industrial1003
Member
- Location
- Easton Pennsylvania
Hello,
Does anyone have a calculation example for sizing the branch circuit fuses (Class RK-5 Time Delay FRS-R-xxx) when the control transformer is tapped from the load side of the branch circuit overcurrent protective device? For anyone with the 2008 Handbook my circuit is exactly like the one in Exhibit 430.15 (correct version). I have not found a example calculation in the Handbook or my Stallcups books. Everyone shows the calculation for just the motor load not including the transformer load.
I have been over section 430.53 C, and my question is how to size the fuses in the safety disconnect switch with consideration that the other load is a transformer (all be it a small one). 430.52 indicates the max allowable fuse at 175% of the full load current from table 430.250, and then my understanding is I would add the rating of the transformer load per 430.53C4 (125% of primary amps). My concern lies it that a transformer also has high inrush as a percentage of its normal load, and when pulling in the coil while the motor is starting will I exceed the value? My control transformer is only 100VA with a primary that is less than 2 amps, so section 430.72C4 allows the transformer primary OCPD to be up to 500%. Since the motor is only a 3 phase 1/2hp 460V I am talking the difference between using a nonstandard 2.5 amp fuse and a standard 3 amp fuse (I know it is really splitting hairs since I can use the next largest standard and be in compliance). I would just like to make sure I am accounting for the transformer correctly in my design calculations.
My system components are:
1/2 HP 3 phase motor 460VAC Continuious duty 1.15SF Nameplate 0.9A Table Value 1.1A Nema Design Code J
100VA single phase transformer 480/240 Primary (using 480) 240/120 Secondary (using 120)
Class RK5 time delay fuses for branch short circuit protection.
My calculations look like:
1.1A X 175% = 1.925 which is approximate 2A max fuses if it were stand alone
100VA / 460 = 0.217 A primary current on transformer
1.925 + (0.217 X 125%) = 2.19A (a 2.5A fuse)
or
1.925 + (0.217 X 500%) = 3.01A (a 3A fuse) Not sure transformer would ever inrush 500%
Any help would be appreciated.
Does anyone have a calculation example for sizing the branch circuit fuses (Class RK-5 Time Delay FRS-R-xxx) when the control transformer is tapped from the load side of the branch circuit overcurrent protective device? For anyone with the 2008 Handbook my circuit is exactly like the one in Exhibit 430.15 (correct version). I have not found a example calculation in the Handbook or my Stallcups books. Everyone shows the calculation for just the motor load not including the transformer load.
I have been over section 430.53 C, and my question is how to size the fuses in the safety disconnect switch with consideration that the other load is a transformer (all be it a small one). 430.52 indicates the max allowable fuse at 175% of the full load current from table 430.250, and then my understanding is I would add the rating of the transformer load per 430.53C4 (125% of primary amps). My concern lies it that a transformer also has high inrush as a percentage of its normal load, and when pulling in the coil while the motor is starting will I exceed the value? My control transformer is only 100VA with a primary that is less than 2 amps, so section 430.72C4 allows the transformer primary OCPD to be up to 500%. Since the motor is only a 3 phase 1/2hp 460V I am talking the difference between using a nonstandard 2.5 amp fuse and a standard 3 amp fuse (I know it is really splitting hairs since I can use the next largest standard and be in compliance). I would just like to make sure I am accounting for the transformer correctly in my design calculations.
My system components are:
1/2 HP 3 phase motor 460VAC Continuious duty 1.15SF Nameplate 0.9A Table Value 1.1A Nema Design Code J
100VA single phase transformer 480/240 Primary (using 480) 240/120 Secondary (using 120)
Class RK5 time delay fuses for branch short circuit protection.
My calculations look like:
1.1A X 175% = 1.925 which is approximate 2A max fuses if it were stand alone
100VA / 460 = 0.217 A primary current on transformer
1.925 + (0.217 X 125%) = 2.19A (a 2.5A fuse)
or
1.925 + (0.217 X 500%) = 3.01A (a 3A fuse) Not sure transformer would ever inrush 500%
Any help would be appreciated.