# Department of Electrical and Computer Engineering

Topics: Trigraph, Transformer, Electric motor Pages: 11 (1260 words) Published: March 12, 2013
The Islamic University of Gaza

Faculty of Engineering

Department of Electrical and Computer Engineering

Final Exam

Electrical Machines (EELE 3351)

Time: 2 Hours

Student’s Name
Student’s Number

Problem #

Problem 1

20

Problem 2

20

Problem 3

20

Problem 4

20

Good Luck
Tuesday May 27, 2008

All kinds of calculators are allowed

1.

The coil on the magnetic core shown has 500 turns. The core has a depth of 5 cm with an airgap length of 0.07 cm. The remaining dimensions are shown in the figure Fringing and leakage flux are to be neglected. The core is composed of steel having a relative permeability of μr=1000.

a)
b)
c)
d)
e)

Calculate the reluctance of the air-gap [3 pts]
Calculate the reluctance of the core [5 Pts]
Sketch the analogous equivalent circuit of this magnetic arrangement [4 Pts] What current is required to produce a flux density of 0.5 T in the air-gap? [4 pts] What flux density is produced in the right leg if the air-gap is removed? [4 Pts]

0.07102
g 

 222817 A.t/Wb
o Ag 4 107 55104
l r .leg
(40  0.07)102
r .leg 

127101 A.t/Wb
r o Ar .leg 1000 4 107 55104
ltot
115102
l ,u ,d .leg 

 61009 A.t/Wb
r o A 1000 4 107 105104
core 127101 61009 188110 A.t/Wb
total core g 188110  222817
total  410927 A.t/Wb
g  Bg  Ag  0.50.050.05  0.00125 Wb
g 1.25 mWb
Ni
500i
g 

1.25103
total 410927
4109271.25103
i
1.027 A
500
l r .leg
40102
127324 A.t/Wb
r .leg 

r o Ar .leg 1000 4 107 55104
core 127324  61009 188324 A.t/Wb
total core 188324 A.t/Wb
Ni 5001.027
core 

 0.00273 Wb
total
188324
core  2.73 mWb

0.00273
Br .leg  core 
1.09 T
Ar .legg 25104
lg

2.

A 100-kVA, 1000/100-V, 60-Hz single-phase transformer gave the following test result: Test
Open-circuit test (HV side open)
Short-circuit test (LV side shorted)
a)
b)
c)

Voltage
100 V
50 V

Current
6A
100 A

Power
400 W
1800 W

Find the parameters of the approximate equivalent circuit referred to the HV side [10 Pts] Determine the voltage regulation at full-load, 0.8 leading power factor [6 Pts] Calculate the efficiency of the transformer [4 Pts]

Open circuit test
400
 0.667
oc  48.19
100  6
Qoc V oc I oc sinoc  100  6  0.745  104.5VA sinoc  0.745
2
2
V
V 2 1002
100
RcL  oc 
X ML  oc 
 25 
 95.72 
Poc 400
Qoc 104.5
1000
a
RcH  102  25  2500 
X MH  102  95.72  9527 
 10
100
Poc V oc I oc cosoc

cosoc 

Short circuit test
1800
sc  68.9
 0.36
50 100
Qsc V sc I sc sinsc  50 100  0.933  4665VA sinsc  0.933
Psc 1800
Q
4665
ReqH  2 
 0.18  X eqH  sc 
 0.4665 
2
I sc 1002
I sc 1002
Psc V sc I sc cossc

cossc 

100000
 100 A
1000
V 1 V 2'  I 2' (ReqH  jX eqH )  1000 0100 36.87(0.18  j 0.4665) V 1  986.41 j 48.12  987.58 2.793 V
V 2'nl V 2'fl
987.58 1000
VR % 
100 
100  1.242%
'
V 2fl
1000
Pout  100  0.8  80 kW
V 12 987.582

Pcore 
 390.13 W  Pnl  400 W
RcH
2500
Pin  Pout  Pcu  Pcore  80000 1800  390.13  82190.13W P
80000
100  97.34%
  out 100 
Pin
82190.13

V 2'  100 10 0  1000 0 V

I 2' rated 

3.

A three-phase, 400-hp, 480-V, 60-Hz, 8-pole, ∆-connected synchronous motor has a synchronous reactance of 1.0 Ω and negligible armature resistance. Ignore friction, windage, and core losses of this motor

a)

What are the magnitudes and phase angles of EA and IA if the motor is supplying 400 hp at 0.8 lagging power factor? [8 pts]
b) Calculate the torque produced by the motor and the torque angle δ [6 pts] c) If EA is increased by 15 percent while the power output remains constant, what is the new magnitude of the armature current? What is the new power factor? [6 pts]

(a)
Pin  Pout ...