**Magnetizing Inrush Current Point(s)**
One or more inrush current points may be plotted on a TCC. Inrush currents are expressed in peak amps. The most common point is 8-12 times rated FLA at 0.1 seconds. Another less common point is 25 times rated FLA at 0.01 seconds.
**Example 1**
Plot the characteristic landmarks for a 1000kVA, 65°C, 4160-480/277V, Δ-YG, oil-filled, substation transformer with an impedance of 6.0%. Consider both the frequent and infrequent fault cases for this application.
Solution
Step 1 – Calculate the FLA
FLA = 1000kVA / (1.732 x 4.16kV) = 139 amps
Step 2 – Determine the Applicable Category
This is a Category II transformer based on the nominal rating of 1000kVA
Step 3 – Calculate the infrequent fault data points from Table 3
I 1800 sec = 2 x 139 amps = 278 amps
I 300 sec = 3 x 139 amps = 417 amps
I 60 sec = 4.75 x 139 amps = 660 amps
I 30 sec = 6.3 x 139 amps = 876 amps
I 10 sec = 11.3 x 139 amps = 1571 amps
I 2 sec = 25 x 139 amps = 3475 amps
Since the transformer is connected Δ-YG a separate set of data points must be calculated for primary-side protective devices. Primary-side devices will only see 58% of a secondary-side, single-line-to-ground fault.
I 1800 sec = 0.58 x 2 x 139 amps = 161 amps
I 300 sec = 0.58 x 3 x 139 amps = 242 amps
I 60 sec = 0.58 x 4.75 x 139 amps = 383 amps
I 30 sec = 0.58 x 6.3 x 139 amps = 508 amps
I 10 sec = 0.58 x 11.3 x 139 amps = 911 amps
I 2 sec = 0.58 x 25 x 139 amps = 2016 amps
Step 4 – Calculate the frequent fault data points from Table 3
I 2 sec = 139 amps / Z = 139 amps / 0.06 = 2316 amps
I 4.08 sec = 0.7 x 139 amps / Z = 97.3 amps / 0.06 = 1622 amps
t 1622 amps = 2551 (0.06)2 = 9.2 seconds
Again, shift the data points by 0.58.
I 2 sec = 0.58 x 139 amps / 0.06 = 1344 amps
I 4.08 sec = 0.58 x 97.3 amps / 0.06 = 941 amps
Step 5 – Calculate Inrush Points
12 x Inrush = 12 x 139 amps = 1668 amps
25 x Inrush = 25 x 139 amps = 3475 amps
The results are plotted in figure 1.
**Example 2**
Repeat Example 1 but now assume the secondary is high-resistance grounded (HRG).
Solution
Step 1 – Same as Example 1
Step 2 – Same as Example 1
Step 3 – Same as Example 1
No shifting of the damage curve is required with a HRG secondary. In this case the primary-side protective devices will not see a ground fault on the secondary-side. Ground fault magnitudes will always be much lower than load current levels.
Step 4 – Same as Example 1
Again, no shifting of data points is required.
Step 5 – Same as Example 1
The results are plotted in figure 2. |