Download EE457 M1 Fall 06-07 - faraday - Eastern Mediterranean University

Eastern Mediterranean University
Department of Electrical and Electronic Engineering
EE 457 POWER SYSTEM ANALYSIS I
MIDTERM EXAM I
: 20 Nov 2006
SOLVE ALL 3 QUESTIONS
: 90 min.
Q1=35 pts, Q2=35 pts, Q3= 30 pts
Date
Duration
1-) A three-phase load draws 200 kW at a power factor of 0.707 lagging from a 380-V source.
In parallel with this load is a three-phase capacitor bank which draws 80 kVA.
(a) Find the total line current drawn from the source. (20 pts)
(b) Find the resultant power factor at the terminals of the source. (10 pts)
(c) Find the capacitance per-phase-Y of the capacitor bank. (5 pts)
Hint: You may assume that the load and the capacitor bank are both Y-connected.
2-) Consider the single-phase transformer circuit shown below. Transformer ratings and reactances are:
Transformer A
: 250 kVA , 250 / 800 V, X = 0.10 p.u.
Transformer B
: 200 kVA , 1000 / 400 V, X = 0.15 p.u.
The transmission line can be represented as a series impedance Zline = 0.2 + j 0.4 . Choose a base
of 300 kVA, 250 V in the voltage source ( Vs ) circuit.
(a) Draw the impedance diagram for this circuit, with all impedances in per unit. (15 pts)
(b) The load voltage is given as VL =3500 V. Find the source voltage required in per unit
and in volts.(12 pts)
(c) Find the source current (Is) in per-unit and in amperes. (8 pts)
Is
transmission line
(Zline =0.2 + j0.4 Ω)
A
B
IL
+
ZL=0.8 + j0.4 Ω
Vs
3-) A 50-Hz three-phase transmission line is 250 km long. The sending end voltage is kept constant at 380
kV. The line is open-circuited at the receiving end, and the receiving end voltage was measured as VR
= 231.4 kV. Afterwards, the receiving end is short-circuited and the short-circuit current was measured
as IR = j 4473.15 A. Calculate the per-unit-length inductance L (H/km/ph.) and capacitance C
(F/km/ph.) of the line. (30 pts)
Useful Information :
1-) Change of per unit basis
Z
''
pu
 S ''
 Z .  b'
 Sb
'
pu
  Vb' 
 .  '' 
  Vb 
2
2-) Lossless transmission line equations
V ( x)  VR cos(  x)  jZ c I R sin(  x)
I ( x)  I R cos(  x)  j
VR
sin(  x)
Zc
  j   zy
Zc  z / y
SOLUTION
1-) (a) Assume that the load and the capacitor bank are both Y-connected.
IT
IL,a
IC,a
380 V
(a)
200 103
 429.8 A
3  380  0.707
 I L,a =429.8-45 A since power factor is lagging
Load current magnitude, I L 
For the capacitor,
S  3 Vll IC
where S  80 kVA, Vll  380 kV
80 103
 IC 
 121.55 A
3  380
 IC ,a  121.5590 A
Total line current,
IT  I L ,a  IC ,a  303.87  j303.87  j121.55  303.87  j182.32 A
 354.37  30.96 A
(b) Resultant power factor at the source terminals = cos(30.96º ) = 0.857
(c) The line-to-neutral voltage is
Vph  X C .I C
Vph  380 / 3  219.4 V
 X C  219.4 /121.55  1.805 Ω 
1
2 fC
 C  1.763 mF
2-) (a)
Vb, B  800 V
Base voltages:
, Vb,C  320 V
Base impedances:
Zb, B
8002
3202

 2.1333  , Zb ,C 
 0.3413 
300  103
300 103
Impedances in per-unit
0.8  j 0.4
Z load 
 2.344  j1.172 p.u.
0.3413
0.2  j 0.4
ZTL 
 0.09375  j 0.18750 p.u
2.1333
2
XT ,A
 300   250 
 0.1 

  0.12
 250   250 
X T ,B
 300   1000 
 0.15  

  0.35156
 200   800 
p.u
2
p.u
Impedance diagram,
j 0.12
0.09375+ j 0.18750
+
j 0.35156
I
Vs
VL
2.344+j 1.172
(b)
I
VL
1.093750

 0.4174  26.57 pu
Z L 2.344  j1.172
 Vs  I (0.09375  j 0.65906)  VL  1.25185  j0.2286  1.2726  10.35 pu
 Vs  1.2726  250  318.15 V
(c) The source current was already found in part (b):
I s  0.4174  26.57 pu
Base current in the source circuit,
S
300kVA
Ib  b 
 1200 A  I s  0.4147  1200  497.64 A
Vb
250V
3- ) (a) When the line is open-circuited
219.93
 0.9504
231.4
 l  cos1 (0.9504)  0.31617 rad.    1.2647 103 rad./km
Vs  VR cos  l
 cos l 
Since the line is lossless,
j   zy

zy   2  1.59945 106 (1)
When the line is short-circuited,
Vs  jZ c I R sin  l
Zc 
Using (1) and (2),
z
y

 Zc 
219.93 103
 158.129 
j ( j 4473.15)sin(0.31617rad.)
z
 Z c2  25004.78 2
y
(2)
1.59945 106
 6.3966 1011  y  j8.0 10 6 S/km/ph.
25004.78
y  jC  C  2.5465 108 F/km/ph.
y2  
z  25004.78 y  j 0.2 /km/ph.  j L
 L  6.3662 104 H/km/ph.