ptonsparky
Tom
- Occupation
- EC - retired
Along with Kwires suggestion, every project you are on at work should bring about a scenario to work with.
I’m going to look this over in more detail tomorrow. I’ll see what I can come up with.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'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.I’m going to look this over in more detail tomorrow. I’ll see what I can come up with.
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.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.
ok. I calculated a total load of 59 amps. Is this correct?[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.
Your formula is right, to determine minimum feeder ampacity, the values you input are not.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.
No it wont be a 60a on the feeder. See 430.52 or click on my hint above.8 wire 60 amp breaker.
8 wire 60 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.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.
My first calculation was a total of 47 amps1 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
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?