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December 09, 2005

Math Geekery Help Needed.

Actually, maybe not math geekery. Possibly simply somone who is properly numerate. Whenever I try and do these things I get lost in a welter of units and orders of magnitude and end up proving that Alpha Centauri is just past Basingstoke, on the right. And pink.

Now, what I want to do is work out how much energy it takes to grow 1 kg of wheat.

Leave aside all the oil and fertiliser that goes in. Just the sunlight that we use. This comes in two forms.

1) Insolation on the actual plants themselves. From here some numbers for the US. In the wheat growing areas looks like an annual average of 4 or 5 kilowatt hours per square metre per day. We want the annual number because wheat fields give one crop a year, right? We are devoting that much incoming energy to growing our wheat.

Wheat yields per hectare vary widely but according to Lester Brown something like 3 tonnes would be a useful average.

A hectare is 10,000 m sq, so our 1,642.6 kilowatt hours gives us 300 grammes of wheat, yes?


2) Water. It’s said that 1,000 tonnes of water are required to grow 1 tonne of wheat. (We’re talking not irrigated, this is rainfall, so that’s also why we’re happy to use the low yield number above). So we need 300 kg of water to grow our 300 grammes of wheat.

1 calorie is the energy required to elevate the temperature of 1 g of water by 1 oC, yes? Assuming average ocean temperature of 4 oC the water must therefore have been heated by 96 oC to become fresh water in the form of rain. 300,000 x 96 gives us 28,800,000 calories of energy in the production of the wheat purely from evaporation of the water that allowed it to grow.

Right. Kilowatt hours to calories. From here, kilowatt hour to calories is  1:820,000 odd.

So, we have energy going in to our wheat.

820,000 x 1,642 + 28,800,000 calories, yes?

1,346,440,000 + 28,800,000 or 1,375,240,000 or 1.4 billion calories as near as damn it.


Whats the caloric energy we get from eating the wheat? From here, for flour (which I’ll assume is the same as wheat) 88 calories per ounce. Those are kilocalories (which is how we usually measure food) so in fact 88,000 calories per ounce, there’s roughly 30 g to an ounce so our 300 g is 880,000 calories. Or a million say.

So, just to be very roundy offy about this, we use a billion calories of energy to get 1 million (ie a thousand times less) in the form that we can actually use.

The point of all this? Not a lot really. Just think about this the next time someone tells you that obviously we’d never put more energy into a system than we get out of it. I mean obviously.

December 9, 2005 in The Blogger Himself | Permalink


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Aren't you forgetting the energy expended in labour as well?


Posted by: Unity | Dec 9, 2005 3:03:15 PM

Solar radiation is about 1.4KW per m squared this far from the sun. (without even checking the wiki).
Of which its often stated that 1.040 KW m2 reaches the surface.(air absorption .. don't know about clouds .. going from memory here).
Halve that for nightime, and knock off about one quarter for latitude (USA is not in tropics).

So, energy falling on USA wheat approx 400 W/m2

So original assumption one order of magnitude out.

Posted by: johnny bonk | Dec 9, 2005 5:10:32 PM

Please disregard previous post, I should read more carefully.

I mistook Kilowatts per meter squared per day

for Kilowatts per meter squared.

I apologise and will be more careful in future.

Posted by: johnny bonk | Dec 9, 2005 5:14:15 PM

Well, 0.1% efficiency doesn't sound completely insane for photosynthesis, but the precipitation bit of your calculation is off.

First, I'm going to do everything in SI units, because we get rid of all the pesky unit conversions that way.

Insolation: take your value for 1646 kWh/year -- that's about 6e9J (6 billion joules) for each 0.3kg wheat. That compares well to this map.

Irrigation: you've computed the cost of raising the water needed to 100°C, but what you're actually interested in is the cost of turning it into water vapour -- there is always water vapour in equilibrium with water at any temperature, and the cost of making it is something like 2.26e6 J/kg. So your 300kg of water require 0.7 billion J to turn into water vapour. That gives us a total energy requirement of 7e9J.

Compare that to the energy obtained from eating the wheat: that gives us 4.2e6 J, or 0.06%.

That's added to insolation of about 5.25 billion J, for about 6 billion J per 300g wheat, or about 23 billion J for each kg of wheat.

(For comparison, suppose that we irrigate rather than waiting for rain, and have to raise water h = 100m from some aquifer to spread it on our fields. The energy required to do that is E = mgh; for m = 300kg, g = ~10m/s², h = 100m. That gives us a total energy of 300,000J for each 300g of wheat -- much less than waiting for rain. But of course that's a false comparison, because you have to renew the aquifer somehow; in any case, the water cost is pretty small compared to the incident solar energy.)

So, should we be surprised that the "efficiency" is so low? Well, quoted values for energy conversion in photosynthesis are 3% to 6%, so 0.06% does seem pretty low. But: not all of the insolation strikes the leaves of the wheat; not all the energy obtained from photosynthesis is turned into grain anyway; not all the incident solar energy is usable for photosynthesis (as anybody who has ever seen a field of wheat is implicitly aware); and wheat doesn't actually grow all year round.

Here's another calculation, which estimates 20g biomass/m²/day for a cornfield in Iowa, or about a factor of twenty better than your number; he then gets an efficiency about twenty times higher. However, you and I don't have appendices, and so can't metabolise most of that "dry mass", so it's not a whole lot of good to us; however, it does suggest that the useful bit of the corn plant is ~5% of the total.

Posted by: Chris Lightfoot | Dec 9, 2005 5:23:53 PM

But you can't eat sunshine...

If only the moon were made of cheese.

Posted by: Sam | Dec 9, 2005 5:37:46 PM

Well of course we put more energy into systems then comes out (as work) that's just simple thermodynamics. And of course we don't put more energy into systems them comes out in total. That's conservation of energy.

I'm not certain what a person who said such a thing would be getting at.

Posted by: David | Dec 9, 2005 5:44:06 PM

1. Work in joules plleassssee

2. Remember yer economics. The sun shines whether or not we've put some wheat here, so I don't see how it can be counted as an input into the production of wheat.

Posted by: dsquared | Dec 9, 2005 5:47:57 PM

The energy input required to turn water to vapour has to take into account enthalpy of vaporisation = 40.7 kJ/mol = 2.26 MJ/kg.

Posted by: David Gillies | Dec 9, 2005 6:36:04 PM

dsquared -- I suspect the relevant point is that it's an opportunity cost. We can't cover the same places in wheat *and* solar panels so sunlight we use for photovoltaics can't be used for wheat. Actually you probably can -- I don't know what the coverage factor for wheat is -- but it's unlikely to be convenient.

Posted by: Chris Lightfoot | Dec 9, 2005 6:53:34 PM

Next you're going to have us work out how much energy it took to create Uranium, aren't you?

Posted by: Rub-a-dub | Dec 9, 2005 7:57:15 PM

Uranium's easy, isn't it? Assume we begin with hydrogen ("The two commonest elements in the universe are hydrogen and stupidity", and we can't make uranium from stupidity). Uranium has a binding energy of something like 1,800 MeV (2.9e-10 J), so it takes about 733 GJ or about 200,000 kWh to make one gram of uranium from hydrogen, compared to about 20,000 kWh (thermal) which would be obtained from fission of that uranium; that translates to (say) 8,000 kWh (electric).

Posted by: Chris Lightfoot | Dec 9, 2005 9:21:18 PM

***Please disregard previous post, I should read more carefully.

I mistook Kilowatts per meter squared per day

for Kilowatts per meter squared.

I apologise and will be more careful in future.***
Am I a plonker or what ?
Kilowatt HOURS per meter squared per day. I really must be more careful. Now then, have misread either the post or my own reply or my correction to my reply or my correction to my correction to my reply?

Posted by: johnny bonk | Dec 13, 2005 11:42:42 PM