JBD,
Please review NEC Articles 240.4(F) and Section 450.
This will explain the reason to define a System's Load characteristics per "Single" or "Dual" Voltage SDS.
It has to do with Overloading the Secondary side Feeders, when only Primary side Over Current Protection is provided.
It has nothing to do with Ground Faults on the Secondary side.
If an SDS with Dual Voltages has only Primary side OCPD, there is a high possibility to Overload the Secondary Feeders with L-N Load Current, without the Primary Feeders' OCPD tripping.
Example:
45 KVA 480V x 208Y/120V 3P 4W Wye Transformer.
Primary Full-Load Amperes (FLA) = 54.1 Amps
Secondary FLA = 125.0 Amps
Primary OCPD only - max. rating = 125% Primary FLA.
(54.1 A * 1.25 = 67.6 A)
Primary Feeder OCPD = 60/3
Primary Feeders: 2#6 THHN cu.
Secondary Feeders: 4#1/0 THHN cu.
Load on Secondary Feeders between Line A and Grounded Neutral Conductor exceeds 125 Amps (15 KVA), goes up to 200 Amps (24 KVA)
Primary Feeder OCPD sees only 50 Amps, while 200 Amps flows on the Secondary side's Feeder.
Saturation will not affect output voltage at this point.
Even with Primary OCPD at 100% (54 Amps), there will only be 50 Amps flowing.
As you can see, there is a possibility to overload Secondary Conductors, on a Dual Voltage SDS having Primary Over Current Protection only.
If the SDS had only One Voltage output, a 208V L-L Secondary Load of 26 KVA (125 Amps L-L) would be reflected back to the Primary, and an L-L Primary Current of 54.2 Amps will be flowing through the Primary side OCPD.
If the Secondary L-L Load increases to 31.2 KVA (150 Amps), the Primary L-L Load will increase to 65 Amps, which will trip the OCPD - likely within 30 Minutes.
A solidly grounded wye secondary can have a line to ground fault that is not reflected back to the delta primary simply by the turns ratio of transformer.
480x208Y/120V Transformer.
Winding ratio = 4:1
L-G Short Circuit Amperes (SCA) on Secondary = 100% value of the Secondary L-L-L Bolted Fault Value.
L-G Secondary Fault SCA reflected to Primary Windings = 0.25 of the L-L-L Secondary SCA.
L-L-L SCA = 12000 Amps
L-G Fault = 12000 Amps x 1.0 = 12000 Amps
Fault Current reflected back to Primary = 3000 Amps (12000 * 0.25 = 3000)
Do the math. Given a single L-G fault of 2.0PU on a wye secondary, what is the corresponding Line current on the delta primary. Will a primary side protective device clear this L-G fault in the required time frame (assume a NEMA AB-1 characteristic)?
Using the example I provided (3000 Amps on Primary side), the OCPD should trip within 0.2 Seconds.
Scott
