NEC2011 - 310.15(B)(2)(a) or 310.15(B)(2)(b)

[curious101]

Hi,
How do I determine if I have to go by 30C or 40C ambient temperature table (tables in the thread line) if I have 37C ambient temperature?
For an outdoor installation of photovoltaic DC conductors, how do I determine the ambient temperature? Do I take the average of year-around temperatures in the specific city?
Thank you!

 

[wwhitney]

How do I determine if I have to go by 30C or 40C ambient temperature table (tables in the thread line) if I have 37C ambient temperature?

Which table of temperature correction factors you use is based on what the assumed ambient temperature is for the table from which you selected your starting ampacity. For example, 2011 NEC Table 310.15(B)(16) is based on 30C ambient, while Table 310.15(B)(20) is based on 40C ambient.

Or you can just use equation 310.15(B)(2) directly instead of the tables.

For an outdoor installation of photovoltaic DC conductors, how do I determine the ambient temperature?

Generally you would consider a sufficiently worst case scenario, so use something like the 99% or 99.7% percentile high outdoor temperature, e.g. from ASHRAE. I think using the past record high temperature would be unreasonably conservative, although if the project is being planned for the next 20+ years, using an inflator to account for global warming would be reasonable.

Cheers, Wayne

 

[curious101]

Which table of temperature correction factors you use is based on what the assumed ambient temperature is for the table from which you selected your starting ampacity. For example, 2011 NEC Table 310.15(B)(16) is based on 30C ambient, while Table 310.15(B)(20) is based on 40C ambient.

Or you can just use equation 310.15(B)(2) directly instead of the tables.


Generally you would consider a sufficiently worst case scenario, so use something like the 99% or 99.7% percentile high outdoor temperature, e.g. from ASHRAE. I think using the past record high temperature would be unreasonably conservative, although if the project is being planned for the next 20+ years, using an inflator to account for global warming would be reasonable.

Cheers, Wayne

Thank you so much, Wayne, for your clear response! I've been trying to find the ASHRAE temperatures, but had no luck. Any links you could suggest? Thanks again!

 

[wwhitney]

Any links you could suggest?

A google search turns up https://ashrae-meteo.info/v2.0/places.php?continent=North America but I have no idea if that's official or reliable.

It says that for the 2021 ASHRAE data for Atlanta, GA airport, for example, the 0.4% dry bulb (what you want) temperature is 34.3 C, while the 1% dry bulb is 33.1 C. It also says the mean extreme annual dry bulb temperature is 35.8 C, with a standard deviation of 2.0 C.

So those numbers mean that if for example the weather station logs temperature once a minute, that's 1440 numbers per day, or 525,500 numbers per year. Of those data points, in the average year 1% are at least 33.1C ; 0.4% are at least 34.3C; and the average largest number seen is 35.8C. Using the 35.8C mean extreme temperature and the 2.0 C standard deviation, it also computes the 5, 10, 20, and 50 year expected extremes, meaning for each time period the temperature for which (if I understand correctly) you'd have a 50% chance of seeing it at least once over that time period.

Having said all that, i have no idea which of those temperatures it would be appropriate to select. It presumably would depend on what you need the value for and what the consequences are if you are off. If your equipment dies if the temperature is exceeded, and you want the equipment to last 20 years, and you're willing to accept a 1% chance of equipment death due to temperature over 20 years, then you'd want the 20 year 1% likelihood extreme value (which is not any of the numbers I listed above, but is computable from the extreme annual mean and standard deviation).
Versus if you're just doing conductor ampacity temperature correction, maybe the 0.4% annual number is fine. Not sure if there's any engineering consensus on which number to use for that purpose.

Cheers, Wayne

 

[curious101]

A google search turns up https://ashrae-meteo.info/v2.0/places.php?continent=North America but I have no idea if that's official or reliable.

It says that for the 2021 ASHRAE data for Atlanta, GA airport, for example, the 0.4% dry bulb (what you want) temperature is 34.3 C, while the 1% dry bulb is 33.1 C. It also says the mean extreme annual dry bulb temperature is 35.8 C, with a standard deviation of 2.0 C.

So those numbers mean that if for example the weather station logs temperature once a minute, that's 1440 numbers per day, or 525,500 numbers per year. Of those data points, in the average year 1% are at least 33.1C ; 0.4% are at least 34.3C; and the average largest number seen is 35.8C. Using the 35.8C mean extreme temperature and the 2.0 C standard deviation, it also computes the 5, 10, 20, and 50 year expected extremes, meaning for each time period the temperature for which (if I understand correctly) you'd have a 50% chance of seeing it at least once over that time period.

Having said all that, i have no idea which of those temperatures it would be appropriate to select. It presumably would depend on what you need the value for and what the consequences are if you are off. If your equipment dies if the temperature is exceeded, and you want the equipment to last 20 years, and you're willing to accept a 1% chance of equipment death due to temperature over 20 years, then you'd want the 20 year 1% likelihood extreme value (which is not any of the numbers I listed above, but is computable from the extreme annual mean and standard deviation).
Versus if you're just doing conductor ampacity temperature correction, maybe the 0.4% annual number is fine. Not sure if there's any engineering consensus on which number to use for that purpose.

Cheers, Wayne

This is all great info! Thank you, Wayne!
I am checking:
1. the DC string voltage calculations which involve the ASHRAE high and low temperatures, and
2. the DC conductor ampacity calculations which involve the maximum ambient temperature.
Is the formula to calculate the 20 year 1% likelihood extreme value somewhere on the internet? Sorry to bother you.... I did a quick search, but many complicated options came up. Not that I don't like complicated things, but I'm already spending probably too much time on this.... Thank you!

 

[wwhitney]

Is the formula to calculate the 20 year 1% likelihood extreme value somewhere on the internet?

On your DC string voltage calculation, what happens if the spec is say 600V max and due to extreme temperature the string OCV is 601V? Does the inverter immediately die, or is it likely that nothing happens? I would think the 20 year 1% likelihood extreme value would only be the appropriate temperature to use if the inverter would die immediately at 601V.

Anyway, to answer your statistics question, let's say that the annual extreme temperature is normally distributed with mean M and standard deviation S, then we are going to sample that distribution 20 times. We want to find the threshold temperature T where if we say that "success" is that each sample is no more than T, and failure is if any sample exceeds T, we have a 1% chance of failure. That means we need a 99% chance of "success" on 20 consecutive samples.

If each sample is independent, then overall success requires 20 individual successful samples in a row. So the chance of overall success is the 20th power of the chance of success on one sample. So we want to know how many standard deviations from the mean do you need to go to get (.99)^(1/20) = 0.9995 of the distribution to be below that value. [These probabilities p are small enough that the approximation (1-p)^n = 1-p*n holds.]

So now we can check a CDF (Cumulative Distribution Function) table for the standard normal distribution and find that the answer is 3.27 standard deviations from the mean. For the Atlanta, GA data where the annual extremal distribution is 35.8 C with a 2.0C standard deviation, that means our 1% likelihood 20 year extreme temperature would be 35.8 + 3.27*2.0 = 42.3C.

But this is not likely to be the appropriate temperature to use, unless your system is extremely sensitive to temperature, and exceedance would be bad, even briefly.

Cheers, Wayne

 

[curious101]

On your DC string voltage calculation, what happens if the spec is say 600V max and due to extreme temperature the string OCV is 601V? Does the inverter immediately die, or is it likely that nothing happens? I would think the 20 year 1% likelihood extreme value would only be the appropriate temperature to use if the inverter would die immediately at 601V.

Anyway, to answer your statistics question, let's say that the annual extreme temperature is normally distributed with mean M and standard deviation S, then we are going to sample that distribution 20 times. We want to find the threshold temperature T where if we say that "success" is that each sample is no more than T, and failure is if any sample exceeds T, we have a 1% chance of failure. That means we need a 99% chance of "success" on 20 consecutive samples.

If each sample is independent, then overall success requires 20 individual successful samples in a row. So the chance of overall success is the 20th power of the chance of success on one sample. So we want to know how many standard deviations from the mean do you need to go to get (.99)^(1/20) = 0.9995 of the distribution to be below that value. [These probabilities p are small enough that the approximation (1-p)^n = 1-p*n holds.]

So now we can check a CDF (Cumulative Distribution Function) table for the standard normal distribution and find that the answer is 3.27 standard deviations from the mean. For the Atlanta, GA data where the annual extremal distribution is 35.8 C with a 2.0C standard deviation, that means our 1% likelihood 20 year extreme temperature would be 35.8 + 3.27*2.0 = 42.3C.

But this is not likely to be the appropriate temperature to use, unless your system is extremely sensitive to temperature, and exceedance would be bad, even briefly.

Cheers, Wayne

Hold on, I am a detailed person, but didn't take enough courses in statistics... Sorry, where do I find the table where the 3.27 comes from?

 

[wwhitney]

Hold on, I am a detailed person, but didn't take enough courses in statistics... Sorry, where do I find the table where the 3.27 comes from?

No problem, you followed well enough to single out the point I didn't explain. : -)

In Excel parlance this value would be NORM.INV(0.9995,0,1), where 0 is the mean and 1 is the standard deviation. This gives us the value for which 99.95% of the time, the standard normal distribution when sampled will be <= that value. LibreOffice tells me it is actually 3.29, so maybe I misread the chart I was looking at, or didn't properly consider the rounding that was used in the chart.

NORM.INV will do the scaling for you, so I could have just asked for NORM.INV(0.9995,35.8, 2.0) to get 42.4 C.

As for what NORM.INV is doing, the standard normal distribution's probability distribution function is exp(-0.5*x^2)/sqrt(2*pi), so NORM.INV is finding the value y such that the integral from -infinity to y of exp(-0.5*x^2)/sqrt(2*pi) = 0.9995.

Normal distribution - Wikipedia

en.wikipedia.org

Cheers, Wayne

 

[curious101]

No problem, you followed well enough to single out the point I didn't explain. : -)

In Excel parlance this value would be NORM.INV(0.9995,0,1), where 0 is the mean and 1 is the standard deviation. This gives us the value for which 99.95% of the time, the standard normal distribution when sampled will be <= that value. LibreOffice tells me it is actually 3.29, so maybe I misread the chart I was looking at, or didn't properly consider the rounding that was used in the chart.

NORM.INV will do the scaling for you, so I could have just asked for NORM.INV(0.9995,35.8, 2.0) to get 42.4 C.

As for what NORM.INV is doing, the standard normal distribution's probability distribution function is exp(-0.5*x^2)/sqrt(2*pi), so NORM.INV is finding the value y such that the integral from -infinity to y of exp(-0.5*x^2)/sqrt(2*pi) = 0.9995.

Normal distribution - Wikipedia

en.wikipedia.org

Cheers, Wayne

So, can I use the same logic for the extreme minimum too? My project is actually in MS, although I'm in Atlanta. The ASHRAE info of the project's location is attached here. With your logic inputting into Excel, I got 44.08 for max temp and -3.018 for the min. Is this correct? Thank you!

 

[wwhitney]

So, can I use the same logic for the extreme minimum too?

If you really want the 1% likelihood 20 year extreme temperatures, then based on the data you posted, the low temp would be NORM.INV(0.0005,-9.6,2.2) and the high temp would be NORM.INV(0.9995,37.5,2.0).

[Of course, if the above data are just based on past observations, the high temp you get will be non-conservative because temperature data almost everywhere shows a warming trend, and there is every reason to expect that trend to continue for the next 20 years.]

1% likelihood in 20 years seems quite conservative. You might ask in the PV forum what likelihood and return period people generally use.

Cheers, Wayne

 

[curious101]

If you really want the 1% likelihood 20 year extreme temperatures, then based on the data you posted, the low temp would be NORM.INV(0.0005,-9.6,2.2) and the high temp would be NORM.INV(0.9995,37.5,2.0).

[Of course, if the above data are just based on past observations, the high temp you get will be non-conservative because temperature data almost everywhere shows a warming trend, and there is every reason to expect that trend to continue for the next 20 years.]

1% likelihood in 20 years seems quite conservative. You might ask in the PV forum what likelihood and return period people generally use.

Cheers, Wayne

Okay, thank you!