Just for fun Motor design calc

[Wgoodrich]

You have three motors fed by one feeder, one being a 3 phase 10 horse power 208 volt motor. The second motor being a single phase 3 horse power motor 208 volt and one 3/4 horse power motor single phase 120 volt.

What is the minimum copper THHN feeder size required for this 3 phase 4 wire wye feeder serving these three motors from a main service rated panel each on a dedicated branch circuit with individual branch circuit overcurrent device and overloads.

Two hots, one neutral, and one grounding wire in the feeder conduit. No voltage drop concern.

Question, what size feeder is required.

Calculate best you can and provide answer you think to be right. After three replies we will dissect this question and its answer as usual. This again is an off the cuff quest written by me.

Have fun and we all can learn. Get involved to best gain from these fun quests.

Good Luck

Wg

[mdshunk]

Feeder... #4 THHN, 100A inverse time breaker

10HP Branch Circuit.... #8 THHN, 80A inverse time breaker

3HP Branch Circuit... #12 THHN, 50A inverse time breaker

3/4HP Branch Circuit... #14 THHN, 40A inverse time breaker

overloads can only be calculated with actual nameplate data. Assumed 75degree terminations.

[rlfrazee]

Fla for all three motors 63.3 amps plus 25% of largest motor in the branch.
63.3A plus 7.7A equals 71.0 Amps. Minimum ampacity would be 85 amps from 75 degree column considering motor terminations. I am in agreement we need to derate because of four conductors so #4 from seventy five degree column gives 85 amps derated to 68 amps.This would call for us to go to #3 THHN and derate 100 amps to 80 amps. #3 THHN as minimum feeder.
Keeping everything on one post by editing, made a math mistake earlier and corrected to 71 amps.......RL

[volt]

This is a "trick" question sometimes found on certification exams. One needs to consider the load balance before answering this question. Using the FLC's from Tables 430.148 and 430.150, the 3 phase 10 HP 208 volt motor puts 30.8 amps on each phase. Since this is the largest motor, 125% x 30.8 = 38.5 amps on each phase.

For the sake of discussion, assume that the single phase 208 volt 3 HP motor (18.7 FLC) is connected across phases "A" and "B" and the single phase 120 volt motor 3/4 HP motor (13.8 FLC) is connected to the "C" phase.

So the max amps on any one phase will be 38.5 amps + 18.7 amps = 57.2 amps

The "trick" is that since the 3/4 HP motor FLC is less than the 3 HP motor _and_ it will be connected to separate phase from the 3 HP motor, the 3/4 HP FLC does not factor into the calculation, per se.

I assume that the "two hots" is a typo, as a 3 phase feeder requires 3 "hots," each on a separate phase.

Using THHN conductors, a #6 THHN derated from the 90 degree C column (75 amps) for 4 conductors (A Wye connected system requires the neutral to be considered a current carrying conductor), the #6 THHN conductor is then good for 60 amps using 75 C terminals. 75 amps x 80% = 60 amps.

The minimum (THHN) feeder conductors are then a #6 THHN.

Without getting into further discussion at this time, I'll just note that WG "conveniently" selected a resulting FLC number that is 2.2 amps above the 55 amps from the 60 degree C column for a #6.

[mdshunk]

This is a "trick" question sometimes found on certification exams. One needs to consider the load balance before answering this question.

There's no provisions that I can find in Art 430 that would allow the consideration of how the load will be balanced. True, you would balance the load out as you suggest. I just don't see how that fact is permitted to be considered in the feeder calc.

[volt]

There's no provisions that I can find in Art 430 that would allow the consideration of how the load will be balanced. True, you would balance the load out as you suggest. I just don't see how that fact is permitted to be considered in the feeder calc.

That's what makes it a trick _question_. I believe the key word that WG used (and many certification exams will also) was the word _minimum_.

WG wrote:
"What is the minimum copper THHN feeder size required for this 3 phase 4 wire wye feeder serving these three motors from a main service rated panel each on a dedicated branch circuit with individual branch circuit overcurrent device and overloads."

If the 3 HP (208V) motor were connected across the "A" and "B" phases and the 3/4 HP (120V) motor was connected to either the "A" or "B" phases, then yes, the feeder would need to be a #4 THHN.

However, since the _minimum_ wire size was asked for, putting the 3/4HP motor on the "C" phase allows for the smaller #6 THHN conductor.

While the NEC doesn't explicitly require load balancing, the wording of the _question_ itself (asking for the minimum wire size) does require, implicitly, that the load be balanced as much as possible.

[Wgoodrich]

Volt, I would say you have taken a test or two of this calaber. Congrats.

Yes there was many tricks involved in this questions.

Yes there was a typo in the 2 hot conductors that should have been 3 hot conductors.

The word minimum would require you to balance the load on this panel. However the NEC does require the load to be balanced best as you can in a service.

You would place the 120 volt motor on one hot leg and hte 208 single phase motor on the other two hot conductors omitting the 120 volt motor load from the calculation. The calculation demands you use the feeder with the largest load on any of the three hot conductors to size you equipment.

The largest being the 3 phase motor using three hot conductors increased 25" then add the full load current of the one single phase 208 volt motor omitting the single phase motor from your equation.


10 hp 3 ph = 30.8 flc Table 430.150
3 hp 1 ph = 18.7 Table 430.148

(30.8 x 125%) + 18.7 = 57.2 amps on the largest loaded of the three phase hot wires.

Yes there is no unbalanced load applied only being one single phase motor using that white wire making that white wire a grounded leg not a true neutral. Per 310.15.B.2.A rulesm, If it were a true neutral that conductor would be omitted in that conduit count of current carrying conductors yet this case not unbalanced load applies but only one 120 volt component. The grounding wire per 310.15.B.2.A rules the grounding wire is not counted as a current carrying conductor. This makes 4 current carrying conductors requiring a 20% deration in ampacity.

Using the 90 degree column for the deration calculation to start with we find 6 awg copper Table 310.16 / 90 degree column we would find ampacity rate at 75 amps. 75 x 80% = 60 amps rating derated. Now remember 110.14.C calls for 60 degree rating to be used. However if you read the rest of that rule motors normally use 75 degree connections as well as panel breaker and panel connections. This would allow per NEC 110.14.C to use 75 degree column compared to your calculation answer using the lower ampacity rating of the two columns. 75 degree column ampacity in TAble 310.16 would be 65 amps compared to the derated calculation of 60 amps. Therefor the 60 amp is the ampacity of 6 awg wire with 4 current carrying conductors in a raceway.

Now volt caught that next part of the tricks. If you will check in NEC 240.4.B if that ampacity is less than 800 amps then you may increase to next higher number. Your calculation landed on a breaker size listed. 240.4.B is only if the ampacity lands not on a breaker size. This derated ampacity calculation of 60 amps is listed in 240.6 as a normal breaker size. Therefor you must use the 60 amp rating.

Now our load on that feeder is calculated as 57.2 amps and a 6 awg THHN wire calculated ampacity for a motor circuit is 60 amps.

6 awg THHN copper is the answer.

Yes it was set to invite you to be mislead down the path of the 55 ampacity of a 60 degree column only as a rougue further muddying the water.

Volt you did great.

Mdshunk, the load is calculated as connected load only. Only two motors are on the heaviest loaded feeder conductor. The third motor would not be a connected load on that largest loaded feeder conductor.

Mdshunk, you stated actual motor name plate rating is used on flc., check NEC 430.6.A.1 calling for table values to be used for a feeder. You would be correct using name place per NEC 430.6.A.2 to size overload data.

You guys did great !

Did you guys gain from this excercise? Was this exercise worth doing again in question form on other subjects of motors or commercial demand load etc?

Wg

[mdshunk]

Mdshunk, you stated actual motor name plate rating is used on flc., check NEC 430.6.A.1 calling for table values to be used for a feeder. Wg

No, I didn't. I said nameplate data needed for choosing overloads. Stop accusing me! Just because I think the world is out to get me, doesn't make me paranoid! ;)

[Wgoodrich]

Must have been my speed reading again jumping in and not reading what you wrote but what I thought you wrote. Word assume sits in on that one I guess. Sorry Mdshunk.

Wg

[rlfrazee]

Well at least my feeder wasnt too small. I really enjoyed the excercise though dont think at my level of knowledge I would have thought of the balancing of loads. However, I did follow thru the solution and am working on understanding all that was discussed. It will certainly speed up my learning curve on industrial motor circuits. I hope we will continue time to time to do these exercises it cant do anything but make us all more knowledgeable......RL

[mdshunk]

While I agree that there is a general requirement for the load of a service to be reasonably balanced, how does this apply to a feeder? I would agree that a balanced load on a feeder is good practice, I can't seem to find even a general requirement that the load on a feeder to a sub panel be balanced. I can see this feeder coming out of a huge MDP, and an unbalanced load on it would't make one bit of difference on the service balancing. I don't have one of those fancy NEC Cd's to do more elaborate searches... only the book and the handbook. Someone help me out.

[Wgoodrich]

MDshunk,

Take a piece of paper. Forget single phase service think three phase service.

Draw a column for line one , line two, line three, then a neutral wire. Having 5 columns the first being name of motor load. Make 3 copies of that form of 5 columns.

Now take 9 - 120 volt motors [lets say each 120 volt motor pulls 9 amps each] and show them all on line one. Watch what happens to the neutral conductor when you get done calculating total load on each of the three hot and the neutral.

Answer with all 120 volt motors on line one of three phase feeder would call for 90 amp service with a minmum of a 90 amp rated neutral wire.

Now take the second sheet and place those motors evenly on each hot leg of that three phase service. This would put 3 motors pulling on each hot leg of the three hot conductors. Now look at the neutral wire. If you showed one motor on each hot wire evenly then the service neutral instead of being 90 amp rated demand is now zero rated load on the neutral service conductor requiring only a 8 awg copper neutral wire only because the NEC requires neutral service wire being no smaller than the equipment grounding wire if no load is currently on that four wire three phase neutral service conductor.

Now look at total demand load of that second BALANCED sheet. you now have 18 amps on each hot service conductor requiring a minimum of 18 amp service but per code upgrades as a minimum of 60 amp service as a minimum service size allowed if more than one breaker in a panel.

Now for giggles take the third sheet keep the 9 - 120 volt motors balanced and add 1 - 220 volt 9 amp motor load three phase. This would add 9 amps to each hot service conductor but still will not increase the neutral service conductor because it does not add load no neutral load used.

It is very common in an industrial setting to have a 1000 amp service with a 50 amp load total on the neutral service conductor. This is true only if the service is a balanced load.


Same rules apply to a feeder of a sub panel serving many motors. If the load is perfectly balanced with even number of 120 volt motors no load is on the neutral feeder conductor size.

The three sheets I had you make up is called a load balancing sheet. If you do not balance hte load of an industrial service or sub panel you will end up per NEC rules with an outragous sub panel or service size required with two hot conductors doing little of nothing.

This is not true in a residential setting. The breakers are alternated between line one and line two pretty well auto balancing the service. You would have to think to make an unbalanced load in a dwelling and just about only if you planned on doing so. Good electricians would not purposely cause an imbalanced load in a dwelling but if they did they would have to work at it.

Dwellings is much different than three phase services in this nature.

Does this help ?

Wg

[rlfrazee]

Wg thanks for taking the time to explain this balancing of loads. I have pretty well worked thru the rest of the solution and see where I went wrong on all of it except load balancing. I couldnt tie together what was happening to the neutral as far as sizing/ampacity. Biggest problem for me that I can see isnt that I cant understand the solution its that I dont know all the code articles that factor into the solution of the minimum feeder size. Anyway hope the exercises continue......RL

[mdshunk]

No... It doesn't clear up why feeders to subs are required to be balanced. If one feeder heavy on A phase, another feeder heavy on B, and another heavy on C, the net service load can still be balanced. You drug in neutral load, which had nothing to do with the original motor feeder question (which had only one 120 volt motor). Of course you'd want to do your best to balance the neutral load between phases, so that as much as possible "cancels itself out". I'm talking about phase load balancing. Don't mean to beat a dead horse, but it's my nature.

[rlfrazee]

Mdshunk, your not beating a dead horse only meant to say I didnt see the fact the neutral being counted as a conductor and that its ampacity could change depending on load balancing or the fact that you needed to balance loads to determine feeder size minimum. Some of the code that was part of this solution I took a look at but couldnt determine if it applied so steered myself in the wrong direction thinking it didnt. After looking at the solution it does make sense to me.....RL ;)

[Wgoodrich]

Ron

See if this helps concerning you seeking rules of balancing the load in the NEC.

210.11 Branch Circuits Required.
Branch circuits for lighting and for appliances, including motor-operated appliances, shall be provided to supply the loads computed in accordance with 220.3. In addition, branch circuits shall be provided for specific loads not covered by 220.3 where required elsewhere in this Code and for dwelling unit loads as specified in 210.11(C).

(B) Load Evenly Proportioned Among Branch Circuits. Where the load is computed on a volt-amperes/square meter or square foot basis, the wiring system up to and including the branch-circuit panelboard(s) shall be provided to serve not less than the calculated load. This load shall be evenly proportioned among multioutlet branch circuits within the panelboard(s). Branch-circuit overcurrent devices and circuits shall only be required to be installed to serve the connected load.

II. Feeders and Services
220.10 General.
The computed load of a feeder or service shall not be less than the sum of the loads on the branch circuits supplied, as determined by Part I of this article, after any applicable demand factors permitted by Parts II, III, or IV have been applied.


COMMENTS;

As far as i can pick up what you seek is where the NEC requires the load to be balanced. This is not just one rule sentence in the Code. The entire section of 220 per calculation of demand is based on the load being balanced. When you perform a demand load calculation whether commercial, industrial, or residential you start that calculation using Table 220.3.B.2.A. This is the rule mentioned in the above rules of per square foot calculation.

Best I got for you

Wg

[volt]

Volt, I would say you have taken a test or two of this calaber. Congrats.

Yes, I have, thanks. This sort of problem was first brought to my attention some 15 years ago when I was preparing for an exam using Tom Henry's books.

Sorry for the delayed response, but I recently had surgery on my right shoulder and while I did pop in a few times, just didn't feel like typing. The 600 MCM finally whooped me after 27 years.


Did you guys gain from this excercise? Was this exercise worth doing again in question form on other subjects of motors or commercial demand load etc?

Wg

Yes, it's fun, plus it will help people who may be preparing for exams. I see I missed the xfmr quiz. The box quiz was good too.