7/28/2019 18-05 Power System Analysis (EL)

1/2

ID.No./Seat No.

MEHRAN UNIVERSITY OF ENGINEERING AND TECHNOLOGY,JAMSHORO.

FIRST TERM FINAL YEAR (7TH TERM) B.E.(ELECTRICAL) REGULAR

EXAMINATION 2009 OF 06-BATCH.

Dated: 18-05-2009. Time Allowed: 03 Hours. Max.Marks-80.NOTE. ATTEMPT ANY FIVE QUESTIONS. ALL QUESTIONS CARRY EQUAL MARKS.

Q.No.

1 Calculate the symmetrical components of the following set of unbalanced currentsanalytically and graphically.

Ia = 100 300 Ib = 60 3000 Ic = 30 1900

2 The one line diagram of an unloaded power system is shown in figure 1.Q.2.

The reactances of the two sections of transmission line are shown in the diagram.The generators and transformers are rated as follows:Generator 1 : 20 MVA 13.8 kV X = 0.20 p.uGenerator 2 : 30 MVA 18 kV X = 0.20 p.uGenerator 3 : 30 MVA 20 kV X = 0.25 p.uTransformer 1 : 25 MVA 13.8 /220Y kV X = 0.12 p.uTransformer 2 : 30 MVA 18 /220Y kV X = 0.12 p.uTransformer 3 : 35 MVA 22Y/220Y kV X = 0.10 p.uChoose a base of 50 MVA, 13.8 kV for generator 1 circuit and draw impedancediagram of the power system. Mark the letters to indicate the points corresponding tothe one line diagram.

3 Discuss the symmetrical component theory. Derive relation between phase voltages and theirsymmetrical components and vice-versa.

4. (a) Define per unit values. What are advantages of using per unit values. Derive expression for baseimpedance from the chosen KVA and KV base values.

(b) Three generators A, B and C are connected to a common bus.Generator KVA rating Voltage KV Reactance X %

A 7500 11 10B 1250 11 10C 1750 11 10

Calculate per unit reactances of all three generators on a common 15 MVA base.

Contd on P/-2

POWER SYSTEM ANALYSIS

7/28/2019 18-05 Power System Analysis (EL)

2/2

( - 2 - )

5. (a) Derive an expression for the fault current for a single line to ground fault at the terminal of analternator.

(b) A solidly grounded generator, rated 10 MVA, 13.8 KV has X = X2 = 15% and X0=6%is operating at rated voltage and is disconnected from the rest of the system. Calculate

the fault current in Amperes for a single line to ground fault at generator terminals.What will be change in fault current for the same fault if neutral is grounded through areactor of 0.3

6. A simple power system is shown in figure 2.Q.6. A fault occurs at the mid point of thetransmission line. Find the per unit fault current for the following faults(a) Single Line to Ground (b) Line to Line (c) Double Line to Ground.Reactances in per unit on a common MVA base are tabulated below. Generator isgrounded through a reactance ofj0.06 p.u. and motor is solidly grounded.Item Positive Seq. Negative Seq. Zero seq.Generator 0.30 0.22 0.05

Motor 0.25 0.18 0.06Transformer T1 0.10 0.10 0.10Transformer T2 0.10 0.10 0.10Line 0.15 0.15 0.45

7. For a load flow solution, the per unit line admittances areBus code Admittance

1 2 2 j 7.01 3 1 l 4.52 3 0.6 j 2.62 4 1 j 4.13 4 2 j 7.0

The Schedule for per unit active and reactive power isBus code P Q V Remarks

1 - - 1.05 Slack 2 0.5 0.2 100 PQ3 0.4 0.3 100 PQ4 0.6 0.2 100 PQ

Find the bus admittance matrix and determine the voltage at all buses after first iteration ofGauss Seidal method

8. (a) Define Stability, steady state stability limit and transient-state stability limit.

(b) Derive an expression for the steady-state power limit of a transmission line of resistance R andreactance X ohms per phase. If voltages at sending and receiving ends are equal and thereactance of transmission line can be varied with resistance remaining constant then themaximum steady-state power can be transmitted over the line is greatest when X = 3 R.

-----------------THE END---------------