Transformers and Induction Motors

Assignment - 2
Q1. Explain the constructional details of core and shell type of three-phase transformers.
Sol: Generally, power is generated and distributed in three phase system. Hence it is obvious that three phase
transformers are required to step up and step down voltages. It is practically possible to use single unit of
three phase transformers rather than using bank of three single phase transformers of same kVA rating. It
is more popular and widely used as it consumes less space, more economical and lighter in contrast to the
bank of three single phase transformers.[2][3] There are 2 types of three phase transformers that are classied
based on the magnetic circuit, core type and shell type, just like single phase transformers.
The constructional details of core type as well as shell type three phase transformers are as follows:
1. Core type three phase transformer construction:
Figure 1: Core type 3 - φ transformer
(a) The core consists of three legs or limbs, each provided with both the windings of one phase.
(b) Each limb has primary and secondary windings in cylindrical shape (previously wound), arranged in
concentric manner.
(c) The magnetic ux of each limb uses the other two limbs for its return path with the three magnetic ux's
in the core generated by the line voltages diering in time-phase by 120◦
. So the ux in the core remains
nearly sinusoidal, producing a sinusoidal secondary supply voltage.[5]
(d) The magnetic frame of the transformer is laminated in order to reduce the eddy current loss that takes
place in core and yoke.
(e) Low voltage (LV) windings are easier to insulate from the core than the high voltage (HV) windings.
1

Hence, LV winding is placed nearest to the core with suitable insulation and oil ducts between core and LV
winding.
(f) The HV windings are placed over LV windings with proper insulation and oil duct.[3]
2. Shell type three phase transformer construction:
Figure 2: Shell type 3 - φ transformer
(a) The core consists of ve legs or limbs.
(b) Five-limb cores are generally used for very large power transformers as they can be made with reduced
height.
(c) Three phases are more independent than they are in core type. Each phase has its individual magnetic
circuit.
(d) The construction is similar to that of three single phase shell type transformers kept on the top of each
other.[2]
(e) Steps b, d, e and f from previous section are exactly same for the construction of shell type three phase
transformer.
The core type three phase transformers are more widely used compared to shell type three phase transformers
as they are more economical, lighter and easier to manufacture. But for high power rating and high voltage
applications, shell type is preferred because of its better short circuit strength characteristic. It also exhibits
better power to weight ratio as compared to core type.[2][3]
In order to achieve three phase operation and enable the conduction of currents in both shell and core type
transformers, the windings are wound and connected either as Y or on each side. These shapes form as a
2

result of the way the three conductors inside the transformer get connected. There are 4 standard ways of
connection of the windings:
1. Y - Y:
Figure 3: Phase wiring for Y-Y arrangement
The winding ends marked with dots are connected to their respective phases A, B, and C. The non-dot ends
are connected together to form the centers of each Y. Having both primary and secondary winding sets
connected in Y formation, this will allow to use neutral conductors (N1 and N2) in each power system.
2. Y - :
Figure 4: Phase wiring for Y - arrangement
3

The secondary windings (bottom set) are connected in a chain, the dot side of one winding connected to
the non-dot side of the next, forming the Δ loop. At every connection point between pairs of windings, a
connection is made to a line of the secondary power system (A, B, and C).
3. - Y :
Figure 5: Phase wiring for - Y arrangement
This type of conguration would allow for the provision of multiple voltages (line-to-line or line-to-neutral)
in the secondary power system (secondary winding). This comes from the primary power system (primary
winding) that does not have neutral conductors.[4]
4. - :
Figure 6: Phase wiring for - arrangement
4

The - arrangement is restricted to applications in which neither the primary nor the secondary side
requires a 3 - φ neutral connection. It is generally used in moderate voltage systems because full line-to-line
voltage exists across the windings, and in heavy current systems because the windings need to carry only
1√
3
of the line current. This arrangement of connection is also unique as when one of the transformers are
removed for repairing or maintenance, the other 2 transformers forms open that can provide 3 - φ voltages
at a reduced level in contrast to the original 3 - φ transformer. [1]
For economical purpose, reliability and safety, Y - connection is generally used on the H.V. side while -
connection is used on the L.V. side. In this case the common point in the Y - connection, also called the
neutral, is grounded. With this, the phase voltage on the Y side is reduced to
1√
3
times the voltage had the
connection been a . [3]
Q2. A three-phase Induction Motor has an eciency of 85% when the load is 60 H.P. At this load both the
stator copper loss and the rotor copper loss are equal to the core losses. Mechanical Losses are (
1
4 )
th
of the
No-load losses. Calculate the slip.
Hint:
• 1 H.P=746 Watts
• No-load loss = Mechanical Losses + Core Loss
Sol:
η = 0.85, Pout = 60 × 746 = 44, 760W = 44.76kW
1) Pin =?
η =
Pout
Pin
(1)
∴ Pin =
Pout
η
=
44.76
0.85
= 52.66
∴ Pin = 52.66kW
2) Let Stator copper loss = Rotor copper loss = Core losses = Pc and Mechanical loss = Friction and windage
loss = PF W
∴ No − loadlosses = Pc + PF W (2)
Mechanical Losses are (
1
4 )
th
of the No-load losses. Hence:
5

No − loadlosses = 4 × PF W
∴ 4PF W = Pc‘ + PF W
∴ Pc = 4PF W − PF W = 3PF W (3)
∴ PF W =
1
3
Pc (4)
3) Total losses of power (Pl) is given by the formula:
Total losses of power = Stator copper loss + Rotor copper loss + Core losses + Mechanical losses = Input
power - Output power
∴ Pl = Pc + Pc + Pc + PF W = Pin − Pout (5)
∴ Pc + Pc + Pc +
1
3
Pc = 52.66 − 44.76 = 7.9
∴
10
3
Pc = 7.9
∴ 10Pc = 3 × 7.9 = 23.7
∴ Pc =
23.7
10
= 2.37kW
4) According to power ow diagram, mechanical power developed (Pmech) given by formula:
Pmech = Pout + PF W (6)
∴ Pmech = 44.76 +
2.37
3
= 45.55kW
Relationship between mechanical power (Pmech) and rotor input power (Pg) is given by:
Pmech = Pg(1 − s) (7)
where s = slip
6

∴ Pg =
Pmech
(1 − s)
(8)
Relationship between rotor copper loss (Pc), slip and rotor input power (Pg) is given by:
Pc = sPg (9)
Plugging the equation 8 into equation 9 gives:
Pc = s[
Pmech
(1 − s)
]
∴
Pc
Pmech
=
s
(1 − s)
(10)
5) Finally to nd the slip (s):
Pc = 2.37kW and Pmech = 45.55kW. Therefore using equation 10:
∴
2.37
45.55
=
s
(1 − s)
∴ 0.05203(1 − s) = s
∴ 0.05203 − 0.05203s = s
∴ s + 0.05203s = 0.05203
∴ 1.05203s = 0.05203
∴ s =
0.05203
1.05203
= 0.04946
Final answer: slip (s) = 0.04946 ≈ 0.0495
Hence % slip (%s):
∴ %s = 0.0495 × 100 = 4.95%
%s = 4.95%
7

Q3. A 3-phase, 50 Hz, 415V, Star-connected Induction motor has the following characteristics:
The stator is constructed as 1 pole pair with per phase impedances referred to stator as (0.2 + j3) Ω/phase
for the stator and (0.2 + j3) Ω/phase for the rotor. The magnetising reactance is j50 Ω/phase.
The stator iron losses are 250 W/phase. Using the approximate equivalent circuit determine the following
when the slip is 0.04
a. Rotor speed
b. Frequency of the rotor current
c. The supply current and power factor
d. The shaft power output and the torque developed by the motor if the friction and windage loss is 500 W.
Sol:
Data:
f = 50Hz, P = 2 (1 pole pair), Xm = 50Ω, s = 0.04, Iron or Core loss = 250 W per phase
Vph = 415√
3
= 239.6V
For stator:
Stator resistance per phase = R1 = 0.2Ω
Stator leakage reactance per phase =X1 = 3Ω
For rotor:
Rotor resistance per phase referred to the stator =R2 = 0.2Ω ∴ R2/s = 0.2
0.04 = 5Ω
Standstill rotor reactance per phase referred to the stator = X2 = 3Ω
Figure 7: Approximate Equivalent Circuit
8

a) Rotor speed (N ) = ?
Synchronous speed (Ns) is given by:
Ns =
120f
p
(11)
∴ Ns =
120 × 50
2
= 3, 000rpm
Slip is given by:
s =
Ns − N
Ns
(12)
Therefore rearranging equation 12 gives:
N = NS − sNs = Ns(1 − s) (13)
∴ N = 3000(1 − 0.04) = 2, 880rpm
N = 2, 880rpm
b) Frequency of rotor current = ?
Frequency of the rotor current (f ) is given by:
f = sf (14)
∴ f = 0.04 × 50 = 2
f = 2Hz
c) Supply current and power factor = ?
Supply current (I) is given by the formula:
I = Is = I0 + I2 (15)
since Supply current (I) = Stator current (Is)
where I0= No load current and I2= Rotor current per phase referred to stator
I0 = Iw + Iµ (16)
where Iw= Core loss current and Iµ= Magnetizing current
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Total Core loss = 250 × 3 = 750 W
Coreloss =
3(|Vph|)2
R0
(17)
∴ R0 =
3(|Vph|)2
Coreloss
=
3 × (239.6)2
750
= 229.64Ω
Iw =
Vph
R0
(18)
∴ Iw =
239.6∠0◦
229.63
= 1.04∠0◦
A
Iµ =
Vph
jXm
(19)
∴ Iµ =
239.6∠0◦
50∠90◦
= 4.792∠ − 90◦
A
Hence from equation 16:
∴ I0 = 1.04∠0◦
+ 4.792∠ − 90◦
= 1.04 + j0 + 0 − j4.792
∴ I0 = 1.04 − j4.792 = 4.903∠ − 77.7◦
A
To calculate rotor current per phase referred to stator (I2):
I2 =
Vph
impedances
(20)
impedances are referred to the stator.
impedances = R1 + X1 +
R2
s
+ X2 (21)
impedances = 0.2 + j3 + 5 + j3 = (5.2 + j6)Ω = 7.94∠49.1◦
Ω
∴ I2 =
239.6∠0◦
7.94∠49.1◦
= 30.18∠ − 49.1◦
A
From equation 15, supply current is calculated as:
Is = 4.903∠ − 77.7◦
+ 30.18∠ − 49.1◦
10

∴ Is = 1.04 − j4.792 + 19.76 − j22.81
∴ Is = 20.8 − j27.60 = 34.56∠ − 53◦
A
∴ Supply current = 34.56∠ − 53◦
A
To calculate power factor( cos φ):
Since it is 3-φ Y - connected induction motor, Supply current = Phase current. Therefore:
Is = Iph = 34.56∠ − 53◦
A
Zph =
Vph
Iph
(22)
∴ Zph =
239.6∠0◦
34.56∠ − 53◦
= 6.93∠53◦
Ω
From the obtained phase impedance value, the phase angle of the impedance equals to the power factor angle.
Hence:
φ = 53◦
Hence the power factor:
Powerfactor = cos φ (23)
∴ cos φ = cos(53◦
) = 0.6018
Powerfactor ≈ 0.602
d) To nd the shaft power (Pout) and torque developed by the motor (T) :
Total Core loss (Pw)= 250 × 3 = 750 W, Rotor resistance per phase = R2 = 0.2Ω
Stator copper loss = 3 × (|I2|)2
R1 and Rotor copper loss = 3 × (|I2|)2
R2
Therefore Total Copper loss (Pct) = Stator copper loss + Rotor copper loss:
Pct = 3 × (|I2|)2
R1 + 3 × (|I2|)2
R2 (24)
∴ Pct = (3 × 30.182
× 0.2) + (3 × 30.182
× 0.2) = 1, 093W
11

Therefore total losses (Pl) = Pw + Pct + PF W = 1, 093 + 750 + 500 = 2, 343W
Stator input or Pin:
Pin = 3VphIphcosφ (25)
∴ Pin = 3 × 239.6 × 34.56 × 0.6018 = 14, 949.75W
Hence shaft power is given by:
Pout = Pin − Pl (26)
∴ Pout = 14, 949.75 − 2, 343 = 12, 606.75W
∴ Shaft Power =Pout ≈ 12.607kW
Therefore torque (T) developed by the motor:
T =
Pout
ω
(27)
∴ ω =
2πN
60
=
2π × 2, 880
60
= 301.59rad/sec
∴ T =
12, 606.75
301.59
= 41.80
T = 41.80Nm
References
[1] Charles K. Alexander and Matthew N.O. Sadiku. Fundamentals of Electric Circuit. Tata McGraw-Hill
Education, fth edition, 2013.
[2] Kiran Daware, 2014.
[3] V.N. Mittle and Arvind Mittal. Basic Electrical Engineering. Tata McGraw-Hill Education, second
edition, 2005.
[4] Tony R. Kuphaldt. Lessons In Electric Circuits, volume 2. Digital Science License, 2000.
[5] Wayne Storr, feb 2015.
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