The Einstein-Szilard Refrigerator

This article looks at the refrigeration system patented by Albert Einstein and Leo Szilard and explores the underlying scientific principles.

The refrigerator is depicted in the patent filed by Einstein and Szilard, a copy of which is available here:  http://opensourceecology.org/w/images/6/63/Einstein_Fridge.pdf.  It’s copied below, but you’ll want to go ahead and click that link so you’ll have the image as a reference.

Einstein-Szilard Refrigerator Patent Drawing

As a quick orientation, the little box on the right, labeled “1,” is the part that gets cold.  All the other things in the drawing are the parts that are necessary to make that little box get cold.  Also note: if you wanted to use this system to, say, keep your food cold, you wouldn’t put the food in that little “1” box.  No, the “1” box would go inside of a larger insulated box, and your food would go in there.  The “1” box would get cold, heat would conduct from the inside of your insulated box through the walls of box “1” into its interior, and that in turn would make your food cold.

The fundamental principle by which this device makes cold is this: when liquids turn into gasses, they suck up heat from their surroundings.  A mass of water vapor at 212 degrees Fahrenheit has more heat energy than the same mass of liquid water at 212 degrees Fahrenheit.  One term for this concept is “enthalphy of vaporization.”  It’s an interesting and powerful scientific principle.  Just for example: it takes 5.4 times as much heat energy to convert water at a given temperature to water vapor at the same temperature as it does to heat the same water from 32°F to 212°F.  This is why sweat is so effective at cooling you off: every ounce of sweat that evaporates draws in enough heat energy to cool three quarts of water by 10°F.  [Can’t help adding: the specific heat of flesh is around 3,470 J/kg*K, so if you weigh 200 pounds, every 2.7 ounces of evaporated sweat is enough to cool your body 1°F.]

The Einstein-Szilard refrigerator, instead of using water as the evaporating substance, uses butane.  Here’s how butane stacks up against water in some key properties:

  Water Butane
ΔHvaporization (kJ/mol) 40.66 21
ΔHvaporization (kJ/kg) 2257 320
Boiling Point 100֗֠ºC -1ºC
Density (liquid) 1,000 kg/m3 600 kg/m3
Condensation pressure at 100ºF ~38 psi / 2.6 atm ~1 psi / 0.068 atm
chemical formula H2O C4H10

So butane has lower enthalpy of vaporization than water, meaning you need to evaporate more of it to produce temperature change.  How much more?  Seven times as much by weight, and almost 12 times as much by volume.  However, it has a couple of other key properties that make it more useful than water for this purpose.  Mainly, its boiling point is much lower than water’s.  Back to box “1;” the liquid butane is evaporated by the addition of gaseous ammonia, which is bubbled through the liquid butane by the porous stone “31.”  The gaseous ammonia causes the butane to evaporate much like water will evaporate into air – even well below its boiling point – until it reaches 100% relative humidity.  As the butane evaporates, it pulls in from its environment the energy required to overcome its enthalpy of vaporization, which makes its environment colder.  Butane’s low boiling point and condensation pressure make it well suited for refrigeration temperatures.

As we’ve now established, all we need is a steady supply of liquid butane and a steady supply of gaseous ammonia, and we’ll have a steady supply of cold.  The purpose of all of the other components of the Einstein-Szilard refrigerator apart from box “1” is to recycle the butane and ammonia so that you don’t have to keep supplying your cooler with fresh butane and ammonia.  That would get expensive.  Einstein and Szilard figured out a series of chemical processes by which the butane and ammonia could be recycled, requiring only a heat source and an environmental heat sink (such as the atmosphere, a lake, or a well).

Separating the Butane Vapor from the Ammonia Vapor

After the ammonia gas evaporates the butane in box “1,” the ammonia and butane gasses are combined.  The first step in recycling them is to separate them.  This is done by exploiting the fact that ammonia dissolves readily in water, whereas butane does not.  This happens in box “6” on the diagram.  The ammonia-butane gas mixture flows to box “6” via pipe “5.”  [Note that pipe “30” passes through pipe “5;” we’ll get to that later.]  In box “6,” water is sprayed through the ammonia-butane gas mixture using sprayer head “35.”  The ammonia dissolves into the water and the ammonia-water solution falls to the bottom of the tank (“26″ on the drawing) because liquids are heavier than gasses.  Removing the ammonia from the butane has the helpful side effect of condensing the butane (re-liquefying it), so the butane also falls to the bottom of the box, but it’s less dense than water so it floats above the water.  This actually completes the recycling of the butane, and the butane is allowed to flow back into box “1” via pipe “11.”

Of course the butane warms back up as it condenses, releasing as much heat energy as it absorbed when it evaporated.  This heat has to be dissipated before the butane goes back into box “1,” and that’s what the heat exchanger “12” is for.

Recycling the Ammonia

After the process in box “6,” the butane has been recycled, but the ammonia is now in solution with water.  It must be returned to its gaseous state in order to be useful.  To achieve this, the Einstein-Szilard refrigerator exploits another property of ammonia: more ammonia can dissolve in cool water than in hot water.  In other words, heating the water-ammonia solution will force out the ammonia, which will return to its gaseous state.  This heating is done in box “29.”  The heat supplied to box “29” is the primary energy required to make this heat pump work.

The hot ammonia gas travels through pipe “30” to box “1.”  On the way, it passes through pipe “5,” where it exchanges heat with the chilled ammonia-butane mixture that is exiting box “1,” so it enters box “1” pre-chilled and does not introduce too much heat to box “1.”

Recycling the Water

The byproduct of the ammonia recovery is hot water, which must be cooled so that it can again be useful for absorbing ammonia.  In the original Einstein-Szilard design, this is done using the heat exchangers at 28 and 12.  Heat exchanger 28 uses this hot, low-ammonia water to pre-heat the high-ammonia water en route from box 6 to box 29 while simultaneously cooling the hot water.  Heat exchanger 12 requires an external heat sink such as a source of cool water – maybe water pumped to a geothermal heat sink.  Air cooling could work, too, with some coils, a pump to move the cooling water through the coils, and a fan to move air over the coils.

This water cooling step is where a couple of common myths often claimed about the Einstein-Szilard refrigerator break down:

Myth #1: It’s a device that turns a heat source into a cold source!

Not entirely true.  It’s impossible to make a system colder by adding heat.  What the Einstein-Szilard refrigerator actually achieves is a heat pump, which is a device that moves heat energy from one place to another.   Specifically, this one moves heat from the surroundings of box “1” into box “1,” then from box “1” to box “6,” then from box “6” into the environment via heat exchanger “12.”  It is very interesting and useful that it does this without needing a motor, just a heat source, but of course it does not defy the laws of conservation of energy.

Myth #2: It’s a refrigerator with no moving parts!

Not quite.  It’s true that it’s a heat pump with no moving (solid) parts, but heat pumps just move heat from point “A” to point “B” (or in this case from box “1” to heat exchanger “12”).  Unless you have some mechanism in place to dissipate the heat from point “B,” the heat pump will overheat the environment at point “B” and stop working.  In order to have what we think of as a household refrigerator, we need a fan or pump or some other way to keep the environment around heat exchanger “12” from getting hot.

Likewise on the cold side, some air circulation around box “1” would be needed so that it doesn’t get too cold right next to box “1” and not cold enough in the rest of your insulated box.

Once the hot water is cooled down, the cycle is complete and we are back to our starting products: gaseous ammonia, liquid butane, and unheated water.  The Einstein-Szilard refrigerator continuously runs this process, cooling the surroundings of box “1” and heating the heat sink around heat exchanger “12.”

 

I intend to come back later and take a longer look at this style of heat pump, performing a somewhat more in-depth analysis.  For now, I have enjoyed considering the key concepts that make the machine work.  It will be a useful thing to consider for using waste heat from industrial processes to cool other processes.  I hope I get a chance to use this concept.

Hydroelectric Power – Part One: Is your stream viable?

This Part One discusses how to assess the viability of a site for hydroelectric power, and simultaneously provides a basic understanding of the physics and math behind hydroelectric power generation. Future articles are planned to address other aspects, such as turbine and generator design, but it may be a while. This here is the fun part.

The first thing to understand is that virtually all of the electricity generated in the world comes from using some physical process to create spin. If you can spin a turbine, that turbine can turn a generator, and that creates electricity. Spin the turbine against greater resistance, and you get more electricity. Hydroelectric power is about using water to spin the turbine. Pretty easy; just drop the water through the turbine. If you have a lot of water or if the water is under pressure – for example, if it has a lot of water above it pushing downward – then it will spin the turbine against greater resistance and generate more power. So hydroelectric power is fundamentally about harvesting the energy difference between water held at different elevations under the force of gravity. If you have a source of water at some elevation and you can control the water’s ability to flow to a lower elevation, then you can steal that energy and make it do work for you.

How much energy can you get? That depends on how much water you have and how much of an elevation change you can produce. Let’s walk through the equations:

  1. The equation we’re looking for defines the relationship between water, elevation, and energy production in familiar units we can use to understand the relationship. As Americans, we understand energy in terms of kilowatt-hours because that’s how our energy utility companies bill us for them, and we understand water in terms of gallons because that’s what we use when we cook, shower, and flush. So let’s try to build an equation that relates Watts to gallons.
    1. A kilowatt-hour (kWh) is a kilowatt (kW) (1,000 watts) of power exerted for one hour. Or half a kW (500 watts) exerted for two hours. Or two kW (2,000 watts) exerted for half an hour. So let’s say you have a one-kW microwave oven; if you run it for five minutes, you’ll use one-twelfth of a kilowatt-hour.  A typical house uses about 900 kWh per month (cite).
    2. Your bathtub probably holds about 40 gallons of water during a typical bath.  There are 7.48 gallons of water in a cubic foot of water.  A keg of beer is a little over two cubic feet of beer.  A gallon of water weighs about eight pounds near its boiling point, or about 8.3 pounds at room temperature.
  2. Since we all understand kilowatt-hours and gallons, the goal is to relate kilowatt-hours to gallons of water under force of gravity.  So, step one, let’s break down the kilowatt-hour into its component units until we find something that makes sense in terms of water and gravity.
    1. A Watt is a unit of power (i.e., how hard you’re pushing), and a Watt-hour is a unit of energy (how hard you’re pushing for how long).  The Joule is also a unit of energy, equal to Watt-second.  There are 3,600 seconds in an hour, so there are 3,600 Joules in a Watt-hour.  Then, because there are 1,000 Watt-hours in a kilowatt-hour, there are 3,600,000 Joules in a kilowatt-hour.  (Keeping all this relatable: a Calorie is 4.187 Joules, and is the heat energy required to heat one gram of water one degree Celsius – so a kilowatt-hour would be enough energy to heat about 3 gallons of water from room temperature (20 C) to boiling (100 C) – do bear in mind, however, that the “calories” that appear on your food’s nutrition labels are actually kilocalories – each one is 1,000 of the SI unit, the Calorie… therefore in order to burn off a 250-calorie Krispy Kreme doughnut, you’ll need to do about 0.30 kilowatt-hours of work).
    2. A Joule is a Newton-meter, which is starting to get us somewhere because Newtons are a measure of force, and the action of gravity on water is a force.  So the only piece of the puzzle still missing is: how many Newtons of force does gravity exert on a gallon of water?
    3. A Newton is a kilogram-meter per second per second.  Gravity pulls on objects at Earth’s surface with a force equal to their mass times 9.8 meters per second per second.  That means a kilogram of water weighs 9.8 Newtons… and because a kilogram weighs 2.2 pounds, that means a gallon of water at room temperature weighs about 37 Newtons.
    4. So then in order to get to Joules, we have to factor in meters.  If we have a gallon of water with two meters of drop, that’s 74 Joules, which is… only about 0.00002 of a kilowatt-hour.  So in order to get any meaningful power out of hydroelectric generation, it takes either a lot of water or a lot of height.
  3. Solving the equation we have set up, it turns out that:
    Wh = 0.01028 * gallons of H2O * meters of drop


    … or, since there are 7.48 gallons in a cubic foot of water (and it’s useful to know cubic feet since the flow rates of most bodies of water are measured in cubic feet per second):
    Wh = 0.07688 * feet3 of H2O * meters of drop


    … or, to calculate wattage expressed in terms of flow rate (gallons or cubic feet per minute):
    W = 0.62 * (gallons of H2O / minute) * meters of drop


    W = 4.6 * (feet3 of H2O / minute) * meters of drop

Now let’s play with those equations some…

First off, let’s say you have a license to dam the Yadkin River for purposes of hydroelectric power generation. Let’s say you dam the Yadkin, giving you the Yadkin’s entire flow rate of about 2,000 cubic feet per second to work with. And let’s say you build your dam to create a 60 foot elevation drop (about 19 meters). That means every second, you can generate about three kWh… which means that every hour, you can generate about eleven megawatt-hours. However, if you really wanted to maximize your profits, you’d probably build in about 40 MW worth of generating capacity, so that you could let the water level rise during non-peak times of day, and then generate like crazy during peak usage hours when energy is selling for a high price. How much money could you make? Well, if you could sell power to the grid for $0.05/kWh, you could make $4.8 million annually.

As another scenario, let’s do a basic analysis to see if it would be worthwhile to install a home hydro system. Say you’ve measured your flow rate and found that a stream on your property has a flow rate of 60 gallons per minute, and it undergoes a 20′ change in elevation on your property. Equation says this can produce somewhere around 67 Watts. Because this is a home setup, figure some losses to inefficiency, so maybe it will actually produce about 57 Watt-hours per hour. That’s about 41 kWh per month… which comes to about four bucks a month if you pay $0.10/kWh for energy. So you probably have better things to do than try to produce energy off that stream.

How much water or drop do you need to make it worthwhile? Well, I’d be pretty happy to have a system that could both erase my power bill and supply me with energy in the event of a major blackout. For such a system, I’d want to generate about 1000 kWh per month. There are about 720 hours in a month, so the system needs to generate about 1.4 kW. Applying an 85% efficiency factor, that means I need about 1.65 kW worth of water and drop. So, at a drop of 6 meters, I’d need a flow rate of 444 gallons per minute. Or, at a drop of 12 meters, I’d need a flow rate of 222 gallons per minute. Either way, that’s a pretty good-sized waterfall.


Measuring your flow rate

To measure your flow rate, grab a five-gallon plastic bucket, divert your stream’s water into it, and see how long it takes to fill. You could also measure the cross-sectional area of your stream and measure the water’s speed and then run those numbers, but that’s probably more work than the plastic bucket. You’re going to have to build some sort of diverter to get all the water into the hydro system anyway. Remember that flow rates can vary a lot with weather and time of year, so it may be advisable to take a few measurements.


Common Questions

Q: Do I need a drop?
A: Yes. Now, I know what your next question is: what about water wheels on old mills? True, those work. The thing is, to get any kind of meaningful electricity generation out of one, it would have to be huge.

Q: Can I keep my waterfall and also use the water and drop to generate electricity?
A: Not really, no. You have to channel the water from the higher elevation to your turbine, which sits at a lower elevation. It can’t just drop; it has to be channeled so that the water at the turbine is under pressure. Of course, you could just channel part of the water from the waterfall, and let the rest cascade over. That works.

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