Calculating power needs

[warmonger0]

Hi and thanks for taking the time to read my post.

I'm trying to calculate my power and battery needs. If anyone can check my math and formulas or offer advice on where I go now to decide on batteries I would appreciate it.

I would like to power/charge the below items for up to 3 hours continuously and I would like to the system to run in 12V so that I can use a 12v to 110v inverter.

1. Laptop (from wall outlet charger)- AC input: 100-240V ~2.5A
DC Output: 19v 4.74A

2. Iphone (from wall outlet charger)- AC input: 100-240v ~0.15A
DC Output: 5v 1A

3. Ipad (from wall outlet charger)- AC input: 100-240v ~0.15A
DC Output: 5.2v 2.4A

I would like to power this item 24h a day

1. Arduino (from Arduino specifications) - the Arduino takes 9v DC @ 2.1A

I'm pretty sure i've calculated the power requirements correctly (below) but I wanted to be sure. Before I get to this I want to include that i'm thinking to plug these into an inverter that will draw off the batteries so that I have a 110v AC outlet and a USB port. I was looking at this one in the 15Amp version for my needs (i'm open to suggestions for a different one). here

I feel like there is a lot of power loss converting from DC (the batteries) to AC (through the inverter) and back to DC (through the respective components adapter) but i'm not sure of a better way to insure I can power as universal a set of components as possible.

I'm not certain how i'm going to wire the Arduino in at this point so lets put that aside I guess.

AC to DC calculations: I found the formulas on this site to be
1A AC = 10A DC. to find DC Amps necessary for the system.
A DC X 12V = DC Watts

So here are the numbers I got for DC Watts


Laptop - 2.5A AC X10 = 25A DC. 25A DC x12V = 300 DC W
Iphone - 0.15A AC X10 = 1.5A DC. 1.5A DC x12V = 18 DC W
Ipad - 0.15A AC X10 = 1.5A DC. 1.5A DC x12V = 18 DC W

I'm pretty sure these are the numbers I need to start with but correct me if I'm wrong.

Okay, so now I add the total DC Watts together to get the maximum load I would draw per hour.
300 DC W + 18 DC W + 18 DC W = 336 W / H

So if I want to run these for 3 hours maximum with all 3 devices plugged in that would be
336W/H * 3 = 1080 W total
lets add 10% for the loss from the inverter and that's 1188 W total

Here is where I get a bit stuck and I'm looking for advice.
How many amps should my inverter be to handle the devices I laid out.
How many and what type of batteries would I need? I was looking to go with a couple 6v Golf Cart batteries in series mostly because of the size of the area I need to put the batteries in. How many Amps out should I look for in batteries given the inverter I specified and the draws of the devices.

If i'm missing anything just let me know or if there are any considerations I should make when choosing power please feel free to advise.

Thank you one and all for taking the time to look at this.

[Mike90250]

sorry - on the road tonight working from a tablet

Just need the AC watts, and the hours for each: Watts * Hour = watt hours
TV 79w 4 hr = 316wh
toaster 800w 0.06h = 48wh and so on.

[daz]

AC to DC calculations: I found the formulas on this site (hyperlink removed) to be
1A AC = 10A DC. to find DC Amps necessary for the system.
A DC X 12V = DC Watts

Dont work it out this way. Watts is watts, regardless of whether it it DC or AC. Your solar system will also be calculated in watts (actually WH), so we keep everything simple by just using watts...
Wattage is voltage multiplied by current, which is why watts AC = watts DC (minus losses for our purposes).
So 100W AC is 100W/220VAC = 0.45A
100W DC is 100W/12V = 8.3A
So regardless of whether DC or AC watts is watts...

So here are the numbers I got for DC Watts

Take all the items that you want to power, and have a look at the ratings (should be a sticker or plate on the device). If they mention a wattage , use that. If they do not mention a wattage, take the current and multiply it by the voltage (eg 1A x 115VAC = 115W).

Take all the wattages and add them together, then multiply them by the amount of hours they will be used (this will give you WH). This WH figure is what you will use to calculate the size solar system you need.

lets add 10% for the loss from the inverter

The actual losses will be much higher than 10%. On top of inverter losses you have battery losses, controller losses, wiring losses, etc.
The actual figure you use will depend on the type of charge controller you use. For PWM add 100% to your required WH figure, and for MPPT add 50% to the required WH figure. This gives you a good ballpark to start from.

Here is where I get a bit stuck and I'm looking for advice. How many amps should my inverter be to handle the devices I laid out.

The inverter needs to be able to handle the peak power draw of all your equipment. So if all your equipment is mainly resistive (eg no motors), and the peak power demand is 400W, double up and go for a 1KW inverter. If you are using motors, due to their large startup power requirements, you will need an inverter rated at 6 - 10 times higher than your actual running power draw.

How many and what type of batteries would I need? I was looking to go with a couple 6v Golf Cart batteries in series mostly because of the size of the area I need to put the batteries in. How many Amps out should I look for in batteries given the inverter I specified and the draws of the devices.

Your battery bank is sized according to your load and power requirements. The battery bank sizing is calculated by taking your required wattage and dividing by your battery bank voltage, then adjusting for discharge depth.

EXAMPLE:
You require 1000WH @ 12V....
1000WH / 12V = 83.33A
The maximum battery discharge depth for lead acid batteries should be no more than 50% DOD. This means we have to double our required battery capacity to achieve a 50% DOD. So...83.33A x 2 = 166.6A battery bank @ 12V. 166AH 12V is not a standard battery, so round up to whatever is the next closest capacity.

For a 30% DOD, you take your required capacity and divide by 30, then multiply by 100.
Example: 1000WH / 12V = 83.33A
83.33A / 30 = 2.78 x 100 = 277.77AH @ 12V (round up to next closest capacity).


The above is the basics to get you started....also, we will need to know the solar insolation for your area, as that determines the solar panel array size that is required.

[warmonger0]

Dont work it out this way. Watts is watts, regardless of whether it it DC or AC. Your solar system will also be calculated in watts (actually WH), so we keep everything simple by just using watts...
Wattage is voltage multiplied by current, which is why watts AC = watts DC (minus losses for our purposes).
So 100W AC is 100W/220VAC = 0.45A
100W DC is 100W/12V = 8.3A
So regardless of whether DC or AC watts is watts...




Take all the items that you want to power, and have a look at the ratings (should be a sticker or plate on the device). If they mention a wattage , use that. If they do not mention a wattage, take the current and multiply it by the voltage (eg 1A x 115VAC = 115W).

Take all the wattages and add them together, then multiply them by the amount of hours they will be used (this will give you WH). This WH figure is what you will use to calculate the size solar system you need.




The actual losses will be much higher than 10%. On top of inverter losses you have battery losses, controller losses, wiring losses, etc.
The actual figure you use will depend on the type of charge controller you use. For PWM add 100% to your required WH figure, and for MPPT add 50% to the required WH figure. This gives you a good ballpark to start from.



The inverter needs to be able to handle the peak power draw of all your equipment. So if all your equipment is mainly resistive (eg no motors), and the peak power demand is 400W, double up and go for a 1KW inverter. If you are using motors, due to their large startup power requirements, you will need an inverter rated at 6 - 10 times higher than your actual running power draw.




Your battery bank is sized according to your load and power requirements. The battery bank sizing is calculated by taking your required wattage and dividing by your battery bank voltage, then adjusting for discharge depth.

EXAMPLE:
You require 1000WH @ 12V....
1000WH / 12V = 83.33A
The maximum battery discharge depth for lead acid batteries should be no more than 50% DOD. This means we have to double our required battery capacity to achieve a 50% DOD. So...83.33A x 2 = 166.6A battery bank @ 12V. 166AH 12V is not a standard battery, so round up to whatever is the next closest capacity.

For a 30% DOD, you take your required capacity and divide by 30, then multiply by 100.
Example: 1000WH / 12V = 83.33A
83.33A / 30 = 2.78 x 100 = 277.77AH @ 12V (round up to next closest capacity).


The above is the basics to get you started....also, we will need to know the solar insolation for your area, as that determines the solar panel array size that is required.

awesome response. thank you for all the information and I will recalculate my needs. According to National Renewable Energy Laboratory, the solar insulation for my area is historically between 6 and 6.5 kWH/m^2/day.

[russ]

For a location in your general area - a 1 kW DC system - a bit more accurate accounting of the insolation available

4.76	113	18
4.82	104	16
5.73	137	22
6.81	157	25
5.93	141	22
6.32	144	23
6.75	157	25
6.59	151	24
5.79	128	20
4.85	114	18
5.12	114	18
4.75	112	18
5.68	1,571	$ 248

[daz]

awesome response. thank you for all the information

You are welcome. Glad to be able to help!

According to National Renewable Energy Laboratory, the solar insulation for my area is historically between 6 and 6.5 kWH/m^2/day.

That sounds a bit on the high side? Have a look at Russ's post above this, and calculate using those figures, as they seem to be more accurate. Rather work on lower insolation figures and have more power available, than the other way round, and then run into problems.

Also, always use the lowest insolation figure for your area, as this way you are covered for winter! In this case, according to Russ's table above, this would be December @ 4.75.

Let us know how your calculations go!

[warmonger0]

You are welcome. Glad to be able to help!



That sounds a bit on the high side? Have a look at Russ's post above this, and calculate using those figures, as they seem to be more accurate. Rather work on lower insolation figures and have more power available, than the other way round, and then run into problems.

Also, always use the lowest insolation figure for your area, as this way you are covered for winter! In this case, according to Russ's table above, this would be December @ 4.75.

Let us know how your calculations go!

Power Needs
Watt useage
Laptop – 110v x 2.5 = 275w
Iphone – 110v x 0.15 = 16.5w
Ipad – 110v x 0.15 = 16.5w
Arduino – 2.1v x 9 = 18.9w

275w +16.5w +16.5w +18.9w = 326.9W
326.9W x3 hours = 980.7Wh (3hours is the time I want to charge and that's the maximum load I anticipate is seeing all 3 devices charging at once and the arduino will essentially always be on. It's more likely that I'll only charge one at a time but I figured I would calculate bigger rather than smaller).
980.7Wh (all 4 items for 3 hours)
396.9w for 21 hours (for the rest of the time the Arduino is on)
1377.6WH for 24 hours(total wattage for 24 hour period)
1377.6wh x .5 = 688.8wh (50% for MPPT charge controller)
1377.6wh + 688.8wh = 2066.4wh (total wh needed to handle)
Peak power draw is 980.7wh
980.7wh x2 = ~ 2kwh so I need at least a 2kw inverter

Battery Bank Sizing
Required wattage/battery bank voltage
2066.4wh/12v = 172.2A
172.2A x2 for lead acid batteries with a depth discharge of no more than 50% is =
344A @ 12v.
34Ah/battery @ 6v
68Ah/2 in series @12v
So I need 12 batteries or 6 pairs in series, in parallel to give me 12v?

This feels like a lot of battery to run a laptop, iphone, ipad for 3 hours and an Arduino for 24h. Did I miscalculate something? I tried to show all my work so it’s easier to follow.

[daz]

Did I miscalculate something? I tried to show all my work so it’s easier to follow.

Yes, it looks like you have made a slight error with the calculations...

Power Needs
1377.6WH for 24 hours(total wattage for 24 hour period)

Ok, so this figure is your daily power consumption (1337.6WH)

1377.6wh x .5 = 688.8wh (50% for MPPT charge controller)
1377.6wh + 688.8wh = 2066.4wh (total wh needed to handle)

...and this is how much power you need to produce a day to cover battery charging, losses, etc (2066.4WH)

Peak power draw is 980.7wh
980.7wh x2 = ~ 2kwh so I need at least a 2kw inverter

You are on the right track here...now you will just need to see if you can get away with using a mod sine inverter, or if you have to use a true sine inverter...

Battery Bank Sizing
Required wattage/battery bank voltage
2066.4wh/12v = 172.2A
172.2A x2 for lead acid batteries with a depth discharge of no more than 50% is =
344A @ 12v.

This part is fine so far.....344.4AH capacity @ 12V for a DOD of 50%
For a DOD of the 30%, you would need a capacity of 574AH @ 12V.

34Ah/battery @ 6v
68Ah/2 in series @12v
So I need 12 batteries or 6 pairs in series, in parallel to give me 12v?

I am not sure what you are trying to do here? From what it looks like, you want to use 12 x 34A 12V batteries to make up the 345AH 12V.
If this is the case, then it is not recommended! Always aim to use the least amount of paralleled batteries, as it will make life easier, and the batteries will last longer than using lots of batteries paralleled together.

For example, why not just use two 6V 365AH batteries in series for going with the 50% DOD?

You can also parallel batteries to get your required capacity, but you there are a few rules that need to be followed to help balance them in the string. One of the rules is using identical interconnects, and connecting the charge/discharge connections at opposite ends (if you search the forum, you will find a few good posts on the advantages and disadvantages of parallel as well as helpful hints on installation).

So if you decided to parallel two batteries together...
You would need two 12V 200AH batteries for 50% DOD,
or for the 30% DOD, you would need four 6V 305AH (or whatever is your local common capacity)...the reason for four batteries is...two sets of two batteries in series, with them paralleled together.

Before you go ahead paralleling batteries together though, search the forum for the threads on paralleled batteries, as you will get a good idea of what is in store!

This feels like a lot of battery to run a laptop, iphone, ipad for 3 hours and an Arduino for 24h.

It is! Unfortunately on solar...the energy has to be stored somewhere!!....just wait till you see how much the batteries weigh!

[Mike90250]

Since this is a "trial" system, I'd go with a cheap pair of 200ah, 6v golf cart batteries and run them hard. They should last 2 years, and then you will have a better idea of what the system is capable of.

To really learn, chose a charge controller that can give you some data logging (monitor battery voltage and charge amps) and most of the small wall adapters have switching supplies in them now, which always run fine on sine wave power, I'd suggest the Morningstar SureSine 300w inverter with pretty low losses.

[PNjunction]

Laptop – 110v x 2.5 = 275w

You may be able to cut this requirement in half or possibly more. The ratings for laptops are usually for when the hard drives are spinning AND it is also recharging the battery. If you can operate without the battery so that it is not charging, so much the better. Yes, eventually battery charging will stop, but then you are sitting with a fully charged internal li-ion battery all the time - which is not really good for the laptop battery either. See if you can run the laptop without the battery.

Of course use power-savings features. The P3-Kill-A-Watt meter will show the wattage savings over this maximum 275w spec as you dial in as much power saving as you can tolerate on the laptop. This meter will also show how ineffective most screen-savers are, and that they can draw considerable power on their own for entertainment purposes. In this case, I just use the simple drop-to-black screensaver.