Dennis Adams's posts

Household garbage contains high quantities of zinc, cadmium, lead, magnesium, copper, potassium, barium, and calcium, not to mention whatever gets included in the vast quantites of plastic containers and wrappings we use, much of which is poisonous when burned.  Fortunately much of this can be also be recycled and most of which can't should definitely not be burned.  burning increases the concentration of the  unburnable materials (lead, copper, cadmium) in the ash, so the ash is probably going to be more toxic than leaving those elements where they were in the first place. If recycling is not an option, bury it.  Granted landfills are not the ideal solution.  Recycling is obviously preferred, however many modern landfills are made on tightly compacted clay layers which make migration of harmfull elements into water supplies much more difficult than in past years.  Some landfills can be tapped for the methane they generate after certain areas are filled in and capped off.

I would have thought composting might be an attractive option, especially in rural areas where the space for composting is more than likely available and the composted products can generally be readily used for improvement of existing soils and growing a really nice vege patch.

For some tips on composting see,
http://users.imag.net/~lon.trea/level1.htm

What turbine manufacturer and model do you have?

The rated speed, voltage and current should be marked on the equipment.

Larger diameter wiring will reduce the voltage loss and thereby increase the amperage.  The amount of the increase may or may not be significant depending on the voltage output by the generator and the length of the wires.

For efficient use, yes.  A charge controller charges the batteries in a controlled manner, permitting greater current to be sent to the batteries when they are low and reducing the current to the batteries when the charge level comes up.  It isn't a necessity to have charge control, but if not, high currents would be sent to the battery even though they lack only 1 mA from being full.  The idea is that a good portion of those high currents could be used elsewhere whenever the batteries do not need to be charged at full rate.

The Xantrex C Series for example can be used as solar charge controllers or load controllers or load diversion controllers. I haven't been looking closely, so if these aren't, I believe there are multiple input models out there.

I'm not much of an electrical engineer, but as I like to think of it, there are basically two types of generators  used in renewable energy, PV and batteries are one and then the induction type Wind/Hydro generators, which are really just induction motors spinning in reverse.  That's leaving synchronus generators aside.  PVs and Batteries do not derive their voltage from the circuit they are supplying (within limits), whereas induction generators do.  Induction generators need an electric field in which to spin in order to generate their voltage.  The strength of the field is determined by the circuit they are supplying.  Therefore changes in the circuit load, change the field of the generator and consequently change its output.

Claudio, you have a very nice website there, clean and with nicely done drawings. 

Not to put a spanner in the works, but you might do better spending less money on your website and paying your engineer a little bit more such that he can finish the calculations with enough detail, including ALL friction factors and a thermodynamic model of something more realistic than the ideal adiabatic process you have chosen to "prove" it works.  I do suspect that you will be able to see for yourself that the laws of thermodynamics are quite troublesome to circumnavigate.  That said, good luck in minimizing those adverse effects.

S'poze it depends on how much that horse drinks. :-)
Looks like we're not going to find out either.

provided you can keep the equipment from freezing up and bursting during the night.

Actually, I suppos the best way is to make an even simpler solar thermal collector installation.

John, I'm wondering first, why you think its expensive and secondly, why you think its more expensive than installing and maintaining a piped supply to and a pump itself in freezing conditions.  If you did go that route, then you still have to power the pump, so solar panels (and batteries) would still be needed.  A simple panel (and maybe batteries) connected to a small resistance heating element (maybe more like heat tracing) might be just the right solution.  Certainly much simpler. If the temperature difference isn't much and there is some insulation available, power reqirements could be quite small. 

I had a look at the AirX cut sheets and they have a speed cotrol that they say kicks in when the batteries have charged and stops the turbine.  When the batteries are charged, apparently the turbine will not start.  That means that, while the batteries are receiving all current, a load "diversion" is not needed as the battery takes 100% of load, and when they are reaching charge, speed control begins and current is limited to less than what can be produced without a diversion load, when charged, breaking kicks in and the stopped turbine (of course) produces nothing and needs no load diversion.  Therefore the AirX turbine does need to accept full load and needs load diversion capability, OR needs speed control while actively spinning.  My previous post above wasn't clear as I did not mention that I was assuming that the case being discussed was limited to fully spinning-free conditions, under which all turbo-machines need load diversion, of which the AirX is no exception.

Can I ask a couple of questions first?  First, how much water is in the trough and what's the lowest temperature you want to have ice-free water available?  What temperature do you want to try to keep the water at?  (Just thinking 33º might be a little too cold to drink.)  Assuming you want to keep water available for a number of cloudy days, you'll need batteries, so how many days to you want to guarantee icefree water?  More days = more batteries.

Wait a minute.  Did you say that you think diversion loads are optional?  Diversion loads are not optional with hydro or wind.  These devices create electricity from spinning a driver which spins a generator.  Once you have something spinning, its not easy to stop.  If you immediately remove or reduce the braking action (in this case that's the electrical load that they have been ramped up against), they can sometimes take off.... sometimes all the way off.  The diversion load is needed to hold them in place and in one piece, and cool, until they can be ramped back to the new power level output required for the new load.  That's if you have control over the new rpm needed.  Changing the flowrate to a hydro, for example (but you can't change the windspeed so easily).  Until then you have acceleration and a possibility for very high voltages, if the load is ramping down, or brownouts from equipment coming on line, but trying to operate on too low power levels as the load is building, but supply remains relatively low.

There are three coefficients of temperature, one for current, one for potential and one for power.  When used with the I-V curve for the reference temperature, such as you find on the cell data sheets, as this BP SX120 S has,

http://www.bp.com/liveassets/bp_internet/solar/bp_solar_north_america/STAGING/local_assets/downloads_pdfs/pq/product_data_sheet_bp_sx_120s_01_4016_v1_en.pdf

you can convert a voltage and current from the reference curve at 25ºC and get the actual current and voltage at the actual operating temperature and draw the rest of the curves that BP has so kindly furnished, or a curve for an intermediate value, such as their listed supposed operating temperature of 47ºC, (strange they don't give that curve).

To do such a calculation, I have made an example, converting the Isc and Voc values of the maximum power point at the reference temperature to the maximum power point at the supposed standard operating condition 47ºC (117ºF), do this,

T2   47   ºC
Tref   25   ºC
dT ºC   22   Cº
      
NEW POWER FROM TC power      
P at mp   119.6   W
TC power   -0.5    /Cº
dT ºC   22   Cº
dW   -11   W
P   119.6   W
Pref@47ºC   108.616   W

New Power Calculated from α and β, which shows the interrelation between all the coefficients,      

NEW CURRENT      
TC current = α   0.065   %/C
 /100   0.00065   /Cº
dT   22   Cº
dI   0.0143   A
Isc @= mp   7.12   A
Isc * dI   0.101816   A
Isc_75   7.221816   A

NEW VOLTAGE      
Voc at mp   16.8   V
TC voltage = β   -80   mV/Cº
 /1000   -0.08   V/Cº
dT ºC   22   Cº
dV   -1.76   V
Voc   15.04   V

NEW POWER      
Power   108.6   W

Same answer as we got from the power coefficient.
 
You have to be very careful with these coefficients, because they can also change according to how the panels are wired, in parallel and series and if there is a different temperature of one panel in the group, which might affect the output of all.  Use it as a relatively good estimate only.  Its not usually exact.

Check out all these articles on the National Renewable Energy Lab site,

http://search.nrel.gov/query.html?col=nrel&qt=hybrid+wind+solar&charset=utf-8&qp=site%3Awww.nrel.gov+site%3Awww.sst.nrel.gov+site%3Arredc.nrel.gov&qc=nrel&ws=0

80 ft is relatively tall.  Are you sure it needs to be that high?

One thing that would make a major difference in cost is if the tower could be guyed, or if it must be free standing.  Guys will take a considerable area.  You should have enough area for at least 3 guy points 2/3 as far away from the tower as the tower is high, 2/3 * 80 = 60 ft, so that would be a triangle 90 ft across the bottom x 90 ft in height.  The guyed tower structure itself would be much cheaper than a free standing tower and the foundation needed to support it.  Can you fit a guyed tower in the area available? 

On the ground yes, but not "grounded" electrically.  The panels need to be bonded to electrical ground.

Hi Steve,

You might try looking at this page.  A bunch of things, maybe not exactly what you're looking for, but pleanty good enough to give you all kinds of ideas.

http://www.otherpower.com/otherpower_experiments.html

When you get bored with that one (should take you awhile), just Google "solar power experiments" and a bunch more will pop up.

Happy Experimenting!

In the US the National Electric Code is followed.  I assume you are in the U.K., so you must really check the U.K. regulations.  To attempt to answer your question, without knowing anything about your local code, I think if each has a separate earth, you should bond the separate earthings together.  The idea is that no components should ever have the possibility of developing differential voltages between them.

Here are some guidelines for inspecting U.S./ N.E.C. compliant installations, within which you will see the above recommendation.  Maybe this document will be of help until you can find out exactly how the system should be wired according to the U.K. electric code.

http://www.irecusa.org/fileadmin/user_upload/NationalOutreachPubs/InspectorGuidelines-Version2.1.pdf

Sorry, here's the page with the PV grounding diagram and info,
http://discoverpower.com/shop/item.asp?PID=80

Hi Thomas,
The correct choice of prop is super critical for the average wind velocity at any given site.  Grabbing at all that air running past at faster than the average speed your prop was designed for won't get you any more power. You'd have to be willing to go up there and change the prop every time the airspeed went higher.  Then, if you do that, you'll lose power production when the wind fell back to normal.  Check this propeller efficiency chart.  Your prop is probably peaking out on power production at +/-28 mph, so ratioing that propeller curve there, its efficiency is falling at around 33-35 mph to Zero.  If you changed to a prop with Hi eff at 35 mph, when the wind is back around 25 mph, you'd only be at 55-60% of optimum juice making capacity, so you'd be up there changing the prop again.  I believe the gen has an internal brake that starts kicking in at 33 to 35 mph, otherwise the prop would vibrate and maybe fly apart when it hit 45-50 or the voltage would get too high at those rpms and start burning through the insulation.

I'm a rotating equipment engineer and don't know much about grounding such that I'd want to give you possibly bad advice, but I have found this information for what may cover some of your questions.  Read this page on lightning arrestors, check the diagram for the PV installation, and see you think any of it might apply to your situation.  I can point you to an engineering Q/A site where you might be able to get a good answer.  http://www.eng-tips.com  Register there and go to the Electrical Engineering and Motor Forum.  Post your question.  Usually the advice is excellent ..and reasonably fast too, although you might want to wait to see if some "alternative" suggestions come up.  Then come back here and tell me what they told you.

but maybe not getting to the full 27.24V

John, You still there?

I think you need to leave it right there at 28.2V.

Power = Voltage x Current and say the wind gen is putting out say 5 amps (?), that would be 5A x 28.2V =  141 W.  So find your current coming from the generator and substitute that current for the 5A above and get your power output.

Continuing,

If 141 W is running to a 27.24 V battery rack, it would charge at 141/27.24 = 5.2A.  Substitute your power output for the 141W number and divide that by 27.24V battery V to get your charge current (A).  There will probably be some line loss from the gen to the bats, so including that might get your gen output equal to or a little lower than the battery voltage, in which case it would only tend to increase the charge current to the battery a little more.  I'll bet you'd be charging at just a bit higher than 5 amps if your wind gen put out 5 amps like this example.

The equipment should be able to survive on the roof without coolant flows, although its life (esp. of any plastic parts) might be compromised somewhat.  Really I'd guess its just much easier to valve some heat off to a rejection point than climb up to the roof every time you want to adjust the water outlet temp.

Hi Wally,

You know it sounds to me like you've already (and quite admirably) answered your own question.  The most important part of the decision to invest in alternative energy first involves a good site analysis.  I think you've done that and have even included some valid moral issues as well.  It is also interesting to note that saving fuel use and expense and even "Going Green", in terms of benefiting the entire planet, for most people involves economic sacrifice in that it requires them to make a rather large investment.  In your case it sounds like you can accomplish all of those things just by letting Mother Nature handle things on her own.  And isn't that the real idea behind all of this and your best alternative too?   

I live in Spain, so I don't know what if any alternatives exist in the US, but here there are a few companies that would allow you to participate in solar power projects, even if you don't have your own rooftop.  They are building PV power stations of 500kW to 5MW and even 20MW, in which they are selling shares of each power station, perhaps to people just like yourself.  Each investor receives his portion of the power sold into the national grid.  In fact, I am planning to start just such a venture.  In that manner, you could be free to "do the right thing" for your site and generate solar power as well.  If you can't find a similar option locally, I could give you the names of some companies in Germany and Spain that already have such projects in operation.  Just a thought.

160ºF is a reasonable temperature coming off solar collectors and appears reasonable for radiant heating systems as well.

My thought was that if you shoot for 170 as a design temperature, it will give you too much heat to reject when solar radiation is strong and require you to shut-down and drain the collectors, or add a heat rejection radiator and/or otherwise complicate the system and its control.

If a little is good, you generally think more is better, but that's only true up to the amount you actually need.

Select a blower with the air flowrate and output pressure you need.  Make a note of the RPM and the Power that the blower needs to put out that air.  Find a motor that runs at the same RPM of the blower and has 10 to 20% more power than what the blower needs.

Sounds pretty hot to me.  How well does your system perform at its existing tempertures?  If it is comfortable, there should be no need to increase those temperatures.

James,

Right, my numbers don't make sense and don't matter anyway, as its only the method used on the page (link) that I'm interested in clarifying.

As you have indicated, a 20% increase factor appears in the calculation, which I don't think was noted.  I was just making sure there wasn't another consideration that might have influenced that increase over the calculated value.

Thanks,
D. Adams

Hi All,  (I'm new here)

I was looking at the inverter calculation on page,
http://store.altenergystore.com/calculators/on_grid_calculator/

and I can't seem to figure how they get from a monthly usage of 100 kW, avg sun hours of 4.13 to Watts needed = 969

There is some kind of loss factor in there.

Any ideas about what the Watts formula contains?
Thanks