ECE 3614
HW # 1
3.4) A single-phase source supplies a load consisting of a resistor R = 20 Ω, which are connected in
parallel. The instantaneous voltage of the source is given by v(t) = 120 cos(ωt+45o) V
Find the following:
a) Phasor voltage V of the source
V=
√
cos(
+ 45°) V = 84.85< 45° V
b) Phasor Current I suppied by the source
Z1 = R = 20 Ω, Y1 = 0.05 Ʊ
Z2 = -j10 Ω (capacitor), Y2 = j0.10 Ʊ
Y = Y1+Y2 = (0.05 + j0.10)
I = YV = (0.05 + j0.10) x (84.85< 45°) = -3+j9 = 9.5< 108.4° A
c) Instantaneous current i(t) supplied by the source
I(t) = 9.5√2 cos(
+ 108.4°) = 13.4 cos(ωt+108.4°) A
3.5) Repeat 3.4 if the resistor and capacitor are connected in series.
a) Phasor voltage V of the source
V=
√
cos(
+ 45°) V = 84.85< 45° V
b) Phasor Current I suppied by the source
Z1 = R = 20 Ω, Z2 = -j10 Ω (capacitor)
Z = Z1+Z2 = (20 - j10)
I= =
(
.
(
°)
)
= 1.2+j3.6 = 3.8< 71.6° A
c) Instantaneous current i(t) supplied by the source
I(t) = 3.8√2 cos(
+ 71.6°) = 5.4 cos(ωt+71.6°) A
3.10) A single-phase source has a terminal voltage V = 120< −15°. It supplies a current of 15< 45° to
an electric load.
a) Find the complex power supplied by the source.
S = V x I* = (120<-15°) x (15<45°)* = (120<-15°) x (15<-45°) = 1800<-60° VA
S = 900 – j1559 VA
b) Determine the real power. State whether the source is delivering or absorbing.
P = +900 W (supplied because it is positive value)
c) Determine the reactive power. State whether the source is delivering or absorbing.
Q = -j1559 VAR (absorbed, because it is negative value)
3.13) A single-phase electrical load draws 10 MW at 0.6 pf lagging.
a) Find the real and reactive power absorbed by the load.
P(real power) = 10 MW, pf = 0.6 lagging
ϴ=cos-1(0.6) = 53.1°
Q(reactive power) = P x tan(ϴ) = 10 tan(53.1°) = 13.33 MVAR
b) Draw the power triangle.
S= (
+
) = (10 + 13.33 ) = 16.66 MVA
c) Determine the kVAR of a capacitor to be connected across the load to raise the power factor
to 0.95.
ϴnew = cos-1(0.95) = 18.2°
Qnew = P x tan(ϴnew) = 10 tan(18.2°)
= 3.29 MVAR
= Qold + Qcap
Qcap = Qnew - Qold
Qcap = 3.29 – 13.33 = -10 MVAR (capacity of the required compensation capacitor)
3.15) An industrial plant consists of several induction motors. The plant absorbs 300 kW at 0.6 pf
lagging from the substation bus.
a) Compute the required kVAR rating of the capacitor connected across the load to raise the
power factor to 0.9 lagging.
PL = 300 KW, pf = 0.6 lagging, ϴ = cos-1(0.6) = 53.1°
QL = PL tan(ϴ) = 300 tan(53.1°) = 400 kVAR
Qnew = PL tan(ϴnew) = 300 tan(cos-1(0.9)) = 145.3 kVAR
Qcap = Qnew – QL = 145.3 – 400 = -254.7 kVAR
b) A 200-hp, 90% efficient, synchronous motor is operated from the same bus at rated conditions
and 0.8 pf leading. Calculate the resulting power factor.
PM’ = 200 hp, η = 90%, pf = 0.8 leading, ϴM = cos-1(0.8) = 36.9°
PM =
.
W = 165800 W = 165.8 kW
QM = PM tan(ϴM) = 165.8 tan(36.9°) = -124.35 kVAR
PT = PL + PM = 300 + 165.8 = 465.8 kW
QT = QL + QM
QT = 400 + (-124.35) = 275.65 kVAR
Tan (QT) =
=
.
.
QT = 30.6°
Pftot = cos(30.6°) = 0.86 lagging