ฝัง
- เผยแพร่เมื่อ 2 มิ.ย. 2024
- Learn how to calculate the battery capacity, or battery size, for your solar electric system.
⏱️Timestamps:
0:06 Intro
0:53 --- Why are batteries needed?
1:10 --- What batteries are available?
1:40 --- What is the useable energy?
2:16 --- What is the battery longevity?
3:06 How to assess your energy demand
5:57 How to estimate system losses
6:23 --- round trip efficiency
7:07 --- wiring losses
7:32 --- conversion efficiency
8:20 How to calculate battery size
8:28 --- select a battery voltage
10:44 --- select days of autonomy
11:51 --- consider thermal losses
12:25 --- specify depth of discharge
13:42 --- calculate battery capacity
14:19 --- calculate number of batteries
15:14 Other videos and resources
For more information, visit: go.umd.edu/FarmSolar
Procedures for determining the performance of stand-alone photovoltaic systems. [www.nrel.gov/docs/fy99osti/27...]
IEEE Guide for the Characterization and Evaluation of Lithium-Based Batteries in Stationary Applications 1679.1-2017. [ieeexplore.ieee.org/document/...]
Exceeding 3% voltage drop can result in cost vs. performance benefit (solar cell arrays). [ieeexplore.ieee.org/document/...]
By far the best breakdown on how to calculate power needs; how to play with battery voltage and solar panel sizing to accommodate said power needs; I’m working on an OTG cabin and wish I could have seen this video last year. Thank you for producing. Consider another video on examples when you increase the solar panel array voltage or the battery voltage capacity, to a fixed power consumption need. Great work!
By far the best video I have seen on the topic. Precise and to the point.
Then you are an idiot - OR - you haven't watched many videos at all.
A very well explained and thorough video, far better than most others out there. Thanks for this and others you have presented.
However, unless I have misunderstood, there seems to be an error in one of the calculations:
At 7.00 minutes you said you would assume using lead-acid batteries at a round trip efficiency of 85%, and the use of these batteries was confirmed by later calculations using a depth of discharge of 50%.
But, at 8 minutes you calculated the combined system losses using a round trip efficiency of 0.95 (matching lithium batteries) not 0.85. This throws off the Load Subsystem Efficiency by quite a bit.. it should be 0.76 not 0.85.
In turn, that throws off the calculation of final battery capacity: it should be around 7645 Ah, not the 6835Wh given.
Most clear explanation about solar system set-up I've ever seen ❤ Good job 👍
Outstanding presentation- thank you
Thanks for the video. Easy to follow. All complexities broken to simpler, digestible chunks.
South Africa.
The best video that analyze this topic. Thanks
Fantastic videos dude 👍🏼❕
I’ve been dealing with solar for about 2 years now part time in Maine on my total off grid cabin and at my house in CT running the essentials in the house TV Wi-Fi freezers fridge raider
Keep the awesome videos coming
Thanks 👍🏼👍🏼👍🏼👍🏼
Благодарю Вас, очень познавательно. Мне понравилось.
THANK YOU VERY MUCH FOR YOUR VIDEO.
Good info, learned alot.
Excellent presentation- thank you very much for the outstanding info
I know this is old but excellent. I will be returning to this and I've subscribed.
Thanks my brother your doing good
Very well explained .thanks Prof..I dont see one done for battery bank recharge rate.. the time a system takes to recharge esp one using solar by day and wind by night.
Thanks a lot, I appreciate your knowledge-sharing session. got a lot of things to my knowledge base.
Thank you again best wishes
Thank you so much.😊
Dear Mr. Schiavone, I have a doubt at last two formula. Temperature conpensation factor 1.19 should be at lower part of the formula, so the result is: (19.528*3)/(24*0.85*0.5*1.19)=4826Ah. The last formula, the battery voltage is 12V, not 24V, so the quantity should be: 4826*2/215=44.9, roundup to 45 pcs. Just for your consideration. Thanks.
Very well explained!! but the number of batteries is double the quantity you get, cuz you are using 12V batteries, it means you will need 64 batteries.
Another very good tutorial...ib the case of a hybrid system what are you solutions..there is continuous charge when the sun goes down and the wind turbines kick in..
thank you so much
Thanks 😁👍🏻
Thanks so much for that information, but I still got a question. If the voltage system selected is 24V, why do you use 12V batteries? that only means that you will have to set more batteries in paralel which, as far as I know, it is not recommended. I'd really appreciate your answer. Thx
Great explanation with the general math, but the math application is very very bad😂. Math formula says 24 volts but got 12v batteries, choose lead batts with 50% round trip but used 95% lithium round trip. Number of batteries should be x2 since they are 12V and not 24Vdc, etc. But the general logic is sound with much details without the insignificant details that usually don't matter but cause a lot of confusion.
What about the inverter efficiency ? Inverters have a stanby consumption,
Meaning at night even with zero load they will draw some amount of watts from your battery , shouldnt this be in the calculation?
I like it
Can you have too much battery?
I think my daily usage is like 30kwh, but i have an EV and will plan on getting another one. Was thinking about building a 200kwh battery storage for fun, seems excessive.
please make a video for battery sizing of a utility PV plant with 50 MWp installed capacity for example.. since there is no load in this case, how we can size the battery
Why would there be no load?
My question is can a 5KW hybrid inverter say a Growatt SPF 5000 ES charge a 15KWH batter; 3 (48V, 100AH)
In the calculations given, you said the total of 6,835 Ah of battery storage. Isn’t this calculation for Watt hours? Given the daily electrical load began at 19,528 Wh?
Well since amps x volts = watts and Ah x volts = watthours and the quation he used had battery voltage in denominator it's been accounted for.
If you leave the battey voltage out of the equation you get watthours.
Or obviously if you multiply by battery volts again at the end you get watthours.
@@ghz24So you’re saying he has 69 AGM 100Ah or 35 Lithium Ion batteries? He can consume 285 amperes per hour per day! I don’t know how any residential home can use that much energy a day. The cost of such a battery bank of storage is astronomical and not even feasible to recharge with only 6~7 hours of solar energy. The 19,528 Wh of daily usage is 585,840 Wh per month.
@@deanhenthorn1890 The average US house uses 886 kWh a month (almost 30 kWh per day) so 585.840 kWh per month is hardly extreme.
You say consumes amps like it measures energy when it's current. Without voltage amps means almost nothing.
After watching more carefully I can see a few issues with this.
First is nobody would set up a 24 volt system for a load of 19kWh per day.
Second the turn around efficiency of the battery should not be included because it is about energy in vs energy out and is relevant to how much electricity it takes to charge the battery not it's fully charged capacity.
Like wise for the wire losses because the battery should be very close to the inverter and the wire loss between the panels and the charge controller are irrelevant to the battery capacity needed.
Finally this high of a capacity battery bank won't use 12 volt pre made battle born type 100 amp-hour batteries.
If I were doing this I'd simplify 19.5 kWh rounded is 20 x 3 days =60. Give 15% for temperature and 15% (even though depth of discharge is 100% of rated capacity for lfp) as a future hedge against aging. So 60 kWh x 1.3 is 78 kWh divide by inverter efficiency of 94% so 78 ÷.94 is 83 kWh.
That is the battery you need for this application.
83 kWh is 83,000 Wh ÷ 50 volts of a 16s lfp battery and get 1660 amphours.
1660 Ah ÷ 280 Ah per cell gives 5.9 cells in parallel.
So a 16s6p battery will give 20 kWh for 3 days and will for 25-30 years since loosing 20% after it's 6000 cycle life is already accounted for.
Lfp prismatic cells can be had for about $100 per kWh so a 83 kWh bank would be about $8-10k.
I gave $1400 for 14.3 kWh plus $300 for shipping but shipping would have been the same for 6 times more battery if I had room and money for more battery.
I'm wondering why 12V battery was used in the end if the calculations were done for 24V battery system. Did I miss something?
Also wondering the same. That would need twice the number of batteries he calculated.
@@senogamoses5319 i guess you wire the batteries in series
It was just for demonstration the point was the “Ah” capacity of the battery shown. Because the voltage can be increased by connecting it in a series connection.
@@AO00720 in which case you would need x2 the number of batteries.
64 batteries needed not 32
@@Cattywampus555 can you be more specific
7:05 Selected (0.85) Lead acid battery
but in
8:13 you used 0.95 (Lithium ion) why?
Lithium batteries have an efficiency of 95%
@@mikerukwava9574 But we are using lead acid batteries.........
I also noted that. An error of commission I suppose.
thanks a lot. it was the only that I think was totally useful for improving the basic knowledge.. the rest was just a blah blah
LFP - lithium-Ferro-Phosphate, not lithium-ion-phosphate! Otherwise, good basic calculations, thanks.
you have a book
Start undersized and live with it. You'll figure out how to manage loads by time of day and season. You can always add more later. If you go too big, it's money down the rat hole.
@6:20 round trip battery efficency has zero to do with battery capacity and should be removed from this calculation.
It's a one way trip not a round trip.
It measures how much more energy you need to charge the battery than it can deliver.
This is added to the charge energy but not subtracted from the capacity.
You rate capacity on energy delivered not energy to recharge.
Likewise line losses don't belong here there are no long runs of wire here.
The inverter should be only a couple feet from the inverter.
Wire from the panels to the charge controller are only involved in charging not discharging.
You are so right , round trip battery efficiency should be used to calculate how many extra panels you need to charge the batteries.
The inverter efficiency factor ( standby consumption ) should have been used instead. Since inverters with zero load also draw an amount of watts from the batteries.
yea,, this shi**t is not easy to explain, and you gave it a good shot. You could have stated some simple perspectives first to help people question 'why its done a certain way' instead of maping it out like biology class. he he. ex. 2 days power is the norm. 2 days is usually 60kwh. max charge rate is normally .5c . Just use lithium batteries, rofl. Use hybrid inverters unless you like financial bdsm with solar. Surge will have to be Calculated and explained to the homeowner if off grid, period. etc etc... loved your video though.. usually i just click to the next one.. cheers.
That electric bill must be from a hot place
11:30 is the only guarantee solution..... a gasoline ⛽️ generator.
Good quality, Flooded lead acid batteries will last a MINIMUM of 15 years to over 50 years or MORE, IF they are maintained and Charged CORRECTLY, as per Dr. Peter Lindermann and John Bedini. And furthermore, John Bedini has tested and observed that LA batteries are in excess of 200% efficient, when Pulse charged.
200% wow who needs solar we can just charge our battery for power.
Lead acid batteries are a rip off and the mist expensive batteries to own. Pulse charging at certain frequencies can restore function to sulfated batteries but only to a point.
Even if they lasted 50 years instead of the 5 that most get it's 45 more years of sucky.
Just the fact you get 85 watthours back for every 100 put in instead of 95 like a lfp is lead killer.
Lead is slow to charge and hates being partially charged and needs 10% more energy to charge it.
Just because you can float a lead acid battery for 50 years in perfect conditions doesn't make it a good battery.
@@ghz24 Don't just talk from the top of your hat, because you THINK you know it all !!! Because the great Nikola Tesla has also worked with Lead Acid batteries and observed most of what I mentioned. Dr. Peter Lindemann and John Bedini we're working in their lab, based on what Nikola Tesla has recorded. So, if something is beyond YOUR understanding, don't be STUPID and exhibit the same in public. And let me further mention, I have LA batteries, purchased in 2014, which I have been maintaining, based on the methods mentioned, and they are STILL in TOP condition.
Is it possible to power the whole world with only solar energy? For this, the device we will produce must be efficient, widespread and economical. What would happen if we covered the Sahara desert with solar panels? Or should we make a dyson sphere around the sun 😝😝🤪
8:03 wrong. you said earlier (at timestamp 6:57) that you were designing this example system with lead acid batteries which you claimed has a round-trip efficiency of .85, not .95. what is your major malfunction? oh, i can tell you're an alcohol drinker.
10:26 wrong, higher voltage begets higher current. the difference between a 12v system and a 24v system is the wiring size, which has different resistances. oy vey.
No he's right because the watts are limiting. As an example lets use a simplified system of two 120 watt panels giving 12 volts and 10 amps to get that 120 watts . If I put the two of them in parallel I get 12 volts still but have 20 amps but, if I wire them in series I get 24 volts and 10 amps to give the same 240 total watts.
So higher voltage results in lower amperage to provide the same power.
Since resistance losses are related to current, lower current and higher voltage results in less loss and/or thinner wires being needed.
In addition charge controllers cost substantially more for higher amperage ratings so a 40 amp controller for a 12 volt battery can handle 480 watts the same 40 amp controller for a 48 volt battery will handle 1920 watts.
@@ghz24his solitary statement devoid of caveats is incorrect and misleading.
@@h7opoloYeah not really. What caveats would you have included? Maybe "to provide the same power"?
I still say he's right and I'll give another example. Say I have a high intensity discharge lamp. The ballist (transformer) has multiple taps for use with different voltages if I use the set for 120 volts it will draw twice the amps as if I use the set of wires (taps) for 240 volts.
If I have 10 panels and they are 325 watt panels they each produce 33 volts and 10 amps at max pp.
If I wire them in series I have 330 volts at 10 amps and 10 guage wire is adequate for a 100' run.
If I wired them all in parallel I'd have 33 volts and 100 amps. What size wire would one need to handle that for a 100 foot run?
His statement is spot on for the subject being discussed considering the understood subject is solar power.
Even batteries follow this the watts a group of batteries can produce doesn't change depending on how they are wired. Series you add voltage and parallel you add amps but watts remain the same.
Just like a transformer if you step voltage up the current goes down and if you step the voltage down the current rises because conservation of energy.
@@ghz24 resistance losses can also be depend on voltage. V^2/R
@@MuhammadTalha-by4ee No the loss due to resistance is I^2R.
It doesn't matter what the voltage is it's not even in the equation.
10 amps through a 10 ohm resistor is 1000 watts loss regardless of if it is hundred volts or a thousand because volts is not mentioned, the difference is the power delivered to the load not the power lost in the wire.
You are using power delivered not power lost. Current is THE determining factor in power loss.
Higher voltage allows the use of lower current to get the same power through the wore but the loss is only dependant on the current and the resistance aka wire size.
Voltage only effects the current needed to deliver a set power but the calculation for loss is only resistance and current.
Tell straight way not teaching math.