138power Systems

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No part of this publication may be reproduced or distributed in any form or any means, electronic, mechanical, photocopying, or otherwise without the prior permission of the author. GATE SOLVED PAPER Electrical Engineering POWER SYSTEMS Copyright © By NODIA & COMPANY Information contained in this book has been obtained by authors, from sources believes to be reliable. However, neither Nodia nor its authors guarantee the accuracy or completeness of any information herein, and Nodia nor its authors shall be responsible for any error, omissions, or damages arising out of use of this information. This book is published with the understanding that Nodia and its authors are supplying information but are not attempting to render engineering or other professional services. NODIA AND COMPANY B-8, Dhanshree Tower Ist, Central Spine, Vidyadhar Nagar, Jaipur 302039 Ph : +91 - 141 - 2101150 www.nodia.co.in email : [email protected]

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Transcript of 138power Systems

No part of this publication may be reproduced or distributed in any form or any means, electronic, mechanical, photocopying, or otherwise without the prior permission of the author.GATE SOLVED PAPERElectrical EngineeringPOWER SYSTEMSCopyright By NODIA & COMPANYInformation contained in this book has been obtained by authors, from sources believes to be reliable.However, neither Nodia nor its authors guarantee the accuracy or completeness of any information herein, and Nodia nor its authors shall be responsible for any error, omissions, or damages arising out of use of this information. This book is published with the understanding that Nodia and its authors are supplying information but are not attempting to render engineeringor other professional services.NODI A AND COMPANYB-8, Dhanshree Tower Ist, Central Spine, Vidyadhar Nagar, J aipur 302039Ph :+91 - 141 - 2101150www.nodia.co.inemail : [email protected] Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount*GATE SOLVED PAPER - EEPOWER SYSTEMS www.nodia.co.inYEAR 2013TWO MARKSQ. 1Forapowersystemnetworkwithnnodes,Z33ofitsbusimpedancematrixis . j05 per unit. The voltage at node 3 is. 13 10c -per unit. If a capacitor having reactance of. j35 -per unit is now added to the network between node 3 and the reference node, the current drawn by the capacitor per unit is(A). 0325 100c - (B). 032580c(C). 0371 100c - (D). 043380cStatement for Linked Answer Questions: 2 and 3In the following network, the voltage magnitudes at all buses are equal to 1 pu, the voltage phase angles are very small, and the line resistances are negligible. All the line reactances are equal to1 j WQ. 2The voltage phase angles in rad at buses 2 and 3 are(A). 012q =- ,. 023q =- (B)02q = ,. 013q =-(C). 012q = ,. 013q = (D). 012q = ,. 023q =Q. 3If the base impedance and the line-to line base voltage are 100ohms and 100kV respectively, then the real power in MW delivered by the generator connected at the slack bus is(A)10 - (B) 0(C) 10(D) 20YEAR 2012ONE MARKQ. 4The bus admittance matrix of a three-bus three-line system isY j1310510181051013=---RTSSSSVXWWWWIf each transmission line between the two buses is represented by an equivalentpnetwork, the magnitude of the shunt susceptance of the line connecting bus 1 and 2 is(A) 4(B) 2(C) 1(D) 0GATE SOLVED PAPER - EEPOWER SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inQ. 5Atwo-phaseloaddrawsthefollowingphasecurrents:( ) ( ), sin i t I tm 1 1w f = -( ) ( ) cos i t I tm 2 2w f = - . These currents are balanced if 1fis equal to.(A) 2f - (B) 2f(C) ( / ) 22p f - (D) ( / ) 22p f +Q. 6Thefigureshowsatwo-generatorsystemapplyingaloadof40MW PD = , connected at bus 2.The fuel cost of generators G1 and G2 are :( ) 10000 / Rs MWh C PG 1 1=and( ) 12500 / Rs MWh C PG 2 2=and the loss in the line is. , P P 05( ) ( ) loss pu G pu 12=where the loss coefficient is specified in pu on a 100MVA base. The most economic power generation schedule in MW is(A), P P 20 22G G 1 2= = (B), P P 22 20G G 1 2= =(C), P P 20 20G G 1 2= = (D), P P 0 40G G 1 2= =Q. 7Thesequencecomponentsofthefaultcurrentareasfollows:1.5 , pu I jpositive=0.5 , pu I jnegative=- 1pu I jzero=- . The type of fault in the system is(A) LG(B) LL(C) LLG(D) LLLGYEAR 2012TWO MARKSQ. 8For the system below,and S SD D 1 2 are complex power demands at bus 1 and bus 2 respectively. If1pu V2= , the VAR rating of the capacitor ( ) QG2 connected at bus 2 is(A) 0.2pu(B) 0.268pu(C) 0.312pu(D) 0.4puQ. 9A cylinder rotor generator delivers 0.5pu power in the steady-state to an infinite bus through a transmission line of reactance 0.5pu. The generator no-load voltage is 1.5pu and the infinite bus voltage is 1pu. The inertia constant of the generator is 5MW- / s MVA and the generator reactance is 1pu. The critical clearing angle, in degrees, for a three-phase dead short circuit fault at the generator terminal is(A) 53.5(B) 60.2(C) 70.8(D) 79.6GATE SOLVED PAPER - EEPOWER SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inYEAR 2011ONE MARKQ. 10A nuclear power station of 500 MW capacity is located at 300 km away from a load center. Select the most suitable power evacuation transmission configuration among the following optionsQ. 11A negative sequence relay is commonly used to protect(A) an alternator(B) an transformer(C) a transmission line(D) a bus barQ. 12For enhancing the power transmission in along EHV transmission line, the most preferred method is to connect a(A) Series inductive compensator in the line(B) Shunt inductive compensator at the receiving end(C) Series capacitive compensator in the line(D) Shunt capacitive compensator at the sending endYEAR 2011TWO MARKSQ. 13A load center of 120 MW derives power from two power stations connected by 220 kV transmission lines of 25 km and 75 km as shown in the figure below. The three generators, G G1 2 and G3 are of 100 MW capacity each and have identical fuel cost characteristics. The minimum loss generation schedule for supplying the 120 MW load is(A) MW lossesMWMWPPP802020123= +==(B) MWMW lossesMWPPP603030123== += GATE SOLVED PAPER - EEPOWER SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.in(C) MWMWMW lossesPPP404040123=== +(D) MW lossesMWMWPPP304545123= +==Q. 14Thedirectaxisandquadratureaxisreactancesofasalientpolealternatorare 1.2p.uand1.0p.urespectively.Thearmatureresistanceisnegligible.Ifthis alternatorisdeliveringratedkVAatupfandatratedvoltagethenitspower angle is(A) 30c(B) 45c(C) 60c(D) 90cQ. 15A three bus network is shown in the figure below indicating the p.u. impedance of each element.The bus admittance matrix, Y -bus, of the network is(A) ..0.....j0302020120080008002--RTSSSSVXWWWW(B)....j1550575125012525---RTSSSSVXWWWW(C) .......j01020020120080008010 --RTSSSSVXWWWW(D)... j1050575125012510--RTSSSSVXWWWWStatement For Linked Answer Questions : 13 & 14Q. 16Two generator units G1 and G2 are connected by 15 kV line with a bus at the mid-point as shown below250MVA G1= , 15 kV, positive sequence reactance25% XG1=on its own base100MVA G2= , 15 kV, positive sequence reactance10% XG2=on its own base L1 and10km L2= , positive sequence reactance0.225 / km XLW =GATE SOLVED PAPER - EEPOWER SYSTEMS www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inQ. 17In the above system, the three-phase fault MVA at the bus 3 is(A) 82.55 MVA(B) 85.11 MVA(C) 170.91 MVA(D) 181.82 MVAYEAR 2010ONE MARKQ. 18Power is transferred from system A to system B by an HVDC link as shown in the figure. If the voltage VAB and VCD are as indicated in figure, and I 0 2 , then(A)0, 0, V V V V < >AB CD AB CD1 (B)0, 0, V V V VAB CD AB CD2 2 1(C)0, 0, V V V V >AB CD AB CD2 2 (D)0, 0 V V