Renewables, or solar, hopefully explained in simple terms.
Feb 9, 2021 18:46:09 GMT 10
hd1340, Beno, and 1 more like this
Post by malewithatail on Feb 9, 2021 18:46:09 GMT 10
Renewables, or solar, hopefully explained in simple terms.
Im not going into the theory of how solar works, look it up if u are that interested. And as this is a very simplified explanation and may not be to the liking of a purist, get over it and write it yourself.
There is a concept that you really do need to think about, and that is voltage drop, or indirectly, efficiency.
The calculations are the same at 110 volts, 240 v or 1,000 volts, only the current changes, and hence the losses. Bear in mind that power (watts) = current {squared} X resistance. And, as the resistance is simply the load we need to run, we can simplify the expression to P=I (squared). A little more math's. Say your load is a refrigerator and it draws 220 watts, remember, voltage determines the current needed. At 110 v, 220 watts is 2 amps (Power=Amps X volts). At 240 volts, its slightly less than 1 amp. Remember that power (loss) is proportional to the current squared, 1/2 the current, 1/4 the losses. Yes, taking it further, at 500 volts, the current is 1/2 amp, so the losses are 1/16 those at 110 volts. Now do you see why we use high voltage for transmission ? The higher the voltage, the less loss there is, but the harder it is to stop things flashing over in big flashes, zaps and noises. The same efficiency "rule" applies when selecting a battery voltage, which we will get to.
Step 1-You need to decide what loads (appliances etc) that you wish to run from the system solar.
Simply list everything you want to run from the system, don't worry about small things like your mobile phone charger, or a rechargeable torch, the system, if properly designed, will run those without an issue, and the amount of time they will run per day.
Example
Laptop 100 watts 8 hour per day=800 watt hours per day
Refrig 200 watts 12 hours per day=2,400 watt hours per day
Freezer 150 watts 10 hours per day=1,500 watt hours per day
Lights (Led types)100 watts total 8 hours per day=800 watt hours per day
etc {fill in your own loads}
Total watt hours per day in this example=5,500 watt hours per day. This is at whatever mains voltage you have, be it 240, 110, 500 etc, doesn't matter as we have only considered the actual power. Remember, the current will vary according to voltage, but power is the load you need to run.
Im not too sure what an 110 v inverters efficiency is (look it up for the model u want to use, and remember, the smaller the load, the worse the efficiency, due to magnetic losses, etc within it). I will assume its around 90 % efficient average throughout the day. This means our actual load that we need to supply is now 5,500X 1.1 or 6,050 watt, approx. 6 Kw hours per day.
Our Australian Standard states that for loads less than 1,200 watt hrs per day, 12 volts is OK, under 4,800 watthrs, 24 volts is a good choice, and greater than 4.8 kwhrs, then32 or more commonly, (These figures are approx). 48 volts.
Step 2- Determine the size of battery you will, need to supply that load, with a reasonable (10 years +) lifespan.
The watt hours per day now need to be turned into amp hours, as battery's are rated in amps per hour, either at the 100 hour rate, or something similar. (there are some traps in battery ratings that we will discuss shortly).
6kwh at 12 volts is 6,000/12=500 amp hours per day, at 24 volts its 6,000/24 or 300 amp hours per day and at 48 volts its 150 amp hours per day. Remember our efficiency calculation, amp hours can also be related to battery efficiency and the higher the voltage, the lower the current and the less the loss, as a squared function. Therefore assuming to be 12 volts, ,at 24 volts is 1/4 the loss and 48 volts 1/16 etc. We now need to account for battery efficiency, and amp hour efficiency is different to watt hour efficiency, but the difference isnt to great in a new battery, and gets worse as the battery ages, so we will assume a battery efficiency of 90 % overall. Ie:You will need to put another 10% more in than what u take out. This assumes a lead acid type deep cycle battery, Car battery's are about 80 %, Lithium approaches 96%. Nicads or Nife types are less than 70% efficiency.
The amount of charge you have to put back into the battery is 110% what you took out, or in the case above, 6,600 watts, or at 12 volts, 550 amphrs (6600/12), at 24 v,275 amp hrs, and at 48 volts, 117.5 amp hrs.
This is the raw power that the battery's have to deliver daily.
Step 3-Determine the size (capacity) of the battery.
Once again I'm assuming a lead acid type as they are still the easiest to use. (Lithium tends to catch fire if a fault happens and has yet to prove itself under cycle conditions). This is where most come unstuck, as there are several ratings for a battery and you need to get the right one.
Once again, referring to the Australian Standard, it states that the system must have 5 days of autonomy, i.e.: it must be able to supply the daily load for a minimum period of 5 days with little or no sun. Your area may be different in relation to sunless days, but 5 days without sun here is doesn't happen very often.
At 12 volts, this is 2,772 amp hours, at 24 volts its 1,386 amp hours, and at 48 volts its 693 amp hours. Most people assume that they now look at the 100 hour (5 day) rating of the battery. But...This is a daily load, averaged over 5 days, so you need to select the battery size based on the 20 hour, or 24 hour, rating, multiplied by 5.
Example, a certain battery has a 100 hour rate of 1,100 amp hours, but its 20 hour rate is only 780 amp hours, so the quicker you draw power out, the lower the rating, conversely, the slower you draw it down, the more amp hrs are available, and for the battery above, its 120 hour rate is 1,275 amp hours.
Practically, for our example above, this means you must look for a battery with a 20 hour rate of at least 693 amp hours (for 48 volts system). I like to go one or two sizes bigger to allow for extra loads, such as wtshtf and the kids turn up with their families.
This 20 hour rating is the trap most, even professional installers, get caught out with, and it results in a system prematurely failing and the installer getting bitten on the arse with a potential warranty claim.
Step 4-Working out how many solar panels are needed.
This has a few traps in it as well. Obviously, the angle that the panel faces is quite important, and for maximum output, the sun needs to be at 90 degrees to the panels surface. There are obviously compromises to be made here as the suns position varies thorough the day, and from season to season the ideal angle for your solar panels can vary depending on your solar objectives. For example, if you are seeking to maximize solar generation during winter months, be it for an off grid system or higher energy consumption levels, you would ideally want to angle your panels at a greater tilt for maximum exposure to the low winter sun. The general rule of thumb is that panels are angled at the latitude angle, plus 15°, is best to maximize winter sun exposure. In reality though, your panels will more than likely be positioned at the angle of your rooftop, unless your roof is dead flat, as the cost to mount the panels at the optimum angle usually outweighs the benefits associated with additional solar generation.
So what percentage of solar energy do you lose from not having panels mounted at the optimum angle? Well, it depends based on the orientation and angle of the panels! Generally, if your roof pitch is anywhere between 10° – 15° either side of the latitude angle, you will only lose between 1 – 1.5% of the maximum solar output possible at your location.
If you do have a flat rooftop, do not despair, usually a mounting of some kind is installed so as to deflect rainwater and allow the rain to clean the panels. Dust and grime will reduce the generation capacity of your solar panels, so this is an important consideration in the design and install process. The added benefit is that the panels will then be angled, likely at around 10°, which will expose them to more sun and allow them to generate more energy.
Now, to some more maths, groan..........
There are two ways to calculate the number of panels needed, one is to use the amp hour calculation, the other is to use the watt hours, I like to use watt hours as solar panels are usually rated in peak watts.
Our daily example load is 6,600 watts, and the solar panels we want to buy are rated at 250 watts. Now for some tricky math's. We need to know how many peak sun hours occur at your location. This info should be available in the planning department of the local Council, but if it isn't, the CSIRO or National Weather Beauru should have it. In our location, peak sun hours are 6 in summer and 5 in winter, so we will use the winter figure to base our calculations on. 6,600 watts/5 = 1,320 watts of peak solar power needed. Dividing the power rating of each panel we come up with a figure of 1,320/250=5.2 panels. As its a bit harder to cope with 0.2 of a panel, we will assume 6.
How will they be arranged ? This depends on the type of regulator you will be using.
Step 5-Regulators.
This is perhaps the hardest part to get the head around, but Einstein once said that "a brain once expanded never returns to its previous state", so lets get going.
There are basically three types of modern regulators, disregarding the old relay operated types that thankfully have all but disappeared from the scene.
There is firstly the switching type that uses a semiconductor device to alternately switch the solar array to the battery, then disconnect it when the battery is charged, reconnecting to keep the battery topped up. This is a simple, cheap way of regulating, but might not fully charge a battery due to it switching off to soon as the battery voltage rises quickly in full sun, especially if you have a large array connected, or as the battery's age. There are not many of these types around today, but some regulators do allow this type of regulation on small systems as an option.
The most common type, and cheapest, is the PWM, or pulse width modulation type. This switches the solar panels to the battery on and off very quickly, 100's times per second, monitoring the battery voltage between switching's. This results in a better charge acceptance by the battery, and less ripple that can cause heating of the cells. but there is obviously some loss in the switching devices.
The newer type, and becoming more common, is the switching type of regulator.
These rely on following the maximum power point of the panel, which represents the bias potential at which the solar cell outputs the maximum net power. The MPP voltage can drift depending on wide range of variables including the irradiance intensity, device temperature, and device degradation. The MPP of a panel is usually stated in the specs, and is always lower than the maximum voltage of the panel. As solar cells are a constant voltage device , ie they output voltage when exposed to light, and the amount of power (amps) they output depends on the amount of illumination they receive, and as this changes through the day, MPPT regulators follow this point and change their switching parameters, sometimes 1,000's times a second. The advantage is that as the cells get hot and their output falls, the regulator can compensate and increase the current.
The other advantage of switching regulators is that they are able to switch a much higher input voltage, to the lower battery voltage, and remembering what was above re efficiency and how its related to voltage. Most of the Victron range of regulators can take an open circuit voltage of 150 v dc in, and switch to the battery voltage, say 48 volts. If you are using a nominal 24 volt panel, with a maximum open circuit voltage of around 46 volts, then this means that 3 panels can be connected in series, saving 2 sets wiring, conduit etc and gaining some extra output. Most modern systems these days uses switching regulators as they are much better than other types for other reasons as well. The disadvantage is that the input voltage is over 60 volts dc and under our electrical rules, unlicensed persons arnt allowed to work on it, and also if there is a wiring fault, a lot more energy is available to cause mischief than a single panel wired to a PWM regulator. There is a switching loss that gets worse as the illumination increases, but with modern semi conductor devices, the loss in efficiency is more than made up for in a gain due to being able to do MPP tracking.
There is another kind of regulator that is seldom used today, except in very small systems, and that's the Linear regulator. It uses a semi conductor to limit the voltage into the battery's, and it is very inefficient and tends to get very hot under load as its passing all the current from the panel to the battery's. I've designed and used several of these in small systems, such as that which charges the small battery that opens our chook trailer doors in the mornings to let the little dears out at dawn. Efficiency isnt an issue here as they are fitted with 80 watt 12 volt panels and the motor operates for about 3 seconds twice a day, not particularly onerous duty and plenty of time to charge back up, even in rain.
A typical 250 watt panel, operating at a MPPT voltage of typically, 37 volts, will deliver about 6.7 amps, so 6 of them will deliver 40 amps, at 24 volts, but we are doing it at 150 volts, so can use 3 panels in series, reducing the number of parallel groups to 2, instead of 6, and the input current to 13 amps, saving cable, conduit and time. And gaining a more efficient system as well. Check that you wont exceed the regulator rating in full sun, as obviously that's when the maximum power will be delivered.
I don't have a solution, but I admire the problem.
Mash forehead on keyboard to continue.
Im not going into the theory of how solar works, look it up if u are that interested. And as this is a very simplified explanation and may not be to the liking of a purist, get over it and write it yourself.
There is a concept that you really do need to think about, and that is voltage drop, or indirectly, efficiency.
The calculations are the same at 110 volts, 240 v or 1,000 volts, only the current changes, and hence the losses. Bear in mind that power (watts) = current {squared} X resistance. And, as the resistance is simply the load we need to run, we can simplify the expression to P=I (squared). A little more math's. Say your load is a refrigerator and it draws 220 watts, remember, voltage determines the current needed. At 110 v, 220 watts is 2 amps (Power=Amps X volts). At 240 volts, its slightly less than 1 amp. Remember that power (loss) is proportional to the current squared, 1/2 the current, 1/4 the losses. Yes, taking it further, at 500 volts, the current is 1/2 amp, so the losses are 1/16 those at 110 volts. Now do you see why we use high voltage for transmission ? The higher the voltage, the less loss there is, but the harder it is to stop things flashing over in big flashes, zaps and noises. The same efficiency "rule" applies when selecting a battery voltage, which we will get to.
Step 1-You need to decide what loads (appliances etc) that you wish to run from the system solar.
Simply list everything you want to run from the system, don't worry about small things like your mobile phone charger, or a rechargeable torch, the system, if properly designed, will run those without an issue, and the amount of time they will run per day.
Example
Laptop 100 watts 8 hour per day=800 watt hours per day
Refrig 200 watts 12 hours per day=2,400 watt hours per day
Freezer 150 watts 10 hours per day=1,500 watt hours per day
Lights (Led types)100 watts total 8 hours per day=800 watt hours per day
etc {fill in your own loads}
Total watt hours per day in this example=5,500 watt hours per day. This is at whatever mains voltage you have, be it 240, 110, 500 etc, doesn't matter as we have only considered the actual power. Remember, the current will vary according to voltage, but power is the load you need to run.
Im not too sure what an 110 v inverters efficiency is (look it up for the model u want to use, and remember, the smaller the load, the worse the efficiency, due to magnetic losses, etc within it). I will assume its around 90 % efficient average throughout the day. This means our actual load that we need to supply is now 5,500X 1.1 or 6,050 watt, approx. 6 Kw hours per day.
Our Australian Standard states that for loads less than 1,200 watt hrs per day, 12 volts is OK, under 4,800 watthrs, 24 volts is a good choice, and greater than 4.8 kwhrs, then32 or more commonly, (These figures are approx). 48 volts.
Step 2- Determine the size of battery you will, need to supply that load, with a reasonable (10 years +) lifespan.
The watt hours per day now need to be turned into amp hours, as battery's are rated in amps per hour, either at the 100 hour rate, or something similar. (there are some traps in battery ratings that we will discuss shortly).
6kwh at 12 volts is 6,000/12=500 amp hours per day, at 24 volts its 6,000/24 or 300 amp hours per day and at 48 volts its 150 amp hours per day. Remember our efficiency calculation, amp hours can also be related to battery efficiency and the higher the voltage, the lower the current and the less the loss, as a squared function. Therefore assuming to be 12 volts, ,at 24 volts is 1/4 the loss and 48 volts 1/16 etc. We now need to account for battery efficiency, and amp hour efficiency is different to watt hour efficiency, but the difference isnt to great in a new battery, and gets worse as the battery ages, so we will assume a battery efficiency of 90 % overall. Ie:You will need to put another 10% more in than what u take out. This assumes a lead acid type deep cycle battery, Car battery's are about 80 %, Lithium approaches 96%. Nicads or Nife types are less than 70% efficiency.
The amount of charge you have to put back into the battery is 110% what you took out, or in the case above, 6,600 watts, or at 12 volts, 550 amphrs (6600/12), at 24 v,275 amp hrs, and at 48 volts, 117.5 amp hrs.
This is the raw power that the battery's have to deliver daily.
Step 3-Determine the size (capacity) of the battery.
Once again I'm assuming a lead acid type as they are still the easiest to use. (Lithium tends to catch fire if a fault happens and has yet to prove itself under cycle conditions). This is where most come unstuck, as there are several ratings for a battery and you need to get the right one.
Once again, referring to the Australian Standard, it states that the system must have 5 days of autonomy, i.e.: it must be able to supply the daily load for a minimum period of 5 days with little or no sun. Your area may be different in relation to sunless days, but 5 days without sun here is doesn't happen very often.
At 12 volts, this is 2,772 amp hours, at 24 volts its 1,386 amp hours, and at 48 volts its 693 amp hours. Most people assume that they now look at the 100 hour (5 day) rating of the battery. But...This is a daily load, averaged over 5 days, so you need to select the battery size based on the 20 hour, or 24 hour, rating, multiplied by 5.
Example, a certain battery has a 100 hour rate of 1,100 amp hours, but its 20 hour rate is only 780 amp hours, so the quicker you draw power out, the lower the rating, conversely, the slower you draw it down, the more amp hrs are available, and for the battery above, its 120 hour rate is 1,275 amp hours.
Practically, for our example above, this means you must look for a battery with a 20 hour rate of at least 693 amp hours (for 48 volts system). I like to go one or two sizes bigger to allow for extra loads, such as wtshtf and the kids turn up with their families.
This 20 hour rating is the trap most, even professional installers, get caught out with, and it results in a system prematurely failing and the installer getting bitten on the arse with a potential warranty claim.
Step 4-Working out how many solar panels are needed.
This has a few traps in it as well. Obviously, the angle that the panel faces is quite important, and for maximum output, the sun needs to be at 90 degrees to the panels surface. There are obviously compromises to be made here as the suns position varies thorough the day, and from season to season the ideal angle for your solar panels can vary depending on your solar objectives. For example, if you are seeking to maximize solar generation during winter months, be it for an off grid system or higher energy consumption levels, you would ideally want to angle your panels at a greater tilt for maximum exposure to the low winter sun. The general rule of thumb is that panels are angled at the latitude angle, plus 15°, is best to maximize winter sun exposure. In reality though, your panels will more than likely be positioned at the angle of your rooftop, unless your roof is dead flat, as the cost to mount the panels at the optimum angle usually outweighs the benefits associated with additional solar generation.
So what percentage of solar energy do you lose from not having panels mounted at the optimum angle? Well, it depends based on the orientation and angle of the panels! Generally, if your roof pitch is anywhere between 10° – 15° either side of the latitude angle, you will only lose between 1 – 1.5% of the maximum solar output possible at your location.
If you do have a flat rooftop, do not despair, usually a mounting of some kind is installed so as to deflect rainwater and allow the rain to clean the panels. Dust and grime will reduce the generation capacity of your solar panels, so this is an important consideration in the design and install process. The added benefit is that the panels will then be angled, likely at around 10°, which will expose them to more sun and allow them to generate more energy.
Now, to some more maths, groan..........
There are two ways to calculate the number of panels needed, one is to use the amp hour calculation, the other is to use the watt hours, I like to use watt hours as solar panels are usually rated in peak watts.
Our daily example load is 6,600 watts, and the solar panels we want to buy are rated at 250 watts. Now for some tricky math's. We need to know how many peak sun hours occur at your location. This info should be available in the planning department of the local Council, but if it isn't, the CSIRO or National Weather Beauru should have it. In our location, peak sun hours are 6 in summer and 5 in winter, so we will use the winter figure to base our calculations on. 6,600 watts/5 = 1,320 watts of peak solar power needed. Dividing the power rating of each panel we come up with a figure of 1,320/250=5.2 panels. As its a bit harder to cope with 0.2 of a panel, we will assume 6.
How will they be arranged ? This depends on the type of regulator you will be using.
Step 5-Regulators.
This is perhaps the hardest part to get the head around, but Einstein once said that "a brain once expanded never returns to its previous state", so lets get going.
There are basically three types of modern regulators, disregarding the old relay operated types that thankfully have all but disappeared from the scene.
There is firstly the switching type that uses a semiconductor device to alternately switch the solar array to the battery, then disconnect it when the battery is charged, reconnecting to keep the battery topped up. This is a simple, cheap way of regulating, but might not fully charge a battery due to it switching off to soon as the battery voltage rises quickly in full sun, especially if you have a large array connected, or as the battery's age. There are not many of these types around today, but some regulators do allow this type of regulation on small systems as an option.
The most common type, and cheapest, is the PWM, or pulse width modulation type. This switches the solar panels to the battery on and off very quickly, 100's times per second, monitoring the battery voltage between switching's. This results in a better charge acceptance by the battery, and less ripple that can cause heating of the cells. but there is obviously some loss in the switching devices.
The newer type, and becoming more common, is the switching type of regulator.
These rely on following the maximum power point of the panel, which represents the bias potential at which the solar cell outputs the maximum net power. The MPP voltage can drift depending on wide range of variables including the irradiance intensity, device temperature, and device degradation. The MPP of a panel is usually stated in the specs, and is always lower than the maximum voltage of the panel. As solar cells are a constant voltage device , ie they output voltage when exposed to light, and the amount of power (amps) they output depends on the amount of illumination they receive, and as this changes through the day, MPPT regulators follow this point and change their switching parameters, sometimes 1,000's times a second. The advantage is that as the cells get hot and their output falls, the regulator can compensate and increase the current.
The other advantage of switching regulators is that they are able to switch a much higher input voltage, to the lower battery voltage, and remembering what was above re efficiency and how its related to voltage. Most of the Victron range of regulators can take an open circuit voltage of 150 v dc in, and switch to the battery voltage, say 48 volts. If you are using a nominal 24 volt panel, with a maximum open circuit voltage of around 46 volts, then this means that 3 panels can be connected in series, saving 2 sets wiring, conduit etc and gaining some extra output. Most modern systems these days uses switching regulators as they are much better than other types for other reasons as well. The disadvantage is that the input voltage is over 60 volts dc and under our electrical rules, unlicensed persons arnt allowed to work on it, and also if there is a wiring fault, a lot more energy is available to cause mischief than a single panel wired to a PWM regulator. There is a switching loss that gets worse as the illumination increases, but with modern semi conductor devices, the loss in efficiency is more than made up for in a gain due to being able to do MPP tracking.
There is another kind of regulator that is seldom used today, except in very small systems, and that's the Linear regulator. It uses a semi conductor to limit the voltage into the battery's, and it is very inefficient and tends to get very hot under load as its passing all the current from the panel to the battery's. I've designed and used several of these in small systems, such as that which charges the small battery that opens our chook trailer doors in the mornings to let the little dears out at dawn. Efficiency isnt an issue here as they are fitted with 80 watt 12 volt panels and the motor operates for about 3 seconds twice a day, not particularly onerous duty and plenty of time to charge back up, even in rain.
A typical 250 watt panel, operating at a MPPT voltage of typically, 37 volts, will deliver about 6.7 amps, so 6 of them will deliver 40 amps, at 24 volts, but we are doing it at 150 volts, so can use 3 panels in series, reducing the number of parallel groups to 2, instead of 6, and the input current to 13 amps, saving cable, conduit and time. And gaining a more efficient system as well. Check that you wont exceed the regulator rating in full sun, as obviously that's when the maximum power will be delivered.
I don't have a solution, but I admire the problem.
Mash forehead on keyboard to continue.
