One of the most mis-calculated current rating calculations on a grid tied system, is the feeder calculation for unbalanced grid interactive inverters on a Delta system.
If you are reading this, you more than likely have a basic knowledge of Delta and Wye 3 phase electrical systems. Most systems installers are interacting with, is the Wye system, that is the most common of the 3 phase systems out there and are the systems that typically incorporate the voltages of 120/208V and 277/480V. These are pretty straight forward connections, even on an un-balanced system. But there are still many Delta systems out there that are typically 120/240V with the High Leg, 240V systems and 480V systems.
The difference between a Wye system and a Delta system is how the voltage and current act. In a Wye system current simply adds up from leg to leg and the systems windings all share a common connection point making all windings 120º apart. The calculation to figure the voltage between legs is the winding voltage times the square root of 3 (√3= 1.732). This makes sizing for current easier as it simply adds up.
What is different about the Delta is that the currents use the same √3 or 1.732 factor only when the currents are equal between a-b, b-c, and c-a. So when it is unbalanced, sizing the legs can be challenging and tricky, as not all calculations for each leg (L1, L2, L3) are always the same.
What is an unbalanced system? An unbalanced system is pretty much exactly that, an unbalance of current between the A, B, and C legs. On a 3 phase inverter (an inverter that connects to leg A B and C) you do not need to worry about unbalanced supply currents as the inverters typically balance this internally. Most utilities have a requirement that allows for a certain percentage of un-balance for each service, and ideally they prefer a perfectly balanced system.
Unbalanced systems typically only occur when the inverters incorporated into a 3 phase system are 1 phase using only 2 legs. If for example you use 3 inverters all equal in size and have an equal amount of Wattage supply from the panels at the same orientation, tilt, and shading. You may still have a balanced system as all legs are receiving the same amount of current.
- The things you will need to do the calculation is:
- The Voltage of the system you are connecting to (this is for Delta systems only)
- The size of the inverters
- The quantity of the inverters
- The “Rated” Continuous AC current output of each inverter (found on the data sheet or specs on the unit)
- Which inverters are connected to what legs (this is important and needs to be relayed for installation)
In my example I will use a system we recently did for a winery:
- Voltage- 120/240V (we only used the 240V in this instance)
- Inverter(s)- 5000W (KACO 5002xi) All of ours were the same in this instance
- Number of Inverters- 4
- Continuous Current- 24A
- Connections to Phases-
- a-b= 24A
- b-c= 24A
- c-a= 48A
As a note it may be best to connect the larger load between a-c phases as that is often where the center tap is and those legs often use more electricity. This generally will help balance the load better to the utility, and they like you better for it.
This is used for any unbalanced legs:
Iphase = √((Iconnected-phases²) + (Iconnected-phases²) + ((Iconnected-phases x Iconnected-phases))
This is used for any balanced legs (both winding feeding leg must be equal):
Iphase = √3 x Ibalanced-phases
The size of the feeder or conductor must be sized off the largest required current flow of one of the 3 legs.
Ia= Current on leg A
Ib= Current on leg B
Ic= Current on leg C
Ia-b= Inverter current between A and B phase
Ib-c= Inverter current between B and C phase
Ic-a= Inverter current between C and A phase
Depending on your connection setup, a variation of the Unbalanced and Balanced leg calculations would be used.
Unbalanced leg calculations-
Ia= √((Ia-b²) + (Ia-c²) + ((Ia-b x Ia-c))
Ib= √((Ib-a²) + (Ib-c²) + ((Ib-a x Ib-c))
Ic= √((Ic-a²) + (Ic-b²) + ((Ic-a x Ic-b))
Balanced leg calculations the connection currents must be equal-
Ia= √3 x balanced current
Ib= √3 x balanced current
Ic= √3 x balanced current
Our Winery Example- (4) KACO 5002xi inverters:
Unbalanced Delta system= Each inverter has a maximum output of 20.83A but due to the NEC 690.8(A)(3) it shall be calculated with the continuous current rating on the inverter which was 24A.
Phase a-b= (1) inverter 24A
Phase b-c= (1) inverter 24A
Phase c-a= (2) inverters 48A
Ia= √ (482 + 242 + (48 x 24)) = 63.49 A
Ib= √ 3 x 20.83 = 41.56 A
Ic= √ (482 + 242 + (48 x 24)) = 63.49 A
This makes the maximum available current on the feeders of 63.49 A and using the Continuous factor of 1.25A per 690.8(B). The conclusion is that a #2 wire installed for the PV feeder is sufficient to supply the calculated continuous current rating of 79.37A as the #2 AL rating in 310.15(B)(16) is 90 A at 75ºC.
Be prepared to be challenged by the inspector as most do not readily do calculations like this and it would be good practice to have this calculation at the ready during inspections.