Need Assistance on Electrical/Water system plan

mtang45mtang45 Registered Users Posts: 8
Hi folks!

New here but have been reading some very informative threads for awhile. I am planning a simple second home that will be completely off grid for water and electricity. What I am trying to determine is the "total system" and don't know how to go about figuring what I need.

The drilling company tells me that I will need to go about 450' deep for my water source and I would need around 8 gpm.

The house electrical needs will encompass lights, small appliances and unfortunately AC. I am leaning toward a wind turbine for power as I have a hilltop location that gets a pretty constant breeze. What I am struggling with is;

1) What kind/size pump is needed to support 8 gpm from 450'
2) How much power needs to be generated to handle the house and pump.
3) What wind (or solar) setup would be necessary and would you go with 12, 24 or 48 volt?

Thanks!

Comments

  • CariboocootCariboocoot Banned Posts: 17,615 ✭✭
    Re: Need Assistance on Electrical/Water system plan

    Welcome to the forum.

    First things first is to size that water pump. 450' is pretty deep, and 8 GPM is pretty high flow. Your well installer should be able to determine the proper pump size. This is no doubt going to be something around 2 HP deep well submerged pump. Do yourself a favour and get one with "soft start" ability or the "3 wire" wiring (not including ground).

    If you want you could have a dedicated solar pump such as the Grundfos: panels drive the pump as per what power is available, filling a reservoir to give you a reserve. Lots of options here and I'm sure others will have ideas too.

    In any case you have to know something about your target power requirements in order to size a system properly. The more power you need, the better off you are with higher system Voltage. See this thread: http://forum.solar-electric.com/showthread.php?15989-Battery-System-Voltages-and-equivalent-power

    If you can, get a Kill-A-Watt meter and actually measure the power use of anything and everything you want to use. Trying to judge by the numbers given by manufacturers and guestimating usage time will not be very accurate.

    As for wind power, set up an anemometer and make sure you have the wind first. Often what seems like a lot to us is nothing to power needs. In general, turbines need really good installations; up high and out of the turbulence. Usually PV will return more power per $ spent than wind will, and you're better off thinking of the PV as primary power for charging and the wind as back-up.

    Best thing you can do is not try to jump to the finish before you know the track, or as Icarus says: "Avoid Ready, Fire, Aim!"
  • westbranchwestbranch Solar Expert Posts: 4,860 ✭✭✭✭
    Re: Need Assistance on Electrical/Water system plan

    First thing is to find out how much power in Kwh you use to live , right now.... Purchase one of these, http://www.p3international.com/products/special/p4400/p4400-ce.html, and measure everything you can in the house!!

    FYI Our minimalist new house projected consumption is for 3 Kwh to 5 Kwh per day excluding pump , NO AC with no furnace or HVAC either. Current house uses 16 Kwh per day...

    hth
     
    KID #51B  4s 140W to 24V 900Ah C&D AGM
    CL#29032 FW 2126/ 2073/ 2133 175A E-Panel WBjr, 3 x 4s 140W to 24V 900Ah C&D AGM 
    Cotek ST1500W 24V Inverter,OmniCharge 3024,
    2 x Cisco WRT54GL i/c DD-WRT Rtr & Bridge, Hughes1100 Sat Modem
    Eu3/2/1000i Gens, 1680W & E-Panel/WBjr to come, CL #647 asleep
    West Chilcotin, BC, Canada
  • CariboocootCariboocoot Banned Posts: 17,615 ✭✭
    Re: Need Assistance on Electrical/Water system plan
    westbranch wrote: »
    First thing is to find out how much power in Kwh you use to live , right now.... Purchase one of these, http://www.p3international.com/products/special/p4400/p4400-ce.html, and measure everything you can in the house!!

    FYI Our minimalist new house projected consumption is for 3 Kwh to 5 Kwh per day excluding pump , NO AC with no furnace or HVAC either. Current house uses 16 Kwh per day...

    hth

    Those numbers are eerily similar to our off-grid cabin vs. our on-grid house. Except we use even less at the cabin than your minimalist house.
  • vtmapsvtmaps Solar Expert Posts: 3,738 ✭✭✭✭
    Re: Need Assistance on Electrical/Water system plan
    mtang45 wrote: »
    What I am trying to determine is the "total system" and don't know how to go about figuring what I need.

    You cannot design a system without knowing your loads. You need to know your daily energy use (kilowattHours) and your peak power use at any one time (watts).

    Step 1: determine loads
    Step 2: choose your battery (most difficult step)
    Step 3: buy enough electronics and solar panels to keep batteries happy

    I cannot emphasize enough the importance of step 1. Its like buying a pickup truck... you must know your loads... do you want a one ton or a half ton truck? If you buy the wrong truck you cannot convert it to the one you should have bought.

    In your case the greatest peak power use (watts) will probably be the water pump. The greatest daily energy use (kwatthours) will probably be the A/C (around here 'AC' usually means alternating current).

    I'm guessing here... but I suspect that your system will be very large and expensive 48 volt system. Where is your system located? We need to know the climate and solar insolation to help design your system.

    By the way, why do you need 8 gallons per minute? That sounds like a lot for a 'simple second home'. How much water do you need per day? There are many ways to have that water without using such a large (2 hp) pump.

    --vtMaps
    4 X 235watt Samsung, Midnite ePanel, Outback VFX3524 FM60 & mate, 4 Interstate L16, trimetric, Honda eu2000i
  • mtang45mtang45 Registered Users Posts: 8
    Re: Need Assistance on Electrical/Water system plan

    Thanks for all the great advice, related threads and suggestions. Since the house is not built yet, I am speculating as to the power/water requirements. If 4 gpm water sounds more reasonable I'm good. "Cabin" will be approximately 1300 sqft with the biggest draw being an A/C unit to keep it cool in the summer plus whatever the requirement for the well pump. Location is Henry County Missouri which is west central part of the state. Summer peak temperatures and humidity make it very uncomfortable.

    So at a minimum, whats the best route to go to get electrical needs started with the above criteria, keeping in mind that expansion may be necessary in the future. I'm willing to tailor the scope of the house based on cost to get started.
  • CariboocootCariboocoot Banned Posts: 17,615 ✭✭
    Re: Need Assistance on Electrical/Water system plan

    The pump is going to be an issue. It is an unknown, and potentially a fairly sizable one. Not only do you need to know how much power is needed to start and run it, but also how long it will run for.

    The other loads you could at least approximate by measuring the consumption of similar items you use now. May not be perfect, but it's better than nowt. It may even identify some items and/or habits you want to change to make things better. ;)

    Expansion of systems is tricky. If you start too small you have to basically start over. If you start too big you spend money you don't need to. The biggest problems are changing system Voltage and availability of compatible components at a later date (the panels you buy today may not exist tomorrow). You can fall back on the "system expansion by duplication" trick, wherein when you need more power you simply build an entirely new system to handle the new demand. Sucks money like a Hoover, but sure to work. :roll:

    It never hurts to run some PV Watts simulations either: http://www.nrel.gov/rredc/pvwatts/grid.html
    Although designed for grid-tie, it will give you some idea of the kind of power you get at different times of year for your location. Keep in mind that off-grid power is only about 52% efficient over-all. And take a look at your existing power bills to get a preliminary idea of your usage.

    Without a target number, you'll be guessing wildly and spending money wrongly.
  • westbranchwestbranch Solar Expert Posts: 4,860 ✭✭✭✭
    Re: Need Assistance on Electrical/Water system plan

    we stated with our floor plan , refined it over several years, talking it out, a LOT, about as many details as we could think of and then put in windows, doors, lighting, appliances, no micro, heating, etc then measures what we have now , as a worst case, and plan on buying some new as we go... then built a spread sheet/work book with time of day use for every item on the list>>> total load estimate. then designed PV and inverter and charger, gen set to match, with expectation that some might have to be beefed up if not robust enough...

    its a long process!

    hth
     
    KID #51B  4s 140W to 24V 900Ah C&D AGM
    CL#29032 FW 2126/ 2073/ 2133 175A E-Panel WBjr, 3 x 4s 140W to 24V 900Ah C&D AGM 
    Cotek ST1500W 24V Inverter,OmniCharge 3024,
    2 x Cisco WRT54GL i/c DD-WRT Rtr & Bridge, Hughes1100 Sat Modem
    Eu3/2/1000i Gens, 1680W & E-Panel/WBjr to come, CL #647 asleep
    West Chilcotin, BC, Canada
  • solarvicsolarvic Solar Expert Posts: 1,043 ✭✭✭
    Re: Need Assistance on Electrical/Water system plan

    You might not need to pump water from 450 ft if the static water level in the well is a lot higher than 450 feet. Probably need to see what the GPM output is and how mich water you can pump without lowering the static water level.
  • BB.BB. Super Moderators Posts: 26,837 admin
    Re: Need Assistance on Electrical/Water system plan

    100 kWH per month/3.3 kWH per day is probably about the minimum you would want to run your home one (deep well is going to add to your power requirements).

    200-300 kWH per month is probably where you will end up if it is more than just that two of you (round guess work--At least you can start sizing a system and check out the costs).

    The other way, start a small system (1-3.3 kWH per day)... Just enough to run some lights, radio, computer, TV, pressurize the home/temp camp water system, perhaps a fridge (3.3 kWH per day system). And use a moderately sized genset for day time/evening power (it would be great if you could use a Honda eu2000i 1.6 kWH 120 VAC genset) and measure those loads (including the genset). (gas and generator noise is going to be your "cost" to figure out your power requirements).

    And use a larger genset (as needed) to run the power tools, table saws, big well pump to fill a cistern (may be a good idea for fire suppression--if an issue in your area, a requirement in some of the southwest areas of the country), etc.

    Once you have your loads measured, you can retire the small system (or move it to a granny/guest flat) and install a system that meets your (now better known) load requirements better.

    For an example system... 3.3 kWH per day or 100 kWH per month, using PV Watts for Columbia (you can use Kansas City if closer to your weather conditions), fixed array, 0.52 system derating (off grid), and tilted to latitude 2kW array (just for discussion):
    "Station Identification"
    "City:","Columbia"
    "State:","Missouri"
    "Lat (deg N):", 38.82
    "Long (deg W):", 92.22
    "Elev (m): ", 270
    "PV System Specifications"
    "DC Rating:"," 2.0 kW"
    "DC to AC Derate Factor:"," 0.520"
    "AC Rating:"," 1.0 kW"
    "Array Type: Fixed Tilt"
    "Array Tilt:"," 38.8"
    "Array Azimuth:","180.0"

    "Energy Specifications"
    "Cost of Electricity:"," 7.0 cents/kWh"

    "Results"
    "Month", "Solar Radiation (kWh/m^2/day)", "AC Energy (kWh)", "Energy Value ($)"
    1, 3.69, 121, 8.47
    2, 4.73, 138, 9.66
    3, 5.10, 157, 10.99
    4, 5.71, 165, 11.55
    5, 5.58, 159, 11.13
    6, 6.04, 164, 11.48
    7, 5.97, 164, 11.48
    8, 6.03, 166, 11.62
    9, 5.47, 152, 10.64
    10, 5.08, 152, 10.64
    11, 3.56, 105, 7.35
    12, 3.23, 103, 7.21
    "Year", 5.02, 1745, 122.15

    So--during November thru January, you will be around 103-121 kWH per month--Probably need a generator during stretches of bad weather.

    Battery wise... 1-3 days of no sun, 50% maximum discharge for longer battery life. A 2 day + 50% load at 3.3 kWH per day:
    • 3,300 kWH * 1/0.85 inverter losses * 1/24 volt battery bus * 2 days no sun * 1/0.50 max discharge = 647 AH @ 24 volt battery bank

    To double check, a battery bank should be charged (very rough rule of thumb) around 5% to 13% rate of charge:
    • 647 AH * 29 volts charging * 1/0.77 panel+controller losses * 0.05 rate of charge = 1,218 watt array minimum
    • 647 AH * 29 volts charging * 1/0.77 panel+controller losses * 0.10 rate of charge = 2,437 watt array nominal
    • 647 AH * 29 volts charging * 1/0.77 panel+controller losses * 0.13 rate of charge = 3,168 watt array "cost effective maximum"

    So a 2,000 Watt to 2,437 Watt array would be a nice fit with a 3.3 kWH per day minimum output and a 647 AH @ 24 volt battery bank (your load requirements may be different than my first guess--and excludes any AC usage).

    To get an idea of the battery bank "useful" output... Very roughly, a C/8 discharge rate is a good continuous maximum (don't over heat a flooded cell battery bank) and C/2.5 is a good number for maximum useful surge (starting a well pump, etc.):
    • 647 AH * 24 volts * 0.85 inverter eff * 0.85 inverter eff * 1/8 rate of discharge= 1,650 watts nominal max continuous rated AC power
    • 647 AH * 24 volts * 0.85 inverter eff * 0.85 inverter eff * 1/2.5 rate of discharge= 5,280 watts max recommended AC surge power

    So, your maximum "useful" rated inverter would be around 1.7 kW to roughly 2.5 kW (good inverters can surge about 2x rated output). Getting a 4kW or larger inverter may be a bit much for this size system. Note that even a 1,650 Watt AC load (like a hair drier, microwave oven, well pump with a failed shutoff float switch) would drain your battery bank in less than 8 hours--So, you have to be very careful about your loads to make sure that nothing bad happens to your battery bank.

    Note the above numbers are conservative (AGM batteries can supply much higher discharge rates). And new/fully charged batteries will do better than 1/2 charged, multiple year older batteries (again, being conservative). Also, the above numbers are not "that accurate". I am just carrying out digits so you can repeat my math, avoid round off error, and trace when I use a calculation result in the next equation.

    Choices like battery bank voltage (12/24/48 volts) can be fudged a bit (just picked 24 volts as a mid-range number).

    Generator rating--A typical battery charger may have a 0.67 power factor (a whole set of threads by itself), and an 80% efficiency. Assuming a rate of charge from 5% to 25% (max not to exceed rule of thumb) would look like:
    • 647 AH * 0.05 rate of charge = 32 amp minimum charger
    • 647 AH * 29 volts charging * 1/0.80 charger eff * 1/0.67 PF * 0.05 rate of charge = 1,750 VA rating min rating for ~32 amp @ 24 volt charger
    • 647 AH * 0.10 rate of charge = 65 amp nominal charger
    • 647 AH * 29 volts charging * 1/0.80 charger eff * 1/0.67 PF * 0.10 rate of charge = 3,501 VA rating min rating for ~65 amp charger
    • 647 AH * 0.13 rate of charge = 84 amp healthy sized charger
    • 647 AH * 29 volts charging * 1/0.80 charger eff * 1/0.67 PF * 0.13 rate of charge = 4,551 VA rating min rating for ~84 amp charger
    • 647 AH * 0.25 rate of charge = 162 amp not to exceed charger
    • 647 AH * 29 volts charging * 1/0.80 charger eff * 1/0.67 PF * 0.25 rate of charge = 8,751 VA rating min rating for ~162 amp charger

    Note--Our rules for thumb for system design are very much designed around "keeping the battery bank happy"...

    If your power needs are not too large and you mostly use power at night, you may get away with a 5% rate of charge (as long as you meet the minimum solar array requirement for your daily loads/battery charging).

    If you power needs are heavy and you use power during the day--You are probably looking at a 10% or so rate of charge minimum.

    Anyway--None of the above is in stone--Just to give you some ideas of what you are looking at for system design, how load drive battery sizing. And battery sizing drives the rest of the system design.

    It is very difficult to "grow" a solar PV system... Generally, past a certain point, you need to increase battery bank votlage to keep the overall size of wiring "reasonable" and other system costs down (like solar charge controllers). Remember power=voltage*current--so doubling the voltage will cut the current by 1/2 (and so on).

    At this level, looking at the daily/monthly power levels--If you need 3x the amount of power, simply multiple the battery bank/solar array/etc. by 3x (you will probably need to up the battery bank voltage to 48 volts for that large of system).

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • mtang45mtang45 Registered Users Posts: 8
    Re: Need Assistance on Electrical/Water system plan
    BB. wrote: »
    100 kWH per month/3.3 kWH per day is probably about the minimum you would want to run your home one (deep well is going to add to your power requirements).

    200-300 kWH per month is probably where you will end up if it is more than just that two of you (round guess work--At least you can start sizing a system and check out the costs).

    The other way, start a small system (1-3.3 kWH per day)... Just enough to run some lights, radio, computer, TV, pressurize the home/temp camp water system, perhaps a fridge (3.3 kWH per day system). And use a moderately sized genset for day time/evening power (it would be great if you could use a Honda eu2000i 1.6 kWH 120 VAC genset) and measure those loads (including the genset). (gas and generator noise is going to be your "cost" to figure out your power requirements).

    And use a larger genset (as needed) to run the power tools, table saws, big well pump to fill a cistern (may be a good idea for fire suppression--if an issue in your area, a requirement in some of the southwest areas of the country), etc.

    Once you have your loads measured, you can retire the small system (or move it to a granny/guest flat) and install a system that meets your (now better known) load requirements better.

    For an example system... 3.3 kWH per day or 100 kWH per month, using PV Watts for Columbia (you can use Kansas City if closer to your weather conditions), fixed array, 0.52 system derating (off grid), and tilted to latitude 2kW array (just for discussion):



    So--during November thru January, you will be around 103-121 kWH per month--Probably need a generator during stretches of bad weather.

    Battery wise... 1-3 days of no sun, 50% maximum discharge for longer battery life. A 2 day + 50% load at 3.3 kWH per day:
    • 3,300 kWH * 1/0.85 inverter losses * 1/24 volt battery bus * 2 days no sun * 1/0.50 max discharge = 647 AH @ 24 volt battery bank

    To double check, a battery bank should be charged (very rough rule of thumb) around 5% to 13% rate of charge:
    • 647 AH * 29 volts charging * 1/0.77 panel+controller losses * 0.05 rate of charge = 1,218 watt array minimum
    • 647 AH * 29 volts charging * 1/0.77 panel+controller losses * 0.10 rate of charge = 2,437 watt array nominal
    • 647 AH * 29 volts charging * 1/0.77 panel+controller losses * 0.13 rate of charge = 3,168 watt array "cost effective maximum"

    So a 2,000 Watt to 2,437 Watt array would be a nice fit with a 3.3 kWH per day minimum output and a 647 AH @ 24 volt battery bank (your load requirements may be different than my first guess--and excludes any AC usage).

    To get an idea of the battery bank "useful" output... Very roughly, a C/8 discharge rate is a good continuous maximum (don't over heat a flooded cell battery bank) and C/2.5 is a good number for maximum useful surge (starting a well pump, etc.):
    • 647 AH * 24 volts * 0.85 inverter eff * 0.85 inverter eff * 1/8 rate of discharge= 1,650 watts nominal max continuous rated AC power
    • 647 AH * 24 volts * 0.85 inverter eff * 0.85 inverter eff * 1/2.5 rate of discharge= 5,280 watts max recommended AC surge power

    So, your maximum "useful" rated inverter would be around 1.7 kW to roughly 2.5 kW (good inverters can surge about 2x rated output). Getting a 4kW or larger inverter may be a bit much for this size system. Note that even a 1,650 Watt AC load (like a hair drier, microwave oven, well pump with a failed shutoff float switch) would drain your battery bank in less than 8 hours--So, you have to be very careful about your loads to make sure that nothing bad happens to your battery bank.

    Note the above numbers are conservative (AGM batteries can supply much higher discharge rates). And new/fully charged batteries will do better than 1/2 charged, multiple year older batteries (again, being conservative). Also, the above numbers are not "that accurate". I am just carrying out digits so you can repeat my math, avoid round off error, and trace when I use a calculation result in the next equation.

    Choices like battery bank voltage (12/24/48 volts) can be fudged a bit (just picked 24 volts as a mid-range number).

    Generator rating--A typical battery charger may have a 0.67 power factor (a whole set of threads by itself), and an 80% efficiency. Assuming a rate of charge from 5% to 25% (max not to exceed rule of thumb) would look like:
    • 647 AH * 0.05 rate of charge = 32 amp minimum charger
    • 647 AH * 29 volts charging * 1/0.80 charger eff * 1/0.67 PF * 0.05 rate of charge = 1,750 VA rating min rating for ~32 amp @ 24 volt charger
    • 647 AH * 0.10 rate of charge = 65 amp nominal charger
    • 647 AH * 29 volts charging * 1/0.80 charger eff * 1/0.67 PF * 0.10 rate of charge = 3,501 VA rating min rating for ~65 amp charger
    • 647 AH * 0.13 rate of charge = 84 amp healthy sized charger
    • 647 AH * 29 volts charging * 1/0.80 charger eff * 1/0.67 PF * 0.13 rate of charge = 4,551 VA rating min rating for ~84 amp charger
    • 647 AH * 0.25 rate of charge = 162 amp not to exceed charger
    • 647 AH * 29 volts charging * 1/0.80 charger eff * 1/0.67 PF * 0.25 rate of charge = 8,751 VA rating min rating for ~162 amp charger

    Note--Our rules for thumb for system design are very much designed around "keeping the battery bank happy"...

    If your power needs are not too large and you mostly use power at night, you may get away with a 5% rate of charge (as long as you meet the minimum solar array requirement for your daily loads/battery charging).

    If you power needs are heavy and you use power during the day--You are probably looking at a 10% or so rate of charge minimum.

    Anyway--None of the above is in stone--Just to give you some ideas of what you are looking at for system design, how load drive battery sizing. And battery sizing drives the rest of the system design.

    It is very difficult to "grow" a solar PV system... Generally, past a certain point, you need to increase battery bank votlage to keep the overall size of wiring "reasonable" and other system costs down (like solar charge controllers). Remember power=voltage*current--so doubling the voltage will cut the current by 1/2 (and so on).

    At this level, looking at the daily/monthly power levels--If you need 3x the amount of power, simply multiple the battery bank/solar array/etc. by 3x (you will probably need to up the battery bank voltage to 48 volts for that large of system).

    -Bill

    Bill,

    Thank you for the very thought out explanation and examples; you definitely gave me a solid starting point. I have received another suggestion to put a standard mechanical windmill on the well and keep a tank filled. Then I could downsize my pump needs to just keeping the house pressurized, especially if I could put the tank at a higher elevation and let gravity do part of the job.

    Anyway thanks all for the great suggestions. I'm off to work on the plan and this weekend will be clearing trees and moving dirt with my dozer to prep the site. I'm sure I will be back seeking more great advice!!

    Everyone have a great Thanksgiving!

    Mark
  • vtmapsvtmaps Solar Expert Posts: 3,738 ✭✭✭✭
    Re: Need Assistance on Electrical/Water system plan
    mtang45 wrote: »
    I have received another suggestion to put a standard mechanical windmill on the well and keep a tank filled. Then I could downsize my pump needs to just keeping the house pressurized, especially if I could put the tank at a higher elevation and let gravity do part of the job.

    You could also achieve that end with a separate solar panel system directly connected to a DC submersible pump. No batteries or inverters needed. When the sun shines you fill the tank. That may be more reliable than using wind to fill the tank.

    --vtMaps
    4 X 235watt Samsung, Midnite ePanel, Outback VFX3524 FM60 & mate, 4 Interstate L16, trimetric, Honda eu2000i
  • TnAndyTnAndy Solar Expert Posts: 249 ✭✭✭✭✭
    Re: Need Assistance on Electrical/Water system plan
    BB. wrote: »
    100 kWH per month/3.3 kWH per day is probably about the minimum you would want to run your home one (deep well is going to add to your power requirements).

    200-300 kWH per day is probably where you will end up if it is more than just that two of you (round guess work--At least you can start sizing a system and check out the costs).


    -Bill


    Bill, I think you meant to say "per month", not "per day" in the above.

    I suggest with the pump and AC load, the load may well run in the 400-500kw/hr/month range in the summer.

    One way ( as previously suggested ) to reduce your water pumping load is to use a Grundfos SQ series pump ( that would DEFINITELY be the pump you ought to put in no matter what, as it can run off DC or AC ) with enough dedicated panel watts to run it at 1/2 to 1 gal/min into a large storage tank ( 1000-1500gallons.....about 50 cents/gallon tank price at Tractor Supply ), then use a small booster pump from the tank to the house supply.

    I have a friend I set up an off grid cabin for that basically does this ( he uses rain water collection, no well ) into a 2500 gallon plastic tank, and the booster pump with a small pressure tank on top uses very little power. He only uses the cabin on weekends, but a 1500w PV system with 800amp/hr battery bank does him fine ( No AC ).....batteries never dip below 10% discharge. He runs a full sized fridge, microwave, coffee make, lights and water pump.

    ry%3D400

    ry%3D480
  • BB.BB. Super Moderators Posts: 26,837 admin
    Re: Need Assistance on Electrical/Water system plan
    TnAndy wrote: »
    Bill, I think you meant to say "per month", not "per day" in the above.
    Thank you Andy... I will fix.

    -Bill :blush:
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • CariboocootCariboocoot Banned Posts: 17,615 ✭✭
    Re: Need Assistance on Electrical/Water system plan

    TnAndy;

    Nice set-up. :D
    But that battery bank looks like it's "laddered"; as though both the positive and negative connections to equipment are on the string closest to the wall. If so you need to move one of those wires forward to the other string for best results.

    If that's not how it is ignore me.
  • TnAndyTnAndy Solar Expert Posts: 249 ✭✭✭✭✭
    Re: Need Assistance on Electrical/Water system plan

    Not sure what you mean by laddered.....the front four are one string, the back four are the other string.
    The negative terminals connect together on the left side, and feed the lead to the inverter......the positive ends connect on the right side, and feed the longer, looped wire on the wall to the inverter.
  • CariboocootCariboocoot Banned Posts: 17,615 ✭✭
    Re: Need Assistance on Electrical/Water system plan
    TnAndy wrote: »
    Not sure what you mean by laddered.....the front four are one string, the back four are the other string.
    The negative terminals connect together on the left side, and feed the lead to the inverter......the positive ends connect on the right side, and feed the longer, looped wire on the wall to the inverter.

    Yes, but it appears the positive and negative wires are both attached to the back four. One of them should be on the front four; the "diagonal wiring" that helps keep current flow even. As per Smart Gauge method #2 http://www.smartgauge.co.uk/batt_con.html
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