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John Baez
John D Cook
New Year, New World

Things do look a bit bleak right now but human beings can sometimes perform miracles, the COVID vaccines and the James Webb telescope are proof of that. Nothing is ever perfect and a new year inspires new ideas for making the world a better place and life a little more pleasant. So here are some of our ideas.

Let's start with something simple like the calendar. Why are there 7 days in a week when we have 365 days in a year? If you divide 365 by 7 you get 52 with one day left over, we have 52 weeks plus one day in a year. This means dates do not fall on the same day of the week every year. New Year's Eve this year is on Friday but next year it will be on Saturday. How do we keep it on Friday every year?

The solution is simple, we change the number of days in a week to 5 instead of 7. We can safely get rid of Monday and Tuesday since nobody really likes those days anyway. Saturday and Sunday will still be the weekend i.e. days where you can do any damn thing you want. The work week will only be three days: Wednesday, Thursday, and Friday. There will be 73 weeks in a year. That's 73 weekends, surely we can all get behind this!

The one possible objection to this new scheme is the fact that a year is actually about 365.25 days long. So we need leap days or we'll eventually be celebrating Christmas in shorts and tank tops the way the Australians do. But we don't need to make leap days a part of the regular calendar. We'll just call them free days, a special holiday where politicians have to clean public toilets on live TV and homeless people get to sleep and eat in the White House.

Now let's look at clocks. There are 24 hours in a day so why not use a 24 hour clock. Forget this AM and PM baloney. 1 PM is simply 13:00, 6 PM is 18:00, and so on. Isn't that simpler? If there's 24 hours then number them that way.

But the most confusing thing about clocks is that I may know the time where I am but I have no idea what the time is on Pitcairn Island. We need one universal time for the whole world. Yes that means some people will be getting up at 7:00 while others will be going to bed at that time but everyone will get used to it. Also, if you go to some other country you don't have to do any calculations to figure out what your people back home are doing right now.

Another more radical solution is to do away with clocks altogether and replace them with a device that shows the longitude where solar noon is occurring at any given moment. This provides a universal time and it can tell you if it's night or day time at any given place if you know it's approximate longitude.

One of the great insanities of this world is the belief that we need economic growth at all costs. If GDP goes down it's seen as some kind of great catastrophe, the Federal Reserve starts printing money like crazy, making bankers ever richer, politicians are voted out of office, and businesses are urged to start hiring and ramp up production. Does it matter what's produced? No, anything will do. A million new bobble head dolls is great as long as it adds to GDP. Most of this new economic "growth" is just crap that will end up in a landfill in a few years. This kind of attitude will eventually destroy this beautiful planet we live on. It also consigns so many people to boring, meaningless lives which is probably the biggest tragedy. We can change the world by buying less crap and making more beautiful things.

Stand on a street corner in any American town of a few thousand people or more and watch the cars go by. The vast majority of them will have a single occupant. Very few will be small and fuel efficient, or better yet, electric. A significant number of them will be monstrously large pickup trucks pumping out clouds of black diesel exhaust. Do we really need giant machines that weigh tons to haul one human being maybe a mile or so down the road? Why can't more people use bikes for short trips around town or to get to work? At least in cities, we should give the same priority to bicycle transportation as we do to automobile transportation.

If you haven't tried bike riding recently, you're missing out on lots of fun. There's something exhilarating about moving along under your own power outside in the fresh air. It's good exercise and will help with that New Year's resolution to lose some weight. It also gives you a whole new perspective on the world. You will see things you never noticed before. Give it a try.

Do you like to save money? Do you like getting something for free? If so then get yourself a clothes line and stop using your electric or gas powered clothes dryer. It doesn't take much to put up a clothes line in your backyard. It's the simplest way to take advantage of free solar energy. Your clothes will smell great and they will last longer, plus you get more fresh air and sunshine. And since most electric power in the US is produced with fossil fuels, you're likely to be adding less CO2 to the atmosphere.

There's good evidence that a plant-based diet can prevent many of the diseases that cause so much human suffering. Cancer, diabetes, and heart disease can be prevented and reversed in people who already have them. It reduces not only human suffering but the suffering of animals too. The industrialized farming of animals is a horrifying thing to see. The conditions under which these animals are forced to live out their short lives before they end up on somebody's dinner plate is a moral and ethical transgression of the first order. There are videos of animals being funneled into a slaughterhouse. You can see the absolute terror on their faces when they get to the point where they finally realize what's going on. Please, let's stop this insanity.

We'd love to see more people have jobs that they enjoy, reducing traffic accidents caused by the mad rush to get home after work to have a little enjoyment at the end of the day.

And last but not least, we'd love to see kindness and compassion increase among humanity.

We wish everyone a happy, healthy and safe New Year.


Making an Irreversible Process Reversible

The concepts of reversible and irreversible processes are very important in thermodynamics but they are usually explained in rather abstract terms that take some getting used to. We will give what we think is a more intuitive example of such processes in the form of the simple problem of transferring liquid from one container to another.

We have two identical cylindrical liquid tanks sitting at the same level next to each other. One of them is full and the other is empty. The problem is to transfer as much liquid as possible from the full tank to the empty tank without moving either tank and keeping them both on the same level, and without any input of energy in the form of a pump.

If we just connect the bottom of the two tanks with a hose then liquid will flow from the full to the empty tank, until they both have the same liquid level. In this way, half of the liquid is transferred from the full to the empty tank, and this is clearly an irreversible process. You can't get all the liquid back into the full tank without the input of additional energy.

Is there any way to transfer more than half the liquid? The answer is yes, if we first partition the empty tank into two parts, one on top of the other, as shown in the following figure. The top part is filled first by connecting its bottom to the bottom of the full tank. The bottom part is then filled in the same way. If the height of the partition is correctly placed we can get more than half of the liquid into the empty tank.

So where exactly should we put this partition? To make things simple let's set the level of the full tank equal to 1. We will call the level of the partition y, and the final height of the liquid above the partition x. The following equation must then be satisfied.

\[1 - x = y + x\]

If we impose the condition that the bottom part must be filled completely then the following equation must also be satisfied

\[1 - x - y = y\]

Solving these equations for \(x\) and \(y\) we get \(x=y=1/3\). This means that if we place the partition at a height of \(1/3\) we can fill the top part to a height of \(1/3\) and fully fill the bottom part for a total height of \(2/3\) which is more than the \(1/2\) we get without partitioning.

It is fairly easy to show that this value for \(y\) is optimal. No other value of \(y\) will let you transfer more liquid. The way to transfer more liquid is to use more partitions. You can partition the second tank into 3 parts by placing the partitions at \(y_1=1/2\) and \(y_2=1/4\) which will allow you to transfer \(3/4\) of the liquid. If you partition into 4 parts by placing partitions at \((y_1,y_2,y_3)=(3/5,2/5,1/5)\) you can transfer \(4/5\) or \(80\%\) of the liquid.

In general partitioning the empty tank into \(N\) parts will allow \(N/(N+1)\) of the liquid to be transferred. By making \(N\) large enough you can transfer as much of the liquid as you want and in the limit as \(N\to\infty\) all of the liquid is transferred and the process becomes completely reversible. You can turn around and transfer all the liquid back to the original tank.

The question of reversibility is all about change in entropy and the loss of useful energy so let's look at the energies involved in the above processes. The gravitational potential energy stored in a cylindrical column of liquid is proportional to the square of the height of the column. For simplicity we'll assume the proportionality constant is equal to 1.

The full tank then starts out with energy \(E=1\). In the case of no partitioning the final height of the liquid in both tanks is \(1/2\) so the final energy is \(1/4+1/4=1/2\). Half the initial energy has been lost but where does it go? It must eventually be dissipated as heat into the surroundings causing a net increase in entropy. In the case of a partition into two parts, the two tanks have final liquid heights of \(1/3\) and \(2/3\) for a final energy of \(1/9+4/9=5/9\). In this case only \(4/9\) of the energy has disappeared and there is correspondingly a smaller increase in entropy.

In general for a partition into \(N\) parts the height of the liquid in the two columns will be \(1/(N+1)\) and \(N/(N+1)\) so that the energy is

\[E=\frac{N^2+1}{(N+1)^2}\]

In the limit \(N\to\infty\) the energy is equal to the initial energy, \(E=1\). No energy is lost, no entropy is created and the process is completely reversible.

Another example of an irreversible process is the transfer of heat from a quantity of water at temperature \(T_1\) to another quantity of water at temperature \(T_2 \lt T_1\). If you split up the \(T_2\) water and put each piece in contact with the \(T_1\) water sequentially you can lower the entropy production and transfer more of the heat i.e. the process becomes less irreversible. This is the basic principle used in heat exchangers. For an interesting take on this idea see the paper by Mischenko and Pshenichka listed in the references below.

References

Eugene G. Mishchenko, Paul F. Pshenichka, "Reversible temperature exchange upon thermal contact", American Journal of Physics, Vol 85, No 1, January 2017, p23-29.


Efficiently Heating a House III

A heat pump needs a source of energy and most of the modern ones run on electricity. You just plug them into your house's electrical system and they'll start chugging away. Another interesting way to power a heat pump is with a heat engine. A thermodynamic diagram of this process is shown in the following figure.

There are three heat reservoirs, the outside environment at temperature \(T_0\), the inside of the house at temperature \(T_1\), and a high temperature heat source at \(T_2\). The high temperature source could come from the combustion of some fuel such as natural gas. There is a heat engine operating between \(T_2\gg T_1\) that produces work \(W\). This work is used to drive the heat pump that operates between \(T_1\gt T_0\) pumping heat into the house.

Does such a setup make sense? The answer is yes, at least from a theoretical perspective. It greatly improves on the efficiency of the heat pump alone, providing a much greater quantity of heat to the house for the same expenditure of energy. The method was first proposed by Lord Kelvin, one of the founders of thermodynamics, way back in 1853. Only a few thermodynamics books mention it, such as the one by Franzo H. Crawford published in 1963. A more modern reference is a paper by E. T. Jaynes published in 2003. Links to these references are given below.

So let's analyze this system to see how good it really is. The energy we have to pay for is \(Q_2\) which comes from the combustion of fuel. The energy that we get going into the house is \(Q_1 + Q_1^{\prime}\) where \(Q_1\) is the heat engine waste and \(Q_1^{\prime}\) is the heat coming from the pump. So the efficiency, or heat gain, is the ratio of \(Q_1 + Q_1^{\prime}\) to \(Q_2\). This is what we will calculate. The goal is to express the ratio in terms of the temperatures of the three reservoirs.

First we'll assume that both the pump and engine are reversible. This means they are as efficient as possible and produce no net entropy. Starting with the engine, we see from the diagram that the work it produces is given by (apply the first law of thermodynamics for a cyclic process)

\[W = Q_2 - Q_1\]

Since the engine produces no net entropy we must have (apply the second law of thermodynamics)

\[\frac{Q_2}{T_2} = \frac{Q_1}{T_1}\]

Using this equation, the work produced by the engine can be expressed as

\[W = Q_2\left(1 - \frac{T_1}{T_2}\right)\]

This work is used by the pump to deliver heat \(Q_1^{\prime}\) to the house so we have (again, apply the first law of thermodynamics for a cyclic process)

\[Q_1^{\prime} = Q_0 + W\]

where \(Q_0\) is the heat extracted from the outside environment. Since the pump produces no net entropy we have (apply the second law of thermodynamics)

\[\frac{Q_0}{T_0} = \frac{Q_1^{\prime}}{T_1}\]

With this equation we can write the heat delivered to the house as

\[Q_1^{\prime} = Q_1^{\prime}\frac{T_0}{T_1} + W\]

Using the above expression for \(W\) and solving for \(Q_1^{\prime}\) we get

\[Q_1^{\prime} = Q_2\frac{T_1}{T_2}\frac{T_2-T_1}{T_1-T_0}\]

The total heat delivered to the house is then

\[Q_H=Q_1^{\prime}+Q_1=Q_2\frac{T_1}{T_2}\frac{T_2-T_0}{T_1-T_0}\]

so the heat gain is

\[G(T_0,T_1,T_2)=\frac{Q_H}{Q_2}=\frac{T_1}{T_2}\frac{T_2-T_0}{T_1-T_0}\]

which can also be written as

\[G(T_0,T_1,T_2)=\frac{1-T_0/T_2}{1-T_0/T_1}\]

As an example, assume the fuel we are using is methane, which has a combustion temperature in air of 1950 ℃ = 2223 K. Assume the outside temperature is 0 ℃ = 273 K, and assume the inside temperature is 20 ℃ = 293 K. Plugging these values into the heat gain formula we get and incredible \(G(273,293,2223) = 12.85\). So for each unit of heat purchased this setup produces almost 13 units of heat in the house. It is important to remember however that we are assuming an ideal heat engine and pump that produce no entropy. With a real engine and pump we would probably only get about half this value.

We can analyze the system without the assumption of a ideal engine and pump by focusing on the overall change in entropy. Note, first of all, that the net result of a cycle is that \(Q_2\) leaves the \(T_2\) reservoir, and \(Q_0\) leaves the \(T_0\) reservoir, so the heat entering the house is \(Q_H=Q_2+Q_0\). The total entropy change of the system is

\[\Delta S = \frac{Q_H}{T_1} - \frac{Q_0}{T_0} - \frac{Q_2}{T_2}\]

In general, this must be greater than or equal to zero (it is only equal when the engine and pump are reversible). Therefore, we have

\[\frac{Q_H}{T_1} \ge \frac{Q_0}{T_0} + \frac{Q_2}{T_2} = \frac{Q_H - Q_2}{T_0} + \frac{Q_2}{T_2}\]

Rearranging this expression gives us

\[Q_H \le Q_2\frac{T_1}{T_2}\frac{T_2-T_0}{T_1-T_0}\]

When engine and pump are reversible, this becomes an equality and we get the previous result. The general expression for the heat gain is then

\[G(T_0,T_1,T_2) \le \frac{1-T_0/T_2}{1-T_0/T_1}\]

This shows that the result we calculated earlier is the maximum possible heat gain, for real engines and pumps it will be smaller.

How could you actually realize such a system? One way is to use a natural gas powered electric generator to power an electric heat pump. Electric generators are generally very efficient and you can also get very efficient electric heat pumps. You won't get anywhere near the ideal heat gain we calculated above, but it should be better than running just a heat pump alone off the electric grid. An additional benefit is that you have an electric backup generator in case the power ever goes out.

References

Sir William Thomson, Baron Kelvin, "Power required for the thermodynamic heating of buildings", Cambridge and Dublin Mathematical Journal, November, 1853. A copy of this paper is found in Kelvin's Volume V of Mathematical and Physical Papers, p124-133.

Franzo Hazlett Crawford, "Heat, Thermodynamics and Statistical Physics", 1963, p217-219.

Edwin Thompson Jaynes, "Note on thermal heating efficiency", American Journal of Physics, Vol 71, No 2, February 2003, p180-182.


Efficiently Heating a House II

Out of all the conventional ways to heat a house using a heat pump is by far the best. The reason is that it takes less energy to move a quantity of heat than it does to generate that heat in the first place. A heat pump can take heat from the cold outside air, yes there is still heat in that air, and move or pump it into the warmer house. Heating a house with a heat pump is like having an inside out refrigerator where you're heating the inside and cooling the outside.

The basic thermodynamic operation of a heat pump is shown in the following figure.

We have two heat reservoirs, the outside of the house at temperature \(T_0\) and the inside at temperature \(T_1\gt T_0\). With an input of energy \(W\) (this is the part you have to pay for) the pump removes a quantity of heat \(Q_0\) from the outside air and deposits a quantity of heat \(Q_1=Q_0+W\) into the inside air. The efficiency of this process for a given heat pump is called its coefficient of performance \(\mathbf{COP}\). It is defined as follows

\[\mathbf{COP}=\frac{Q_1}{W}\]

In other words, it's the ratio of what you get to what you pay for. If \(\mathbf{COP}=2\) for example, then one unit of input energy will provide two units of heat energy to the inside of the house.

There is no single value for \(\mathbf{COP}\) since it is generally a function of the temperatures \(T_0\) and \(T_1\). Ideally there should be a plot of \(\mathbf{COP}\) for different values of \(T_0\) and \(T_1\) but we have not seen a single heat pump manufacturer that does this. Instead they use (for the case of heating performance) something called a heating seasonal performance factor \(\mathbf{HSPF}\). I should mention here that there are different types of heat pumps. Some use the air as a heat source while others use ground or water as a heat source. The HSPF mainly applies to air source heat pumps.

\(\mathbf{HSPF}\) is expressed as the heat output over an entire heating season measured in BTUs (British thermal units) divided by the total electricity used measured in watt-hours. This gives it the somewhat bizarre units of BTU/watt-hour. You can convert \(\mathbf{HSPF}\) to the average \(\mathbf{COP}\) over a heating season, which is unitless, by multiplying \(\mathbf{HSPF}\) by \(0.2931\), the number of watt-hours in a BTU.

High efficiency heat pumps have an \(\mathbf{HSPF}\) of at least 8 or an average \(\mathbf{COP}\) of \(8\cdot 0.2931=2.3448\). By contrast, an electrical resistance heater, which can do no more than turn all of its input energy into heat, would have \(\mathbf{COP}=1\) or \(\mathbf{HSPF}=1/0.2931=3.4118\).

It is interesting to compare a real heat pump with an ideal heat pump which produces no entropy so that

\[\frac{Q_1}{T_1}=\frac{Q_2}{T_2}\]

with the temperatures measured in Kelvin. The \(\mathbf{COP}\) can then be expressed in terms of the temperatures as

\[\mathbf{COP}=\frac{T_1}{T_1-T_0}\]

No heat pump operating between the temperatures \(T_1\gt T_0\) can have a greater \(\mathbf{COP}\) than this. As an example, if the outside temperature is 0°C and the inside temperature is 20°C then \(T_0=273.15 K\), \(T_0=293.15 K\) and the maximum \(\mathbf{COP}\) that any heat pump can have is 14.6575.

Currently, in the United States, the minimum HSPF set by the government is 8.2, but will rise to 8.8 in 2023. You will pay more for higher HSPF. For example, a Senville 12,000 BTU mini split heat pump with HSPF of 8.5 costs 799.99 USD, while a Fujitsu 12,000 BTU mini split heat pump with HSPF of 12.5 costs 1539.00 USD. For this price difference of 739.01 USD the \(\mathbf{COP}\) rises from 2.49 to 3.66.

Of course, reliability is another factor to consider when buying a heat pump. Consumer Reports has a reliability rating for numerous brands.

A blog post we found worthwhile reading on the topic of installing a heat pump yourself is "Our DIY Heat Pump Install – Free Heating and Cooling for Life?".



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