Given a calculated load that is at least partially continuous, this updated four-part test assures that the conductors are correctly chosen.
Selecting conductors is probably the most fundamental task facing installers; certainly no part of this industry is far removed from these decisions. Most loads, especially at the feeder level of the distribution, are at least partially continuous in character. Recent NE Code changes have complicated the selection process.
This opening paragraph is very close to the one that appeared here in the May 1992 issue, and it is as true today as it was then. At that time, we published a four-part test to ensure proper sizing of conductors carrying loads that are at least partly continuous. That article was Code-perfect and definitive, and it caused major overcurrent device manufacturers to realize that the way they were testing their devices did not reflect some Code applications.
Rather than rewrite the product standards, and then face the possibility of a major redesign and an extensive retesting program, which would have cost millions of dollars, these manufacturers proposed major revisions to the 1996 NEC to prevent that from being necessary. As part of their substantiation, copies of the May 1992 issue of EC&M were sent to each member of the Code-making panel. Those proposed revisions for the 1996 NEC were accepted, which means the earlier article is no longer valid. This article replaces the earlier one, which should no longer be used in jurisdictions under the 1996 NEC.
NEC changes, the history
The key change was to partially reconnect the concept of a 125% oversizing for the overcurrent device with a 125% oversizing for circuit conductors. Nevertheless, there is a subtle difference between the size requirement for an overcurrent protective device and the size requirement for the conductor connected to it. We'll start with Sec. 220-10(b) in the 1984 NEC:
(b) Continuous and Noncontinuous Loads. Where a feeder supplies continuous loads or any combination of continuous and noncontinuous load, neither the ampere rating of the overcurrent device nor the ampacity of the feeder conductors shall be less than the noncontinuous load plus 125 percent of the continuous load.
As a transitional period in the 1987 NEC, service conductors were treated the same as in the 1984 code, but feeders followed a new rule that was subsequently extended to both feeders and service conductors. This was Sec. 220-10(b) in the 1990 NEC:
(b) Continuous and Noncontinuous Loads. Where a feeder supplies continuous loads or any combination of continuous and noncontinuous load the rating of the over-current device shall not be less than the noncontinuous load plus 125 percent of the continuous load.
The result was to remove the 125% allowance for continuous loading for the conductor and leave it in place for the overcurrent device. There was a solid technical basis for that distinction, which remained unchanged in the 1993 NEC. In general, overcurrent protective devices will malfunction if subjected to more than 80% of their rating for continuous periods of time (note that 125% is simply the reciprocal of 80%). The UL Electrical Construction Materials Directory (the "Green Book") contains the following rule under the product category "Circuit Breakers, Molded Case, and Circuit Breaker enclosures":
Unless otherwise marked, circuit breakers should not be loaded to exceed 80 percent of their current rating, where in normal operation the load will continue for 3 hours or more.
A similar rule applies to fused switches. These requirements are conditions of the product listing and as such they are enforceable under Sec. 110-3(b). They also are covered directly in the Code, in Sec. 220-10(b) for feeders and, per Sec. 23042(a), for services by reference. Sec. 2203(a) and Sec. 210-22(c) incorporate this principle for branch circuits. This rule relates solely to the mechanical functioning of the devices to which the conductors are connected. It has nothing to do with protecting the conductors themselves.
This is because the ampacity of a conductor is inherently a continuous property due to the way it's defined in Art. 100:
Ampacity: The current in amperes that a conductor can carry continuously under the conditions of use without exceeding its temperature rating.
This was the reason conductors were treated differently from overcurrent devices in the 1987, 1990, and 1993 NEC. Although those earlier rules were reaffirmed on 25 separate occasions by the Code-making panel, they were substantially reversed in the 1996 NEC. The 1996 NEC seemingly went back to 1984, but not completely. The new wording is:
(b) Continuous and Noncontinuous Loads. Where a feeder supplies continuous loads or any combination of continuous and noncontinuous loads, the rating of the over-current device shall not be less than the noncontinuous load plus 125 percent of the continuous load. The minimum feeder circuit conductor size, without the application of any adjustment or correction factors, shall have an allowable ampacity equal to or greater than the noncontinuous load plus 125 percent of the continuous load.
Exception: Where the assembly including the overcurrent devices protecting the feeder(s) are listed for operation at 100 percent of their rating, neither the ampere rating of the overcurrent device nor the ampacity of the feeder conductors shall be less than the sum of the continuous load plus the noncontinuous load.
Remember, overcurrent devices, in fact devices in general, do not have ampacities. Only conductors have ampacities. Therefore, as we will stress over and over in this article, you must always be prepared to do a series of independent calculations, some related to conductor ampacities, some related to allowable terminations, and some related to how an overcurrent device will respond to continuous loads. Each of these calculations has an independent technical basis, and therefore each must stand alone.
This article shows you, step by step, how to apply the new language. You have to size the overcurrent device, and then figure the conductors not one, not two, but three different ways to be sure you are correct. Only one of these calculations is a true ampacity calculation. The other calculations make certain that the overcurrent device to which the conductors are connected functions within tested parameters. Then, and only then, you can select a conductor that will be as small as your competition is likely to bid and still survive the conditions of use and have overcurrent protection within Code limits. This principle of making and then comparing independent calculations has never been more important than in the 1996 NEC.
Test 1
The first test, although not actually a conductor rule, does influence the choice of conductors under other tests:
Will the overcurrent device accommodate the continuous portion of the load? Unless listed for operation at 100% of a continuous load, the overcurrent protective device must be rated at least 125% of the continuous portion of the load, plus 100% of the noncontinuous portion. The overcurrent protective device may be rounded up to the next higher standard size, subject to the usual 800 A limitation.
Example 1: On a 120/240V system, there is 16,200VA of continuous load and 12,200VA of noncontinuous load. What is the minimum size overcurrent protective device that could be used? Note that these loads are taken from Example 3 in Chapter 9 of the NEC, the only example to combine continuous and noncontinuous loads.
Answer:
Step 1: (16,200VA x 1.25) + 12,200VA = 32,450VA
Step 2: 32,450VA/240V = 135A
Step 3: Rating (next larger size per Sec. 240-6) = 150A
This rule is familiar to the industry; it has been a requirement in the Code for generations. This rule has not applied directly to the conductor since the 1984 NEC; however, in the 1996 NEC, it now does. This new rule, as we will see however, differs from the 1984 NEC in an important way.
Test 2
Will the conductor terminations make adequate allowance for the heating effects, both at the terminals and within the over-current device, created by all elements of the connected load, both continuous and noncontinuous? Qualified testing laboratories make assumptions about the sizes of conductors that will be used and the maximum temperature that their device terminations will reach when testing overcurrent devices, and these become restrictions on the use of these products.
The first assumption is a worst-case assessment of the maximum temperature the termination will reach, even without continuous loading. This has been an enforceable part of test lab requirements for many years, and is now in as Sec. 110-14(c).
A second assumption is that a conductor will be a partial heat sink for heat generated within a continuously loaded device. This is the basis for the major change in these procedures for the 1996 NEC. The Code now requires building in an additional 25% of the continuous portion of a load when sizing conductors. This provides "headroom" at the termination so that heat can be dissipated more easily.
This is potentially far more restrictive than previous requirements that only considered the absolute temperature of a termination. The temperature limitations in Sec. 110-14(c) still apply in two ways, however. First, if the load is noncontinuous, these limitations apply directly. Secondly, if you are dealing with continuous loads, you still must start there to establish the maximum temperature rating you'll use to figure the minimum conductor sizes that will allow your terminations to function in accordance with the test lab assumptions:
The temperature thresholds in this section (normally 60 [degrees] C for 100A and below and No. 1 terminations or smaller; 75 [degrees] C for higher rated equipment) set the column in Table 310-16 (or other applicable ampacity table) that you need to refer to in selecting a conductor with respect to terminations. In the case of purely noncontinuous loads, you use this section directly.
Here's how. Open your Code Book and look at the insulation temperature columns in Table 310-16. Normally these are used to decide the maximum current a conductor can safely carry. That isn't true in this case. Here, we use the table values in reverse, to predict how hot a conductor will run under a specified load. Consider a No. 2 THW (or other 75 [degrees] C) conductor with an ampacity of 115A. That entry in the table means we can infer that a No. 2 conductor, regardless of insulation type, will run at some temperature over 75 [degrees] C if it is loaded beyond 115A, and conversely that the same piece of copper will not exceed 75 [degrees] C if the loading doesn't exceed 115A.
In the case of continuous loads, there's an additional step. We need to build in additional conductor sizing to provide enough "headroom" for the overcurrent device to properly dissipate heat. This is figured as an additional 25% of the continuous portion of the load, and read from the same ampacity column as would be used for noncontinuous loads, i.e., the column corresponding to the temperature thresholds in Sec. 110-14(c).
Although nominally referring to ampacity, these requirements have nothing to do with ampacity in the pure sense. They are saying, in effect, don't take a slug of copper (or aluminum), bolt it to a device, and cause it to exceed a certain temperature. Remember, a metal-to-metal electrical connection is also a pretty good thermal connection. Comparatively very few devices function normally with a thermally coupled heat source approaching the boiling point of water, as could be the case with THHN or other 90 [degrees] C conductors. In addition, the 1996 NEC says we need to add some more mass at the terminations so the conductor will work as a heat sink in conjunction with the overcurrent device.
This rule stands apart from rules covering conductor derating (for conductor fill and/or ambient temperature) in raceways or cable assemblies. Judge the termination, then independently judge the raceway or cable assembly, and only then determine and meet the worst case. Never begin derating calculations (which follow in the third test) with the conductor ampacity determined by this test. Here's an example:
Example 2: Given the rating of the over-current protective device from Example 1 (150A), what conductors would be permitted to terminate on such a device?
Answer:
Step 1: Determine the basic temperature limitation for the device based on Sec. 11014(c) and Sec. 110-3(b): 75 [degrees] C.
Step 2: Using the appropriate temperature column, base the minimum size of the conductors on the actual load as adjusted if required for continuous operating conditions. In this case, per Example 1, use 135A. Column: 75 [degrees] C [Sec. 110-14(c)]
No. 1 Copper: 130A
No. 1/0 Copper: 150A
No. 2/0 Aluminum: 135A
At 135A, No. 1 would run hotter than 75 [degrees] C, requiring rejection; No. 1/0 Copper (or 2/0 Aluminum) must be used. This is the minimum termination size regardless of insulation type and other conditions of use. Be careful, however. Not all of these conductors necessarily comply with other rules that must be followed, as explained in other tests.
Note that the actual current flowing through the conductors is: (16,200VA + 12,200VA) / 240V = 118A. The additional 17A of capacity (for the total of 135A in Step 2 above) is phantom load. The No. 1 Copper (75 [degrees] C Ampacity = 130A) will not overheat this termination of its own accord. The conductor itself would not be damaged at all. However, it will not provide an adequate heat sink so that the over-current device will operate properly. This is an example of the 1996 NEC change; a No. 1 conductor could have been used under the prior three Codes because it would be protected by the 150A overcurrent device per Sec. 240-3(b). The minimum conductor size for a 3-conductor feeder to one of the stores has gone up from No. 1 to 1/0.
Test 3
Will the conductors overbear under the assumed loading and conditions of use? The conductors must have sufficient current-carrying capacity so they will carry the calculated load without exceeding their ampacity, as taken from the table and adjusted for numbers of conductors in a raceway or cable, and adjusted for ambient temperature.
Note that the definition of ampacity, as given in Art. 100, is the continuous current carrying rating of a conductor, under the conditions of use. Therefore, make no adjustment for continuous loading now. As required in Sec. 220-10(a), make sure that the conductor ampacity after derating will carry the load. No next-higher-standard-size allowance applies at this stage.
The Code cannot possibly give a wire size for every load under every condition of use; it can only rate the available sizes of conductors, as it does in Table 310-16. Where no more than three conductors are in a raceway or cable, and where the ambient temperature does not exceed 30 [degrees] C, the table can be used directly. With many conductors or high ambient temperatures, the table values require adjustment, as follows.
Multiple conductors: If there are more than three conductors in a raceway or cable assembly, then this condition will decrease the current that can be carried in a conductor, because there will be more sources of heat within the same confined area. This condition is covered in Note 8(a) to the ampacity tables. The requirement is expressed as a series of factors by which conductor ampacities are to be multiplied. We usually want to go the opposite way, and take a given load and find the required conductor size. Since this is opposite to the table, users should remember to divide the specified load by these factors, which is opposite to multiplication.
Example 3: Suppose the calculation in Example 1 applied to each of two stores, and their feeders were run together in the same raceway. What conductors would not overheat in the combined raceway?
Answer: This would be a condition of 4 to 6 conductors within a raceway, calling for a 0.8 derating factor per Note 8(a) to the tables. As explained, we divide the required load rating (118A) by this factor: 118A / 0.8 = 148A. Any conductor with a table ampacity equal to or greater than 148A could be used, such as 3/0 TW, 1/0 THW, 1 THHN, or 3/0 THW aluminum. Note that we didn't include the 17A of phantom load from Ex. 2. Therefore, some of these conductors may not comply with other rules that must be followed, as explained in Tests 2 and 4. At this point, we are only concerned with how a given conductor will behave in the cable or raceway under the conditions of use.
There simply isn't any substitute to doing a series of independent calculations. Sometimes the termination rules will be the limiting factor, and sometimes the conductor ampacity will be limiting. They must be figured independently, and then compared at the end to determine the worst case.
This is where the new provision is more technically correct and superior to the 1984 NEC rule. The 1984 NEC required an ampacity adjustment of 125% generally, even along the length of the feeder where that increase wasn't relevant. In this example, we would have had to include that phantom load, resulting in 135A / 0.8 = 169A. This would have added a wire size and accomplished nothing in terms of safety.
High temperature ambients: If the conductors are located in a higher temperature ambient, then the allowable ampacities are reduced by multiplying by the factors provided in the lower section of the table. Again, for practical reasons the table refers to loading of given conductors; we usually want to go the opposite way. That is, we want to know what conductor could be used for a given load. Since this is opposite to the table, users should remember to divide the specified load by these factors in this case, which is opposite to multiplication.
Example 4: If the conductors in Example 3 were to run in a 40 [degrees] C ambient, what size would they have to be in order to not be damaged by overheating? Note once again that not all of the conductors shown in the answer comply with other rules that must be followed, as explained in Tests 2 and 4.
Answer: Divide the required load rating (118A) by the correction factor that applies to the insulation category in question:
For TW, 118A/0.82 = 144A;2/0 permitted
For THW, 118A/0.88 = 134A; 1/0 permitted
For THHN, 118A/0.91 =130A;2 permitted
For Al THW, 118A/0.88 = 134A; 2/0 permitted
Multiple conductors and high temperature at the same time: If more than three conductors are run together in high ambient temperatures, then both derating factors will apply. As in Examples 3 and 4, when the load is known, we divide the load by the derating factors, one after another. The order doesn't matter; in general x/y/z = x/z/y.
Example 5: If the conditions in Examples 3 and 4 were both true (4 to 6 conductors in a common raceway at 40 [degrees] C), now how large would the conductors have to be? As with the previous three examples, at this point we are only looking at how the conductors will behave in the raceway under the conditions of use, and the results may not comply with other rules explained later. See Example 7.
Answer: Divide the required load rating (118A) by the correction factor that applies to the category of insulation in question, and then by the factor for multiple conductors in a raceway from Note 8(a):
For TW, 118A/0.82/0.8 = 180A; 4/0 permitted
For THW, 118A/0.88/0.8 = 168A; 2/0 permitted
For THHN, 118A/0.91/0.8 = 162A; 1/0 permitted
For Al THW, 118A/0.88/0.8 = 168A; 4/0 permitted
Multiple ampacities on a given circuit run. Sec. 310-15(c) applies to the condition where more than one ampacity would apply to a given circuit. This could happen if a raceway, generally at normal temperatures, passed through a boiler room or other high temperature area on the way to its final destination. In addition, multiple circuits could be present at the beginning of a conduit ran, and then exit through tee fittings leaving only the circuit being calculated.
In these cases, the lowest calculated ampacity would be used, subject to an exception that recognizes the ability of conductors to function as a heat sink, in this case for adjacent segments of the conductor and not for a connected device:
Exception: Where two different ampacities apply to adjacent portions of a circuit, the higher ampacity shall be permitted to be used beyond the point of transition, a distance equal to 10 feet (3.05 m) or 10 percent of the circuit length figured at the higher ampacity, whichever is less.
Example 6: Suppose the combined raceway in Example 3 were 8 ft long into a junction box or tee conduit body. From that point, a 5-ft length of EMT dropped into a panelboard for the first store, and the other raceway extended an additional 80 ft to a panelboard for the second store. Would the conductor ampacities need to be reduced in consideration of the number of conductors in the common raceway in this case?
Answer: The exception would not help the first store since 10% of the port,on of its circuit figured without derating (5 ft x 0.1 =0.5 ft) is far less than the length subject to derating, or 8 ft. The second store would benefit, however. Although the first store would need conductors with an acceptable rating after derating for the number of conductors in the raceway, the second store would be permitted to use the basic table value.
This is the only test concerned with ampacity in the true sense. It applies to conductors within their raceway or cable assembly. It is entirely separate from and has no relation to making-an allowance for continuous loading on an overcurrent device, or the temperature rating of a device termination, or whether the overcurrent device is properly protecting them.
Test 4
Will the overcurrent protective device selected in the first test protect the conductors selected in the third test, within the parameters of See. 240-3? The over-current protective device must always protect its conductors under their actual conditions of use.
Sec. 215-3 requires overcurrent protection for all feeders, and in general, Sec. 240-3 requires this rating to not exceed the ampacity of the feeder conductors. However, Sec. 240-3(b) permits the next higher standard size overcurrent device to be used on circuits below 800A.
This principle also applies to branch circuits except those supplying multiple receptacle outlets. Testing laboratory requirements also track these requirements, recognizing the termination of, for example, 4/0 THW (ampacity 230A) as a 75 [degrees] C termination on a 250A device. This means that as long as the final ampacity of a conductor exceeds, even by only one ampere, the rating of a standard size of overcurrent protective device, it may be protected by the next higher standard size (assuming not over 800A, etc.).
Example 7: What would be the smallest size conductor rated 90 [degrees] C or lower that could supply the first store in Example 6, if the ambient temperature were 40 [degrees] C? This, in effect, duplicates the conditions in Example 5. The load will remain the same as described in Example 1.
Answer:
Step 1: Determine the minimum adjusted ampacity:
Next lower standard size: 125A 125A + 1A = 126A.
(If the final ampacity after all derating is 126A or more, the 150A overcurrent device, being the next higher standard size, will protect the conductors.)
Step 2: Determine the derating factors:
Temperature, 0.91; Conductor fill, 0.8.
Step 3: Divide Step 1 by Step 2
126A/0.91/0.8 = 173A
Result: No. 2/0 THHN, ampacity 195A
NOTE: No. 2/0 THW (table ampacity 175A) would not work because its temperature derating factor is larger than for the THHN. Here's the calculation:
126A/0.88/0.8 = 179A required, or 3/0 THW, ampacity 200A.
Normally, the artificially inflated loads that have to be included to account for required heat sinking for continuous loading result in coincidental compliance with this test. That isn't necessarily so, however. In this case, the 150A overcurrent device will not properly protect and cannot be used with 1/0 THHN, even though:
(1) 1/0 THHN has sufficient ampacity to accommodate the conditions of use (see Example 5); and
(2) 1/0 THHN is large enough to meet all the termination requirements (see Examples 1 and 2).
Example 8: Recalculate Example 3 assuming no continuous load, and determine suitable conductors.
Answer: The load is still 118A, but the overcurrent device could be 125A. A No. 1 conductor (75 [degrees] C column temperature of 130A) will not overheat the terminal. The raceway is subject to Note 8(a), however.
No. 1 THW: 130A x 0.8 = 104A, which must be rejected, but No. 1 THHN: 150A x 0.8 = 120A, OK
This is a good example of the utility of higher temperature insulation in spite of termination rules. Although high temperature insulation won't help with heating at terminations, it can be very helpful in countering derating if that is required over the length of the raceway or cable. There are many instances where the careful use of this principle will allow a raceway size reduction as well as smaller conductors.
Summary
The four rules for conductor sizing can be summarized as follows. They will apply, in some form, to virtually every circuit covered by the NEC:
1. The overcurrent device must handle the continuous part of the load, figured at 125%, plus the noncontinuous part.
2. The conductors must be sized with enough headroom so they are able to fulfill their heat sinking function at the terminations, and without overheating those terminations under the applicable temperature limitations. This is true even if the conductor itself will not be damaged.
3. The conductor size and insulation type must be coordinated with the actual conditions of use so the insulation will not degrade over time.
4. The overcurrent protective device selected under Rule 1 must properly protect the conductors selected under the Rule 3.
The most important thing to remember is not to mix up apples and oranges. The termination rules are one, and the raceway or cable rules are another. The rules that apply to the raceway or cable have their own technical validity without affecting the terminations, and vice-versa. As a concluding exercise, we can verify that the 2/0 THHN shown in Example 7 does indeed meet all the rules. Note that in this case we are determining the ampacity of a given conductor, and therefore we begin with the Table value, multiply by the required factors, and then look at the load:
Verification, Test 1: The 150A Overcurrent device (determined in Example 1) is correct per Sec. 220-10(b).
Verification, Test 2: The operating temperature of 2/0 THHN does not reach 75 [degrees] C until it is carrying 175A. Therefore, even with a 135A load allowance for continuous loading, it will not overheat the terminals or impede necessary heat dissipation, per Sec. 110-14(c) and Sec. 110-3(b).
Verification, Test 3: The Table ampacity of 2/0 THHN is 195A. This must be adjusted for the conditions of use:
195 x 0.91 x 0.8 = 142A
This exceeds the actual loading of 118A.
Verification, Test 4: The 150A overcurrent device will protect a 142A conductor because it is the next higher standard size. Note that 1/0 THHN (Table Ampacity = 170A) would violate Sec. 240-3(b):
170A x 0.91 x 0.8 = 124A
The 150A overcurrent device is more than the next higher standard rating.