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JohnWW
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Extra Large Capacitors For Power Storage
Extra Large Capacitors For Power Storage
At a recent auction in my city in New Zealand, I purchased ten extra-large-capacity electrical capacitors (for $NZ10 each - someone else was also
after them and bidded me up from $1 each), which had been made surplus (due to being obsolete) to the requirements of the local Fonterra Northland
dairy factory, which processes most of the milk produced in my region, following a control system upgrade. Judging by other surplus electronic
equipment also sold in the same auction by the same vendor, they appear to have been made in about 1989 in Japan, and were part of the factory's
automatic process control system. As a chemical engineer, I was familiar with the equipment that was up for sale. The capacitors were probably
formerly used as voltage surge protectors. I am going to see if I can get more of them from other Fonterra process plants, when they become
redundant.
They each weigh 1.35 Kg, are cylindrical with a diameter of 90 mm and length 165 mm, and are cased in black PVC and polypropylene plastic. They each
have a rated capacitance of 10,000 millifarads (MFD) = 10 farads, which for the whole 10 capacitors would be 100 farads connected in parallel (but
only 1 farad connected in series), and are rated for use at a voltage of up to 350 v with allowable surge capacity up to 400v. A photograph of one of
them is attached.
The metal band around the circumference is a mounting bracket. The two 8mm diameter x 10 mm long aluminium cylindrical protuberances (with internal
threads) on top are the terminals.
I am interested in using them for power storage, in connection with a solar power system that I am going to install on my roof, to avoid using grossly
overpriced mains electricity (the power supply companies are "ripping off" everyone in New Zealand), and in conjunction with a set of eight newly
reconditioned 12v deep-cycle lead-acid storage batteries I also bought at the same auction-house. (However, even the best deep-cycle lead-acid
batteries need replacing after 10 years of frequent use at the most, hence my looking at capacitors). To use them, I would probably connect the solar
panels sufficiently in series to deliver power at about V = 350v DC (12v panels x 30 in series would do) to the capacitors so as to fully charge them,
BUT there would have to be some sort of automatic switching transistor arrangement to prevent their discharge through the solar panels at night (when
the solar cells would be unpowered).
The total maximum charge on each capacitor, using the maximum rated voltage of 350v, would then be Q = VC = 3,500 coulombs each or 35,000 coulombs for
the ten, connected in parallel, representing energy of E = ½CV² = ½QV = 612,500 watt-seconds each, or 6,125,000 watt-seconds = 6,125
kilowatt-second = 1.701 kilowatt-hours for all 10 connected in parallel. (1 coulomb = 1 ampere-second, 1 watt = 1 ampere-volt). This is probably
adequate for most ordinary domestic use, at least other than in cold winters, but may exceed the daily generation capacity of a typical domestic
solar-panel installation. Besides, at 350v it would be not particularly safe to have around, and minimum 400v-rated electrical cable would have to be
used with them instead of the regular 250v cable used in New Zealand.
If, however, nominal 12v solar panels (the same voltage as used by DC to 110v or 230v AC inverters, which are usually run off 12v car batteries) were
connected all in parallel to charge the capacitors at only 12v combined, the 10 capacitors in parallel would only be able to store 120 coulomb each or
1,200 coulombs combined, representing energy of only 21,000 watt-seconds each, or 210,000 watt-seconds = 0.0583 kilowatt-hours for the 10 in parallel.
This would be much less than adequate to store typical daily domestic power.
There are, however, two major problems to be overcome:
(1) Provision of some sort of automatic switching transistor arrangement, between the solar panels and the capacitors, to prevent the capacitors from
discharging through the solar panels at night when the solar cells are unpowered; and:
(2) Certainly as far as a 350v arrangement would be concerned, provision of some sort of voltage regulator, between the capacitors and the inverter,
to deliver electric power from the capacitors at a constant approximately 12v (±1v) to a DC-AC inverter during discharge of the capacitors at night
when the solar panels are not working, during which discharge the voltage delivered by the capacitors would also drop exponentially. The voltage
supplied would otherwise steadily decrease during discharge; after less than 1/10 discharged from a 12v arrangement, it would drop below 11v, reducing
the effective storage capacity by a factor of about 10. This loss of effective storage capacity could be partly overcome by arranging the solar panels
to deliver power at say 24v to the capacitors instead of 12v or 350v, about twice that required by the inverter but still fairly safe.
Not myself being an electrical engineer, can anyone here suggest solutions to these two problems?
(BTW, If I could make a 350v arrangement of the solar panels and capacitors safe, and could afford a sufficiently large bank of efficient solar panels
and the required switchgear and a transformer-inverter at economic prices, I would be looking at supplying surplus power to the national grid, stepped
down to 230v AC).
P.S. Some of the info in the "Exploding Wires" thread (using power from capacitors) in this section has also been helpful. For further general info on
the subject, see http://en.wikipedia.org/wiki/Capacitor and http://en.wikipedia.org/wiki/Types_of_capacitor and especially http://en.wikipedia.org/wiki/Supercapacitor . Can someone please supply this restricted paper referenced therein:
Zorpette, Glenn (2005). "Super Charged: A Tiny South Korean Company is Out to Make Capacitors Powerful enough to Propel the Next Generation of
Hybrid-Electric Cars". IEEE Spectrum 42 (1): 32. doi:10.1109/MSPEC.2005.1377872. ISSN 0018-9235 .
(Edit: Since posted: http://www.sciencemadness.org/talk/files.php?pid=162141&... ).
P.S. Two revisions: -
(a) In the light of the Wikipedia "Supercapacitor" article, I now think that that the 10 large "10,000 MFD" capacitors I have are, in fact 10,000
millifarads, or 10 farads, not microfarads which would be µFD, as I previously thought.
(b) In the light of comments subsequently made on my original posting, it appears I wrongly calculated the possible KW/hr storage by multiplying
instead of dividing.
I have revised my original calculations as above, to incorporate both corrections.
[Edited on 13-9-09 by JohnWW]
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not_important
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1 Ws = 2.7777777777778⋅10-7 kWh, thus your 122500 watt-seconds is about 0,034 kWh.
Capacitors suck at energy storage, except when you need to dump out a lot of watts in a short time.
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watson.fawkes
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Oops. You multiplied by 3600 s/hr here rather than dividing by
it. What's seven orders of magnitude between friends?
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JohnWW
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BTW, By searching on Google for this search string:
"HCG F4A" capacitor OR condenser
I obtained 58 results for this particular type of capacitor, mainly places offering them for sale mostly second-hand, although none of them seem to
specify their 10,000 millifarad capacity.
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woelen
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JohnWW, these are not millifarad capacitors, but microfarad capacitors. So, your total capacity when all are connected in parallel is 100000 micro
Farad, which is 0.1 Farad. For storing energy this is totally useless, but it can make a big bang or a nice pulse. But your capacitors most likely are
not constructed for pulsing and may be destroyed or damaged if used as pulse capacitor frequently.
But if you want to be sure whether these are milli- or microfarad capacitors, try the following:
- Charge one of them on a 12 V power supply.
- Take a 12 V lamp for a car, use a low power one, not one of the main lights.
- Connect the lamp to the capacitor.
If it is a 10 Farad capacitor, then it will give off light for quite some time (several seconds at least, maybe one minute or so). If it is a 0.01
Farad capacitor, you'll just see a brief flash of light, or nothing at all.
[Edited on 13-9-09 by woelen]
[Edited on 13-9-09 by woelen]
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JohnWW
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Quote: Originally posted by woelen  | (cut) But if you want to be sure whether these are milli- or microfarad capacitors, try the following:
- Charge one of them on a 12 V power supply.
- Take a 12 V lamp for a car, use a low power one, not one of the main lights.
- Connect the lamp to the capacitor.
If it is a 10 Farad capacitor, then it will give off light for quite some time (several seconds at least, maybe one minute or so). If it is a 0.01
Farad capacitor, you'll just see a brief flash of light, or nothing at all. |
Charging one of the capacitors up using a 12v supply would, from the above calculations, be quite inadequate for any sort of experiment. I would do
that sort of experiment, after charging with a much higher voltage up to 350v DC, ONLY if I could get my hands on, as noted in my first post, a
suitable voltage regulator which would allow the capacitor to discharge at a nearly constant 12v DC. Where could I get one? Besides, from the
Wikipedia "Supercapacitor" article, I am fairly convinced that they are 10,000 millifarad, in the light of their size.
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not_important
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You can determine the capacitance with 12 VDC quite easily. Short the capacitor for some time to make sure it is discharged. Remove the short, place
a volt meter across the terminals, and charge the capacitor from a 12 VDC supply through a 10K resistor. A 100000 microfarad capacitor will charge
to around seven and a half volts in about ten seconds, a 100000 millifarad unit will take almost 7 hours to reach the same voltage. No highh voltages
needed, no risk of giving your heart a jolt.
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woelen
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JohnWW, the size does not tell that much. I have capacitors of 600 uF/350 V which have a size of 10 cm height, 7 cm diameter and weigh a few hundreds
of grams. I am almost 100% sure that yours are 10000 uF and not 10000 mF. Now you have two methods (not_important's and mine) to test their capacity.
Both methods are equally suitable and only require a 12 V DC power supply and which method you select depends on what other components and measuring
gear you have at hand.
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franklyn
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The capacitors you have are aluminum electrolytics and are rated at 10,000 - M I C R O farads.
If it were millifarads just as this it would say so ( it is never abbreviated ). By ( Capacitance X Volts X Volts )
divided by 2 , (.01 farad x 350 x 350 ) / 2 , comes to ~ 613 joules , which if discharged in one second
is the equivalent number of watts , 613 , According to NIST: power, radiant flux, watt, W = J / s .
http://physics.nist.gov/cuu/Units/units.html
( a resistance of 20 ohms would do so but rather violently unless it had at least 100 watt rating ,
3 light bulbs in series would work ). Ten of these fully charged would power a hair blower/dryer for
no more than 4 seconds. They can serve to provide very high surge demand when parallel with a
battery bank. For example a 30,000 BTU air conditioner may draw ~ 24 Amps operating , but may
require 100 amps for 2 seconds when starting.
A L L cable insulation is good for 400 + volts and actually even likely good for 600 volts. A megger
insulation tester can tell if there is insulation breakdown , these test in steps of 500 , 750 , 1000 volts.
I gather you have some notion of harnessing solar power to mitigate your electric utility expenditure.
The very first thing you need do is decide how much you want to budget for this , only you know
how deep your pockets are. After some study of the available ways and means you must know
exactly what you need to obtain , rather than randomly acquiring things which you believe may have
some vague utility or application. You will then have some clear idea of just how much return this
project will net you , in effect is it worth it long term.
A good place to start is this thread http://www.sciencemadness.org/talk/viewthread.php?tid=10515
Installations can be modest but an expenditure comparable to a car purchase is called for if significant
substitution of grid power is being sought. Apart from the PV ( Photovoltaic ) Solar Panels , hardware
required is as you guessed , a DC to AC Inverter , and Storage Batteries. Car batteries may be
procured very cheaply from automobile wrecking and salvage yards. The inverter is not something
that can be cobbled together and it will be a significant cost of the total outlay. Most commonly
available in 12 and 24 volt units for RV ( recreational vehicles ) motor homes and marine boat use.
Note that if you plan to link photovoltaics to output at a.c. voltage level , then you will need to
have 27 ( twenty seven ) 12 volt batteries in series to take the charge , something in excess of ~ 325
volts ( since the batteries actually output nearer to 14 volts each ) this corresponds to the peak value
of the line voltage and must be present to provide a facsimile of the actual utility provided a.c..
You then must obtain a high voltage photovoltaic inverter which do cost more because there are few
manufactured. http://www.solarelectricsupply.com/Inverters/inverters.html
Better to keep to the more conventional lower voltage.
Note that MOT ( microwave oven transformers ) are rated at over 1000 watts and have a turns ratio
of around 18 and that quite closely matches the a.c. output from the 12 to 14 input. One or more
transformers in parallel can adequately distribute power and step up the voltage to the required a.c
service level , if you really , really , really want to adapt your own makeshift inverter.
A charge controller ( voltage regulator / battery charger ) is used to interface the Photovoltaic array
with the batteries from which power is supplied to a bidirectional inverter , one that converts DC power
to AC or rectifies AC power to DC to charge the batteries from the mains. The inverter has an H-bridge
comprised of four electronic power switches usually Insulated Gate Bipolar Transistors (IGBT's) each
with a diode parallel and opposite in polarity forming the full wave rectifier. S E E
http://www.dprg.org/tutorials/1998-04a , and , http://www.modularcircuits.com/h-bridge_secrets1.htm
The diagonally opposite switches ( Q1, Q4 ) and ( Q2, Q3 ) are switched on by pulses , pulse-width
modulated to produce sinusoidal output voltage with a transformer. S E E below
Called a switch mode chopper circuit , in a mains connected system it acts as an interface that converts
DC current produced by the solar cells into utility grade ac current. The inverter must produce good
quality sine wave output and be in phase with the grid voltage. A typical grid connected inverter may
use a pulse-width modulation PWM scheme and operate in the range of 2 kHz up to 20 kHz.
DC-AC inverter Pulse Width Modulation
http://encon.fke.utm.my/courses/notes/inverter-2002.pdf
A Guide to PV System Design and Installation
http://www.energy.ca.gov/reports/2001-09-04_500-01-020.PDF
Photovoltaic Power Systems
http://www.polarpower.org/static/docs/PVWhitePaper1_31.pdf
Photovoltaic Power Systems - National Electrical Code NEC Suggested Practices
http://www.altestore.com/store/media/pdfs/photovoltaic_NEC_c...
Power Electronics Handbook
S E E -> 23.2 Power Electronics for PhotovoltaicPower Systems page 540 ( pdf pg 530 )
T H I S . I S . A . PDF . F I L E . N O T . A N . M O V . ( change the extension to view 
http://www.megaupload.com/?d=2E9DLZI8
Solaris Power
Extensive Detailed How To ( Get this one ! )
T H I S . I S . A . PDF . F I L E . N O T . A N . M O V . ( change the extension to view
http://www.megaupload.com/?d=PKE9PFJM
Additionally attached below
Guerrilla Solar A modest plug into outlet contrivance
Attachment: Guerrilla-Solar.pdf (335kB)
This file has been downloaded 890 times
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JohnWW
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Thank you very much for your info, Franklyn, particularly regarding voltage regulators for the storage, and photovoltaic cells-storage interfaces to
prevent discharge through the panels.
As well as replacing overpriced mains electricity in my principal house, I am looking at buying a rural farm property with no present mains
electricity connection, and for which several tens of thousands of dollars would be charged by the local lines company (Northpower) to connect to the
national grid because of the distance of possible house sites from existing reticulation. I am fairly sure that this cost would be substantially
greater than installing a solar panel (and possibly wind) array, with power storage as described using capacitors (as well as batteries), especially
as cooking, and any required heating (hardly ever required in my part of New Zealand in a properly insulated house, the temperature hardly ever
falling below 5ºC), could be accomplished using bottled gas and wood.
[Edited on 14-9-09 by JohnWW]
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12AX7
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Actually, that's a megafarad (MF), but those don't exist either. They are fairly standard 10,000uF 350V electrolytics.
Put a couple together and you can make an excellent photoflash supply / wire exploder.
Edit: datasheet for the next series up.
http://www.hitachiaic.com/docs/products/Screw_Terminal_Elect...
Tim
[Edited on 9-13-2009 by 12AX7]
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densest
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Photovoltaic charge controllers come up for sale on surplus electronic sites fairly often in the $US 50-350 range for 300-1KW units. Try www.goldmineelec.com, www.bgmicro.com, www.halted.com, www.alltronics.com, www.allelectronics.com and others I don't know or remember at this time. They do all the work of mating lead acid batteries with photovoltaic
arrays.
It's old hat, but it's worth remembering that replacing all lighting with either compact fluorescent (for area lighting) or LED (for spot/directed
lighting) saves a lot of power.
NZ is a hotbed (heh) of geothermal work. Perhaps you should check with the local geological office?
A lot of people have worked on 12V/24V appliances and lighting (see above). If you completely avoid any electrically powered heating and cooling you
can reduce your power demands a lot. I don't know of any domestic freezers that work on a heat cycle, but half a century ago I lived with a kerosene
powered refrigerator in Liberia, West Africa. It barely kept the freezing compartment below 0C, but it used no electricity, which was a good thing as
the power went off regularly and for long periods. I think it ran on the ammonia/hydrogen cycle - very old technology.
Tens of thousands of dollars just might be enough to fit a usable photovoltaic system. Be sure that a hydroelectric (got a stream on the property?)
system is likely to be much more powerful. A 6-foot (2m) head is about the minimum for an amateur installation. Also consider wind power if you have a
nice slot between two mountains.
Since you're talking about a farm, I assume you need pumps and refrigeration as well as lighting. Both can be driven from gas or petrol if needed -
the capital costs are high but the running costs are relatively low, especially for a large installation because the efficiency goes up with larger
units.
Yes, lead acid batteries have a limited life. There are two *I think* similar but incompatible chemistries: Pb (hardened with Sb) and Pb/Ca. (Subject
to correction by more knowledgeable people). The "float" voltage and normal discharge voltages differ by a couple tenths of a volt. So if you parallel
a lot of batteries, make them all the same type, auto/truck or alarm/backup, but don't mix them. The higher voltage batteries will never charge fully
and die young or the lower voltage batteries will overheat and die young. Auto batteries are designed to deliver huge amounts of energy over a very
short time but never fully discharge - their plates are thin. So if you use auto batteries expect to replace them often or use 5-10X as many as you
think you might need so they never ever discharge more than 20% or so.
If you have any motors which start and stop frequently, consider using ultracapacitors in parallel with your batteries to supply a 1-10 s huge surge.
Single phase induction motors are very close to a short circuit across the power line while starting. Three phase motors are much more benign (but
still draw power corresponding to the acceleration of masses). It could be cost effective to put converters (VFD variable frequency drives are very
good 1-3 phase converters, or rotating converters (an ingenious setup with a large 3-phase motor driving the 3rd leg)) where large motors must be
started frequently and use 3-phase motors. VFDs have the advantage of allowing a soft (slow) startup with a correspondingly smaller surge draw.
Using heat pumps with ground reservoirs for heating and cooling might be worth the capital cost. Drill a borehole into rock with an average
temperature of 5-10C and use that to dump excess heat from a chiller greatly increases the amount of heat you can pump if the air temperature is 20C
or above. I'm not familiar with the climate of New Zealand - does it ever get very cold or very hot in the area where your intended farm is located?
[Edited on 13-9-2009 by densest]
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not_important
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While I am a proponent of wind and solar power, there are things you need to consider if you are involved with putting together a system.
You must remember that the power rating of a photovoltaic panel is it's rating under standard illumination conditions, effectively full noon day
sunlight. The actual average power output, the availability or capacity factor, is much less and varies depending on your location. Values for good
locations generally run around 0,2, with temperate climate values being more in the range of 0,15 to 0,18.
This means that a panel with a rating of 100 W is going to on the average deliver 15 to 18 watts, or 360 to 431 watt-hours per day. Obviously the
actual power peaks in the midday, the average or daily total is used for calculating how large a PV array is needed to meet the daily power demand and
assumes some sort of power storage.
To put enough power into storage to meet some specific base level of delivered power, your average output rating must be several times larger than the
baseload, and your nameplate or peak rating even larger - scaled up by the availability factor. For large scale wind from turbine farms the nameplate
generation capacity with CAES storage, a farm with a rated output of 2,5 to 3 times the steady amount of power to be delivered results in around 2/3
of the power coming from wind and 1/3 from the natural gas used by the CAES. That is, for every kW to be supplied continuously 24 hours a day, you
need 2,5 to 3 kW of nameplate wind capacity, and will derive 650 watts from wind and 350 watts from natural burned in the CAES turbine.
For solar the overrating is even higher, in part because of the lower availability factor ( 0,15 to 0,2 vs 0,3 to 0,4 for wind), and because solar is
'noisier' than wind, the power generated fluctuates more quickly over a wider range than wind does. The actual rated power capacity of the PV array
should be on the order of 5 to 7 times the steady state power output.
Now with battery systems all of the power is coming from the PV array, so the excess capacity must be even higher for reliable power. One the other
hand, it is likely that the demand pattern isn't a flat load, but varies with peaks in the morning and early evening and a much smaller value late at
night. On top of that batteries are a more efficient means of storing power than is compressed air; CAES is targeted for applications where hundreds
of kilowatts to 10s of megawatts are being supplied and battery arrays are impractical.
Solar PV output can fluctuate greatly on the scale of 10s of seconds. A ultracapacitor based smoothing storage plus batter based bulk storage might be
the best combination. Overrating the battery array so as to avoid wide charge-discharge swings can prolong battery life.
Small scale wind power can be of lower quality than solar PV, if you're not using a tower of at least 20 meter height then wind may be more expensive
than solar per reliable watt. Wind quality, both speed and how non-turbulent the wind is, is reduced the near the ground the turbine is, and the
closer are buildings, trees, hedgerows, and so on. Not only does turbulence reduce power output, but the increased stresses and shocks decrease the
lifespan of the equipment. Small 'rooftop' wind turbines produce very little power on the average, in one case a turbine with a rating of 500 watts
produced an average power output of 6 watts; the power output was taking for winds on the borderline of being gale force winds, far in excess of what
is available on most rooftops.
The image below is 3 graphs of solar PV output. The top shows some daily outputs from a 55 kW array in Australia, with summer and winter maximums and
minimums. The next two are for a 4,5 MW array in the U.S. The 1st graph shows 7 days output at one minute resolution, the day to day variation is
clearly visible. The last graph is the same site showing a single day's output at ten second resolution, the rapid fluctuations of the high desert
near ideal location are quite evident.
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chemoleo
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| Quote: |
You can determine the capacitance with 12 VDC quite easily. Short the capacitor for some time to make sure it is discharged. Remove the short, place a
volt meter across the terminals, and charge the capacitor from a 12 VDC supply through a 10K resistor. A 100000 microfarad capacitor will charge to
around seven and a half volts in about ten seconds, a 100000 millifarad unit will take almost 7 hours to reach the same voltage. No highh voltages
needed, no risk of giving your heart a jolt. |
Not_important, how did you calculate this ? Charging will be an exponentional rise to maximum function, so I assume you defined >0.95 out of 1
(actual V divided by max V) as the full charge? Would be interesting to know!
Also, how are capacitors designed that do NOT die when shortcircuiting them (exploding wire etc)? I've had exactly that problem in the past, a few
rapid discharges and the capacitors (taken from a TV) died...
Never Stop to Begin, and Never Begin to Stop...
Tolerance is good. But not with the intolerant! (Wilhelm Busch)
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watson.fawkes
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| Quote: Originally posted by chemoleo | | Charging will be an exponentional rise to maximum function, so I assume you defined >0.95 out of 1 (actual V divided by max V) as the full charge?
Would be interesting to know! |
See http://hyperphysics.phy-astr.gsu.edu/HBASE/electric/capdis.html for a short exposition with further links.
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watson.fawkes
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| Quote: Originally posted by chemoleo | | Also, how are capacitors designed that do NOT die when shortcircuiting them (exploding wire etc)? I've had exactly that problem in the past, a few
rapid discharges and the capacitors (taken from a TV) died... |
To think about this sanely, you need to focus
on ways that a physical capacitor departs from the ideal model of a capacitor. Non-ideal capacitors can be modeled with parasitic components,
typically resistors and inductors.
Resistors generate heat loss. The faster the discharge, the more power dissipated as heat. If you generate heat faster than the capacitor's ability to
shed it, you'll do Bad Things to the inside, such as, say, boiling the electrolyte or melting the dielectric. Another kind of failure comes from
thermal stress and warping, bringing the plates closer to each other somewhere, which lowers the voltage rating of the capacitor, so that a subsequent
pulse causes failure. Therefore, one technique is to reduce heat loss by lowering internal resistance. This means thicker internal conductors, more
conductive conductors (copper instead of aluminum, for example), and lower loss dielectrics. For even more extreme applications, there are design
changes to lower the thermal resistance of the interior of the capacitor to the outside.
A capacitor that makes a good pulse requires low inductance, but this is a different concern than the one you asked about.
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12AX7
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High pulse caps are also build like tanks (and live inside heavy tanks!). The compressive electric force of full charge is quite strong, though it
holds the dielectric in compression fairly simply. Along comes a couple kiloamp discharge and the magnetic force pushes everything apart (think
railgun), while at the same time, the electric force drops to zero (or goes to zero, then goes back up again, then down some more, since discharge is
usually underdamped slightly).
If the capacitor isn't constructed properly, it may also discharge unevenly. A simple wound capacitor has two leads stuck somewhere in the middle of
the winding. So the middle does all this first, then towards the ends. I suppose if the velocity happens to match the speed of sound in the
material, some ugly shock wave type action could cause interesting mechanical problems.
Tim
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not_important
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| Quote: Originally posted by chemoleo |
Not_important, how did you calculate this ? Charging will be an exponentional rise to maximum function, so I assume you defined >0.95 out of 1
(actual V divided by max V) as the full charge? Would be interesting to know!
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The standard definition of R-C time constant, the time to reach 63.2 percent of full charge voltage or
to discharge to 36.8 percent of the starting charged voltage. T (seconds) = R (Ohms) × C (Farads)
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franklyn
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30 year old assessment of solar applications
http://www.osti.gov/accomplishments/documents/fullText/ACC01...
A wealth of information is available on line , what I googled
http://rapidshare.com/files/161735022/Stand-alone_Photovolta...
http://rapidshare.com/files/164116389/Planning_and_Installin...
http://rapidshare.com/files/119677336/Solar_House_-_A_Guide_...
Photovoltaics are only perhaps 12 % efficient at conversion of incident radiation
because it only captures a narrow fraction of the spectrum. Passive collectors
capture almost all but as low temperature heat with few prospects for efficient
use beyond space heating. Still the ultra low cost and high latent heat available
from brine ponds makes this an area for continued research.
http://www.solarponds.com
Theoretical Analysis of Closed Rankine Cycle Solar Pond Power Generator
http://ccsenet.org/journal/index.php/mas/article/download/26...
Experimental Study and Modeling of a Low Temperature Rankine Cycle for Small Scale Cogeneration
http://www.labothap.ulg.ac.be/cmsms/Staff/QuoilinS/TFE_SQ010...
conclusion ( in part ):
"with the use of closed Rankine cycle solar pond power generator. It has been concluded that
the generation of maximum thermal efficiency of 27.94 % and specific power of 16.852 kW/m2
can be achieved by using the effective closed Rankine cycle solar pond power generator."
The only other low temperature device known I previously discussed here _
http://www.sciencemadness.org/talk/viewthread.php?tid=5080#p...
Proceedings of the Nitinol Heat Engine conference
http://handle.dtic.mil/100.2/ADA108973 , redirects to _
http://www.dtic.mil/cgi-bin/GetTRDoc?AD=ADA108973&Locati...
Heat Engine Driven by Shape Memory Alloys: Prototyping and Design
http://scholar.lib.vt.edu/theses/available/etd-09252002-1707...
- *.mpg video -
http://scholar.lib.vt.edu/theses/available/etd-09252002-1707...
Thermoacoustic devices as yet still operate at too high a temperature to be powered
by the heat of a brine pond , though it can operate by concentrated sunlight.
http://www.city.ac.uk/sems/research/research_structure_and_a...
http://www.score.uk.com/research/default.aspx
http://www.lanl.gov/mst/engine
Thermoacoustic Engines
http://www.lanl.gov/thermoacoustics/Pubs/ICSV9.pdf
Also known as "Laminar Flow Stirling Engine"
http://www.youtube.com/watch?v=cAyw_dOioMU
Thermoacoustic Magneto thermodynamic generator , U.S. patent 4599551
http://www.pat2pdf.org/pat2pdf/foo.pl?number=4599551
This last device above uses the resonant wave produced to directly generate electric
current with a fluid metal. I wonder if a gas of Xenon seeded with Cesium could be
operable at lower temperature differential.
.
[Edited on 15-9-2009 by franklyn]
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not_important
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| Quote: Originally posted by franklyn | ...
Photovoltaics are only perhaps 12 % efficient at conversion of incident radiation
because it only captures a narrow fraction of the spectrum. ...
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"High-power module using 156mm square poly-crystalline silicon solar cells with 13.74% module conversion efficiency"
http://www.solarhome.org/224wattmodule.aspx
And that polycrystalline cells, the high end of the commercial lines runs about 18% efficient.
Silicon photovoltaic cells capture over half the solar spectrum, having a bandgap of 1,1 eV. The problem is that the excess energy of an absorbed
photo, energy that puts the photon above the bandgap energy, goes into making heat. Thus as te photo energy moves out of the very near infrared
towards the blue end of the spectrum, an increasing percentage of the light goes into heat.
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JohnWW
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Thanks again for your info, Franklyn. I am aware that there have been substantial improvements in the efficiencies of solar photovoltaic cells in
recent years, as the result of which I have not yet been in any particular hurry to buy them. Large drops in their prices per KW generated in the near
future can be expected.
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DerAlte
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Look at it this way, JohnWWW:
A 40Ah car battery, 12v, stores 40x12= 480 Wh or 1.728E6 joules of energy.
An equivalent capacitor of C Farads at 360 volts stores ½*C*V^2 joules if you could gainfully extract it). Thus the equivalent capacitor on the basis
of energy storage is 2*1.728E6/(360^2) = 27 F at rating 360V. Huge! Especially compared with the battery. Electrical storage cannot remotely compete
with electrochemical. If Li ion cells become markedly cheaper and reliable due to vehicle use, maybe they would be the way to go.
Here I use 1MWh/month in summer = average of 1.39Kw rate/hr. or 33kWh/ day. The house is all electric. Air conditioning is a must and the most
expensive item – outside temps are around 93F and humidity typically 50% RH or greater, which feels like 100F. Full sunshine averages perhaps
6hr/day. Assuming optimum vertical angle of panels for latitude, and a cosine factor for illumination angle, this could be equivalent to full
illumination of perhaps 4 hours. Hence the panels would have to be rated at 24/4 = 6 times my average usage, i.e. 8.34 kW. At 15% overall efficiency
the required insolation at 1kWh/m^2 is 55.6 kW requiring 55.6 m^2 of panel. If I covered my south facing roof area not shaded by trees I actually have
this available.
Sizing the storage requirement: In the worst weather (hurricane season) a one day supply – 33kWh reserve - should be adequate. About 70 typical
car batteries. However, one ought to allow for peak usage of about 12 kW, the all-equipment-on level. I have no idea what the allowable draw on lead
acid batteries is, for a reasonable life, but suppose it is ¼ capacity. This needs a 12x4 kWh rating = 48 kWh (100 batteries).
I have no idea how much a plant of this size would cost. I’d guess ~$30,000 US. At this rate it would take about 20 years for payoff (but ignoring
maintenance). Since this represents a return of about 5%/an. I can consistently do better investing (bar the current crisis!) and paying the
14cent/kWh, exorbitant as it is.
Now consider your rural farmhouse. Here the situation differs. Solar water heating is very efficient and reasonably trouble free (I had it at my
previous house. 2m^2 of panel provided overkill for a family of 4.5 all year). Heating and cooling could be provided by an ammonia cycle reversible
machine using solar heat as the source. Also very trouble free, if inefficient. Alternatively, solar water heating panels might provide enough winter
heat, depending on climate, and perhaps air conditioning isn’t necessary in NZ in summer. The electrical plant could then be minimized to a fraction
of that needed by my Florida example, and far more cost effective.
It seems that solar energy is on the edge of being practical in some areas of this globe. The chief problem is amortization of the capital cost, as in
nuclear energy production.
Incidentally, I thought cost/kWh was low in NZ due to extensive hydroelectric production.
Regards, Der Alte
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merrlin
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Maxwell Laboratories produces ultracapacitors with energy densites of up to 5.5 Wh/kg. Their largest MC device is 3000 farads at 2.7 volts.
Attachment: MC_Cell_Power_1009361_rev9.pdf (521kB)
This file has been downloaded 799 times
Attachment: energy_buffers.pdf (568kB)
This file has been downloaded 1182 times
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JohnWW
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| Quote: Originally posted by DerAlte | Look at it this way, JohnWW:
A 40Ah car battery, 12v, stores 40x12= 480 Wh or 1.728E6 joules of energy. An equivalent capacitor of C Farads at 360 volts stores ½*C*V^2 joules if
you could gainfully extract it). Thus the equivalent capacitor on the basis of energy storage is 2*1.728E6/(360^2) = 27 F at rating 360V. Huge!
Especially compared with the battery. Electrical storage cannot remotely compete with electrochemical.(cut)
Incidentally, I thought cost/kWh was low in NZ due to extensive hydroelectric production. |
Thanks for your observations, Der Alte. Yes, for water-heating, I could use some large solar water panels, besides bottled-gas and/or wood heating for
hot water.
In spite of extensive hydro-electric and geothermal generation, and now wind generation, with tidal generation planned, the cost of electricity in New
Zealand is exorbitant due to lack of competition in generation and retailing, inadequate regulations, and inadequate enforcement of regulations, in
the partly-privatized system that arose in a few years of Thatcher-inspired right-wing free-market madness about 20 years ago. No government since has
had the courage to do anything about it, except indirectly by introducing new building regulations requiring insulation, and subsidize this and
retro-fitting of insulation in houses.
[Edited on 16-9-09 by JohnWW]
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JohnWW
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| Quote: Originally posted by merrlin |
Maxwell Laboratories produces ultracapacitors with energy densites of up to 5.5 Wh/kg. Their largest MC device is 3000 farads at 2.7 volts.
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Thanks for your effort, Merrlin. However, that largest Maxwell Labs ultracapacitor, at its low rated voltage of only 2.7v which is suitable for
electronic circuits but not large-scale power storage, holds only ½CV² = 10,935 watt-second = 0.00304 KW-hr. That would be quite inadequate for my
purposes. Maxwell should come back here only when they have a capacitor of that size that can withstand up to 350v DC; this would store 51.04 KW-hr,
more than ample to power an all-electric house at any time of the year.
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