ADAMSON UNIVERSITY
COLLEGE OF ENGINEERING
ELECTRICAL ENGINEERING DEPARTMENT
STUDENT NO:
SECTION:
NAME:
PROFESSOR:
865. In symmetrical components, what is the vector sum of 1+a+a2?
a = 1<120 = -0.5+j0.866
a2 = 1<240 = -0.5-j0.866
1+a+ a2 = 1 + 1<120 + 1<240
= 1 + (-0.5+j0.866) + (-0.5-j0.866)
=0
866. If the load of a wye-connected transformer are:
IA = 10 cis (-30)
IB = 12 cis 215
IC = 15 cis 82
What is the phase b positive sequence component?
IA1 = (1/3) (IA + a IB + a2IC)
= (1/3) [10<-30 + (1<120)(12<215) + (1<240)(15<82)]
= (1/3) [10<-30 + 12<335 + 15<322]
= (1/3) [8.66-j5 + 10.875-j5.071 + 11.820-j9.235]
IA1 = 10.45-j6.435 = 12.27<-31.624
IA1 = a2 IA1 = (1<240) (12.27<-31.624)
IA1 = 12.27<208.376
867. Given three unbalanced three-phase voltages:
VA = 150+j0 V
VB = -90-j120 V
VC = -120+j90 V
Determine VA1
VA = 150+j0 = 150<0
VB = -90-j120 = 150<-126.869
VC = -120+j90 = 150 <143.13
VA1 = (1/3) (VA +a VB + a2 VC)
= (1/3) [150<0 + (1<120)(150<-126.869) + (1<240)(150<143.13)]
= (1/3) [150<0 + 150<-6.869 + 150<383.13]
= (1/3) [150+j0 + 148.923-j17.94 + 137.942+j58.923]
VA1 = 145.62 + j 13.66
868. The three-phase unbalanced currents are:
IA = 10 cis (-30)
IB = 0
IC = 10 cis 150
Find the negative sequence current of phase a.
IA2 = (1/3) (IA + a2 IB + a IC)
= (1/3) [10<-30 + (1<240) (0) + (1<120) (10<150)]
= (1/3) [10<-30 + 0 + 10<270] = (1/3) [8.66-j5 + 0 –j10]
IA2 = 2.886-j5 = 5.77<-60
IA2 = 5.77 cis (-60)
869. Given the following currents:
IA = 60+j0 A
IB = -36-j48 A
IC = -48+j36 A
Solve for the negative sequence component of IA.
IA = 60+j0 = 60<0
IB = -36-j48 = 60<-126.869
IC = -48+j36 = 60 <143.13
IA2 = (1/3) (IA +a2 IB + a IC)
= (1/3) [60<0 + (1<240) (60<-126.869) + (1<120) (60<143.13)]
= (1/3) [60<0 + 60<113.13 + 60<263.13]
= (1/3) [60 + (-23.569+j55.177) + (-7.177-j59.569)]
IA2 = 9.751-j1.464
870. The three-phase unbalanced currents are:
IA = 10 cis (-30)
IB = 0
IC = 10 cis 150
Find the zero sequence current.
IA0 = (1/3) (IA1 + IA2 + IA0)
= (1/3) [10<-30 + (0) + (10<150)]
= (1/3) [8.66-j5 + 0 + (-8.66+j5)]
IA0 = 0
871. Given the following currents:
IA = 60+j0 A
IB = -36-j48 A
Solve for zero sequence component of IA.
IC = -48+j36 A
IA0 = (1/3) (IA1 + IA2 + IA0)
= (1/3) [60+j0 + (-36-j48) + (-48+j36)]
IA0 = -8-j4
872. The sequence currents of phase a current are as follows:
Zero sequence current =12.13 <17.34
Positive sequence current =2.98 <10.06
Negative sequence current =708.26 <-31
Determine phase a current
Ia = Ia1 + Ia2 + Ia0
= [2.98<10.06+ 708.26< - 31 + 14.13<17.34)]
= 2.934 +j 0.52 + 607.097 – j364.78 +13.487 + j 4.211
= 623.518 – j 360.05
Ia = 720<30
873. The sequence currents of phase a current are as follows:
Zero sequence current =0.47 + j1.49
Positive sequence current =18.4 cis (-31.6)
Negative sequence current =3.23 cis 168.2
Determine phase a current
Ib1 = a2Ia1 = 1<240 * 18.4<-31.6 = 18.4<208.4 = -16.185 – j 8.75
Ib2 = aIa2 = 1<120 * 3.23<168.2 = 3.23<288.2 = 1.009 – j 3.068
Ib0 = aIa0 = 0.47 + j 1.49
Ib = Ib1 + Ib2 + Ib0
= -15.185 – j 8.75 + 1.009 – j3.068 + 0.47 + j1.49
= -14.706 – j10.328 = 17.97<-145 = 18<(360 -145)
Ib = 18cis 215
874. The sequence currents of phase a current are as follows:
Zero sequence current =0.47 + j1.49
Positive sequence current =18.4 cis (-31.6)
Negative sequence current =3.23 cis 168.2
Determine phase c current
Ic1 = aIa1 = 1<120 * 18.4<-31.6 = 18.4<88.4 = 0.513 – j 18.39
Ib2 = a2Ia2 = 1<120 * 3.23<168.2 = 3.23<408.2 = 2.153 + j 2.407
Ic0 = Ia0 = 0.47 + j 1.49
Ic = Ic1 + Ic2 + Ic0
= 0.513 + j 18.39 2.153+ j 2.407 + 0.47 + j1.49
= 3.136 + j22.287 + 22.5<82
Ic = 22.5 cis 82
875. Determine the symmetrical components of the line current in line a if one of
the in phase impedance of its delta connected load connected across lines ca is
removed. The delta load with impedance of 10<0 ohms per phase is supplied from a
220 volts, 6- cycle, 3 phase. Assume a phase sequence of a-b-c.
Iab = Vab /Z = 220<0/10<0 = 22<0 =22 A
Ibc = Vbc /Z = 220<-120/10<0 = 22<0 =22<-120 A = -11- j19.052
Ia = Iab = 22<0 A
Ib = Ibc – Iab = -11 –j19.052 -22
= -33 – j 19.052
= 38.104< -150 A
Ic = -Ibc = - (-11 – j19.052) = 22<60 A
Ia1 = 1/3 (Ia + aIb + a2Ic)
= 1/3 [22<0 + (1<120*38.104<-150) + (1<240*22<60)]
Ia1 = 22 – j12.7 = 25.4<-30 A
Ia2 = 1/3 (Ia + a2Ib + aIc)
= 1/3 [22<0 + (1<240*38.104<-150) + (1<120*22<60)]
Ia2 = j 12.7 = 12.7<90 A
Ia0 = 1/3 (Ia1 + Ia2 + I0)
= 1/3 [22<0 + 38.104<-150 + 22<60]
Ia0 = 0 A
876. A star connected balanced load takes 75 A from a balanced 3 phase, 4wire
supply. If the two supply lines of the fuses are removed determine the symmetrical
components of the line currents after the fuses are removed.
Ia = 75<0 A
Ib = 0 A
Ic = 0 A
Ia1 = 1/3 (Ia + aIb + a2Ic)
= 1/3 [75<0 + (1<120*0) + (1<240*0)]
Ia1 = 25<0 A
Ib1 = a2Ia1 = (1<240)*(25<0) = 25<240 A
Ic1 = aIa1 = (1<120)*(25<0) = 25<120 A
Ia2 = 1/3 (Ia + a2Ib + aIc)
= 1/3 [75<0 + (1<240*0) + (1<120*0)]
Ia2 = 25<0 A
Ib2 = aIa1 = (1<240)*(25<0) = 25<120 A
Ic2 = a2Ia2 = (1<120)*(25<0) = 25<240 A
Ia0 = 1/3 (Ia1 + Ia2 + I0)
= 1/3 [75<0 +0 + 0]
Ia0 = 25<0 A
Ib0 = Ic0 = 25<0 A
877. A 50 MVA, 33kv/ 11kv, 3 phase, wye delta connected transformer has a 3%
impedance. What is the percent impedance at 100 MVA base and 34.5kv base?
Zpu2 = Zpu1 (Vbase1 / Vbase2)2 (Sbase2/Sbase1) = 0.03 *(33/34.5)2 * (100/50) = 0.054896
Zpu2 = 5.49%
878. At a certain location of an electrical system, the available short circuit MVA is
10 at 110 kV while its Thevenin’s equivalent reactance is 0.05 pu. Determine the per
unit reactance of this point using a base 20 MVA and 115 KV?
Solution:
Z pu2 = (0.05)(110/110)²(20/10)
Z pu2 = 0.09
879. A 13.8 kV/440v, 50 KVA single phase transformer has a leakage reactance of
300 ohms referred to the 13.8 kv side. Determine the per unit value of the leakage
reactance for the voltage base.
Solution:
X pu = (300x50,000)/(13,800)
X pu = 0.079
880. A 5 KVA, 2400-120/240 volt distribution transformer when given a short circuit
test has 94.2 volts applied with rated current flowing in the short circuited wiring.
What is the per unit impedance of the transformer?
Solution:
Isc = 5000/2400 = 2.083 A
Z = 94.2/2.083 = 45.22 Ω
ZPU = (2.083X45.22)/(2400)
ZPU = 0.0392
881. A 3 phase, 375 KVA, 480 V, 50 HZ, wye connected alternator has an equal positive
and negative sequence reactance of 10%. Find the symmetrical fault current if a 3
phase fault occurs at the alternator terminals.
Solution:
IPU = 1/j.01 = -j10
Ibase = (375X10ᶾ)/(√3 X 480) = 451.05 A
IF = 10(451.05) = 4510.5 A
KVA = 375/0.1 = 3750 KVA
IF = 3750/(√3X0.48)
IF =4510.5 A
882. A generator rated 600 kVa, 2400v, 60 cycles, 3 phase, 6 poles and wye
connected has 10% synchronous reactance. If a three phase fault occurs, what will
be the short circuit current?
Solution:
IPU = 1/j.01 = -j10
Ibase = (600X10ᶾ)/(√3X2400) = 144.34 A
IF = 10X 144.34
IF =1443.4A
883. A 20 MVA, 13.8 KV, 60 HZ, three phase synchronous turbo alternator has a
positive, negative and zero sequence reactance of 0.25, 0.35 and 0.15 respectively. If
lines b and c are short circuited, determine the sub transient fault current in line
b. Assume the alternator is operating at no load and at rated voltage with the
neutral of the alternator solidly grounded.
Solution:
Ibase = (20000000)/(√3X13800) = 836.7 A
Ia1 = 1.0/(j0.25+j0.35)
Ia1 = -j1.667 pu
Ia2 = j1.667 pu
Ia0 = 0
Ic = (1<120)(1.667<-90)+(1<-120)(1.667<90)+0
Ic = 2.886 pu
Ic fault = 2.886(836.7)
Ic fault =2414.7 A
884.) At a certain location in an electric system, the available fault MVA is 400
MVA. A 15 MVA, 34.5 kV/6.24 kV, 2.5% impedance, wye-wye grounded transformer is
installed at the location. Determine the short circuit MVA at the secondary side of
the transformer.
Solution:
Sf = √3Vf [(Vt / √3) / (Zf)]
Zf = (6.24 x 103)2 / (400 x 106) = 0.097 Ω
ZL = 0.064896 Ω
Sf new = Vf2 / Zf = (6.24 x 103)2 / (0.16224)
Sf new= 240 MVA
885.) two three-phase transformers are connected in parallel at the primary as
well as at the secondary sides. One is rated 10 MVA, 34.5/13.8 kV and 4% impedance
while the other is rated 7.5 MVA, 34.5/13.8 kV and 5% impedance. The primary
tapping point has a 3-phase short circuit MVA of 1000 MVA. Determine the fault
current delivered to a 3-phase fault at the secondary side bus bars.
Solution:
MVAf = MVAc / Z1 pu = 10 / 0.035 = 285.714 MVA
If = 285.714 x 106 / √3(13.8 x 103)
If = 11,953.41 A
886.) a 100 MVA, 22 kV synchronous turbo generator has positive and negative
reactances of o.20 p.u and a zero sequence reactance of 0.05 p.u The neutral of the
generator is grounded through a reactor of 0.242 ohm. Determine the ratio of the
subtransient current for a single line to ground fault to the subtransient current
for a three-phase fault. Assume the generator is operated without a load and at
rated voltage.
Solution:
Xbase = Vbase2 / Sbase = (22,000)2 / (100,000,000)
Xbase = 4.48 Ω
Ratio = -j5.001 / -j5
Ratio = 1.0
887.) a generator is rated 100 MVA, 15 kV, it is Y-connected, solid grounded and is
operated at rated voltage at no-load and is disconnected from the rest of the
system. Its positive and negative sequence reactances are each 0.10 p.u and its zero
sequence reactance is 0.05 p.u Calculate in ohms of inductive reactance to be
inserted in the neutral connection of the generator to limit the fault current for a
single line to ground fault to the fault current of a symmetrical three-phase fault.
Solution:
Xbase = (15,000)2 / (100,000,000)
Xbase = 2.25 Ω
Xneutral = (0.0167)(2.25)
Xneutral = 0.0376 Ω
888.) the reactance of a three-phase alternator is 8 percent. If the alternator is
rated 25 MVA at 13.25 kV output voltage, line to phase, solve for the magnitude of
the fault current generated when a short circuit occurs between 2 phases at the
terminals.
Solution:
kVAf = KVAc / Xpu = 25000 / 0.08 = 312500 kVA
If 3Ф = kVAf / √3 (kVf) = 312,500 / √3 (√3 (13.25)
If 3Ф = 7861.6 A
IfLL = (√3 / 2) (If 3Ф)
IfLL = 6808 A
889.) a 3-phase, 375 kVA, 480 V, 50 Hz, wye connec ted alternator has an equal
positive and negative sequence reactance of 10%. Find the unsymmetrical fault
current, if a 2-phase fault (line to line) occurs at the alternator terminals.
Solutions:
Ia1 = 1.0 / j0.1 + j0.1
Ia1 = -j5 pu
Ia2 = -Ia1 = j5 pu
Ia0 = 0
Ib fault = Ib pu (Ibase) = 8.66(451.05)
Ib fault =3906.09 A
890. A 5 MVA, 13.8 kVA , 480 V. 5% impedance transformer is tapped at 13.8 kV line
where the Thevenin’s equivalent impedance is 0.5 ohm. Determine the fault current
at the primary for a three phase fault at the secondary.
%IZ = SZ/V2
Z = (%IZ)(V2) / S = ((0.05)(13.8 x 103)2) / (5x106)
Z = 1.9044 ohm
Zt = Zth + Z = 0.5 +1.9044 = 2.4044 ohm
If = Vf / Zt = (13800/sqrt3) / 2.4044 = 3313.68 A
891. A 10 kVA, 110V, 3 phase, 4 wire, 60 Hz alternator generates 70 volts per phase
when excited to give rated at full load. The armature synchronous impedance per
phase is 0.2 ohm. What is the current in each phase if the generator terminals are
short circuited?
If = Vp / Zp = 70 / 0.2 = 360 A
892. At a certain point of the system network the positive, negative and zero
sequence impedances are 0.25 pu, 0.25 pu, and 0.3 pu, respectively. The base MVA is
100. The voltage level at that point is 34.5 kV. Determine the zero sequence current
for a one line to ground fault.
Ia0 pu = Ia1 pu = Ia2 pu
Ia0 pu = 1 / 0.25 + 0.25 + 0.3 = 1.25 pu
Ibase = 100x106 / sqrt(3) (34.5x103) = 1673.48 A
Ia0 = Ia0 pu (Ibase ) = 1.25(1673.38)
Ia0 = 2091.81 A
893. In a short circuit analysis, the positive, negative and zero sequence impedances
are 0.20 pu, 0.20 pu, and 0.25 pu respectively, using a base MVA of 50. Estimate the
fault current on the faulted lines if a double line to ground fault at the 138 kV level
occurs.
Ia1 = 1 / j(0.2 + (0.2x0.25 )/(0.2+0.25))
Ia1 = -j3.214
Va1 = 0.3572 pu
Ia0 = j1.4288
Ia2 = j1.786
Ib = -4.3297 + j2.1428 = 4.831˂153.67 pu
Ibase = (50x106) / sqrt(3) (138x103) = 209.184 A
If = 4.831(209.183) = 1010.568 A = 1011A
894. A 15 MVA, 6600 V, 60 Hz, y connected synchronous alternator has a positive,
negative and zero sequence per unit reactance of 0.20, 0.20, and 0.10 respectively.
The neutral of the generator is grounded through a reactor with a per unit
reactance of 0.05 based on the generator rating. If a double line to ground fault
occurs at the alternator terminals, estimate the current that flows in the reactor.
Ia1 = - j3.214
Va1 = 0.3572 pu
Ia0 = j1.4288
Ia2 = j1.786
Ib = -4.3297 + j2.1428
Ic = 2.7983 + j1.1607 + ( 1.5467 – j0.892) + j1.4288
Ic = 4.3297 +j2.1428
If = 4.2856(1312.16) = 5623.39 A
If = 5623 A
895. In a short circuit analysis, the positive, negative and zero sequence per unit
reactance of 0.25, 0.25, and 0.3 respectively. The base MVA is 100. Determine the
fault current for a three phase fault at the 115kV level.
KVAf = KVAc / Xpu = 100000/0.25 = 1312.18 A
If = 400000 / sqrt(3)(115) = 2008 A
896. A 3 – phase 220 – V, Y- connected alternator has a synchronous impedance of
0.15 + j2 ohms per phase. The alternator is connected through a short transmission
line whose impedance per wire is 2 + j1 ohms to a Y-Y connected transformer bank
whose total equivalent impedance referred to the secondary is 50 + j86.6 ohms. A
three – phase load is connected from the secondary side of the bank through a
transmission line whose impedance is 50 + j75 ohms per wire. If a symmetrical three
– phase fault occurs at the side. How much is the current will flow to the alternator
windings. Assume the bank has a transformation ratio of 1:10.
Total impedance per phase at the secondary side of the transformer bank:
Z2 = (50 + j86.6) + (50 + j75)
Z1 = 3.15 + j4.616 = 5.588 cis 55.69o
If = E/Zt = 220 sqrt. 3 /5.588
If =22.73 A
897. Three single-phase transformers each rated 50 kVA, 2400 volts primary. 277
volts secondary, four (4) percent impedance have their primaries connected in
delta and their secondaries in wye. Calculate the fault drawn if a short circuit
occurs between two phases at the secondary terminals.
%IZ = SZ / V2
Z = 0.06138 Ω
If = VL / 2Z = 277 / 2x0.0138
If = 2256.43 A
898. La Tondena Distillery inc. located in canluabng calamba. Laguna installed an
emergency 3-phase generator GM generator is rated 350kVA, 460 volts, 60 Hz with a
reactance of 8%. Solve for the symmetrical 3-phase short circuit.
KVA = KVA / Xpu = 350/ 0.08 = 4375 kVA
If = KVA / sqrt. 3 (KVf) = 4375 /sqrt. (0.46)
If = 5491 A
899. A 15 MVA. 34.5 kV / 6.24 kV transformer is connected at an infinite bus. The
percent impedance of the transformer is 2.5. What is the current at the 34.5 kV side
for a three phase short at the 6.24 kV side?
KVAf = KVA / Xpu = 15000 / 0.025 = 600000 kVA
If = kVA / sqrt. 3 x kVf = 600000 /sqrt. 3 x 34.5
If = 10040.87 A
900. A three-phase, 3-wire generator Is rated 325kVA, 480 volts, 60 Hz with a
reactance of eight percent. Solve for the fault current delivered by the generator
during a 2-phase short circuit at the terminals.
KVAf = KVA / Xpu = 325 / 0.08 = 4062.5 kVA
If = kVA / sqrt. 3 x kVf = 4062.5 /sqrt. 3 x 0.08
If = 4232 A
901. The secondary side of 5 MVA, 345 / 13.8k, 3-phase transformer is connected to a
bus bar served by a 20 MVA, 13.8 kV, 3-phase alternator. The primary side of the
transformer is tapped from an infinite bus. Estimate the fault current delivered
to a 3-phase fault near the common bus bar. The transformer and alternator have
4% and 7% reactance, respectively.
Ztrans.pu = Ztrans.pu X (MVAc / MVAtran) = 0.16
Refer to the equivalent circuit diagram:
Ztrans.pu = Ztrans.pu x Zalt.pu / Ztrans.pu +Zalt.pu =0.16(0.075) / 0.16 + 0.075 = 0.051
Ztrans.pu =0.235
Strans = MVAc / Ztrans.pu = 20 /0.051 = 392.157 MVA
If = 16,406.66 A
902. Determine the minimum rupturing capacity of the circuit breaker as shown, if
a three-phase symmetrical fault occurs at a point P.
Use the 10 MVA as the common MVA base.
Xa = Xpu (MVAc / MVAbase)
Xg = 0.03 (10 / 25) = 0.12 pu
Xtransfomer = 0.20 (10 / 20 =0.10
Refer to the equivalent circuit diagram:
X1 = 0.12 + 0.15 + 0.15 + 0.10
X1 =0.52 pu
MVAf = MVAc / X1 =19.23
Rupturing capacity od CB =MVA
=20 MVA
903. Given the one line diagram as shown. Determine the current delivered by
generator. A when a three-phase fault occurs at point P.
Use the 75 MVA as the common MVA base.
Xa = Xpu (MVAc / MVAbase)
Xa = 0.20 (75 / 50) = 0.30 pu
Xtransfomer = 0.20 (10 / 20 =0.10
Refer to the equivalent circuit diagram:
X1 = (0.3)(0.15) /0.3 + 0.15 =0.25
X1 =0.35 pu
Refer to the equivalent circuit diagram:
Sf = kVAc / X Ipu = 214285 kVA
If =8965 A
Refer to circuit diagram of the transformer bank :
Ia =172.5 A
904. Given the one-line diagram as shown, determine the reactance of the limiting
reactor x in order to limit the three-phase symmetrical fault power at point P to
400 MVA.
three-phase fault occurs at point P.
Use the 20 MVA as the common MVA base.
Xa = Xpu (MVAc / MVAbase)
Xa = 0.20 (20 / 25) = 0.16 pu
Xtransfomer = 0.25 (20 / 50 ) =0.10
Refer to the equivalent circuit diagram:
X1 = 1/ (1 /0.15) + (1 /0.16) + (1 / 0.1 + X)
X1 =0.1+X / (2.2916 + 12.916X EQ. 1
Refer to the equivalent circuit diagram:
MVAf = MVAc / X Ipu
Xf= MVAc / MVAf =20 /400
Xf=0.05
Substitute X1 in EQ. 1:
0.05 =0.1+X / (2.2916 + 12.916X)
0.11458 + 0.6458X = .1 + X
X = 0.04116 pu
Xbase = Vbase 2/ Sbase =9.522 Ω
X = Xpu (Xbase ) =0.04116 (9.522)
X = 0.392 Ω