3 phase load question

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B

Bill Turax

Member
Location
MD
Hello, this is my first post.

I have a question that I am hoping you all can help me with.

I am looking at a job that involves replacing (5) EVSEs. The existing (5) EVSEs are being fed from a 208 volt/3 phase panel. The panel is fed from a 3 phase, 100 amp breaker at the main source.

My question is this:
Each EVSE is 2 pole and will have a 32 amp maximum draw, and a 40 amp calculated load (125% of the maximum draw). Assuming the 100 amp panel is evenly loaded, 2 of the phases would have 120 amps and 1 phase would have 160 amp calculated load. So a 100 amp feeder is to small to handle this.

However the buildings electrical supervisor is telling me that since it is a 3 phase system each lines total Kva (208*160 = 33.2kva) can be divided over 360 volts because of 3 phase (208x1.72), bringing the highest loaded phase to a total of 92.4 amps. Making the 100 amp feed adequate.

Is this correct? I thought I could only apply that 360 voltage if the device loads were 3 phase loads, these are only using 2 legs of the 3 phase.

See link to 1 line diagram of this question.
https://drive.google.com/open?id=14rmxaSFbhigNRo2fvnd5LstEGlJA0flj
 
jumper

jumper

Senior Member
Location
3 Hr 2 Min from Winged Horses
You do not add amps for 160 and since the load is not balanced the way you drew it will show highest phase load.

Your diagram is correct, but use kva. place one half of a load on each phase. 208 x 40/2 per phase.

Then add up kva on the phase.

I did it quickly and got about 140A on highest leg using 40A per rec. A phase per your diagram.
 
charlie b

charlie b

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Location
Lockport, IL
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Semi-Retired Electrical Engineer
First of all, I strongly recommend that you always do the math in units of power (i.e., KVA), not in units of amps, and only convert to amps as the last step in the calculation. Working in amps can, if you are careful, get you (perhaps accidentally) the right answer. But working in KVA will get you the right answer every time. So I would call it a better strategy.

Here is how I would handle this design:

I will first mention that I don’t know whether to use 32 amps or 40 amps. That is, I do not know whether EVSEs run at full load for more than 3 hours. But for the present, I will use 40.

  • (40 amps) times (208 volts) times (5 EVSEs) equals 41.6 KVA.
  • Pretend for a moment that the loads are evenly balanced.
  • Then (41.6 KVA) divided by (208 * 1.732) equals 116 amps.
  • Therefore, given that at least one leg will be loaded higher than at least one other, you will have to accommodate a load of more than 116 amps when picking your conductor.
  • I would change the problem to say there were six EVSEs, and balance the load amongst the three phases.
  • (40 amps) times (208 volts) times (6 EVSEs) equals 49.9 KVA.
  • Then (49.9 KVA) divided by (208 * 1.732) equals 139 amps (which by the way confirms Derek's answer, but not his method, as I don't know what his method was).
  • I would select 1/0 copper, with its ampacity of 150 amps.

 
jumper

jumper

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Derek used the same method Mr b., Derek’s problem is that he is a moron and when he is in a hurry he writes like a hyper 4 year with dyslexia.:ashamed1:

Just looked at what I actually wrote.::slaphead::slaphead::slaphead:
 
augie47

augie47

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Location
Tennessee
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State Electrical Inspector (Retired)
jumper said:
Derek used the same method Mr b., Derek’s problem is that he is a moron and when he is in a hurry he writes like a hyper 4 year with dyslexia.:ashamed1:

Just looked at what I actually wrote.::slaphead::slaphead::slaphead:
Click to expand...

and when he's not in a hurry.................................................
 
Jraef

Jraef

Moderator, OTD
Staff member
Location
San Francisco Bay Area, CA, USA
Occupation
Electrical Engineer
Bill Turax said:
... since it is a 3 phase system each lines total Kva (208*160 = 33.2kva) can be divided over 360 volts because of 3 phase (208x1.72), bringing the highest loaded phase to a total of 92.4 amps. Making the 100 amp feed adequate. ...
Click to expand...
This part of his (the building electrical supervisor) theory by the way is just wrong. Weirdly concocted, and wrong. It's funny (sadly) how some people just create math to suit their predetermined outcome...
 
B

Bill Turax

Member
Location
MD
Thanks you so much guys! I knew that the electrical supervises math seemed funny.

Considering the fact that these new EVSE units are replacing (5) old ones I will inform my customer that I can only install (4) new units on the existing 100 amp panel, not (5). If I understand the math above correctly, that will leave me with a balanced load of 93 amps per phase (40*208*4=33,280va / 208x1.72 = 92.4). If they want 5 (or 6 as that makes more sense) They can either have me replace the 100 amp feeder with a 150 amp feeder and 150 amp breaker at the main source. or run a separate 40 amp circuit from the main power source to the 5th EVSE.
 
B

Bill Turax

Member
Location
MD
Let me know if I got this straight:
you got to 104 by balancing the loads, then doing the calculation. so 4x40x208 = 33,280va, however this is unbalanced so you added 1/2 of one 40 amp load to the unbalanced line to get to 37,440. from there you can divide by 208x1.72 to get ~ 104A.

So to make the 100 amp panel code complaint, as is, I can only install (3) new EVSE units. this load would be (40x3x208 = 24,960va) that would be 70 amps per phase.

Also, if I understand this concept now, then the below will be true. (3 phase, 208v system)
The customer also wants to install (2) 100 amp EVSE units (80 amp load each). so if I wanted to install a sub-panel to handle the 200 amp load plus the (2) 40 amp loads that will be deleted from the 100 amp panel then my feeder calculation would look like this:

100 x 208 x2 = 41,600va
40 x 208 x 2 = 16,640va
this makes L1=20,800va, L2&L3 = 18,720va
so I need to add 2,080 to both L2 and L3 to balance the loads.
so I add 41,600+16,640+4,160 = 62,400va
62,400va / 208x1.72 = 174A
So the code minimum OCPD would be 175 amp with 2/0 CU feeders. Correct?
 
jumper

jumper

Senior Member
Location
3 Hr 2 Min from Winged Horses
Bill Turax said:
Let me know if I got this straight:
you got to 104 by balancing the loads, then doing the calculation. so 4x40x208 = 33,280va, however this is unbalanced so you added 1/2 of one 40 amp load to the unbalanced line to get to 37,440. from there you can divide by 208x1.72 to get ~ 104A.

So to make the 100 amp panel code complaint, as is, I can only install (3) new EVSE units. this load would be (40x3x208 = 24,960va) that would be 70 amps per phase.
Click to expand...

Yes to both.

I do not have time to do your additional calc, but you seem to get it now.
 
I

Ingenieur

Senior Member
Location
Earth
load 5 x 40 x 208 = 41600 va
a-b 2 units 40% x va = 16640 per winding / 120 = 138.7 A
b-c 2 units 40% 16640, 138.7 A
a-c 1 unit 20% x va = 8320, 69.3 A
 
M

MyCleveland

Senior Member
Location
Cleveland, Ohio
My question is this:
Each EVSE is 2 pole and will have a 32 amp maximum draw, and a 40 amp calculated load (125% of the maximum draw). Assuming the 100 amp panel is evenly loaded, 2 of the phases would have 120 amps and 1 phase would have 160 amp calculated load. So a 100 amp feeder is to small to handle this.

I put together yet another spreadsheet for this type of calculation. I am posting a PDF of the PHASORS along a three phase feeder with TAPS serving each point of five with a 32 amp load L-L.
Phasors are in light yellow cells. I understand this is not the exact application but the SUM of all riser Phasors, or currents at the serving panel are at the top of page. View attachment 3Phase Main - 1Phase Branches.pdf
 
jumper

jumper

Senior Member
Location
3 Hr 2 Min from Winged Horses
MyCleveland said:
My question is this:
Each EVSE is 2 pole and will have a 32 amp maximum draw, and a 40 amp calculated load (125% of the maximum draw). Assuming the 100 amp panel is evenly loaded, 2 of the phases would have 120 amps and 1 phase would have 160 amp calculated load. So a 100 amp feeder is to small to handle this.

I put together yet another spreadsheet for this type of calculation. I am posting a PDF of the PHASORS along a three phase feeder with TAPS serving each point of five with a 32 amp load L-L.
Phasors are in light yellow cells. I understand this is not the exact application but the SUM of all riser Phasors, or currents at the serving panel are at the top of page. View attachment 19397
Click to expand...

Chart looks great but too complicated for me.

I used 40A per EVSE and my numbers match Charlie and Iggy.
 
I

Ingenieur

Senior Member
Location
Earth
total load 5 x 32 x 208 = 32280 va

a-b 2 units 40% of total load
ph a = (20+20)% x total va ~ (1/2 a-b + 1/2 b-c)
= 13322 per ph winding / 120 = 110.9 A

b-c 1 unit 20% of total load
ph b = (20+10)% x total va ~ (1/2 a-b + 1/2 b-c)
= 9984/120 = 83.2 A

c-a 2 units 40% of total load
ph c = (10+20)% x total va ~ (1/2 b-c + 1/2 a-c)
= 9984/120 = 83.2 A
 
B

Bill Turax

Member
Location
MD
thanks all, I appreciate the responses. I thought I had this figured out, however I am more confused than before:?.

I am looking to find the leg with the largest load so I can size my feeder and breaker properly.

so is a 100 amp 3 phase panel and feeder to small for (4) 32 amp (40 amp calculated) load? my math below says yes, however it is not nearly as involved as the last few posts.

(4) 2 pole 40 amp loads * 208 volts = 12,480va
to make the calculation balanced I need to use the line with the highest va. so 1248*3 = 74,440 va
37,440 va / (1.72x208) = ~104A - TO LARGE FOR THE EXISTING FEEDER

however (3) 40 amp EVSE's would be ok:
(3) 2 pole 40 amp loads * 208 volts = 8,320va
the calculation is already balanced, so 8,320*3 = 24,960va
24,960 va / (1.72x208) = ~70A - OK TO USE WITH EXISTING FEEDER
 
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