IGEE 402 – Power System Analysis MID-TERM EXAM – Sample

IGEE 402 – Power System Analysis MID-TERM EXAM – Sample

Joós, G.
IGEE 402 – Power System Analysis
MID-TERM EXAM – Sample
Instructions:
1.
- Duration: 75 minutes.
- Material allowed: a crib sheet (double sided 8.5 x 11), calculator.
- Attempt all questions. Make any reasonable assumption.
- Return the completed answer sheet attached.
QUESTION 1 (25 points)
A single phase 600 V, 60 Hz feeder supplies 2 loads in parallel: (a) a heating load with a 10 Ω
resistance; (b) a 70 kVA motor load, with a power factor of 0.70 (lagging). Compute the total
current, power, reactive power and power factor. Draw the V-I vector diagram. A capacitor is
connected in parallel with the loads to increase the power factor to 0.95. Compute the reactive
power supplied by the capacitor and the value of the capacitor bank in µF.
2.
QUESTION 2 (25 points)
A short circuit test on a single phase 2400/240 V, 50 kVA, 60 Hz transformer yields the
following results seen from the high side: 66 V, 20.8 A, 750 W. The open circuit test done from
the low voltage side gives: 240 V, 5.97 A, 213 W. Give the approximate equivalent circuit
referred to the high side, ignoring the magnetizing branch. For a 50 kVA inductive load (0.75
power factor) fed at 240 V, compute the current and voltage on the high side. Using as a base
the transformer rated values, find: (a) the transformer parameters in pu for the approximate
equivalent circuit; (b) the current and voltage on the high side; (c) the voltage regulation; (d)
the efficiency (include core losses).
3.
QUESTION 3 (25 points)
Three identical single-phase 69 kV/6.9 kV, 60 Hz, 33 MVA distribution transformers, with a
leakage reactance of 0.10 on the transformer base, are fed from the 69 kV bus. They supply
two three phase 12 kV loads connected in parallel: a 60 MW, 0.8 power factor (lagging) load
(assume a Y connection) and a capacitor bank of three elements (-j15 Ω) connected in ∆ (hint:
convert into Y). Compute: (a) total line current; (b) the total power and total reactive power
drawn from the feeder; (c) the resulting power factor. Draw the single line pu circuit, using a
100 MVA, 12 kV base.
4.
QUESTION 4 (25 points)
A 500 km, 500 kV, 60 Hz transmission line has a series impedance of z = 0.033 + j0.35 Ω/km
and a shunt reactance of y = j4.42 x 10-6 S/km. Assume the line is lossless. Calculate: (a) the
ABCD parameters; (b) the nominal π equivalent circuit parameters; (c) the surge impedance
loading. The line delivers 1200 MW at a unity power factor and at a 0.95 pu voltage. Using the
ABCD parameters, compute: (a) the sending end voltage and δ angle; (b) the no load receiving
end voltage for the rated sending end voltage. Compute the sending end voltage and δ angle
using the nominal π equivalent and ignoring the shunt elements.
QUESTION 1
Vl := 600
600
R := 10
Ir :=
S := 70000
pf := 0.70
φ⋅ 
180 
Pr := Vl⋅ Ir
10
4
= 45.573 Pm = 4.9 × 10
 π 
Ir = 60
Pr = 3.6 × 10
Pm := S⋅ pf
φ := acos( pf )
Im :=
Im = 116.667
Qm := S⋅ sin( φ)
Irc := Ir
Imc := Im⋅ cos( φ) + Im sin( φ) ⋅ i
Imc = 81.667 + 83.317i
Itc := Irc + Imc
Itc = 141.667 + 83.317i
Itc = 164.351
 180  ⋅ arg( Itc) = 30.461

 π 
a :=
4
S
Vl
4
Qm = 4.999 × 10
180
π
cos( arg( Itc) ) = 0.862
4
Pt := Pr + Pm
Pt = 8.5 × 10
Qt := Qm
Qt = 4.999 × 10
Stc := Pt + Qt⋅ i
Stc = 8.5 × 10 + 4.999i × 10
pft :=
4
Pt
Pt
Xc := Vl⋅
Vl
Qcap
Vl
Xc⋅ i
Itcc := Itc + Icc
C :=
1
(60⋅ 2⋅ π ⋅ Xc)
4
Stc = 9.861 × 10
a⋅ acos( pfc) = 18.195
Qtcor := Stcor ⋅ sin( acos( pfc ) )
pfc
Qtcor = 2.794 × 10
4
a⋅ arg( Stc) = 30.461
pft = 0.862
pfc := 0.95
Stc
Stcor :=
Icc :=
4
4
Qcap := Qt − Qtcor
Qcap = 2.205 × 10
4
Xc = 16.325
Icc = −36.753i
Itcc = 141.667 + 46.564i
−4
C = 1.625 × 10
a⋅ arg( Itcc) = 18.195
QUESTION 2 - Transformers
Calculations in A and V
Vpr := 2400
Vsr := 240
Sr := 50000
Vpsc := 66
Ipsc := 20.8
Psc := 750
Vpsc
Zpsc :=
Zpsc = 3.173
Ipsc
Rpsc :=
Psc
Xpsc :=
( Zpsc⋅ Zpsc − Rpsc⋅ Rpsc)
Sl := 50000
Vlp :=
Ilp :=
Rpsc = 1.734
Ipsc⋅ Ipsc
Xpsc = 2.658
pf := 0.75
 Vpr  ⋅ Vl

 Vsr 
 Vsr  ⋅ Ils

 Vpr 
Ils :=
a⋅ acos( pf ) = 41.41
Sl
φ := acos( pf )
Vl := 240
Ils = 208.333
Vl
Ilpc := Ilp⋅ ( cos( φ) − i⋅ sin( φ) )
Ilpc = 15.625 − 13.78i
Ilpc = 20.833
3
Vp := Vlp + Ilpc⋅ ( Rpsc + i⋅ Xpsc)
Vp = 2.464 × 10 + 17.638i
Vp = 2.464 × 10
Vp
3
Vpr
= 1.027
Calculation on a pu basis
Vpbase := Vpr
Ipbase :=
r :=
3
Sbase := Sr
Sbase
Zpbase :=
Vpbase
Rpsc
x :=
Zpbase
Vpbase
Ipbase
Xpsc
Zpbase
4
Vpbase = 2.4 × 10
Sbase = 5 × 10
Ipbase = 20.833
Zpbase = 115.2
r = 0.015
x = 0.023
ip := 1 ⋅ ( cos( φ) − i⋅ sin( φ) )
ip = 0.75 − 0.661i
vp := 1 + ip⋅ ( r + i⋅ x )
vp = 1.027 + 7.349i × 10
Vreg :=
efficiency :=
Vreg = 0.027
1

pin := Re vp⋅ ip + pcore
(
pout
pin
−3
vp = 1.027
( vp − 1 )
pout := 1 ⋅ pf
ip = 1
)
pin = 0.769
efficiency = 0.975
pcore :=
213
50000
QUESTION 3 - Three-phase circuits
Vline := 12000
Plphase := 60⋅
10
6
pf := 0.8
3
φ := −acos( pf )
Illine :=
Vlphase :=
a⋅ φ = −36.87
Slphase :=
 Slphase  ⋅ ( cos( φ) + i⋅ sin( φ) )

 Vlphase 
Vline
3
Vlphase = 6.928 × 10
3
Plphase
7
Slphase = 2.5 × 10
pf
3
Illine = 2.887 × 10 − 2.165i × 10
3
3
Illine = 3.608 × 10
Icphase :=
Vlphase
3
Icphase = 1.386i × 10
15 
 −i⋅

 3
Icline := Icphase
3
Iline := Illine + Icline
Iline = 2.887 × 10 − 779.423i
3
Iline = 2.99 × 10
a⋅ arg( Iline) = −15.11
cos( arg( Iline) ) = 0.965
Pt := 3 ⋅ Slphase ⋅ cos( φ)
Pt = 6 × 10
Qt := 3 ⋅ Slphase ⋅ sin( −φ) − 3 ⋅ Icline ⋅ Vlphase
Qt = 1.62 × 10
Sbase := 100 ⋅
Load
Capacitor
10
6
Vbase :=
3
12000
p := 20⋅
10
p = 0.6
Sbase
xc :=
15
3 ⋅ Zbase
Ibase :=
3
6
xc = 3.472
7
pf = 0.8
7
a⋅ atan
Qt 
 Pt 
Sbase
Vbase
= 15.11
Zbase :=
Vbase
Ibase
Zbase = 1.44
QUESTION 4 - Transmission lines
−6
rl := 0.033
xl := 0.35
X := ixl⋅ l
X = 175i
Y := iyl⋅ l
Y = 2.21i × 10
 xl 

 yl 
Zc :=
β :=
yl := 4.42⋅ 10
l := 500
Vl := 500
Vl
= 288.675
3
−3
Zc = 281.399
SIL :=
xl⋅ yl
( Vl⋅ Vl)
SIL = 888.418
Zc
β = 1.244 × 10
−3
Lossless ABCD
A := cos( β ⋅ l)
A = 0.813
B := i⋅ Zc⋅ sin( β ⋅ l)
B = 163.936i
C := i⋅
sin( β ⋅ l)
C = 2.07i × 10
Zc
A := 1 + Y⋅
Medium line
−3
X
A = 0.807
2
B := X
B = 175i
C := Y⋅  1 + Y⋅

3
Vrated := 500 ⋅ 10
X
4
6
Vrl := 0.95⋅ 500 ⋅
Vsl := A⋅ Vrl + B⋅ Ir
10
3
3
5
Ir = 1.459 × 10
5
5
Vsl = 2.212 × 10 + 2.552i × 10
5
a arg( Vsl) = 49.086
Vrated
3
Vrnl := A⋅ Vsr
Vregl :=
Vrnl = 2.329 × 10
( Vsl − Vrl )
Vrl
5
Vrl = 2.742 × 10
3
Vrl = 2.742 × 10
Vsl = 3.378 × 10
Vsr :=
−3
Srated := 1200⋅ 10
pf := 1
Srated   1 
Ir := 

 3   Vrl 
C = 1.996i × 10
Vregl = 0.232
5