Any grounded or ungrounded systems mainly depend on the customer demands.
The ungrounded electrical systems are used where the customer and the design engineer do not want the overcurrent protection device acting on the ground fault line. However, there are some ungrounded elements such as building steel or iron and, water pipeline, which are intentionally grounded. In an ungrounded system, there is no internal connection between any line (including the neutral) and ground.
The ungrounded system is basically grounded through distributed capacitance.
There are some advantages and disadvantages of ungrounded systems. The ground fault current in this system is very low (5 A or less) and provides more reliability during fault conditions. The voltage between the healthy lines and ground is very high. The effect of harmonics in the ungrounded system will die out itself within the system. The outside source interferences are usually neglected as there is no con-nection with the soil. The ungrounded system is very poor to protect the electrical appliances due to transient voltages. Sometimes, these transient voltages propagate or elongate to the nearby equipment which can destroy the insulation of those equipment. In this system, it is very difficult to locate the line to ground fault. An ungrounded wye-connection is shown in Fig.4.2, where a ground fault occurs in line R. Each line has distributed earth capacitance. The capacitance of line R will discharge through the faulty path to the capacitance of lines Y and B. The charge and discharge of capacitance of line R will continue due to the healthy lines Y and B. The currents in the lines Y and B can be written as,
icY ¼ vYG
1=jxCY
¼ jxCYvYG ð4:1Þ
icB¼ vBG
1=jxCB¼ jxCBvBG ð4:2Þ
But, the distributed capacitance of each line is the same and it can be expressed as,
CR¼ CY¼ CB¼ C ð4:3Þ
The expression of fault current is,
if ¼ icYþ icB ð4:4Þ
Substituting Eqs. (4.1)–(4.3) into Eq. (4.4) provides,
if ¼ jxCðvYGþ vBGÞ ð4:5Þ
The vector diagram of the wye-connection is shown in Fig.4.3. The current icB
leads the voltage vBGby 90° and The current icY leads the voltage vYGby 90°. From Fig.4.3, the following voltage expression can be written as,
vBG¼ vBnþ vn ð4:6Þ
vYG¼ vYnþ vn ð4:7Þ
Substituting Eqs. (4.6) and (4.7) into Eq. (4.5) yields,
if ¼ iR ¼ jxCðvYnþ vnþ vBnþ vnÞ ð4:8Þ if ¼ jxCðvYnþ vBnþ 2vnÞ ð4:9Þ
R
Y B
n
icB
icY
if
CB
CR
CY
G
G
G Fig. 4.2 Wye-connection
with faulted line R
156 4 Grounding System Parameters and Expression of Ground Resistance
Resolve the voltage vBn into vBncosh in the vn direction and vBnsinh is per-pendicular on the vn. This can be expressed as,
vBn¼ vBncosh þ vBnsinh ð4:10Þ Again, resolving the voltage vYninto vYncosh in the vndirection andvYnsinh in the perpendicular direction of the vn, the following expression can be found,
vYn¼ vYncosh vYnsinh ð4:11Þ Adding Eqs. (4.10) and (4.11) yields,
vYnþ vBn¼ vBncosh þ vBnsinh þ vYncosh vYnsinh ð4:12Þ icY
icB
if
θ θ
vYG
vYn Bn
v vBG
G
Rn Bn Yn
v =v =v vn
n Rn
v = −v vRn
icY
icB
if
vYn
vBn
vRn
90°
90°
vBR
vYR
(a)
(b) Fig. 4.3 aVector diagram of
faulted wye-connection system. b Vector diagram of faulted wye-connection system
But, vBnsinh is equal and opposite to vYnsinh, then Eq. (4.12) can be modified as,
vYnþ vBn¼ vBncosh þ vYncosh ¼ vn ð4:13Þ Substituting Eq. (4.13) into Eq. (4.9) yields,
if ¼ jxCðvnþ 2vnÞ ¼ j3xC vn ð4:14Þ Alternative approach:
If the line to ground fault occurs at line R, then the line to ground voltage becomes zero i.e.,vRG¼ 0. Applying KVL between R-G-Y in Fig.4.2yields,
vRG vYGþ vRY ¼ 0 ð4:15Þ
0 vYGþ vRY ¼ 0 ð4:16Þ
vYG¼ vRY ð4:17Þ
Again, applying KVL between B-G-R yields,
vRG vBGþ vBR¼ 0 ð4:18Þ
vBG¼ vBR ð4:19Þ
Equations (4.1) and (4.2) can be modified as, icY ¼ jxCvYR¼ jxC ffiffiffi
p3
vph ð4:20Þ
icB¼ jxCvBR¼ jxC ffiffiffi p3
vph ð4:21Þ
From Fig.4.3b, the expression of the fault current is, if ¼ 2icBcos 30¼ 2icYcos 30¼ ffiffiffi
p3
icB¼ ffiffiffi p3
icY ð4:22Þ
Substituting Eq. (4.21) into Eq. (4.22) yields, if ¼ ffiffiffi
p3
jxC ffiffiffi p3
vph¼ 3jxC vph ð4:23Þ
A delta-connection with the ungrounded system is shown in Fig. 4.4. This system is basically grounded through an earth capacitance. Applying KVL between R-G-Y yields,
vRG vYG vRY ¼ 0 ð4:24Þ
158 4 Grounding System Parameters and Expression of Ground Resistance
vRY ¼ vRG vYG¼ 1 0j ð4:25Þ
vRG¼ vYGþ 1 0j ð4:26Þ
Again, applying KVL between Y-G-B yields,
vYG vBG vYB¼ 0 ð4:27Þ
vYB¼ vYG vBG¼ 1 120j ð4:28Þ
vBG¼ vYG 1 120j ð4:29Þ
Applying KVL between B-G-R yields,
vBG vRG vBR¼ 0 ð4:30Þ
vBR¼ vBG vRG¼ 1 120j ¼ 1 240j ð4:31Þ The earth capacitance will be equal for equal length and transposed of the conductor. This can be written as,
CR¼ CY¼ CB¼ C ð4:32Þ
In this condition, the expressions of the current are, iRG¼vRG
1 jxC
¼ jxC vRG ð4:33Þ
iYG¼ jxC vYG ð4:34Þ
G
G
G
B
R
Y CB
CR
CY
iBG
iRG
iYG
Fig. 4.4 Ungrounded delta-connection system
iBG¼ jxC vBG ð4:35Þ The sum of the currents at node G can be expressed as,
iRGþ iYGþ iBG¼ 0 ð4:36Þ
Substituting Eqs. (4.33)–(4.35) into Eq. (4.36) yields,
jxC vRGþ jxC vYGþ jxC vBG¼ 0 ð4:37Þ jxC ðvRGþ vYGþ vBGÞ ¼ 0 ð4:38Þ
vRGþ vYGþ vBG¼ 0 ð4:39Þ
Substituting Eqs. (4.26) and (4.29) into Eq. (4.39) yields,
vYGþ 1 þ vYGþ vYG 1 120j ¼ 0 ð4:40Þ vYG¼1
3ð1 120j 1Þ ð4:41Þ
vYG ¼ 1ffiffiffi
p 1503j ð4:42Þ
Substitute Eq. (4.42) into Eq. (4.26) provides, vRG¼ 1ffiffiffi
p 1503j þ 1 0j ð4:43Þ
vRG¼ 1ffiffiffi
p 303j ð4:44Þ
Again, substituting Eq. (4.42) into Eq. (4.29) yields, vBG¼ 1ffiffiffi
p 1503j 1 120j ð4:45Þ vBG¼ 1ffiffiffi
p 903j ð4:46Þ
Again, consider that the line to ground fault occurs in line R as can be seen in Fig. 4.5. The impedances due to line to earth are considered to be small and the impedance due to the fault is considered to be zero as shown in Fig.4.6. The current in the ground to line B can be written as,
160 4 Grounding System Parameters and Expression of Ground Resistance
iBG¼ iB¼vBG ZB
ð4:47Þ
The current in the ground to the line Y is, iYG¼ iY ¼vYG
ZY
ð4:48Þ
The voltage of line R to the ground is,
vRG¼ 0 ð4:49Þ
Applying KVL between the Y-G-R yields,
vYG vRG vRY ¼ 0 ð4:50Þ
Fig. 4.5 Ungrounded delta connection with fault at line R
G equivalent circuit when fault at line R
Substitute Eq. (4.49) into Eq. (4.50) provides,
vYG¼ vYR ð4:51Þ
Applying KVL between the B-G-R yields,
vBG vRG vBR¼ 0 ð4:52Þ
Substituting Eq. (4.49) into Eq. (4.52) yields,
vBG¼ vBR ð4:53Þ
The expression of the fault current can be written as,
if ¼ iYGþ iBG ð4:54Þ
Substituting Eqs. (4.47) and (4.48) into Eq. (4.54) yields, if ¼vYG
ZY þvBG
ZB ð4:55Þ