I received a comment to my post on the Moller Skycar Volantor suggesting that I "look up the Wankel rotary engine (Mazda RX8) - 1.3L = 200+ HP." I'm not completely sure what the commenter was getting at, but I think he or she was saying that it's possible to get the power claimed by Moller in a package of the type he claims to have it in. Let me be clear (quoting our President): I agree. There's no question in my mind that that's possible. I didn't discount that possibility in my post.

What I did say isn't possible is to develop that amount of power and still achieve 20 miles per gallon of fuel at the speed claimed. That is, the combination of airspeed, fuel economy, and engine power claimed is not realistic. I'd like to delve into this in a little more detail. An internal combustion engine works by taking some combustible fuel into an enclosed space and igniting it. This results in a large temperature rise, and the consequent increase in pressure is used to push down a piston, turn a rotor, or turn a turbine thus converting the potential energy in the chemical bonds of the fuel into mechanical energy and using it to do work.

Any given fuel has a fixed amount of energy available for release by oxidation for any given mass, volume, or number of "moles" of substance. In a perfect world, unencumbered by the second law of thermodynamics, we could harness 100% of this energy to do useful work. Even in such a perfect world, no more could be had. And once the time over which this chemical potential energy is converted by oxidation into internal energy or "heat" is noted, we can divide the energy in the quantity of fuel (energy available is the same as work that can be done) by the time, we have work divided by time, or power.

For example, my Piper Saratoga PA32R-301T will burn about 18 gallons of fuel per hour to go about 170 knots. A "knot" is one nautical mile per hour, and a nautical mile is 6080 feet or about 1.152 statute miles. So, the plane burns 18 gallons in an hour to go 170*1.152 or 195.6 miles. Thus it's achieving 10.9 m.p.g., but more to the point of this post, it's burning 18 gallons per hour. There are a variety of sites with somewhat differing values for the energy density of the 100LL avgas I burn in the Saratoga (see here and here for example) so I'll use an average of 32.6 megajoules/liter or 123.4 (easy to remember) megajoules/gallon.

So, I'm releasing 18*123.4 megajoules/hour of chemical potential energy. Since an hour is 3600 seconds that's 18*123.4/3600 megajoules per second or 617,000 joules per second. By definition, a joule/second is a watt so this is 617 kilowatts or 827 horsepower. This is interesting, my pilot's operating handbook says that the power setting producing this fuel flow is 70% of the 300 maximum continuous horsepower available, or 210 horsepower. Thus, I'm using (210/827)*100% or 25.4% of the heat released by burning the avgas. This is pretty close to the type of efficiency we've come to expect of internal combustion engines in typical applications.

The maximum possible efficiency allowed by the laws of thermodynamics for a "heat engine" was determined in the 19th Century beginning with the work of
Sadi Carnot and is determined solely by the temperature of the working fluid (fuel/air mixture as it oxidizes in this case) and the cold reservoir (the atmosphere) and for reasonable temperatures, as stated in my post on the Moller Skycar, is about 82%. For a working internal combustion engine, the maximum efficiency is related to the compression ratio and, for a typical compression ratio of 10.5:1, works out to be about 61%. Various factors involving friction, the operating points of the engine, inefficiencies in the fuel delivery and exhaust, and many other factors cause the actual efficiency to be as low as it is. These considerations apply equally to the Wankel or rotary engines in the Moller Skycar.

But, if the rate of fuel burn in volume/time and the type of fuel are known (or can be calculated) then the maximum available power can be calculated. An estimate of engine efficiency can then give the power available to turn wheels, turn a propellor, turn a ducted fan, etc. This is what I did in the Moller Skycar post, using the miles per hour divided by miles per gallon to give gallons per hour and thus the heat energy available. This enabled me to demonstrate that the claims made by Moller for the Skycar are not feasible.

A look at energy use in my life and how it applies to others' lives

## Tuesday, March 31, 2009

## Sunday, March 29, 2009

### "Exponentially" really does mean something

In my never ending quest to understand what people believe about the world, and specifically about energy and fuel economy, I run across repeated misunderstanding of the word "exponentially." This or that "increases exponentially." Now some things do, in fact, increase exponentially. Money in a bank account at a fixed rate of interest comes to mind. The population of bacteria in a petri dish or humans on a finite planet (for a while) are other examples.

But it means more than "gets big quickly." People who should know better, or at least should learn better before using it this way frequently use it to mean this. An example is an article at the Planet Green web site entitled "Become a "Hypermiler," Save Even More Gas" by one Collin Dunn of Corvallis, OR. As usual, driving more slowly is first on his list. OK, I agree completely, but then he states that "The amount of drag your vehicle generates increases exponentially with each increase in speed; that is, driving a little faster generates a lot more drag, which requires more gas to overcome."

Well, it's just some guy writing an article for a Cable Channel web site, right? But he got this gem from the Toyota Open Road Blog where they go into detail to precisely state their error: "The amount of drag your vehicle generates is not linear – it does not increase at the same rate as your vehicle’s speed does. Instead, drag is more or less proportional to the square of speed. It increases exponentially." NO, NO, NO!

"More or less proportional to the square of speed" is correct. That is, if you know the drag at some speed, and want the drag at some other speed, take the ratio of the second speed to the first, square it, multiply the drag at the first speed by the squared ratio and there's your approximate drag at the second speed. For example, doubling the speed would approximately increase aerodynamic drag by a factor of four. This is NOT exponential, the mathematically astute refer to it as a power function. The variable (speed) is the base, the power (2 in this case) is the exponent. The exponent is fixed.

An exponential function has a fixed base, and the variable is the exponent. Now, depending on the ranges of the variables and the fixed numbers and the proportionality constants, the exponential function may be smaller than the power function at some values of the variable, but the exponential function is always ultimately larger for large values of the variable. So while something that increases exponentially really does get large very fast, at least after a while, it's not true that anything that increases quickly at any point increases exponentially. This most assuredly does include aerodynamic drag.

And don't even get me started on "mega."

But it means more than "gets big quickly." People who should know better, or at least should learn better before using it this way frequently use it to mean this. An example is an article at the Planet Green web site entitled "Become a "Hypermiler," Save Even More Gas" by one Collin Dunn of Corvallis, OR. As usual, driving more slowly is first on his list. OK, I agree completely, but then he states that "The amount of drag your vehicle generates increases exponentially with each increase in speed; that is, driving a little faster generates a lot more drag, which requires more gas to overcome."

Well, it's just some guy writing an article for a Cable Channel web site, right? But he got this gem from the Toyota Open Road Blog where they go into detail to precisely state their error: "The amount of drag your vehicle generates is not linear – it does not increase at the same rate as your vehicle’s speed does. Instead, drag is more or less proportional to the square of speed. It increases exponentially." NO, NO, NO!

"More or less proportional to the square of speed" is correct. That is, if you know the drag at some speed, and want the drag at some other speed, take the ratio of the second speed to the first, square it, multiply the drag at the first speed by the squared ratio and there's your approximate drag at the second speed. For example, doubling the speed would approximately increase aerodynamic drag by a factor of four. This is NOT exponential, the mathematically astute refer to it as a power function. The variable (speed) is the base, the power (2 in this case) is the exponent. The exponent is fixed.

An exponential function has a fixed base, and the variable is the exponent. Now, depending on the ranges of the variables and the fixed numbers and the proportionality constants, the exponential function may be smaller than the power function at some values of the variable, but the exponential function is always ultimately larger for large values of the variable. So while something that increases exponentially really does get large very fast, at least after a while, it's not true that anything that increases quickly at any point increases exponentially. This most assuredly does include aerodynamic drag.

And don't even get me started on "mega."

## Thursday, March 26, 2009

### A potpourri of cluelessness

In my reading of various blogs, news articles, forum posts, etc. I've encountered a large number of writings indicative of the strange and distorted ideas people have about the subjects of this blog, that is, of fuel economy, energy use, physics, and life. I've decided to periodically quote some of them along with, where it's relevant, my own comments.

The following is from Physics Forums where the question was "Does RPM affect gas mileage?" An answer was:

"Assume that you travel a set distance at the same speed, first in 3rd gear, then in 4th. Can you see that the engine will turn more times in the lower gear. Each revolution of the engine will "consume" the same amount of air/fuel mixture. Therefore you must consume more air/fuel mixture at the lower gear.

Now a lot of modern engines are getting smarter about feeding fuel so, the assumption of a constant a/f mixture may not be valid. With intelligent fuel metering the millage difference may be small.

The biggest difference to fuel consumption is your rate of acceleration. If you like to feel some acceleration and take pride in your ability to get to 60mph (100kph) then you will see improvement by playing "old lady" for a while.

You do more work when doing 5s to 60 vs 10s to 60. This increased work MUST be reflected in fuel consumption."

Well. The fact is that RPM will affect gas mileage but this guy's argument is full of errors. He (I assume it's "he") correctly states that the lower gear will require more revolutions to travel a given distance than a higher gear, but then claims that each revolution will consume the same amount of fuel/air mixture regardless of the gear. This is false. The lower gear will travel a lesser distance and therefore do less work for each revolution, thus requiring less throttle since the gear ratio will give a greater mechanical advantage. But you'll need to apply this lesser throttle through more revolutions of the engine. Now, due to throttling and frictional losses, you'll be less efficient in the lower gear but it's certainly not true that "Each revolution of the engine will "consume" the same amount of air/fuel mixture." And this was true before cars included computerized controls, it's only a matter of the throttle position.

Next, "Integral" (his screen name, complete with integral symbol avatar) is completely off base with his concept of acceleration and work. Going from 0 to 60 adds the same amount of kinetic energy to the car, and thus takes the same amount of work, regardless of whether it's done in 5 seconds or 10. It's true that taking 10 seconds is more fuel efficient, since the energy is added over a longer distance but Integral confuses work with power. And this is from a Physics Forum where he has enough posts to be a mentor. Further, someone points out his errors and he stridently defends them, even insulting the corrector.

Moving on, "Josh," in reply to a tip at Daily Fuel Economy Tip that suggested minimizing use of electrical accessories said: "This is simply not true …. and alternator is not operated on a clutch therefore it spins at the same speed no matter what is being used in the car. It is not on a clutch like the AC that just kicks in when the compressor comes on."

Of course, this is false. Raising the electrical load causes an increased magnetic field in the alternator field coil, resulting in a larger torque to be overcome by the alternator drive belt, thereby using more fuel.

These are merely misunderstandings of fundamental physical principles and don't reach the heights of looniness achieved by some of the more "outro" world wide web denizens. I'll cover more of each type in subsequent posts.

The following is from Physics Forums where the question was "Does RPM affect gas mileage?" An answer was:

"Assume that you travel a set distance at the same speed, first in 3rd gear, then in 4th. Can you see that the engine will turn more times in the lower gear. Each revolution of the engine will "consume" the same amount of air/fuel mixture. Therefore you must consume more air/fuel mixture at the lower gear.

Now a lot of modern engines are getting smarter about feeding fuel so, the assumption of a constant a/f mixture may not be valid. With intelligent fuel metering the millage difference may be small.

The biggest difference to fuel consumption is your rate of acceleration. If you like to feel some acceleration and take pride in your ability to get to 60mph (100kph) then you will see improvement by playing "old lady" for a while.

You do more work when doing 5s to 60 vs 10s to 60. This increased work MUST be reflected in fuel consumption."

Well. The fact is that RPM will affect gas mileage but this guy's argument is full of errors. He (I assume it's "he") correctly states that the lower gear will require more revolutions to travel a given distance than a higher gear, but then claims that each revolution will consume the same amount of fuel/air mixture regardless of the gear. This is false. The lower gear will travel a lesser distance and therefore do less work for each revolution, thus requiring less throttle since the gear ratio will give a greater mechanical advantage. But you'll need to apply this lesser throttle through more revolutions of the engine. Now, due to throttling and frictional losses, you'll be less efficient in the lower gear but it's certainly not true that "Each revolution of the engine will "consume" the same amount of air/fuel mixture." And this was true before cars included computerized controls, it's only a matter of the throttle position.

Next, "Integral" (his screen name, complete with integral symbol avatar) is completely off base with his concept of acceleration and work. Going from 0 to 60 adds the same amount of kinetic energy to the car, and thus takes the same amount of work, regardless of whether it's done in 5 seconds or 10. It's true that taking 10 seconds is more fuel efficient, since the energy is added over a longer distance but Integral confuses work with power. And this is from a Physics Forum where he has enough posts to be a mentor. Further, someone points out his errors and he stridently defends them, even insulting the corrector.

Moving on, "Josh," in reply to a tip at Daily Fuel Economy Tip that suggested minimizing use of electrical accessories said: "This is simply not true …. and alternator is not operated on a clutch therefore it spins at the same speed no matter what is being used in the car. It is not on a clutch like the AC that just kicks in when the compressor comes on."

Of course, this is false. Raising the electrical load causes an increased magnetic field in the alternator field coil, resulting in a larger torque to be overcome by the alternator drive belt, thereby using more fuel.

These are merely misunderstandings of fundamental physical principles and don't reach the heights of looniness achieved by some of the more "outro" world wide web denizens. I'll cover more of each type in subsequent posts.

## Saturday, March 21, 2009

### The Moller Skycar

Since I was a little boy, there has been talk of flying cars. And since I was a little boy, Paul Moller has been a short few years away from going into production on such a vehicle. He still is, though now it's called the Skycar Volantor. It's reminiscent of the wag's remark about fusion energy: "fusion is the energy source of the future, and always will be."

But what about the M400 Skycar? It's "specifications" can be found here. What a fine way to get around! 275 m.p.h. cruise at better than 20 m.p.g. It's stated that the production model will employ eight rotary engines rated at 150 h.p. per engine burning any of a variety of fuels, but ethanol is suggested. This is the power required for the vertical take off capability. At this site, there are a variety of videos, including a hover test of the M400, it apparently will get off the ground. There are also some specifications wherein it's indicated that the "nominal continuous power" is 720 horsepower and the fuel is ethanol. This would be running at 60% power and makes sense. It's not stated whether the engines are turbocharged, but they must be since the operational ceiling is stated to be 36,000 feet (!).

So let's take a look at a vehicle utilizing internal combustion engines to get 20 m.p.g. (the specs. say ">20" so this should be conservative) at 275 m.p.h. This means it's burning 13.75 gallons of ethanol per hour. OK, we find here that ethanol has an energy density of 6,100 watt hours/liter, or about 83,100,000 joules/gallon. So the total available energy in 13.75 gallons of ethanol is 1,143,000,000 joules. Burning this in one hour or 3600 seconds at 100% efficiency will produce 317,395 watts or 426 horsepower. This means the Skycar's engines are using fuel with an efficiency of 169%. I rather doubt it.

Let's suppose, then, that the figures come from burning gasoline. Gasoline has a significantly higher energy density, and would mean the claim is only an efficiency of 106%. Still higher than the figures one typically sees for an internal combustion engine. All right, suppose that the 720 horsepower only applies to the "top speed," listed as 360 m.p.h. Now, with gasoline, we're looking at an efficiency of 81%. Well, at least it's no longer in the category of the "over unity" nut cases, but I've never seen a real engine with such a specification and, unless it's operating at a very high temperature and dumping into a very cold reservoir, thermodynamics won't allow it. If a heat engine is operating from 1500 K into a reservoir of 273 K (about 2240 degrees fahrenheit into 32 degrees fahrenheit) the absolute theoretical maximum efficiency is 81.8%. No real engine comes close.

Is 720 horsepower sufficient to produce a speed of 360 m.p.h.? It's likely that it is. The Piper Meridian, for example, utilizes a Pratt & Whitney PT6A-42A turboshaft engine running at 500 horsepower to cruise at 260 knots, or about 300 m.p.h. It burns on the order of 40 gallons of Jet A fuel each hour to do it though. Speed available goes up approximately with the cube root of power, so 720 horsepower should be able to give an increase of about 12.9% over the Meridian, all else being equal. This would be about 339 m.p.h. Maybe the Skycar is a little aerodynamically cleaner, maybe its ducted fans are slightly more efficient than the Meridian's propeller. I'll call it plausible.

So if the Skycar were to be a real product, what are its claimed advantages? Well, it's driveable at low speed from your garage to an approved takeoff location where it can take off vertically. It's being designed, ultimately, to be fully automated, no pilot intervention necessary, and thus able to be used by those with no flying skills of any kind. You were previously able to buy a place in line for a production Skycar at a cost for delivery of about $400K to $1M (according to some dated web sites) depending on where in line you were. You could put only part of that down to reserve your place. It seems that that's no longer the case though.

Dr. Moller is now stating that, in limited production, the M400 Skycar Volantor will sell for about $500,000 and be available in "about three years." In mass production (i.e., when everyone you know is buying one) they'll sell for $60,000 to $80,000. If you'd rather own the company than the Skycar, it's sold over the counter as MLER.OB on the OTCBB ("Over the Counter Bulletin Board"). You can peruse the financials here. Note the negative book value. No wonder they're not taking deposits. Its market capitalization values the company at $8.27M, with 54% owned by insiders and 5% holders. Pretty thin. You just can't beat the laws of thermodynamics.

But what about the M400 Skycar? It's "specifications" can be found here. What a fine way to get around! 275 m.p.h. cruise at better than 20 m.p.g. It's stated that the production model will employ eight rotary engines rated at 150 h.p. per engine burning any of a variety of fuels, but ethanol is suggested. This is the power required for the vertical take off capability. At this site, there are a variety of videos, including a hover test of the M400, it apparently will get off the ground. There are also some specifications wherein it's indicated that the "nominal continuous power" is 720 horsepower and the fuel is ethanol. This would be running at 60% power and makes sense. It's not stated whether the engines are turbocharged, but they must be since the operational ceiling is stated to be 36,000 feet (!).

So let's take a look at a vehicle utilizing internal combustion engines to get 20 m.p.g. (the specs. say ">20" so this should be conservative) at 275 m.p.h. This means it's burning 13.75 gallons of ethanol per hour. OK, we find here that ethanol has an energy density of 6,100 watt hours/liter, or about 83,100,000 joules/gallon. So the total available energy in 13.75 gallons of ethanol is 1,143,000,000 joules. Burning this in one hour or 3600 seconds at 100% efficiency will produce 317,395 watts or 426 horsepower. This means the Skycar's engines are using fuel with an efficiency of 169%. I rather doubt it.

Let's suppose, then, that the figures come from burning gasoline. Gasoline has a significantly higher energy density, and would mean the claim is only an efficiency of 106%. Still higher than the figures one typically sees for an internal combustion engine. All right, suppose that the 720 horsepower only applies to the "top speed," listed as 360 m.p.h. Now, with gasoline, we're looking at an efficiency of 81%. Well, at least it's no longer in the category of the "over unity" nut cases, but I've never seen a real engine with such a specification and, unless it's operating at a very high temperature and dumping into a very cold reservoir, thermodynamics won't allow it. If a heat engine is operating from 1500 K into a reservoir of 273 K (about 2240 degrees fahrenheit into 32 degrees fahrenheit) the absolute theoretical maximum efficiency is 81.8%. No real engine comes close.

Is 720 horsepower sufficient to produce a speed of 360 m.p.h.? It's likely that it is. The Piper Meridian, for example, utilizes a Pratt & Whitney PT6A-42A turboshaft engine running at 500 horsepower to cruise at 260 knots, or about 300 m.p.h. It burns on the order of 40 gallons of Jet A fuel each hour to do it though. Speed available goes up approximately with the cube root of power, so 720 horsepower should be able to give an increase of about 12.9% over the Meridian, all else being equal. This would be about 339 m.p.h. Maybe the Skycar is a little aerodynamically cleaner, maybe its ducted fans are slightly more efficient than the Meridian's propeller. I'll call it plausible.

So if the Skycar were to be a real product, what are its claimed advantages? Well, it's driveable at low speed from your garage to an approved takeoff location where it can take off vertically. It's being designed, ultimately, to be fully automated, no pilot intervention necessary, and thus able to be used by those with no flying skills of any kind. You were previously able to buy a place in line for a production Skycar at a cost for delivery of about $400K to $1M (according to some dated web sites) depending on where in line you were. You could put only part of that down to reserve your place. It seems that that's no longer the case though.

Dr. Moller is now stating that, in limited production, the M400 Skycar Volantor will sell for about $500,000 and be available in "about three years." In mass production (i.e., when everyone you know is buying one) they'll sell for $60,000 to $80,000. If you'd rather own the company than the Skycar, it's sold over the counter as MLER.OB on the OTCBB ("Over the Counter Bulletin Board"). You can peruse the financials here. Note the negative book value. No wonder they're not taking deposits. Its market capitalization values the company at $8.27M, with 54% owned by insiders and 5% holders. Pretty thin. You just can't beat the laws of thermodynamics.

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