Analysis of Power Systems

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Show how the connections must be modified to allow parallel operation with a Ydll transformer

Solution

           

Draw a phasor diagram for the modified connections

Solution

           

3 (a) Two single-phase transformers have the following impedances referred to the secondary side

Determine the load taken by each machine as a percentage of its rating when they share a common load of 500 kvA at 0.73 p.f lag

                        Solution

            Given the data of the transformers

            Transformer A

            Rating = 300 KVA

            Impedance = 0.09 + j 0.18 ohms

Let us consider,

            Z1 = 0.09 + j 0.18 ohms

            Z2 = 0.09 + j 0.23 ohms

KVA showed by transformers 1 is given as

SA = ST * (Z2/Z1+Z2)

= 500* [(0.06 + j 0.23)/(0.09 + j 0.23) + (0.09 + j 0.23)

=275.8 KVA

SB = ST * (Z1/Z1 +Z2)

=500 * [(0.09 + j 0.18)/ (0.09 + j 0.18) + (0.09 + j 0.23)

=227.72 KVA

b) In the parallel load of 500 KVA, the transformer B getting over loaded.

Percentage of the over load on B = [(227.72 – 200)/ 200] * 100

=12.36 %

4 (a) where the power limit and the MVA limit coincide

i. power factor

power factor = cos 45.81 = 0.697 leading

ii. the load angle

tan α = 0.72 / 0.7

 angle = 45.81 degrees

iii. The MVA output

Vt Ia Sin @ = 0.72 p

Vt Ia Cos @ = 0.7

                                                            = 1.0042

            b. Where the power limit and the stability margin coincide

a) The p.u

MVA output =  (0.72 + 0.42)^2 + (0.7)^2

= 1.34

b) The p.u

0.7^2 + 0.72^2

5 Solution

Maximum margin = 1 p.u

Excitation limit = 2.8 p.u

Ex/Xs = 1

V = 12/28

= 0.714

Generator current input = 10/28

                                    = 0.357

September 25, 2023
Category:

Life Science

Subcategory:

Hero Physics

Subject area:

Power Energy Electricity

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Number of words

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