Apply the Cec, Regulations and Standards

competency G1

APPLY THE CEC, REGULATIONS AND STANDARDS

Student Activities

Student Activity In the table depicted in Figure 1, indicate (as per examples) if the conductor insulation ratings are permitted for the given termination temperature ratings, and whether any de-rating is required.

Solution: CEC Rule 4-006 Temperature limitations (see Appendix B)

Termination rating (°C) / Conductor insulation rating
60°C / 75°C / 90°C
60°C / OK / OK (at 60°C rating) / OK (at 60°C rating)
75°C / NOT OK / OK (at 60°C or 75°C rating) / OK (at 60°C or 75°C rating)
60/75°C / OK (at 60°C rating) / OK (at 60°C or 75°C rating) / OK (at 60°C or 75°C rating)
90°C / NOT OK / NOT OK / OK

Figure 1 Maximum conductor termination temperature chart

Student Activity If the four conductors in Example #4 were installed where the ambient temperature could rise to 40°C, would a Table 5A de-rating be required in addition to the Table 5C de-rating?

Example 4 Find the ampacity of four #1/0 AWG copper conductors with RW75 insulation when installed in conduit.

Solution:

Where the surrounding temperature is over 30° C, sub CEC Rule (8) of Rule 4-004 requires the use of a correction factor from Table 5A.

As already determined, Table 5C required a correction factor for 4 conductors of 0.80, and Table 5A requires an additional correction factor of 0.88.

Therefore, the ampacity of each conductor = 150 A x 0.80 x 0.88 = 105.6 A.

Student Activity If the four conductors in Example #4 were feeding a 120/208 V three phase load, would a Table 5C de-rating be required?

Example 4 Find the ampacity of four #1/0 AWG copper conductors with RW75 insulation when installed in conduit.

Solution:

In this example, since the fourth conductor is a neutral, its’ current (if any) will only be the unbalanced current from the other three conductors. Sub CEC Rule (3) of Rule 4-004 does not require you to count the neutral for de-rating purposes. Therefore, we would consider 3 current carrying conductors for determining conductor size and ampacity. Therefore, the ampacity of each conductor = 150 A.

Student Activity If one of the metric 78 (3”) conduits in Example #20 (Figure 18) was installed opposite the other metric 78 (3”) conduit, resulting in a straight pull, while the remaining five conduits were left as is on the existing walls, would a different size pull-box be required?

Figure 18 Straight and angle pulls

Solution:

In this example, since one conduit produces a “straight” pull, while the other two conduits remain as “angle” pulls, we require the use of CEC Rules 12-3036 (2)(b) and (c).

Length A = 8 x 78

= 624 minimum inside dimension

Length B = 6 x 53 + (27)

= 345 minimum inside dimension

Minimum distance between conduit entries (C):

• for 53 (2") conduit = 6 × 53 = 318 mm

• for 27 (1") conduit = 6 × 27 = 162 mm

Student Activity If the transformer rated over 750 V in Figure #28 was a 225 kVA, 14.4 kV: 25 kV single phase, step-up transformer, what difference (if any) of maximum overcurrent protection would there be on a fuse versus a circuit breaker?

Figure 28 Individual transformer overcurrent protection

Solution:

IL = VA ÷ EL

= 225 000 VA ÷14 400 V

= 15.625 A

CEC Rule 26-252 (1)(a) indicates a rating of not more than 150% of the rated primary current for fuses, therefore 15.6 x 1.5 = 23.4 A.

CEC Sub-rule (2) permits the next higher standard rating, should 150% of the primary current not correspond to a standard rating. Therefore, a 25 A fuse is permitted.

CEC Rule 26-252 (1)(b) stipulates a rating of not more than 300% of the rated primary current for breakers, therefore 15.6 x 3 = 46.9 A. Therefore, a 45 A maximum breaker size is allowed.

Student Activity If the other than dry type of transformer in Figure #30 was a 37.5 kVA, 600 V: 240 V single phase, step-down transformer, what difference (if any) of maximum overcurrent protection would there be on a fuse versus a circuit breaker?

Figure 30 Individual transformer overcurrent protection

Solution:

IL = VA ÷ EL

= 37 500 VA ÷ 600 V

= 62.5 A

CEC Rule 26-254 (1) indicates a rating of not more than 150% of the rated primary current, therefore 62.5 x 1.5 = 93.75 A.

CEC Sub-rule (2)(a) permits the next higher standard rating, should 150% of the primary current not correspond to a standard rating. Therefore, a 100 A overcurrent device (either a fuse or a breaker) is permitted.

Student Activity If Motor 1 in Example 27 (Figure #46) was for use in a Varying duty service with a 30 minute rating, what difference (if any) would be required of its’ branch circuit minimum conductor size and/or its’ maximum overcurrent protection?

Figure 46 Motor 1 branch circuit riser

Solution:

Minimum ampacity of the branch circuit conductors

CEC Rule 28-106 (2) applies here. The branch circuit conductors shall have an ampacity of the applicable percentage given in Table 27 of the full load current rating. Table 27 specifies that the conductor size for a Varying duty service with a 30 minute rating shall be not less than 150% of the full load current rating of the motor.

1.5 x 96 A = 144 A

Therefore, Table 2 requires AWG No.1/0 TW75 (150 A) copper conductor as the minimum that are to be used for the branch circuit conductors.

Note, there is no change to the branch overcurrent protection.

Student Activity If Motor 1 in Example 30 (Figure #51) was for use in a Varying duty service with a 30 minute rating, what difference (if any) would be required of the feeder circuit minimum conductor size and/or the feeder circuit maximum overcurrent protection?

Figure 51 Riser for motors 1, 2 and 3.

Solution:

Minimum ampacity of the feeder conductors

CEC Rule 28-108 (1)(c) applies here because the feeder supplies both continuous and non-continuous duty service motor loads. The three full-load currents (FLA) of the motors are:

Motor 1 96 A
Motor 2 27 A
Motor 3 156 A

According to the CEC Rule, you must add 125% of the largest continuous duty service motor full load current to the full load current of the other continuous duty service motor, and the calculated current of the non-continuous duty motor being fed by the same feeder.

(156 x 1.25) + (96 x 1.5) + 27 = 366 A

The minimum feeder conductor ampacity is 366 A. According to Table 2, a 500 kcmil TW75 (380 A) conductor is required.

Note, there is no change to the feeder overcurrent protection.

Student Activity All of the overcurrent devices for Example #1 are marked with a maximum conductor termination temperature of 60/75°C and also marked for continuous operation at 80%. Would the total suite feed calculated need any adjustment for the circuit breaker and wire sizing if NMD90 copper was used as the feeder?

Solution:

CEC Rule 8-202 (2) indicates that the loads in dwelling units shall not be considered to be a continuous load for the application of Rule 8-104 (6). Therefore, we are permitted to load the circuit to 100% of its’ 60 A minimum rating required by CEC Rule 8-202 (1)(b). As a result, we require a 2P60A circuit breaker.

Maximum ampacity of the NMD90 conductors would be based on the 75°C column of Table 2. Therefore, we require a #6/3 NMD90 (65 A).

Would the total house load sub-feed calculated need any adjustment for the circuit breaker and wire sizing if NMD90 copper was used as the feeder?

Solution:

CEC Rule 8-202 (2) indicates by its’ omission of sub-rule (4) that loads not located in dwelling units shall be considered to be a continuous load for the application of Rule 8-104 (6). Therefore, we are required to load the circuit to a maximum of 80% of its’ rating. As a result, we require a 3P70A (69.4 A calculated) circuit breaker.

Maximum ampacity of the NMD90 conductors would be based on the 75°C column of Table 2. Therefore, we require a #4/4 NMD90 (85 A).

Would the total main service feed calculated need any adjustment for the circuit breaker and wire sizing if R90XLPE 600V unjacketed copper was used as the feeders?

Solution:

CEC Rule 8-202 (2) indicates that the total load calculated according to CEC Sub-rule (1) and Sub-rule (3)(a), (b), and (c) shall not be considered to be a continuous load for the application of Rule 8-104 (6). Therefore, we are permitted to load this portion of the circuit to 100% of its’ rating.

However, Rule 8-202 (2) indicates by its’ omission of sub-rule (3)(d) and (e) that loads not located in dwelling units shall be considered to be a continuous load for the application of Rule 8-104 (6). Therefore, we are required to load this portion of the circuit to a maximum of 80% of its’ rating.

As a result, we require a 3P200A (184.9 A calculated) circuit breaker.

Maximum ampacity of the R90XLPE conductors would be based on the 75°C column of Table 2. Therefore, we require a #3/0 R90XLPE (200 A).

Student Activity All of the overcurrent devices for Example #2 are marked with a maximum conductor termination temperature of 60/75°C and also marked for continuous operation at 80%. Would the total suite feed calculated need any adjustment for the circuit breaker and wire sizing if TW75 copper was used in electrical non-metallic tubing as the feeder?

Solution:

CEC Rule 8-202 (2) indicates that the loads in dwelling units shall not be considered to be a continuous load for the application of Rule 8-104 (6). Therefore, we are permitted to load the circuit to 100% of its’ rating. As a result, we require a 2P100A (89.2 A calculated) circuit breaker.

Maximum ampacity of the TW75 conductors would still be based on the 75°C column of Table 2. Therefore, we require a #4 TW75 (85 A).

NOTE: Refer to 4-004 (23) and Table 39.

Would the total house load sub-feed calculated need any adjustment for the circuit breaker and wire sizing if TW75 copper was used in electrical non-metallic tubing as the feeder?

Solution:

Maximum ampacity of the TW75 conductors would still be based on the 75°C column of Table 2. Therefore, we require a #1 TW75 (130 A).

Would the total main service feed calculated need any adjustment for the circuit breaker and wire sizing if TW75 copper was used as the feeders?

Solution:

As a result, we require a 3P400A (361 A calculated) circuit breaker.

Maximum ampacity of the TW75 conductors would still be based on the 75°C column of Table 2. Therefore, we require a 500 kcm TW75 (380 A).