Question

Explain the reasons for applying system grounding in electric power systems. [4 points]

Consider the industrial plant distribution bus which is scpplied by a utility source with per-unit sequence reactances as shown in Figure (4-a). Assume that all reactances are given on a \(5,000-\mathrm{kVA}\) base, and that the plant"s bus voltage is \(4,160-\mathrm{V}\).

b- Determine the fault current for a single line to ground fault on phase A of the bus. \([4\) points]

Assume now that an ungrounded \(5,000 \mathrm{kVA}\) generator is added at the distribution bus as shown in Figure (4-b). [4 points]

c- Determine the fault current for a single line to ground foult on phase A of the bus. [4 points]

Assume now that the utility transformer is grounded through a \(8 \Omega\) grounding resistor as shown in Figure (4-c.)

d- Determine the value of the grounding resistor in per unt. [4 points]

Determine the fault current for a double line to ground fault on phases \(B\) and \(C\) of the bus. [4 points]

Answer 1

advantage of neutral grounding During unbalanced condition, floating neutral point will take plance if the system is ungrounded. That will cause unbalaned phase nollage. Grounding will eleminate this problem by adding reference zero potential point. During fault (unsymmetrical fault) arcing ground phenomenon will arise if the system is ungrounded. This is caused by charging current due to einbalanced fault. To eleminate this grounding is necessary So that fault current will gets a return path to flow. ground fault cet bust (b) Case single line to 1 (4-0) the sequence N/W j0.15 VTH = 1 pu (thevenin uoltage) X'TH = j*' (Heverin reactance at ) X'TH = 0.15 pu Jx1 fault point ,

the sequence N/W jo.15 (the sequence the venin equivalent ) all reactance at fault point , XTH=x" XTH = 0.15 pu zero sequence N/W jo 1 mm jxo Lew sequence thevenin eq. reactance, a at fault point X TH = xo XTH = 0.1 pu fault current for L-G fault at Bus If = BUTH X'TH+X"TH+XTH 3x1 0:15 +0:15 +0:1 fault current in per unit If = 7.5 eu = 693.93 Amp Bare current at bus SB 5000X10 IB = 53 VB = J3 x 4160 fault current in Amp If = (amp) If X IB = 7.5X693.93 = 5204.40 Amp. (PU)

case 2 (4-6) (5000kVA gen. is added into system) the sequence N/W joob Ejxg n E jxt ť VTH = 1 pu X'TH = (xģ11 X'=) | x TH = (0-06 || 9:15) X'TH = 0.06) (0.95) 0.06 +0.15 X'TH = 0.04286 pu (the seq. thevenin eq. reactance aut fault point - ue sequence N/W X'TH = (x + 11x"g) = (0-06 || 0:15) D jx'g 10.15 Jx -ul sequence theverin eq. I reactance at fault point) X TH= 0.0429 pu Lero sequence w/w XºTH = (x°1 || xg) - (0-1110-04) XTH = 0·0206 pu hoor Er det

Pu fault current Ic = 3 UTH Xu+ 'T + XTH 3x2 OOH 29 + 0 042 9 + 0-0286 0 149 If = 26.22 38 pull PU) fault current in amp + = If camp) If X (pod IB = 26.2230X 693.93 If = 20.20 K Amp Il Camp) d Per unit value of grounding resistor Rn = ou Rn = Rocactual) (VBuse)? Sbase (PU) 2 Base ZB = (4160)2 SOOO X103 ZB = 3.4611 r Rn = B 3.4611 = 2.3114 PU Rn = 2.3114 pu /

e L-G fault when utility transformer grounded through on resistor the sequence N/W UTH = 1 pu og sd jou Ejxg. Eixt XTH = 0.0429 pu F { same as previous case) - he sequence N/W 90.01 x"TH= 0.0429 pu (same as previous cases JX"g Sl.or ww jx'T zero sequence N/W į T 3 274 = (ixg || (Uxq + 3Rn)) jos irt = | j0.04 116.9342 +30-2)] 3x12.314) { 3Rn ZOTH = 0.04-09.67 °

Per unit -G fault current - 3.UTH 3×(120) | XTH )x"TH + 2TH 0-0 929 + 30-0429 + 0-04 /0941 3 0.1250 (09.09 IRL = 23.0482 pu Fault current in amp IB = 693.93 Amp If amp If amp - If & IB (PU) = (23.8482)XL693.93) If = 16.55 K Amp amp