PE power Q4

[yaoyaodes]

The minimum size THWN copper conductors rated at 75 C installed in conduit required to serve a continuous duty 230-V 10-hp single-phase induction motor and noncontinuous 1-kW resistance heater on a circuit operating at 240-V single phase are:

Can not really follow the solution, can anyone explain to me a little bit.

 

[akyip]

Off the top of my head, since I don't have this problem right in front of me:

The current needed for the heater is simply 1 KW divided by 240 V. So that is:

1000 W / 240 V = 4.167 A

For the motor, you need to use the Motor Tables at the end of NEC Section 430 to find the FLC (full load current) needed to properly size the conductors. It should be Table 430.248 for 1-phase AC motors, off the top of my head. Look for 1-phase, 230 V, 10 HP in this table to find the required FLC. Off the top of my head, I believe it is 50 A.

Then still for the motor, you need to multiply this motor FLC by 125% as per NEC Section 430.22 and 430.24.

50 A x 1.25 = 62.5 A

So the total amps required for both the motor and the heater is:

62.5 A + 4.167 A = 66.667 A

Then you need to go to NEC Table 310.15(B)(16) (if I remember right off the top of my head) to find the correct minimum size conductor that has an ampacity greater than or equal to this total current (66.667 A).

I will double-check my work I did for this problem when I have the time.

 

[akyip]

I should also mention that for this problem, you need to look primarily at NEC Section 430.24. This section specifically covers branch circuits consisting of a combination of motor loads and/or non-motor loads, which this problem is all about.

NEC 430.24 for combination motor and non-motor loads: The required ampacity is the total sum of:

-125% of the FLC (full load current) of the highest-rated motor
-100% of the total sum of FLCs of all other motors
-125% of sum of ampacities of continuous non-motor loads
-100% of sum of ampacities of non-continuous non-motor loads

The given motor is the only motor here, so it is treated as the highest-rated motor. Hence for the motor:

50 A x 125% = 62.5 A

The heater is stated to be non-continuous. So for the heater:

100% x (1000 W / 240 V) = 4.167 A

So the total amps that the conductor is required to serve is:

62.5 A + 4.167 A = 66.667 A

Then use Table 310.15(B)(16) to find the correct minimum conductor size whose ampacity is greater than or equal to this total amps.

 

[yaoyaodes]

Off the top of my head, since I don't have this problem right in front of me:

The current needed for the heater is simply 1 KW divided by 240 V. So that is:

1000 W / 240 V = 4.167 A

For the motor, you need to use the Motor Tables at the end of NEC Section 430 to find the FLC (full load current) needed to properly size the conductors. It should be Table 430.248 for 1-phase AC motors, off the top of my head. Look for 1-phase, 230 V, 10 HP in this table to find the required FLC. Off the top of my head, I believe it is 50 A.

Then still for the motor, you need to multiply this motor FLC by 125% as per NEC Section 430.22 and 430.24.

50 A x 1.25 = 62.5 A

So the total amps required for both the motor and the heater is:

62.5 A + 4.167 A = 66.667 A

Then you need to go to NEC Table 310.15(B)(16) (if I remember right off the top of my head) to find the correct minimum size conductor that has an ampacity greater than or equal to this total current (66.667 A).

I will double-check my work I did for this problem when I have the time.

you are absolutely correct, thanks. I am not so familiar with NFPA 70

 

[yaoyaodes]

Off the top of my head, since I don't have this problem right in front of me:

The current needed for the heater is simply 1 KW divided by 240 V. So that is:

1000 W / 240 V = 4.167 A

For the motor, you need to use the Motor Tables at the end of NEC Section 430 to find the FLC (full load current) needed to properly size the conductors. It should be Table 430.248 for 1-phase AC motors, off the top of my head. Look for 1-phase, 230 V, 10 HP in this table to find the required FLC. Off the top of my head, I believe it is 50 A.

Then still for the motor, you need to multiply this motor FLC by 125% as per NEC Section 430.22 and 430.24.

50 A x 1.25 = 62.5 A

So the total amps required for both the motor and the heater is:

62.5 A + 4.167 A = 66.667 A

Then you need to go to NEC Table 310.15(B)(16) (if I remember right off the top of my head) to find the correct minimum size conductor that has an ampacity greater than or equal to this total current (66.667 A).

I will double-check my work I did for this problem when I have the time.

Thanks, really appreciated. Yeah, I checked. you are correct. how can you remember all these?

 

[akyip]

Thanks, really appreciated. Yeah, I checked. you are correct. how can you remember all these?

Only because I did a lot of repetitive/repeat problem solving with the NCEES practice exam and other sample practice exams I had. I studied for about a year due to my original exam dates being pushed back because of CoViD.

 

[adamcarlisle]

I have a question about this problem. The problem states that " the minimum size THWN copper conductors rated at 75 °C installed in conduit required to serve....". Because the current is less than 100A, and according to 2017 NEC 110.14(C)(1)(a) wouldn't we use the 60 deg Celsius column on Table 310.15(B)(16)? Both come out with the same answer of 4 AWG, but if they didn't would you go with the 60 deg column or the 75 deg column?

 

[adamcarlisle]

I have a question about this problem. The problem states that " the minimum size THWN copper conductors rated at 75 °C installed in conduit required to serve....". Because the current is less than 100A, and according to 2017 NEC 110.14(C)(1)(a) wouldn't we use the 60 deg Celsius column on Table 310.15(B)(16)? Both come out with the same answer of 4 AWG, but if they didn't would you go with the 60 deg column or the 75 deg column?

The question also does specifically say to use the THWN wire. The 60 deg C column is not for the THWN wire type so I think I answered my own question here but if it didn't specifically say THWN maybe you do use the 60 deg C??? who knows lol