Solution Assignment 2 - Chapter 4

Solution Assignment 2 - Chapter 4 - Assignment #2 Chapter 4...

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1 Assignment #2– Chapter 4 Prof. S. Santoso (instructor) Example 4.3 and 4.4 Please refer to the textbook for the solution Problem 4.1 From Eq (4.4) and the problem statement , the single-phase short-circuit capacity given as three- phase quantity and single-phase quantity Three-phase quantity are the following: * 1 * 1 3 _ 1 ) ( 3 ) ( 3 sc LL sc LN sc quan I kV I kV MVA φ = = Single-phase quantity 3 / ) ( ) ( * 1 * 1 1 _ 1 sc LL sc LN sc quan I kV I kV MVA = = It can be seen from the above two equations that 1_ 3 1 3 sc sc quan quan MVA MVA φφ = (Eq A) From Eq (4.9), the zero-sequence impedance given as three-phase quantity = * 3 * 3 _ 1 3 _ 0 ) ( 2 ) ( 3 sc sc quan base eq quan MVA MVA MVA Z is (Eq B) Substituting Eq A into Eq B , the zero-sequence impedance given as single-phase quantity = * 3 * 1 _ 1 1 _ 0 ) ( 2 ) ( 1 sc sc quan base eq quan MVA MVA MVA Z is
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2 Problem 4.2 At node ‘A’, MVA 3 Φ = 5000MVA and MVA 1 Φ = 3000MVA Taking base as 100MVA, p.u 06 . 0 5000 2 3000 3 100 . p.u 02 . 0 5000 100 0 0 1 j Z Z j Z eq eq eq = × = = = Now for 30MVA, 138/12.47kV transformer, p.u 1332 . 0 04 . 0 30 100 1 j j Z T = × = Computation of Thevenin Impedance: For positive sequence, Positive sequence diagram (Thevenin’s equivalent impedance) Z 1 eq(new) = j0.02 + j0.1333 = j0.1533 p.u. Negative sequence impedance is the same as above. For zero sequence, the diagram is as follows: Zero sequence diagram
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3 Z 0 eq(new) = j0.1333 p.u. For calculation of short circuit capacity, we need to compute the 3-phase symmetrical fault and single line to ground fault short circuit currents. A 4630 10 47 . 12 3 10 100 3 3 6 = × × × = × = base base base V MVA I Let the pre-fault voltage at fault location (transformer terminal) be 1 p.u. Fault current in case of three phase symmetrical fault at transformer terminals is given by, p.u. 52 . 6 1533 . 0 1 1 1 j j Z I eq SC = = = kA 202 . 30 4630 52 . 6 3 = × = kA sc I φ The three-phase short circuit capacity is given by: MVA 652.32 202 . 30 47 . 12 3 3 = × × = new MVA Fault current in case of single phase fault is given by, p.u. 82 . 6 4399 . 0 3 3 2 0 1 j j Z Z Z I eq eq eq SC = = + + = kA 577 . 31 4630 82 . 6 1 = × = kA SC I The single phase short circuit capacity is given by: MVA 682.012 577 . 31 47 . 12 3 1 = × × = new MVA
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4 Problem 4.3 Load is not connected to the service transformer. Hence no load current is present in the system. The p.u values are calculated as in example 4.3. Let us assume that the pre-fault voltage at fault location F is ° = 0 1 F V p.u. The positive sequence diagram is given by Figure 4.6 but with no load. (It is a open circuit at node D). From the above diagram, ° = = = 0 1 F C B V V V p.u. Since the service transformer is wye-grounded delta connection, there will be a -30° phase shift from the primary to he secondary. In other words, ° = 30 1 D V p.u. The pre-fault voltages are given as follows V B(abc) V F(abc) V C(abc) V D(abc) ° ° ° 120 0000 . 1 120 0000 . 1 0 0000 . 1 ° ° ° 120 0000 . 1 120 0000 . 1 0 0000 . 1
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Solution Assignment 2 - Chapter 4 - Assignment #2 Chapter 4...

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