# Need help with Off-Grid Batteries, Inverters and Panels

AmazonWiz
Registered Users Posts:

**12**✭✭
Hello.

I am currently also working on an Off-Grid project in Riyadh, Saudi Arabia. The facility currently, completely runs on a diesel generator. The average power of the connected loads is 6580 W.

The facility needs 24/7 operation. Therefore, the Watt-Hour per day would be 6580 x 24 = 157920 WH.

Should the solar array size be calculated as follows:

157920 * 1/0.77 * 1/6.92 sunny hours = 29,637 W solar array size.

Therefore the number of panels (assuming we use 280W panels) would be 29,637/280 = 106 panels.

Now whats confusing me is how to choose batteries and inverters.

For batteries what I did is follows:

Now the inverter chosen is one 30kva inverter. I believe that should be sufficient, with 23 panels connected in series and 5 parallel strings.

Does this look fine ? Which charge controller should I use ?

I am currently also working on an Off-Grid project in Riyadh, Saudi Arabia. The facility currently, completely runs on a diesel generator. The average power of the connected loads is 6580 W.

The facility needs 24/7 operation. Therefore, the Watt-Hour per day would be 6580 x 24 = 157920 WH.

Should the solar array size be calculated as follows:

157920 * 1/0.77 * 1/6.92 sunny hours = 29,637 W solar array size.

Therefore the number of panels (assuming we use 280W panels) would be 29,637/280 = 106 panels.

Now whats confusing me is how to choose batteries and inverters.

For batteries what I did is follows:

Daily Energy Requirement from Battery | 157,920 | Watt-Hours |

Minimum Size of Battery | 3290 | Amp-Hours |

Days of Autonomy (DoA) | 2 | Days |

Adjusted Battery Capacity for DoA | 6580 | Amp-Hours |

Maximum Depth of Discharge (DoD) | 90 | Percent |

Adjusted Battery Capacity for DoD | 7311.111111 | Amp-Hours |

Coulombic efficiency of Battery (Ec) | 90 | Percent |

Adjusted battery size for Ec | 8123.45679 | Amp-Hours |

AH of one battery | 200 | Ampere Hours |

Total No. of batteries required | 41 |

Does this look fine ? Which charge controller should I use ?

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## Comments

30,942adminIs this a different system/installation than we have been discussing in the other thread?

For Off Grid/Backup Power systems, you do need to size the inverters+battery bank to supply surge power too.

OK, the basic math. I like to start with the Battery Bank sizing first--It is the "heart" of your power system. Assuming 2 days of autonomy and 50% maximum discharge (for longer battery life). Note, assuming 48 volt battery bank--But for larger systems, other (higher) battery bank voltages may be available/make sense (total battery cost is about the same P=V*I -- Higher V means lower I for same P):

- 6,580 Watt load * 24 hours per day = 157,920 Watt*Hours per day
- 157,920 WH per day * 1/0.85 AC inverter Eff * 1/48 volt battery bank * 2 days storage * 1/0.50 max discharge = 15,482 AH @ 48 volt battery bank

For example, using large fork lift batteries--A battery bank may look like (note, you will buy batteries "locally", not from US--Your choices will vary):Crown Industrial Battery - 24 Volts, 1560 Amp-hours

2 batteries in series * 10 parallel strings = 20 batteries total (just an example--This is a very large battery bank)

Next, we size a battery bank with 5% to 13% or so solar charging... For a full time off grid system, 10% rate of charge minimum is highly recommended (and especially for industrial/fork lift type batteries).

- 15,482 AH battery bank * 59 volts charging * 1/0.77 solar panel+controller deratings * 0.10 minimum rate of charge = 118,628 Watt array minimum (based on battery bank capacity)

Note, you have a 6,580 Watt 24 hour x 7 day a week load--So, you need to add some solar panels to "carry the load" too while the battery bank is charging--So, the solar array needed to do this would be:- 6,580 Watt load * 1/0.77 panel+charge controller derating * 1/0.85 AC inverter eff = 10,053 Watt array to carry loads durign daytime charging
- 10,053 Watt array (loads) + 118,628 Watt for charging = 128,681 Watt array nominal (in my fist pass opinion)

There there is sizing the array based on loads and hours of sun per day...http://solarelectricityhandbook.com/solar-irradiance.html

## Average Solar Insolation figures

Measured in kWh/m2/day onto a solar panel set at a 65° angle from vertical:(For best year-round performance)

## Riyadh

Measured in kWh/m2/day onto a solar panel set at a 50° angle from vertical:Average Solar Insolation figures

(Optimal winter settings)- 157,920 WH per day * 1/0.52 typical off grid system eff * 1/4.75 hours of sun = 63.935 Watt array (December, break even on "average" weather)

So, based on suggested battery bank size, a 128,681 Watt array would be suggested.And based on load+sun, a 63.935 Watt array would be minimum.

You can talk with your battery supplier about their recommendations--But be aware that they may not have much in the way of solar power experience--They are familiar with a fork lift (discharge 8 hours during the day and recharge 8-16 hours every night). Solar does not give you the ability for the "perfect" charging cycle.

Also forklift batteries may not have much "extra water" capacity vs batteries designed for solar with more water above the plates. There are automatic watering systems (and fork lift batteries can use more distilled water)--So maintenance may be an issue too (how often is somebody on site, etc.).

Lots of questions--Hope this helps.

-Bill

12✭✭Yup, this is a different project that on the previous thread.

I would like to show you some of my calculations regarding battery sizing and array sizing.

So what I did was

6580 Watt load * 24 hours operation = 157,920 Watt-Hours per day.

I then sized the battery, assuming a 48V system Voltage.

I then want to size the panels, though unsure how to do that, since the calculations I read seem too confusing. What I did was:

However, 738 panels seems to be a lot. What do you think?

30,942adminFirst, I would disagree with this assumption about your battery:

Maximum Depth of Discharge
(DoD)
90
Percent

For Lead Acid deep cycle type Batteries--Generally you would never go below 20% State of Charge (80% Depth of Discharge). Lead Acid Batteries tend to have a 20% spread in Amp*Hour capacity from Mfg. variations, "unbalanced" cell, etc. Also, as the batteries age, they lose capacity--You would have little "spare" capacity as the batteries age (some battery vendors have used 20% loss of capacity as end of battery life).If you were thinking of LiFePO4 or some other battery type--Perhaps 90% DOD would be OK. I typically use 50% DOD as the planned discharge level for better battery life/unplanned issues.

- Adjusted battery size for Ec 8123.45679 Amp-Hours
- AH of one battery 200 Ampere Hours
- Total No. of batteries required 41

This appears to assume a 48 volt @ 200 AH battery--Can you double check this? Lead Acid Batteries tend to be in 2/4/6/8/12/24 volt size--And in the smaller AH capacity, 12 volt would seem to be the maximum. For example, your 41 batteries are 12 volt @ 200 AH, then you would have a string of 4 batteries in series and 41 parallel string--Giving you 164x 12 volt @ 200 AH batteries.In general, I would suggest 2/4/6 volt Lead Acid Cells/Batteries (cell is single 2 volt cell, multiple cells become a "battery"). I do not like a lot of parallel battery strings. But, prefer large AH cells (or battery assemblies) for various reasons. You need to see what is available from your local suppliers (or what you can reasonably import). Also, you need to figure out what size battery you can move around (forklift, crane, hard concrete floor, etc.). Larger batteries can easily be >1,000 kg in weight.

Regarding solar array--You appear to be using some sort of website (possibly) to do that calculations--I do not like to use this as I have to back out the underlaying math to figure out what they are doing/assumptions made/etc.

I cannot figure out exactly what your table is telling me and what information you may have supplied (panel wattage, Vmp, Imp, etc.).

In general, I like to work with paper numbers (like array wattage) first--Then after the basics of the paper design is accepted, then look at actual panel+charge controller+battery bank configuration options (ie., 200 or 300 watt panels, 60 cell, 72 cell, brand/model of charge controllers, etc.).

Take a look at my math--You can certainly make different assumptions (deratings, DoD for battery bank, etc.). But usually my basic math gets you close enough (say within 20% of final system design) to let you do some quick pricing/availability/paper configurations to see if the project makes sense as requested.

-Bill

33✭✭In your Panel calculation:

"Next, we size a battery bank with 5% to 13% or so solar charging... For a full time off grid system, 10% rate of charge minimum is highly recommended (and especially for industrial/fork lift type batteries).

- 15,482
AH battery bank * 59 volts charging * 1/0.77 solar panel+controller
deratings * 0.10 minimum rate of charge = 118,628 Watt array minimum
(based on battery bank capacity)

Note, you have a 6,580 Watt 24 hour x 7 day a week load--So, you need to add some solar panels to "carry the load" too while the battery bank is charging--So, the solar array needed to do this would be:I would like to ask is that the figure of 128,681 Watt Array nominal, is it the total Watts required in a day by the system?

If it is the total watt array nominal required then that means we have to divide this by the number of sun hours available and panel watts to determine the amount of panels required?

If this is the correct way of estimating number of panels then what does this mean:

"Now the math. You have lots of winter sun (relatively) too--So the "worst" average month is 4.75 Hours of sun per day (December):

- 157,920
WH per day * 1/0.52 typical off grid system eff * 1/4.75 hours of sun =
63.935 Watt array (December, break even on "average" weather)

So, based on suggested battery bank size, a 128,681 Watt array would be suggested.And based on load+sun, a 63.935 Watt array would be minimum."

I am just highly confused right now between these calculations. It would be very kind if you could list down the entire calculation step by step with description so we can understand the entire process more accurately without confusion.

Also would like to thank you for being so helpful on this website it really means a lot!

30,942adminFor the battery bank, we have two things we need to satisfy... The first is the minimum rate of charge for the battery bank. For Flooded Cell Lead Acid batteries, 5% is really the minimum rate of charge that anyone should have--More or less, it is the minimum current needed to equalize the battery bank (5% of 20 hour rated battery bank capacity--Or 200 AH*5%=10 amps minimum). The minimum "stirs the electrolyte" (important for "tall battery cases") and keeps the pores of the battery plates "open" (clear of sulfated lead--which is an insulator). Plus make up for self discharge (older batteries and especially forklift/traction/industrial batteries will tend towards 2% rate of self discharge before batteries need to be replaced--Above 2% rate of self discharge, batteries can overheat/fail/even catch fire). Below that 5% rate of charge level, you really only have enough current to "float" the battery bank.

For systems that charge during the day and discharge at night--Then the current form the solar array is 100% available for charging the loads. For AmazonW's installation, he has daytime loads (fueling station) and 24 hour x 7 days a week loads. That current is drawn from the solar array directly and supplied to the loads--And (in theory) is never available to recharge the battery bank. So, the "base load" needs to be "added to the solar array" to ensure the minimum 5% (or 10%) rate of charge is available for recharging the batteries during day time.

The second set of calculations is simply the 24 hour load needs to be replaced the next day. So that is simply the load divided by the system efficiency divided by minimum hours of sun per day.

Then comes the "justifications and rules of thumbs"--As you saw, for a very sunny area (like Saudi Arabia)--You can get get an array that is "way over sized" for the loads, if you size the array for the minimum rate of charge.

For systems in the far north/poor weather locations, there the "driving factor" is simply enough array to recharge the battery bank (where you have 2-3 hours per day, or less, of minimum sun--And want to avoid genset/fuel costs).

After you have made this "rough sizing" battery bank and array sizing estimates based on rules of thumb (say 10% rate of charge, hours of sun vs loads, sizing battery bank for 2 days of "no sun", etc.)--You go back and review these assumptions.

As you see, much of the cost of a system is the battery bank itself. You can reduce the size of the battery bank if you live in a very sunny region and/or are willing to use a backup genset more often (bad weather, winter, during periods of heavy electrical demand, etc.). Also, you can go back and review your electrical loads/assumptions. Perhaps you can delay loads during bad weather (i.e., less irrigation pumping during cloudy weather, less fuel pumping on weekends). Perhaps you can search for more energy efficient loads, limiting peak usage, pump to cistern with pump+solar panels (less expensive system), and pump from cistern to point of use with smaller pressure pump (less battery power needed for pumping from cistern above ground vs deep well).

If you can reduce battery size, that reduces the 5% to 10%+ minimum charging current--Reducing the size of solar array needed. And, of course, if you can reduce/delay power usage, that reduces solar array size.

As you reduce battery bank size--Your hours of charging become critical too... A "small battery bank" (say 1 day and 50% discharge) system needs about all of the hours per day of available sun to recharge (5 hours of 10% rate of bulk charge+2-6 hours of absorb charge). You now need to think about a tracking system (mechanical tracker) or a "virtual tracker" (a somewhat oversized array with 1/2 of array pointing south east, other 1/2 pointing south west or so). For an RV--That may mean AGM or LiFePO4 batteries (higher charging current, less hours in Bulk/Absorb (larger array, more efficient charging, etc.).

Besides the above, having an "oversized" solar array gives you "extra power" for relatively low costs (solar panels are at historic lows for pricing). You have to do much less load/sun/battery management. Some people really enjoy daily/hourly system management--Others want an install and (to the extent possible) forget power system.

Lots of stuff to juggle here. The rules of thumbs are there to let us quickly size a paper system that will be "pretty much" guaranteed to work based on what was presented by the questioner.However, if the costs/size of the system exceed what the poster can justify--It gives a bases for a "cost vs benefits" system redesign. What is the owner willing to give up, replace with more efficient loads, replace with propane/other fuel for heating/cooking, what loads can be delayed/not used (cloudy weather, no irrigation during winter), etc.

-Bill

33✭✭30,942adminTake care,

-Bill

8✭✭I think 7KW inverter is enough with in-rush current supporting.

I am working @ jeddah in solar field.

[email protected]

1,547✭✭✭✭This thread is almost 2 years old. You might not get a reply from the OP. Suggest you start your own thread.

2.1 Kw Suntech 175 mono, Classic 200, Trace SW 4024 ( 15 years old but brand new out of sealed factory box Jan. 2015), Bogart Tri-metric, 700 ah @24 volt AGM battery bank. Plenty of Baja Sea of Cortez sunshine.