Power generation system. Part 2

Whenever I have contemplated something like this, I have always wondered why no one has produced a small gen set that runs off town gas ? Interruptions to the gas supply are very rare.
Would it be feasable to convert a small petrol motor to run off the bayonet point a barbecue or heater connects to ?
Dave Goldfinch
Reply to
Dave Goldfinch
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**All very interesting and unreasonably confusing. I did a quick search and came up with more than 20 references to ENERGY payback for Silicon Solar Cells. The WORST figure I came up with was 5 years, whilst the best was 1.25 years. Naturally, the financial payback time is in the order of 20-30 years, at present energy prices.
To use some of your own data, it would take 11 years to refine one kg of silicon, using a single 115 Watt panel.
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Trevor Wilson
www.rageaudio.com.au
Reply to
Trevor Wilson
If you want a quiet RF environment, some "pure" sinewave inverters put out a bit of RF. Mine (SEA Voyager) puts an S5 signal on some of the amateur bands. Can be suppressed of course - but that's more work. It is very efficient though - even at low levels (
Reply to
Alan Peake
1. It's simple math Trevor: find out how much solar energy is typical for your location in watts per metre square, multiply that by the efficiency for your solar cells and you get a figure for power production for the area of solar cells you have in Watts. Assume for simplicity that you get that wattage as long as the sun is above the horizon so you multiply the wattage by the time to get Watt-Hours of energy you make for a DAY.
2. Work out how many Kgs of silicon you need per square metre of silicon and multiply that by 2130 KiloWatt/Hours per kilo of silicon.
3. So you have how much energy your cells make and how many watt hours of power it took to make the silicon used in them. Divide the second by the first to get how many DAYS it's going to take for the energy made by your cells to equal the energy taken to make the silicon in them.
One Kilo of silicon can make a lot more than a single 115 Watt panel. Depends on how thin you slice it.
Regards Mark Harriss
Reply to
Mark Harriss
at least get the units right - kW*hrs not kW/hrs. the difference is proportional to hours^2.....
Reply to
Terry Given
2130kW-hrs.
1 kWhr = 1kW * 3.6ks = 3.6MJ
2130kW-hrs = 7.668GJ per kg of silicon?
the specific heat capacity of silicon is 700J/(kg*K). For 1 kg, m*cp = 700J/K.
7.668GJ/[700J/K] = 10,954,286 K
thats enough to heat 1kg of Si to 11 million degrees C!!!
the melting point of Si is about 1700K so that could melt 1kg of Si about 6,444 times.
Clearly this number is wrong.
Its out by about three orders of magnitude - if it were 2130W-hrs, that would be enough to melt 1kg of Si six-and-a-half times, a thoroughly believable proposition.
dividing your numbers by 1000 changes the answer somewhat, although the methodology is sound.
which does not alter the fact that at 115W takes 66,678,260 seconds to use 7.668GJ, IOW 18,522 hrs = 2.1 years. Add in the various efficiency factors, and Trevors calc is about right.
he said nothing about how many solar panels that 1kg of Si will make.
Cheers Terry
Reply to
Terry Given
Hi Terry, I pulled that figure off the net as the total energy cost to make and refine silicon to semiconductor grade, which is where cell makers buy their single crystal silicon: I would imagine with constant remelting of silicon from zone refining that it would approach that energy figure.
I always have trouble with my kilowatt hours terminology, thanks for the advice.
Mark Harriss
Reply to
Mark Harriss
What features would you look for in a "pure" sinewave inverter? what features would you like?
Here's what I think:
- true sinewave (even at no load) - line interactive - fully regenerative (eats motor loads etc) - low (ideally no) EMI. Run an AM radio sitting on top of it... - doesnt care which wires go to the grid, which to the load (alas, it does care about which go to the battery) - IP66/IP68 (run the sucker underwater...) - drop it from 1m and give it a good boot while its running (wont die) - programmable input PF -1...+1 (IOW can do VAR compensation) - input can act as harmonic filter - failsafe (fully fused, thermal modelling/monitoring etc) - no inrush current - sub-cycle brownout/dropout detection - bumpless power transfer - optically isolated comms link - user serviceable cooling fan - liquid cooling option - 3:1 peak to average power ratio - > 20kHz switching frequency (inaudible) - single- & three-phase models - wide range of battery voltages (12-24-48V) - built-in battery condition monitoring - simple yet useful user interface (small LCD, a few buttons) - optional bypass contactor
Cheers Terry
Reply to
Terry Given
I dont buy it. They melt the Si a few times, not a few thousand times.
The physics dont lie.
Thats one problem with the web, bullshit abounds. When in doubt find (2N+1) different sources, and make a majority decision. Seriously, it looks like either they have screwed up the units (W-hr cf kW-hr) or have included the energy involved in building the entire infrastructure - not an uncommon trick with greenies, but desperately unfair - where do you stop, pretty soon one ends up calculating the energy required to build our entire civilisation from scratch.....
According to Goodge, Semiconductor Device Technology, ch.3 chemical reduction is first used to purify SiO2. It then undergoes zone refining several times (pulled thru a heater, making a molten segment that travels along the Si rod, essentially pushing the contaminants to the end). The end(s) are lopped off, and the resultant pure Si is re-melted, and a single crystal is pulled (Czochralski process)
So the Si would certainly be melted several times, which is what your number indicate assuming its W-hr not kW-hr
I'm a pedant.... but the "/" symbol really confuses things.
Cheers Terry
Reply to
Terry Given
Yep I think so too: so I'm madly trying to find that 2130 KW-Hr figure again, or any figure for that matter.
Regards Mark
Reply to
Mark Harriss
Hi Terry, here's a different site:
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with a powerpoint presentation that has the same figure for silicon production (page 12) of 2130 KWH per kilo. This one, though does specify that this is the TOTAL cost to get from sand to a finished silicon WAFER per kilo, not just a kilo as I stated.
Also the author does cite references of studies for these figures so I'm inclined to believe him. As solar cells are made from wafers, I think it may still be a valid comparison.
At any rate see what you think.
Regards Mark Harriss
Reply to
Mark Harriss
Most of those are features I am used to seeing in a commercial (i.e. not domestic) UPS.
However review for a moment your 3:1 peak:average capability. Peak is determined largely by current handling in the semis. Average is more determined by thermal factors.
I can turn a 2:1 into a 3:1 very cheaply - by removing part of the heatsink. One needs to be rather careful specifying such a ratio unless the average power output capability is also defined adequately.
Reply to
budgie
some of them you wont find in commercial UPS either :) Yet...
determined
yep.
ROTFLMAO! nicely put. Say Pcont = 3kW.
Cheers Terry
Reply to
Terry Given
Hi Mark, thanks for that.
Complete life-cycle energy costings....
the numbers in the table on p. 16 dont stack up:
multiply the % columns: 0.9*0.9*0.42*0.50*0.56 = 0.095
divide energy by % and add: 123/.9 + 50/.9 + 250/.42 + 250/.50 + 240/.56 = 1594kWh
Hmm, thats not quite right, probably a typo on their part, the order of magnitude didnt change.
But clearly the erroneous figure 2130kWh is for 1kg of *processed* wafer, and should be 1594kWh = 5.7GJ. Typical sneaky bastards, present data in the way that serves their purpose best....because the yield is 9.5%, that 5.7GJ is "spent" on 10.5kg of raw Si, enough to melt it (1700K) 458 times, a hell of an improvement on 6,444 times.
So if we multiply the 1600kWh by the weight of a solar cell in kg (much less than 1kg) that will give us a fair estimate for the total energy cost of the solar cell.
If we assume the wafer is 200mm diameter and 0.5mm thick (a guess), its volume is 15.7e-6 m^3, so weighs about (15.7e-6 m^3)*(2330kg/m^3) = 0.0366kg, so per wafer its about 58kWh = 210MJ.
If you have some figures on Si area, thickness and power output, we can work out the total payback time.
which pays back *ALL* of the energy used in the entire manufacturing process.
Fun with physics.
Oh yeah, then do the same for the diesel generator. That steel didnt smelt itself....
Cheers Terry
Reply to
Terry Given
Can't believe you didn't quote toaster specs. Should know them all. 1 room heater = 20 toasters, just you have to keep pushing to toast button down all the time.
If loss of power is a concern then use an inverter for essential lighting and power and use gas for the other essential items like the oven and hotwater. Forget about using stuff like microwaves and electric jugs for a few days while power is out.
Gas fridges are available, old man has a large domestic size fridge running off BBQ gas bottles on his boat and then there's always the Engel type smaller ones which are normally used for camping. Maybe power the fridge a number of times a day off a genset used for that purpose only.
Even in areas where town gas is not available there might be bottled gas service available where they'll come and top up your tank or supply two large tanks with a change over switch and swap them over when you call and tell them you have an empty.
Trying to power electric hot-water and oven off an inverter is ridiculous and a waste of money. Use the BBQ and install a back-up instantaneous hot-water system off a 9kg BBQ gas bottle if needed for a back-up. The later can be picked up cheap second-hand.
Reply to
David Sauer
Not to mention CO2 produced from electricity used and carbon burnt for the first step of the process.
I'm happy with the 0.5 mm thickness of a single cell, it may well be thinner now, I still haven't found any processing costs for making the wafer into a cell either which would be handy. I did find a commercial cell efficiency of 11.5% and I would think using a solar flux figure multiplied by the efficiency would be ok for power output.
It'd be interesting to compare the CO2 made for both paths also.
Not only that but the genny is not used all the time, what are the figures for continuous use. Hmmm... then there's battery costs as well.
Regards
Mark
Reply to
Mark Harriss
this total-cost-of-ownership thing gets pretty tricky pretty fast eh?
look at the mic preamp gain post on a.b.s.e. I've got a link there to a company that sells die. A typical die is 11mil thick, about 0.28mm hence my guess of 0.5mm
thats probably beyond me, my chemistry is not so great. But I'd be happy to learn how...
Reply to
Terry Given
Yes, I'd like all of those Terry :) The Voyager is pretty good on most except the RFI. It's rated at 1700W continuous, 2400W for 30 minutes and 3900W for 5 seconds. Starts all my 240v motors OK (biggest is half horse)and the 19" CRT monitor doesn't dim the lights either at switch-on. (The 10HP/3.5 KVA petrol genny coughs a bit under this load). The THD is quoted at 14MHz. Alan
Reply to
Alan Peake
"Alan Peake"
** A simpler calc is how long to pay for themselves, compared to mains power.
A more realistic hours figure at 50 watts output is 1000 per year - takes into account winters and cloudy days.
So 50 kWH = $ 7.5 worth of power per year.
Break even happens at 500/7.5 = 67 years.
Shame the panels will have died after 15 to 20 years.
........... Phil
Reply to
Phil Allison
Perhaps the simpler maths would be to just look at the cost of buying panels. 50W panels used to cost about $500. How much of that is for the energy required to manufacture them? $50? How much is a KWH? Say $0.15 - then a 50W panel takes about 50/.15 or about 333KWH. So the 50W panel has to make 333KWH to break even. At 5Hrs/day, this takes 333K/(5*50) or 1333 days or 3.65 years. Alan
Reply to
Alan Peake

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