3 Phase

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Zyb

Zyb

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Maine
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Design Engineer
Do I need to divide by 1.732 in sizing the fuse in disconnect or not?

IQ7 Max. Continuous Current = 1.0 A

Total No. of Inverter = 42

1.0 x 42 x 1.25 = 52.5A = 60A Fuse

1.0 x 42 x 1.25 / 1.732 = 30.3 A = 30A Fuse

30A or 60A Fuse?

Thank You!
3-Phase.png
 
Zyb said:
Do I need to divide by 1.732 in sizing the fuse in disconnect or not?

IQ7 Max. Continuous Current = 1.0 A

Total No. of Inverter = 42

1.0 x 42 x 1.25 = 52.5A = 60A Fuse

1.0 x 42 x 1.25 / 1.732 = 30.3 A = 30A Fuse

30A or 60A Fuse?

Thank You!
3-Phase.png
Click to expand...
Your line current is (14.0A)(sqrt3) = 24.25A, and (1.25)(24.25A) = 30.3A, so you are almost right, but you cannot round down that 0.3A. Your minimum fuse size is 35A.
 
ggunn said:
Your line current is (14.0A)(sqrt3) = 24.25A, and (1.25)(24.25A) = 30.3A, so you are almost right, but you cannot round down that 0.3A. Your minimum fuse size is 35A.
Click to expand...

I finally found my answer

breaker at load Center : 20A (2 Pole Breaker) each
1 x 14 x 1.25 = 17.5A

Fuse at disconnect : (3) 45A Fuse
IQ7 continuous power = 280W
(42 x 280) / (208 x 1.732) = 32.64A continuous
32.64 x 1.25 = 40.80A

3.1-Phase.png
 
Where are you getting the 280W continuous output from. The data sheet for the IQ7 states max continuous current of 1.15A when operating at nominal 208V. 42 x 1.15 x 1.25 / 1.732 = 34.9A. You can still use a fused disconnect with 45A fuses in it if you want as long as the feeder conductors between the disconnect and the AC combiner are rated for it, but you only need to install 35A fuses.
 
Be careful with the Watt or KW ratings with this calculation. These values may not necessarily match the Amps of maximum output current. Some inverters have headroom on their Amp ratings above the value you'd expect from P=I*V or P=I*V*sqrt(3). The Amp rating (maximum continuous output current) on the datasheet overrides the amp rating you'd calculate from the power or KVA rating.

When phase-to-phase inverters are balanced among the phases (which is what you should generally aim to do), you can calculate the total current by the following method.
42*1.15A * 208V = 10046.4 VA

10046.4 VA/120V/3 = 27.9A per phase
27.9A * 1.25 = 35A OCPD

If you have an unbalanced setup, you could either pick the pair of phases with the most current, and conservatively multiply that by 3*208V to get a virtual total VA that you then divide by (3*120V). Or you'd use the square root formulas to combine phase-to-phase connected currents.

Ia = Ia0 + sqrt(Iab^2 + Ica^2 + Iab*Ica)
Ib = Ib0 + sqrt(Iab^2 + Ibc^2 + Iab*Ibc)
Ic = Ic0 + sqrt(Ica^2 + Ibc^2 + Ica*Ibc)
 
Carultch said:
I thought 220.5(B) allows you to round down to the nearest amp.
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You are correct, sir! Otherwise, my calculation is correct, but I didn't look up the inverter maximum current at 208V, which is what I assume his 3P voltage is. He stated it as 1.0A, which is what I used. BTW, you cannot use the module or inverter rating in Watts to set the rating of the OCPD; you must use the inverter Imax.

Another thing: 690.9 sets the minimum OCPD rating; you can use higher rated OCPD as long as your conductors are sized appropriately and as long as you do not exceed any published maximum OCPD from the inverter manufacturer.
 
These inverters are connected in a 3ph delta configuration. So the current between the inverters and the inverter CB in the panel is just the inverter Imax and sizing the CB is Imax*1.25. But leaving the 3ph panel the currents combine into line currents using the 3ph rules. Since these are balanced it makes it easy, the line current out of the 3ph panel is the inverter Imax *SQRT(3). The size of the fuse in the disconnect is then Imax *SQRT(3)*1.25.
If the inverters were not balanced then things get tricky. There was an old SolarPro article that went into how to calculate the line currents in an unbalanced combination of inverters in a 3 phase system.
 
Forget the 1.732.
We are looking at currents here.
 
Last edited:
Besoeker3 said:
Forget the 1.732.
We are looking at currents here.
Click to expand...
If you have a 3 phase system with (3) 2-wire loads/sources with power factor 1 connected in a delta as A-B, B-C, and C-A, and if each load/source has a current of 1 amp, then with two loads/sources connected to each line conductor, the current on A, B, or C will be 1.732 amps. Which applies to the OP.

Cheers, Wayne
 
wwhitney said:
If you have a 3 phase system with (3) 2-wire loads/sources with power factor 1 connected in a delta as A-B, B-C, and C-A, and if each load/source has a current of 1 amp, then with two loads/sources connected to each line conductor, the current on A, B, or C will be 1.732 amps. Which applies to the OP.

Cheers, Wayne
Click to expand...
OK.
 
pv_n00b said:
These inverters are connected in a 3ph delta configuration. So the current between the inverters and the inverter CB in the panel is just the inverter Imax and sizing the CB is Imax*1.25. But leaving the 3ph panel the currents combine into line currents using the 3ph rules. Since these are balanced it makes it easy, the line current out of the 3ph panel is the inverter Imax *SQRT(3). The size of the fuse in the disconnect is then Imax *SQRT(3)*1.25.
If the inverters were not balanced then things get tricky. There was an old SolarPro article that went into how to calculate the line currents in an unbalanced combination of inverters in a 3 phase system.
Click to expand...
I have a hard copy of it. It's pretty good, and I've not been able to find it or anything comparable on line since SP went away.
 
ggunn said:
I have a hard copy of it. It's pretty good, and I've not been able to find it or anything comparable on line since SP went away.
Click to expand...
People can still download issues at solarprofessional.com. If you post the month and year of the article people can find it easier.
 
jaggedben said:
People can still download issues at solarprofessional.com. If you post the month and year of the article people can find it easier.
Click to expand...
The hard copy I have doesn't have a publishing date on it. It was written by Marvin R Hamon, PE, if that helps.
 
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