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.
Fig. 1 1000kVA, ∆-YG, liquid-immersed transformer damage curves
Fig. 2 1000kVA, ∆-YG (HRG), liquid-immersed transformer damage curves Example 3
Plot the characteristic landmarks for a 1500kVA, 150°C, 13800-480/277V, ∆-∆, dry-type, substation transformer with an impedance of 5.75%. Consider the infrequent fault case for this application. Solution Step 1 – Calculate the FLA FLA = 1500kVA / (1.732 x 13.8kV) = 62.8 amps Step 2 – Determine the Applicable Category This is a dry-type, Category II transformer based on the nominal rating of 1500kVA Step 3 – Calculate the infrequent fault data points from Table 7 I 100 sec = 3.5 x 62.8 amps = 220 amps I 10 sec = 11.2 x 62.8 amps = 703 amps I 2 sec = 25 x 62.8 amps = 1570 amps Since the transformer is connected ∆-∆ a separate set of data points must be calculated for primary-side protective devices. Primary-side devices will only see 87% of a secondary-side, line-to-line fault.
I 100 sec = 0.87 x 3.5 x 62.8 amps = 191 amps I 10 sec = 0.87 x 11.2 x 62.8 amps = 612 amps I 2 sec = 0.87 x 25 x 62.8 amps = 1366 amps Step 4 – Calculate Inrush Points 12 x Inrush = 12 x 62.8 amps = 754 amps 25 x Inrush = 25 x 62.8 amps = 1570 amps The results are plotted in figure 3.
Fig. 3 1500kVA, ∆- ∆, dry-type transformer damage curves References
Other Application Guides offered by SKM Systems Analysis at www.skm.com ABB Protective Relaying Theory and Application, 2nd Edition, 2004
The latest revision of:
IEEE Std 242, IEEE Recommended Practice for Protection and Coordination of Industrial and Commercial Power Systems (IEEE Buff Book) IEEE Std C57.12.00, IEEE Standard General Requirements for Liquid-Immersed Distribution, Power and Regulating Transformers IEEE Std C57.12.01, IEEE Standard General Requirements for Dry-Type Distribution and Power Transformers Including Those with Solid-Cast and/or Resin-Encapsulated Windings IEEE Std C57.12.59, IEEE Guide for Dry-Type Transformer Through-Fault-Current Duration IEEE Std C57.109, IEEE Guide for L iquid-Immersed Transformer Through-Fault-Current
Duration
Insulating materials Insulation Maximum Class Temperature Y
90°C
A
105°C
Hybrid A
110°C
E
120°C
B
130°C
F
155°C
H
180°C
C
>180°C
>H
220°C
Insulating Materials Cotton, silk, paper, wood, cellulose, fibre without impregnation or oil-immersion Class Y impregnated with natural resins, cellulose esters, insulating oils, etc., also laminated wood, varnished paper TInsuldur® InsulationT Kraft paper with epoxy binders activated under pressure Synthetic-resin enamels, cotton and paper Laminates with formaldehyde bonding Mica, glass fibre, asbestos, etc., with suitable bonding substance; built-up mica, glass-fibre and asbestos laminates The materials of Class B with more thermally-resistant bonding materials Glass-fibre and asbestos materials, and built-up mica, with appropriate Silicone resins Mica, ceramics, glass, quartz, and asbestos without binders or with silicone resins of superior thermal stability TNOMEX® insulation, varnish dipped and vacuum pressure impregnated (VPI)
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