Physics Help Forum Efficiency of converting sunlight into chemical energy?
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Feb 20th 2018, 02:56 AM   #11
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 Originally Posted by ltlmissmegan He probably is, and we've been learning about solar thermal energy, and how PVs work.
What's a PV? Pressure/Volume ideal gas law?

By the way. Solar energy is not the same as thermal energy. In fact they're often taught in different sections of courses and texts.

 Feb 20th 2018, 03:00 AM #12 Senior Member   Join Date: Oct 2017 Location: Glasgow Posts: 362 I don't think a consideration of wet/dry leaves or anything like that is necessary. Efficiency, $\displaystyle e$, is defined as: $\displaystyle e = \frac{total \, used \, energy}{total \, input \, energy} = \frac{E_{use}}{E_{input}}$ So we need to evaluate the two energies for the ten year period and then divide one by the other to get the efficiency. The total input energy is going to be the total solar radiation captured by the tree. You know the incident solar irradiance (168 W/m2) so if you multiply this value by the capture area (the area of a circle of radius 8m) then you'll get the total input power, P: $\displaystyle P = A I_{solar} = \pi r^2 I_{solar}$ Now, power is the rate of change of energy, so to get the total energy gathered over a year, you need to multiply the power by the time (in seconds). $\displaystyle E_{input} = P t$ where $\displaystyle t$ is the duration of 10 years (in seconds). You might also want to consider the dormancy period here... since the time spent growing is actually less than ten years, you can calculate the duration spent actually growing and use that for $\displaystyle t$ instead. What about day-time and night-time? Maybe we can ignore this for now and then investigate it later To get the useful energy, $\displaystyle E_{use}$, we need to consider how much energy is gathered and actually put into chemical energy. There's different ways of calculating this, but whatever you do, don't use $\displaystyle E=mc^2$... you'll get a massive number that's very unrealistic. I found a book that said that glucose stores 16000 J of chemical energy per gram. This is similar to wood, which is about 15000 J per gram (from wikipedia). It's up to you which value you want to use. If we take 15000 J of chemical energy per gram, then that's 15 MJ per kg of energy. It should be straightforward to then get the energy wrapped up in 540 kg of wood. Once you have the two energies, you can get the efficiency Last edited by benit13; Feb 20th 2018 at 04:18 AM.
Feb 20th 2018, 04:26 AM   #13
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 Originally Posted by benit13 I don't think a consideration of wet/dry leaves or anything like that is necessary.
I don't know where this comes from. Perhaps you confused what I meant about water? What I said was
 You're not taking into account the water, carbon, etc that makes up the tree.
This has nothing whatsoever to do with wet/dry leaves. I hope you didn't think I was talking about wet trees?

I specifically spoke only about the water that the tree is partially composed of. Haven't you ever tried to start a fire using wood from a freshly cut live tree? Its almost impossible to do because of the water in the wood. The wood has to be dried out first. A large percentage of the mass of a tree is water. I never realized how much until this thread was started and I looked it up. It turns out that you cannot dismiss the amount of carbon and water in a tree because it you do then you basically don't have a tree. See

 It varies by species and other factors; however, it is often reported that live trees are approximately 50% water by weight and 50% carbon (oven-dried weight). ...

In fact a survival skill is how to get water from trees. E.g. see https://www.outdoorlife.com/blogs/su...nd-syrup-trees

The mass increase in trees is due to the addition of a variety of molecules such as carbon and oxygen. Cellulose is a component of trees and is made of carbon and oxygen.

See: https://serc.carleton.edu/eslabs/carbon/1a.html

That page might help.

benit13 - I doubt that an efficiency can be found using the information given. However, I have two friends who might help. One is a biologist and the other a chemist. And they're both extremely knowledgeable and intelligent. The biologist replied
 You can figure out how much of the weight gain is water by using the formula for photosynthesis. There are also minerals NPK... To consider but if you assume they are negligible you could determine how much of the weight is water and how much is carbs including cellulose by calculating in the co2 and solar energy.
Another aspect of the energy in a tree is the gravitational potential energy in a tree. Anybody who has ever cut down a tree is intimately familiar with this concept.

Last edited by Pmb; Feb 20th 2018 at 04:45 AM.

Feb 20th 2018, 06:55 AM   #14
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 Originally Posted by Pmb What's a PV? Pressure/Volume ideal gas law? By the way. Solar energy is not the same as thermal energy. In fact they're often taught in different sections of courses and texts.
PV - Photovoltaics

Feb 20th 2018, 07:07 AM   #15
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I was referring to Oz's post about wet leaves.

 benit13 - I doubt that an efficiency can be found using the information given. .
By using the information given, the equations I specified and assuming that

1. the tree gained 540 kg of "wood" over the 10 years
2. the solar irradiance is constant (no consideration of day or night cycles)
3. the tree spends 2/3 of its time actively converting incident solar irradiance to chemical energy wrapped up in "wood"
4. the chemical energy wrapped up in 1 gram of "wood" is 15000 J (this is the only additional quantity not provided in the question)

I get an efficiency of 0.00358. That's the right order of magnitude for the efficiency of photosynthesizing plants (~0.1% to 1%).

Sure, the real problem is much more complex than is suggested by the OP's original question, but the information provided can be used for a back-of-the-envelope calculation.

Last edited by benit13; Feb 20th 2018 at 07:30 AM.

Feb 20th 2018, 07:58 AM   #16
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 Originally Posted by benit13 I was referring to Oz's post about wet leaves.
I see. Sorry. I don't read 99% of his posts for obvious reasons.

 Originally Posted by benit13 4. the chemical energy wrapped up in 1 gram of "wood" is 15000 J (this is the only additional quantity not provided in the question)
As I said. It can't be calculated from the information given. This information wasn't given. Had it been then the answer would have been easy to arrive at , just as you showed. Bravo!

Feb 20th 2018, 05:32 PM   #17
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 Originally Posted by benit13 I was referring to Oz's post about wet leaves. By using the information given, the equations I specified and assuming that 1. the tree gained 540 kg of "wood" over the 10 years
Nowhere is the word wood used

OP says " the MASS OF THE TREE INCREASED BY 540Kg" ...

This mass increase will be leaves , wood and bark ...

Searching this I find Quote "..by weight wet wood has less than half - 42% - of the energy of oven-dry wood "

Calorific values are usually for oven dried wood 0% water , so figures must be more than halved ...

Feb 21st 2018, 12:54 AM   #18
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 Originally Posted by oz93666 Nowhere is the word wood used OP says " the MASS OF THE TREE INCREASED BY 540Kg" ... This mass increase will be leaves , wood and bark ... Searching this I find Quote "..by weight wet wood has less than half - 42% - of the energy of oven-dry wood " Calorific values are usually for oven dried wood 0% water , so figures must be more than halved ...
No worries, that's just one of my assumptions for the calculation. If you want to go and add additional complexity by taking into account the composition of the tree in more detail, then go ahead. I just wanted to say that it's not necessary in the first instance for a quick calculation; the bare minimum is to find the amount of chemical energy per unit mass, so I just looked up a value for a substance that I thought made sense for a tree.

 Feb 21st 2018, 01:28 AM #19 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 1,035 I suspect this question was posed as a talking point, not a fine engineering calculation. Thus it would be OK to state assumptions and offer back of envelope ball park calculations. The value of the question is the discussion which would then centre round the presentation. It is a recognised teaching technique to ask a question with insufficient detailed data supplied to widen horizons and stimulate thought. It is suprising how how close even the average human can get with guesses of the missing data.

 Tags chemical, converting, efficiency, energy, sunlight

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