School Work

Web for Power Systems II B.tech 6th Sem

Description
Nature of Faults in Electrical System
Categories
Published
of 16
18
Categories
Published
All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you.
Share
Transcript
   Web Answers for Power Systems - II Chapter 2: Nature of Faults in Electrical System Question 5  Answer: By percentage reactance we mean the percentage of the total phase voltage dropped in the circuit when full load current is flowing i.e. % 100  IX  X V  ….(1)  where I = full load current, V = phase voltage and X = reactance in ohms per phase. It can also be expressed in terms of KVA or KV as under: From equation (1) ……….(2) where  X   is the reactance in ohms. If  X   is the only reactance element in the circuit then short circuit current is given by SC  V  I  X  . By putting the value of X from equation (1), 100% SC   I I  X   i.e. short circuit current is obtained by multiplying the full load current by 100/%  X  . In view of what have been stated above it is worthwhile to mention here the advantage of using  percentage reactance instead of ohmic reactance in short circuit calculations. Percentage reactance values remain unchanged as they are referred through transformers, unlike ohmic reactances which  become multiplied or divided by the square of transformation ratio. Further, as the various equipments used in the power system have different KVA ratings, it is necessary to find the percentage reactance of all the elements on a common KVA rating which is known as base KVA and the conversion is effected by using the relation as %age reactance of base KVA = BaseKVARatedKVA   %age reactance at rated KVA. These all led to conclude that it is preferable to express the reactances of various elements in  percentage values for short circuit calculations.    Question 6  Answer: Whatever may be the value of base KVA, short circuit current is the same i.e., the value of short circuit current does not change if different base KVAs are taken. In order to explain the answer, a single line diagram of the system is shown below in which a 3 phase transmission line operating at 66 KV and connected through a 1000 KVA transformer with 5% reactance to a generating station bus bar is considered. The generator is of 2500 KVA with 10% reactance. i) If 2500 KVA is chosen as the common base KVA, the reactance of the various elements in the system on this base value will be: Reactance of transformer at 2500 KVA base 25005 12.5%1000  Reactance of generator at 2500 KVA base =  250010 10%2500  Total percentage reactance on the common base KVA % 12.5 10 22.5%  X   The full load current corresponding to 2500 KVA base at 66 KV is given by: 2500 100021.87   3 66 1000  I A     Short circuit current 100 10021.87 97.2% 22.5 SC   I I A X   ii) Now if 5000 KVA is chosen as the common base KVA, then: Reactance of transformer at 5000 KVA base 500010 25%1000  Reactance of generator at 5000KVA base 500010 20%1000  Total percentage reactance on the common base KVA %  X   = 25 + 20 = 45% ~ 2500 KVA 10% 1000 KVA 5% 11/66 KV 66 KV Line Single line diagram of the system    Full load current corresponding to 5000 KVA base at 66 KV is 5000 100043.743 66 1000  I A     Short circuit current, 100 10043.74 97.2% 45 SC   I I A X   If the two cases as shown above are examined, it may be observed that for two different base KVA, short circuit current is the same. Example 8 Solution: Let base KVA be 35,000 KVA % Reactance of alternator A at base KVA 35,000% 30 70%15,000  A  X   % of reactance of alternator B at base KVA 35,000% 50 87.5%20,000  B  X   Line current corresponding to 35,000 KVA at 12 KV  A , I   1684101231000035 33  Fig. (ii) shows the reactance diagram of the network at the selected base KVA. Total % reactance from the generator neutral upto fault point, %  A B  X X X   70 87.538.89%70 87.5  A B A B  X X  X X      Short circuit current 100 1001684 4330% 38.89 SC   I I A X  . Example 9   Solution: Assuming base MVA as 1200, the percentage reactance of one generating station is 100% and that of the other is % 1501008001200  The % reactance of the cable is 0.5 1200100 496%11 11      When a 3-  fault takes place at 1200 MVA capacity plant the equivalent circuit will be as follows. When the fault is F, fault impedance between F and the neutral bus will be 86.59% The short circuit MVA of this bus will be as follows 1200100 138686.59  MVA  For fault at the other station, the equivalent circuit will be as follows: The equivalent fault impedance between F and neutral bus will be 119.84%   The short circuit MVA will be 1200100 1001119.84  MVA . Example 10   Solution: Let 10,000 KVA be the base KVA % Reactance of alternator A at base KVA 3 10,000% 10 10%10 10  A  X   % Reactance of transformer or base KVA 3 10,000% 5 10%5 10 T   X   Since the line impedance is given in ohms, conversion into % impedance is made by using the following relation. % impedance = ~ ~ 10MVA 10% Load 5MVA 5% F 1   1Ω 4Ω F 2  Fig: 1 ZPB 150% 100% 496% F 150% 496% 100% F
We Need Your Support
Thank you for visiting our website and your interest in our free products and services. We are nonprofit website to share and download documents. To the running of this website, we need your help to support us.

Thanks to everyone for your continued support.

No, Thanks
SAVE OUR EARTH

We need your sign to support Project to invent "SMART AND CONTROLLABLE REFLECTIVE BALLOONS" to cover the Sun and Save Our Earth.

More details...

Sign Now!

We are very appreciated for your Prompt Action!

x