Proper Load calcs on 120/208V side of 480:120/208V transformer

[msimms]

i am verifying that i am calculating my single phase loads properly, on the load side of 3-phase 480:120/208V xfmr. right now, i have only single phase 120 and single phase 208V loads, and thus far, i have stayed under 38kVA, so i am assuming that a 45kVA XFMR will work. im basically adding all 120V loads together with their amp loads, and then all 208V loads with their amp loads, and summing the total kVA (120*A + 208*A) together. am i doing this right? the code indicates that a 45kVA three phase transformer on the load side should be protected by a 125A breaker, but that appears to be for 3phase loads.

again, there are zero three phase loads, all single.

thanks.

 

[charlie b]

I'm basically adding all 120V loads together with their amp loads, and then all 208V loads with their amp loads, and summing the total kVA (120*A + 208*A) together. Am I doing this right?

Yes, that is correct. The next step would be to take the 38 KVA load, divide by 208 volts, and divide again by the square root of 3. That tells you the total load will result in a current of about 105 amps. So a 125 amp breaker does not seem out of place. The full load current of the 45KVA transformer is 125 amps. You can have a secondary breaker set at 125% of that value. I would typically use a 150 amp breaker in this application.

 

[augie47]

charlie b, I am thinking that, since we are dealing with single phase loads, he would need to make sure he does not load any one phase to over 15 kva. Is that correct ?

 

[charlie b]

I am thinking that, since we are dealing with single phase loads, he would need to make sure he does not load any one phase to over 15 kva. Is that correct ?

Yes. It is commonly presumed that the loads will be spread out evenly amongst the three phases. When I design an electrical distribution system, I fill in panel schedules with the KVA per load shown on the relevant phases. The spreadsheet adds up the loads, and tells me if any one phase is significantly more heavily loaded then the others. I issue the schedules with my design package. But I have no way of knowing whether the installer will connect the loads exactly as shown in my schedule. I tend to doubt it. For my part, I will have done my due diligence if I can show that a solution exists for obtaining a balanced loading condition.

 

[mull982]

Yes. It is commonly presumed that the loads will be spread out evenly amongst the three phases. When I design an electrical distribution system, I fill in panel schedules with the KVA per load shown on the relevant phases. The spreadsheet adds up the loads, and tells me if any one phase is significantly more heavily loaded then the others. I issue the schedules with my design package. But I have no way of knowing whether the installer will connect the loads exactly as shown in my schedule. I tend to doubt it. For my part, I will have done my due diligence if I can show that a solution exists for obtaining a balanced loading condition.

When you fill in the loads on the individual phases you would use the current with the L-N voltage for that particular phase correct? I can see how this works for three phase loads but what about a single phase 208 or 240V load?

 

[Besoeker]

Yes. It is commonly presumed that the loads will be spread out evenly amongst the three phases.

It's not a presumption I would make. It's too risky.

 

[mull982]

When you fill in the loads on the individual phases you would use the current with the L-N voltage for that particular phase correct? I can see how this works for three phase loads but what about a single phase 208 or 240V load?

Let me re-phrase that using somewhat of an example.

If I have a 120/208V 3-phase panel and have a 3-phase load of 10A I know that each phase in the panel will have 10A * 120V = 1200VA. This 1200VA is added for each of the three phases giving the total VA of the load as 3600VA. This is confirmed with the equation 1.73 * 208 * 10 = 3598VA. Therefore determining the kVA per phase is a simple task as mentioned.

However if I have a single phase 208V load I do not see the same results. Again using the 10A example, this would give me a total load VA of 208 * 10A = 2080VA. However if I went to compute the individual phase VA I would arrive at 120V * 10A = 1200VA. Multiplying this result by 2 for the two different phases that are used I get 2400VA. This 2400VA does not match the 2080VA that I calculated. What am I doing wrong in this case?

 

[winnie]

Again using the 10A example, this would give me a total load VA of 208 * 10A = 2080VA. However if I went to compute the individual phase VA I would arrive at 120V * 10A = 1200VA. Multiplying this result by 2 for the two different phases that are used I get 2400VA. This 2400VA does not match the 2080VA that I calculated. What am I doing wrong in this case?

You are missing the 'Oregon Fudge Factor'. (Seriously; search for that phrase to find past threads.)

Because the phase angle of the voltage developed by the transformer coils is different from the phase angle of the voltage applied to the load, there really is a difference in the power factor as measured in the different parts of the circuit.

Imagine for a moment that the load was a purely resistive, unity power factor load, with 10A flowing through it. The 10A is in phase with the 208V applied to the load. This 10A must therefore _not_ be in phase with the 120V developed by the two coils feeding the load.

-Jon

 

[mull982]

You are missing the 'Oregon Fudge Factor'. (Seriously; search for that phrase to find past threads.)

Because the phase angle of the voltage developed by the transformer coils is different from the phase angle of the voltage applied to the load, there really is a difference in the power factor as measured in the different parts of the circuit.

Imagine for a moment that the load was a purely resistive, unity power factor load, with 10A flowing through it. The 10A is in phase with the 208V applied to the load. This 10A must therefore _not_ be in phase with the 120V developed by the two coils feeding the load.

-Jon

I have heard of this fudge factor before.

I see what you are saying about the phase angles. How do you then determine the individual phase kVA for a 208V single phase load connected L-L?

Does the same rule apply for 120V/240V loads? I would think that these would just be addative and that the individual phase kVA @120V can be added together to give the total 240V kVA.

 

[Smart $]

I have heard of this fudge factor before.

I see what you are saying about the phase angles. How do you then determine the individual phase kVA for a 208V single phase load connected L-L?

Typically, as Charlie outlined above. But as you said, it totals as 2400VA. Calculating in VA does not account for power factor. In the case of a 10A, 208V purely resistive load, the power factor is .866 (cos 30?, where 30? is the current angle off L-N voltage angle). If you mulitply your VA times pf you'll be back to the real power of 2080W (or somewhere close, as both 208 and .866 are rounded numbers).

Does the same rule apply for 120V/240V loads? I would think that these would just be addative and that the individual phase kVA @120V can be added together to give the total 240V kVA.

No. They are as you say.

 

[winnie]

I see what you are saying about the phase angles. How do you then determine the individual phase kVA for a 208V single phase load connected L-L?

I don't think that it is possible without considering the other loads on the transformer. The load on the transformer is the vector sum of all the connected loads. The 'phase angle' of A-B line to line loads is 'balanced' on phase A with the 'phase angle' of A-C line to line loads. So the loading placed on phase A by A-B loads will depend upon the presence of A-C loads.

As a conservative approximation I would just multiply the load current by the L-N voltage and assign that value to the connected phase. This would tend to overestimate the loading.

Another very common approximation is to simply total up the KVA being connected and assume that it is balanced. This would tend to underestimate the transformer KVA required, however normal load calculations tend to seriously overestimate the load, so this usually isn't a problem.

Does the same rule apply for 120V/240V loads? I would think that these would just be addative and that the individual phase kVA @120V can be added together to give the total 240V kVA.

120/240 voltages from a single phase center tapped system are all 'in phase', so there is not fudge factor to contend with. In such a system, line-line loads are in phase with the line-neutral loads.

-Jon

 

[mull982]

I don't think that it is possible without considering the other loads on the transformer. The load on the transformer is the vector sum of all the connected loads. The 'phase angle' of A-B line to line loads is 'balanced' on phase A with the 'phase angle' of A-C line to line loads. So the loading placed on phase A by A-B loads will depend upon the presence of A-C loads.

As a conservative approximation I would just multiply the load current by the L-N voltage and assign that value to the connected phase. This would tend to overestimate the loading.

Another very common approximation is to simply total up the KVA being connected and assume that it is balanced. This would tend to underestimate the transformer KVA required, however normal load calculations tend to seriously overestimate the load, so this usually isn't a problem.

-Jon

I see completely now what you are saying about the phase angles, and how the load on each phase is a verctor sum of the currents on that phase.

When I design an electrical distribution system, I fill in panel schedules with the KVA per load shown on the relevant phases. The spreadsheet adds up the loads, and tells me if any one phase is significantly more heavily loaded then the others.

Charlie I have seen many of times panel schedules showing what you describe as showing the KVA per load on the relevent phases. Which of the two methods above do you use or are typically used to divide 208V loads between phases?