Sizing Generator

SolarSailor
SolarSailor Solar Expert Posts: 49
On the generation side of a battery array, I've read here that charging capacity needs to be 5%-15% of array size.

For now, let's assume the generator remains unknown (i.e., wind turbine, PV, etc), but that it can provide the needed input.

If my battery array is sized to achieve 100% demand (during night time hours) of, say 2kW for six hours (6pm-midnight, or 5pm-11pm) when no sunlight is available, then the charging mechanism would need be 15% of 2kW, or 300-Watts.

If the generator were comprised of 200-Watt PV panels, then it'd take 2 x 200-Watts = 400-Watts of capacity to recharge the battery array (with a 33% margin).

Now, here's the queston. If the daytime use is also 2kW throughout the day (when charging might occur), then the generation mechanism would need to be able to provide 2kW continuous in addition to the 300-Watts of charging, right?

That'd be a total of 2,300-Watts of generation capacity. And, if 200-Watt PV panels are used, then 2300 / 200 = 11 + 1 = 12 x 200-Watt PV Panels for a system size of 2,400-Watts (2.4kW, or 120% of demand).

First off, are my calcs close to right?

And, would the 300-Watts of charging capacity keep the array charged with use as described above?

Comments

  • SolarSailor
    SolarSailor Solar Expert Posts: 49
    Re: Sizing Generator

    In the example above, with 300-Watts of charging capacity, it would take 1kW / 300 = 3.3 hours to recharge for 1kWH of output, right?

    Then, the six hours of 2kW output during night time hours would require 6 x 2kWh = 12 kWhs of needed ouput from the battery array.

    To get 12kWhs of output, then, I'd need 12kWhs of charge input to the system during daylight hours. Is that right?

    And, to get 12kWhs of charging at 300-Watts (per hour) would require 12,000 / 300 = 40 hours of charge time. Is that right?

    Then, to get my 12kWhs of charging during an 8-hour charging day, I'd need 5X the charging capacity of 300-Watts, or 1,500-Watts to do the job.

    Am I on the right track here?

    Thanks
  • BB.
    BB. Super Moderators, Administrators Posts: 33,433 admin
    Re: Sizing Generator

    Yes, you are correct with the numbers you have posted.

    By the way, Watts is a Rate (Joules per Second) so Watt per Hour is not correct and can get confusing. We use Hours because Watts*Seconds (instead of Watts*Hours) is a very small number and would make our house hold bills 3,600 times larger to use the real units from Physics. 12 kWhrs = 43,200,000 Joules

    Back to sizing your loads... From your other thread, you are talking about doing things efficiently (LEDs for lighting and such) in a smallish home (container). Averaging 12 kWHrs per night + 12 kWHrs during the day = 24 kWhrs per 24 hour period--if I understand you correctly.

    Or 720 kWhrs per month. That is, for an off-grid home, quite a bit of power for somebody going off grid. It is certainly possible to do that--but I would recommend that you aim at a maximum of 100 kWhrs per month (3.3 kWhrs per day total) as a more reasonable starting point. And you would have to accomplish that through conservation (Efficient appliances, turning things off when not needed, lots of insulation, double pane windows, no or little usage of AC, etc.).

    A standard container is not insulated and either you have to insulate from the inside--which would make the container inside dimensions even smaller (5" in the walls and floor, 12" or more for ceiling) and it would not be usable--or adding insulation to the outside--basically building a weather proof and insulated exterior which would make the shipping container redundant.

    Or--look at using refrigerated/insulated shipping containers. Apparently they may use ThermoCor which can be rated up to R45 per inch. At least that is a practical amount of insulation for your "home". Or doing the insulation yourself when finishing the interior.

    I have several times worked out the costs of various forms of power and a rough rule of thumb is (assuming 20 year equipment life, good sun, and battery replacement every 4-8 years or so):

    $0.10-$0.30 per kWH--utility power
    $0.10-$0.35 per kWH--Grid Tied Solar Power (no batteries)
    $0.40-$0.70 per kWH--Grid Tied with Off Grid backup (Hybrid System)
    $0.50-$1.00+ per kWH--Generator Power
    $1.00-$2.00+ per kWH--Off Grid power

    So--your 720 kWH per month utility power bill for ~$100 or less would cost you (real--out of pocket money) ~$700-$1,200+ a month for an Off-Grid system. You have to pay for the system up front--plus replace the batteries every ~4-8-10 years or so.

    Another reason an Off-Grid system is expensive is because of the extra losses due to Battery Losses and Inverter Losses. So, the end to end efficiency works out for a Grid Tied system of 0.77 vs for an off-grid system of 0.52 ...

    We can use the PV Watts web program to estimate the size of system you would need for your home. Start with 0.77 derating (GT) and 1kW array (easy number) for Tulsa OK:
    "Station Identification"
    "City:","Tulsa"
    "State:","Oklahoma"
    "Lat (deg N):", 36.20
    "Long (deg W):", 95.90
    "Elev (m): ", 206
    "PV System Specifications"
    "DC Rating:"," 1.0 kW"
    "DC to AC Derate Factor:"," 0.770"
    "AC Rating:"," 0.8 kW"
    "Array Type: Fixed Tilt"
    "Array Tilt:"," 36.2"
    "Array Azimuth:","180.0"

    "Energy Specifications"
    "Cost of Electricity:"," 7.7 cents/kWh"

    "Results"
    "Month", "Solar Radiation (kWh/m^2/day)", "AC Energy (kWh)", "Energy Value ($)"
    1, 4.01, 97, 7.47
    2, 4.46, 95, 7.32
    3, 5.33, 122, 9.39
    4, 5.88, 128, 9.86
    5, 5.61, 122, 9.39
    6, 5.77, 118, 9.09
    7, 6.06, 126, 9.70
    8, 5.95, 125, 9.62
    9, 4.99, 105, 8.09
    10, 5.41, 122, 9.39
    11, 4.11, 92, 7.08
    12, 3.70, 88, 6.78
    "Year", 5.11, 1339, 103.10
    You can see in the 3rd column that you will get 88-128 kWH per month per 1kW of solar panels. If you want 720 kWHrs per month, you would need (assuming no generator use 9 months out of the year):

    720 kWH per month / 97 kWH per 1kW panel for January = 7.4 kW of solar panels

    An GT system, roughly, costs $6-$10 per Watt installed. Or:

    7,400 Watt System * $7 per Watt = $51,800

    Now, you can get a 30% federal tax break ($15,000 off), and you may have other local breaks and rebates too (www.dsireusa.org)

    Now do the same calculation with Derating=0.52:
    "Station Identification"
    "City:","Tulsa"
    "State:","Oklahoma"
    "Lat (deg N):", 36.20
    "Long (deg W):", 95.90
    "Elev (m): ", 206
    "PV System Specifications"
    "DC Rating:"," 1.0 kW"
    "DC to AC Derate Factor:"," 0.520"
    "AC Rating:"," 0.5 kW"
    "Array Type: Fixed Tilt"
    "Array Tilt:"," 36.2"
    "Array Azimuth:","180.0"

    "Energy Specifications"
    "Cost of Electricity:"," 7.7 cents/kWh"

    "Results"
    "Month", "Solar Radiation (kWh/m^2/day)", "AC Energy (kWh)", "Energy Value ($)"
    1, 4.01, 64, 4.93
    2, 4.46, 63, 4.85
    3, 5.33, 81, 6.24
    4, 5.88, 85, 6.54
    5, 5.61, 80, 6.16
    6, 5.77, 78, 6.01
    7, 6.06, 83, 6.39
    8, 5.95, 82, 6.31
    9, 4.99, 69, 5.31
    10, 5.41, 81, 6.24
    11, 4.11, 61, 4.70
    12, 3.70, 58, 4.47
    "Year", 5.11, 884, 68.07
    Same thing, take Feburary 64 kWhrs per month and (again very roughly) $12-$20 per Watt installed for Off Grid system (need to include backup genset too):

    720 kWHr per month / 64 kWhr per month per 1kW of panels = 11.25 kW of panels

    Assume ~$15 per Watt for parts:

    11,250 Watts of panels * $15 per watt = $168,750 for Off-Grid solar

    Plus you will need another (very roughly) $10,000 worth of batteries every 10 years or so...

    Now, you can save a bit of energy by forgoing the inverter (15% losses) and using AGM batteries which are 10% loss instead of 20% losses for Flooded Cell, and such... But that is still a fraction of overall price... And AGMs can cost 2x the cost of flooded cell batteries and DC equipment can cost more too...

    Unless you have a great desire to live 100% off grid while using a lot of power--something is going to have to give.

    Realistically, using less electricity (through conservation and life style changes) and accepting that a genset for 10 days every 5 years (get a quiet / fuel efficient one and everyone will be happier) + Grid Tied solar is probably better for both return on investment and the environment too.

    The above is a very rough estimate of the scale of your system based on rules of thumbs and retail costs. I am not in the solar business and not a contractor and do not know your specifics... The above is an "order of magnitude" estimate of the costs involved for what you appear to be asking for. (i.e., SWAG -- Scientific Wild A$$ Guess :roll: )

    Please ask questions and do your own research before spending any money.

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • Cariboocoot
    Cariboocoot Banned Posts: 17,615 ✭✭✭
    Re: Sizing Generator
    On the generation side of a battery array, I've read here that charging capacity needs to be 5%-15% of array size.

    Nope. Charging current, regardless of source, needs to be 5%-15% of battery bank Amp/hr capacity.

    Generator should be able to supply as much power as the array, as it substitutes for the PV's when the sun doesn't shine. Also remember that most inverter/chargers switch the AC load to the generator when acting as a charger. This means the generator should have capacity to supply the load requirements plus the charging requirements. Since loads are intermittent, however, there is some 'fudge factor' to this as charging current can go down periodically to make up for increased loads (without over-taxing the generator).

    The basic formula is: calculate loads first. The inverter is sized for the maximum cumulative loads. The battery bank is sized to handle the loads over time between charges. The array and generator are sized to charge the batteries.

    If you don't know your load requirements you either spend too much buying too large of a system or else you run out of power before you want to.
  • icarus
    icarus Solar Expert Posts: 5,436 ✭✭✭✭
    Re: Sizing Generator

    SS,


    Here is my back of the napkin calc, and it is pretty close, given how easy it is to do.

    You can use it to figure from either end.

    To find the net/net amount of energy a system will reasonably produce OUT THE OUTLETS) taking in to account all system efficiencies (PV/wiring/controller losses/ wiring losses/ battery charging ef/ inverter ef/ wiring losses) try this:

    Take the name plate rating of the PV array, divide this by 2 to allow for said losses, then multiply that number by the average number of hours of GOOD sun one might expect. (seldom more than 4).

    So, let's take 1000 watts of Pv for example. 1000/2X4=2000watt/hours/day on average (In most locations,, see Bill's Pv watts link for yours!)

    So the maximum reasonable amount of energy you could expect to harvest with a battery based system is (1kw) is ~ 2kwh/day. Do realize of course that in addition to battery based systems costing more up front, they run far LESS efficiently mostly due to charge efficiency losses.

    In addition, there are many practical battery concerns, all of which effect (affect?) the longevity of the battery.

    http://www.windsun.com/Batteries/Battery_FAQ.htm#Lifespan%20of%20Batteries
    http://www.batteryfaq.org/

    The above links will provide a great deal of info regarding the care and feeding of batteries, and are a must read for any battery based system.

    Finally, most people do two things wrong when they start out. They overestimate the amount of solar power they will get (net/net) from a system, and they underestimate the loads they will use, resulting in significant charging deficits.

    Good luck,

    Tony
  • niel
    niel Solar Expert Posts: 10,300 ✭✭✭✭
    Re: Sizing Generator

    actually 5-15% is wrong as it is 5-13% rate because trojan batteries and those similar to them are not recommending beyond a 13% charge rate. many other batteries can take a higher charge rate, but with much higher maintenance for the fla types. as was mentioned, the agm style of battery can take higher charging rates with concorde's lifeline and sunxtenders for example able to take on a 100% charge rate and would make generator charging a much shorter time frame if able to deliver the higher power to the batteries from both the generator's and charger's abilities (can be pvs and cc as charging source). i cannot say other agms will be able to do 100%, but rest assured it would most likely be at least around a 30% charge rate capability.

    as with any battery you also don't want to drain it past 50% or the lifetime of the battery suffers and adding it generally improves too if not allowed to be even half drained, but will have a peak before 100%.

    btw, battery capacities are usually written in amp hours (ah).

    you are essentially correct that if you supply the load power to the batteries and the amount of power needed to maintain the batteries too that the batteries will stay at the level that they currently hold. if they started out full they would stay full. if not full, they would stay not full. remember, solar is not a constant everyday exacting source as it varies considerably so much higher levels of pv power should be used and can be 2x or more what you initially thought the pv power should be.
  • SolarSailor
    SolarSailor Solar Expert Posts: 49
    Re: Sizing Generator

    I'm having a little trouble making the jump from watts to amp-hour ratings.

    My electric bill is always in kW-Hours, so Watt-Hours seems normal.
    That is, 300 Watts per hour = 0.3 kWh.

    But, when we get to batteries, suddenly it becomes Amp-Hours.

    Any helpful analogies or rules of thumb here?
  • BB.
    BB. Super Moderators, Administrators Posts: 33,433 admin
    Re: Sizing Generator

    Power = Volts * Amps = V^2 / R = I^2 * R = Watts

    So:

    300 Watts / 120 volts = 2.5 amps
    300 Watts / 12 volts = 25 amps

    By the way--don't use Watts per hour... It is just Watts.

    When you look at your utility bill--it is in kWH (kilo Watts * Hours).

    100 watt light bulb running for 20 hours:

    100 Watts * 20 Hours = 2,000 Watt*Hours = 2 kWH

    Watts is like Gallons per Hour. Watts is how much power you are using (rate)

    Watt*Hours is like Gallons. Watt*Hours is how many gallons you would have to buy at the pump at the end of the trip to refill your tank.

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • SolarSailor
    SolarSailor Solar Expert Posts: 49
    Re: Sizing Generator
    BB. wrote: »
    Power = Volts * Amps = V^2 / R = I^2 * R = Watts

    So:

    300 Watts / 120 volts = 2.5 amps
    300 Watts / 12 volts = 25 amps

    By the way--don't use Watts per hour... It is just Watts.

    When you look at your utility bill--it is in kWH (kilo Watts * Hours).

    100 watt light bulb running for 20 hours:

    100 Watts * 20 Hours = 2,000 Watt*Hours = 2 kWH

    Watts is like Gallons per Hour. Watts is how much power you are using (rate)

    Watt*Hours is like Gallons. Watt*Hours is how many gallons you would have to buy at the pump at the end of the trip to refill your tank.

    -Bill

    LOL, still, when you use 100 watts for twenty hours, it's 100 x 20 = 2000 watts, or 2 kWhs.

    The 2.5 Amps is constant, for use in sizing wire. While watts remains rate.
    I get watt*hours as gallons, volume. There seems to be a fundamental piece of electricity which I'm missing.

    Since all the Ohm's Law formulas are directly relational, the conversion from kWhs to Amp-Hours is also relational. But, then, voltage comes to play. 10 Amp-Hrs at 12-vdc is way different than 10 Amp-Hrs at 120-vac. And, batteries don't put out AC current.

    Just have to keep messing with it until it makes sense.
  • BB.
    BB. Super Moderators, Administrators Posts: 33,433 admin
    Re: Sizing Generator

    For a first cut, DC and AC current/voltage/resistance calculations are the same.

    And AH is a simplification that was applied to batteries because "everybody working on the battery bank knows the voltage". And, if this was talking about modern cars--pretty much it is 12 volts. For our solar systems--the answer can be 12/24 or 48 VDC. So--People keep asking us about systems with XXX AH--and we will keep having to ask what the battery bank voltage is. ;)

    AH is handy when working with batteries. Typically Lead Acid batteries are near 100% efficient when dealing with Amp*Hours... When dealing with Watts, batteries will vary between 80-90% (sometimes higher) in efficiency.

    Because P=V*I -- You can see the simple relationship between V and I... If V is 10x larger, power is 10x larger.

    If you want the same power, if Voltage is 10x larger, you take 1/10 the current. One reason that 120 Volts is better for distributing power than 12 Volts.

    Regarding the complication of dealing with AC... AC volts and Current is normalized to the DC power equivalent using a function called the Root Mean Square voltage (or current).

    So, 120 VDC and 120 VAC RMS have the same "power potential" (the actual AC sine wave peak is 1.41x the RMS voltage or 120VACRMS*1.41=169 volts peak). The 1.41 factor is based on the sine wave shape. A Modified Square Wave has a different factor. A square wave the factor is 1.0 ... etc.

    And a simplification for dealing with AC power is the "Power Factor" or PF. So:

    DC: P=V*I
    AC: P=V*I*PF

    PF can be calculated with sine waves and an oscilloscope measuring the voltage and current... Or just use a speciallized meter, like a Kill-A-Watt and read PF from the display (typically from ~0.5 for motors and CFL bulbs to 1.0 for filament lamps and electric heaters).

    Effectively, for AC, a "bad PF" (less than 1.0) causes more current to flow in the wiring (which gives waste heat):

    AC: P=V*I*PF

    Or:

    I=P/(V*PF)

    Generally "good PF" >0.9 and really bad PF is <0.6

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • niel
    niel Solar Expert Posts: 10,300 ✭✭✭✭
    Re: Sizing Generator

    simply, if you have a 12v battery rated at 100ah, this is 100x12=1200wh or 1.2kwh.
    now a 24v battery bank at 100ah (2 of the same 12v batteries in series) now becomes 100x24=2400wh or 2.4kwh.
    paralleling those same 2 100ah 12v batteries would become 200x12=2400wh or 2.4kwh.
  • SolarSailor
    SolarSailor Solar Expert Posts: 49
    Re: Sizing Generator

    By saying not to use Watts per Hour (i.e., "W/hr"), and from the article someone linked, it's really to be stated as Watt-Hours, or "W-Hr", which meaning remains as intended originally. Slight distinction, but important, obviously, to those close to the industry.

    Still not getting the Amp-Hr translations, though.
    Neil, you're multiplying Volts * Amp-Hrs and getting Wh's and kWh's.
    I'm going to my Ohm's Law page and work on this one.
  • BB.
    BB. Super Moderators, Administrators Posts: 33,433 admin
    Re: Sizing Generator

    Some basic electrical equations are (both for AC or DC):

    V=I*R
    P= V*I = V^2 / R = I^2 * R

    Power is a rate (Watts, Horse Power, miles per gallon, etc.). Now the real Power from Physics is typically in Seconds. However, for home sized to nuclear power stations, Hours is used instead (to keep the numbers a bit more "human sized"). The conversion is:

    60 Sec/min * 60 Min/Hr = 3,600 Sec/Hr

    So when we talk about work regarding electrial appliances--we take the Power (rate) and multiply it by Time--or Hours in our case:

    Work (energy) = P*Time = P*Hours = V*I*Hours

    Rewrite the equation to get Amp*Hours:

    Work = P*Hours = Watt*Hours = V*I*Hours = V * (I*Hours) = V * (Amp*Hours)

    So, when working with battery current flow and time---the Voltage term is just being ignored. For example a 100 Amp*Hour battery is a "large sized" 12 volt car battery. But we don't know how much work it can do.. For that we need to add the "Volts" back to the equation:

    Capacity = 100 AH at what voltage? (12 volts):
    Work = Watt*Hours = V*I*Hours = 12 volts * 100 AH = 1,200 Watt*Hours

    The difference between Watt*Hours and kWatt*Hours is:

    1 kWH = 1,000 Watt*Hours (k=x1,000 or "kilo" or 1,000)

    So, if you see a "K" in kWatts, kWH, kVolts, kAmps, etc... Just multiply the kX number by 1,000 to remove the "K" and deal with the "real number" with the prefix removed. Divide a number by 1,000 to get the "k" prefix back.

    Now, all of the above applies for DC circuits and AC values measured in RMS (Root Mean Square), and the AC voltage/current are "in phase". V and I and R (really Z as in impedance) are Vectors (have both magnitude and direction). The math gets very complex very quickly... A simple way around the math is to use a term called Power Factor (PF). This turns equations like power (and work) into:

    P= V * I * PF (for "real" power)
    Work=P=V*I*PF*Hours (for real work)
    VAR=V*I= Volt*Amp or VA or VAR (total current for wire sizing, etc.)

    VAR is important as PF ranges from 0.5-1.0 ... Which means that circuits that run motors and some electronics and electronic lights can use up to twice as much current as may be assumed from just knowing the Wattage. For small systems, not usually an issue...

    But if your system includes a lot of motors (such as well pump) or electronics (computers, CFL lighting)--can require you to get a larger inverter/wiring for proper and safe operation.

    Note that when PF is near 1.0 -- AC and DC circuit math is the same (PF=1 and drops out of the equations).

    OK--I will shut up now. :roll:

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • SolarSailor
    SolarSailor Solar Expert Posts: 49
    Re: Sizing Generator

    I saw in another posting where Power Factor PF is really just a vector associated with the Sine wave, and when 0 or 90-degs it varies from 1.0 to zero as the amount of forward work being done. Seems equivalent to the physics pulling problem of straight forward requiring the least energy, while trying to move a block forward from a 90-deg angle to it is nearly impossible, wastes lots of energy.

    Something in between is, as I understand it, like a motor starting up, drawing lots of amps, then tapering off to a straight line pull.

    Not sure why a motor does this, but I know it does.
  • BB.
    BB. Super Moderators, Administrators Posts: 33,433 admin
    Re: Sizing Generator
    I saw in another posting where Power Factor PF is really just a vector associated with the Sine wave, and when 0 or 90-degs it varies from 1.0 to zero as the amount of forward work being done.
    Yes, you can calculate the PF of the offset between voltage and current sine wave using Sine(phase angle)=PF

    When you have funky current wave forms, the math becomes more complex (funky=non-linear) and it is just easier to use a meter that can measure Power Factor. (motors tend to be linear in that current is an offset sine wave; computers/CFLs/dimmer-switches/etc. tend to be non-linear current wave forms).
    Seems equivalent to the physics pulling problem of straight forward requiring the least energy, while trying to move a block forward from a 90-deg angle to it is nearly impossible, wastes lots of energy.
    The analogy is close... Work is the effort done in the direction of motion. So while pulling on the rope from 90 degrees really hard is a lot of force--since none of the effort is in the direction of motion, then the "PF=0" (cosine90=0.0) applies.

    Note the the force on the rope is real, and will break the rope if pulled hard enough--but none of the force is used to move the car forward.
    Something in between is, as I understand it, like a motor starting up, drawing lots of amps, then tapering off to a straight line pull.
    There is a very clever way of looking at why a single phase AC Induction motor works the way it does. I can go into the hand waving--but basically it ends up that the inductance of the field following the rotation of the armature reduced the inductance of the field coils--allowing the current to flow sort as if the field coil was a fairly low inductance coil/inductor. So the current lags the voltage by some 45 degrees or so (sine45=0.70).

    The interesting part of the model... Imagine two field coils and looking at the end of the motor. Top and Bottom coil produce an alternating North/South field going up and down--with no rotation. That can be thought of as one field rotating counter clockwise, and one field rotating clockwise--the some of the two rotating field vectors is a single UP/Down North/South field with zero rotation.

    Now, put an armature in there and rotate it clockwise near the frequency of the line (near 3,600 RPM.

    The counter clockwise field "sees" a clockwise rotating armature... The mount of magnetic coupling is very low--so that "direction" of counter clockwise field has high impedance (and little AC current).

    The clockwise field however is turning very close to the speed of the armature--so it has very strong magnetic coupling with the armature (actually the magnetic field in the iron armature that is generated by the rotating field "slip" -- basically turning the armature into a north/south magnet to be pulled by the field coils). The rotating armature also create a back EMF (Electro Magnetic Force) that is generated electricity that prevents too much current from flowing and overheating the motor.

    So (if I explained it correctly) that is how you get a rotating field in a single phase induction motor.

    To start a single phase motor, you need a second set of fields offset from the first. The idea is you need to create a rotating vector from a single phase of 120 VAC at 60 Hz. To to this, is to insert a capacitor into the second field which offsets the current flow to the winding--now you get a, sort of, rotating field to provide starting torque. The capacitor can either be put in temporarily (start cap with some sort of switch) or permanently (run cap). There are other ways of doing this too (i.e., shaded pole motors). They all have there advantages and disadvantages.

    You can even "start" an AC Induction motor with a blown cap or bad centrifugal switch (for motor start cap type) by giving the motor a spin and then applying AC power (I used to do this with a pull cord like a lawnmower gas engine for a rotary phase converter before making a more sophisticated installation).

    Now that electronics are so "cheap" and flexible--they can now take other styles of motors and run them with electronics that generate the multiphase wave forms (i.e., a three phase inverter) and variable frequency required to make for much more efficient motors (such as those than are now used in many of the very efficient refrigeration/heat pumps today).

    -Bill

    If you are studying Power Engineering (4 year degree)--they go into this type of stuff very deeply--enough to make your (or at least my) eyes cross... The math and the physical really fit together nicely.
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • SolarSailor
    SolarSailor Solar Expert Posts: 49
    Re: Sizing Generator

    ....I just wanted to plug in my PlayStation while waiting on the laundry to finish.:confused:

    Now, I don't know if I'm washing or hanging out.