Electrical Power Engineering (2)

Faculty ofEngineering Tanta University

Dr. Ahmed Mohamed Azmy

Department of Electrical Power and Machine EngineeringTanta University - Egypt

Code: EP2207

Lecture: 4 Tutorial: 4 Total: 8

valuebase valueactualvalue unitPer

base

basebase V

SI

base

2base

base

basebase S

VIVZ

Impact of TransformersI1

V1

-

+V2

-

+ I2

base11

2base2 V

NNV

basebase2base1 SSS

base12

1base2 I

NNI

pu2pu1 VV

ExampleA transformer is rated at 13.8kV/138kV/4.157kV with rated power of 50MVA/40MVA/10MVA. Terminations are as followings:13.8 kV winding: 13.8 kV Source138 kV winding: 35 MVA load, pf = 0.866 lag.4.157 kV winding: 5 MVA load, pf = 0.866 leading

Using Sbase = 10 MVA, and voltage ratings as bases, (a) Solve for the primary current, power, and power factor (b) Draw the pu equivalent circuit

Example

1

3

2

---

+

++

V1

V2

V3

+

-

I1

I2

I3

Solution

...... up 30 53s up 531035s 22

...... up 30 50s up 50105s 33

......... up 00 01VVV up 01813813V 3211

...*

up 30 53VSI

2

22

Solution

.. 30 5.0*

3

33 up

VSI

..... up4233.775 30 5030 53III 321

....* up423 7753IVS 111

lagging 0.9177pf MVA7537107753S1 ,..

AI 27360138.010775.31

Solution

1

3

2

--

+

+V1

V2

V3

+

-

I1

I2 =

I3 =

3053.

3050.

001.

Steps of converting actual single-phase system to the

per unit quantities:

• Choose MVA base MVAbase for the whole system

• Choose a kilo voltage base kVbase for one section

• Calculate the voltage base in other sections in the

network using the transformation ratio of transformers

• Calculate the pu value using the general equation

valuebase valueactual valueunitPer

• Calculate the impedance base and current base using MVAbase and kVbase in each section

base

2base

base

basebase S

VIVZ

base

basebase V

SI

To convert the actual three-phase system to the per unit quantities, perform the following steps:

• Choose an apparent power base "VA3-base" for the

three phase for all parts in the system

• Choose a line-to-line voltage base VL-base for one

section in the circuit

• Calculate the impedance base and current base in

terms of VAbase and Vbase in each section of the power

system using the following equations:

baseL

basebase V

SI

3

3

• Calculate the voltage base in other sections in thenetwork using transformation ratio of transformersconsidering the method of winding connection

For star connection:

base3

2baseL

base SVZ

For delta connection:

base

baseLbase S

VZ

3

2

3

• Calculate the pu value using the general equation

valuebase valueactualvalue unitPer

ZZY 31

YZZ 3

Changing the base

If the values of the parameters are given for certainbases and it is required to calculate the values forother bases, a certain relation can be usedThis relation modifies the old value according to theration between new and old bases according to thefollowing relation:

Example:For the network shown, the data are:

Draw the impedance diagram for the networkassuming bases of 20 MVA and 11 kV in the G1 area.

G1 G2

G3

J 10 WTr1 Tr2

Tr3

Tr1: 20 MVA, 11/110 kV, x=J0.12 p.u.G1: 20 MVA, 11 kV, x=J0.4 p.u.Tr2: 10 MVA, 6/110 kV, x=J0.15 p.u.G2: 10 MVA, 6 kV, x=J0.35 p.u.Tr3: 5 MVA, 3/110 kV, x=J0.1 p.u.G3: 5 MVA, 3 kV, x=J0.3 p.u.

SolutionMVAb = 20 MVA

kVb-1 = 11 kV

kVb-II = 110 kV

kVb-III = 6 kV

kVb-IV = 3 kV

XL = 10 * 20 / (110)2 = j 0.017

Xg1 = 0.4 * 20 / 20 (11/11)2 = j 0.4

Xg2 = 0.35 * 20 / 10 (6/6)2 = j 0.7

G1 G2

G3

J 10 WTr1 Tr2

Tr3

III

IV

III

Solution

B

J 0.017

J 0.3

C

J 0.4

A

J 0.12

J 0.017 J 0.017

J 0.4 J 1.2 J 0.7

Xtr2 = 0.15 * 20 / 10 (110/110)2 = j 0.3

Xtr3 = 0.1 * 20 / 5 (110/110)2 = j 0.4

2.1j33

5203.0X

2

3g

12.0j110110

202012.0X

2

1tr

Generator G1 200 MVA, 20 kV, Xd = 15%Generator G2 300 MVA, 18 kV, Xd = 20%Generator G3 300 MVA, 20 kV, Xd = 20%Transformer T1 300 MVA, 220Y/22 kV, Xd = 10%

Transformer T2Three single-phase units each rated 100 MVA, 130 Y / 25 D kV, X = 10%

Transformer T3 300 MVA, 220/22 kV, X = 10%

Example: The power system shown in the figure has the following ratings:

G1 G2

G3

J 75 WT1

T2

T3

Y Y

J 75 W

Y YJ 50 W

Y YY

The transmission linereactances are asindicated in thefigure. Draw thereactance diagram inper unit choosing thegenerator 3 circuit asthe base.

MVAbase = 300 MVAH.V.S. of transformer T2 is Y connectedIts rated line to line voltage is √3 x 130 = 225 kVL.V.S. is connected in Its line voltage is 25 kVVb, III = 20 kV, Vb, IV = 20 x 220/22= 200 kV

G1 G2

G3

J 75 WT1

T2

T3

Y Y

J 75 W

Y Y

J 50 W

Y YY

I II

III

IV

22 220

22 220 225 25

Vb, I = 200 x 22/220= 20 kV

Vb, II = 200 x 25/225= 22.22 kV

Vb, I = 200 x 22/220= 20 kV

Vb, II = 200 x 25/225= 22.22 kV

Generator G1

Generator G2

Generator G3

Transformer T 1

Transformer T 2

Transformer T 3