Studying

hoody32

Member
Location
norwalk, ct
Are you taking a test for a license that is limited to mostly residential work? If so you may not find as many if any motor circuit related questions.

Experience does help , even more if you are involved in design stages. You could spend a lot of time installing motor circuits but if you only did what someone else told you to do or work off their design, you won't learn as much about how to make selections of materials.

Practice with test preparation materials or other mock scenarios is next best thing. If you don't want to spend $$ on providers of such things you can make up your own scenarios, do your calculating the present things to us on this site (show how you came up with your results) and we can help you figure out what you did right or wrong.

Simple problem to start with might be:

supply volts 480 volts three phase, 30C ambient, all terminations are 75C and all copper conductors having 90C insulation

Motors (all three phase motors) 1 HP, 1HP, 5HP 10 HP.
All motors have 1.15 SF, nameplate ratings are 1.9, 1.9, 6.9 and 13 amps.

One other continuous heating load of 15kW, 3 phase 3 wire 480 volts.

Calculate minimum Feeder size, and feeder overcurrent protection to supply these loads.

For sake of more code usage exercises assume we need to tap the feeder to each branch circuit device and tell us what size feeder tap each one needs (assume all are in close proximity and 10 foot tap rule would apply).

Calculate minimum size circuit conductors for each individual load.

Calculate minimum size conductors needed each branch circuit if all branch circuits were run in a single raceway and then calculate minimum size GRC would be needed to contain them plus the minimum size EGC that would be required to be run with them, also state what size the EGC needs to be.

Determine max allowed overcurrent device using time delay fuses.

For more practice also determine max allowed overcurrent devices using inverse time breakers.

Determine maximum motor overload setting for each motor.

ADD: OK maybe not extremely simple, but needed to get enough variety of things that can apply into one scenario. Look at it as several simple questions all coming from one scenario.
I’m going to look this over in more detail tomorrow. I’ll see what I can come up with.
 

kwired

Electron manager
Location
NE Nebraska
I’m going to look this over in more detail tomorrow. I’ll see what I can come up with.
I'll tell you right now I don't have an already made up answer key for this as did not copy it nor have I calculated any of this, just made it up as I entered it - so when you have some results tell us why you came up with the results you have and we will either pat you on the back or steer you in the right direction.
 

hoody32

Member
Location
norwalk, ct
I'll tell you right now I don't have an already made up answer key for this as did not copy it nor have I calculated any of this, just made it up as I entered it - so when you have some results tell us why you came up with the results you have and we will either pat you on the back or steer you in the right direction.
that’s perfectly fine. I should be able to come up with something later today. I’m headed into work now but I went over some of the information so I can think about how to tackle this problem while I’m at work.
 

hoody32

Member
Location
norwalk, ct
[QUOTE="kwired, post: 2581745, member: 96907"



Simple problem to start with might be:

supply volts 480 volts three phase, 30C ambient, all terminations are 75C and all copper conductors having 90C insulation

Motors (all three phase motors) 1 HP, 1HP, 5HP 10 HP.
All motors have 1.15 SF, nameplate ratings are 1.9, 1.9, 6.9 and 13 amps.

One other continuous heating load of 15kW, 3 phase 3 wire 480 volts.

Calculate minimum Feeder size, and feeder overcurrent protection to supply these loads.

For sake of more code usage exercises assume we need to tap the feeder to each branch circuit device and tell us what size feeder tap each one needs (assume all are in close proximity and 10 foot tap rule would apply).

Calculate minimum size circuit conductors for each individual load.

Calculate minimum size conductors needed each branch circuit if all branch circuits were run in a single raceway and then calculate minimum size GRC would be needed to contain them plus the minimum size EGC that would be required to be run with them, also state what size the EGC needs to be.

Determine max allowed overcurrent device using time delay fuses.

For more practice also determine max allowed overcurrent devices using inverse time breakers.

Determine maximum motor overload setting for each motor.

ADD: OK maybe not extremely simple, but needed to get enough variety of things that can apply into one scenario. Look at it as several simple questions all coming from one scenario.
[/QUOTE] ok. I calculated a total load of 59 amps. Is this correct?
I have to finish the rest when I get home from work
 

kwired

Electron manager
Location
NE Nebraska
[QUOTE="kwired, post: 2581745, member: 96907"



Simple problem to start with might be:

supply volts 480 volts three phase, 30C ambient, all terminations are 75C and all copper conductors having 90C insulation

Motors (all three phase motors) 1 HP, 1HP, 5HP 10 HP.
All motors have 1.15 SF, nameplate ratings are 1.9, 1.9, 6.9 and 13 amps.

One other continuous heating load of 15kW, 3 phase 3 wire 480 volts.

Calculate minimum Feeder size, and feeder overcurrent protection to supply these loads.

For sake of more code usage exercises assume we need to tap the feeder to each branch circuit device and tell us what size feeder tap each one needs (assume all are in close proximity and 10 foot tap rule would apply).

Calculate minimum size circuit conductors for each individual load.

Calculate minimum size conductors needed each branch circuit if all branch circuits were run in a single raceway and then calculate minimum size GRC would be needed to contain them plus the minimum size EGC that would be required to be run with them, also state what size the EGC needs to be.

Determine max allowed overcurrent device using time delay fuses.

For more practice also determine max allowed overcurrent devices using inverse time breakers.

Determine maximum motor overload setting for each motor.

ADD: OK maybe not extremely simple, but needed to get enough variety of things that can apply into one scenario. Look at it as several simple questions all coming from one scenario.
ok. I calculated a total load of 59 amps. Is this correct?
I have to finish the rest when I get home from work
[/QUOTE]
I think I know how you got 59 amps, but that is not correct, is higher than minimum ampacity needed so you would still have sufficient conductor size to pass inspection.
 

hoody32

Member
Location
norwalk, ct
Per 430.24 the sum of:
Highest rated motor 13 amps@ 125%= 16.25
Sum of all others 1.9 1.9 6.9= 10.7
15kw/480=31.25@ 125%=39
I’m calculating a load of 66 amps.
 

kwired

Electron manager
Location
NE Nebraska
Per 430.24 the sum of:
Highest rated motor 13 amps@ 125%= 16.25
Sum of all others 1.9 1.9 6.9= 10.7
15kw/480=31.25@ 125%=39
I’m calculating a load of 66 amps.
Your formula is right, to determine minimum feeder ampacity, the values you input are not.

Take a look at 430.6 then decide what values to plug in for motor current.

15kW 480 volts 3 phase is 15000/480/1.732 = 18 amps is also a mistake on input values.
 

mikeames

Senior Member
Location
Germantown MD
Occupation
Teacher - Electrician - 2017 NEC
1 HP = 2.1 Amps (FLC) ------------------------ Overloads Service factor 1.15 - 430.32 = 1.9 x 1.25


1 HP = 2.1 Amps (FLC) ------------------------ Overloads Service factor 1.15 - 430.32= 1.9 x 1.25


5 HP = 7.6 Amps (FLC) ------------------------ Overloads Service factor 1.15 - 430.32= 6.9 x 1.25


10 HP = 14 Amps (FLC) ------------------------ Overloads Service factor 1.15 - 430.32= 13 x 1.25


15KW = 15000/ (1.732*480) = 18.04 Amps

125% Largest Motor

14X1.25 = 17.5 Amps


Plus Others
17.5
2.1
2.1
7.6
18.04

Total = 47.34 Amps

30C ambient = 100% Table 310.15 B 2A

8 AWG CU is good for 50 amps =- 75c Column of Table 310.15 B 2B

GFP- Short Protection Table 430.52

47.34 X 250% = 118.35 Amps

Round down to standard size in Table 240.6 A


110 Amp breaker
 

kwired

Electron manager
Location
NE Nebraska
8 wire 60 amp breaker.
I come up with needing minimum ampacit of 51.8, which is a little too much for 8 AWG so 6 AWG is minimum conductor needed.

430.6 tells you you need to use NEC tables at end of 430 to determine motor current - guessing you figured that out after I told you to look there.

430.24 says:

Conductors supplying several motors, or a motor(s) and other load(s), shall have an ampacity not less than the sum of each of the following:

125 percent of the full-load current rating of the highest rated motor, as determined by 430.6(A)

Sum of the full-load current ratings of all the other motors in the group, as determined by 430.6(A)

100 percent of the noncontinuous non-motor load

125 percent of the continuous non-motor load.

So you should have largest motor of 14 amps x 1.25 = 17.5
the two 1 hp motors @ 2.1 amps each = 4.2
the five hp motor @ 7.6 amps
the continuous heat load of 15kW should be 18 amps x 1.25= 22.5

sum of all those is 51.8, that is minimum ampacity needed for the feeder.

Feeder overcurrent device is trickier as there is really no code mandated minimum. You probably do want it to be more than 51.8 amps and the 60 amp breaker you chose would probably work in many cases, there is some risk it won't hold while starting the 10 HP motor if all the other loads are running at same time. NEC would allow you to go with up 70 amp feeder breaker on this circuit (with 6 AWG conductor) because of the motor loads (2.5 times largest motor plus all other loads then next standard size up) but may still allow higher if that still doesn't hold while starting.
 

kwired

Electron manager
Location
NE Nebraska
After looking again, I’m going to change my answer to awg 6 and an 80 amp breaker.
I was composing my last reply when you posted this. I had 80 amp breaker initially but that still may be right, need to double check this, depends if we need to count the 15 kw load at 100% or 125% in this part of the process.
 

kwired

Electron manager
Location
NE Nebraska
I was composing my last reply when you posted this. I had 80 amp breaker initially but that still may be right, need to double check this, depends if we need to count the 15 kw load at 100% or 125% in this part of the process.
430.63: Where a feeder supplies a motor load and other load(s), the feeder protective device shall have a rating not less than that required for the sum of the other load(s) plus the following:

15kW load in our example should be calculated at 100% according to this - so the 70 amp breaker would be the correct maximum feeder breaker on a 6 AWG conductor. If you chose to run larger feeder conductor then you would need need to protect it at it's ampacity or next size up if no more than 800 amps, so if you did increase to 4 AWG 90 amp breaker would be allowed.
 

mikeames

Senior Member
Location
Germantown MD
Occupation
Teacher - Electrician - 2017 NEC
I thought feeder calcs got rounded up as well but then I was taught that its rounded down because going above 250% is only allowed as an exception if breaker is tripping. My calcs are above in post #29
 

hoody32

Member
Location
norwalk, ct
Few questions. Where does the service factor come into play? To find the current of the heater how do we know we have to decide by 1.732? What table should I be using for the breaker?
 
Last edited:

hoody32

Member
Location
norwalk, ct
1 HP = 2.1 Amps (FLC) ------------------------ Overloads Service factor 1.15 - 430.32 = 1.9 x 1.25


1 HP = 2.1 Amps (FLC) ------------------------ Overloads Service factor 1.15 - 430.32= 1.9 x 1.25


5 HP = 7.6 Amps (FLC) ------------------------ Overloads Service factor 1.15 - 430.32= 6.9 x 1.25


10 HP = 14 Amps (FLC) ------------------------ Overloads Service factor 1.15 - 430.32= 13 x 1.25


15KW = 15000/ (1.732*480) = 18.04 Amps

125% Largest Motor

14X1.25 = 17.5 Amps


Plus Others
17.5
2.1
2.1
7.6
18.04

Total = 47.34 Amps

30C ambient = 100% Table 310.15 B 2A

8 AWG CU is good for 50 amps =- 75c Column of Table 310.15 B 2B

GFP- Short Protection Table 430.52

47.34 X 250% = 118.35 Amps

Round down to standard size in Table 240.6 A


110 Amp breaker
My first calculation was a total of 47 amps
 

mikeames

Senior Member
Location
Germantown MD
Occupation
Teacher - Electrician - 2017 NEC
Few questions. Where does the service factor come into play? To find the current of the heater how do we know we have to decide by 1.732? What table should I be using for the breaker?
430.32 is the overloads. That's where service factor comes in. You take the name plate (FLA)

Its 3 Phase heater so to find current you have to remember that the current is supplied and returned on all three phases unlike a single phase. 1.732 is the square root of three. Memorize it. We could explain why that is but that's a different topic. Additionally after a while you will memorize 360 and 831 those numbers are 1.732 X the common three phase voltages such as 208 and 480.
 
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