I further, I see how efficiency is NOT a factor. Your factor of 0.9/STC_irradiance is essentially a conversion from 6.8kWh/(m**2*day) to effective full sun hours/day.
No, I see my mistake. Reading formulas in text boxes is an error trap, I should have double checked. Your units work out, my apologies.
So if you are saying your 0.9 factor along with the/STC_irradiance value does account for conversion efficiency, and also the difference between the value of 6.8kWh/(m**2*day) and hours/day equivalency, then I can see how your formula can work. In which case, efficiency does matter. Your insistence that conversion efficiency was not a factor was a point of confusion.
I see what you did now, my apologies. Thanks for staying with me on it.
^NO. Check your units. Your answer is actually 22 kWH-cubed/day. Which makes no sense.
That is because you have W rated * irradiance * irradiance, all three with W as a unit numerator... so you get Watts cubed.. But the 1000W/m2 radiance is a lab test value for standardizing comparisons, it is not applicable in this analysis, and like I said the 6.8 kWh/m2.day doesn't apply.
Like I said, units never lie. To get kWh instead of kWh-cubed, you'll need to divide somewhere in your conversion by a kW factor twice.
6.8kWh/m2.day represents the available power of the sunlight hitting a square meter of surface in a day, on average.
To obtain KWH electrical per day using 6.8kWh/m2.day, you must;
1) Convert solar to electrical (0.15 efficiency, or whatever you assume the solar panel conversion efficiency is) to get kWh-e/m2.day
2) Multiply by the surface area of the panel. to get kWh-e/day terminal output
3) Factor any electrical losses from panel output to load. kWh-e/day system output.
The 3.6 KW-e/Panel is a panel rating. If you start with panel rating, you can
1) Multiply by effective full sun hours per day to get KWh-e/Day/panel (this is typically about 5 to 7 hours/day)
2) Multiply by the number of panels for total KWh-e/Day terminal
3) Apply losses to get kWh-e/day system output.
In this case, you can combine 2 and 3 in your "90%" factor if you want, but you cannot use the 6.8kWh/m2.day, it does not apply, as it is surface area dependent.
Part of the confusion is due to the calculation in your original post by erroneously and needlessly calculating "2250kWh/(kWp.y) of electricity." Which made now sense since we were looking for output/day and we already had units in days. Then you switched method entirely and started with a panel rating you found by googling, and just calculated straight from that value with really documenting it very well. If somehow you used 6.8kWh/m2.day, then your units are wrong, and your number is wrong. It looks reasonable purely out of luck.
If you don't follow, just show your calculation including all units in factors and results for each step. Units never lie.
That is utterly ridiculous. You just changed the size to compensate for the change in efficiency. That is totally meaningless, and not related to the original calculation, which was based on an assumed carport size (area). If you assume a size (area), then efficiency matters. If you assume it can be any size, then you can't calculate the output.
Simply put, if you use 6.8kWh/m2/day, the only way you can calculate the electrical output is to use the conversion efficiency, which change the units to electrical, and to multiply by area to cancel out the square meter in the denominator of the value.
If you ignore panel area and just assume a panel rating, then the 6.8kWh/m2/day is not what you use in your calculation. Its quite easy to see just by following units.
You must assume cell efficiency if you are calculating based on the 6.8kWh/(m.day) extracted from the map/chart. It is absolutely relevent, hence the error in
6.8kWh/(m.day) in Arizona on a tilted plane gives you about 2500kWh/(m.y) With a performance ratio of 90% for your PV installation, you can get 2250kWh/(kWp.y) of electricity.
Which is more correctly 0.15 x 2500kWh/(m.y) = 375 kWh/(m.y)
You can ignore that method, and begin with a different starting point and instead calculate from a given cell rating, then the 90% performance ratio makes sense and yes, physical size does not matter, only the equivalent full solar hours. Which often ranges between 5 and 7, but is different from the 6.8kWh/(m.day) from the chart. You just got lucky that 6.8 was still a reasonable value.
I agree with that, so lets just clarify; You stated;
6.8kWh/(m.day) in Arizona on a tilted plane gives you about 2500kWh/(m.y)
With a performance ratio of 90% for your PV installation, you can get 2250kWh/(kWp.y) of electricity.
I see that as 6.8 x 365 = 2500. 2500*0.9=2250
6.8kWh/(m.day) = solar irradiance/(m.day)
You seem a assume 20% cell efficiency is already factored into 6.8kWh/(m2.day). It is not. A typical cell produces more like 1 kWh/m/day, not 6 kWh/m/day.
With the performance ratio, you can convert solar irradiance (in kWh/m2.year) directly into specific yield (in kWh/kWp.year).
You state performance ratio of 90%. However, kWh/m2 electrical output = more like 0.15 kWh/m2 Solar irradiance.
I don't see how you included the 0.20 conversion solar to electrical efficiency in your performance ration. 90% only accounts for losses from electrical output of panel to load, plus shadowing and other physical installation factors as you stated.
In other words, you do not get 90% of 6.8 kwh/m2/day in electrical output. You get more like 15%.
6.8kWh/(m.day) in Arizona on a tilted plane gives you about 2500kWh/(m.y)
With a performance ratio of 90% for your PV installation, you can get 2250kWh/(kWp.y) of electricity.
Your answer is about in the right ballpark, but the method is in error.
The chart which shows up to 6.8kWh/m2/day (for SE US) reflects actual solar energy, not electrical. Solar cell conversion efficiency is about 20%, resulting in about 1.36 kWh electric/m2/day. Considering the 6.8 is really that max possible, and there are electrical losses through the inverter, you could expect about 1 kwh-e/m2 If the panel area is 440f2=40m2, you could get as much as 40 Kwh-e/day. That is a generously high number. If you start with a more realistic 5.8kWh/m2/day and less than optimal conversion efficiency, the result is closer to 35 Kwh-e/day.
If you assume the setup is made of 24 x 250W panels, that's 6KW capacity. Assuming 5 equivalent "full sun" hours per day average (a generous number), you can get about 30KWh-e. I'd go with 30Kwh-e/day. One could easily come up with a much loser estimate if one wanted to be conservative in that direction.
I have begun to despise the term "hacked". As anything that can be used in any manner other than its purest fundamentally intended purpose, is considered to be hackable.
Not everything needs to be secure. My mailbox in not secure. I have photos printed at by others. When I start taking nude selfies, I'll make sure wifi is turned off.
I think he's saying. "It's too hard, lets just ignore it an assume its not a problem".
It certainly is a interesting question, but since development is so far from being that advanced, there is little benefit in trying to answer it now. Its good to keep in mind.
Yes, in my mind it was too easy. Its relative in this sense, but if you are an experienced operator those steps are not difficult to accomplish and there are relatively few of them. Simply the step of disabling ECCS without any consequences is an example. I am not saying that was the only cause or even the primary cause. Controls in place both organizationally, procedurally, and via design were inadequate and too easy to circumvent. All in the name of performing an unneeded test.
But yes, if you were to remove all safety systems from operation, including somehow overriding the added interlocks that are in place, violate the protocols which doesn't allow a single authority to control everything, and perform those same maneuvers, you'd get a similar result. Given the changes made since Chernobyl's accident, that would be very difficult to accomplish and essentially would have to be intent on causing a major accident.
They just need to embrace the P2P and filesharing world. Simply sell a license to own a particular movie in any format, obtain it any way you want. They could even sell movie player boxes that automatically download any content you license.
Slashdot Mirror
Silly String has the original patent on 3D printing. They are just waiting to strike.
Ok, I will correct myself again.
I further, I see how efficiency is NOT a factor. Your factor of 0.9/STC_irradiance is essentially a conversion from 6.8kWh/(m**2*day) to effective full sun hours/day.
No, I see my mistake. Reading formulas in text boxes is an error trap, I should have double checked. Your units work out, my apologies.
/STC_irradiance value does account for conversion efficiency, and also the difference between the value of 6.8kWh/(m**2*day) and hours/day equivalency, then I can see how your formula can work. In which case, efficiency does matter. Your insistence that conversion efficiency was not a factor was a point of confusion.
So if you are saying your 0.9 factor along with the
I see what you did now, my apologies. Thanks for staying with me on it.
^NO. Check your units. Your answer is actually 22 kWH-cubed/day. Which makes no sense.
That is because you have W rated * irradiance * irradiance, all three with W as a unit numerator... so you get Watts cubed.. But the 1000W/m2 radiance is a lab test value for standardizing comparisons, it is not applicable in this analysis, and like I said the 6.8 kWh/m2.day doesn't apply.
Like I said, units never lie. To get kWh instead of kWh-cubed, you'll need to divide somewhere in your conversion by a kW factor twice.
The factor you should be using rather than 6.8kWh/m2.day can be found here;
.95 (loss factor) = 20.5 kwh/day
http://www.wholesalesolar.com/...
If we have a panel rating of 3.6 kw,
3.6KWp x 6 hrs/day x
Notice the units work out.
6.8kWh/m2.day represents the available power of the sunlight hitting a square meter of surface in a day, on average. To obtain KWH electrical per day using 6.8kWh/m2.day, you must;
1) Convert solar to electrical (0.15 efficiency, or whatever you assume the solar panel conversion efficiency is) to get kWh-e/m2.day
2) Multiply by the surface area of the panel. to get kWh-e/day terminal output
3) Factor any electrical losses from panel output to load. kWh-e/day system output.
The 3.6 KW-e/Panel is a panel rating. If you start with panel rating, you can
1) Multiply by effective full sun hours per day to get KWh-e/Day/panel (this is typically about 5 to 7 hours/day)
2) Multiply by the number of panels for total KWh-e/Day terminal
3) Apply losses to get kWh-e/day system output.
In this case, you can combine 2 and 3 in your "90%" factor if you want, but you cannot use the 6.8kWh/m2.day, it does not apply, as it is surface area dependent.
Part of the confusion is due to the calculation in your original post by erroneously and needlessly calculating "2250kWh/(kWp.y) of electricity." Which made now sense since we were looking for output/day and we already had units in days. Then you switched method entirely and started with a panel rating you found by googling, and just calculated straight from that value with really documenting it very well. If somehow you used 6.8kWh/m2.day, then your units are wrong, and your number is wrong. It looks reasonable purely out of luck.
If you don't follow, just show your calculation including all units in factors and results for each step. Units never lie.
That is utterly ridiculous. You just changed the size to compensate for the change in efficiency. That is totally meaningless, and not related to the original calculation, which was based on an assumed carport size (area). If you assume a size (area), then efficiency matters. If you assume it can be any size, then you can't calculate the output.
Simply put, if you use 6.8kWh/m2/day, the only way you can calculate the electrical output is to use the conversion efficiency, which change the units to electrical, and to multiply by area to cancel out the square meter in the denominator of the value.
If you ignore panel area and just assume a panel rating, then the 6.8kWh/m2/day is not what you use in your calculation. Its quite easy to see just by following units.
Which is more correctly 0.15 x 2500kWh/(m.y) = 375 kWh/(m.y) electricity
6.8kWh/(m.day) in Arizona on a tilted plane gives you about 2500kWh/(m.y) With a performance ratio of 90% for your PV installation, you can get 2250kWh/(kWp.y) of electricity.
Which is more correctly 0.15 x 2500kWh/(m.y) = 375 kWh/(m.y)
You can ignore that method, and begin with a different starting point and instead calculate from a given cell rating, then the 90% performance ratio makes sense and yes, physical size does not matter, only the equivalent full solar hours. Which often ranges between 5 and 7, but is different from the 6.8kWh/(m.day) from the chart. You just got lucky that 6.8 was still a reasonable value.
How about doing it to all campaign sites for anyone who supports it?
forgive my units.... all m = m2
6.8kWh/(m.day) in Arizona on a tilted plane gives you about 2500kWh/(m.y) With a performance ratio of 90% for your PV installation, you can get 2250kWh/(kWp.y) of electricity.
I see that as 6.8 x 365 = 2500. 2500*0.9=2250
6.8kWh/(m.day) = solar irradiance/(m.day)
You seem a assume 20% cell efficiency is already factored into 6.8kWh/(m2.day). It is not. A typical cell produces more like 1 kWh/m/day, not 6 kWh/m/day.
With the performance ratio, you can convert solar irradiance (in kWh/m2.year) directly into specific yield (in kWh/kWp.year).
You state performance ratio of 90%. However, kWh/m2 electrical output = more like 0.15 kWh/m2 Solar irradiance.
I don't see how you included the 0.20 conversion solar to electrical efficiency in your performance ration. 90% only accounts for losses from electrical output of panel to load, plus shadowing and other physical installation factors as you stated.
In other words, you do not get 90% of 6.8 kwh/m2/day in electrical output. You get more like 15%.
The can buy Monster cables next.
22kWh/day really is your best case scenario.
6.8kWh/(m.day) in Arizona on a tilted plane gives you about 2500kWh/(m.y) With a performance ratio of 90% for your PV installation, you can get 2250kWh/(kWp.y) of electricity.
Your answer is about in the right ballpark, but the method is in error. The chart which shows up to 6.8kWh/m2/day (for SE US) reflects actual solar energy, not electrical. Solar cell conversion efficiency is about 20%, resulting in about 1.36 kWh electric/m2/day. Considering the 6.8 is really that max possible, and there are electrical losses through the inverter, you could expect about 1 kwh-e/m2 If the panel area is 440f2=40m2, you could get as much as 40 Kwh-e/day. That is a generously high number. If you start with a more realistic 5.8kWh/m2/day and less than optimal conversion efficiency, the result is closer to 35 Kwh-e/day.
If you assume the setup is made of 24 x 250W panels, that's 6KW capacity. Assuming 5 equivalent "full sun" hours per day average (a generous number), you can get about 30KWh-e. I'd go with 30Kwh-e/day. One could easily come up with a much loser estimate if one wanted to be conservative in that direction.
If there were ever a satellite network that would deliver sports only (ESPN, FS. etc), it would rip the fabric of the cable TV world.
I have begun to despise the term "hacked". As anything that can be used in any manner other than its purest fundamentally intended purpose, is considered to be hackable.
Not everything needs to be secure. My mailbox in not secure. I have photos printed at by others. When I start taking nude selfies, I'll make sure wifi is turned off.
Exactly. My wife is free to read my email any time she wants, and vice versa. Can't imagine needing to hide anything.
I've also learned there are two sides to every story. Be very careful judging if you've only heard one.
^based on his background, I'll consider his knife & kitchen organization advice. Outside that, its just another dude's pondering.
I think he's saying. "It's too hard, lets just ignore it an assume its not a problem".
It certainly is a interesting question, but since development is so far from being that advanced, there is little benefit in trying to answer it now. Its good to keep in mind.
That's too complicated, we've simplified it. All we need to know is which Party initiated it.
Finally, the fabled future of the feature phone is found to be fabulous.
Is that spring unique to RBMK? I can't recall seeing it on other designs. Typically just gravity is sufficient (for top fed rods).
Yes, in my mind it was too easy. Its relative in this sense, but if you are an experienced operator those steps are not difficult to accomplish and there are relatively few of them. Simply the step of disabling ECCS without any consequences is an example. I am not saying that was the only cause or even the primary cause. Controls in place both organizationally, procedurally, and via design were inadequate and too easy to circumvent. All in the name of performing an unneeded test.
But yes, if you were to remove all safety systems from operation, including somehow overriding the added interlocks that are in place, violate the protocols which doesn't allow a single authority to control everything, and perform those same maneuvers, you'd get a similar result. Given the changes made since Chernobyl's accident, that would be very difficult to accomplish and essentially would have to be intent on causing a major accident.
They just need to embrace the P2P and filesharing world. Simply sell a license to own a particular movie in any format, obtain it any way you want. They could even sell movie player boxes that automatically download any content you license.