Page 177 - Electrical Safety of Low Voltage Systems

160 Chapter Nine

The occurrence of a ground fault causes the system capacitance to
become an unbalanced load (i.e., the impedances of the three branches
1, 2, and 3 are no longer equal). The earth electrode resistance R G ,in
fact, is in parallel to the capacitance of the faulty phase (e.g., line 1
in Fig. 9.5). Upon the first fault the system evolves from a balanced
three-phase capacitive load, with no neutral wire, to an unbalanced
capacitive-resistive load.
Because of this unbalance, a potential difference V NG = V N −V G ,
also referred to as neutral potential rise, appears between the point
of neutral N at the source and the ground G at the faulty ECP. The
presence of V NG changes the voltage between the line conductors and
theground(i.e.,V 1G ,V 2G ,andV 3G )withrespecttothesystemsvoltages
(i.e., V 1N , V 2N , and V 3N ). The two sets of vector quantities identical in
normal conditions will now differ.
By applying Kirchhoff’s voltage law to each one of the closed loop
in Fig. 9.5, we obtain

V = V + V (9.4)
1G 1N NG
V 2G = V 2N + V NG (9.5)

V = V + V (9.6)
3G 3N NG
Let us calculate V NG by applying Millman’s theorem to the three-
phase circuit in Fig. 9.5 (see App. B for more details):

−V 1N [(1/R G ) + j C 0 ] − V 2N j C 0 − V 3N j C 0
V =
NG
(1/R G ) + j3 C 0

−V 1N (1/R G ) − j C 0 V 1N + V + V 3N
= 2N 0
(1/R G ) + j3 C 0
V 1N
=− (9.7)
1 + j3 C 0 R G
The above result has been obtained by knowing that the vectorial
summation (V 1N + V 2N + V 3N ) is zero, since these vectors are equal
◦
in magnitude and equally displaced by 120 .
We can now rewrite Eqs. (9.4) through (9.6), as follows:
V 1N j3 C 0 R G V 1N
V 1G = V 1N − = (9.8)
1 + j3 C 0 R G 1 + j3 C 0 R G
V 1N
V = V − (9.9)
2G 2N
1 + j3 C 0 R G
V 1N
V = V − (9.10)
3G 3N
1 + j3 C 0 R G