SECTION 1 / AFDELING 1 - MEMO Question 1: Give the following ...

SECTION 1 / AFDELING 1 - MEMO
FW LEUSCHNER
Xformers & Electrical Machines / Xformators & Elektriese Masjiene
Question 1: Give the following equations for: Vraag 1: Gee die
volgende vergelykings vir:
a. Transformer: The output voltage as a function of the winding ratio and the
input voltage.
Transformator: Die uitgangspanning as ‘n funksie van die
windingsvehouding en die insetspanning
4
Vout = Vin x N2 / N1
Pout = Pin = Vout x Iout = Vin x Iin
b. Transformer: The output impedance as viewed from the input voltage, as
a function of the output impedance and the winding ratio
Transformator: Die uitgangsimpedansie soos gesien vanaf die
insetspanning, as ‘n funksie van die uitgangs-impedansie en die
windingsverhouding
Z’L = ZL x (N1 / N2) 2
c. Machines: The ratio of Mechanical Torque and the Armature Current of a
separately excited rotating DC machine.
Masjiene: Die verhouding tussen die Meganiese Draaimoment en die
Ankerstroom van ‘n apart-opgewekte GS masjien.
d. Machines: The ratio of induced EMF EA and the rotating speed ω of a
separately excited rotating DC machine.
Masjiene: Die verhouding tussen die opgewekte EMK EA en die
rotasiespoed ω van ‘n apart-opgewekte GS masjien.
e. Machines: For calculating the synchronous speed in r.p.m. of an
Induction Motor as a function of Pole Pairs and Supply Frequency.
Masjiene: Vir die berekening van die sinchrone spoed in o.p.m. van ‘n
Induksiemotor as ‘n funksie van Poolpare en
Toevoerfrekwensie.
ns = (60 x f) / Pp
V2 / I2 = ZL + 15.51 & 15.56
Tdev / IA = KΦ
EA / ωm = KΦ
Field rotates ns in r.p.m. / 60 gee o.p.s. Hz
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4
4
4
[20]

Question 2 / Vraag 2
An electric vehicle (battery powered) is running uphill at constant speed and
constant field current (separately excited DC machine).
What happens when the vehicle runs down the other side of the hill (no
change in field current or speed)?
‘n Elektriese voertuig (aangedryf deur batterye) ry op teen ‘n bult teen ‘n
konstante spoed en konstante veldstroom (apart opgewekte GS masjien.
Wat gebeur wanneer die voertuig teen die anderkant van die bult afry (geen
verandering in veldstroom of spoed)?
Uphill / opdraend:
Battery voltage = VB and EA = VB - ( IA x RA)
EA = KΦωm ∴ωm = constant & Φ = constant ∴ EA = constant
and < VB IA = constant ∴ Tdev = constant & P = constant
Down hill / afdraend:
IF = constant ∴ Φ = constant and K = constant. Assume ωm =
constant ∴ same speed and wind losses
∴ Tdev = constant in opposite direction (generator) i.e. Tinput to
Rotor. Tin = KΦIA ∴ IA = Tin / KΦ in opposite direction but
constant.
∴ VB = EA + ( IA x RA) ∴ Battery gets charged at VB and IA ⇒
NOTE: I A is now negative (∴ EA > VB )
Assume constant battery voltage ???)
Question 3 / Vraag 3
For a three-phase squirrel cage induction motor, the following is known:
Supply voltage 400V (line voltage). Frequency 50 Hz. Delivered output
power 10kW at 20A input current and 0.75 power factor.
What is? i. The electrical input power into the motor. ii. The efficiency of
the motor in %. iii. The values of three capacitors to be connected across
the three phase connections on the motor – namely A, B and C to
improve the power factor to 0.90
i.
PIn = √3 VLine x I x cosθ = 10.39kW
(3)
ii.
η = ( Pout / Pin ) x 100 = 96.2%
(3)
iii. Change PF from .75 to 0.90 by adding one capacitor across
each line voltage. Total Qin = √3 VLine x I x sinθ = 9.17kVAr (9)
For PF to be 0.90 we need cosθ = 0.90 or θ = 25.84° and Qin’ =
√3 VLine x I x sinθ = 6.04kVAr
∴ Add Q =3.13kVAr Capacitive or QEach = 1.04kVAr per phase.
Q = VLine
[15]
2 / XC ∴ XC = VLine 2 / Q = 153.8Ω and
XC = 1 / 2πfC ∴ C = 1 / 2πf XC = 20.7 µF
[10]