Deleted

ABrass

Your advice is only valid* for a very narrow target audience (off grid systems) and you don't understand the numbers you're using.

If you're trying to work things out based on instantaneous power then you're going to keep getting it wrong. Especially if you have a battery.

I can believe that a 2kW array might only generate 1.5kWh on a particular day in October. Saying it generated 1500W that day is meaningless, or wrong.

*arguably valid

ABrass

You are describing watt hours and kWh. You may not have typed in the correct units but that is what you're trying to describe.

30 watt current draw (instantaneous), running for an hour is 30wH (0.03kWh). 24 of those gives 720wH or 0.72kWh.

ABrass

Deleted_User said:

This seems to be the point of contention - yes?

KWh = KW x Hours

So take a theoretical example where a solar array generates 1kw of power an hour over a period of 4 hours.

You could correctly say either of these two things:

1. The array produced 4kwh of energy (so kw x 4 hours = 4kwh)
2. The array produced 4kw of power in a 4 hour period.

I'm sticking to power in -> power out to illustrate the impact of using your system efficiently. To my mind converting to KWh complicates things and is likely to lead to incorrect results when doing calculations:

1. On an average December day my 2kw array will generate a TOTAL 500 watts of power. 
2. That could equally be expressed as 0.5 kwh of energy
3. However, given that KWh = KW x hours then it follows that you would be able to calculate how much power the panels produced per hour using the formula kw = KWh / time. So kw = ( 0.5 / 24 = 20 watts). Which is definitely incorrect as the sun is only up for a few hours in winter!

So  that's why I've missed out the conversion to KWh. Instead I'm working with power and explicitly expressing the time component.

You wrote:

"30 watt current draw (instantaneous), running for an hour is 30wH (0.03kWh). 24 of those gives 720wH or 0.72kWh."

But also if a product consumes 30 watts an hour for 24 hours it consumes 720 watts of power.

And I'm interested in power in - > power out. (or more importantly the difference between the two).

But setting all that aside - do you understand what I'm saying? Or do you want me to explain by converting to KWh then back to KW?

No, you can't say the array generated 4kW over a period of 4 hours, because that would actually be 16kWh.

These are not units invented by this forum, these are scientific and engineering terms that have clearly defined meanings.

As to your 0.5kWh example, you probably had 4 hours generating energy and it would have averaged at 125W over that time, but in reality it would have started at zero then ramped up to 200W and then ramped down.

Martyn1981

Deleted_User said:

This seems to be the point of contention - yes?

KWh = KW x Hours

So take a theoretical example where a solar array generates 1kw of power an hour over a period of 4 hours.

You could correctly say either of these two things:

1. The array produced 4kwh of energy (so kw x 4 hours = 4kwh)
2. The array produced 4kw of power in a 4 hour period.

The first is correct, stating the amount of energy, the second isn't as it talks about power.

The second suggests either that the max power reached during the 4hr period was 4kW, or that perhaps 16kWh of energy was produced. The second does not really explain the amount of energy, without some additional info.

Much better to use energy when talking about energy (Wh's) and power when talking about power (W's).

But also if a product consumes 30 watts an hour for 24 hours it consumes 720 watts of power.

Nope. At no point does it consume 720W's (power) if it's consuming 30W's (power), but it will consume 720Wh's (energy) over a 24hr period.




Quote from the American comedian Steven Wright - I was going 70 miles an hour and got stopped by a cop who said, “Do you know the speed limit is 55 miles per hour?” “Yes, officer, but I wasn’t going to be out that long…”

70sbudgie

If I have understood your point correctly, this is how my system works. It is a different set up, so the detail is different, but the principles that I believe you are trying to share, are the same.

I have a 4.35kWp E-W solar array, grid tied without batteries. 

On this sunny November day, my panels have so far generated 4.4kWhrs, but have peaked at 900W. If my inverter is 95% efficient then I would have had a maximum of 855W to power my appliances.

The efficiency of my inverter is irrelevant to me as every kWhr that comes from my PV system is free energy as far as I am concerned. However, I do like to optimise the use of my solar panels. 

Ways that I can do that include not running high load appliances at the same time. I have the option to operate my microwave at 1000W, 600W or 300W and choose according to the output of my panels. Operating the dishwasher at the same time as the washing machine exceeds the output of the panels. I don't know the exact figures, but lets say the dishwasher peaks at 800W and the washing machine at 1.2kW. I'm never going to cover the washing machine during a November day, but I can get perhaps 55W of its peak power for free if I run it at the same time as the dishwasher. If I am really clued up about the operating cycles of my appliances, I can time it so their peaks don't coincide.

Another way I optimise my panels is from their E-W orientation. My inverter is 3.6kW, but this summer my generation peaked at 2.7kW. But I had enough generation for my import to be zero from about 8am to about 18:45. (On that particular day, I generated 22.9kWhrs, over nearly 18 hours - I generated less than my baseload for a long time)

My base load (fridge, freezer, tv box, children's nightlight etc) is about 150W. So this means that from about 8am my solar output, including inverter losses is at least 150W.

This is also the same principle that means that E7 is currently more cost effective for me than Octopus Go. The extra 3 hours more than make up for the slightly higher rate.

Going back to your original point, yes there probably is a diminishing rate of return for increasing array size. However, that is usually offset by economies of scale. Increasing the storage capacity of batteries (which uses units of kWhrs) is a way to mitigate against too much generation wastage. It is also worth being aware of the charge rate of batteries (measured in kW) when optimising a system with energy storage.

ABrass

Deleted_User said:

ABrass said:

Deleted_User said:

This seems to be the point of contention - yes?

KWh = KW x Hours

So take a theoretical example where a solar array generates 1kw of power an hour over a period of 4 hours.

You could correctly say either of these two things:

1. The array produced 4kwh of energy (so kw x 4 hours = 4kwh)
2. The array produced 4kw of power in a 4 hour period.

I'm sticking to power in -> power out to illustrate the impact of using your system efficiently. To my mind converting to KWh complicates things and is likely to lead to incorrect results when doing calculations:

1. On an average December day my 2kw array will generate a TOTAL 500 watts of power. 
2. That could equally be expressed as 0.5 kwh of energy
3. However, given that KWh = KW x hours then it follows that you would be able to calculate how much power the panels produced per hour using the formula kw = KWh / time. So kw = ( 0.5 / 24 = 20 watts). Which is definitely incorrect as the sun is only up for a few hours in winter!

So  that's why I've missed out the conversion to KWh. Instead I'm working with power and explicitly expressing the time component.

You wrote:

"30 watt current draw (instantaneous), running for an hour is 30wH (0.03kWh). 24 of those gives 720wH or 0.72kWh."

But also if a product consumes 30 watts an hour for 24 hours it consumes 720 watts of power.

And I'm interested in power in - > power out. (or more importantly the difference between the two).

But setting all that aside - do you understand what I'm saying? Or do you want me to explain by converting to KWh then back to KW?

No, you can't say the array generated 4kW over a period of 4 hours, because that would actually be 16kWh.

These are not units invented by this forum, these are scientific and engineering terms that have clearly defined meanings.

As to your 0.5kWh example, you probably had 4 hours generating energy and it would have averaged at 125W over that time, but in reality it would have started at zero then ramped up to 200W and then ramped down.

A 1kw solar array will generate 1kw of power per hour, therefore over 4 hours it will generate 4kw of power.

KWh = KW x hours

KW = KWh / time

You can calculate KWh by multiplying KW (1kw) x Time (4 hours) = 4kwh

In this example you can also calculate KW by dividing KWh (4kwh) / Time (4 hours) = 1kw.

But in the example we're talking about where we know 500 watts of power was generated over a 24 hour period it's not strictly correct to refer to it as KWh. This is because you don't know the time element of the formula.  

You're partially right in what you're saying, but KWh become meaningless for calculations without a time reference. 

If you won't believe me, have a read at this article and it may become clearer:

"Energy consumption expressed in terms of kWh doesn't often mean much unless you also know the length of the period that the kWh were measured over. And it's difficult to make fair comparisons between kWh figures unless they are all from periods of exactly the same length. Figures expressed in terms of power (e.g. kW) make many things more straightforward..."

https://www.energylens.com/articles/kw-and-kwh

A 1kW array is not going to generate 1kW over winter at any point.

Yesterday my 8kW array generated 10kWh, it had a peak power of 3kW.

ABrass

This is really a thread about batteries and when to use them btw. The PV is almost irrelevant.

ABrass

Deleted_User said:

ABrass said:

This is really a thread about batteries and when to use them btw. The PV is almost irrelevant.

No. The thread is about diminishing returns on investment from increasing the size of your solar array.

Its also about checking POWER IN from your solar array and POWER OUT via your inverter. If OUT is much smaller than IN you need to look at why. 

Reason being that if you have an efficiency problem throwing more solar panels at it won't necessarily improve things. In the example I gave above, doubling the number of panels HALVED the usable power from solar. 

The thing about solar is it is unpredictable. Houses are wired in a way that let us have energy on demand. But solar isn't like that. If you try to make solar seemlessly merge with your mains system then your solar is going to be working very inefficiently.

Solar systems need to be used very differently from mains. Suppliers of home back up systems like to advertise it as being seamless - and it may appear that way. But look under the hood and it's very wasteful. 

You're wrong in your units, your method and unsurprisingly your conclusions.

For example, if you double your PV then any fixed losses from the inverter become a smaller fraction of the total power. You get more than double the usable power.

You also keep ignoring your battery in your descriptions and talking about PV. But with a battery you have the 10% round trip losses involved as well.

And I'm still not sure you understand how PV power actually varied through the day. How many kWh did you generate yesterday?

Magnitio

ABrass said:

Deleted_User said:

ABrass said:

This is really a thread about batteries and when to use them btw. The PV is almost irrelevant.

No. The thread is about diminishing returns on investment from increasing the size of your solar array.

Its also about checking POWER IN from your solar array and POWER OUT via your inverter. If OUT is much smaller than IN you need to look at why. 

Reason being that if you have an efficiency problem throwing more solar panels at it won't necessarily improve things. In the example I gave above, doubling the number of panels HALVED the usable power from solar. 

The thing about solar is it is unpredictable. Houses are wired in a way that let us have energy on demand. But solar isn't like that. If you try to make solar seemlessly merge with your mains system then your solar is going to be working very inefficiently.

Solar systems need to be used very differently from mains. Suppliers of home back up systems like to advertise it as being seamless - and it may appear that way. But look under the hood and it's very wasteful. 

You're wrong in your units, your method and unsurprisingly your conclusions.

For example, if you double your PV then any fixed losses from the inverter become a smaller fraction of the total power. You get more than double the usable power.

You also keep ignoring your battery in your descriptions and talking about PV. But with a battery you have the 10% round trip losses involved as well.

And I'm still not sure you understand how PV power actually varied through the day. How many kWh did you generate yesterday?

Ever feel like you're banging your head against a brick wall?

ABrass

Your point is that you should turn your inverter off when it's dark or when there are clouds. Fine, it's an interesting concept. Utterly impractical but maybe it will save you a few joules.

Then you should buy a battery, work out the maximum output of that and the inverter and run everything off the battery at once. Fine, utterly impractical but maybe it'll save you a few joules.

If your time is worthless then it might make sense. And it completely ignores the normal situation where people get paid for export or have no batteries. It's advice for a very niche use case