Page 581 - Handbook of Electrical Engineering

574 HANDBOOK OF ELECTRICAL ENGINEERING

q) Simplified solution

In the proposed system, used as the example above, it is acceptable to ignore the cable impedance
R cm + jX cm and the transformer impedance R c + jX c for approximate calculation purposes. This
is only allowable for the following reasons:-

• The power system distribution cables or overhead lines are well rated for their current duty
and are short in length.
• The series impedances for cables are usually small in comparison with the transient reactances
and load impedances, but this is not always the case with low voltage situations where for
example the route lengths of cables can be relatively long.
Figure G.3 shows a simplified form of Figure G.2, where Z o is the equivalent impedance
of Z ol in parallel with Z og and is calculated as follows:-

1
R o = = 2.2321 pu
1 1
+
R ol R og
1
X o = = 4.1903 pu
1 1
+
X ol X og
r) Initial conditions
The motor starter is open. The motor terminal V mo equals the generator terminal V go voltage,
which is 1.0 per-unit.
Hence,
V mo = V go = 1.0 + j0.0pu.

Find the initial values of I c and V l ,i.e. I co and V lo , noting that I m = 0.0
At the MCC the parallel load is R o in parallel with X o .

+ j X ol .
Convert the parallel load into a series load of R ol
The conversions are:-
2
R o X o
R ol = pu per phase
2 2
R o + X o
2
X o R o
X ol = pu per phase
2 2
R o + X o
Where R o and X o were found above.
Hence

Z ol = R ol + jX ol = 1.7388 + j0.9262 pu
1.0 + j0.0
V mo
I oo = I go = = = 0.448 − j0.2386 pu
Z ol 1.7388 + j0.9262