Lecturer: Dr. J B E AL-ATRASH No. of Pages: 4

Transcription

1 Philadelphia University Faculty of Engineering Dept. of Electrical Engineering Student Name: Student Number: Final Exam Course Title: Design of T&D System Date: 15 th June 2016 Course No: / Time Allowed:120 Minutes Lecturer: Dr. J B E AL-ATRASH No. of Pages: 4 Question 1: (10 Marks) Objectives: (Fundamental information in T& D design) Choose the correct answer? (2 Marks x 5) 1. The SIL of transmission line working at 750 KV (double circuit) is a) 750 MW b) 1875 MW c) 3750 MW d) 1500 MW Answer: c 2. The height of H.V transmission working on 400 KV is a) 28 m b) 35 m c) 48 m d) 75 m Answer: c 3. The short circuit level at any bus in the power system is given by MVAs.c=MVAb/Xs.c Where a) MVAb in MVA and Xsc in p.u from that bus up to the Xs of the source b) MVAb in MVA and Xsc in ohm from that bus up to the Xs of the source c) MVAb in MVA and Xsc in p.u from that bus up to and including the Xs of the source Answer: c 4. Typical area of land needed to build indoor substation 132/33 KV is a) 2600 m 2 b) 500 m 2 c) between m 2 Answer: a 5. The characteristic impedance for a transmission line is equal to a) LC b) L/C c) LC d) L/C Answer: b Question 2: (10 Mark) Objectives: (Tie connection between two neighboring power system) If two neighboring power systems are to be connected together to exchange power during MD of 500 MW (unity power factor) at a distance of 500 km. Calculate a) The level of voltage at which this transmission line should operate on to permit the required power exchange?(4 Marks) Solution SIL 400KV =(400 KV) 2 /300=533.3 MW Then choose operating voltage of 400 KV 1

2 b) Discuss the design and configuration (draw) of both the towers and conductors to be used in this transmission line? Give the approximate height of the tower, the insulator level voltage and the distance between successive towers?(6 Marks) Double circuit Bundled twin conductor Single earth wire The tower should be designed as double circuit (bundled) with single earth wire. The approximate height is 48 m for each tower. The insulators voltage level is 15 KV The distance between successive towers=500 m. I L =S/( 3xV L )=500 MVA/( 3x400KV)=721.7 Amps Then choose c.s.a of 650 mm 2 X T.L = 500x(0.05+j0.18)=25+j90 Ω For D.L X T.L =+J45 Ω Question 3: (10 Marks) Objectives: (Calculation of annual energy, MD and load factor) 2

3 Fig 1: Daily load curve of the national grid. The load measurements are given in Table 1 Time (hrs) P (MW) Time (hrs) P (MW) Table 1 (A) The daily load curve of the Jordan national power system is given in Fig 1 shows that the MD was 3300 MW during the day between a.m at the 9 th August Calculate the daily load factor for that day? (6 Marks) (B) Calculate the energy consumed in the year 2015 given that the annual energy factor (ALF) = 0.62? (4 Marks) Solution 3

4 DLF= Actual energy consumed/ Energy consumed in that period assuming the maximum demand prevailing all the time = [2600x1+2x ( ) +2700x1]/3300 x 24 = /79200 =0.84 B) Calculate the energy consumed given that the annual energy factor (ALF) = 0.65? (4 Marks) Energy consumed = ALF x MD X 8760=0.62 x 3300 x 8760 = GWh Question 4: Objectives: (Design transmission power system) (10 Marks) If it is required to supply a new project which has Maximum Demand of 400 MW at unity power factor at the end of the next 10 years; and the nearest point of connection is at distance of 35 km, 400 KV and at fault level of MVA Then calculate 1. The number and the rating of the transformers near the sight of the project? (2 marks) 2. The rating of different circuit breakers (in MVA and KA) used to supply this new load? (4 marks) 3. The short circuit level at the high voltage side and low voltage side of the suggested substation? (4 marks) 400 kv 2 x 400 MVA 132 kv The MVA SC at 400 KV = MVA I L = (400/1.0) /( 3 x 400)= 577 A 4

5 Then choose 600 mm 2 X of 10 km = 0.22 x 10=2.2 Ω Z base = (400) 2 /100 = 1600 Ω X p.u = (2.2/1600) x ½ = p.u Xsource= 100/15000 = The fault level at 400 kv (the point of connection after 10 km) is MVA s.c = 100/( ) = 100/( ) = MVA In this case C.B s at 400 kv are MVA, 30.3 ka The fault level at 132 kv is MVA s.c = 100/[ (100/400)0.14// (100/400)0.14] =100/[ // 0.035] =100/(0.0247) = MVA Then use MVA C.B, 25 ka Table 1 Switchgear Fault Ratings System Voltage BS 116 Rating LE.C Rating kv MVA ka Table 2: Electrical characteristics of bare aluminum conductors steel-reinforced c.s.a(mm 2 ) R (Ω/km) X (Ω/km) Ampers Table 3: Three phase transformer group Ratio KV Rating MVA Reactance % 5

6 33/11 132/33 400/