Need help with series or parallel
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AmazonWiz
Registered Users Posts: 12 ✭✭
Hello.
I am looking at a solar grid tied project. The total AC load at the site is around 62250 W. According to my calculations, I have selected three 25KW inverters. I am however confused as to how to calculate how many panels should be connected in series or parallel. Any tips would be greatly appreciated.
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Welcome to the forum AmazonWiz.
I am not quite sure I understand your power needs... Is this an off grid or grid tied solar system (battery bank or just solar panels + GT inverter tied to utility power)?
Are you saying you have 62,250 Watt "average" or "peak" loadsOr is this the energy usage per day (Watt*Hours of energy used per day). I will do some quick mathBut with a system this large, you will need a professional engineer to size/design/help install the systemThis is a big installation.
First, you need to figure out the basics of your loads. Lets say you are looking at a 62,000 Watt average installed AC load. And it runs 10 hours per day at a 50% duty cycle (compressors are on 1/2 of the time). Then: 62,000 Watts * 10 hours per day * 0.50 duty cycle = 310,000 Watt*Hours per day
 310,000 WH per day * 1/0.85 AC inverter eff * 1/48 volt battery bank * 2 days storage * 0.50 maximum discharge = 30,392 AH @ 48 volt battery bank (this is a "huge" battery bank and you may run at higher than 48 volts for your installationBut the overall cost of the bank will be about the same as batteries are in different series/parallel combination)
 30,392 AH * 59 volt charging * 1/0.77 panel+controller deratings * 0.10 rate of charge = 232,873 nominal solar array (based on battery bank size).
http://solarelectricityhandbook.com/solarirradiance.htmlRiyadh
Measured in kWh/m2/day onto a solar panel set at a 65° angle from vertical:
Average Solar Insolation figures
(For best yearround performance)
Lets guess that your actual loads vary with the amount of sunSunny months need more cooling that less sunny months. Lets use ~6.00 hours of sun for the hot months:Jan Feb Mar Apr May Jun 4.61
5.46
5.75
6.04
6.54
6.92
Jul Aug Sep Oct Nov Dec 6.79
6.76
6.59
6.36
5.12
4.43
 310,000 WH per day * 1/0.52 average off grid system eff * 1/6.00 hours of sun per day = 99,359 Watt array minimum (based on loads and sun)...
And, if you have utility power and want to run a Grid Tied solar system, the size of the array would be roughly (to "break even" with your utility power usage): 310,000 WH per day * 1/0.77 GT system efficiency * 1/6.92 hours of sun for peak AC month = 58,179 Watt solar array minimum
Bill
Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset 
Thank you Bill for the math. I do have a lot of questions.
The system shall be an on grid system, without any batteries. The total load at the site is 62250 W. We plan to use 290 panels, each of 280W. For this, we plan to use three 25kva inverters. Now what I'm confused at is how do I design the system to see how many panels should be connected in series and how many in parallel. 
Remember your AC / Utility Mains run the AC loads (the size of the solar array and gt inverters do not affect how the loads will work)... The solar panels + GT inverter will essentially "recharge" the AC Grid. During the middle of the day you will be pumping energy back to the utility (more solar power than loads)... At morning/evening/night you will be drawing energy from the utility.
SoYour average loads * hours per day = Watt*HOurs per day (usually kWH per day is what your utility meter measures1kWH=1,000WH).
What I do not understand just yet is your 62250 Watt a "load" (average power draw) or is it energy used per day (62250 WH per day or 6225W*10hours per day = 62250 WH per day)...
So, the calculation would look like:
Watt*Hours per day * 1/0.77 GT system efficiency * 1/x.x hours of noon time sun energy per day = Nominal solar array rating
Sothe load math looks like:
100 watt load * 1 hour per day = 100 WH per day
100 watt load * 8 hours per day = 800 WH per day
100 watt load * 24 hours per day = 2400 WH per day
Your sizing of the array to the GT invertersYou have the rough idea of XXXX Watts of solar panel = 2x 25,000 Watt GT inverters is correctBut way you picked 290 panels * 280 Watts per panelI do not have a good idea.
Series/parallel panels (wattage, current, Vmparray series votlage, etc.) are all based on the GT inverter you pick. What is it input requirements?
Bill
Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset 
Thank you Bill. That cleared up many confusions that I was having.
A question though. If I want to power up an air conditioner using solar. Do I need to consider surge power/in rush current of the air conditioner ?
Also, the 62250 W is the average power load. 
You will do two different designs.
First, ignore solar, size the utility drop, the main panel, branch circuits to power each device (lights, AC, elevator, etc.).
The utility power runs everything.
Next, you figure out how many Watt*Hours per day you want to offset. Seasons and even your utility billing plan will affect your calculations and decisions (more ac in summer, more sun available in summer, etc.)
Next you design your solar power system. Size the array to "turn your meter backwards during the 6 sunny hours part of the day". And your meter runs forward the other 18 hours of the day.
Note, your main panel/peak load on your wiring system may be the generation by your solar power system (utility drop, main panel have to be larger because your solar system may output 180,000 Watts [6 hours hours of solar production vs 24 hours for steady state load of 62,000 watts].
Learning the basics will be very helpful to you. However a system this big will probably require at least one power engineer to do the actual design and over see the installation.
I do not know near enough to help you make detailed design decisions.
BillNear San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset 
Thanks Bill. That was very helpful.
Just to make sure. The utility power would run everything? The solar power system would just essentially recharge the AC grid ?
I had thought that the solar power system would be in use during peak sunny hours, and if there's excess, it would recharge the grid. That is not the case ?
Also what if this was an off grid system ? 
Are you sure your average load is 60kW? A standard grid service here is only 15kW. 63Amp main fuse, 230V. In order to get anything larger you have to get three phase.
Are you sure you have you decimal place in the right place or mixing units, or do you have some kind of heavy industrial application?
1.8kWp CSUN, 10kWh AGM, Midnite Classic 150, Outback VFX3024E,
http://zoneblue.org/cms/page.php?view=offgridsolar 
60Kw is an industrial plant, and your plant engineer should coordinate with the power company on the design.
Powerfab top of pole PV mount  Listeroid 6/1 w/st5 gen head  XW6048 inverter/chgr  Iota 48V/15A charger  Morningstar 60A MPPT  48V, 800A NiFe Battery (in series) 15, Evergreen 205w "12V" PV array on pole  Midnight ePanel  Grundfos 10 SO59 with 3 wire Franklin Electric motor (1/2hp 240V 1ph ) on a timer for 3 hr noontime run  Runs off PV 
 Midnight Classic 200  10, Evergreen 200w in a 160VOC array 
 VEC1093 12V Charger  Maha C401 aa/aaa Charger  SureSine  Sunsaver MPPT 15A
solar: http://tinyurl.com/LMRSolar
gen: http://tinyurl.com/LMRLister , 
The facility is an accommodation. The following loads are to be connected:
Airconditioner 13 units .... 3000W / unit
Refrigerator 13 units... 200W / unit
TV  13 units... 100 W / unit
Washing Machine  3 units... 600W / unit
Drinking Water Cooler  1 unit.... 100W / unit
Water Heater  3 units.... 4500W / unit
Lamps  35 units... 20W / unit
Misc Devices (blenders, etc)  65 units... 50W / unit
That gives a total of 62250W. It will indeed be three phase.

Technicallythe utility looks like a "voltage source" (or an "AC Battery"Which keeps the 230 VAC voltage "stable" regardless of load).
A typical Grid Tied Inverter looks like a "current course" (or constant current devicecurrent level based on the amount of sun hitting the panels at any point in time).
As far as your "billing" is concerned, yes, the solar panels+GT Inverter will "spin the power meter backwards" when loads are less than GT inverter outputAnd "recharge" the utility Grid.
Unless you have electrical or electronics experience"Normal" people rarely experience a "current source" and how it functions.
https://en.wikipedia.org/wiki/Current_source
https://en.wikipedia.org/wiki/Voltage_source
We can discuss it more if you are interestedCurrent sources are actually quite interesting devices and are very heavily used in integrated circuits (typically analog devices). Arc Welders and Florescent Ballasts are also examples of current sources.
As far as your A/C systems, lights, etc... their starting surge and the 230 VAC line voltage is "stabilized" by the utility itself (also know as a voltage source/sink). Having a "current source" recharge a "voltage source" is very common and work well together.
I am not sure how far you want to take this discussion (your choicejust ask).
Bill "I type too much sometimes" B.Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset 
Thanks Bill. That's great to hear that I would not need to consider the starting surge of A/Cs.
What is confusing me now is that I added the average power from all the loads. Next I chose the losses to be around 1.3, and multiplied the load of 62250 by 1.3 to get 80925W and therefore chose 290 x 280W panels to run this 80925W.
Is this way wrong? Do I have to use watthours instead ? 
Watts is a "rate"Like Liters per minute
Watt*Hours is an "amount'Like total Liters pumped
So, if you have an "average load" of 62250 Watt and run it 10 hours per day: 62,250 Watts * 10 Hours per day = 622,500 Watt*Hours = 622.5 kWH per day
The typical number we use is 77% of solar panel name plate roughly equals the average GT inverter output on a typical warm day (in cold winter weather, below freezing, the solar panel Vmp rises, and P=V*I  So rising V gives more power with cold solar panels). 622,500 WH per day * 1/0.77 GT system efficiency * 1/6.92 hours of sun for peak AC month = 116,827 Watt solar array
in the end, you need to understand your power needs pretty closely. And how that may change over the seasons (no AC in winter?).
Also power company billing will have a large effect on your eventual final design.
Bill
Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset 
Thanks Bill.
This clears up a lot of my confusion.
Also do I need to consider the power factor of each load ? In that case the average load would be much higher.
Also since according to the calculations you did, the solar array should be 116,827W, the number of panels required (if choosing 280W), would be
116,827/280 = 417.2 panels ? 
Yes, your # of solar panel math is correct (based on my guesses/examples).
Power Factor is a large discussion in itselfYou need to size the branch circuits and main panel+utility on Power Factor for two reasons.
First reason is that the wiring/fuses/transformers (more or less) use VA (Volt*Amps) for sizing the circuits. For example if you have a typical induction motor with poor PF~0.7, then a 4,000 Watt load would look like (made up numbers for example): 4,000 Watts / 230 VAC = 17.4 Amps (based on Watts)
 17.4 amps / 0.70 PF = 25.6 Amps (actual wire current based on poor PF)
 230 Volts * 25.6 Amps = 5,888 VA
Then there is the other side of the discussion... First, most GT Inverters output near 1.0 PF  And some newer models can actually be programmed to output +/ 0.8 PF or soTo offset power PF loads to a certain extent.
Second, for motors, it is possible to add (special) capacitors to the motor (or the utility can do it on their network) to bring the PF closer to 0.95 or so (correct bad PF locally, and not have the bad PF run all the way back to the generator/power station). In California, summer means lots of A/C and water pumping (irrigation of crops)So the power company will switch in capacitor banks during summer.
ThirdThere are A/C systems that are designed to have 0.95 or better power factorSo it reduces your problems. Choosing efficient/modern hardware is getting much easier these days. And, it is generally better (cost less overall) to start by looking for very efficient electrical appliances. The new stuff can be much better than what was available 20 years ago. Add insulation to walls and ceilings, double or triple pane widows, etc. are all (usually) worth the time and moneyInstead of being tied to high electric bills for years/decades in the future.
NoteThis is a basic PF discussionPoor PF has other causes and effects that can cause more headaches and complexities... Motors+Capacitors are "simple" solution that (usually) do not apply to many other types of loads (computer power supplies, CFL and LED fixtures, etc.). These other "issues" may or may not apply to you (depends on your specific loads and how many of each you haveI.e., if you have a large computer center to go with your A/C system, there are "Power Factor Corrected" type computer power supplies available).
Bill
Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset  4,000 Watts / 230 VAC = 17.4 Amps (based on Watts)

This is a shot off my TED5000, you can see each zone has its own Power Factor.
Plus the shot is of selling to the grid, here they give kWh credits within our TOU and a credited true up at year end of about $0.025 per kWh for any excess.
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