design

arslan

i want to install a solar system at my home due to high load shedding effect in my area. for this i need some expert consultation for designing and wiring it.

my load is 3x fans of 100w each and 3x CFL's of 24W each i want to run these appliance on solar in sun hours plus 6 hours of backup. if i am not wrong total watts for load comes to 375W and watthours comes to 6000Wh.

now i need following things to know
1. PV panels selection? as per my reserch my solar insolation is 5.8 and so i need 9 panels of 180W, is it correct? please guide me
2. what would be rating of charge controller for this?
3. what would be rating of inverter for this ?
4. what would be rating of batteries for 6 hours backup? and which batteries do i need?
5. how many panels will i connect in series and parrallel? i mean how will i get to know that how much current and voltage do i need for my controller.

dear admin, i am requesting to make me understand these with the load that i have mentioned. i shall be very thank full to you if i could get rid of darkness by installing solar at my home.
regards.

niel

Re: design

before everybody gets to the meat of the details i have to ask why not charge the batteries from the grid when the grid is up? you can still add solar to it if you wish, but i'd take advantage of when the grid is present.

BB.

Re: design

Welcome to the forum "arslan",

In general, the first step is measuring your loads and doing as much conservation as you can. In your case, if you can reduce the amount of energy used by the fans, that would be a big help. 6,000 Watt*Hours per day is not a small system--So the costs to build such a system will not be small--And remember that batteries will last somewhere around 3-8 years--And that adds more costs (electronics such as the AC inverter, charge controllers, etc. will have around ~10+ year life and need repair/replacement too).

Also, there are other choices you can make too. If you lose power on summer afternoons for 4-8 hours per day--You could just install a battery bank+AC charge controller+AC Inverter and just use the utility power to recharge your battery bank and skip the whole Solar Array--This is a very common solution in many parts of the world (especially if you do not have room to install the array or cannot justify the price of the solar panels).

Based on what you asked for, and using basic design rules to sketch out a system design, it would look something like this.

• 300 Watts of fans + 3x*24 watts of lighting = 372 Watt average load
• 372 watt load * 6 hours per day = 2,232 Watt*Hours per day

My first question is do you have other loads that add up to 6,000 WH per day, or are you making some other "adjustments" to your AC power loads? I will go ahead and use 6,000 WH as your loads for now--If you need less, just plug in the numbers for a smaller system.

First the battery bank. Normally we suggest 1-3 days of backup power and 50% maximum discharge. An off grid system, I would start with 2 days of backup and 50% maximum discharge or 4x your daily load (this would be assuming you lose AC power for days or weeks at a time). If you mostly just lose power in the afternoon/evenings, you could go with 1 day of storage and 50% maximum discharge--This keeps the battery costs/size lower (you may have to replace batteries more often--But the operational costs will be similar--I.e., replace 4 batteries every 3 years vs replacing 8 batteries every 8 years).

• 6,000 WH * 1/0.85 AC inverter eff * 2 days of storage * 1/0.50 maximum discharge * 1/48 volt battery bank = 588 AH @ 48 volt battery bank

Next, calculate the size of the solar array--I do two calculations. One based on the size of the battery bank and minimum charging current (larger battery bank needs larger solar array), and the second based on the "hours of sun" you get per day.

First, size of solar array based on battery bank. Nominal rates of charge for a solar power battery system are around 5% to 13% per day. Usually 10% or more is the best solution for a system that is cycled daily.

• 588 AH * 59 volts charging * 1/0.77 panel+controller deratings * 0.05 rate of charge = 2,253 Watt array minimum
• 588 AH * 59 volts charging * 1/0.77 panel+controller deratings * 0.10 rate of charge = 4,505 Watt array nominal
• 588 AH * 59 volts charging * 1/0.77 panel+controller deratings * 0.13 rate of charge = 5,857 Watt array "cost effective maximum"

Second is sizing the array on the amount of sun you get per day... Using the Solar Electricity Handbook for Islamabad, with fixed array titled 56 degrees from vertical, facing south, we get:

(For best year-round performance)

Jan
Feb
Mar
Apr
May
Jun


4.81

5.06

5.69

6.35

6.63

6.56



Jul
Aug
Sep
Oct
Nov
Dec


5.77

5.52

6.25

6.76

5.91

4.79



Your region appears to have lots of sun--But we don't want to use "average numbers", but the minimum amount for when you plan on using the system, plus you need some extra power to allow for variations in daily sun (you normally would only plan on using ~66-75% of your system "predicted output" per day unless you plan on making of the difference with AC Grid or Backup Generator. Lets use 5 hours of sun minimum per day:

• 6,000 WH * 1/0.52 system AC end to end efficiency * 1/5.0 hours of sun per day = 2,308 Watt array minimum

Notice that I picked a 48 volt battery bank--That is to keep your DC copper wiring "reasonable" in size and keep the solar charge controller costs down. A 588 AH battery bank would need ~10% rate of charge or ~58.8 amps maximum charging current. Many solar charge controller have an output limit of around 60-80 amps--If you used a 24 volt battery bank, you would would have a 1,176 AH battery bank and ~118 amp charging current--And would need 2 charge controllers instead (another $700 or so in US).

Your AC inverter would only need to supply 372 watts continuously, and fans+lighting do not have a whole lot of starting surge (refrigerator needs a lot of surge current to start the compressor--5x rated current is a good guess). So, you should be very happy with a 800-1,200 watt AC inverter.

You also have the choice of MSW (modified square/sine wave) or TSW (true or pure sine wave) AC inverter. MSW inverters cost much less. TSW inverters have an output AC wave form that is the same as utilty power. A few appliances/electronics can have problems on MSW (overheating, buzzing noise, etc.). But MSW inverters cost much less.

Also, for 48 volt battery banks, finding "smaller" AC inverters can be difficult. Usually 48 volt AC inverters are used for larger AC power loads.

If your daily power usage is closer to 2,232 WH per day--All the above numbers are 1/2.7 smaller--And you could use a much smaller battery bank+solar array+charge controllers... And use a 24 volt battery bank (a 12 volt battery bank is still a bit low of voltage for such a large battery bank).

Anyway--That is how the math looks to me. Any questions/corrections to my guesses before we start talking about hardware?

-Bill

arslan

Re: design

thank you very much bill for your kind cooperation. thanks indeed.
bill to be frank, i didnt get all the details you told me for the systems, so lets go step by step.
i repeat my objecive, i.e i want my 3x fans of 100W each and 3x lights of 24W each Run on solar during sun hours, and after sun hours only 6 hours of back up.

please correct me if i am wrong with these calculations
fans = 100W x 3 x 18hours = 5400Wh (i have made 18 hours as 12hours of sun hours at summer and 6hours for backup charging)
lights= 24W x 3 x 8hours = 576Wh
total Wh = 5976wh

now for this Wh at 5.8 solar insolation how much wattage of panels i need to make PV array?
arslan

BB.

Re: design

Ok--My mistake--Yes, 6,000 WH is what you are looking for.

And everything above applies to your 6 kWH per day worth of loads.

There is a secondary sizing issue (I did not want to get into it as I figured my first post was big enough to cause confusion)... But basically, the batteries "want" around 5% to 13% rate of charge (and many battery vendors recommend 10% or more).

If you have "loads during the day", we need to account for them too. Say we have a 588 AH battery bank and you have:

• 372 Watts * 1/0.85 AC inverter eff * 1/48 volt nominal battery voltage = 9.1 amp 48 VDC AC inverter input current

And if we choose 10% of 588 AH battery capacity as our "charging current", we would have 58.8 average charging current---But since you have a day time load of 9.1 amps DC (at 48 volts), then the total solar charging would need to support Battery Charging + AC Loads:

• 58.8a + 9.1a = 68.9 Amps solar array current
• 67.9amps * 59 volts charging * 1/0.77 panel+charger deratings = 5,203 Watt Array

As you can see, at 10% of battery bank capacity, the extra ~9 amp DC load is not really a big amount... If you were to charge at 5% rate of charge (588*0.05=30 amps), then the 9 amp load is very significant and needs to be accounted for (battery would only get 21 amps of charging or ~3.6% rate of charge).

Lets look at a "minimum" sized hardware system. Assume that you only need the fans when it is a sunny day and that you only lose power on a warm sunny day (when you need the fans). So the solar array supplies Fan+Light power during daylight and the batteries supply power after sunset (6 hours). And that you will will want 2 "nights" of backup battery power for fans+lights (and never discharge the battery by more than 1/2 -- for longer battery life)...

Battery would be:

• 372 Watts*6 hours * 2 days of storage * 1/0.50 maximum discharge * 1/0.85 inverter efficiency * 1/24 volt battery bank = 438 AmpHour @ 24 volt (a much smaller battery bank allows lower DC bus voltage)

The size of solar array based on power used per day is still your 6 kWH per day. However, only part of the power is used from the battery bank (flooded cell batteries are about 80% efficient)

• 372 Watt * 6 hours * 1/0.52 system efficiency = 4,292 Watt*Hours of "raw solar power" for nigh-time loads
• 6,000 WH total AC load - (372 Watts * Hours night-time loads) = 3,768 Watt*Hours "day time" Loads from solar array
• 3,768 WH * 1/0.85 inverter eff * 1/0.77 charge+panel deratings = 5,758 Watt*Hours "day time" loads driving by solar array
• (3,768W+5,758W daily loads after all losses) * 1/5.8 hours of sun = 1,642 Watt "break even" solar array based on loads

And double check that we meet our 5% to 13% recommended rate of charge:

• 438 AH * 29 volts charging * 0.05 rate of charge = 635 Watt minimum array
• 438 AH * 29 volts charging * 0.05 rate of charge = 1,270 Watt nominal array
• 438 AH * 29 volts charging * 0.05 rate of charge = 1,651 Watt "cost effective maximum" array

But we need to account for the AC loads during the day:

• 372 Watts * 1/0.85 AC inverter eff * 1/0.77 panel+controller deratings = 568 Watts "day time" solar panel power

Add the two together and we get:

• 635 watt + 568 Watt = 1,200 Watt array for day+night loads minimum
• 1,270 Watt + 568 Watt = 1,838 Watt array for D+N loads nominal
• 1,651 Watt + 568 Watt = 2,219 Watt array for D+N loads cost effective maximum

Based on what you asked me and some simple division of day + night load usage, your battery bank would be 438 AH @ 24 volts (approximately 8x "golf cart" 6 volt @ 220 AH sized batteries--4 in series for 24 volts and two parallel strings for ~440 AH).

And your solar array would range from a minimum of 1,642 Watts (to support your 6kWH per day+night loads at 5.8 hours of sun per day) to ~1,838 watt for a "nominal rate of charge" for battery bank to ~2,219 Watt Array for "cost effective" maximum charging (plus loads) battery bank support.

And just to be clear--When we talk about battery AH capacity (amp*hours)--We are talking about the "20 Hour Discharge Rate" (i.e., you can draw 440 Amps for 20 hours until the battery is dead). There are other discharge rates--But this 20 hour rate usually works best for estimating off grid solar power system usage (discharging a battery "faster" than 20 hour rate makes the battery "appear" smaller because of higher internal battery losses at higher discharge current).

Please feel free to ask point by point questions--I wrote a lot here and I did not want to add detailed descriptions to each step--It would make the post 3x larger.

I did use a lot of "magic" derating numbers here---They are fairly conservative and found to give reliable results for sizing your system. We could play with them to get them more exact for your needs (perhaps you have a 90% efficient AC inverter instead of 85%, etc.)--But this is "close enough" for the first paper designs. Anything within 10% of the results is pretty much in the margin of error for solar (i.e., a 440 AH vs a 464 AH battery bank is pretty much within the margin of error).

-Bill

niel

Re: design

arslan wrote: »

thank you very much bill for your kind cooperation. thanks indeed.
bill to be frank, i didnt get all the details you told me for the systems, so lets go step by step.
i repeat my objecive, i.e i want my 3x fans of 100W each and 3x lights of 24W each Run on solar during sun hours, and after sun hours only 6 hours of back up.

please correct me if i am wrong with these calculations
fans = 100W x 3 x 18hours = 5400Wh (i have made 18 hours as 12hours of sun hours at summer and 6hours for backup charging)
lights= 24W x 3 x 8hours = 576Wh
total Wh = 5976wh

now for this Wh at 5.8 solar insolation how much wattage of panels i need to make PV array?
arslan

i can respect that you want to do this with solar, but there's no need to diss me to the point of being rude and you didn't answer the question as it is a legit question. this is a large endeavor for 3 lights and 3 fans. a thorough understanding on our parts does help you to better fulfill your goals.

SolInvictus

Re: design

Within an hour of sunrise and sunset, the PV array probably will not output enough power for your loads due to the direction of the sun and shading. Therefore, 8 hours running from the batteries and 10 hours running from the PV panels is a safer assumption.

Given:
PV daily capacity factor: 5.8 hours of sun
load: 372 W

{[(372 W * 10 hr) + ((372 W * 8 hr) / .75 (efficiency of batteries))] / .75 (efficiency of inverter, charge controller & wiring)} / 5.8 hr = 1,767 rated watt PV array minimum

Because this minimum size is not enough to deal with even partial cloudiness, you should choose more power for the PV array.

Assuming a 24 V wet lead-acid battery array is discharged to 75% state of charge each day, (372 W * 8 hr) / (1 - .75) / 24 V = 496 Ah battery array at the 20 hour rate of discharge.

arslan

Re: design

thank you very much for guidance.
i have another question could you please make me understand that.
the thing is i have 2HP electric submersible water pump and i want to run only 2 hours a day on solar.
let say if i use 7 panels of 150W and 3Kw inverter to drive this pump only 2 hour in day time (NO BACKUP).
now i am confused that in solar time which 2 hours i will drive the pump? as i designed the system only for two hours.
regards
arslan

BB.

Re: design

First, I would look at the 2 hp pump--that is a large pump for home use, but for industrial/agricultural use, may make sense (depends on how deep the well is, how high a pressure at the top of the well to point of use, volume, etc.).

So, first thing to check is the sizing/type of motor for your needs (and the pump). If you can use a smaller/more efficient pump, you win because of less power used and smaller solar power system.

Second--When to pump... many people will run such loads after the battery bank is fully charged. And use the "excess" solar power to run their pumps.

You could also put "2 hp" worth of extra panels on your array (plus charge controller) and simply just pump when the sun is up.

Lastly, there are other pumping alternatives. There are smaller pumps that are designed to run directly from solar panels. No batteries, no inverters, no maintenance (battery charging, watering, replacement) required. These are not usually cheap pumps, but overall can be significantly cheaper to install and run vs full off grid solar power.

And there are other options too just hitting the market. You change to a three phase motor and run a VFD (Variable Frequency Drive). You have to find a VFD designed (or can be converted) to run from solar panels and the VFD will generate the 3 phase power needed to run the pump (and you can adjust the frequency to run the pump slower/faster when needed).

BB. wrote: »

Some discussions about VFD (Variable Frequency Drives)... Basically a variable frequency inverter with (typically) three phase output. Used to soft start motors (handy for 3 phase well pumps, or pumps with well head starting capacitor) and can also turn an AC motor into a variable speed motor (very handy for pumping applications).

WELL PUMP and Inverter QUESTION
Wind/solar for large scale pumping etc (out of my depth!)
could use knowledge - using Gould jet pump - transfering from 230vAC to ? DC (new link/thread 10/27/2012)
Help required to design off grid system (information on possibilities to connect "standard VFDs direct to solar panels) (new link 1/13/2013)

May may not makes sense to use a solar+VFD for your situation--But certainly worth looking at if water pumping is a major issue for you.

-Bill

SolInvictus

Re: design

2 hp is 1491 W and starting power could be 5 to 10 times the operating power. I am not sure a 3,000 W inverter with a 6,000 W surge could start it. Seven 150 W PV panels with direct sunlight would provide 7 * 150 W * .75 = 788 W which means batteries must source the remainder. Start this pump around 11 am on a sunny day when you have maximum power from the PV panels and the batteries are fully charged. The battery array would need to be 48 V and the capacity high (how high?) to handle the starting power. As BB asks, can you use a water pump that uses less power?