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OK. That is different watt hour usage then your previous post.
Based on that data your math is correct for either the 24V or 48V battery.
The 5 daily multiplier makes an assumption that you will only use 20% of the battery system each day but you still have multiple days of drainage if you do not get any charging from the sun for a couple of days. You just have to figure using additional charging system besides the sun to get the batteries back to 100% SOC or you will damage them if you use them 3 days straight or 60% DOD.Leave a comment:
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OK. Going back to your "Dude, you gotta quit using attachments for text."
I changed to different browser and it seems to be working. Thank fk!
I'm reading and re-reading over your posts and formulas.
You said, "low voltage = low power." "Voltage is pressure." which is similar to my understanding of voltage being the force behind of the current (Amp). Without voltage energy wont create an Amp (current). Energy (charged particle) needs force/pressure (voltage), to flow in a current (Amp) which provides us with the power/consumption (watt) to make something work.
Amp is a unit of current or rate of flow and watt is a measure of power/rate of consumption that moves energy (charged particles/electricity).
The formulas work on 3 principles mostly by pressure moving energy through a current at what rate of consumption... Then the correct size wiring comes into play by way of limiting resistance to maximise efficiency and lower cost and size.
Am I on the right layman's track?
All my watt hours were taken directly from the power metre. However, the metres have a Hi and Lo Watt rating as well as normal (not average) usage. Do i need to calculate the Hi ratings?
I will write it all down again
AC for 4hrs @ 106w = 424w
Fridge 24hrs @ 32w = 768w
2 lights 6hr @ 20w = 120w
TV 4hr @ 63w = 252w
Tools 1hr @ 750w = 750w
Charger 2hr @ 90w = 180w
Washer 90min @ 50w = 75w Cold wash
Radio 2hr @ 50w = 100w
Total 2669
We can round it up to 2700Wh.
OK Battery Bank must be 5x Daily Usage = 13,500Wh 13.5kWh
13500Wh / 24V = 562.5 AH
13500Wh / 48V = 281.25 AH It would appear 48V be the best option for this size. Does that factor in any depth %?
I see how 7 100Ah 12V batteries will cook in no time. He needs 4 12V 282Ah in series.
Based on that data your math is correct for either the 24V or 48V battery.
The 5 daily multiplier makes an assumption that you will only use 20% of the battery system each day but you still have multiple days of drainage if you do not get any charging from the sun for a couple of days. You just have to figure using additional charging system besides the sun to get the batteries back to 100% SOC or you will damage them if you use them 3 days straight or 60% DOD.Leave a comment:
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Yes Sir.
I will write it all down again
AC for 4hrs @ 106w = 424wh
Fridge 24hrs @ 32w = 768wh
2 lights 6hr @ 20w = 120wh
TV 4hr @ 63w = 252wh
Tools 1hr @ 750w = 750wh
Charger 2hr @ 90w = 180wh
Washer 90min @ 50w = 75wh Cold wash
Radio 2hr @ 50w = 100wh
Total 2669 wh
We can round it up to 2700Wh.
There is no way to make 7 batteries work except in parallel. 7 is a prime number. You cannot use prime numbers of panels or batteries because that only leaves you two possible configurations of all in parallel or all in series. 7 12-volt batteries in series is 84 volts. In parallel leaves you with a 12 volt battery. Neither are workable. You can work with 2, 4, 6, and 8.
Leave a comment:
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I see how 7 100Ah 12V batteries will cook in no time. He needs 4 12V 282Ah in series.
Leave a comment:
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OK. Going back to your "Dude, you gotta quit using attachments for text."
I changed to different browser and it seems to be working. Thank fk!
I'm reading and re-reading over your posts and formulas.
You said, "low voltage = low power." "Voltage is pressure." which is similar to my understanding of voltage being the force behind of the current (Amp). Without voltage energy wont create an Amp (current). Energy (charged particle) needs force/pressure (voltage), to flow in a current (Amp) which provides us with the power/consumption (watt) to make something work.
Amp is a unit of current or rate of flow and watt is a measure of power/rate of consumption that moves energy (charged particles/electricity).
The formulas work on 3 principles mostly by pressure moving energy through a current at what rate of consumption... Then the correct size wiring comes into play by way of limiting resistance to maximise efficiency and lower cost and size.
Am I on the right layman's track?
All my watt hours were taken directly from the power metre. However, the metres have a Hi and Lo Watt rating as well as normal (not average) usage. Do i need to calculate the Hi ratings?
I will write it all down again
AC for 4hrs @ 106w = 424w
Fridge 24hrs @ 32w = 768w
2 lights 6hr @ 20w = 120w
TV 4hr @ 63w = 252w
Tools 1hr @ 750w = 750w
Charger 2hr @ 90w = 180w
Washer 90min @ 50w = 75w Cold wash
Radio 2hr @ 50w = 100w
Total 2669
We can round it up to 2700Wh.
OK Battery Bank must be 5x Daily Usage = 13,500Wh 13.5kWh
13500Wh / 24V = 562.5 AH
13500Wh / 48V = 281.25 AH It would appear 48V be the best option for this size. Does that factor in any depth %?
I see how 7 100Ah 12V batteries will cook in no time. He needs 4 12V 282Ah in series.Leave a comment:
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Leave a comment:
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Look at the numbers you got from the solar link.
In Winter you have 2.36 kwh/m^2
Average you have 4.54 kwh/m^2
In Summer peak you have 6.72 kwh/m^2
For an off-grid system you must use Worse case solar insolation. For you that is winter 2.36 kwh/m^2 which is 2.36 Sun Hours in a day. How long you have day light has little to do with Sun Hours.
So to find panel wattages we have to know how many watt Hours you need in a day. Example lets say 3 Kwh/day A battery system is only 66% efficient and we have to account for those losses and make an adjustment To do that we take your daily watt hour consumption and mulitply it by 1.5 or divide it by .66. Either way means to have 3 Kwh usable the panels must generate 3000 wh x 1.5 = 4500 wh or 4.5 Kwh.
Now that we know that and the minimum Sun Hours we can determine what the panel wattage requires by factoring the hours of of watt hours to leave us with WATTS.
Remember Watt Hours = Watts x Hours.
So we automatically know that Watts = Watt Hours / Hours
So now we have the numbers.
4500 watt hours / 2.36 Sun Hours = 1906 wats is required. 1906 is a queer number so in this example 2000 watts works and can use 8 250-watt panels will work. Now in Summer months with 6.72 Sun hours we have a surplus that is never used unless we know how and when to use it. That matters not in winter with only 2.36 Sun Hours. If you used Summer Sun hours would calculate only a 600 watt panel which would leave you dark all winter long without power or running the generator 9 months out of a year.Leave a comment:
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You might get 10 hours of sunlight but your panels will only produce close to their nameplate wattage during the insulation hours of the day which varies but is probably not more than 6 hours. The other hours will produce some power but it is hard to get a good number on what that is due to the changing amount of sunlight each day.Leave a comment:
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When performing a daily watt hour usage estimate it does not matter when you use the load, it still all adds up to what a battery system may have to provide if you do not get any sun that day. If you don't size your battery on your total usage you run the risk of killing that battery so while the peak load is important your battery size must be based on total usage.Leave a comment:
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Gotcha again. You do not understand the difference between power and energy. You have to get your head wrapped that. Watts and Watt Hours are not the same thing.
A 1000 watt Inverter can deliver up to 24 Kwh in a day.
Here is what you just said and asked put in a different way. Read it and you will see how silly it is.
My car goes 100 Kilometers per hour. How far did you travel?
You cannot answer because it is nonsense. Now you could answer if you said and asked: My car is going 100/Kph for 2 hours. How far did I travel? Easy peasy 200 kilometers.
Or; I need to go 200 kilometers in 2 hours. How fast do you need to go? 100/Kph
Watt Hours is energy, Watts is Power.
Watt Hours = Watts x Hours.
So you are on the right track, but making the wrong conclusions because you do not understand and confusing yourself. For the Inverter there are 3 possible answers, answers and 2 are BAD.
1. If you estimate 300 watt is peak, and the real number is 500 watts, your Inverter is not going to work and trip off-line. If you keep trying it will burn up.
2. If you estimate 500 watts and buy a 500 watt inverter it will work for a short period of time. But like a car, it may go 200/Kph, bu tif you get in, floor the throttle and keep it there you know what is going to happen. The motor will burn up.
3. If peak is 500 watts, multiply by 1.5, buy 750 Watt Inverter, then life is good and long lived.
Then I am assuming he has to know what he will use, when he will use it. He has to program what he does, when he uses air con with TV, or when he uses tools with air con etc. Either of those combinations will draw different amounts of peak power. For him to build a 5.8Kw (5800Wh) system will be laughable to him, as he wont afford it.
So it comes down to a good program, sticking to it. Setting up a solar system which includes some future additions (TV, Air Con, Washer) and sticking to a program. I don't even know if for sure he will ever get a TV or air con.
All I know is his total Wh usage will be up to 2669Wh each day, including future addition of household items. This is what I asked before. 2669Wh is 2.669kWh over the whole day. Why do my kWh figures total 5.8kWh and not equate with 2669Wh? I used a power meter for each device. The power metres have a Hi W rating, usage & Lo W rating. Did I need to factor in the Hi W rating.
For eg. the Washer has a Hi rating of 2134W, but usage ranging from 5-50w during the wash cycle. A power tool is easy, its continuous when you use it.Leave a comment:
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Battery = 5 x 1000 wh / 12 volts = 416 AH 12 volt battery
Panel Wattage = [1000 wh x 1.5] / 2.36 Sun Hours = 635 watts. You can get by with 600 watts. Or 3 200-watt panels wired in series.
MPPT Charge Controller Min Amps = 600 watts / 12 volts = 50 amps
Inverter min Req = 100 watts x 1.5 = 150 watts.
But I can work out the total with that formula.
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A 1000 watt Inverter can deliver up to 24 Kwh in a day.
Here is what you just said and asked put in a different way. Read it and you will see how silly it is.
My car goes 100 Kilometers per hour. How far did you travel?
You cannot answer because it is nonsense. Now you could answer if you said and asked: My car is going 100/Kph for 2 hours. How far did I travel? Easy peasy 200 kilometers.
Or; I need to go 200 kilometers in 2 hours. How fast do you need to go? 100/Kph
Watt Hours is energy, Watts is Power.
Watt Hours = Watts x Hours.
So you are on the right track, but making the wrong conclusions because you do not understand and confusing yourself. For the Inverter there are 3 possible answers, answers and 2 are BAD.
1. If you estimate 300 watt is peak, and the real number is 500 watts, your Inverter is not going to work and trip off-line. If you keep trying it will burn up.
2. If you estimate 500 watts and buy a 500 watt inverter it will work for a short period of time. But like a car, it may go 200/Kph, bu tif you get in, floor the throttle and keep it there you know what is going to happen. The motor will burn up.
3. If peak is 500 watts, multiply by 1.5, buy 750 Watt Inverter, then life is good and long lived.Last edited by Sunking; 02-19-2017, 04:58 PM.Leave a comment:
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Sunking. I wish I did not have to use attachments.
The cant do long posts for some reason.... MAJOR PIALeave a comment:
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