tutorial8.pdf

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EEL303: Power Engineering I - Tutorial 8

1. The line currents in a 3- system in phases a,b & c are 1030o, 20-45o & 3060o

respectively referred to the same reference vector. Find the symetrical components ofthe currents. [Ans: I

(0)a =13.7924.01o A, I

(1)a =9.63 -3.29o A, I

(2)a =13.55180.2486o A ]

Solution:

I(0)a =13

(Ia + Ib + Ic) = 12.6 + j5.61 = 13.7924.01oA

I(1)a =1

3(Ia + aIb + a

2Ic) = 9.6j0.55 = 9.63 3.29oA

I(2)a =1

3(Ia + a

2Ib + aIc) = 13.55j0.0588 = 13.55180.2486oA

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 ofvoltage. [Ans: V

(0)a =21.3 -28.18o kV, V

(1)a =52.6515.79o kV, V

(2)a =30.85-7.95o kV ]

Solution: Taking phase a as reference we have Va=100o kV, Vb=33-100

o kV &Vc=38176.5

o kV

V(0)a =1

3(Va + Vb + Vc) = 21.3 28.18okV

V(1)a =1

3(Va + aVb + a

2Vc) = 52.6515.79okV

V(2)a =13

(Va + a2Vb + aVc) = 30.85 7.95okV

3. One conductor of a 3- line is open. The current flowing to the -connected load througline a is 10 A. Assuming that line c is open, find the symmetrical components of linecurrents with

(a) current in line a as reference

(b) current in line b as reference [Ans(a) : I(0)a = 0A; I

(1)a = 5.7735 30oA;

I(2)a = 5.7735 + 30oA; (b) : I

(0)b = 0A; I

(1)b = 5.7735 + 30

oA; I(2)b = 5.7735

30oA]

Solution: (a) Taking line a as reference, we have Ia=100oA and since line c is

open Ic=0 Ia + Ib = 0 Ib = Ia = |Ia|180o=10180oA

I(0)a =1

3(Ia + Ib + Ic) = 0A

Electrical Engineering Dept - IIT Delhi


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I(1)a =1

3(Ia + aIb + a

2Ic) = 5.7735 30oA

I(2)a =1

3(Ia + a

2Ib + aIc) = 5.7735 + 30oA

(b) Taking line b as reference, we have Ib=0, and since line c is open Ic=0 Ia + Ib = 0 Ib=100oA and Ia=10180oA

I(0)b =

1

3(Ib + Ic + Ia) = 0A

I(1)b =

1

3(Ib + aIc + a

2Ia) = 5.7735 + 30oA

I(2)b =

1

3(Ib + a

2Ic + aIa) = 5.7735 30oA

4. Three identical Y-connected resistors form a load bank with a 3- rating of 2300 V and500 kVA. If the load bank has applied voltages |Vab| = 1840V, |Vbc| = 2760V&|Vca| =2300V, find the line voltages and currents in p.u into the load. Assume that the neutralof the load is not connected to the neutral of the system and select a base of 2300V,500kVA. [Ans : Vab = 0.882.82

o;Vbc = 1.2 41.41o;Vca = 1.0180o;Ia = 0.783135.814

o; Ib = 1.0264 63.2535o; Ic = 1.1888 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 inload is 1.0 p.u. & p.u line voltages with Vbase = 2300V are

|Vab| = 18402300

= 0.8, |Vbc| = 27602300

= 1.2, |Vca| = 23002300

= 1.0

Assuming an angle of 180o for Vca, the phasor diagram for line voltages is shown inFigure 1.



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where, angle made by voltage Vab with reference angle made by voltage Vbc with referenceThese angles can be found using law of cosines as

cos =|Vab|2 + Vca|2 Vbc|2

2|Vab||Vca| = 0.125 = 82.82o

cos=|Vbc|2 + Vca|2 Vab|2

2|Vbc||Vca| = 0.75 = 41.41o

Therefore Vab = 82.82o & Vbc = 41.41o (since Vbc lags reference by 41.41o)

V(0)ab =

1

3(Vab + Vbc + Vca) = 0

V(1)ab =1

3(Vab + aVbc + a2Vca) = 0.985773.55op.u

V(2)ab =

1

3(Vab + a

2Vbc + aVca) = 0.2347220.24op.u

We have V(1)ab =

330oV

(1)an & V

(2)ab =

3 30oV(2)an

The absence of neutral connection make zero sequence current absent. Therefore, thephase voltages at load contain +ve and -ve sequence components only. The aboveequations get modified to V

(1)ab = V

(1)an 30o & V

(2)ab = V

(2)an 30o if line voltages are

expressed in terms of line to line base voltages and if phase voltaes are expressed interms of line to neutral base voltage.

Thus,V

(1)

an = 0.9857

(73.55

o

30o

) = 0.9857

43.55

o

&V(2)an = 0.2347(220.24o + 30o) = 0.2347250.24o

Since each resistor has an impedance of 1.0 p.uI(1)a = 0.985743.55o & I

(2)a = 0.2347250.24o

Ia = 0.783135.814o, Ib = 1.0264 63.2535o & Ic = 1.1888 22.674o

5. Using symetrical components, calculate the power absorbed in the load of above problemusing symetrical components. [Ans: S3=513.34 kW]

Solution:

S3 = V(0)a I(0)a + V(1)a I(1)a + V(2)a I(2)a

With the given base of 2300V, 500kVA we have

S3 = 0 + 0.985743.55o 0.985743.55o + 0.2347250.24o 0.2347250.24o

= (0.9857)2 + (0.2347)2 = 1.0267p.u

= 1.0267 500 = 513.34kW


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