Application of the Floating Wye Connection
Floating WYE Capacitor
Much has been said and written regarding the benefits and advantages of connecting banks in a WYE configuration with the neutral point grounded (grounded WYE) or ungrounded (floating WYE).
High Fault Current
One of the concerns with a capacitor bank is when the available fault current is of high magnitude. When one capacitor unit fails and just before its fuse clears, all of the available fault current will "dump" into this faulted unit. This could result in a violent rupture of the faulted unit, possible damage of other equipment, and costly outages. One possible solution is to use current limiting fuses which have interrupting ratings as high as 50kA. However, normally capacitor substation blocks (outdoor, open structures) are of larger (KVAR) size and this could be expensive. Another more effective method is to float the neutral (ungrounded WYE) in a floating WYE capacitor configuration. The fault current is held to threetimes the line current eliminating the concern of high fault current damaging other good capacitors and possibly rupturing the faulted capacitor.
Floating Neutral
A little mathematical analysis can show why this is so. Figure 1 shows a balanced floating capacitor bank. Let 1 PU (per unit) equal V_{a}N, V_{b}N, and V_{c}N.



Figure 1 

Figure 2 
Assume a capacitor is failing in Leg (a). Just before the fuse attached to the capacitor clears, the neutral (N) shifts to Va. (See Figure 2). There is an increase in voltage across the capacitors in lines VcN' and VbN'. This increase is intuitively 1.73 x VN or 1.73 x (1 PU). However, in pure mathematics, we cannot accept intuitive logic. This will need to be proven.
There are a few methods of proving this. One is mathematical and the other is analytical. The mathematical solution uses the Law of Cosines.
Mathematical Solution
When the neutral shifts from N to N' (again see Figure 2), N' is at the same potential as Va. This happens as a capacitor in this leg starts to fail and prior to the fuse of the failing capacitor clearing.
The Law of Cosines is (b')^{2} = a^{2} + b^{2 }+ 2abCosØ.
Using Figure 2, we have:
(b')^{2} = (V_{b}N')^{2}
a^{2} = (N'N)^{2} or (1 PU)^{2}
b^{2} = (V_{b}N)^{2} or (1 PU)^{2}
CosØ = (0.5), Ø = 120°
Then (V_{b}N')^{2} = (1)^{2} +(1)^{2} + 2(0.5)
(V_{b}N')= 2 + 1 = 3
b'= 1.73
(V_{b}N')= 1.73 (1 PU)
Analytical Solution
If an analytical approach is preferred, we can solve the voltage problem this way, which allows us to use the same assumptions we used for the mathematical solution and let the same capacitor fail, we can develop Figure 3. The triangles N', N, Vb: N', N, Vc are equal. Also, the center angles are all equal (120°). Therefore, triangle N', Vb, Vc is an equilateral triangle.



Figure 3 

Figure 4 
A perpendicular line is drawn from N to E. Based on the fact that the triangle is equilateral, this line not only bisects the angle(Vb, N, Va), it segments the line b' into two equal parts (VbE and N'E).
Isolating triangle E, Vb, N, we have the following (see Figure 4).
Then (EN) = sin 30°. Therefore, EN = 0.5
And V_{b}E  cos 30°.Therefore, V_{b}E= 0.866
Proving Line V_{b}E is 0.866, then Line EN' = 0.866 + 0.866 = 1.73 and voltage across leg V_{b}N' is 1.73 x 1PU.
The shift in the capacitor current has the same effect as the voltage shift. Of course, the current is leading the voltage by 90°.
The Phasor Diagram for the current in this condition is shown in Figure 5. (Capacitors are shown for reference only.)



Figure 5 

Figure 6 
Using Kirchoff's Law
In = IØa + IØb + IØc = 0
Again, if a unit fails and before the fuse clears, we have a neutral shift in current.
When this occurs, we then have the condition shown in Figure 6.
In Figure 5, In = 0. However, in Figure 6, IØa is now considered In'; and because of Kirchoff's Law, we have IN = IØb + IØc.
Under normal conditions, IØa = IØb = IØc, but as shown in the voltage calculation when the unit in leg (a) begins to fail, Vb = Vc = (1.73 x 1 PU). The same calculations apply in the current calculations. Therefore, IØb = IØc = (1.73 x normal line current).
Using Polar and Complex calculations (see Figure 6), we have the following
Therefore, IN' = 3 times normal line current.
A more graphic approach would be by using vectors (see Figure 7).

Figure 7

A = B = 1.73. Therefore, Figure 7 is a rhombus. The diagonals of a rhombus intersect at right angles. Also, the diagonals divide the rhombus into four equal and congruent triangles.
Then Vectorially
A + B=2C
C=(cos 30°)(1.73)
C=(0.866)(1.73)
C=1.5
Therefore
2C=3
Summary
There are advantages in both grounded and floating WYE banks. However, floating WYE banks can be used on both 3 phase, 3 wire and 3 phase, 4 wire systems. As shown in the analogy, a major advantage in using a floating WYE bank is when the fault current is of a high magnitude. The floating WYE bank will limit the fault current to three times the line current.
