Energy consumption problem

Abdullahx227 · February 2017

Welcome guys.
i have been installed a solar system, after calculating the energy consumption and running the system, the system have unexpected consumption happened, it supposed to be 2800Wh capacity but some day we consume only 1813Wh and the inverter shut down
first i will show you how i calculate the consumption.

i have a
40W halogen Lamp 1
30W LED 1
45W LED 1
12W LED 12PCS
total = 259W x 7hr = 1813Wh
all these working together for 7hr that day

the batteries i have two in parallel 12V,200Ah lead-acid (not deep cycle)
the total capacity should be:10V x 400Ah = 4000Wh ( i use 10V because the inverter cutoff voltage at (10 ~ 10.5V))
and assuming system loss with be (25% ~30%)
so the total become 0.7 x 4000 = 2800Wh
inverter 2000W, 12V(modified sine wave)...too much
charger controller 50A( the common on)
the batteries that day before running the system were at 13.7 volts (fully charged)

my question is where are the rest of the capacity goes ? 2800Wh - 1813Wh = 987Wh ??
are there any mistakes in my calculations ?

Thank you

animatt · February 2017

Is the inverter on 24x7? What is the power consumption of the inverter to just stay powered on?
Controller power consumption?

How are you getting 200 ah value per battery if they are not deep cycle batteries. I believe non deep cycle do not usually get giving an AH value.

Also is your amp hour rating. Is it a 20 hour rate or 100 hour rate?

Apart from that lead acid batteries if you care for them should not be fully discharged.

Estragon · February 2017

Also, how long has the system been running?

How old are the batteries?

What are your charging sources? Solar - how many, size, configuration of panels. Other sources - generator?

Did this happen suddenly or does it seem to be getting slowly worse?

Where are you located?

westbranch · February 2017

I believe there is a mathematical error, you should only use 50% of the batteries capacity when calculating the loads that can be powered... so you would only have 2000Wh not 4000Wh. Then you can apply system losses etc.

Abdullahx227 · February 2017

animatt

the system installed two months ago
everything new

Is the inverter on 24x7? What is the power consumption of the inverter to just stay powered on?
for the second question i think the consumption was taken into account from the system losses( 20% ~ 30%),
Controller power consumption? also included

How are you getting 200 ah value per battery if they are not deep cycle batteries.
I believe non deep cycle do not usually get giving an AH value.
there are two different batteries used in solar system in my area, deep cycle the efficient and usual way( expensive)
or using car batteries i use two of them (not recommended i know, but less price than the previous.

Also is your amp hour rating. Is it a 20 hour rate or 100 hour rate?
not sure.

thank you for your reply

Aguarancher · February 2017

" i use 10V because the inverter cutoff voltage at (10 ~ 10.5V)) and assuming system loss with be (25% ~30%) " you can't use those figures. here is the battery state of charge with no loads: the battery voltage @ 11.51V = 10% remaining charge. you are killing your batteries

Abdullahx227 · February 2017

Estragon said:

Also, how long has the system been running?
two months ago
How old are the batteries?
new
What are your charging sources? Solar - how many, size, configuration of panels. Other sources - generator?
solar (off grid system), 4 panels each 150W, see the picture
Did this happen suddenly or does it seem to be getting slowly worse?
some time we use the system but without running all lights , one day we did and this happened
Where are you located?
average sun hours for my area about 7.5hr/day
now in winter, the worst case 6 hrs/day
the panels located at 23 degrees fixed all the year

Abdullahx227 · February 2017

Aguarancher said:

" i use 10V because the inverter cutoff voltage at (10 ~ 10.5V)) and assuming system loss with be (25% ~30%) " you can't use those figures. here is the battery state of charge with no loads: the battery voltage @ 11.51V = 10% remaining charge. you are killing your batteries

what would you recommend ?

BB. · February 2017

Abdullahx227 said:

Welcome guys.
i have been installed a solar system, after calculating the energy consumption and running the system, the system have unexpected consumption happened, it supposed to be 2800Wh capacity but some day we consume only 1813Wh and the inverter shut down
first i will show you how i calculate the consumption.

i have a
40W halogen Lamp 1
30W LED 1
45W LED 1
12W LED 12PCS
total = 259W x 7hr = 1813Wh
all these working together for 7hr that day

My first question is to double check the LED ratings... Those pretty high wattage numbers (i.e., those would be very bright lights for interior usage--And if some are used for exterior/security lighting--I would suggest that you look for motion activated lights. That can reduce power usage a lot--Plus, many times, always on security lighting just highlights what people are interested in stealing. Motion activated lights keep your place in the dark and hidden--until the lights are activated--Which then attracts attention and can scare off intruders.

In either case--The numbers do add up and are close to a "medium sized" off grid power system (1,000 WH per day or less, small... Approximately 3,300 WH per day is a medium system that can run a very efficient off grid home.

the batteries i have two in parallel 12V,200Ah lead-acid (not deep cycle)
the total capacity should be:10V x 400Ah = 4000Wh ( i use 10V because the inverter cutoff voltage at (10 ~ 10.5V))
and assuming system loss with be (25% ~30%)
so the total become 0.7 x 4000 = 2800Wh
inverter 2000W, 12V(modified sine wave)...too much
charger controller 50A( the common on)
the batteries that day before running the system were at 13.7 volts (fully charged)

I am not a fan of paralleled 12 volt batteries for a 12 volt system. It is very difficult to monitor the health of the batteries (both batteries have the same "bus voltage" and you cannot with a simple volt meter easily monitor the health of each 12 volt battery). My preference would be (for example) 2x 6 volt @ 200 AH batteries in series (in the states, we call these "golf cart" batteries). Now you can use a volt meter to check each battery to make sure they are sharing charging/loads equally.

If you can--I would suggest getting a DC Current Clamp DMM (digital multi meter). They run from ~$60 USD to >$300 USD. You can simply put the "current clamp" around one wire and measure the current flow--And figure out what is happening quite easily (and safely) in your system.

my question is where are the rest of the capacity goes ? 2800Wh - 1813Wh = 987Wh ??
are there any mistakes in my calculations ?

Thank you

You also asked about inverter "self consumption" (energy used just when the inverter is turned on). That can run from 6 Watts to 20+ Watts pretty easily for a 2,000 Watt PSW inverter. There are inverters with remote on/off and a "sleep mode" that can reduce power usage when the AC power is not needed. As a worst case example:

20 Watt Tare Load (AC inverter) * 24 hours = 480 WH per day -- Or almost 1/3rd more power than your loads.

Here is an example of a very nice/small 300 Watt 12 volt AC inverter with remote on/off and power save mode:

http://www.morningstarcorp.com/products/suresine/

Both 120 VAC @ 60 Hz and 230 VAC 50 Hz versions are available.

-Bill

BB. · February 2017

In general, the first step for designing an off grid power system--Conservation is key. It is almost always more cost effective to conserve energy than to generate it with solar.

Based on your power estimates, here is an example of how I would suggest designing your off grid system using our "rule of thumbs" for a full time off grid system.

First the battery bank size. In general, 2 days without sun and 50% maximum discharge gives us a cost effective system with good performance and life:

1,813 WH per day * 1/0.85 typical AC inverter eff * 1/12 volt battery bank * 2 days storage * 1/0.50 max discharge = 711 AH @ 12 volt battery bank

This AH rating of battery bank (near 800 AH) is large enough, that it may be worth looking at a 24 volt battery bank @ 356 AH. Same amount of stored energy, but the current for charging/loads is 1/2 that of a 12 volt system.

Also--I would suggest getting a hydrometer to measure the specific gravity of electrolyte of your flooded cell batteries--That will help you figure out how your batteries+charging system are performing.

If this is a weekend system, or your just don't want that large of a battery bank, you could go with 1/2 * 311 AH @ 12 volts or a 356 AH @ 12 volts--But I would not recommend this for a full time off grid system.

Next, two solar array calculations. The first is based on the size of the battery bank--A larger battery bank simply needs more current to properly charge the battery bank. Recommend 5% rate of charge for weekend/seasonal cabin/home/camp. And 10% to 13% or so for a full time off grid system:

711 AH * 14.5 volts charging * 1/0.77 panel+controller eff * 0.05 rate of charge = 669 Watt array minimum
711 AH * 14.5 volts charging * 1/0.77 panel+controller eff * 0.10 rate of charge = 1,339 Watt array nominal
711 AH * 14.5 volts charging * 1/0.77 panel+controller eff * 0.10 rate of charge = 1,741 Watt array cost effective maximum

Next is sizing the array based on your loads and hours of sun per day... For example:
http://www.solarelectricityhandbook.com/solar-irradiance.html

Ad Dammam Saudi Arabia
Average Solar Insolation figures

Measured in kWh/m2/day onto a solar panel set at a 64° angle from vertical:
(For best year-round performance)

Jan	Feb	Mar	Apr	May	Jun
4.51	5.31	5.53	5.89	6.41	6.77
Jul	Aug	Sep	Oct	Nov	Dec
6.49	6.64	6.78	6.32	4.99	4.25

This region is blessed with lots of sun year round--So the array needed to power the loads would not be that large:

1,831 AH per day * 1/0.52 off grid system eff * 1/4.25 hours of sun (December) = 829 Watt array

Of course, these are long term average numbers--If you have some periods of poor sun/weather--Then you would want a bit larger array if you cannot turn off some of your loads after a few days of bad weather. For example, you may want to only count on 75% of the system predicted performance to run your base loads (especially if you do not have a backup generator):

829 Watt array (December "nominal") * 1/0.75 "derating for poor weather at times" = 1,105 Watt array for "reliable" base loads

Anyway--I have not gone into details of battery/charge controller selection just yet--I wanted to show a "back of the envelope" set of calculations to figure out what you are expecting (energy wise) from your system.

-Bill

Estragon · February 2017

Is it sunny *every* day?

You are using around half or more your capacity daily using batteries which aren't designed to be deeply cycled. For long life, deep cycle batteries shouldn't be taken below 50% regularly, and need to be fully charged about once/wk. On top of your 1800w estimated loads, your inverter could be using 30w or so just to be on. 30x24=720w/day if you leave it on. Using a 2000 watt inverter for ~300w of loads won't be very efficient so more losses there.

A sulfated battery will appear to charge (fairly quickly) but will also discharge quickly under load. Two months is plenty of time for this to have happened. Your panels will likely be giving you 2.5kw/day or so which may just cover load. Any cloudy days or higher use would make it hard to catch up with no other charging source. If the batteries are flooded, I would try equalizing the batteries in the hope the sulfation can be at least partly reversed.

Abdullahx227 · February 2017

BB. said:

In general, the first step for designing an off grid power system--Conservation is key. It is almost always more cost effective to conserve energy than to generate it with solar.

Based on your power estimates, here is an example of how I would suggest designing your off grid system using our "rule of thumbs" for a full time off grid system.

First the battery bank size. In general, 2 days without sun and 50% maximum discharge gives us a cost effective system with good performance and life:
1,813 WH per day * 1/0.85 typical AC inverter eff * 1/12 volt battery bank * 2 days storage * 1/0.50 max discharge = 711 AH @ 12 volt battery bank
This AH rating of battery bank (near 800 AH) is large enough, that it may be worth looking at a 24 volt battery bank @ 356 AH. Same amount of stored energy, but the current for charging/loads is 1/2 that of a 12 volt system.

Also--I would suggest getting a hydrometer to measure the specific gravity of electrolyte of your flooded cell batteries--That will help you figure out how your batteries+charging system are performing.

If this is a weekend system, or your just don't want that large of a battery bank, you could go with 1/2 * 311 AH @ 12 volts or a 356 AH @ 12 volts--But I would not recommend this for a full time off grid system.

Next, two solar array calculations. The first is based on the size of the battery bank--A larger battery bank simply needs more current to properly charge the battery bank. Recommend 5% rate of charge for weekend/seasonal cabin/home/camp. And 10% to 13% or so for a full time off grid system:
711 AH * 14.5 volts charging * 1/0.77 panel+controller eff * 0.05 rate of charge = 669 Watt array minimum
711 AH * 14.5 volts charging * 1/0.77 panel+controller eff * 0.10 rate of charge = 1,339 Watt array nominal
711 AH * 14.5 volts charging * 1/0.77 panel+controller eff * 0.10 rate of charge = 1,741 Watt array cost effective maximum
Next is sizing the array based on your loads and hours of sun per day... For example:
http://www.solarelectricityhandbook.com/solar-irradiance.html
Ad Dammam Saudi Arabia ( from where you get this location )
Average Solar Insolation figures
Measured in kWh/m2/day onto a solar panel set at a 64° angle from vertical:
(For best year-round performance)
Jan Feb Mar Apr May Jun
4.51
5.31
5.53
5.89
6.41
6.77

Jul Aug Sep Oct Nov Dec
6.49
6.64
6.78
6.32
4.99
4.25

This region is blessed with lots of sun year round--So the array needed to power the loads would not be that large:
1,831 AH per day * 1/0.52 off grid system eff * 1/4.25 hours of sun (December) = 829 Watt array
Of course, these are long term average numbers--If you have some periods of poor sun/weather--Then you would want a bit larger array if you cannot turn off some of your loads after a few days of bad weather. For example, you may want to only count on 75% of the system predicted performance to run your base loads (especially if you do not have a backup generator):
829 Watt array (December "nominal") * 1/0.75 "derating for poor weather at times" = 1,105 Watt array for "reliable" base loads
Anyway--I have not gone into details of battery/charge controller selection just yet--I wanted to show a "back of the envelope" set of calculations to figure out what you are expecting (energy wise) from your system.

-Bill

Thank you very much Bill,

thanks for everyone helps me in this problem

Abdullah

Abdullahx227 · February 2017

Estragon said:

Is it sunny *every* day?

You are using around half or more your capacity daily using batteries which aren't designed to be deeply cycled. For long life, deep cycle batteries shouldn't be taken below 50% regularly, and need to be fully charged about once/wk. On top of your 1800w estimated loads, your inverter could be using 30w or so just to be on. 30x24=720w/day if you leave it on. Using a 2000 watt inverter for ~300w of loads won't be very efficient so more losses there.

A sulfated battery will appear to charge (fairly quickly) but will also discharge quickly under load. Two months is plenty of time for this to have happened. Your panels will likely be giving you 2.5kw/day or so which may just cover load. Any cloudy days or higher use would make it hard to catch up with no other charging source. If the batteries are flooded, I would try equalizing the batteries in the hope the sulfation can be at least partly reversed.

("Using a 2000 watt inverter for ~300w of loads won't be very efficient so more losses there")
Yes you are right, how much the losses will be ?
how can i differentiate between amount of loss in a 1500W inverter and 2000W inverter when using ~300W instantaneous ?

Thnak you Estragon

-Abdullah

mcgivor · February 2017

@Abdullahx227 asks

("Using a 2000 watt inverter for ~300w of loads won't be very efficient so more losses there")

Yes you are right, how much the losses will be ?

how can i differentiate between amount of loss in a 1500W inverter and 2000W inverter when using ~300W instantaneous ?

All inverters will have a self consumption load on the battery, generally the larger the inverter, the larger the self consumption. Even if there is no load on the inverter but it is turned on, it will still costume power from the battery, therefore always turn it off when not required. Using a smaller inverter, more closely matching you loads will lower the self consumption.
Some of the better inverters have a sleep mode, which goes to sleep and uses only a very small amount of power, when something is turned on it comes to life, but there has to be a load large enough to wake it up, so turning a 5W led light on will unlikely wake it up, all have different wake up ratings depending on manufacturer.
These values are generally in the specifications and can be found online.

Abdullahx227 · February 2017

mcgivor
Ok, thank you very much

Energy consumption problem

Comments

Ad Dammam Saudi Arabia
Average Solar Insolation figures

Ad Dammam Saudi Arabia ( from where you get this location )
Average Solar Insolation figures

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Energy consumption problem

Comments

Ad Dammam Saudi ArabiaAverage Solar Insolation figures

Ad Dammam Saudi Arabia ( from where you get this location )Average Solar Insolation figures

Categories

Ad Dammam Saudi Arabia
Average Solar Insolation figures

Ad Dammam Saudi Arabia ( from where you get this location )
Average Solar Insolation figures