“Be kind, for everyone you meet is fighting a hard battle” - Often attributed to Plato but likely from Ian McLaren (pseudonym of Reverend John Watson)

Saturday, November 20, 2010

When ignorance becomes a movement: The rise of Snookiism - By David Rothkopf | David Rothkopf

When ignorance becomes a movement: The rise of Snookiism - By David Rothkopf | David Rothkopf

The "embarrassed to be conservative" meme of my blog is, I'm afraid, becoming more and more prevalent. Why is it necessary to celebrate ignorance in order to be "conservative?"

C: Did a scientist say it?
S: Yes
C: Did he/she go to a college or university?
S: Yes, of course
C: Then I don't believe a word her or she says. All colleges and universities are nothing more than institutions for brainwashing attendees into believing in one-world government and the socialist agenda.
S: But we're talking about (climate, public health, evolution), not politics.
C: You weren't paying attention, were you? It's all part of the conspiracy for one-world government and the socialist agenda.

I really don't want this blog to become a platform for political commentary, and my belief in preserving the environment, fiscal conservatism, personal responsibility, minimal government intrusion, and the brilliance and timelessness of our Constitution is unshaken. But the TMZ/MTV/"I don't really like reading books"/Jersey Shores/Rush Limbaugh/Marc Morano/Glenn Beck absorption of the American dialogue is filling me with grief for the disappearance of thinking, moral conservatism.

Hello? Is there anybody out there?

Update: I tried my first xtranormal movie to illustrate my point.

Friday, November 19, 2010

What can the wind do?

After my post regarding the wind turbines atop Houston's Hess tower, I thought a very brief and very elementary primer on wind power might be in order. Moving air posesses kinetic energy in accordance with E=1/2*m*v^2. This energy can be harnessed to generate electricity. So how much energy is this?

Let's look at the turbines from the Hess Tower that I discussed in the post.  Here, each turbine intercepts an area of 8.25 m^2. Let's let s be the wind speed (since the turbines have a vertical axis the wind direction doesn't matter, so we don't need the vector quantity) in m/s (meters per second). Then, in one second, each turbine will intercept a volume V of 8.25*s m^3 (meters cubed). Assuming air has a density, d, of 1.16 kg/m^3, the mass,m, of air intercepted in a second is V*d=8.25*s*1.16 kg. This mass, moving at a speed of s m/s has a kinetic energy, E, of 0.5*m^s^2 or 0.5*8.25*s*1.16*s^2.

One thing to note before moving on is that s is present twice - once to the first power and once squared - meaning that the available energy PER SECOND (remember, we are talking about the amount of energy available in one second's worth of wind) is proportional to the cube of the wind velocity. Now, the total available energy per second in the wind is otherwise known as power. In other words, this number is the power that would be produced at a given wind speed, s, if 100% of it could be converted to electrical energy. Of course it can't, and we'll get to that later.

So, where are we? For a speed, s, the kinetic energy, E, per second, or power, P, in watts passing through one of the Hess turbines is 0.5*8.25*1.16*s^3 or, P=4.785*s^3 watts. Like a good physicist, let's check the units: 0.5 is dimensionless; 8.25 is m^2; 1.16 is kg*m^(-3); s is m*s^(-1) but it's cubed so that factor has units m^3*s^(-3). Thus, we have: for m, m^2*m^(-3)*m^3 or m^2; kg is merely kg^1; for s we have  s^(-3). The total is kg*m^2*s^(-3). Now, power is energy per second, or force * distance/second, or (mass*distance/time^2)*distance/second. This is kg^1*m*s^(-2)*m*s^(-1) or kg*m^2*s(-3). The units check. For ANY equation relating physical quantities, the units on each side MUST match, this is a necessary (but not sufficient) condition for the equation to be valid.

Let's stick a number in and see what we get. A 15 m.p.h. wind is fairly robust, how much power is available for the turbine's conversion to electricity? 15 m.p.h. is 6.71 m/s, the total power (through an area the size of a single Hess turbine) is 4.785*6.71^3 or 1,443 watts. If it all could be converted to electricity, it would light a little more than 14 100 watt light bulbs.

Of course it can't all be converted. One excellent intuitive way of understanding this is that the kinetic energy comes from the speed, to extract it all we would need to bring the wind to a complete stop, requiring a solid wall at the "turbine" location, something that obviously cannot work. Albert Betz was the first to calculate the absolute theoretical maximum energy that could ideally be extracted from wind. This is called, appropriately enough, the Betz Limit, and it's 59.3%.

Now a real turbine will not approach this limit, and horizontal axis wind turbines, at least in steady wind, are more efficient than vertical axis wind turbines like those at Hess Tower. From the data available at the
Cleanfield Energy site (the manufacturer of the Hess turbines) I estimate that the claim is that the turbines are about 37% efficient.

I've included a chart showing the total power in wind, as well as the Betz Limit and the power at 37% efficiency for wind from 0 to 30 m.p.h. The actual numbers apply specifically to a single Cleanfield turbine like the ones on the Hess Tower, but the shapes apply to all wind turbines, and keep in mind that the turbines typically don't turn at all until the wind speed reaches somewhere around 8 or 10 m.p.h. (Click to enbiggen).

Here's a graph showing the Betz Limit power in wind for speeds from 2.5 m/s to 20 m/s (about 5.6 m.p.h. to 44.7 m.p.h.) and turbine area of 5 m^2 to 25 m^2. Note the extreme dependence on turbine size and, especially, wind speed.

Thursday, November 18, 2010

Could the boys and girls at the Wall Street Journal please buy Keith Johnson a calculator?

As I read some of the news of the day, concentrating on some of my RSS feeds for wind related stories, I was drawn to this article in the (formerly) esteemed Fox Financial News Wall Street Journal. The thrust of the article is that offshore wind power is expensive to install. I have no reason to think that that's not true, it looks like South Korea expects to pay about $3.32M per megawatt, a lot of money in anyone's book.

But in the linked article we find:

"Take the new proposal for the world’s biggest wind farm by another Texas oil man, peak oil prophet Matt Simmons. His Ocean Energy Institute proposes building a 5,000 megawatt deepwater wind farm in the Gulf of Maine, blessed with some of the world’s strongest sustained winds.
The problem is that, as envisioned, the Maine offshore wind farm would be very expensive—and that vision includes some very optimistic assumptions.
Ocean Energy figures capital costs for the project could go as high as $4.5 billion a megawatt, a lot more than Mr. Pickens projects for his massive Texas wind farm. All in, the costs for the Maine project could come to $25 billion, or $5 billion a megawatt, the Ocean Energy folks told Earth2Tech. That compares to upfront costs of about $600 million per megawatt for old-fashioned coal-fired plants."
Woah. Let's see here. $5B/megawatt for 5,000 megawatts. That's $25T (trillion). But didn't he say that it would cost $25B? Yes, he did. So, which is it? I can assure my panic stricken windpower fans that $25B is correct for an installed cost of $5M/megawatt.
Keith Johnson is the lead writer of the Environmental Capital section of the Wall Street Journal which "provides daily news and analysis of the shifting energy and environmental landscape." It's led by Journal energy reporter Russell Gold.
The first commenter pointed out the three order of magnitude error, saving me the trouble.

I don't have a barometer but I have an iPhone

There's an old apocryphal tale of a high school physics student taking a test who's asked "how would you use a barometer to measure the height of a building?" The expected answer is "measure the barometric pressure at the bottom and the top, use the pressure lapse rate with altitude to determine the height." This was not the answer given by the student. When his teacher marked the answer as incorrect, the student protested, telling the teacher that there are many ways to use a barometer to find the height of a building. The teacher was curious and asked the student to name a few. The student came up with (in addition to the obvious answer above):
  • On a sunny day, set the barometer on a sidewalk and measure the height of the barometer and its shadow. Measure the building's shadow, apply the ratio.
  • Drop the barometer from the roof, time its descent to the street and use s=.5*g*t^2.
  • From the roof, tie the barometer to a string, lower it until it almost touches the street. Set it swinging and measure the period, and apply t=2*Pi*(L/g)^0.5
  • Measure the height of the barometer, use the core stair to count barometer heights to the top of the building, multiply.
  • The easiest? Take the barometer to the building manager and say "I have a fine barometer here, and if you'll tell me the height of your building, I'll give it to you."
Not having a barometer handy, I didn't do any of those, but I used the Pasco SPARKvue program on my iPhone to measure and record the acceleration during an elevator trip from the third floor to the 11th of the Hilton Americas-Houston where I'm staying. The program will email the data in CSV format. I then used Excel to numerically integrate the acceleration and the velocity to find displacement. The distance worked out to  be 100' 1 1/2" for the eight floors, a floor height of about 12.5'. This is a little more than I'd have estimated. The hotel has 19 floors and, I assume, a mechanical penthouse, so I'm going with 19*12.5 + 10 (the mechanical penthouse will be shorter) for a total of 247.5'.

The top (and steady) speed reached by the elevator (a Schindler) was 7.9 m.p.h. The peak acceleration was about 0.8 m/s^2, less than 0.1 "g". And if my iPhone were as smart as it (and Steve Jobs) thinks it is, it could distinguish acceleration from gravity in a closed elevator cab. Oh, wait...

Wednesday, November 17, 2010

A wind blows in Houston

I'm at the ASNT (American Society for Nondestructive Testing) Fall Conference and Quality Testing Show in downtown Houston, TX. I'm staying at the Hilton Houston Americas, with a northward view across Disovery Green Park. I was here in May of 2009 for the Clean Technology Conference and Expo and saw an office building under construction. It's now completed, it had been called Discovery Tower but has now been renamed Hess Tower since the Hess Corporation (formerly Amerada Hess) leased all 872,000 square feet of space in the building.

As I looked out my 11th floor window at the 30 story tower, a spinning motion caught my eye. I looked up at the tower and, sure enough, there appeared to be a series of vertical axis wind turbines adorning the top of the tower. They are certainly eye-catching and, to my uneducated eye, add to the visual appeal of the building. But do they provide enough energy to justify their cost to manufacture, install, and maintain?

Information as to the cost and expected energy output is sparse in the google-verse though I did see that Cynthia Cisneros of KTRK did a brief television piece about it. She exhibited her and her editor's cluelessness by stating that each of the 10 turbines is "designed to generate approximately 3.5 kilowatts per hour." She does state (and I've read at a couple of other sites) that the turbines should generate sufficient energy to light the building at night or to power two office floors.

I was able to finally determine that the turbines are manufactured by Cleanfield Energy, headquartered in Ontario, Canada and that they are Cleanfield's Model V3.5. They're "designed to harness urban wind efficiently and effectively." On their web site Cleanfield provides a specification sheet for the turbine which I would copy here if it weren't for fear of a lawsuit.

But using the "Estimated Energy per Year" chart and an average wind velocity of 4.5 meters/second (downgraded for the urban setting from an 80 meter wind average wind velocity chart here) I'll estimate that each turbine will provide about 1,500 kilowatt hours/year for a total of 15,000 kilowatt hours/year, or 15 megawatt hours/year. Note that this is a capacity factor (assuming 3.5 kilowatts rated power) of 1,500/(3.5*24*365) * 100% or about 4.9%.

Now, I suppose that Hess Corporation gets a better deal on energy than I do, but it's probably not too far off to figure they pay about $0.10/kilowatt hour. So, by avoiding the purchase of 15,000 kilowatt hours they'll save something like $1,500. I don't know what the generators cost but I suspect that the net present value of the investment is negative no matter how small Hess' cost of capital.

I've foundcouple of estimates for electricity use in office buildings on a "per square foot" basis, at 18.9 and 17 kWh/year. If I assume that the Hess building is MUCH more efficient at, say, 15 kWh/year/ft^2, then at 872,000ft^2/30 floors there are about 29,000 ft^2/floor. Two such floors would use 870,000 kilowatt hours/year. Use of a median rather than an average electrical energy usage might cut this number by about a third (very optimistically) so let's say 580,000 kilowatt hours/year might be a reasonable lower bound. I don't think the turbines will do it. They'd supply about 1,250 ft^2, a small office.

A video is here:


Friday, November 05, 2010

$200M/day, 34 naval vessels, raining kittens and puppies

I'm going to have to come up with a new name for my personal political viewpoint, I can't continue to be associated with what are commonly referred to now as "conservatives."

I've mentioned on a couple of occasions that I make it my habit to listen to and read from sources of editorial opinion that span the political spectrum, e.g., I listen to both KPFK, the local Pacifica network outlet where National Public Radio is regarded as a bunch of right wing reactionaries and tools of the bourgouisie, and to KRLA, the Salem Radio Network outlet where Hugh Hewitt, Michael Medved, Dennis Prager, Dennis Miller, Bill Bennett, and Glenn Beck (!) hold forth.

I was listening to Mike Gallagher Wednesday evening to sample the conservative talk radio reaction to Tuesday's election. I was hearing generally what I expected when my ears pricked up as Gallagher started going on about Obama's visit to India. He said he'd read that $200M per day would be spent on the trip, including hotels, security, air transport, 34 naval vessels, etc. He didn't buy into this figure with complete abandon, saying that it was from an Indian newspaper, and prefacing some of his remarks with "if this is true..." But in speaking with subsequent callers he did rail against Obama using that figure.

Now, as it turns out, even the hardcore conservatives have distanced themselves from the story, but just how clueless do you have to be to hear such a thing and not immediately reject it out of hand? What mental processes must have broken down to make a person WITH A NATIONALLY SYNDICATED TALK SHOW so credulous that he even momentarily took such a thing seriously?

Does being a conservative mean that if someone tells you it's raining kittens and puppies you worry about taxes being raised to clean up the poop?