ECE 423, Spring 2005, First Exam: 4:00-4:50pm, February 2 (Wed.), 2005
Name:
Student Number:
Problem
Max Points
1.
0.40 pu
2.
0.40 pu
3.
0.20 pu
Total
1.00 pu
Points
Instructions:
•
Closed book, but you are allowed to have one-page single-sided handwritten notes.
•
Read all the problems first.
•
Attempt to solve first the problems you can, don’t spend too much time on a problem.
•
Read carefully the statement and/or circuits of each problem.
•
Write your solutions neatly and box your answers (and box steps for partial credits)
Problem 1 (0.40 pu) A 3-phase 480 V source supplies two balanced loads in parallel: one in Y-connection
and the other in ∆-connection. The Y-connected load draws 10 kW at pf of 0.866 lagging and the ∆connected load 10 kVA at pf of 0.5 leading.
1. (0.15 pu) Determine the real, reactive, and apparent power delivered by the source and (0.10 pu)
draw the power triangles for the source and two loads.
2. (0.15 pu) Calculate the line (i.e., the source) current and respective impedances of the Yconnected load and ∆-connected load.
Solution Key
1.
SY
QY
PY = 10 kW, (pf)Y = .866 lagging, thus
SY = PY / (pf)Y = 10 k / .866 = 11.55 kVA, and
QY = SY*sin(cos-1(pf)Y) = 11.55 k *sin(30°) = 5.77 kVar.
S∆ = 10 kVA, (pf)∆ = .5 leading, thus
P∆ = S∆ * (pf)∆ = 10 k * .5 = 5 kW, and
Q∆ = −S∆*sin(cos-1(pf)∆) = −10 k *sin(60°) = 8.66 kVar.
PY
P∆
PS
QS
SS
S∆
Q∆
PS = PY + P∆ = 15 kW
QS = QY + Q∆ = −2.89 kVar
SS = (PS ^2+ QS ^2)^1/2 = 15.275 kVA
2.
Iline = SS / (1.732*Vll) ⁄−tan-1(QS / PS) = 15.275k / (1.732*480) ⁄−tan-1(−2.89 / 15) = 18.37 ⁄10.9° A
ZY = (Vll2 / SY ) ⁄cos-1(pf)Y = 4802 / 11.55k ⁄cos-1.866 = 19.95 ⁄ 30° = 17.275 + j9.975 Ω
−cos-1(pf)∆ = 3*4802 / 10k ⁄−
−cos-1.5 = 69.12 ⁄ −60° = 34.56 − j59.9 Ω
Z∆ = (3*Vll2 / SY ) ⁄−
Problem 2 (0.40 pu) A three-zone 3-phase network is shown below. Using base values of 50 kVA and 480
V in Zone 1, draw the pu circuit (0.03 pu) and determine the pu values for the source voltage (0.03 pu) and
all the impedances (0.24 pu). Calculate the load current in pu and in amps (0.10 pu).
Zone 1
Zone 2
Zone 3
T1
T2
G1
3-phase
VS = 480 V
50 kVA
480/2400 V
Xeq= .1 pu
Zline= j1 Ω
Load
30 kVA
2300/460 V
Xeq= .1 pu
Zload= 10+j10 Ω
Solution Key
Xeq1
G1
.1 pu
Zline
Xeq2
Zload
.00868 pu .153 pu
2.17 + j 2.17 pu
o
1∠0 pu
Zone 1
Zone 2
Zone 3
Sbase = 50 kVA
Vbase_1 = 480 V
Vbase_2 = 480 V *(2400/480) = 2400 V
Vbase_3 = 2400 V *(460/2300) = 480 V
No change
Zbase_2 = Vbase_22 / Sbase = 115.2 Ω
Zbase_3 = Vbase_32 / Sbase = 4.608 Ω
Xeq no change in pu
Zline_pu = Zline / Zbase = .00868 pu
Xeq2_pu = Xeq2_old *(4602 / 30k) / Zbase_3
= .153 pu
Zload_pu = Zload / Zbase_3 = 2.17+j2.17 pu
IL_pu = 1 ⁄ 0° / ( j.1 + j.00868 + j.153 + 2.17 + j2.17) = 1 ⁄ 0° / ( 2.17 + j2.43) = .307 ⁄−48° pu
IL_base = Sbase / (1.732* Vbase_3) = 50 k / (1.732* Vbase_3) = 60.14 A
IL_amps = IL_pu*IL_base = 18.46 ⁄−48° A
Problem 3 (0.20 pu) A single-phase 50 kVA, 2300/230 V, 60 Hz distribution transformer has 2 ohm total
equivalent leakage reactance and a 2000 ohm magnetizing reactance referred to the high-voltage side.
Neglect the copper and core losses. Assume that the rated voltage is applied to the high-voltage side and
equal leakage reactances for the primary and referred secondary. Draw the equivalent circuit and determine
its parameters in pu for the transformer (0.10 pu). Calculate the open circuit voltage in pu (0.10 pu).
Solution Key
Sbase = 50 kVA
V1_base = 2300 V
Z1_base = V1_base 2 / Sbase = 105.8 Ω
Xeq1_pu = 1 ohm / Z1_base = .00945 pu (2 ohm equally divided between the primary and referred secondary
leakage reactance)
Xm_pu = 2000 ohm / Z1_base = 18.90 pu
.00945 pu
V1=1 pu
.00945 pu
18.90 pu
V2
V2 = V1 * 18.90 / (18.90+.00945) =.99950 pu or
V2 = V1 * j18.90 / (j18.90+j.00945) =.99950 ⁄ 0° pu , assuming V1 = 1 ⁄ 0° pu.