Solar Powered Sumbermisble Pump

[garman]

I'm trying to run my 300 Gallon Aquarium off of solar power - off the grid.

1200 GPH Mag Drive Pump AC 110 watts -24 Hours
and
a 60" LED Light 46 Watts - 12 Hours

Point me in the right direction.

From an off grid calculator it it showing that the system will use

124.48 Kwh/HR per month
I live in Miami Florida so 4.5 Peak Hours of sun a day and want to run it %100 off the grid , so I need at least 922 watts of power.

I should be looking at 1000+ watt systems?

[inetdog]

Yes. Even more if you want to be able to stay off grid through the winter months.
Why do you want to be totally off grid? It would be more economical to use grid tie inverters to reduce your power cost if grid power is available to where your tank will be.

[mapmaker]

1200 GPH Mag Drive Pump AC 110 watts -24 Hours
and
a 60" LED Light 46 Watts - 12 Hours

Welcome to the forum,

Have you measured the wattage with a kill-a-watt meter? or are those numbers from the labels? (the labels are usually not as accurate as an actual measurement)

From an off grid calculator it it showing that the system will use

124.48 Kwh/HR per month

That makes no sense... 124.48 kilowattHours per hour per month?????

110 watts X 24 hours = 2640 wattHours per day
46 watts X 12 hours = 552 wattHours per day

2640 + 552 = 3192 wattHours per day

3192 watthours per day X 30 days = 95760 watthours per month = 95,760 kilowatthours per month

Since the sun comes up every day (not every month), the number to use for off grid system design is 3.192 kwh per day. The monthly figure is more useful for a grid tie system.

Next step in off grid design is to determine a battery size and system voltage. First take the daily energy consumption (AC watts) and figure your inverter is 85% efficient: 3.192 kwh ÷ .85 = 3.76 kwh.

A typical balanced design will have the battery drawn down to 75% SOC per day, thus your battery should store 4 X 3.76 kwh = 15 kwh.

For this capacity, I suggest a 24 volt system. That would be 15,000 watthours ÷ 24 volts = 625 amphours.

Surrette makes a good 683 ah battery in their "5000 series". It will cost you about $4000 plus shipping, and with careful management will last about 10 years.

How are we doing so far? Next step is to determine the size of the solar array that will keep those batteries happy, choose an inverter and balance of system, and decide what to do (generator?) if you have more than two cloudy days in a row.

--mapmaker

[paulcheung]

I have a similar setup, I have a pond that use 125 watts pump run 24/7. and another one run 8 hours in the day from 7am for the water fall.

I have 3 225 watts solar panels wired in series and 60 amps CC for 12 volts batteries, 1000 watts pure sine wave inverter and 4 6volts 225 amps deep cycle batteries. It run quite well since I installed the third panel. You can have the same but try to put in 24 volt batteries in series, or 2 400 amps batteries in 12 volts. also you can have a smaller inverter like 300 watts pure sine wave will do the work. I bought mine so large when I don't know better.

[Johann]

Welcome to the forum,

Have you measured the wattage with a kill-a-watt meter? or are those numbers from the labels? (the labels are usually not as accurate as an actual measurement)



That makes no sense... 124.48 kilowattHours per hour per month?????

110 watts X 24 hours = 2640 wattHours per day
46 watts X 12 hours = 552 wattHours per day

2640 + 552 = 3192 wattHours per day

3192 watthours per day X 30 days = 95760 watthours per month = 95,760 kilowatthours per month

Since the sun comes up every day (not every month), the number to use for off grid system design is 3.192 kwh per day. The monthly figure is more useful for a grid tie system.

Next step in off grid design is to determine a battery size and system voltage. First take the daily energy consumption (AC watts) and figure your inverter is 85% efficient: 3.192 kwh ÷ .85 = 3.76 kwh.

A typical balanced design will have the battery drawn down to 75% SOC per day, thus your battery should store 4 X 3.76 kwh = 15 kwh.

For this capacity, I suggest a 24 volt system. That would be 15,000 watthours ÷ 24 volts = 625 amphours.

Surrette makes a good 683 ah battery in their "5000 series". It will cost you about $4000 plus shipping, and with careful management will last about 10 years.

How are we doing so far? Next step is to determine the size of the solar array that will keep those batteries happy, choose an inverter and balance of system, and decide what to do (generator?) if you have more than two cloudy days in a row.

--mapmaker

So.....how many watts of panels does this member need to do the job ?

[mapmaker]

So.....how many watts of panels does this member need to do the job ?

I usually, as a starting point, design for a 10% charge rate. The units are a bit idiomatic... What I'm getting at is that the battery's capacity, in amphours at the 20 hour rate (C20) is 683 amphours. Ten percent of that number is 68.3 amphours... but we want to be able to charge at 68.3 amps (not amphours). That, by the way, may be expressed as C20/10 (the 20 hour Capacity divided by 10).


The solar array needs to provide 68.3 amps, and you must account for loads and controller losses and cable losses. This system has small, but constant loads. I will somewhat arbitrarily call those loads and losses 10% (the controller, solar array string voltages, temperature and many other factors make up that 10%). In the OPs system the loads are small enough to fit within the 10%).


So now we need 68.3 amps + 6.83 amps = 75 amps.


You need to be able to make that 75 amps at about 60 30 volts (that 60 30 volts is somewhat near your absorb and or equalization voltage)
75 amps X 60 30 volts = 4500 2250 watts.


Of course solar panels rarely produce their nameplate power, so you must derate them. For a first approximation, 75%


4500 2250 watts ÷ 75% = 6000 3000 watts


So to answer your question... first approximation is 6000 3000 watt solar array.


In any real system, the numbers can be worked out more precisely, depending on many factors, including some that I have mentioned.


--mapmaker
------ Edit: should have used numbers for a 24 volt system. Correct numbers in bold. Thanks, Paul

[Robert1234]

I ran 500 gallons of aquaponics (600 gph of pumps, two air pumps, two T8 fixtures, etc) on a 3 kwatt array but that was marginal. Much happier to be expanded to a 5 kwatt array and grid tied now.

Based on experience (without doing the calcs) I'd say a 6 kwatt array need is realistic based on what's been presented. Bravo, mapmaker . Not a bad guess at all. Takes a lot more panels than what people might first feel is necessary.

[Johann]

I usually, as a starting point, design for a 10% charge rate. The units are a bit idiomatic... What I'm getting at is that the battery's capacity, in amphours at the 20 hour rate (C20) is 683 amphours. Ten percent of that number is 68.3 amphours... but we want to be able to charge at 68.3 amps (not amphours). That, by the way, may be expressed as C20/10 (the 20 hour Capacity divided by 10).


The solar array needs to provide 68.3 amps, and you must account for loads and controller losses and cable losses. This system has small, but constant loads. I will somewhat arbitrarily call those loads and losses 10% (the controller, solar array string voltages, temperature and many other factors make up that 10%). In the OPs system the loads are small enough to fit within the 10%).


So now we need 68.3 amps + 6.83 amps = 75 amps.


You need to be able to make that 75 amps at about 60 volts (that 60 volts is somewhat near your absorb and or equalization voltage)
75 amps X 60 volts = 4500 watts.


Of course solar panels rarely produce their nameplate power, so you must derate them. For a first approximation, 75%


4500 watts ÷ 75% = 6000 watts


So to answer your question... first approximation is 6000 watt solar array.


In any real system, the numbers can be worked out more precisely, depending on many factors, including some that I have mentioned.


--mapmaker

Thank you very much for breaking it down for me and the OP.
At least next time I should be able to calculate it for myself.

[paulcheung]

Seriously? Use 6KW PV array to run 3.2kwh daily load? Am I missing something?

[paulcheung]

A typical balanced design will have the battery drawn down to 75% SOC per day, thus your battery should store 4 X 3.76 kwh = 15 kwh.

For this capacity, I suggest a 24 volt system. That would be 15,000 watthours ÷ 24 volts = 625 amphours.

Surrette makes a good 683 ah battery in their "5000 series". It will cost you about $4000 plus shipping, and with careful management will last about 10 years.


--mapmaker

I guess I see what is wrong. You use 24 volts here. and you use 48 volt on the next reply.

[paulcheung]

Next step in off grid design is to determine a battery size and system voltage. First take the daily energy consumption (AC watts) and figure your inverter is 85% efficient: 3.192 kwh ÷ .85 = 3.76 kwh.

A typical balanced design will have the battery drawn down to 75% SOC per day, thus your battery should store 4 X 3.76 kwh = 15 kwh.

For this capacity, I suggest a 24 volt system. That would be 15,000 watthours ÷ 24 volts = 625 amphours.

Surrette makes a good 683 ah battery in their "5000 series". It will cost you about $4000 plus shipping, and with careful management will last about 10 years.

How are we doing so far? Next step is to determine the size of the solar array that will keep those batteries happy, choose an inverter and balance of system, and decide what to do (generator?) if you have more than two cloudy days in a row.

--mapmaker

I usually, as a starting point, design for a 10% charge rate. The units are a bit idiomatic... What I'm getting at is that the battery's capacity, in amphours at the 20 hour rate (C20) is 683 amphours. Ten percent of that number is 68.3 amphours... but we want to be able to charge at 68.3 amps (not amphours). That, by the way, may be expressed as C20/10 (the 20 hour Capacity divided by 10).


The solar array needs to provide 68.3 amps, and you must account for loads and controller losses and cable losses. This system has small, but constant loads. I will somewhat arbitrarily call those loads and losses 10% (the controller, solar array string voltages, temperature and many other factors make up that 10%). In the OPs system the loads are small enough to fit within the 10%).


So now we need 68.3 amps + 6.83 amps = 75 amps.


You need to be able to make that 75 amps at about 60 volts (that 60 volts is somewhat near your absorb and or equalization voltage)
75 amps X 60 volts = 4500 watts.


Of course solar panels rarely produce their nameplate power, so you must derate them. For a first approximation, 75%


4500 watts ÷ 75% = 6000 watts


So to answer your question... first approximation is 6000 watt solar array.


In any real system, the numbers can be worked out more precisely, depending on many factors, including some that I have mentioned.


--mapmaker

I think here is what is wrong. 683 amps is for 24 volt and the last calculation 6000 watts is base on 48 volts.

[mapmaker]

I think here is what is wrong. 683 amps is for 24 volt and the last calculation 6000 watts is base on 48 volts.

You are correct. I will go back and correct it. thanks for catching it. We seem to have lost the OP. --mapmaker

[Robert1234]

Seriously? Use 6KW PV array to run 3.2kwh daily load? Am I missing something?

------ Edit: should have used numbers for a 24 volt system. Correct numbers in bold. Thanks, Paul

Now account for the aquaponic pump and lights having power factors of 0.35 and 0.6, respectively and you may end up back where you started

Like I said earlier, my experience was that a 3 kw array barely kept up with my system which is similar to what the OP appears to be designing (albeit my pumps were lower wattage). He's gonna be adding air pumps into that setup to, so a large array is inevitable. If he has no backup power source, he best overdesign the array or the tanks will go off-line at some point in the first few months of operation. Depending on his stocking density, he could lose many of his fish friends within the first 6-12 hours of the pumps stopping.

[paulcheung]

Now account for the aquaponic pump and lights having power factors of 0.35 and 0.6, respectively and you may end up back where you started

Like I said earlier, my experience was that a 3 kw array barely kept up with my system which is similar to what the OP appears to be designing (albeit my pumps were lower wattage). He's gonna be adding air pumps into that setup to, so a large array is inevitable. If he has no backup power source, he best overdesign the array or the tanks will go off-line at some point in the first few months of operation. Depending on his stocking density, he could lose many of his fish friends within the first 6-12 hours of the pumps stopping.

Now, something is seriously wrong with your setup!!. I have 6400watts array that run my whole house with pond pumps and 2 refrigerator and a freezer, washing machine and other lights etc. I only need to use my generator for few hours when the sun is covered with cloud for the day.

I have a small pond with 2 pumps rated 125VA each, when I use the kill-a-watt meter to check them , it say 65 watts and 120VA on the meter. I use 3 225watts solar panel and a 60 amps CC with 4 6volts 225 amps deep cycle batteries. I have one pump run 24/7 and the other run 8 to 12 hours on the day depend on the weather. so I don't see how you have to use 5000watts to run a little 125watts pump and 2 air pumps with 2 lights.

[mapmaker]

Now account for the aquaponic pump and lights having power factors of 0.35 and 0.6, respectively and you may end up back where you started

Actually, low power factor loads have very little effect on the draw from the battery.

The low power factor load means the inverter must be rated higher than the watt consumption, and there will be a lightly higher cable loss.

Suppose the loads have a watt draw of 150 watts, and a PF of 0.5. The VA will be 300 watts. The inverter must be rated 300 watts, but only the 150 real watts are drawn from the battery. The excess VA watts (over the real watts) is current just "sloshing around" in the wiring. It is not consumed.

If 300 watts from the battery were consumed, and only 150 watts were consumed by the load, the inverter would be turning 150 watts into heat.... and you would know it... the inverter would be running very hot.

--mapmaker