Solar calculations for beginners

The start of the "Roger Writes" series - Jan 2022

With increasing home energy prices, the idea of moving your house/cabin/shed over to Solar power has a lot of appeal.

Often on the forums, you see a lot of "how much solar do I need to move my house off-grid?" type questions. This is really like saying "What car should I buy?", without a lot of addition information it's impossible to say.

So what should you be thinking about?

Power - How much per day? What's the peak power you will need? (Is it a motor/compressor, or a non-inductive load?)
Sunlight - Where you are in the world, what's the average number of sunlight hours per day?
Storage - How much can you use during daylight? How much will be during darkness? Do you need to allow for Winter dark days?

Units

Some people get angry over the wrong units, or the capitalisation, and most predictive text makes it difficult to get right!
The main things we will talk about are:

Watts - 'W' - Named after 'James Watt', so always should be capital 'W'
Volts - 'V' - Named after 'Alessandro Volta', so always should be capital 'V'
Amps - 'A' - Named after 'André-Marie Ampère', so always should be capital 'A'
Ohms - 'Ω'
Hours - 'h' - should always be a lower case, capital 'H' is Henrys, which is a measurement of inductance.
Kilo - 'k' - should be lower case, capital 'K' is Kelvin, which is a measurement of temperature.

What do they mean and how are they used:

Watts or kilo Watts(1kW = 1000W) - Used for instantanious power. Typically used to measure the power from a solar panel, charging a battery, or powering a device.
Volts - Used for potential. Typically from a solar panel, or a battery bank.
Amps - Used for current flow. Typically the current going in or out of a device.
Ohms - Used for resistance. Typically cable resistance, or battery/solar panel internal resistance.
Watt hours - Wh (or kilo Watt hour kWh, 1kWh = 1000Wh) - Used for energy. Typically used to measure battery capacity, power from a solar panel over a time period, power require for a device in a certain time.

Ohms law states: Volts = Amps x Ohms, and Watts = Volts x Amps
From which we can derrive:
Volts=Watts/Amps
Amps=Watts/Volts
Watts=Amps x Amps x Ohms

Examples

Lets start by assuming that everything is 100% effecient, then add some margin at the end....

Example #1

We've got a 10W LED bulb, that's on for half of the day, how much energy will it need? 10W x 24 hours x 50% = 120Wh
So how big a panel will I need to power it, if I average 2 hours of sunlight per day? 120Wh/2h=60W panel.
Assuming you're not using your LED bulb during daylight hours, you will need to store the energy. If we assume that 100% is used in non-daylight hours, then it would need 120Wh of storage.
Typically batteries are in multiples of 12V, for a small system like this, 1 battery will be enough, so 1 battery of 12V, so 120Wh/12V=10Ah.
So for a 10W bulb, on a 12V system, we would need at least a 10Ah battery, and 60W solar panel.

Example #2

We've got a 1000W water pump, that's on for 10 minutes per hour, how much energy will it need? 1000W x 24 hours x (10 minutes / 60 minutes) = 4000Wh
So how big a panel will I need to power it, if I average 2 hours of sunlight per day? 4000Wh/2h=2000W of panels.
Assuming you'll want this to run outside of daylight, then you will need to store the energy. If we assume that 75% is used in non-daylight hours, then it would need 4000Wh x 75% = 3000Wh of storage.
For this sort of power, you would go to 24V (or higher), so 2x12V battery in series. 3000Wh/24V=125Ah
We would need at least 3000Wh=2x 125Ahi@12V batteries, and 2000W of solar panels.
This would probably be an AC motor, so the startup power will be at least 2x, so we would need at least a 2000W inverter.

Margins

Do you need to allow dark/overcast day when there will be next to no solar? If so you will need 2x the battery capacity. It is also common to add more solar to catch up for the darker days.
How often do you want to replace the batteries? If you charge/discharge to 100% then batteries won't last very long (Lead acid might survive as few as 160 cycles to 100% Depth of Discharge, Lithium might be 1000 cycles). If you reduce the depth of discharge (by adding more batteries), then you can significantly increase the battery life. For some lead-acid, if you have 2.5x the battery capacity (so 40% DoD), then you increase the battery lifetime by about 4x.
Solar chargers. PWM controllers are only about 70% efficient, so you will need to add more solar to get the required solar output. MPPT controllers are about 97%+ effecient, so for larger systems, this can make a big difference.
If you can power the device from DC, then direct to battery (for 12V/24V devices) is 100% effecient, if it's another DC voltage, then DC:DC convertors, can be 95%+ effecient. Otherwise if it's AC only, then you'll need an inverter, cheap ones aren't very effecient, and can have significant standby power consumptions. More expensive ones, are more effecient, but don't expect much more than 70...80%.
Wiring losses, battery heating, solar panel angle to sun, and lots of other factors exist, so this will just be a baseline

So for our two examples:
A 10W bulb, on a 12V system, we would need at least a 10Ah battery, and 60W solar panel. Assuming we want to allow for one dark day (Battery 2x), and recharge in a reasonable time (Solar panel 1.5x) and we want our battery to last a long time (Battery 3x), and we'll be using a cheap PWM controller (solar 1.5x), then our battery would need to be at least 60Ah, and the solar panel at least 135W.
For a 1000W pump that's on 10 minutes per hour, then we start with 3000Wh=2x125Ah@12V batteries, and 2000W of solar panels. But if we need to allow for 2 dark days (Battery 3x), and recharge in a reasonable time (Solar Panel 1.5x), and we want our batteries to last (Battery 2x), we'll be using a MPPT controller (so solar + 5%), but using a 75% effecient inverter (battery and solar +25%), we would actually need 25200Wh=12x175Ah@12V of battery and 4200W of solar!

Why 24V or 48V

12V kit is unversal and cheap, so why go to the bother of 24V (or higher)?
Cable size/cost is the main reason. For the same power, if you double the voltage, then you half the amps (as Power=Volts x Amps), Power = Amps x Amps x Ohms, so if the cabling is the same, then the power lost in the cable drops to 25%. Or for the same power loss you can use cable with 4x the resistance, so it can be a lot thinner. Many countries have a "Safe working voltage" which is usually 50V or 60V, above this you need to have special handling or protection, so 48V is a good compromise for this.
For solar, the same thing applies, so putting several panel in series, and reducing the current, can significantly reduce the solar cabling cost.

The Roger Writes series

I research / dabble with lots of things, and figured that if I write my notes here, I can quickly reference them, also, sometimes, they are useful to others!
Here is what I have so far:

Cosplay Arduino programming (RW#13 - June 2024)
ATU-100 modifications (RW#12 - February 2024)
Garage door open warning (RW#11 - January 2024)
Horizontal Halos and Vs (RW#10 - December 2023)
Faking Dell PSUs (RW#9 - December 2023)
Designing NEC2 antennas (RW#8 - November 2023)
4NEC2 antenna models (RW#7 - October 2023)
3-element cubical quad for 2m (RW#6 - September 2023)
Cobwebb Antenna dimensions (RW#5 - August 2023)
WS2812 and similar 'Smart' LEDs (RW#4 - January 2023)
Investigation and details of the AFHDS-2A protocol (RW#3 - June 2022)
Changes we've made to our caravan (RW#2 - January 2022)
Basic solar/battery/off-grid calculations (RW#1 - January 2022)
Credit card size walking robot aka Monster-Chan (RW#0 - Originally 2016 - Moved here 2022)
Why you should avoid certain number of drives with ZFS (aka why isn't my ZFS/raidz array the size I expected)
Simple way to connect a Servo to a PC (RS232 port)
Using ZFS with FreeBSD

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This page was lasted updated on Thursday, 17-Aug-2023 12:40:28 BST