P54x/EN OP/La4 Operation
(OP) 5-16
MiCOM P543, P544, P545 & P546
OP
Since relays A and B are identical, relay B also sends out data messages to end A.
Assume relay B sends out a data message at tB3. The message therefore contains the time
tag tB3. It also returns the last received time tag from relay A (i.e. tA1) and the delay time,
td, between the arrival time of the received message, tB*, and the sampling time, tB3, i.e. td
= (tB3 - tB*).
The message arrives at end A after a channel propagation delay time, tp2. Its arrival time is
registered by relay A as tA*. From the returned time tag, tA1, relay A can measure the total
elapsed time as (tA* - tA1). This equals the sum of the propagation delay times tp1, tp2 and
the delay time td at end B.
Hence,
(tA* - tA1) = (td + tp1 + tp2)
The relay assumes that the transmit and receive channels follow the same path and so have
the same propagation delay time. This time can therefore be calculated as:
tp1 = tp2 = ½(tA* - tA1 - td)
Note: The propagation delay time is measured for each received sample
and this can be used to monitor any change on the communication
link.
As the propagation delay time has now been deduced, the sampling instant of the received
data from relay B (tB3*) can be calculated. As shown in Figure 3, the sampling time tB3* is
measured by relay A as:
tB3* =
(tA* - tp2)
In Figure 3, tB3* is between tA3 and tA4. To calculate the differential and bias currents, the
vector samples at each line end must correspond to the same point in time. It is necessary
therefore to time align the received tB3* data to tA3 and tA4. This can be achieved by
rotating the received current vector by an angle corresponding to the time difference
between tB3* and tA3 (and tA4). For example a time difference of 1ms would require a
vector rotation of 1/20 * 360° = 18° for a 50 Hz system.
As two data samples can be compared with each data message, the process needs to be
done only once every two samples, therefore reducing the communication bandwidth
required.
Note: The current vectors of the three phases need to be time aligned separately.
1.1.1.2 Time alignment of current vectors with GPS input (all models)
The effe
ct of the deployment of switched SDH (Synchronous Digital Hierarchy) networks on
telecommunications circuits used in the application of numerical current differential
protection to transmission lines.
Such telecommunications networks can be deployed in flexible, self-healing topologies.
Typically, ring network topologies are employed and these are characterized by the ability to
self-heal in the event of a failure of an interconnection channel.
Consider a simple ring topology with 6 nodes, A - F, and consider two equipment situated at
nodes B and C. Under healthy conditions equipment at B communicates with equipment at
C directly between nodes B and C and equipment at C communicates with equipment at B
directly between nodes C and B. In this condition the communications propagation time
between nodes B and C will be the same as that between nodes C and B and so the
traditional technique described in could be used to apply numerical current differential
protection (see Figure 4).
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