I have a 120/240 3 phase, 4 wire (center tap) delta system (B stinger). My 3 phase load is 80 kva per phase (A, B, and C). My single phase load is 34 kva phase A and 41 kva phase C. What are the amps for each phase and how is this calculated? What size (amp rating) panel is needed?
unbalanced load calc for 3 phase delta system
- Thread starter rnolan
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Re: unbalanced load calc for 3 phase delta system
Is this an exam question or a real situation?
How about you giving it a shot first, and showing us your results? It would help us to help you if we knew what parts of this process you already understood, and if we could see where you might be going wrong.
Is this an exam question or a real situation?
How about you giving it a shot first, and showing us your results? It would help us to help you if we knew what parts of this process you already understood, and if we could see where you might be going wrong.
Re: unbalanced load calc for 3 phase delta system
I recently had a test question similar to this. I have been tying to find study info on calcs for delta sevices and so far I have not found any real good info. So if someone could go thru the steps involved here it would be greatly appreciated.
Re: unbalanced load calc for 3 phase delta system
rnolan , since,i am new to this site,( and i have been really enjoying the reading) i'll take a shot at it.
First, the 3 phase mulitplier for 240 volt system is 415.(240x1.732),so Phase a-b-c will be 192-192-192.(80000/415).A phase single phase load(34000/120) would be 283.C phase load 41000/120 would be 341.So total a phase load would be 475,B phase 192, C phase 533. 125% of the largest load would be the minimum breaker size 666amps or the next higher 800amp.Hope this helps,or gets you close.
Rick
rnolan , since,i am new to this site,( and i have been really enjoying the reading) i'll take a shot at it.
First, the 3 phase mulitplier for 240 volt system is 415.(240x1.732),so Phase a-b-c will be 192-192-192.(80000/415).A phase single phase load(34000/120) would be 283.C phase load 41000/120 would be 341.So total a phase load would be 475,B phase 192, C phase 533. 125% of the largest load would be the minimum breaker size 666amps or the next higher 800amp.Hope this helps,or gets you close.
Rick
Re: unbalanced load calc for 3 phase delta system
RUWIRED,
I see this problem a bit differently. The phase voltage in this delta system is 240V. I don't see where you get 415.
That being the case, the application of a bit of trig to the problem yields 577A, 700A, and 759A for the line currents. I have assumed that all loads present the same PF to make life easier.
Oops! Only the 577A is right. I will come up with the correct numbers manana.
Rattus
[ February 05, 2006, 11:56 PM: Message edited by: rattus ]
RUWIRED,
I see this problem a bit differently. The phase voltage in this delta system is 240V. I don't see where you get 415.
That being the case, the application of a bit of trig to the problem yields 577A, 700A, and 759A for the line currents. I have assumed that all loads present the same PF to make life easier.
Oops! Only the 577A is right. I will come up with the correct numbers manana.
Rattus
[ February 05, 2006, 11:56 PM: Message edited by: rattus ]
Re: unbalanced load calc for 3 phase delta system
80000/415 = 192 amps per phase. I don't see how you arrived at your figures.
Edit
I read the post again. I thought it was 80 kw total. 3 x 80 = 240 kw/.415 = 578 amps per phase.
[ February 06, 2006, 11:36 AM: Message edited by: bob ]
He is using a multipler of 240 x 1.73 = 415.
80000/415 = 192 amps per phase. I don't see how you arrived at your figures.
Edit
I read the post again. I thought it was 80 kw total. 3 x 80 = 240 kw/.415 = 578 amps per phase.
[ February 06, 2006, 11:36 AM: Message edited by: bob ]
Re: unbalanced load calc for 3 phase delta system
All the figures are given on a per phase basis. If you use 240V to calc. line current you have to multiply the Apparent Power by 3.
I get the following.
A = 80 kVA = 577.4 Amps
B = 34 kVA = 283 + 577.4 = 860.6 Amps
c = 41 kVA = 341 + 577.4 = 919 Amps.
[ February 06, 2006, 11:33 AM: Message edited by: spsnyder ]
All the figures are given on a per phase basis. If you use 240V to calc. line current you have to multiply the Apparent Power by 3.
I get the following.
A = 80 kVA = 577.4 Amps
B = 34 kVA = 283 + 577.4 = 860.6 Amps
c = 41 kVA = 341 + 577.4 = 919 Amps.
[ February 06, 2006, 11:33 AM: Message edited by: spsnyder ]
Re: unbalanced load calc for 3 phase delta system
Don't believe you add individual single phase KVAs to each phase of the 3 phase KVA. I believe you add all KVAs for each item (3 phase and single phase), then calculate current for design load from total KVA.
Not sure about the issue with the trig for the delta.
Don't believe you add individual single phase KVAs to each phase of the 3 phase KVA. I believe you add all KVAs for each item (3 phase and single phase), then calculate current for design load from total KVA.
Not sure about the issue with the trig for the delta.
Re: unbalanced load calc for 3 phase delta system
It is like this gents,
The phase currents in windings AB and BC are
80KVA/240V = 333A.
These currents combine to yield,
Ib = 333Ax1.732 = 577A
For winding CA, 68KVA of the single phase load may be treated as a 240V load and added to the 80KVA load to yield 148KVA of 240V load. Load (not line) current is 148KVA/240 = 616A. To obtain the line current, Ia, we combine vectorially the phase current from winding AB.
Ia = 616A + 333A[cos(60) +jsin(60)] = 783A + j289A yields 835A magnitude.
The remaining 7KVA of single phase load adds 58.3A into node C. Then
Ic = 616A + 58.3A + 333A[cos(60) +jsin(60)] = 841A + j289A yields 890A magnitude.
Phase currents cannot be added algebraically since their phase angles are different. The standard formulas do not work because the imbalance is so great. This is not a simple problem.
Anyone see it differently?
Edited to correct errors.
Rattus
[ February 08, 2006, 11:51 AM: Message edited by: rattus ]
It is like this gents,
The phase currents in windings AB and BC are
80KVA/240V = 333A.
These currents combine to yield,
Ib = 333Ax1.732 = 577A
For winding CA, 68KVA of the single phase load may be treated as a 240V load and added to the 80KVA load to yield 148KVA of 240V load. Load (not line) current is 148KVA/240 = 616A. To obtain the line current, Ia, we combine vectorially the phase current from winding AB.
Ia = 616A + 333A[cos(60) +jsin(60)] = 783A + j289A yields 835A magnitude.
The remaining 7KVA of single phase load adds 58.3A into node C. Then
Ic = 616A + 58.3A + 333A[cos(60) +jsin(60)] = 841A + j289A yields 890A magnitude.
Phase currents cannot be added algebraically since their phase angles are different. The standard formulas do not work because the imbalance is so great. This is not a simple problem.
Anyone see it differently?
Edited to correct errors.
Rattus
[ February 08, 2006, 11:51 AM: Message edited by: rattus ]
Re: unbalanced load calc for 3 phase delta system
Rattus,
"The remaining 8KVA of single phase load"
41KVA-34KVA=7KVA. I would think you should have stated the reamining 7KVA of single phase load and this would change your numbers some.
Interesting with the delta the way the phase, line and load currents are calculated.
Rattus,
"The remaining 8KVA of single phase load"
41KVA-34KVA=7KVA. I would think you should have stated the reamining 7KVA of single phase load and this would change your numbers some.
Interesting with the delta the way the phase, line and load currents are calculated.
Re: unbalanced load calc for 3 phase delta system
I am getting
Ia = 835 A
Ib = 890 A
Ic = 577 A
I added them vectorially, if that is a word. For B phase I had 333.33 Amps + 341.667 Amps at an angle of 0 and 333.33 Amps at an angle of 120.
For A phase I had 333.33 Amps + 283 Amps at an angle of 0 and 333.33 Amps at an angle of -120.
Am I confused?
I am getting
Ia = 835 A
Ib = 890 A
Ic = 577 A
I added them vectorially, if that is a word. For B phase I had 333.33 Amps + 341.667 Amps at an angle of 0 and 333.33 Amps at an angle of 120.
For A phase I had 333.33 Amps + 283 Amps at an angle of 0 and 333.33 Amps at an angle of -120.
Am I confused?
Re: unbalanced load calc for 3 phase delta system
ok, i reworked the math due to thinking single phase loads were 120 volt.
80000/415=192 a-b-c
34000/240=141 a-c
7000/120=58 c-n
Total-A=333
Total-B=391
Total-C=192
Total-N=177
391/.8=488 next higher 600amp main breaker
Rick
ok, i reworked the math due to thinking single phase loads were 120 volt.
80000/415=192 a-b-c
34000/240=141 a-c
7000/120=58 c-n
Total-A=333
Total-B=391
Total-C=192
Total-N=177
391/.8=488 next higher 600amp main breaker
Rick
Re: unbalanced load calc for 3 phase delta system
80KVA/240V = 333A
Add these vectorially to obtain the line current,
Ib = 333Ax1.732 = 577A
Now figure the "phase" currents at A and B:
Ian = 34KVA/120V + 333A = 616A (algebraic)
Ibn = 41KVA/120V + 333A = 675A (algebraic)
Now add vectorially the 333A phase currents from the other windings to obtain,
Ia = 835A
Ib = 890A
Now if we had only the 120V loads on ANC, we would have a quite different problem.
There is no way to get 415V from a 240V delta. The balanced phase currents are,
80KVA/240V = 333A
Add these vectorially to obtain the line current,
Ib = 333Ax1.732 = 577A
Now figure the "phase" currents at A and B:
Ian = 34KVA/120V + 333A = 616A (algebraic)
Ibn = 41KVA/120V + 333A = 675A (algebraic)
Now add vectorially the 333A phase currents from the other windings to obtain,
Ia = 835A
Ib = 890A
Now if we had only the 120V loads on ANC, we would have a quite different problem.
Re: unbalanced load calc for 3 phase delta system
Bob, If i have a 3 phase load drawing 80 kva, i don't say i have a 240kva load, i say i have an 80kva load. amperage is based on 80kva / by voltage-phase at hand. in this case it is 240 3 phase and the multiplier is 415. If i have a 192 amp 240v 3 phase load and want to find kva, i multiply 192x415= 79680kva(80).
Rick
Bob, If i have a 3 phase load drawing 80 kva, i don't say i have a 240kva load, i say i have an 80kva load. amperage is based on 80kva / by voltage-phase at hand. in this case it is 240 3 phase and the multiplier is 415. If i have a 192 amp 240v 3 phase load and want to find kva, i multiply 192x415= 79680kva(80).
Rick
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