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S265 20/08/
TH E MANCH ESTER M ETR O PO LITAN UNIVER SITY
FACULTY O F SCIENCE AND ENGINEER ING
D EPA R TMENT O F ENGINEER ING AND TECH NO LO GY
SESSIO N 2001/
Exam ination for th e BEng (H O NS) ELECTR ICA LAND ELECTR O NIC ENGINEER ING (FULL-TIME/SANDWICH /PA R T-TIME) YEA R TW O
UNIT 64EE2073: ELECTR ICALPO W ER ENGINEER ING
Tuesday 14 May 2002
9 .30 am to 11.30 am
Instructions to Candidates
A nsw er ANY FO UR questions.
A llquestions carry equalm ark s. Graph paper is provided. Break dow n of m ark s for each question is sh ow n in square parenth eses.
- A b ipolar junction transistor sw itch h as th e follow ing particulars:
β min = 25 ; β max= 100 ;
V (^) BEsat = 1. 5 V;and VCEsat = 2 V.
Th e value of th e load resistance R (^) C is 20 O h m s. Th e dc supply voltage is 100 V and th e input voltage to th e b ase circuit VBB is 5 V.
(a) Sk etch th e circuit arrangem ent and use it to calculate: (i) a suitab le value of transistor b ase resistance R (^) B w h ich w illallow saturation of th e transistor w ith an overdrive factor (O D F) of 5; and [16]
(ii) th e totalpow er dissipation in th e transistor. [4]
(b ) Explain, b riefly, th e significance of O D F in designing BJT sw itch es. [5]
- A forward converter is shown in Figure Q2:
(a) W ith th e aid of appropriate equivalent circuit diagram s and w aveform s, sh ow th at th e output voltage is given by V (^) O = DVI w h ere D is th e duty cycle of th e converter. [15]
(b ) Th e converter of Figure Q 2 h as th e follow ing particulars: Input voltage Vin=24 V; O utput voltage Vout=12 V; Sm ooth ing inductor value L= 75 μH ;and O perating frequency, f , of 15 k H z. D e rive an expression for calculating th e m axim um inductor ripple current and use to it to calculate th e value of th is current. [10]
L
D C Vin Vout
Figure Q
- (a) A n alternator is connected to constant voltage, constant frequency busbars and is operating at constant pow er output. Th e equivalent circuit per ph ase is sh ow n in Figure Q 4. Starting from unity pow er factor operation and w ith th e aid of ph asor diagram s, explain h ow lagging or leading pow er factors m ay be ach ieved. [8]
(b) An 11 kV cylindrical rotor alternator is rated at 15 MW at 0.9 pf lagging and has a synchronous reactance of 12 Ω /phase and negligible resistance. It is delivering 10 MW at 0.8 pf lagging to constant voltage constant frequency busbars. By draw ing a ph asor diagram to scale or oth erw ise, determ ine th e induced em f (E) in th e stator w indings due to excitation current and th e load angle (δ) at w h ich th e m ach ine operates. [6]
(c) W ith out ch anging th e excitation current, th e pow er output is increased to 15 MW. D e term ine th e new pow er factor and load angle at w h ich th e m ach ine now operates. [6]
(d) Com m ent on th e new load angle value and state th e action w h ich m ust b e tak en to im prove th e stab ility of th e alternator w ith out reducing th e pow er output. [5]
E G V
Xs I
Figure Q 4
- Part of a pow er system is sh ow n in Figure Q 5. A 3-ph ase sh ort circuit to earth develops close to busbar 2 D e term ine:
(a) th e MVA fault levelat busbar 2. [15]
(b ) th e fault current supplied by th e 3 MVA generator. [5]
(c) th e voltage at busbar 3 during th e fault and before th e fault h as b e e n cleared. [5]
A ppropriate pu and oh m ic reactances are indicated alongside each item of plant;it m ay be assum ed th at allsystem resistance can b e ignored. It is suggested th at you ch oose 5 MVA as b ase.
Figure Q 5
33 kV network 1000 MVA fault level
T
T
T (^) T
T
11
(^31 )
1 2
Busbar 1 Busbar 2
Busbar 3
11 kV
L
L
L
13 23
12
0.9 Ω
0.3 Ω
0.6 Ω
G
2 MVA 2 MVA 11/0.415 kV (^) 11/0.415 kV 0.05 pu (^) 0.05 pu
Load (^) Load
5 MVA 33/11 kV 0.05 pu
3 MVA 0.1 pu
