One of the themes I’ve been exploring for the past few years is computational thinking. It’s an evocative phrase that has led me in a few different directions. One is my intentional use of tagging and syndication as key strategies for social information management. Another is my growing interest in the kinds of uses of WolframAlpha outlined in Kill-A-Watt, WolframAlpha, and the itemized electric bill.
A lot of what I’ve read and heard about WolframAlpha seems to focus on its encyclopedic nature. But it aims to be a compendium of computable knowledge, and as such I think its highest and best use will be to enable computational thinking.
Here’s one small but telling example from my Kill-A-Watt essay:
A: About half the energy released by combustion of one kilogram of gasoline.
Q: ( 1 kilogram / density of gasoline ) / 2
A: Less than a fifth of a gallon.
I was trying to understand what 9 Watts, over the course of a month, means. WA offered the comparison to the amount of energy in gasoline, but reported in kilograms. I still think in gallons. The conversion is:
( 1 kg / .73 kg/L) / 2 = .685L * .264 gallons / L = .18 gallons
If you don’t do that kind of thing on a regular basis, though — as I don’t, and as many of us don’t — it’s hard to get over the activation threshold. Looking up and applying the relevant formulae is a multistep procedure. WA collapses it into a single step:
( 1 kilogram / density of gasoline ) / 2
It knows the density of gasoline, and when you do the computation it reports results in a variety of units, including gallons.
I was feeling a bit guilty about needing this sort of intellectual crutch. But then I heard from a friend who had just read the Kill-A-Watt/WA piece. It reminded him of an Energy Tribune article entitled Understanding E=mc2 which concludes:
A 1000-MW coal plant — our standard candle — is fed by a 110-car “unit train” arriving at the plant every 30 hours — 300 times a year. Each individual coal car weighs 100 tons and produces 20 minutes of electricity. We are currently straining the capacity of the railroad system moving all this coal around the country. (In China, it has completely broken down.)
A nuclear reactor, on the other hand, refuels when a fleet of six tractor-trailers arrives at the plant with a load of fuel rods once every eighteen months. The fuel rods are only mildly radioactive and can be handled with gloves. They will sit in the reactor for five years. After those five years, about six ounces of matter will be completely transformed into energy. Yet because of the power of E = mc2, the metamorphosis of six ounces of matter will be enough to power the city of San Francisco for five years.
This is what people finds hard to grasp. It is almost beyond our comprehension. How can we run an entire city for five years on six ounces of matter with almost no environmental impact? It all seems so incomprehensible that we make up problems in order to make things seem normal again. A reactor is a bomb waiting to go off. The waste lasts forever, what will we ever do with it? There is something sinister about drawing power from the nucleus of the atom. The technology is beyond human capabilities.
But the technology is not beyond human capabilities. Nor is there anything sinister about nuclear power. It is just beyond anything we ever imagined before the beginning of the 20th century. In the opening years of the 21st century, it is time to start imagining it.
Six ounces of matter? Really? My friend wrote:
I remember at the time I tried to run simple order of magnitude calculations in my head to verify the number, but it got messy, I got sidetracked, and forgot.
This time I went to Wolfram-Alpha, and the answer was right there, clear as day, in seconds (and yes, it’s really 6 ounces of matter).
I went back to the article, and the only quantity of energy reported for San Francisco was that Hetch Hetchy Dam “provides drinking water and 400 megawatts of electricity to San Francisco.” That alone would come to:
400MW * 5 years = ~700 grams = ~25 ounces
Or, if Wikipedia is right and the dam yields only about 220MW, then:
220MW * 5 years = ~386 grams = ~14 ounces
Of course since San Francisco has other sources of power, the amount of matter would be more. Still, this doesn’t invalidate the author’s point: we’re talking ounces, not tons.
When I mentioned this to my friend, though, he wrote back:
I went the other way around:
http://www.wolframalpha.com/input/?i=6oz * c^e2 in gw hr
It gives 4247 GWhr which is definitely in the ballpark for San Francisco.
Sweet!
I didn’t actually follow up on that result just now, but over 5 years it comes to:
http://www.wolframalpha.com/input/?i=4247GWh / 5 years = ~100MW. That’s a quarter of what the article reports for Hetch Hetchy, it’s half what Wikipedia reports, and I still don’t know how it relates to San Francisco’s total power draw.
Even so, we’re playing in the kind of ballpark we need to be able to play in if we’re going to have any kind of reasoned discussion about future energy mixes like Saul Griffith’s straw-man proposal of:
2TW Solar thermal, 2TW Solar PV, 2TW wind, 2TW geothermal, 3TW nukes, 0.5TW biofuels
What I find most striking about the energy literacy talks that Saul’s been giving lately is his ability to move fluidly between the personal quantities of energy we experience directly, the city-scale quantities we experience indirectly, and the global quantities that most of us can scarcely imagine.
My point here isn’t to revisit the dispute that Stewart Brand and Amory Lovins are having about the future role of nuclear power. Nor to endorse William Tucker, the author of that Energy Tribune article, who is a journalist not a scientist or an engineer, and whose argument fails to address issues of security and waste disposal.
Instead I want to focus on how mental power tools like WolframAlpha, by making computable knowledge easier to access and manipulate, can augment our ability to think computationally. If we’re going to reason democratically about the energy, climate, and economic challenges we face, we’re going to need those power tools to be available broadly and used well.
On my second try at formulating a query, I found the energy equivalent of 1 oz of matter from Alpha using this query –
e = mc^2 1 oz
It returned 707.8 GWh (Gigawatt-hours)
It’s amazing how Alpha can figure out how to interpret these queries!
There’s something puzzling about these e=mc^2 calculations. Shouldn’t the m in question be the difference in mass between the fuel and the end products of fission?
Wolfram Alpha to the rescue again, in my attempt to understand an example fission reaction:
(mass of U236 – (mass of Ba131 + mass of Kr92))/(mass of U236)
returns
.055974
So only about a 20th of the initial fuel mass is actually converted in this case.
Oops, should have read more carefully–I see that this issue was made explicit above:
“about six ounces of matter will be completely transformed into energy”
http://lmgtfy.com/?q=6*oz*c^2%2F%28GW*hr%29
you can type: 6*oz*c^2/(GW*hr)
in Google. Google calculator is pretty awesome as well.
this on the other hand is pretty darn impressive:
http://www.wolframalpha.com/input/?i=limit+x^x+x-%3E0
click “Show steps”