Благодарю Вас, очень познавательно. Мне понравилось.

Then you are an idiot - OR - you haven't watched many videos at all.

Why would there be no load?

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.

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.

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

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.

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.

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.