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POLYTECHNIC OF NAMIBIA http://www.polytechnic.edu.na
SCHOOL OF ENGINEERING DEPARTMENT OF ELECTRICAL ENGINEERING http://www.polytechnic.edu.na/EEE
BACHELOR OF ENGINEERING (B.Eng.)
EMC510S ELECTRICAL MACHINES 216
TEST 2
DATE: 30 TH^ APRIL, 2010 TIME: 17H15 – 20 H 0 0 (2¾ HOURS)
This paper consists of four (4) questions.
INSTRUCTIONS:
- Attempt all questions.
- Use the given lined paper for your solutions and use both sides of the answer sheet.
- All working and all steps must be shown clearly. Marks will be lost for ‘skipped’ steps.
- All cell phones must be switched off during the test. 10% will be deducted from the
total mark in penalty for disturbances caused by cell phones.
- You're not allowed to leave the room during the test.
- Consultations with fellow students are not allowed during the test. 10% will be
deducted from the total mark in penalty for such consultations.
TOTAL MARKS: 92
EXAMINER: Mr. K. Kanyimba
MODERATOR: Mr. G. Gope
QUESTION ONE (28 MARKS)
(a) Draw a well-labelled equivalent circuit of a two-winding transformer referred to the
secondary circuit. Use the subscript ‘s’ for all secondary quantities and the subscript ‘p’
for all primary quantities. Use the “ ‘ “ (prime) to indicate referred quantities. Assume the
transformer is loaded with a load, ZL. (6 marks)
Solutions:
(a)
(b) A 4000/400 V, 10 kVA two-winding transformer has primary and secondary winding
resistances of 13 Ω and 0.15 Ω, respectively. The leakage reactance referred to the
primary is 45 Ω. The magnetizing impedance referred to the primary is 6 kΩ and the
resistance corresponding to the core loss is 12 kΩ. (i) Determine the total resistance
referred to the primary and the values of all impedances referred to the secondary. (ii)
Determine the input current when the secondary terminals are open-circuited. (iii)
Determine the input current when the secondary load current is 25 A at a power-factor of
0.8 lagging. (22 marks)
Solution:
2 1 '^2 1 2
N 4000
(i) Turns ratio, n = 10 13 0.15 10 N 400
p p eq p s s s
N
R R R R R
N
(^2 2 )
s s s eq p eq s eq p eq p p
N N
R R X X
N N
(^2 2 ) ' '
s s c c m m p p
N N
R R X X
N N
^ ^ ^
' ' From the phasor diagram, for (^) p , cos( ) cos( )
cos( ) i.e. ,
but cos( ) (cos cos sin sin ) cos sin.
( cos sin ) There fore , ...( 1
s p s p p s s p e e s
p e e s
p
e e s e e s e s e s e s
p e s e s
p
V V V V I Z
I Z
VR
V
Z Z R X
I R X
VR
V
(b) The primary and secondary windings of a 40 kVA, 6600/250 V single-phase two-
winding transformer have resistances of 10 Ω and 0.02 Ω, respectively. The leakage
reactance of the transformer referred to the primary is 35 Ω. (i) Calculate the primary
voltage required to circulate full-load current when the secondary is short-circuited.
(ii) Calculate the full-load regulation at unity power-factor. Neglect the no-load current.
(14 marks)
Solutions:
(^2 ) '
(i) Primary voltage required to circulate full-load current on short-circuit:
Referring impedances to the primary side: 0.02 13. 250
The full-load prima
ry c
p s s s
N
R R
N
^
urrent is, 6.061 A 66
p FL p
S
I
V
^ ^
'
On short-circuit the equivalent impedance referred to the primary side is:
Hence the primary voltage requir
ed to c
irculate full-load current on
Z (^) p eq R (^) p Rs jX (^) p eq j
.
short-circuit is:
V (^) p sc I (^) p FL Zp eq 6.061 42.381 256.871 2 57 V ... (3.0)
(ii) Full-load Regulation at unity power-factor:
( cos sin ) (^) 6.061 23.9 1 Voltage Regulation, 100% 100 2.195% 6600
p e s e s
p
I R X
VR
V
QUESTION THREE (2 0 MARKS)
(a) From first principles show that the efficiency of a two-winding transformer is at its
maximum when the variable I^2 R loss is equal to the constant core loss, PC. ( 5 marks)
Solution:
2 .
Output Power Output Power Transformer efficiency, Input Power Output Power + Power losses
..
(. ) ( )
s s
s s C s s eq
I V p f
I V p f P I R
. 2 C s. s-eq.
= ... eqn. 3.1. (. ) ( / )
where P is the core loss, I is the resistive loss and R is the transformer equivalent
resistance referred to the secondary sid
e
s
s C s s s eq
s eq
V p f
V p f P I I R
R
- For a normal transformer, V is approximately constant, hence for a load of givens
power-factor the efficiency is a maximum when the denominator of eq n. 3.1 is a minimum,... (1.
i.e., when
s s
d V dI
2
. 2..
2
C
. 0 0 or
i.e., the efficiency of a transformer is at its maximum when the variable I R loss is equal to
the constant core loss, P.
C C s s eq s eq s s eq C s s
P P
p f I R R I R P I I
^ ^ ^ ^ ^ ^
(b) The required no-load voltage ratio in a 150 kVA single-phase 50 Hz core-type, two-
winding transformer is 5000/250 V. (i) Calculate the efficiency at half-rated kVA and unity
power-factor. (ii) Determine the efficiency at full-load and 0.8 power-factor lagging. (iii)
Find the load kVA for maximum efficiency. The full-load I^2 R loss is 1800 W and the core
loss is 1500 W. (1 5 marks)
Solutions:
2 2
(i) At half-rated kVA (i.e. half of full-load curent) and unity p.f.
Resistive loss = 0.5 1800 0.45 kW 2
1 Core loss = 1.5 kW Power output = 0.5 150 kVA = 7
... (1.0) 5 k 2
cu FL
FL
P
S
^ ^ ^
VA ...(1.0)
Power losses Therefore, efficiency, 1 Power output + Power losses
1.5 0. 1 75 (1.5 0.45)
0.97466 p.u. or 97.5% ... (2.0)
(ii) At full-load kVA and 0.8 p.f. lagging,
Power output = S p.f.=150 0.8 = 120 kW
1.8 1. and 1 0.97324 p.u. or 97.3% 120 (1.
3
p
3
s
(i) Current ratings of the two-winding transformer:
The 7200 V winding: I 13.889 A 7200
The 600 V winding: I
166.667 A... (2.0)
p
s
S
V
S
V
p c s p c
p
In the autotransformer:
I 166.667 A; I 13.889 A I I I 166.667 13.889 180.556 A
kVA Ratings:
Primary: kVA 1300.0026 1300 kVA 1000 100
Sec
nda
o ry
V I p p
p
: kVA 1300.0032 1300 kVA 1000 100
V Is s
Srating ( ) (^1300) (ii) 13 (
rating^0
Autotransformer
S Two windng Transformer
