California State Polytechnic University Pomona
Department of Electrical and Computer Engineering
ECE 3810 Introduction to Power Engineering - Fall 2020
Solution for Homework 1
1-phase, 3-phase circuits, power, power factor, per unit system
Problem 1
A balanced 415.6922-V three-phase source supplies a balanced delta connected load of
ZD = 30+j42  per phase (Fig. 1). The load is connected to the source by a 3-phase line with the
impedance of ZL=2.1+j5.6  per phase.
Fig. 1 Y- connected balanced 3-phase circuit
1) Convert the 3-phase circuit to an equivalent 1-phase circuit, taking the line to neutral Van as
reference.
2) Calculate the current in phase a i.e. the line current Ia.
3) Determine the line-to-line voltage at the load terminal.
4) Calculate the total (i.e. 3-phase) real and reactive power consumed by the load.
5) What is the power factor of the load?
Solution:
1) Equivalent 1-phase circuit
Van = 24000 V,
Vbn = 240 -1200 V = -120 - 2.07.85,
Vcn = 240 -2400 V = -120 + 207.85
ZY = 10.0000 +14.0000i ohm
Equivalent 1-phase circuit
Cal Poly Pomona ECE 3810 Introduction to Power Engineering – Instructor Dr. Ha Thu Le
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2) Line current Ia
Ia = Van / ZY + ZL = 5.4734 - 8.8659i = 10.4193- 58.3110 A
3) Line-to-line voltage at the load terminal
V2 = 1.7886e+02 - 1.2032e+01i = 179.2610 -3.84870 V (Phase-neutral voltage at Y-connected
load terminal)
VL_load = 310.4891( -3.84870 + 300 ) V
4) Real and reactive power consumed by the load
S = 3*V2*conj(Ia) = 3.2569e+03 + 4.5596e+03i = 5.6033e54.46230 kVA
5) Power factor of load
PF_load = cosd(S_ang) = cos(angle(Zdelta))
PF_load = 0.5812 lagging
Problem 2
A 3-phase delta connected capacitor bank is shown in Fig. 2. Given that VL is line-line voltage, IL is line
current, IP is phase current, XC is capacitive reactance per phase. Also, assume that the phase shift
between VL and IP is C.
1) Determine the 3-phase reactive power QC for the delta-connected capacitor bank (i.e. derive a formula
to calculate QC) based on voltage and impedance. Show all your steps to derive the formula.
2) Assume that the delta-connected capacitor bank is now Y-connected equivalent, derive the formula to
calculate QC for this case based on the formula obtained in Part 1.
Fig. 2 3-phase -connected capacitor bank
Students do derivation by themselves.
Final answer is as follows:
For delta-connected capacitor bank: QC = - 3(VL)2 / X
For Y-connected capacitor bank: QC = - (VL)2 / XY
Where X is reactance per phase of delta-connected capacitor bank and XY is reactance per phase of the
Y-connected capacitor bank.
Note that QC has the negative sign as the capacitor delivers reactive power.
Cal Poly Pomona ECE 3810 Introduction to Power Engineering – Instructor Dr. Ha Thu Le
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Problem 3
Three loads are connected in parallel across a 12.47 kV three-phase supply.
Load 1: 135 kW and 1500 kVAR
Load 2: 250 kW at 0.88 power factor leading
Load 3: Resistive load of 90 kW
1) Find the total complex power, power factor, and the supply current.
2) Draw the power triangle for each load and for the combination of the three loads.
3) A Y-connected capacitor bank is connected in parallel with the loads. Find the total kVAR and the
capacitance per phase to improve the overall power factor to 0.98 lagging.
4) Find the new supply current when the capacitor bank is connected.
Solution:
1)
Total complex power, power factor, and the supply current.
S1 = 1.3500e+05 + 1.5000e+06i VA
S2 = 2.5000e+05 - 1.3494e+05i VA
S3 = 90000 W
S = S1 + S2 + S3 = 4.7500e+05 + 1.3651e+06i VA = 1.4453e+06  70.81370 VA
Power factor of the combined load:
PF1 = cosd(Sang) = 0.3286 lagging
The supply current is
I = conj ( S / (sqrt(3)*Vs)) =
2)
21.9921 -63.2013i = 66.9183 -70.81370 A
Power triangle for each load and for the combination of the three loads
3) Capacitance per phase to raise the overall power factor to 0.97 lagging, Y-connected
capacitor bank
theta_new = cos-1(0.98) = 11.47830
Qnew = P tan(11.47830) = 9.6453e+04 VAR
Qc = Qold – Qnew = 1.2686e+06 VAR
Cal Poly Pomona ECE 3810 Introduction to Power Engineering – Instructor Dr. Ha Thu Le
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Xc = Vs^2 / Qc =
122.5757 ohm
C = 1/ (2*pi*60*Xc) = 2.1640e-05 F = 21.64 microF
4)
New supply current
Snew = real(S) + j*Qnew = 4.7500e+05 + 9.6453e+04i VA
Inew = conj( Snew / (sqrt(3) * Vs)) = 21.9921 - 4.4657i = 22.4409 -11.47830 A
Note: The new supply current is less than the supply current before the capacitor is installed.
Problem 4
The one-line diagram of a three-phase power system, its three-phase power and line-line ratings are as
follows.
Fig. 3 One-line diagram of a 3-phase power system
G:
85 MVA 22 kV X = 0.18 per unit
T1:
55 MVA 22/138 kV X = 0.10 per unit
T2:
55 MVA 138/11 kV X = 0.06 per unit
T3:
80 MVA 22/220 kV X = 0.064 per unit
T4:
80 MVA 220/11 kV X = 0.08 per unit
M:
67 MVA 10.45 kV X = 0.182 per unit
Lines 1 and 2 have series reactances of 58.2 and 75.6, respectively. At bus 4, the three-phase load
absorbs 62 MVA at 10.45 kV and 0.65 power factor lagging.
By selecting a common base of 100 MVA and 22 kV on the generator side, draw an impedance diagram
showing all impedances including the load impedance in per-unit.
Solution:
Vbase1 = 138;
% base voltage for Line 1
Vbase2 = 11;
% base voltage for Motor and Load
Vbase3 = 220;
% base voltage for Line 2
Converting impedances to new base
Xg_new = Xg * (MVAbase/85)
= 0.2118 pu
Cal Poly Pomona ECE 3810 Introduction to Power Engineering – Instructor Dr. Ha Thu Le
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X_T1_new = X_T1 * (MVAbase/55)
= 0.1818 pu
X_T2_new = X_T2 * (MVAbase/55)
= 0.1091 pu
X_T3_new = X_T3 * (MVAbase/80)
= 0.0800 pu
X_T4_new = X_T4 * (MVAbase/80)
= 0.1000 pu
Xm_new = Xm * (10.45/11)^2 * (MVAbase/67)
= 0.2452 pu
Line 1 and 2
Zbase_L1 = Vbase1^2 /MVAbase
= 190.44 ohm
Zbase_L2 = Vbase3^2 / MVAbase
= 484 ohm
Xpu_L1 = X_L1/Zbase_L1
= 0.3056 pu
Xpu_L2 = X_L2/Zbase_L2
= 0.1562 pu
theta_load = acos(PF_load)
= 49.4584 deg.
Load
Sload_PQ = P_load + j*Q_load
= 40.3000 +47.1159i MVA
Zload_ohm = kV_load^2 / conj(Sload_PQ) = 1.1449 + 1.3385i ohm
Zbase_load = Vbase2^2 / MVAbase
Zpu_load = Zload_ohm / Zbase_load
= 1.21 ohm
= 0.9462 + 1.1062i pu
One-line diagram showing all impedances
Cal Poly Pomona ECE 3810 Introduction to Power Engineering – Instructor Dr. Ha Thu Le
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Problem 5
The one-line diagram of a 3-phase 4-bus power system is shown in Fig. 4 where the impedances are
marked in per unit on a 100-MVA 138-kV base. The load at bus 2 is
S2 = 86.5 MW +j20.8 MVAR and at bus 4 is S4 = 14 MW +j22.5 MVAR. The voltage at bus 4 is fixed at
131.100 kV.
Fig. 4 One-line diagram of 4-bus power system
1) Working in per unit, determine the voltage at buses 3, 2, 1 in per unit and in kV.
2) Assume that the lower and upper limit for all the bus voltages of the power system are +/-10% (i.e.
all bus voltages must be in the range of 0.9 and 1.1pu of the nominal values). Propose a solution
to bring all the bus voltages within the permissible limits.
3) Show by calculation that your proposed solution works and find the new bus voltages in per unit
after the solution has been implemented.
Solution:
Converting load power and voltage to per unit power and per unit voltage
S2pu = S2/MVAbase
= 0.8650 + 0.2080i pu
S4pu = S4/MVAbase
= 0.1400 + 0.2250i pu
V4pu = (V4kv/Vbase) * (cosd(0)+j*sind(0)) = 0.95 pu
Solving the power system backward from Bus 4 to Bus 1 to find bus voltages
I4pu = conj(S4pu) / conj(V4pu) = 0.1474 - 0.2368i = 0.2789  -58.1092 deg
I34pu = I4pu;
I23pu = I34pu;
V3pu = V4pu + I34pu * Z34pu
= 1.0211 + 0.0442i = 1.0220  2.4793 deg
V2pu = V3pu + I23pu * Z23pu
= 1.1395 + 0.1179i = 1.1456  5.9070 deg
I2pu = conj(S2pu) / conj(V2pu)
= 0.7698 - 0.1029i = 0.7766  -7.6137 deg
I12pu = I2pu + I23pu
= 0.9171 - 0.3397i = 0.9780  -20.3264 deg
V1pu = V2pu + I12pu * Z12pu
= 1.2754 + 0.4847i = 1.3644  20.8111 deg
Cal Poly Pomona ECE 3810 Introduction to Power Engineering – Instructor Dr. Ha Thu Le
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Voltage at Bus 1, 2, 3 in per unit and in kV
V123_pu = [ 1.3644
1.1456
1.0220 ]
It is visible that V1 and V2 violate the limits of +/- 10%
V123_kV = 1.0e+05 * [1.8829
1.5809
1.4104] V
Methods to bring all bus voltages within +/- 10% limits
a) Place a capacitor bank at Bus 4: Sc = 26 MVAR to supply reactive power to Load S4
V123 = [ 1.0995
0.9280
0.9400 ] pu
b) Placing a capacitor bank at Bus 2 Sc2 = 20 MVAR and at Bus 4 Sc4 = 21.5 MVAR to
compensate for load S2 and S4 reactive power demand
V123_pu = [1.0651
0.9656
0.9542] pu
c) Reduce impedance of ALL lines as follows:
Z12pu = j*0.1 pu
Z23pu = j*0.25 pu
Z34pu = j*0.15 pu
V123 =
[1.0950
1.0464
0.9858] pu
Cal Poly Pomona ECE 3810 Introduction to Power Engineering – Instructor Dr. Ha Thu Le
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