Final Exam, Second Semester: 2015/2016 Electrical Engineering Department

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1 Philadelphia University Faculty of Engineering Student Name Student No: Serial No Final Exam, Second Semester: 2015/2016 Electrical Engineering Department Course Title: Power II Date: 21 st June 2016 Course No: / Time Allowed: 2 hrs Lecturer: Dr. J B E AL-ATRASH No. of Pages: 4 Instructions: ALLOWED: Non-programmable calculator, pens and drawing tools (no red colour). NOT ALLOWED: Papers, literatures and any handouts. Otherwise, it will lead to the non-approval of your examination. Shut down Telephones, and other communication devices. Please note: Write your name and your matriculation number on every page of the solution sheets. All solutions together with solution methods (explanatory statement) must be inserted in the labeled position on the solution sheets. Support your answer with diagrams, equations and examples when possible You can submit your exam after the first hour. Question 1: (10 Marks) Objectives: (To examine the general knowledge of the student in power systems) Choose the correct answer?(5x 2 Marks) 1) The protection system is designed to operate a) Within 3 seconds b) more than 3 seconds c) between 2-3 seconds d) Within 2 seconds e) more than 2 seconds Answer: d 2) C.T (Current Transformer) saturation a) Is necessary for the operation of protective relay b) can cause mal operation of protective relays c) can damage the current transformer itself d) Increase the reliability of the protection scheme e) can damage the circuit breaker Answer: b 3) Load flow results can be used in a) In planning of power system b) In contingency studies c) In loss calculations and power factor correction d) In stability studies e) In (a), (b) and (c) Answer: e 4) The generator is defined as stable if retained its stable condition a) After a 3 second of major fault b) within 3 second of major fault c) within 1-3 second after a major fault d) within 1 second after a major fault e) after adjusting the field excitation of the rotor windings Answer: d 1

2 5) Guass-Siedel method a)a linear method which converges in 3-4 iterations b) Non linear method which converges in 3-4 iterations c) A linear method which converges in many several iterations d) Non linear method which converges in many several iterations Answer: c Question 2: (10 Marks) Objectives: (To examine the ability of the student to distinguish between different methods of loads flow) A) For the radial feeder of Fig 1operating on 33 kv; the maximum three phase fault currents at buses 1, 2, 3 are 1 800, and A respectively. Use the attached time current curve and determine the CTS & TDS of each relay. Assume all p.f of loads =1.0 G 300/5 R 3 250/5 R 2 150/5 R 1 Solution The load currents at each of the three buses are: S 1 =5 MVA S 3 =4 MVA S 2 =6 MVA Fig 1 I 1 =I L1 =S1/ ( 3 (33 KV)) = 6 MVA/ ( 3 (33 KV)) = A I 2 = (I 1 + I L2 )= (S1+S2)/( 3x33 KV) = (6+4)/ ( 3x33 KV)=174.9 A I 3 =I 2 + I L3 = (S1+S2+S3)/ ( 3 x 33 KV) = (6+4+5)/ ( 3 x33 KV) = A Using the C.T ratios, the corresponding relay currents due to load currents are I L,R1 = /(150/5)= 3.49 A I L,R2 = I L, R 3 =174.9 /(250/5)= 3.49 A I L,R4 =262.4/(300/5)= 4.37 A The available CTS settings for the O.C relay are 4,5,6,7,8,10 and 12 A (80,100,120,140,160,180,200 and 240 %). Therefore, the CTS for the relays R1, R2 and R3 are selected as: CTS1=4 A (80%) CTS2= 4 A (80%) CTS3=5 A (100%) 2

3 Next; TDS settings for each relay using the maximum fault currents. The relay R1 is at the end of the radial system, and therefore no coordination with any other relay is necessary. The fault current as seen by R1 I f,r1 =1800/(150/5)=60 A As a multiple of the selected CTS (or the pickup value) I f,r1 /CTS1 =60/4= 15 Since the fastest possible operation is desirable, the smallest (TDS) is selected. Then TDS1=1/2 (Fig 2) The operating time for relay R1 (from associated curve) T1=0.07 s. In order to set relay R2 as a back relay, to respond to the balanced three-phase fault at bus 1, it is assumed that the error margin=0.3. Then T2= T =0.07+4=0.47 s. The fault current for a fault at bus 1 as a multiple of the CTS at bus 2 can be found I f,r1 /CTS2 =60/4=15 Therefore from Fig 2 for the relay R2 for 0.45 s operating time and 15 ratios the TDS can be determined as: TDS2 =2.8 The next step is to determine the setting for relay R3. A three-phase fault at bus 2 produces a fault current of 2500 A. Therefore I f,r2 = I f,r3 =( 2500)/(250/5)=50 A And I f,r2 /CTS2 = I f,r3 /CTS3 =50 /4=12.5 A new curve for the time dial setting of 3 can be drawn in Fig 2 between the two curves shown for the time dial settings of 2 and 5. Then the operating time of relay R2 and R3 can be found from this new curve for the associated multiple of the CTS as 0.6 s. Therefore, permitting the same CDT for relay R4 to respond to a fault at bus 2 as for relay R2 responding to a fault at bus 1. T4= =1.0 s. The corresponding current in the relay can be given as a multiple of the relay pickup current. Thus I f,r4 = 3600/(300/5)=60 A I f,r4 /CTS4 =60/5=12 3

4 Therefore, for relay R4 for 1.0 s operating time and a 12 ratio, the TDS is TDS4 =3.9 Question 3: Objectives: (To examine the student ability to understand stability problem) (10 Mark) For synchronous generator J (d 2 θ m /dt 2 ) =T a =T m T e Where J= Moment of inertia θ m = Angular displacement of rotor t = Time T m = Mechanical or shaft torque supplied by the prime mover T e = Electrical or electromagnetic torque T a = Accelerating torque Prove that the swing equation is given by (H/180f)( d 2 δ /dt 2 )=P a =P m -P e? Solution If J is the combined moment of inertia of the prime mover and generator, neglecting frictional and damping torques, from law's of rotation then J(d 2 θ m /d t 2 )= T a =T m - T e (1) Where θ m = is the angular displacement of the rotor with respect to the stationary reference axis on the stator. But θ m = ω sm t +δ m (2) Where ω sm = angular velocity and δ m = is the rotor position before disturbance at time t=0, measured from the synchronously rotating reference frame. Derivative of (2) gives the rotor angular velocity ω m = dθ m /dt = ω sm + dδ m /dt (3) And the rotor acceleration is d 2 θ m /dt 2 = d 2 δ m /dt 2 (4) Substituting (4) into (1), then J(d 2 δ m /d t 2 )= T m T e (5) Multiplying (5) by ω m results in J ω m (d 2 δ m /d t 2 )= ω m T m ω m T e (6) Since angular velocity times torque is equal to the power, then J ω m (d 2 δ m /d t 2 )= P m P e (7) The quantity J ω m is called the inertia constant and is denoted by M. It is related to kinetic energy of the rotating masses, H. 4

5 H=(1/2) J ω 2 m =(1/2) M ω m (8) Or M =(2 H)/ ω m (9) Although M is called inertia constant, it is not really constant when rotor speed deviates from synchronous speed. However, since ω m does not change by a large amount before stability is lost, M is evaluated at the synchronous speed and is considered to remain constant, i.e., M = (2 H)/ ω sm (10) The swing equation in terms of the inertia constant becomes M(d 2 δ m /d t 2 )= P m P e (11) Where P m (p.u) and P e (p.u) are the per unit mechanical power and electrical power, respectively. (2H)/ω s )(d 2 δ/dt 2 ) =P m (p.u) P e (p.u) (12) But ω s =2π f (H)/π f )(d 2 δ/dt 2 ) =P m (p.u) P e (p.u) (13) Where δ = is in electrical radians. If δ is expressed in electrical degrees, the swing equations becomes (H)/180 f 0 )(d 2 δ/dt 2 ) =P m (p.u) P e (p.u) (14) Question 4: (10 Marks) Objectives: (To examine the ability of the student to understand Newton-Raphson method) Start with defining the voltages at two adjacent nodes V i = V i δ i and V j = V j δ j and the admittance between them Y ij = Y ij -θ ij, A) Derive the relation of real power P i and reactive power Q i? (6 Marks) B) Obtain from A) the Jacobian sub-matrices J 1, J2, J3 and J 4? (4 Marks) Solution -Application of Newton-Raphson Method to load flow equations in polar coordinates Let V i = V i δ i And Y ij = Y ij -θ ij =1/(R ij + j X ij ) = G ij - j B ij Therefore I i = n j=1 Y ij V j = n j=1 Y ij V j -θ ij +δ j (1) Thus P i + j Q i = V i I * i or P i - j Q i = V * i I i Then P i - j Q i = V * i n j=1 Y ij V j -θ ij +δ j = V i -δ i n j=1 Y ij V j -θ ij +δ j 5

6 = n j=1 Y ij V i V j -(θ ij + δ i -δ j ) But e -j( ) θ ij + δ i -δ j = cos (θ ij + δ i -δ j ) j sin (θ ij + δ i -δ j ) Then P i = n j=1 Y ij V i V j cos (θ ij + δ i -δ j ) (2) Q i = n j=1 Y ij V i V j sin (θ ij + δ i -δ j ) (3) Note that changes in power are related to the changes in voltage magnitudes and phase angles. ΔP J 1 J 2 Δδ ΔQ J 3 J 4. Δ V (4) Where ΔP = Power mismatch = P specified - P calculated The elements of sub matrices of this Jacobain matrix can be found as follows For [J 1 ] sub matrix, the off diagonal elements ( i j) are J 1(ij) = - P i / δ j = - Y ij V i V j sin (θ ij + δ i -δ j ) (5) J 1(ii) = - P i / δ i = - n j=1 Y ij V i V j sin (θ ij + δ i -δ j ) J 1(ii) = Y ij V i 2 sin (θ ii ) -Q i i=j (6) Similarly, for the sub matrix [ J 2 ], the off diagonal elements ( i j) are J 2(ij) = - P i / V j = - Y ij V i cos (θ ij + δ i -δ j ) (7) J 2(ii) = P i / V i = n j=1 Y ij V j cos (θ ij + δ i -δ j ) J 2(ii) = P i / V i + Y ij V i cos (θ ii ) i=j (8) For [J3] sub matrix, the off diagonal elements ( i j) are J 3(ij) =- Q i / δ j = - Y ij V i V j cos (θ ij + δ i -δ j ) (9) J 3(ii) = Q i / δ i = n j=1 Y ij V i V j cos (θ ij + δ i -δ j ) J 3(ii) = - Y ij V i 2 cos (θ ii ) + P i i=j (10) Similarly, for the sub matrix [ J 4 ], the off diagonal elements ( i j) are 6

7 J 4(ij) = Q i / V j = Y ij V i sin (θ ij + δ i -δ j ) (11) J 4(ii) = Q i / V i = - n j=1 Y ij V j sin (θ ij + δ i -δ j ) J 4(ii) = Q i / V i + Y ij V i sin (θ ii ) i=j (12) 7