Electrical Power Drivers ET3026WB Exam Solutions Jun2009 (1)

Uitwerkingen tentamen ET3026WB 15-6-09

Problem 1 230504

. . 0,75 41, 40o

U Vf HzI Ap f

==== =

a) sin 230 4 sin 41.40 608, 4Q U I VAr= = = b) cos 230 4 0,75 690P U I W= = = c) sin 4 sin 41, 40 2,65QI I A= = = The current to be compensated for by the capacitor. So Ic=IQ

230 86,82,65C C

UXI

= = = 1

2186,8 36,67

2 50

CX f C

C FC

= = =

d) Draw the fasordiagram of the voltage, the current trough the load, the reactive current and the current trough the capacitor.

Ic

U

IIq

1


Problem 2 First simplify the circuit to show only one phase. The line to neutral terminal voltage is

fixed at: 20,8 123KVE K= = V

a) See graph. This is the nominal voltage (no saturation).

b) At no load there is no voltage drop in the synchronous reactance; consequently,

12oE E KV= =

E, Eo c) The power per phase is: 36 / 3 12P M= = WThe full load current is:

612 10 100012000

PI AE

= = = The voltage across Xs equals: 1000 9 9 90ox sE jIX j KV= = = The voltage Eo is the sum of the voltages E and the voltage across Xs.

2 2 2 212 9 15xE E E KV= + = + = The phasor diagram.

Eo=15KV Ex=9KV

E=12KV 1KA

2


d) With Capacitive load of 12Mvar The reactive power per phase is: 12/3=4MVAr The line current is:

64 10 33312000

QI AE

= = = The voltage across Xs is:

333 9 3 90ox sE jIX j kV= = = As before, Ex leads I by 90o phasor diagram

Ex=3kV

E=12kV

Eo=9kV

I=333A

The voltage Eo generated by Ix is equal to the phasor sum of E and Ex 0 12 ( 3) 9E E Ex kV= + = + =

3


Problem 3 a) Solution El=D.EH=0.2x100V=20V because the battery voltage is greater then EL, current flows out of terminal 5 and into terminal 1.

Its average value is: 30 20 52L

I A= =

b) The duration of one cycle is: 1 1 5020000

Tf

s= = = thus S1 is closed for a time 0,2 50 10aT s s = = and S2 is closed for 40s

At i=5A the voltage drop over de inductor, E41 equals the voltage of the battery minus the IR drop over de resistor: 30 (5 2) 20V A = V IL is flowing into terminal 4 and that terminal 4 is (+) with respect to terminal 1, it follows that IL must be increasing. In the 40s that S2 is closed the inductor accumulates 20V. 40s=800V.s

The current increases by an amount of 800 . 0,0810

V sI AmH = =

The peak to peak ripple is 0,08A

4


1. (1 point) Draw the schematic of a two quadrant and a four quadrant electronic DC-DC converter.

5


2. (3 points) Draw the wave shapes EAY and EAB of the DC to AC sine wave converter from figure 21.88

6


3. (3 points) A simple battery charger. Draw the voltage between point 2 and 1, 3 and 1 and 4 and 1and current I of the circuit from fig. 21.11

4. ( 2points) Draw the fasor diagram of a synchronous motor with torque angle 40o.

E

E0

I

Ex

Ex

5. (1 point) What is the relation between reactive power Q, active power P and apparent power S?

2 2S P Q= +

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Electrical Power Drivers ET3026WB Exam Solutions Jun2009 (1)

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