Solution Manual Power System Protection By Paithankar

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MS Electrical Energy Engineering NUST Islamabad ,Pakistan

Q.1 Assume a 250:5 C.T. ratio with a primary current of 1=150 < 0° . Neglect the primary leakage inductance and assuming secondary impedance of 0.03+j0.05 ohms and a total burden of 0.06 ohms and a magnetizing impedance referred to the secondary side as Z’ M=7+j10E, calculate the phase error of the CT.

Q.2 Now consider that the burden on CT has been increased to 0.12 ohms. Calculate the CT phase error?

Q 3) Compare the results of Q1 and Q2 and summarize the conclusion inferred from increasing the value of burden on CT phase angle error ?

chapter 1 Q:1.4

Q 4) The load current in a system is 100 A. The short-circuit current for the smallest fault is 1000 A. An over-current relay capable of carrying a current of 5 A continuously is to be used. (a) Suggest a suitable CT ratio. (b) What will be the relay current for the smallest fault? (c) Will the protective CT be able to faithfully reproduce the fault current?

Q:5 A generator is rated at 500 MVA, 22 Kv. Its Y Connected windings have reactance of 1.1 p.u. Find the ohmic value of the reactance ?

ch:3 Q:3.1

Q 6) For a 5 A CT, the total impedance of the secondary winding and leads is 0.01 ohm. Determine the VA capacity of the CT secondary at (a) the rated secondary current and (b) the maximum fault current of 20 times the rated CT secondary current.

ch:3 Q:3.2

Q 7) In a simple differential scheme, the two CTs have a nominal ratio of 100 : 1. The CT errors for a 'through fault' current of 1000 A are as follows: CT1 CT2 Ratio error +5% +6% Phase angle error +2° -4°

Ch:3 Review Question 8 Q 8) Prove that in case of differential protection the slope of the internal fault characteristics for a single-end-fed system is 200%

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