Need help with series or parallel

AmazonWiz

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.  

BB.

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" loads--Or is this the energy usage per day (Watt*Hours of energy used per day). I will do some quick math--But with a system this large, you will need a professional engineer to size/design/help install the system--This 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

Lets guess you are using a 48 volt Lead Acid battery bank with 2 days of energy storage and 50% maximum discharge (for longer battery life):

• 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 installation--But the overall cost of the bank will be about the same as batteries are in different series/parallel combination)

Next size the solar array for proper battery bank charging... We would normally suggest 5% to 13% rate of charge for a lead acid battery bank. 5% would be OK for weekend/seasonal use. 10%+ is better for full time off grid:

• 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).

And you need to size based on your loads and amount of sun. For a fixed array located in Riyadh Saudi Arabia:
http://solarelectricityhandbook.com/solar-irradiance.html

Riyadh
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)

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

Lets guess that your actual loads vary with the amount of sun--Sunny months need more cooling that less sunny months. Lets use ~6.00 hours of sun for the hot months:

• 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)...

So, you should have a minimum of ~100,000 Watt solar array and ideally closer to 235,000 Watt (for best health of the battery bank).

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

Anyway, that is the basic math--You probably have lots of questions.

-Bill

AmazonWiz

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. 

BB.

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.

So--Your average loads * hours per day = Watt*HOurs per day (usually kWH per day is what your utility meter measures--1kWH=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

So--the 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 inverters--You have the rough idea of XXXX Watts of solar panel = 2x 25,000 Watt GT inverters is correct--But way you picked 290 panels * 280 Watts per panel--I do not have a good idea.

Series/parallel panels (wattage, current, Vmp-array series votlage, etc.) are all based on the GT inverter you pick. What is it input requirements?

-Bill

AmazonWiz

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.

BB.

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.

-Bill

AmazonWiz

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 ?

zoneblue

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?

mike95490

60Kw is an industrial plant, and your plant engineer should coordinate with the power company on the design.

AmazonWiz

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.

BB.

Technically--the 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 device--current 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 output--And "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 interested--Current 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 choice--just ask).

-Bill "I type too much sometimes" B.

AmazonWiz

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 watt-hours instead ?

BB.

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

But you will not have "full sun" 10 hours per day--So you need a larger array that can output more power (watts) in a shorter amount of time so that the WH used ~ WH generated.

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

Note, in my earlier post above, I made an assumption that your AC units were running 50% duty cycle (motor on for 10 minutes, off for 10 minutes, as an example). Here I just took your number as the average load--So the solar array is much larger based on this assumption.

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

AmazonWiz

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 ?

BB.

Yes, your # of solar panel math is correct (based on my guesses/examples).

Power Factor is a large discussion in itself--You 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

Second--The utility itself has to size its wiring/transformers for the VA of your loads, not Watts. So if you have very bad PF, they may apply the 1/0.70 to your billing and charge you more money because of the poor PF.

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 so--To 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.

Third--There are A/C systems that are designed to have 0.95 or better power factor--So 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 money--Instead of being tied to high electric bills for years/decades in the future.

Note--This is a basic PF discussion--Poor 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 have--I.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

solar_dave

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.