RHO derating factors
- Thread starter skeeton
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- Lockport, IL
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- Semi-Retired Electrical Engineer
Re: RHO derating factors
No. You will find a similar statement in B.310.15(B)(3). But that Annex does not include any enforceable requirements. My view is that if you do not install in the exact manner shown in the tables and figures, you need to have a formal calculation performed ?under engineering supervision.? However, I have had state-level inspection authorities assert that Table 310.16 provides acceptable ampacity limits, even if they are applied to ductbanks. Speaking now as a licensed engineer, I would only consider using 310.16 for a ductbank if the load had been calculated in accordance with 220.
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- Mission Viejo, CA
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- Professional Electrical Engineer
Re: RHO derating factors
at my age there are times when creeping senility cause me to overlook the obvious - so, with an otherwise Code compliant design, where would I not calculate a load per 220?
Charlie, I certainly agree with your "engineering supervision" sentiments and I'm equally puzzled with AHJs apparently being unaware of (or ignoring) the additional heating effects of conductors in ductbanks. But-
at my age there are times when creeping senility cause me to overlook the obvious - so, with an otherwise Code compliant design, where would I not calculate a load per 220?
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- Lockport, IL
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- Semi-Retired Electrical Engineer
Re: RHO derating factors
Suppose you design an office building. You perform a service calculation (per NEC 220) for the core and shell and the first four tenants. You get a calculated result of 1650 amps, but you know that the actual load will be lower. The owner tells you there is room for at least two more tenants. You select a 1,500 KVA transformer (480/277V secondary), and a 2000 amp main board. Using Table 310.16, you select five sets of 500 MCM THHN Copper. You calculate the ampacity at 5 x 380 = 1900. You select a 2000 amp main frame breaker, and set the trip at 1900 amps. At the beginning of this project, the cables are sized for 1900 amps, and the calculated service load is 1650.
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A year later, two new tenants prepare to move in. You perform a 30 day load study, and discover the actual peak load is closer to 1200 amps. Add 25% for conservatism, and call the existing load 1500 amps. You perform a service calculation for the two new tenants, and conclude they will add 300 amps to the total load. You add the 1500 to the 300, and your result is still below the 1900 amp setting of the main breaker.
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My answer to the first question is ?yes, but with some reservations.? That is because I have looked at Table B.310.7. Using that table (not a code requirement), I see that 5 sets of 500 MCM THHN Copper in a six-way ductbank, using a RHO of 90, and a Load Factor of 100%, will give me an ampacity of 5 x 273 = 1,365 amps. That is below the calculated result of 1650 amps, but I also know that the calculation method is conservative, and that the actual load will be much lower.
My answer to the second question is ?no.? First, it is clear that the total calculated load is 1650 (original) plus 300 (new tenants), for a total of 1950, and that the conductors are only sized for 1900 amps. You might argue that you have new information, that the actual load is less than the original 1650, so that you should have capacity to add 300 more, and still be less than the 1900 amps. I counter-argue that you would be mixing the concepts of NEC-220 calculated load and the concept of measured load. The measured load is already too close to the 1365 amps that I would get from B.310.7. I don?t see any room for the addition of 300 more amps, even if that is just the 220 calculation, and even if I know the actual new load addition would be less than 300 more amps.
Good question. Forgive my answer being a bit long and complex.
Suppose you design an office building. You perform a service calculation (per NEC 220) for the core and shell and the first four tenants. You get a calculated result of 1650 amps, but you know that the actual load will be lower. The owner tells you there is room for at least two more tenants. You select a 1,500 KVA transformer (480/277V secondary), and a 2000 amp main board. Using Table 310.16, you select five sets of 500 MCM THHN Copper. You calculate the ampacity at 5 x 380 = 1900. You select a 2000 amp main frame breaker, and set the trip at 1900 amps. At the beginning of this project, the cables are sized for 1900 amps, and the calculated service load is 1650.
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- <font size="2" face="Verdana, Helvetica, sans-serif">Question 1: Is this design acceptable?</font>
A year later, two new tenants prepare to move in. You perform a 30 day load study, and discover the actual peak load is closer to 1200 amps. Add 25% for conservatism, and call the existing load 1500 amps. You perform a service calculation for the two new tenants, and conclude they will add 300 amps to the total load. You add the 1500 to the 300, and your result is still below the 1900 amp setting of the main breaker.
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- <font size="2" face="Verdana, Helvetica, sans-serif">Question 2: Is this design change acceptable?</font>
My answer to the first question is ?yes, but with some reservations.? That is because I have looked at Table B.310.7. Using that table (not a code requirement), I see that 5 sets of 500 MCM THHN Copper in a six-way ductbank, using a RHO of 90, and a Load Factor of 100%, will give me an ampacity of 5 x 273 = 1,365 amps. That is below the calculated result of 1650 amps, but I also know that the calculation method is conservative, and that the actual load will be much lower.
My answer to the second question is ?no.? First, it is clear that the total calculated load is 1650 (original) plus 300 (new tenants), for a total of 1950, and that the conductors are only sized for 1900 amps. You might argue that you have new information, that the actual load is less than the original 1650, so that you should have capacity to add 300 more, and still be less than the 1900 amps. I counter-argue that you would be mixing the concepts of NEC-220 calculated load and the concept of measured load. The measured load is already too close to the 1365 amps that I would get from B.310.7. I don?t see any room for the addition of 300 more amps, even if that is just the 220 calculation, and even if I know the actual new load addition would be less than 300 more amps.
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- Mission Viejo, CA
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- Professional Electrical Engineer
Re: RHO derating factors
Whew - well, at least it wasn't obvious
I still submit, it isn't the load calcs you don't like but the ampacity calcs - which we already agreed were problematic.
Whew - well, at least it wasn't obvious
I still submit, it isn't the load calcs you don't like but the ampacity calcs - which we already agreed were problematic.
- Location
- Illinois
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- retired electrician
Re: RHO derating factors
rabalex,
When the duct bank ampacities were moved from Article 310 to the annex, many of the comments in the ROC said that the only reason that there are not more problems with burning up the conductors in underground duct banks is because of the Article 220 calculations. These calculations almost always provide calculated loads that are 30% to 50% higher than the actual load. There have been problems where the ampacities in Table 310.16 are used for duct banks where the actual load is close to the ampacity shown in the table. These types of problems are more common in industrial type ocupancies where the actual load is often very close to the calculated load.
don
rabalex,
When the duct bank ampacities were moved from Article 310 to the annex, many of the comments in the ROC said that the only reason that there are not more problems with burning up the conductors in underground duct banks is because of the Article 220 calculations. These calculations almost always provide calculated loads that are 30% to 50% higher than the actual load. There have been problems where the ampacities in Table 310.16 are used for duct banks where the actual load is close to the ampacity shown in the table. These types of problems are more common in industrial type ocupancies where the actual load is often very close to the calculated load.
don
- Location
- Mission Viejo, CA
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- Professional Electrical Engineer
Re: RHO derating factors
Don,
I was aware of that. In some other thread I took Article 220 - stripped out all the "dwelling" specific material and asked folks to calculate the industrial load under discussion based on whatever "demand/diversity" they could justify from what was left. I'm sure you know there weren't too many opportunities.
I also said Art 220 was the closest thing we electrical folks have to a "safety factor." Where some mechanical / structural codes require safety factors in excess of 20 we are getting off fairly easy. Too easy in some cases.
I may believe the NEC is excessive in a few places, but ampacity ratings in general is not one of them.
Bob
Don,
I was aware of that. In some other thread I took Article 220 - stripped out all the "dwelling" specific material and asked folks to calculate the industrial load under discussion based on whatever "demand/diversity" they could justify from what was left. I'm sure you know there weren't too many opportunities.
I also said Art 220 was the closest thing we electrical folks have to a "safety factor." Where some mechanical / structural codes require safety factors in excess of 20 we are getting off fairly easy. Too easy in some cases.
I may believe the NEC is excessive in a few places, but ampacity ratings in general is not one of them.
Bob
- Location
- Lockport, IL
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- Semi-Retired Electrical Engineer
Re: RHO derating factors
Something like that. Actually, my issue lies in the title block of Table 310.16. It has been discussed in a previous thread, so it need not be taken up again here. But to state my issue, the title block says that the Table applies to ?not more than three current-carrying conductors in raceway. . . . ? The question is whether having a second (or third, or sixth) conduit in ?close proximity? (not a well-defined concept, but I mean within the same ductbank) causes the Table to no longer apply.
Re: RHO derating factors
Charlie B
I agree with most of your comments but I would argue the using a 100% load factor for an office building would be excessive. I think that 75% LF would be more in line with this type of load.
Obviously the 75% LF is not listed. However
B.310.15(B)(7) suggests the method to obtain a rating for installations that do not match the installations shown in the tables.
Charlie B
I agree with most of your comments but I would argue the using a 100% load factor for an office building would be excessive. I think that 75% LF would be more in line with this type of load.
Obviously the 75% LF is not listed. However
B.310.15(B)(7) suggests the method to obtain a rating for installations that do not match the installations shown in the tables.
- Location
- Lockport, IL
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- Semi-Retired Electrical Engineer
Re: RHO derating factors
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I have no idea why the NEC uses an LF of 50% in some of the Tables (e.g., the RHO = 60 columns of B.310.7). You would almost have to have a large peak (such as 4 hours worth of heavy air conditioning load in the early afternoon) and to have the building essentially turned off during the night, to get to an LF of 50%.
I?m sure you are right. However, coming up with a value for LF, and verifying it ?by measurement or calculation,? as B.310.15(B)(7) would require, is no superficial task. For starters, the NEC never defines ?Load Factor.? Here is how one of my old textbooks tells me how to calculate LF:
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- <font size="2" face="Verdana, Helvetica, sans-serif">Suppose the building ran at a constant load of 80 KW for 12 hours a day, and at a constant load of 40 KW for the other 12 hours a day. The average is 60 KW, and the peak is 80 KW. Dividing 60 by 80 gives you a LF of 75%.</font>
I have no idea why the NEC uses an LF of 50% in some of the Tables (e.g., the RHO = 60 columns of B.310.7). You would almost have to have a large peak (such as 4 hours worth of heavy air conditioning load in the early afternoon) and to have the building essentially turned off during the night, to get to an LF of 50%.
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- Mission Viejo, CA
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- Professional Electrical Engineer
Re: RHO derating factors
From IEEE 100 (Dictionary) as derived from IEEE 141 (Industrial Power Systems) and 241 (Commercial Power Systems):
Using the example you gave, I believe the NEC would say you have an 80kW continuous load for ampacity calculation purposes since its more than 3 hours. In this case, the "load factor" would be irrelevant. Of course, "Non-continuous loads" would also be irrelevant since they are fully incorporated in the continuous load.
The apparent rigidity of the NEC unfortunately gives the impression that it is also rigorous. In many cases it is; but the ampacity rating of conductors in the "condition of use" is not one of them.
Bob
[ July 14, 2004, 05:44 PM: Message edited by: rbalex ]
From IEEE 100 (Dictionary) as derived from IEEE 141 (Industrial Power Systems) and 241 (Commercial Power Systems):
Charlie B,
Using the example you gave, I believe the NEC would say you have an 80kW continuous load for ampacity calculation purposes since its more than 3 hours. In this case, the "load factor" would be irrelevant. Of course, "Non-continuous loads" would also be irrelevant since they are fully incorporated in the continuous load.
The apparent rigidity of the NEC unfortunately gives the impression that it is also rigorous. In many cases it is; but the ampacity rating of conductors in the "condition of use" is not one of them.
Bob
[ July 14, 2004, 05:44 PM: Message edited by: rbalex ]
- Location
- Lockport, IL
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- Semi-Retired Electrical Engineer
Re: RHO derating factors
But if I wish to use the measured peak load as the basis for calculating ?Load Factor,? then the concept of continuous versus non-continuous does not enter into the calculation. It is nice to see that the current IEEE definition that you have quoted matches that given by my 20 year old textbook.
I suspect that we are saying the same thing, but in different terms.
I think not. You are looking at a measured load. If I wish to use that measured peak load as the basis for deciding whether I can add more load to the facility, then 220.35(2) would have me add 25% to the measured load. It does not say that the extra 25% is to account for continuous load. Indeed, the measured peak need not have been present for 3 hours. I think the extra 25% is an ?uncertainty factor,? to allow for the possibility that the measured peak might have been higher if I had taken the measurements during some other 30 day period.
But if I wish to use the measured peak load as the basis for calculating ?Load Factor,? then the concept of continuous versus non-continuous does not enter into the calculation. It is nice to see that the current IEEE definition that you have quoted matches that given by my 20 year old textbook.
I suspect that we are saying the same thing, but in different terms.
Re: RHO derating factors
Just for information, My AIEE - IPCEA Manual lists
3 cable 500 kcm 20C earth 75C conductor LF 100 at
320 amps and for 75 LF at 357 amps.
Okonite defines LF as the ratio of the average hourly load to the maximum hourly over 24 hours.
I suppose you could have 200% load for 8 hrs and 50% load for 16 hrs and get a 100% LF. I wonder how long that would last.
[ July 14, 2004, 08:33 PM: Message edited by: bob ]
Just for information, My AIEE - IPCEA Manual lists
3 cable 500 kcm 20C earth 75C conductor LF 100 at
320 amps and for 75 LF at 357 amps.
Okonite defines LF as the ratio of the average hourly load to the maximum hourly over 24 hours.
I suppose you could have 200% load for 8 hrs and 50% load for 16 hrs and get a 100% LF. I wonder how long that would last.
[ July 14, 2004, 08:33 PM: Message edited by: bob ]
- Location
- Lockport, IL
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- Semi-Retired Electrical Engineer
Re: RHO derating factors
Let?s ignore the odds that the system could operate at 200% load for 8 hours without something tripping. Here?s the math:
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That definition is consistent with the other two given above.
No, the average load would be 100%, but the LF would be 50%.
Let?s ignore the odds that the system could operate at 200% load for 8 hours without something tripping. Here?s the math:
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- <font size="2" face="Verdana, Helvetica, sans-serif">200% for 8 hours plus 50% for 16 hours gives a total of 2400 units of load-hours.</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">Divide this by 24 hours, you will have an average load of 100%.</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">LF is defined as ?average divided by peak.?</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">A 100% average divided by a 200% peak gives an LF of 50%.</font>
pierre
Senior Member
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- Westchester County, New York
Re: RHO derating factors
It is amazing how this subject can create such expression - it is too bad I do not know enough to add to this, but this thread has been very informative - thanks guys!
Pierre
It is amazing how this subject can create such expression - it is too bad I do not know enough to add to this, but this thread has been very informative - thanks guys!
Pierre
- Location
- Mission Viejo, CA
- Occupation
- Professional Electrical Engineer
Re: RHO derating factors
Charlie B,
This debate has been both fun and informative.
I don't believe the 25% adder for measured loads has anything to do with a load safety factor. I think it's simply the uncorrelated residue of the 125% multiplier found originally sprinkled throughout the NEC to protect UL listed overcurrent devices.
I say that based on the original text of 215-2 and 220-35, when 220-35 was first introduced in the 1981 Code.:
Note at the time 215-2 said nothing directly about the 125% multiplier nor did it mention 100% rated equipment as it does now in the exception.
In '81, it appears to me that the CMP simply treated the maximum demand as a continuous load whether it actually was or not. Even the current 220-35 exception uses a 15 minute demand maximum which could be an excursion.
220-35 has gone through several revisions, including acceptance of the 30 day demand reading. The 30 day reading is an exception, not an alternate, to the main rule. It has several additional qualifiers, including :
The question left outstanding to me is:
Is the "new load" calculated per the rest of Article 220?; i.e., is the new load 125% of the new continuous load + 100% of the new non-continuous load OR is the new load simply 100% of both the continuous load and non-continuous load? I have written (Non-formal) interpretations of NFPA staff for BOTH methods.
I have genuinely appreciated the thoughtfulness this forum gives to such debates.
[ July 15, 2004, 12:33 PM: Message edited by: rbalex ]
Charlie B,
This debate has been both fun and informative.
I don't believe the 25% adder for measured loads has anything to do with a load safety factor. I think it's simply the uncorrelated residue of the 125% multiplier found originally sprinkled throughout the NEC to protect UL listed overcurrent devices.
I say that based on the original text of 215-2 and 220-35, when 220-35 was first introduced in the 1981 Code.:
Note at the time 215-2 said nothing directly about the 125% multiplier nor did it mention 100% rated equipment as it does now in the exception.
In '81, it appears to me that the CMP simply treated the maximum demand as a continuous load whether it actually was or not. Even the current 220-35 exception uses a 15 minute demand maximum which could be an excursion.
220-35 has gone through several revisions, including acceptance of the 30 day demand reading. The 30 day reading is an exception, not an alternate, to the main rule. It has several additional qualifiers, including :
I think the key word is computed in 215-2 [corrected the reference to 215-2]in both the '81 and current NEC. It usually means "calculated" but it often means "determined by any valid means." I believe "calculated" or "measured" or a proper combination of both are valid, sufficiently conservative, and applicable as the basis for determining the required conductor ampacities.
The question left outstanding to me is:
Is the "new load" calculated per the rest of Article 220?; i.e., is the new load 125% of the new continuous load + 100% of the new non-continuous load OR is the new load simply 100% of both the continuous load and non-continuous load? I have written (Non-formal) interpretations of NFPA staff for BOTH methods.
I have genuinely appreciated the thoughtfulness this forum gives to such debates.
[ July 15, 2004, 12:33 PM: Message edited by: rbalex ]
- Location
- Lockport, IL
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- Semi-Retired Electrical Engineer
Re: RHO derating factors
My answer begins in 220.3(C), which in turn points us back to 220.3(A) and (B). Those sections will lead us to add 25% to the largest motor, but do not require us to add 25% to the continuous loads. So if the question is, ?I have measured this much load, and I want to add this much load, is my service adequate to supply that much load,? then I infer that you do not add 25% to the continuous load.
On the other hand, if the topic is ampacity, then 215.2(A)(1) requires that the feeder conductor must have sufficient ampacity to handle and extra 25% of the continuous loads. So if the question is , ?I have measured this much load, and I want to add this much load, do my feeder conductors have adequate ampacity for that much load,? then I infer that you must add 25% to the continuous load.
Finally, the rules for sizing a service per 230.23(B) simply point back to the load calculations in 220. Those calculations do not add 25% for continuous loads. So if the question is , ?I have measured this much load, and I want to add this much load, do my service conductors have adequate ampacity for that much load,? then I infer that you do not add 25% to the continuous load.
Fun, ain?t it?
Good question. I can?t spend much time analyzing this one, but my initial answer is, ?No, yes, and no,? in the following order:
My answer begins in 220.3(C), which in turn points us back to 220.3(A) and (B). Those sections will lead us to add 25% to the largest motor, but do not require us to add 25% to the continuous loads. So if the question is, ?I have measured this much load, and I want to add this much load, is my service adequate to supply that much load,? then I infer that you do not add 25% to the continuous load.
On the other hand, if the topic is ampacity, then 215.2(A)(1) requires that the feeder conductor must have sufficient ampacity to handle and extra 25% of the continuous loads. So if the question is , ?I have measured this much load, and I want to add this much load, do my feeder conductors have adequate ampacity for that much load,? then I infer that you must add 25% to the continuous load.
Finally, the rules for sizing a service per 230.23(B) simply point back to the load calculations in 220. Those calculations do not add 25% for continuous loads. So if the question is , ?I have measured this much load, and I want to add this much load, do my service conductors have adequate ampacity for that much load,? then I infer that you do not add 25% to the continuous load.
Fun, ain?t it?
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