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Protection of transformers

The following should be observed for HV fuse-link selection:

a) Transformer ratings
– Service voltage (U)
– Rated output (S)
– Relative short-circuit voltage (uk %)
– Inrush current (factor 8…12 IN)

b) Time-current characteristic of HV fuse-links

c) Secondary devices/selectivity

Procedure based on an example:

A 630kVA transformer has a transformer rated current of 18,2A at a service voltage of 20kV. The relative short-circuit voltage is 4% and the inrush current is 12 x lN. The short-circuit currentn on secondary terminal short-circuit is given from the relative short-circuit voltage. The transformer must be designed to withstand this current for 2 seconds. This condition results in point b) in Fig. 3. HV fuse-links must interrupt this current within 2 seconds. In Fig. 3, the fuse link F4 must not be used for this transformer, as the fuse-link will require longer than 2 seconds to melt at this short-circuit current.

Procedure based on an example graphic
F1– F4) Time-current characteristics of HV fuse-links
a) Inrush current
b) Lowest short-circuit current of transformer

The inrush current is plotted for a duration of 0,1 seconds, resulting in point a). This inrush current must not melt the fuselink, for which reason the fuse-link F1 cannot be used for this transformer. The fuse-links F2 and F3 can be used for this transformer, since their time-current characteristics are between the points a) and b). A transformer can thus be assigned several HV fuse-links for various rated currents. Decisive for selection of the correct fuse is the time-current characteristic and not the rated current of the HV fuse-link.