_2,p_2)

(Since these are probability densities, the actual probability is always obtained by multiplying with the differential(s) dqdp).

This means that the logarithm of the composite density is equal to the sum of the logarithms of the individual densities. This the root cause of having logarithms show up everywhere in statistical mechanics.

A mechanical system of particles is characterized by only 7 ‘meaningful’ additive integrals: Energy, momentum and angular momentum – they add up when you combine systems, in contrast to all the other millions of integration constants that would appear when solving the equations of motions exactly. Momentum and angular momentum are not that interesting thermodynamically, as one can change to a frame moving and rotating with the system (LL also cover rotating systems). So energy remains as the integral of outstanding importance.

From counting states to energy intervals

What we want is to relate entropy to energy, so assertions about numbers of states covered need to be translated to statements about energy and energy ranges.

LL denote the probability to find a system in (micro-)state n with energy E_n as w_n – the quantum equivalent of density \rho . w_n has to be a linear function of the energy of this micro-state E_n as per the additivity just mentioned above, and thus LL omit the subscript n for w:

w_n = w(E_n)

(They omit any symbol ever if possible to keep their notation succinct ;-))

A thermodynamic system has an enormous number of (mechanical) degrees of freedom. Fluctuations are small as per the law of large numbers in statistics, and the probability to find a system with a certain energy can be approximated by a sharp delta-function-like peak at the system’s energy E. So in thermal equilibrium its energy has a very sharp peak. It occupies a very thin ‘layer’ of thickness \Delta E in config space – around the hyperplane that characterizes its average energy E.

Statistical weight \Delta \Gamma can be considered the width of the related function: Energy-wise broadening of the macroscopic state \Delta E needs to be translated to a broadening related to the number of quantum states.

We change variables, so the connection between Γ and E is made via the derivative of Γ with respect to E. E is an integral, statistical property of the whole system, and the probability for the system to have energy E in equilibrium is W(E)dE . E is not discrete so this is again a  probability density. It is capital W now – in contrast to w_n which says something about the ‘population’ of each quantum state with energy E_n.

A quasi-continuous number of states per energy Γ is related to E by the differential:

d\Gamma = \frac{d\Gamma}{dE} dE.

As E peaks so sharply and the energy levels are packed so densely it is reasonable to use the function (small) w but calculate it for an argument value E. Capital W(E) is a probability density as a function of total energy, small w(E) is a function of discrete energies denoting states – so it has to be multiplied by the number of states in the range in question:

W(E)dE = w(E)d\Gamma

Thus…

W(E) = w(E)\frac{d\Gamma}{dE}.

The delta-function-like functions (of energy or states) have to be normalized, and the widths ΔΓ and ΔE multiplied by the respective heights W and w taken at the average energy E_\text{avg} have to be 1, respectively:

W(E_\text{avg}) \Delta E = 1
w(E_\text{avg}) \Delta \Gamma = 1

(… and the ‘average’ energy is what is simply called ‘the’ energy in classical thermodynamics).

So \Delta \Gamma is inversely proportional to the probability of the most likely state (of average energy). This can also be concluded from the quasi-classical definition: If you imagine a box full of particles, the least possible state is equivalent to all particles occupying a single cell in phase space. The probability for that is (size of the unit cell) over (size of the box) times smaller than the probability to find the particles evenly distributed on the whole box … which is exactly the definition of \Delta \Gamma.

The statistical weight is finally:

\Delta \Gamma =  \frac{d\Gamma(E_\text{avg})}{dE} \Delta E.

… the broadening in \Gamma , proportional to the broadening in E

The more familiar (?) definition of entropy

From that, you can recover another familiar definition of entropy, perhaps the more common one. Taking the logarithm…

log S = log (\Delta \Gamma) = -log (w(E_\text{avg})).

As log w is linear in E, the averaging of E can be extended to the whole log function. Then the definition of ‘averaging over states n’ can be used: To multiply the value for each state n by probability w_n and sum up:

- \sum_{n} w_n log w_n.

… which is the first statistical expression for entropy I had once learned.

LL do not introduce Boltzmann’s constant k here

It is effectively set to 1 – so entropy is defined without a reference to k. k is is only mentioned in passing later: In case one wishes to measure energy and temperature in different units. But there is no need to do so, if you defined entropy and temperature based on first principles.

Back to units

In a purely classical description based on the volume in phase space instead of the number of states there would be no cell of minimum size, and then instead of the statistical weight we had simply this volume: But then entropy would be calculated in a very awkward unit, the logarithm of action. Every change of the unit for measuring volumes in phase space would result in an additive constant – the deeper reason why entropy in a classical context is only defined up to such a constant.

So the natural unit called R_0 above should actually be Planck’s constant taken to the power defined by the number of particles.

Temperature

The first task to be solved in statistical mechanics is to find a general way of formulating a proper density function small w_n as a function of energy E_n. You can either assume that the system has a clearly defined energy upfront – the system lives on a ‘energy-hyperplane in phase space’ – or you can consider it immersed in a larger system later identified with a ‘heat bath’ which causes the system to reach thermal equilibrium. These two concepts are called the micro-canonical and the canonical distribution (or Gibbs distribution) and the actual distribution functions don’t differ much because the energy peaks so sharply also in the canonical case. It’s that type of calculations where those hyperspheres are actually needed.

Temperature as a concept emerges from a closer look at these distributions, but LL introduce it upfront from simpler considerations: It is sufficient to know that 1) entropy only depends on energy, 2) both are additive functions of subsystems, and 3) entropy is a maximum in equilibrium. You divide one system in two subsystems. The total change in entropy has to be zero as this is a maximum (in equilibrium), and what energy dE_1 leaves one system has to be received as dE_2 by the other system. Taking a look at the total entropy S as a function of the energy of one subsystem:

0 = \frac{dS}{dE_1} = \frac{dS_1}{dE_1} + \frac{dS_2}{dE_1} =
= \frac{dS_1}{dE_1} + \frac{dS_2}{dE_2} \frac{dE_2}{dE_1} =
= \frac{dS_1}{dE_1} + \frac{dS_2}{dE_2}

So \frac{dS_x}{dE_x} has to be the same for each subsystem x. Cutting one of the subsystems in two  you can use the same argument again. So there is one very interesting quantity that is the same for every subsystem – \frac{dS}{dE}. Let’s call it 1/T and let’s call T the temperature.

Spheres in a Space with Trillions of Dimensions

I don’t venture into speculative science writing – this is just about classical statistical mechanics; actually about a special mathematical aspect. It was one of the things I found particularly intriguing in my first encounters with statistical mechanics and thermodynamics a long time ago – a curious feature of volumes.

I was mulling upon how to ‘briefly motivate’ the calculation below in a comprehensible way, a task I might have failed at years ago already, when I tried to use illustrations and metaphors (Here and here). When introducing the ‘kinetic theory’ in thermodynamics often the pressure of an ideal gas is calculated first, by considering averages over momenta transferred from particles hitting the wall of a container. This is rather easy to understand but still sort of an intermediate view – between phenomenological thermodynamics that does not explain the microscopic origin of properties like energy, and ‘true’ statistical mechanics. The latter makes use of a phase space with with dimensions the number of particles. One cubic meter of gas contains ~1025 molecules. Each possible state of the system is depicted as a point in so-called phase space: A point in this abstract space represents one possible system state. For each (point-like) particle 6 numbers are added to a gigantic vector – 3 for its position and 3 for its momentum (mass times velocity), so the space has ~6 x 1025 dimensions. Thermodynamic properties are averages taken over the state of one system watched for a long time or over a lot of ‘comparable’ systems starting from different initial conditions. At the heart of statistical mechanics are distributions functions that describe how a set of systems described by such gigantic vectors evolves. This function is like a density of an incompressible fluid in hydrodynamics. I resorted to using the metaphor of a jelly in hyperspace before.

Taking averages means to multiply the ‘mechanical’ property by the density function and integrate it over the space where these functions live. The volume of interest is a  generalized N-ball defined as the volume within a generalized sphere. A ‘sphere’ is the surface of all points in a certain distance (‘radius’ R) from an origin

x_1^2 + x_2^2 + ... + x_ {N}^2 = R^2

(x_n being the co-ordinates in phase space and assuming that all co-ordinates of the origin are zero). Why a sphere? Because states are ordered or defined by energy, and larger energy means a greater ‘radius’ in phase space. It’s all about rounded surfaces enclosing each other. The simplest example for this is the ellipse of the phase diagram of the harmonic oscillator – more energy means a larger amplitude and a larger maximum velocity.

And here is finally the curious fact I actually want to talk about: Nearly all the volume of an N-ball with so many dimensions is concentrated in an extremely thin shell beneath its surface. Then an integral over a thin shell can be extended over the full volume of the sphere without adding much, while making integration simpler.

This can be seen immediately from plotting the volume of a sphere over radius: The volume of an N-ball is always equal to some numerical factor, times the radius to the power of the number of dimensions. In three dimensions the volume is the traditional, honest volume proportional to r3, in two dimensions the ‘ball’ is a circle, and its ‘volume’ is its area. In a realistic thermodynamic system, the volume is then proportional to rN with a very large N.

The power function rN turn more and more into an L-shaped function with increasing exponent N. The volume increases enormously just by adding a small additional layer to the ball. In order to compare the function for different exponents, both ‘radius’ and ‘volume’ are shown in relation to the respective maximum value, R and RN.

The interesting layer ‘with all the volume’ is certainly much smaller than the radius R, but of course it must not be too small to contain something. How thick the substantial shell has to be can be found by investigating the volume in more detail – using a ‘trick’ that is needed often in statistical mechanics: Taylor expanding in the exponent.

A function can be replaced by its tangent if it is sufficiently ‘straight’ at this point. Mathematically it means: If dx is added to the argument x, then the function at the new target is f(x + dx), which can be approximated by f(x) + [the slope df/dx] * dx. The next – higher-order term would be proportional to the curvature, the second derivation – then the function is replaced by a 2nd order polynomial. Joseph Nebus has recently published a more comprehensible and detailed post about how this works.

So the first terms of this so-called Taylor expansion are:

f(x + dx) = f(x) + dx{\frac{df}{dx}} + {\frac{dx^2}{2}}{\frac{d^2f}{dx^2}} + ...

If dx is small higher-order terms can be neglected.

In the curious case of the ball in hyperspace we are interested in the ‘remaining volume’ V(r – dr). This should be small compared to V(r) = arN (a being the uninteresting constant numerical factor) after we remove a layer of thickness dr with the substantial ‘bulk of the volume’.

However, trying to expand the volume V(r – dr) = a(r – dr)N, we get:

V(r - dr) = V(r) - adrNr^{N-1} + a{\frac{dr^2}{2}}N(N-1)r^{N-2} + ...
= ar^N(1 - N{\frac{dr}{r}} + {\frac{N(N-1)}{2}}({\frac{dr}{r}})^2) + ...

But this is not exactly what we want: It is finally not an expansion, a polynomial, in (the small) ratio of dr/r, but in Ndr/r, and N is enormous.

So here’s the trick: 1) Apply the definition of the natural logarithm ln:

V(r - dr) = ae^{N\ln(r - dr)} = ae^{N\ln(r(1 - {\frac{dr}{r}}))}
= ae^{N(\ln(r) + ln(1 - {\frac{dr}{r}}))}
= ar^Ne^{\ln(1 - {\frac{dr}{r}}))} = V(r)e^{N(\ln(1 - {\frac{dr}{r}}))}

2) Spot a function that can be safely expanded in the exponent: The natural logarithm of 1 plus something small, dr/r. So we can expand near 1: The derivative of ln(x) is 1/x (thus equal to 1/1 near x=1) and ln(1) = 0. So ln(1 – x) is about -x for small x:

V(r - dr) = V(r)e^{N(0 - 1{\frac{dr}{r})}} \simeq V(r)e^{-N{\frac{dr}{r}}}

3) Re-arrange fractions …

V(r - dr) = V(r)e^{-\frac{dr}{(\frac{r}{N})}}

This is now the remaining volume, after the thin layer dr has been removed. It is small in comparison with V(r) if the exponential function is small, thus if {\frac{dr}{(\frac{r}{N})}} is large or if:

dr \gg \frac{r}{N}

Summarizing: The volume of the N-dimensional hyperball is contained mainly in a shell dr below the surface if the following inequalities hold:

{\frac{r}{N}} \ll dr \ll r

The second one is needed to state that the shell is thin – and allow for expansion in the exponent, the first one is needed to make the shell thick enough so that it contains something.

This might help to ‘visualize’ a closely related non-intuitive fact about large numbers, like eN: If you multiply such a number by a factor ‘it does not get that much bigger’ in a sense – even if the factor is itself a large number:

Assuming N is about 1025  then its natural logarithm is about 58 and…

Ne^N = e^{\ln(N)+N} = e^{58+10^{25}}

… 58 can be neglected compared to N itself. So a multiplicative factor becomes something to be neglected in a sum!

I used a plain number – base e – deliberately as I am obsessed with units. ‘r’ in phase space would be associated with a unit incorporating lots of lengths and momenta. Note that I use the term ‘dimensions’ in two slightly different, but related ways here: One is the mathematical dimension of (an abstract) space, the other is about cross-checking the physical units in case a ‘number’ is something that can be measured – like meters. The co-ordinate  numbers in the vector refer to measurable physical quantities. Applying the definition of the logarithm just to rN would result in dimensionless number N side-by-side with something that has dimensions of a logarithm of the unit.

Using r – a number with dimensions of length – as base, it has to be expressed as a plain number, a multiple of the unit length R_0 (like ‘1 meter’). So comparing the original volume of the ball a{(\frac{r}{R_0})}^N to one a factor of N bigger …

aN{(\frac{r}{R_0})}^N = ae^{\ln{(N)} + N\ln{(\frac{r}{R_0})}}

… then ln(N) can be neglected as long as \frac{r}{R_0} is not extreeeemely tiny. Using the same argument as for base e above, we are on the safe side (and can neglect factors) if r is of about the same order of magnitude as the ‘unit length’ R_0 . The argument about negligible factors is an argument about plain numbers – and those ‘don’t exist’ in the real world as one could always decide to measure the ‘radius’ in a units of, say, 10-30 ‘meters’, which would make the original absolute number small and thus the additional factor non-negligible. One might save the argument by saying that we would always use units that sort of match the typical dimensions (size) of a system.

Saying everything in another way: If the volume of a hyperball ~rN is multiplied by a factor, this corresponds to multiplying the radius r by a factor very, very close to 1 – the Nth root of the factor for the volume. Only because the number of dimensions is so large, the volume is increased so much by such a small increase in radius.

As the ‘bulk of the volume’ is contained in a thin shell, the total volume is about the product of the surface area and the thickness of the shell dr. The N-ball is bounded by a ‘sphere’ with one dimension less than the ball. Increasing the volume by a factor means that the surface area and/or the thickness have to be increased by factors so that the product of these factors yield the volume increase factor. dr scales with r, and does thus not change much – the two inequalities derived above do still hold. Most of the volume factor ‘goes into’ the factor for increasing the surface. ‘The surface becomes the volume’.

This was long-winded. My excuse: Also Richard Feynman took great pleasure in explaining the same phenomenon in different ways. In his lectures you can hear him speak to himself when he says something along the lines of: Now let’s see if we really understood this – let’s try to derive it in another way…

And above all, he says (in a lecture that is more about math than about physics)

Now you may ask, “What is mathematics doing in a physics lecture?” We have several possible excuses: first, of course, mathematics is an important tool, but that would only excuse us for giving the formula in two minutes. On the other hand, in theoretical physics we discover that all our laws can be written in mathematical form; and that this has a certain simplicity and beauty about it. So, ultimately, in order to understand nature it may be necessary to have a deeper understanding of mathematical relationships. But the real reason is that the subject is enjoyable, and although we humans cut nature up in different ways, and we have different courses in different departments, such compartmentalization is really artificial, and we should take our intellectual pleasures where we find them.

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Further reading / sources: Any theoretical physics textbook on classical thermodynamics / statistical mechanics. I am just re-reading mine.

And Now for Something Completely Different: Rotation Heat Pump!

Heat pumps for space heating are all very similar: Refrigerant evaporates, pressure is increased by a scroll compressor, refrigerant condenses, pressure is reduced in an expansion value. *yawn*

The question is:

Can a compression heat pump be built in a completely different way?

Austrian start-up ECOP did it: They  invented the so-called Rotation Heat Pump.

It does not have a classical compressor, and the ‘refrigerant’ does not undergo a phase transition. A pressure gradient is created by centrifugal forces: The whole system rotates, including the high-pressure (heat sink) and low-pressure (source) heat exchanger. The low pressure part of the system is positioned closer to the center of the rotation axis, and heat sink and heat source are connected at the axis (using heating water). The system rotates at up to 1800 rounds per minute. The process is explained in this video by ECOP.

A mixture of noble gases is used in a closed Joule (Brayton) process, driven in a cycle by a ventilator. Gas is compressed and thus heated up; then it is cooled at constant pressure and energy is released to the heat sink. After expanding the gas, it is heated up again at low pressure by the heat source.

In the textbook Joule cycle, a turbine and a compressor share a common axis: The energy released by the turbine is used to drive the compressor. This is essential, as compression and expansion energies are of the same order of magnitude, and both are considerably larger than the net energy difference – the actual input energy.

In contrast to that, a classical compression heat pump uses a refrigerant that is condensed while releasing heat and then evaporated again at low pressure. There is no mini-turbine to reduce the pressure but only an expansion valve, as there is not much energy to gain.

This explains why the Rotation Heat Pumps absolutely have to have compression efficiencies of nearly 100%, compared to, say, 85% efficiency of a scroll compressor in heat pump used for space heating:

Some numbers for a Joule process (from this German ECOP paper): On expansion of the gas 1200kW are gained, but 1300kW are needed for compression – if there would be no losses at all. So the net input power is 100kW. But if the efficiency of the compression is reduced from 100% to 80% about 1600kW are needed and thus a net input power of 500kW – five times the power compared to the ideal compressor! The coefficient of performance would plummet from 10 to 2,3.

I believe these challenging requirements are why Rotation Heat Pumps are ‘large’ and built for industrial processes. In addition to the high COP, this heat pump is also very versatile: Since there are no phase transitions, you can pick your favorite corner of the thermodynamic state diagram at will: This heat pump works for very different combinations temperatures of the hot target and the cold source.