ECE 404
Introduction to Power Systems
HW #5
3.14, 3.19, & 3.35
3.14 A single-phase, 50 kVA, 2400/240 Volt, 60 Hz, distribution transformer is used
as a step-down transformer at the load end of a 2400 feeder whose series
impedance is (1.0 + j 2.0) ohms. The equivalent series impedance of the
transformer is (1.0 + j 2.5) ohms. referred to the high-voltage (primary) side.
The transformer is delivering rated load at 0.8 power factor lagging and at rated
secondary voltage.
Neglecting the transformer excitation current, determine:
a) the voltage at the transformer primary terminals, V p ,
b) the voltage at the sending end of the feeder, V S, and
c) the real, PS, and reactive, Q S, power delivered to the sending end of the
feeder.
j :=
First draw a circuit diagram that represents the problem.
R F + jX F
+
VS
-
+
VP
-
kVA := kW
ID
R T + jX T
Is
−1
+
E1
-
T1
+
E2
-
SD
+
VD
-
Givens:
RF := 1.0⋅ Ω
RT := 1.0⋅ Ω
XF := 2.0⋅ Ω
XT := 2.5⋅ Ω
ZF := RF + j ⋅ XF
ZT := RT + j ⋅ XT
ZF = ( 1 + 2j) Ω
ZT = ( 1 + 2.5j) Ω
a :=
2400⋅ V
a = 10
240 ⋅ V
VD := 240 ⋅ e
SD := 50⋅ e
j ⋅ 0⋅ deg
⋅V
j ⋅ acos( 0.8)
C:\JLAW\CLASSES\ECE 404 Fall
2008\Homework\HW05
\solutionR1.xmcd
⋅ kVA
SD = ( 40 + 30j) ⋅ kVA
Page 1 of 6
September 17, 2008
ECE 404
Introduction to Power Systems
HW #5
3.14, 3.19, & 3.35
Solution:
E2 := VD
( )
E2 = 240 V
arg E2 = 0 ⋅ deg
ID = 208.333 A
arg ID = −36.87 ⋅ deg
E1 := a⋅ E2
E1 = 2.4⋅ kV
arg E1 = 0 ⋅ deg
ID
IS :=
a
IS = 20.833 A
arg IS = −36.87 ⋅ deg
VP := E1 + ZT⋅ IS
VP = 2.45⋅ kV
arg VP = 0.683 ⋅ deg
VS := VP + ZF⋅ IS
VS = 2.49⋅ kV
arg VS = 1.151 ⋅ deg
ID :=
⎯
⎛ SD ⎞
⎜ ⎟
⎝ VD ⎠
⎯
SS := VS⋅ IS
( )
PS := Re SS
( )
( )
( )
( )
( )
SS = ( 40.868 + 31.953j ) ⋅ kVA
PS = 40.9⋅ kW
kVAR := kW
( )
QS := Im SS
C:\JLAW\CLASSES\ECE 404 Fall
2008\Homework\HW05
\solutionR1.xmcd
QS = 32⋅ kVAR
Page 2 of 6
September 17, 2008
ECE 404
Introduction to Power Systems
HW #5
3.14, 3.19, & 3.35
3.19 Using the transformer ratings as base quantities, work Problem 3.14 in per-unit.
Sbase := SD
Sbase = 50⋅ kVA
High Voltage Base Quantities
Low Voltage Base Quantities
VbaseH := 2400⋅ V
VbaseX := 240 ⋅ V
Sbase
IbaseH :=
VbaseH
Sbase
IbaseX :=
VbaseX
IbaseH = 20.833 A
IbaseX = 208.333 A
ZbaseH :=
VbaseH
ZbaseX :=
IbaseH
ZbaseH = 115.2 Ω
pu := 1
SD
SDpu :=
Sbase
SDpu = ( 0.8 + 0.6j) ⋅ pu
VD
VDpu :=
VbaseX
VDpu = 1 ⋅ pu
ZFpu :=
ZT
ZbaseH
ZF
ZbaseH
IbaseX
ZbaseX = 1.152 Ω
Converting Given Quantities to Per Unit.
ZTpu :=
VbaseX
(
−3
(
−3
ZTpu = 8.681 × 10
ZFpu = 8.681 × 10
)
+ 0.022j ⋅ pu
)
+ 0.017j ⋅ pu
apu := 1.0
C:\JLAW\CLASSES\ECE 404 Fall
2008\Homework\HW05
\solutionR1.xmcd
Page 3 of 6
September 17, 2008
ECE 404
Introduction to Power Systems
HW #5
3.14, 3.19, & 3.35
Solution in Per Unit
E2pu := VDpu
(
)
(
)
(
)
(
)
E2pu = 1 ⋅ pu
arg E2pu = 0 ⋅ deg
⎜
⎟
⎝ VDpu ⎠
IDpu = 1 ⋅ pu
arg IDpu = −36.87 ⋅ deg
E1pu := apu⋅ E2pu
E1pu = 1 ⋅ pu
arg E1pu = 0 ⋅ deg
IDpu
ISpu :=
apu
ISpu = 1 ⋅ pu
arg ISpu = −36.87 ⋅ deg
IDpu :=
⎯
⎛ SDpu ⎞
(
)
(
)
VPpu := E1pu + ZTpu⋅ ISpu
VPpu = 1.02⋅ pu
arg VPpu = 0.683 ⋅ deg
VSpu := VPpu + ZFpu⋅ ISpu
VSpu = 1.04⋅ pu
arg VSpu = 1.151 ⋅ deg
⎯
SSpu := VSpu⋅ ISpu
SSpu = ( 0.817 + 0.639j) ⋅ pu
(
)
PSpu = 0.817 ⋅ pu
(
)
QSpu = 0.639 ⋅ pu
PSpu := Re SSpu
QSpu := Im SSpu
C:\JLAW\CLASSES\ECE 404 Fall
2008\Homework\HW05
\solutionR1.xmcd
Page 4 of 6
September 17, 2008
ECE 404
Introduction to Power Systems
HW #5
3.14, 3.19, & 3.35
3.35 Consider a bank of three single-phase two-winding transformers whose high-voltage
terminals are connected to a three-phase, 13.8 kV feeder. The low-voltage terminals are
connected to a three-phase substation load rated 2.4 MVA and 2.3 kV.
Determine the required voltage, current, and MVA ratings of both windings of each
transformer, when the high-voltage/low-voltage windings are connected
a) wye-delta,
b) delta-wye,
c) wye-wye, and
d) delta-delta.
Added by Prof. Law: NR and the ratio of N H/HX
MVA := MW
S3φ := 2.4⋅ MVA
S1φ :=
S3φ
S1φ = 800 ⋅ kVA
3
VXLL := 2.3⋅ kV
VXΔ := VXLL
VXY :=
VHLL := 13.8⋅ kV
VXLL
VHΔ := VHLL
VHY :=
NR :=
VHY
VXΔ
3
= 3.464
C:\JLAW\CLASSES\ECE 404 Fall
2008\Homework\HW05
\solutionR1.xmcd
13.8⋅ e
2.3⋅ e
VHΔ
VXY
VHLL
3
VXΔ = 2.3⋅ kV
VXY = 1.328 ⋅ kV
VHΔ = 13.8⋅ kV
VHY = 7.967 ⋅ kV
j ⋅ 30⋅ deg
⋅ kV
j ⋅ 0⋅ deg
⋅ kV
= 10.392
Page 5 of 6
NR = 6
VHY
VXY
=6
( )
arg NR = 30⋅ deg
VHΔ
VXΔ
=6
September 17, 2008
ECE 404
Introduction to Power Systems
HW #5
3.14, 3.19, & 3.35
S1φ
IHY :=
VHY
IHY = 100 A
S1φ
IXY :=
VXY
IXY = 602 A
S1φ
IHΔ :=
VHΔ
IHΔ = 58 A
S1φ
IXΔ :=
VXΔ
IXΔ = 348 A
Connection H‐X Wye‐Delta Delta‐Wye Wye‐Wye Delta‐Delta VX (kV) 2.3 1.33 1.33 2.3 IX (A) 348 602 602 348 SX
(kVA)
800
800
800
800
VH
(kV)
7.967
13.8
7.967
13.8
IH
(A)
100
58
100
58
SH
(kVA)
800
800
800
800
VX (kV)
2.3
1.33
1.33
2.3
IX (A)
348
602
602
348
SX (kVA)
800
800
800
800
NR
NH/NX 6 at 30 deg 3.464 6 at 30 deg 10.392 6
6 6
6 Or more logically set up than in‐class Connection VH H‐X (kV) Wye‐Delta 7.967 Delta‐Wye 13.8 Wye‐Wye 7.967 Delta‐Delta 13.8 IH (A) 100 58 100 58 C:\JLAW\CLASSES\ECE 404 Fall
2008\Homework\HW05
\solutionR1.xmcd
SH (kVA)
800
800
800
800
Page 6 of 6
NR 6 at 30 deg 6 at 30 deg 6
6
NH/NX 3.464 10.392 6 6 September 17, 2008