EEL303: Power Engineering I - Tutorial 8
1. The line currents in a 3-φ system in phases a,b & c are 106 30o , 206 -45o & 306 60o
respectively referred to the same reference vector. Find the symetrical components of
(1)
(2)
(0)
the currents. [Ans: Ia =13.796 24.01o A, Ia =9.636 -3.29o A, Ia =13.556 180.2486o A ]
Solution:
Ia(2)
1
Ia(0) = (Ia + Ib + Ic ) = 12.6 + j5.61 = 13.796 24.01o A
3
1
Ia(1) = (Ia + aIb + a2 Ic ) = 9.6 − j0.55 = 9.636 − 3.29o A
3
1
= (Ia + a2 Ib + aIc ) = −13.55 − j0.0588 = 13.556 180.2486o A
3
2. The line to ground voltages on the high voltage side of a step up transformer are 100kV,
33kV & 38kV on phases a,b &c respectively. The voltage on phase ’a’ leads that of
phase ’b’ by 100o and lags phase ’c’ by 176.5o . Determine the symetrical components of
(1)
(2)
(0)
voltage.
[Ans: Va =21.36 -28.18o kV, Va =52.656 15.79o kV, Va =30.856 -7.95o kV ]
Solution: Taking phase ’a’ as reference we have Va =1006
Vc =386 176.5o kV
o
kV, Vb =336 -100o kV &
1
Va(0) = (Va + Vb + Vc ) = 21.36 − 28.18o kV
3
1
Va(1) = (Va + aVb + a2 Vc ) = 52.656 15.79o kV
3
1
Va(2) = (Va + a2 Vb + aVc ) = 30.856 − 7.95o kV
3
3. One conductor of a 3-φ line is open. The current flowing to the ∆-connected load throug
line ’a’ is 10 A. Assuming that line ’c’ is open, find the symmetrical components of line
currents with
(a) current in line ’a’ as reference
(0)
(1)
(b) current in line ’b’ as reference
[Ans(a) : Ia = 0A; Ia = 5.77356 − 30o A;
(0)
(1)
(2)
(2)
Ia = 5.77356 + 30o A; (b) : Ib = 0A; Ib = 5.77356 + 30o A; Ib = 5.77356 − 30o A]
Solution: (a) Taking line ’a’ as reference, we have Ia =106 0o A and since line ’c’ is
open Ic =0 ⇒ Ia + Ib = 0 ⇒ Ib = −Ia = |Ia |6 180o =106 180o A
1
Ia(0) = (Ia + Ib + Ic ) = 0A
3
Electrical Engineering Dept - IIT Delhi
EEL303: Power Engineering I - Tutorial 8
1
Ia(1) = (Ia + aIb + a2 Ic ) = 5.77356 − 30o A
3
1
Ia(2) = (Ia + a2 Ib + aIc ) = 5.77356 + 30o A
3
(b) Taking line ’b’ as reference, we have 6 Ib =0, and since line ’c’ is open Ic =0 ⇒
Ia + Ib = 0 ⇒ Ib =106 0o A and Ia =106 180o A
(0)
Ib
(1)
Ib
(2)
Ib
1
= (Ib + Ic + Ia ) = 0A
3
1
= (Ib + aIc + a2 Ia ) = 5.77356 + 30o A
3
1
= (Ib + a2 Ic + aIa ) = 5.77356 − 30o A
3
4. Three identical Y-connected resistors form a load bank with a 3-φ rating of 2300 V and
500 kVA. If the load bank has applied voltages |Va b| = 1840V, |Vb c| = 2760V &|Vc a| =
2300V , find the line voltages and currents in p.u into the load. Assume that the neutral
of the load is not connected to the neutral of the system and select a base of 2300V,
500kVA.
[Ans : Vab = 0.86 82.82o ; Vbc = 1.26 − 41.41o ; Vca = 1.06 180o ;
o
Ia = 0.78316 35.814 ; Ib = 1.02646 − 63.2535o ; Ic = 1.18886 − 22.674o ]
Figure 1: Phasor Diagram for line to line voltages
Solution:
The rating of the load is same as specified base, so the value of each resistance in
load is 1.0 p.u. & p.u line voltages with Vbase = 2300V are
1840
2760
2300
|Vab | =
= 0.8, |Vbc | =
= 1.2, |Vca | =
= 1.0
2300
2300
2300
o
Assuming an angle of 180 for Vca , the phasor diagram for line voltages is shown in
Figure 1.
Electrical Engineering Dept - IIT Delhi
EEL303: Power Engineering I - Tutorial 8
where,
α −→ angle made by voltage Vab with reference
β −→ angle made by voltage Vbc with reference
These angles can be found using law of cosines as
cosα =
|Vab |2 + Vca |2 − Vbc |2
= 0.125 ⇒ α = 82.82o
2|Vab ||Vca |
cosβ =
|Vbc |2 + Vca |2 − Vab |2
= 0.75 ⇒ β = 41.41o
2|Vbc ||Vca |
Therefore 6 Vab = 82.82o & 6 Vbc = −41.41o (since Vbc lags reference by 41.41o )
1
(0)
Vab = (Vab + Vbc + Vca ) = 0
3
1
(1)
Vab = (Vab + aVbc + a2 Vca ) = 0.98576 73.55o p.u
3
1
(2)
Vab = (Vab + a2 Vbc + aVca ) = 0.23476 220.24o p.u
3
√
√
(1)
(1)
(2)
(2)
We have Vab = 36 30o Van & Vab = 36 − 30o Van
The absence of neutral connection make zero sequence current absent. Therefore, the
phase voltages at load contain +ve and -ve sequence components only. The above
(2)
(2)
(1)
(1)
equations get modified to Vab = Van 6 30o & Vab = Van 6 − 30o if line voltages are
expressed in terms of line to line base voltages and if phase voltaes are expressed in
terms of line to neutral base voltage.
(1)
Thus, Van = 0.98576 (73.55o − 30o ) = 0.98576 43.55o &
(2)
Van = 0.23476 (220.24o + 30o ) = 0.23476 250.24o
Since each resistor has an impedance of 1.0 p.u
(1)
(2)
Ia = 0.98576 43.55o & Ia = 0.23476 250.24o
⇒ Ia = 0.78316 35.814o , Ib = 1.02646 − 63.2535o & Ic = 1.18886 − 22.674o
5. Using symetrical components, calculate the power absorbed in the load of above problem
using symetrical components.
[Ans: S3−φ =513.34 kW]
Solution:
S3−φ = Va(0) Ia(0)∗ + Va(1) Ia(1)∗ + Va(2) Ia(2)∗
With the given base of 2300V, 500kVA we have
S3−φ = 0 + 0.98576 43.55o × 0.98576 43.55o + 0.23476 250.24o × 0.23476 250.24o
= (0.9857)2 + (0.2347)2 = 1.0267p.u
= 1.0267 ∗ 500 = 513.34kW
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