# Service Size



## FrunkSlammer (Aug 31, 2013)

Call an electrician. :laughing:


----------



## HARRY304E (Sep 15, 2010)

backstay said:


> I have twenty four 8.1kW (at 208)1ph heaters to run a service for. This is continuos. The incoming power is 120/208 three phase. Can I get away with a 600 amp 3 ph service?



That comes out to 930 amps if I did the math right.


220.51 Fixed Electric Space Heating. Fixed electric space-heating loads shall be calculated at 100 percent of the total connected load. However, in no case shall a feeder or service load current rating be less than the rating of the largest branch circuit supplied.
Exception:  Where reduced loading of the conductors results from units operating on duty-cycle, intermittently, or from all units not operating at the same time, the authority having jurisdiction may grant permission for feeder and service conductors to have an ampacity less than 100 percent, provided the conductors have an ampacity for the load so determined.


----------



## backstay (Feb 3, 2011)

HARRY304E said:


> That comes out to 930 amps if I did the math right.
> 
> 220.51 Fixed Electric Space Heating. Fixed electric space-heating loads shall be calculated at 100 percent of the total connected load. However, in no case shall a feeder or service load current rating be less than the rating of the largest branch circuit supplied.
> Exception:  Where reduced loading of the conductors results from units operating on duty-cycle, intermittently, or from all units not operating at the same time, the authority having jurisdiction may grant permission for feeder and service conductors to have an ampacity less than 100 percent, provided the conductors have an ampacity for the load so determined.


But spread out over a three phase service can I get away with a smaller amp service? I should only have 547 amps on any one leg. Or am I confused.


----------



## Dennis Alwon (May 9, 2009)

I think any phase will see a max of 16 * 8.1kw. So If each phase sees 129600 watts then divide by 360 (208 * 3 sq.rt) and you get 360 amps. So yes a 600 amp feeder should be more than adequate but I screw these up alot---


----------



## backstay (Feb 3, 2011)

Dennis Alwon said:


> I think any phase will see a max of 16 * 8.1kw. So If each phase sees 129600 watts then divide by 360 (208 * 3 sq.rt) and you get 360 amps. So yes a 600 amp feeder should be more than adequate but I screw these up alot---


Thanks Dennis, I was thinking I would see 66% of the load on each leg. That's where I can up with just under 600 amps. I like your number better. I will have to pull out my books and double check, not that I doubt you.


----------



## Vintage Sounds (Oct 23, 2009)

I calculated 539.6A at 208 3 phase. 

Are you saying these heaters are on 100% of the time, 24/7? no thermostat?


----------



## Dennis Alwon (May 9, 2009)

Vintage Sounds said:


> I calculated 539.6A at 208 3 phase.
> 
> Are you saying these heaters are on 100% of the time, 24/7? no thermostat?


I did not take that into consideration. How did you get 539.6 amps


----------



## LGLS (Nov 10, 2007)

I want to see that first electric bill.


----------



## HARRY304E (Sep 15, 2010)

Dennis Alwon said:


> I think any phase will see a max of 16 * 8.1kw. So If each phase sees 129600 watts then divide by 360 (208 * 3 sq.rt) and you get 360 amps. So yes a 600 amp feeder should be more than adequate but I screw these up alot---


Yup,see I knew I messed it up...:no::laughing:


----------



## backstay (Feb 3, 2011)

Vintage Sounds said:


> I calculated 539.6A at 208 3 phase.
> 
> Are you saying these heaters are on 100% of the time, 24/7? no thermostat?


No, they are storage heaters. They charge at night for up to 8 hrs. Off peak rate, under 5 cents a kilowatt hour. Only cheaper heat here is ground source heat pump.

That was mine number too. But that was off the top of my head


----------



## HARRY304E (Sep 15, 2010)

kVA x 1000
1.73 x E

24x8100=194,400/1.73x208=359.84=194,400/359.84= 540.24 amps:blink::laughing:


Okay Dennis how did you come up with 129,600 watts?

I'm missing a key point


----------



## stars13bars2 (Jun 1, 2009)

I got 540amps too.
24 x 8.1 = 194.4k / 1.73 = 112370watts / 208v = 540.24amps.
Harry beat me to it.


----------



## backstay (Feb 3, 2011)

HARRY304E said:


> kVA x 1000
> 1.73 x E
> 
> 24x8100=194,400/1.73x208=359.84=194,400/359.84= 540.24 amps:blink::laughing:
> ...


I think because there are 8 heaters across each phase pair. So they overlap, that's how he got 16. If you draw it out it makes sense. The rest of his calcs, I don't know. But I trust him, besides I'll put his name on the calc sheet!


----------



## HARRY304E (Sep 15, 2010)

backstay said:


> I think because there are 8 heaters across each phase pair. So they overlap, that's how he got 16. If you draw it out it makes sense. The rest of his calcs, I don't know. But I trust him, *besides I'll put his name on the calc sheet!*


 Yeah,that's the ticket...:laughing:


----------



## ablyss (Feb 8, 2014)

I didn't see where he mentioned anything abut delta/wye
Correct me if i'm wrong but if it's a delta wouldn't it only be 208 off of one leg?


----------



## backstay (Feb 3, 2011)

ablyss said:


> I didn't see where he mentioned anything abut delta/wye
> Correct me if i'm wrong but if it's a delta wouldn't it only be 208 off of one leg?


It's a 120/208 wye.


----------



## wildleg (Apr 12, 2009)

I get what Dennis got. 16 heaters on any one phase =360.16 amps. 220.51 says calcuate at 100%, so 400 amp service should do ?


----------



## ohiosparky99 (Nov 12, 2009)

HARRY304E said:


> kVA x 1000 1.73 x E 24x8100=194,400/1.73x208=359.84=194,400/359.84= 540.24 amps:blink::laughing: Okay Dennis how did you come up with 129,600 watts? I'm missing a key point


He's figuring each unit single phase " only using 2 legs" so each phase won't have the full load of all 24 heaters, only 16 max


----------



## Tsmil (Jul 17, 2011)

Sorry, but you will need 600 amp service. Each group of 8 heaters will draw 311.5 amps.... 8100 x 8 / 208= 311.5. Each node will be supplying 2 groups of heaters 120 degrees out of phase with each other... 311.5 x 1.73 = 538.9. Or just work it with total wattage as was shown earlier.


----------



## Laroc3 (Jul 20, 2013)

Tsmil said:


> Sorry, but you will need 600 amp service. Each group of 8 heaters will draw 311.5 amps.... 8100 x 8 / 208= 311.5. Each node will be supplying 2 groups of heaters 120 degrees out of phase with each other... 311.5 x 1.73 = 538.9. Or just work it with total wattage as was shown earlier.


My math is the same.


----------



## Dennis Alwon (May 9, 2009)

Tsmil said:


> Sorry, but you will need 600 amp service. Each group of 8 heaters will draw 311.5 amps.... 8100 x 8 / 208= 311.5. Each node will be supplying 2 groups of heaters 120 degrees out of phase with each other... 311.5 x 1.73 = 538.9. Or just work it with total wattage as was shown earlier.


I don't doubt that you are correct but see if I got it straight. Let's say the loads were 8.1kw but 3 phase instead of single phase. We would have 24 heaters at 3 phase. Then we have 24 * 81000 divided by 360 and we get 540 amps??? No....


----------



## Laroc3 (Jul 20, 2013)

Dennis Alwon said:


> I don't doubt that you are correct but see if I got it straight. Let's say the loads were 8.1kw but 3 phase instead of single phase. We would have 24 heaters at 3 phase. Then we have 24 * 81000 divided by 360 and we get 540 amps??? No....


You just did the same thing. So yes a 600 amp


----------



## Tsmil (Jul 17, 2011)

Dennis Alwon said:


> I don't doubt that you are correct but see if I got it straight. Let's say the loads were 8.1kw but 3 phase instead of single phase. We would have 24 heaters at 3 phase. Then we have 24 * 81000 divided by 360 and we get 540 amps??? No....


Yep. If 3 phase. But where you went wrong earlier is you were calculating single phase heater from single phase of supply and still divided by 380. Should have devised by 208.


----------



## Dennis Alwon (May 9, 2009)

Tsmil said:


> Yep. If 3 phase. But where you went wrong earlier is you were calculating single phase heater from single phase of supply and still divided by 380. Should have devised by 208.


But you all are saying that single phase needs 540 amps-- how could both need the same


----------



## Dennis Alwon (May 9, 2009)

Laroc3 said:


> You just did the same thing. So yes a 600 amp


They are not 3 phase loads in the example-- this is my confusion yet you get the same answer


----------



## Tsmil (Jul 17, 2011)

A watt is a watt is a watt. If the heaters can be distributed equally, as in this case, to give a balanced load, it makes things very simple. 24 heaters at 8100 watts each would be 194400 watts. Divide by 360 and tada....539.6.


----------



## Laroc3 (Jul 20, 2013)

Dennis Alwon said:


> They are not 3 phase loads in the example-- this is my confusion yet you get the same answer


I know there not 3 phase loads, but you have to balance the system. So you put 8 heater on each phase. But each load will be shared between two phases which are120 degrees out from each other so you have to use1.73. X 208. Only because the system is balanced it works out both ways.


----------



## Nukie Poo (Sep 3, 2012)

Laroc3 said:


> I know there not 3 phase loads, but you have to balance the system. So you put 8 heater on each phase. But each load will be shared between two phases which are120 degrees out from each other so you have to use1.73. X 208. Only because the system is balanced it works out both ways.


What they have calculated is the load on one leg of a 3-ph 208 service. calculating for single phase service would yield 935-amps


----------



## wildleg (Apr 12, 2009)

Tsmil said:


> A watt is a watt is a watt. If the heaters can be distributed equally, as in this case, to give a balanced load, it makes things very simple. 24 heaters at 8100 watts each would be 194400 watts. Divide by 360 and tada....539.6.


your calculation would be correct if these were 3 phase heaters.

unfortunately, they are single phase, so you are in error.


----------



## Dennis Alwon (May 9, 2009)

Laroc3 said:


> I know there not 3 phase loads, but you have to balance the system. So you put 8 heater on each phase. But each load will be shared between two phases which are120 degrees out from each other so you have to use1.73. X 208. Only because the system is balanced it works out both ways.


So you are saying that 24 heater that are 3 phase which will take up 72 spaces and draw 8.1kw on each phase is the same as 24 single phase breakers take up 48 spaces with 8.1 kw load on each phase.

I see that as 16 - 3 phase loads instead of 24 and I think that is my issue


----------



## Dennis Alwon (May 9, 2009)

Phase A sees 129,600 watts, phase B sees 129,600 watts and same for phase C when we divide the single phase loads evenly and each phase sees the current 16 times (16*8100)= 129,600

Now if they are 3 phase heaters each phase see (24*8100)=194,400 watts and yet the both need the same size service-- that is what I don't see but I never did understand the phase angles and how it affects everything. Probably never will.


----------



## HARRY304E (Sep 15, 2010)

Ha Ha! 

This is a good test question...:laughing:


----------



## backstay (Feb 3, 2011)

HARRY304E said:


> Ha Ha!
> 
> This is a good test question...:laughing:


After seeing all the back and forth, I don't feel so bad that I had to ask.


----------



## Dennis Alwon (May 9, 2009)

I don't trust myself so I would not go with that but it seems 600 amps is fine


----------



## HARRY304E (Sep 15, 2010)

backstay said:


> After seeing all the back and forth, I don't feel so bad that I had to ask.





Dennis Alwon said:


> I don't trust myself so I would not go with that but it seems 600 amps is fine


This would be a good thread for next door where the super electrical engineers hang out..

.


----------



## Nukie Poo (Sep 3, 2012)

Dennis Alwon said:


> Phase A sees 129,600 watts, phase B sees 129,600 watts and same for phase C when we divide the single phase loads evenly and each phase sees the current 16 times (16*8100)= 129,600 Now if they are 3 phase heaters each phase see (24*8100)=194,400 watts and yet the both need the same size service-- that is what I don't see but I never did understand the phase angles and how it affects everything. Probably never will.


If the heaters were 3ph 8.1kW PER PHASE, you would get 14Kw of heat total. A single phase 8.1kW heater is 8.1KW no matter how you slice it


----------



## Dennis Alwon (May 9, 2009)

Nukie Poo said:


> If the heaters were 3ph 8.1kW PER PHASE, you would get 14Kw of heat total. A single phase 8.1kW heater is 8.1KW no matter how you slice it


I see that-- thank you

3 phase = 14kw/360= 38.8 amps

1 phase= 8.1kw/208= 38.8 amps


----------



## Tsmil (Jul 17, 2011)

Nukie Poo said:


> If the heaters were 3ph 8.1kW PER PHASE, you would get 14Kw of heat total. A single phase 8.1kW heater is 8.1KW no matter how you slice it


When adding wattage, there is no 3 phase calculation. A watt is a watt is a watt. 8100 watts on phase A + 8100 watts on phase B + 8100 watts on place C = 24300 watts.


----------



## Rollie73 (Sep 19, 2010)

I have never understood why some people have trouble with these calcs........but math was always my strong subject so....

A quick easy way to figure a 3 phase service that usually works out ok for me (even if it isnt the 100% correct way) is to figure it as a single phase service and then divide it the square root of three.

24 x 8100 (8.1KW)=194400

194400/208=934.61 amps for a single phase

934.61/1.732=539.616 amps for a 3 phase wye

A 600 amp service should do the job just fine but isn't going to leave any room for future expansion.


----------



## Nukie Poo (Sep 3, 2012)

Tsmil said:


> When adding wattage, there is no 3 phase calculation. A watt is a watt is a watt. 8100 watts on phase A + 8100 watts on phase B + 8100 watts on place C = 24300 watts.


Yeah. How stupid can I be - and I went school for EE, ME, and NE. I'll have to chalk it up to the post-op narcotics Im on right now. Double Inguinal hernia repair. If you ever get one (common among tradesmen) , make sure you have the surgeon check both sides; you DO NOT want to do this twice.


----------



## Rollie73 (Sep 19, 2010)

Nukie Poo said:


> Double Inguinal hernia repair. If you ever get one (common among tradesmen) , make sure you have the surgeon check both sides; you DO NOT want to do this twice.


 
Be careful......even if the surgeon checks both sides and fixes you up.....they can come back. A weakened lining is a weakend lining and it can tear again. My uncle went through hell with the same problems and they could never get him fixed quite right.


Want to share some of those post-op narcotics??:whistling2::whistling2:


----------



## backstay (Feb 3, 2011)

Rollie73 said:


> I have never understood why some people have trouble with these calcs........but math was always my strong subject so....
> 
> A quick easy way to figure a 3 phase service that usually works out ok for me (even if it isnt the 100% correct way) is to figure it as a single phase service and then divide it the square root of three.
> 
> ...


That's what I did originally, I just questioned the answer. And by the talent disagreeing on the answer, I'm not alone.


----------



## Nukie Poo (Sep 3, 2012)

Rollie73 said:


> Be careful......even if the surgeon checks both sides and fixes you up.....they can come back. A weakened lining is a weakend lining and it can tear again. My uncle went through hell with the same problems and they could never get him fixed quite right. Want to share some of those post-op narcotics??:whistling2::whistling2:


Oh great. Did they use the mesh method?


----------



## Rollie73 (Sep 19, 2010)

Nukie Poo said:


> Oh great. Did they use the mesh method?


Yup they did. Just be careful with ANY heavy lifting for a while. Talk to your doctor about it when you go in for a check up.


----------



## Nukie Poo (Sep 3, 2012)

Rollie73 said:


> Yup they did. Just be careful with ANY heavy lifting for a while. Talk to your doctor about it when you go in for a check up.


Thanks to my guys, the heaviest thing I lift nowadays is my Kriegoff K80


----------



## CADPoint (Jul 5, 2007)

backstay said:


> That's what I did originally, I just questioned the answer. And by the talent disagreeing on the answer, I'm not alone.


AH, Fixed electrical heaters are considered continious that will push you up to 675 ~ amps...

Call an 800!

Or a 700 Amp breaker in a 800 panel.


----------



## backstay (Feb 3, 2011)

CADPoint said:


> AH, Fixed electrical heaters are considered continious that will push you up to 675 ~ amps...
> 
> Call an 800!
> 
> Or a 700 Amp breaker in a 800 panel.


Only if the OCD and equipment is not rated for 100 %. 

230.42(A)(2)


----------



## Dennis Alwon (May 9, 2009)

Tsmil said:


> A watt is a watt is a watt. If the heaters can be distributed equally, as in this case, to give a balanced load, it makes things very simple. 24 heaters at 8100 watts each would be 194400 watts. Divide by 360 and tada....539.6.


So earlier you said I should have divided by 208. Then you go ahead and divide by 360.

If you look at these loads as three phase then there wont be 24 of them but rather 16 3 phase loads so it seems like you are mixing them up

Yeah I though I had it- doh


----------



## Tsmil (Jul 17, 2011)

The OP said 24 single phase loads of 8100 watts each. To convert to 3 phase, you will need 3 heaters in delta for balanced load for a total of 24300 watts. You will have 8 groups of heaters to use all 24 heaters for a total wattage of 194,400 watts. This is a balanced load allowing for easiest calculation. 194400/(208x1.73). Giving current per phase. 

Diagrams always help me.


----------



## 99cents (Aug 20, 2012)

dennis alwon said:


> so earlier you said i should have divided by 208. Then you go ahead and divide by 360.
> 
> If you look at these loads as three phase then there wont be 24 of them but rather 16 3 phase loads so it seems like you are mixing them up
> 
> yeah i though i had it- doh


208 x 1.73 = 360


----------



## 99cents (Aug 20, 2012)

You guys are making load calcs too complicated.


----------



## ablyss (Feb 8, 2014)

It is my understanding the motors are setup up to run on a wye 208 2 pole breaker. 208 being the maximum voltage supplied is what you should consider in your calculations. Total watts / voltage = amps. This is the total amps on each leg. No further calculations are required.. No need to use 1.73 or 360 because you already have that value based in the voltage.


----------



## Tsmil (Jul 17, 2011)

ablyss said:


> It is my understanding the motors are setup up to run on a wye 208 2 pole breaker. 208 being the maximum voltage supplied is what you should consider in your calculations. Total watts / voltage = amps. This is the total amps on each leg. No further calculations are required.. No need to use 1.73 or 360 because you already have that value based in the voltage.


Better retread original post. 24 single phase heaters rated at 8100 watts at 208 volts fed from 3 phase panel.


----------



## Dennis Alwon (May 9, 2009)

Thanks all for the help-- I see the issue


I read this thread at mike holt and here is the thread but it is closed... http://forums.mikeholt.com/showthread.php?t=140398 Basically they did this as well as what you all stated and got the same answer










I think the problem is I should be using 4.05kw for each phase . It should look like this


----------



## backstay (Feb 3, 2011)

Thanks for everyone who contributed information to this thread! The discussion was eye opening. I had the right answer(I think) but didn't really see the whole thing. This is a very large project for a one man show and if I can get them to go for it beer is on me. Thanks again!


----------



## ablyss (Feb 8, 2014)

You go by what the nameplate says.
If the nameplate say 8.1W at 208v wye then that is what you figure on each leg.


----------



## ablyss (Feb 8, 2014)

936A is the number I'm getting

81000w / 208v = 39 A
24 x 39 = 936 

If you were calculating loads based off 120v or 240v then you would divide by 1.73 but the nameplate is clear it is already figured in the 1.73
That's my understanding at least


----------



## ponyboy (Nov 18, 2012)

ablyss said:


> 936A is the number I'm getting 81000w / 208v = 39 A 24 x 39 = 936 If you were calculating loads based off 120v or 240v then you would divide by 1.73 but the nameplate is clear it is already figured in the 1.73 That's my understanding at least


Incorrect. Go over the previous posts for further explanation


----------



## ablyss (Feb 8, 2014)

ponyboy said:


> Incorrect. Go over the previous posts for further explanation


I'll go by what the nameplate says, you go by what other ppl say.


----------



## Dennis Alwon (May 9, 2009)

ablyss said:


> I'll go by what the nameplate says, you go by what other ppl say.


Fine but think about it.

8.1k at 208V=~ 39 amps if the volt was 240= half 120V= 33.75

8.1 amps at 120V= 67.5 amps

If each leg saw 8100 watts then the 40 amp breaker at 208 V would not hold

Don't you size the breaker based on 208V in this case. Forget the 125% for now.


----------



## Dennis Alwon (May 9, 2009)

Dennis Alwon said:


> Fine but think about it.
> 
> 8.1k at 208V=~ 39 amps if the volt was 240= half 120V= 33.75
> 
> ...



To add to this the amperage on each leg would be 39 amps but not the same for the wattage. If it was we would meed a dp 70 amp breaker and it would defeat the purpose of using 240V in many cases.


----------



## 99cents (Aug 20, 2012)

ablyss said:


> You go by what the nameplate says.
> If the nameplate say 8.1W at 208v wye then that is what you figure on each leg.


This is a load calc. Who gives a rat's ass about legs on a load calc? Balancing loads has nothing to do with a load calc.

These are heaters (no inrush to worry about). Watts divided by volts times 1.73.

Done.


----------



## Tsmil (Jul 17, 2011)

This is getting confusing. I just retread all posts. Supply is 208/120 wye. Loads are all 208v single phase. So all loads would be connected phase to phase. Nothing gets connected to neutral. Load is delta connected. 24 heaters in total. Split heaters evenly across phases and you get 8 heaters on AB, 8 heaters on AB, and 8 heaters on AC. Again, nowhere is anything connected to neutral so it makes no difference whether the supply is 208/120 wye or 208 delta. All heaters connected in a balanced delta configuration would be 194,400 watts. 194400 / (208 x 1.73) = 539.6.


----------



## wildleg (Apr 12, 2009)

balancing the loads on the legs has EVERYTHING to do with a load calc when you are balancing single phase loads on a three phase system. Do you really want to put in a service that is 2 or three times the size you need because you can't figure out the correct way to balance the loads ? I calculated it wrong like Dennis did and I would like to thank everyone for showing the correct calculation. The Holt thread shows it perfectly.


----------



## 99cents (Aug 20, 2012)

Tsmil said:


> This is getting confusing. I just retread all posts. Supply is 208/120 wye. Loads are all 208v single phase. So all loads would be connected phase to phase. Nothing gets connected to neutral. Load is delta connected. 24 heaters in total. Split heaters evenly across phases and you get 8 heaters on AB, 8 heaters on AB, and 8 heaters on AC. Again, nowhere is anything connected to neutral so it makes no difference whether the supply is 208/120 wye or 208 delta. All heaters connected in a balanced delta configuration would be 194,400 watts. 194400 / (208 x 1.73) = 539.6.


Doesn't matter if they're connected phase to phase or phase to neutral, the load calc is the same.

Load calcs are easy. 

Single phase - Watts divided by volts.
Three phase - Watts divided by volts X 1.73.

The only time you complicate those equations is when you have inrush - motors, welders, etc.


----------



## 99cents (Aug 20, 2012)

wildleg said:


> balancing the loads on the legs has EVERYTHING to do with a load calc when you are balancing single phase loads on a three phase system. Do you really want to put in a service that is 2 or three times the size you need because you can't figure out the correct way to balance the loads ? I calculated it wrong like Dennis did and I would like to thank everyone for showing the correct calculation. The Holt thread shows it perfectly.


The guy's question was on the load calc. How can you possibly screw up balancing 24 equal wattage heaters on a three phase system?


----------



## Dennis Alwon (May 9, 2009)

99cents said:


> The guy's question was on the load calc. How can you possibly screw up balancing 24 equal wattage heaters on a three phase system?


Well I did--:blush:--- Question then become what if it isn't balanced? The mike holt thread shows that. The error is basically over thinking the problem.


----------



## Rollie73 (Sep 19, 2010)

99cents said:


> Doesn't matter if they're connected phase to phase or phase to neutral, the load calc is the same.
> 
> Load calcs are easy.
> 
> ...


You can't make it any simpler. :thumbsup:

I guess that Canadian Red Seal apprenticeship training really pays off.:jester::jester:


----------



## 99cents (Aug 20, 2012)

I complicate my life with women. Electricity is easy  .


----------



## Tsmil (Jul 17, 2011)

99cents said:


> Doesn't matter if they're connected phase to phase or phase to neutral, the load calc is the same. Load calcs are easy. Single phase - Watts divided by volts. Three phase - Watts divided by volts X 1.73. The only time you complicate those equations is when you have inrush - motors, welders, etc.


In this case it does matter if loads are connected phase to phase or phase to neutral. The loads were rated at 208 volts and connecting them phase to neutral is 120 volts. 

If the heaters were 8100 watts at 120 volts, they would be connected to neutral but supply current should remain the same. 

The confusion always starts when you look for answers to something different than what you have.


----------



## Dennis Alwon (May 9, 2009)

Rollie73 said:


> I guess that Canadian Red Seal apprenticeship training really pays off.:jester::jester:


It is simple but understanding the theory is another thing. I never took any electrical classes. I learned by doing it but it has its draw backs. It really has not limited me in any way doing resi work.


----------



## ablyss (Feb 8, 2014)

I think I understand now. The question at hand was concerning the service size and because it is 3-phase the total combined load is divided by 1.73. 

Edit: multiply by 1.73


----------



## Tsmil (Jul 17, 2011)

Unbalanced loads can be a real pain.

Unbalanced wye loads are not bad because each branch of the wye is connected to only one phase.

Unbalanced delta loads are a different story. This is something i have to deal with quite frequently. I use a different method that us not 100% accurate but comes so close that the error is virtually irrelevant. Easier to explain with picture.


----------



## Rollie73 (Sep 19, 2010)

Dennis Alwon said:


> It is simple but understanding the theory is another thing. I never took any electrical classes. I learned by doing it but it has its draw backs. It really has not limited me in any way doing resi work.


True enough Dennis.........I'm just trying to be an ass today:thumbsup:


----------



## Rollie73 (Sep 19, 2010)

ablyss said:


> I think I understand now. The question at hand was concerning the service size and because it is 3-phase the total combined load is divided by 1.73.


 
Yes........you have the idea now.


----------



## Tsmil (Jul 17, 2011)

Dennis Alwon said:


> It is simple but understanding the theory is another thing. I never took any electrical classes. I learned by doing it but it has its draw backs. It really has not limited me in any way doing resi work.


Damn, I gotta give you credit for that. Overall, you have a great grasp and after reading so many of your posts, I have learned a lot from you. I have always been a theorist and worked with math in most of what I have done my whole life.


----------



## Dennis Alwon (May 9, 2009)

Now lets go back to the unbalanced load in the mike holt thread. Charlie mentions adding the kva and divide by 3 to get 389 amps. Forget 80% rules etc and say we have a 400 amp breaker-- Is phase B really seeing 416 amps?? Otherwise what happens here. My gut would be to size the feeder to the largest load.


----------



## Tsmil (Jul 17, 2011)

Dennis Alwon said:


> Now lets go back to the unbalanced load in the mike holt thread. Charlie mentions adding the kva and divide by 3 to get 389 amps. Forget 80% rules etc and say we have a 400 amp breaker-- Is phase B really seeing 416 amps?? Otherwise what happens here. My gut would be to size the feeder to the largest load.


Mine too.


----------



## Rollie73 (Sep 19, 2010)

Tsmil said:


> Mine too.


In any load like that, I would always try to size my feeders to the largest of the loads. In this case the B phase which would be seeing the 416 Amps.

By the CEC......I would be using a 500MCM copper conductor.


----------



## 99cents (Aug 20, 2012)

Dennis Alwon said:


> Now lets go back to the unbalanced load in the mike holt thread. Charlie mentions adding the kva and divide by 3 to get 389 amps. Forget 80% rules etc and say we have a 400 amp breaker-- Is phase B really seeing 416 amps?? Otherwise what happens here. My gut would be to size the feeder to the largest load.


I know what you're saying but three phase theory can drive you nuts if you overthink it. This is AC. Everything is in cycles and peaks and valleys. It's in the timing. Three phase is essentially three single phases operating at offsetting times. When one is high, the other two are lower. If, in fact, you are seeing 416 amps, it is for a fraction of a fraction of a second.


----------



## Going_Commando (Oct 1, 2011)

I think after reading this thread I am more confused than when I started, I think I'll stick to my old method, i.e. the 99cents and Tsmil method. Simple math is far easier than complex math.


----------



## Tsmil (Jul 17, 2011)

99cents said:


> I know what you're saying but three phase theory can drive you nuts if you overthink it. This is AC. Everything is in cycles and peaks and valleys. It's in the timing. Three phase is essentially three single phases operating at offsetting times. When one is high, the other two are lower. If, in fact, you are seeing 416 amps, it is for a fraction of a fraction of a second.


You will get 416 amps RMS.


----------



## 99cents (Aug 20, 2012)

Tsmil said:


> You will get 416 amps RMS.


Yeah, you're right. Just when I get three phase settled out in my tiny little brain, you had to throw in RMS...


----------



## Knightryder12 (Apr 4, 2013)

Dennis Alwon said:


> Thanks all for the help-- I see the issue
> 
> 
> I read this thread at mike holt and here is the thread but it is closed... http://forums.mikeholt.com/showthread.php?t=140398 Basically they did this as well as what you all stated and got the same answer
> ...


I have attached panel schedules for everyone to see. Looks like he will need an 800 amp service because there is no way to totally balance the load. It appears one leg will be over 600 amps. But I could be wrong.


----------



## Dennis Alwon (May 9, 2009)

I don't understand. The chart above shows that it is totally balanced. However it does not need to be totally balanced.

Your schedule does not show 24 heaters only 21 but I still don't get what you are getting at. The only possible thing would be if the 125% needs to be added to this. Some think not.


----------



## Dennis Alwon (May 9, 2009)

According to 220.51 100% is all that is needed to calculate the load


----------



## Knightryder12 (Apr 4, 2013)

Dennis Alwon said:


> I don't understand. The chart above shows that it is totally balanced. However it does not need to be totally balanced.
> 
> Your schedule does not show 24 heaters only 21 but I still don't get what you are getting at. The only possible thing would be if the 125% needs to be added to this. Some think not.


Schedule #1 has 20 heaters and schedule #2 has 4 heaters. With a 42 space panel you can't even out the load. Schedule #1 has schedule #2 loads linked to it with a total of 24 heaters. If you look at the schedule #1 connected load area (bottom left) phase A has a connected load of 607amps. i tried to move the load around and when i did one phase was always at 607amps. but Like I said i could be wrong.


----------



## Dennis Alwon (May 9, 2009)

Knightryder12 said:


> Schedule #1 has 20 heaters and schedule #2 has 4 heaters. With a 42 space panel you can't even out the load. Schedule #1 has schedule #2 loads linked to it with a total of 24 heaters. If you look at the schedule #1 connected load area (bottom left) phase A has a connected load of 607amps. i tried to move the load around and when i did one phase was always at 607amps. but Like I said i could be wrong.


You are not limited to 42 space panels...


----------



## Knightryder12 (Apr 4, 2013)

Dennis Alwon said:


> You are not limited to 42 space panels...


I know. I need to create schedules with 54 or 60 spaces. But the 42 space is the most commonly used in commercial work for us here in fla.


----------



## Dennis Alwon (May 9, 2009)

Okay I don't know what I was looking at before. 

So what is the difference between I wrote with the 4.05kva amps and what you have other than you using 4.1kva.

Using two panels would not change anything esp if you feed the second panel from the first. You can put the breakers in their perspective phases. You don't have to start at the A phase if it doesn't work.


----------



## Knightryder12 (Apr 4, 2013)

Dennis Alwon said:


> Okay I don't know what I was looking at before.
> 
> So what is the difference between I wrote with the 4.05kva amps and what you have other than you using 4.1kva.
> 
> Using two panels would not change anything esp if you feed the second panel from the first. You can put the breakers in their perspective phases. You don't have to start at the A phase if it doesn't work.


You are right. On the second panel I would just have to start the loads at C phase and all should be balanced


----------

