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Polyphase Induction Motor
 In electric motor, conversion of electrical power to mechanical power takes place in the rotating
part of the machine.
 Invented by “Nikola Tesla”
Two Main Parts
1. Rotors- rotor winding is the moving part
2. Armature- the stationary parts
Rotor
 Rotor windings receives power through induction which is similar to the secondary winding of the
transformer. When current flows through the secondary winding, it creates an electromagnetic
field causing the rotor to move because of the interaction of magnetic fields between stator and
rotor winding.
Basic Principle of Operation
- The long force on a conductor
 Faraday’s Law of Magnetic Induction
Two General Types of Rotor
1. Squirrel Cage (SCRIM)
2. Wound Rotor (WRIM)
Stator- use whole coiled loop winding which has the following;
1.
2.
3.
4.
There are as many coils in the entire winding as there are slots.
The number of coils per phase is equal to slots per phase.
The number of coils in each pole group per phase equals slots per pole times phase.
Coil in each group is always connected in series.
IGBT’s – Insulated Gate Bipolar Transistor; can operate at frequencies up to 20 kV
GTO’s – Gate-Turn-Off (thyristor)
-
Can handle currents of several flow ampere at voltage of up to 5Kv
Converter- (frequency converter)
-
The modern way of controlling speed and torque of motor drives
Induction Motors
 Pole Pitch- the distance between adjacent poles
 Plugging- to stop quickly the motor by reversing the rotating (through interchanging the two
Single Phasing
-
When one fuse is blown in a 3𝜙 IM, the motor continues to run but the current on the
remaining two lines will double.
Asynchronous Generator
-
When the induction motor rotates faster than synchronous speed, it generates electricity
which is returned to the line.
Dynamic Breaking
-
Is achieve when the armature winding AC source is replaced by a DC source
-
In SCRIM, it indicates the type of (Squirrel cage motor to be change by changing the
number) bars used in the rotors.
Code Letter
Consequent
-
Permits the synchronous speed of squirrel cage motor to be change by changing the
number of stator poles.
-
Starting and stop of motor for inching or to move the motor only for only a few second
for mechanical positioning.
Jogging
Varmeter
-
Used to measured reactive power; the voltage is shift by 90% before it is applied to the
PT coil.
Revolving Field Speed
-
Also known as synchronous speed (𝑁𝑠𝑦𝑛 )
𝑧𝑓
𝑃
𝑁𝑠𝑦𝑛 = 𝑟𝑝𝑠 =
𝑁𝑠𝑦𝑛 = 𝑟𝑝𝑚 =
120𝑓
𝑃
Sample Problems
1. Calculate the synchronous speed of an 8-pole induction motor when supplied with power from;
(a) 60Hz, (b) 50Hz, (c) 25Hz source.
Solution;
120𝑓
𝑃
=
120(60)
8
b.) 𝑁𝑠𝑦𝑛 =
120𝑓
𝑃
=
120(50)
8
=
750 rpm
c.) 𝑁𝑠𝑦𝑛 =
120𝑓
𝑃
=
120(25)
8
=
375 rpm
𝑁𝑠𝑦𝑛 =
a.)
=
900 rpm
Slip and Rotor Speed
-
Although the rotor of induction motor rotates in the same direction as the revolving field,
it cannot do so at synchronous speed. There must be relative motion between the
revolving field and rotating rotor.
-
The difference between the synchronous speed and actual rotor speed, which is usually
specified in terms of synchronous speed as a percentage.
Slip
%slip =
𝑁𝑠𝑦𝑛 −𝑁𝑅
𝑁𝑠𝑦𝑛
× 100%
m Decimal Form
S=
𝑁𝑠𝑦𝑛 −𝑁𝑅
𝑁𝑠𝑦𝑛
Slip Speed
Slip Speed = 𝑁𝑠𝑦𝑛 − 𝑁𝑅
Rotor Speed
120𝑓
𝑃
𝑁𝑅 =
(1 − 𝑠)
Sample Problems
1. The rotor speed of a 6-pole, 50 Hz induction motor is 960 rpm. Calculate the percent slip.
Solution;
𝑁𝑅 =
120𝑓
𝑃
S = 1-
(1 − 𝑠)
𝑁𝑅 ⋅𝑃
120𝑓
=1−
(960)(60)
(120)(50)
S=0.04
%Slip = 4%
2. Calculate the speed of a 60Hz, 14-pole motor if the slip is 0.05
Solution;
120𝑓
𝑃
𝑁𝑅 =
𝑁𝑅 =
(1 − 𝑠)
120(60)
14
(1 − 0.05)
𝑁𝑅 = 488.57 𝑟𝑝𝑚
Rotor Generated Voltage and Frequency
𝐸𝑅 = 𝑆𝐸𝐵𝑅
𝑓𝑅 = 𝑆𝑓
Where;
𝑬𝑹 – generated voltage per phase in the rotor @ slip s
𝑬𝑩𝑹 – blocked rotor generated voltage per phase
𝒇𝑹 – rotor frequency
Equivalent circuit per phase of a wound Rotor Induction Motor (WRIM)
Where;
𝑟1 – stator resistance per phase
𝑟2 – rotor resistance per phase
𝑋1 – stator leakage reactance per phase
𝑋1 – rotor leakage reactance
𝑅𝑚 – resistance corresponding to the iron
𝑅𝑥 – external resistance
losses & friction windage loss
Sample Problem
1. A 3𝜙 60Hz, 6-pole, 220 V wound rotor induction motor (WRIM) has a stator that connected is
star. The rotor is half as many turns as many in the stator. If the speed of the rotor 𝑁𝑅 is 1,110
rpm, calculate the following: (a) slip; (b) 𝐸𝐵𝑅𝜽 ; (c) 𝐸𝑅𝜽 ; (d) 𝐸𝑅 between terminals; (e) frequency
of rotor 𝑓𝑅 .
Solution;
a) 𝑁𝑅 =
S = 1-
𝑁𝑅 ⋅𝑃
120𝑓
S = 0.075
120𝑓
𝑃
(1 − 𝑠)
=1−
(1110)(6)
(120)(60)
𝑁1
𝑁2
b)
𝐸
= 𝐸1
2
Where;
1
𝑁2 = 2 𝑁1
𝐸1 = 220𝑉
𝑁1
1
𝑁
2 1
=
220
𝐸2
𝐸𝐵𝑅𝜽 = 𝐸2 = 110𝑉
c) 𝐸𝑅𝜽 = 𝑆𝐸𝐵𝑅𝜽 = (0.015)(110)
𝐸𝑅𝜽 = 8.25𝑉
d) 𝐸𝑅𝐿 = 8.25(√3)
𝐸𝑅𝐿 = 14.29
e) 𝑓𝑅 = 𝑆𝑓 = (0.075)(60)
𝑓𝑅 = 4.5Hz
2. A 6-pole 3𝜙 squirrel cage rotor induction motor (SCRIM) is connected to 60 cps supply. At full
load, the rotor induced emf make a 72 complete cycle in 1 minute. Find the rotor speed 𝑁𝑅 .
Solution;
𝑐𝑦𝑐𝑙𝑒
1𝑚𝑖𝑛𝑢𝑡𝑒
𝑓𝑅 = 72 𝑚𝑖𝑛𝑢𝑡𝑒 (60𝑠𝑒𝑐𝑜𝑛𝑑𝑠) = 1.2 𝑐𝑝𝑠
S=
𝑓𝑅
𝑓
=
∴ 𝑁𝑅 =
1.2 𝑐𝑝𝑠
60 𝑐𝑝𝑠
120𝑓
𝑃
= 0.02
(1 − 𝑠) =
120(60)
6
(1 − 0.02)
𝑁𝑅 = 1,176 𝑟𝑝𝑚
Rotor Current & Power
𝐼𝑅 =
𝑆𝐸𝐵𝑅
2 +𝑆 2 𝑋 2
√𝑅𝑅
𝐵𝑅
=
𝐸𝐵𝑅
2
2
√(𝑅𝑅 ) + 𝑋𝐵𝑅
𝑆
3. Using the data of problem number 1, calculate the 𝐼𝑅 if 𝑅𝑅 =0.1 Ω & 𝑋𝐵𝑅 = 0.5 Ω.
Solution;
𝐼𝑅 =
𝐸𝐵𝑅
2
2
√(𝑅𝑅 ) + 𝑋𝐵𝑅
=
110
2
√( 0.1 ) + 0.52
0.075
𝑆
𝐼𝑅 = 77.25 A
-
What is the rotor reactance when it is already running?
Solution;
@ stand still; (60Hz)
𝑋𝐿 = 2𝜋𝑓𝐿
𝐿=
0.5
2𝜋(60)
𝐿 = 1.326𝑀𝐻𝑧
@ running; (4.5Hz)
𝑋𝐿 = 2𝜋𝑓𝐿 = 2𝜋(4.5𝐻𝑧)(1.326𝑀𝐻𝑧)
𝑋𝐿 = 0.0375 Ω
Another solution
𝑋𝐿 = 𝑆𝑋𝐵𝑅 = 0.075(0.5)
𝑋𝐿 = 0.0375 Ω
4. The 6-pole wound rotor induction motor is excited by a 3𝜙, 60 Hz source and the full load speed
is 1140 rpm. Calculate the frequency of the rotor and current under the following:
a.) At stand still
b.) Motor is running @ 500rpm the same direction as the revolving field.
c.) Motor is turning @ 500rpm @ different direction as the revolving field.
d.) Motor running at 2000 rpm in the sane direction as the revolving field.
Solution;
𝑁𝑠 =
120𝑓
𝑃
=
120(60)
6
= 1,200 𝑟𝑝𝑚
a. @ Stand still
S=
𝑁𝑠 −𝑁𝑅
𝑁𝑠
=
1200−0
1200
=1
𝑓𝑅 = 𝑆𝑓 = (1)(60) ;
𝑓𝑅 = 60Hz
b. When the rotor turns in the same direction of the field, the motor speed is positive
S=
𝑁𝑠 −𝑁𝑅
𝑁𝑠
=
1200−500
1200
= 0.583
𝑓𝑅 = 𝑆𝑓 = (0.583)(60)
𝑓𝑅 = 35Hz
c. When the rotor turns in the opposite direction of the revolving field, the motor speed is
negative
S=
𝑁𝑠 −𝑁𝑅
𝑁𝑠
=
1200−(−500)
1200
= 1.417
𝑓𝑅 = 𝑆𝑓 = (1.417)(60)
𝑓𝑅 = 85Hz
Note: a slip greater than 1 implies that the motor is operating as a break. (ex. Lowering of load
in a tower crane)
d. The motor speed is positive because the rotor turns in the same direction as the revolving
speed.
S=
𝑁𝑠 −𝑁𝑅
𝑁𝑠
=
1200−2000
1200
= −0.6667
𝑓𝑅 = 𝑆𝑓 = (−0.6667)(60)
𝑓𝑅 = -40Hz
Note: a negative frequency means that the phase sequence of the voltage induce in the
rotor winding is reverse.
Note: a negative slip implies that the motor is operating as a generator.
Induction Motor Losses and Efficiency
Power Input 𝑃𝑡
𝑃2
𝑃𝑚
Rotor
Stator
RPI
Stator copper loss
Core loss
(𝐼2 ⋅ 𝑅𝑠𝑡𝑎𝑡𝑜𝑟 )
(𝐼𝑅2 ⋅ 𝑅𝑅 )
RPD
Rotor copper loss
𝑃2 = 𝑃1 – (stator copper and core losses)
𝑃𝑚 = 𝑃2 – (rotor copper loss)
RPI (Rotor Power Input) – active power supply to the rotor
RPD (Rotor Power Develop) – mechanical power develops
On per Phase Rotor Equivalent Circuit
𝐸𝑅 = 𝑆𝐸𝐵𝑅
Where;
𝑋2 = 2𝜋𝑓2 𝐿
𝑋𝑅 = 2𝜋𝑓𝑅 𝐿
𝑅𝑅 – rotor resistance unaffected by “f”
𝑋2 – reactance per phase at stand still
Friction and Windage loss
Output power of
motor in Hp
𝑋𝑃 – reactance per phase when running & affected by frequency in the rotor
On per Phase Basis
Rotor Power Input (RPI) - 𝐼𝑅2 ⋅
𝑅𝑅
𝑆
Rotor Copper Loss (RCL) - 𝐼𝑅2 ⋅ 𝑅𝑅 = 𝑅𝑃𝐼 ⋅ 𝑆
1−𝑆
)
𝑆
Rotor Power Develop (RPD) - 𝐼𝑅2 ⋅ 𝑅𝑅 (
= 𝑅𝑃𝐼(1 − 𝑆)
Sample Problem
5. Using the data of sample number 1 and 3, calculate, (a) RPI, (b) RCL, (c) RPD in Watts and (d) RPD
in horsepower
Solution;
𝑅𝑅
𝑆
(77.25)2 (0.1)
Given:
a.) RPI = 3𝐼𝑅2 ⋅
𝐼𝑅 = 77.25 A
b.) RCL = 𝑅𝑃𝐼 ⋅ 𝑆 = (0.075)(24.025) =
𝑅𝑅 = 0.1 Ω
c.) RPD = 𝑅𝑃𝐼(1 − 𝑆) = (24.025)(1 − 0.075) =
𝑆 = 0.075
d.) RPD = RPD⋅ 746𝑊 = 22278.625𝑊 746𝑊 =
=3⋅
0.075
ℎ𝑝
24.025 kW
=
ℎ𝑝
1.801875 kW
22,278.625
W kW
29.86 hp
6. A 3𝜙 induction motor having a nominal rating of 100hp and has a synchronous speed 𝑁𝑆 OF 1800
rpm is connected to a 600V source. The two-wattmeter method shows a total power of 70 kW
and an ammeter reading indicates a line current of 78A. Precise measurements give a rotor speed
𝑁𝑅 of 1763 rpm. In addition, the following characteristics are known about the motor; stator iron
loss= 2KN; friction and windage loss= 1.2KN; resistance between two terminals= 0.34 Ω. Calculate;
(a) power supplied to the rotor, (b) rotor copper loss, (c) mechanical power supplied to the load
in hp, (d) efficiency, (e) torque develop at 1763 rpm.
Solution;
𝑃𝑖𝑛 = 𝑃𝑡 = 𝑊1 + 𝑊2 = 70 𝑘𝑊
1
2
1
2
𝑅𝑆 = 𝑅 = (0.34) = 0.17Ω
∴ 𝑃𝑐𝑢𝑟 = 3𝐼 2 𝑅𝑆 = (3)(0.17)(78)2 = 3.1028 𝑘𝑊
a.) RPI = 𝑃𝑖𝑛 − (𝑃𝑐𝑢𝑟 + 𝑃𝑊 ) = 70 − (2 + 3.1028) =
b.) 𝑃𝑐𝑢𝑟 = 𝑅𝑃𝐼 ⋅ 𝑆
Where; S =
𝑁𝑠 −𝑁𝑅
𝑁𝑠
1800−1763
1800
=
= 0.02056
1.3343 kW
𝑃𝑐𝑢𝑟 = (64.8972) ⋅ (0.02056) =
c.) RPD = RPI - 𝑃𝑐𝑢𝑟 = 64.8972 − 1.3343 = 63.563 𝑘𝑊
𝑃𝑜𝑢𝑡 = 𝑅𝑃𝐷 − 𝑃𝑓𝑤 = 63.563 − 1.2
ℎ𝑝
𝑃𝑜𝑢𝑡 = 62.363𝑘𝑊 [0.746𝑘𝑊] =
d.) 𝓃 =
𝑃𝑜𝑢𝑡
𝑃𝑖𝑛
× 100% =
𝑅𝑃𝐼
e.) T = 9.55 𝑁 = 9.55
𝑠
62.363
×
70
82.597 hp
100% =
64.8972×103
1800
=
344.32N.m
Rotor Torque
 The force tending to produce rotation
2𝜋𝑁 𝑇
𝑅
Hp = 33,000
T = 7.04
𝑅𝑃𝐼
𝑁𝑆
𝑅𝑃𝐼
in lb.ft
in lb.ft
T = 9.55 𝑁 = 9.55
𝑆
𝑃𝑜𝑢𝑡
𝑁𝑆
in N.m
89.09%
64.8972 kW
7. Calculate the torque develop by the motor in the last example, where RPI is 24,000W, 𝑁𝑆 =
1200𝑟𝑝𝑚
Solution;
𝑅𝑃𝐼
𝑁𝑆
T = 7.04
2,400
= 7.04 1,200 =
140.8 lb.ft
Maximum Torque
𝑅
 Occurs when the slip is nearly equal to 𝑋 𝑅
𝐵𝑅
𝑅
𝑆𝑚𝑎𝑥𝑇 = 𝑋 𝑅
𝐵𝑅
𝐸2
𝑅𝑃𝐼𝑚𝑎𝑥𝑇 = 2𝑋𝐵𝑅
𝐵𝑅
𝐼𝑅𝑚𝑎𝑥𝑇 =
𝐸𝐵𝑅
√2𝑋𝐵𝑅
Sample Problems
8. Calculate the maximum torque that can develop by No. 1, (b) the speed at which the torque will
occurs.
Solution;
a.
𝐸2
𝑅𝑃𝐼 = 2𝑋𝐵𝑅 =
𝐵𝑅
𝑇𝑚𝑎𝑥 = 7.04
(3)(110)2
(2)(0.5)
𝑅𝑃𝐼𝑚𝑎𝑥
𝑁𝑆
= 36,300𝑊
36,300
= 7.04 ( 1,200 )
𝑇𝑚𝑎𝑥 = 213 𝑙𝑏. 𝑓𝑡
𝑅
0.1
b. 𝑆𝑚𝑎𝑥𝑇 = 𝑋 𝑅 = 0.5 = 0.2
𝐵𝑅
𝑁𝑅 =
120𝑓
𝑃
(1 − 𝑠) = 𝑁𝑆 (1 − 𝑆) = 1200 (1 −
𝑁𝑅 = 960 𝑟𝑝𝑚
0.1
)
0.5
Starting Torque
𝐼𝑅𝑠𝑡 =
𝐸𝐵𝑅
2 +𝑋 2
√𝑅𝑅
𝐵𝑅
𝑅𝑃𝐼𝑠𝑡 =
2
𝐸𝐵𝑅
2
2 )
(𝑅𝑅 +𝑋𝐵𝑅
- per phase
⋅ 𝑅𝑅
9. Calculate the starting torque develop by the motor of example no. 1.
Solution;
𝑅𝑃𝐼𝑠𝑡 =
2
𝐸𝐵𝑅
2 +𝑋 2 ) ⋅ 𝑅𝑅
(𝑅𝑅
𝐵𝑅
(3)(110)2
𝑅𝑃𝐼𝑠𝑡 = (0.1)2 (0.5)2 (0.1) = 13.96 𝑜𝑟 14 𝑘𝑊
14,000
14,000
𝑇𝑠𝑡 = 7.04 ( 1,200 ) = 82 𝑙𝑏. 𝑓𝑡 𝑜𝑟 9.55 ( 1,200 ) = 111.42 𝑁. 𝑚
Induction Motor Efficiency
%𝓃 =
𝑃𝑜𝑢𝑡
𝑃𝑖𝑛
𝑊𝑎𝑡𝑡 𝐿𝑜𝑠𝑠𝑒𝑠
× 100% = (1 − 𝑊𝑎𝑡𝑡 𝑜𝑢𝑡𝑝𝑢𝑡+𝑤𝑎𝑡𝑡 𝑙𝑜𝑠𝑠𝑒𝑠) × 100%
 The efficiency of induction motor as well as another operating characteristic may be determined
by performing three tests.
3. The Stator-Resistance Test
When the induction motor runs @ no-load, the slip is exceedingly small thus making
𝑅𝑅
𝑆
very large and 𝐼𝑅 becomes negligible compared to 𝐼𝑃 . Thus, at no-load, the circuit consist
essentially of the magnetizing branch 𝑅𝑚 & 𝑋𝑚 only.
𝑆𝑛𝑙 = √3 ⋅ 𝐸𝑛𝑙 ⋅ 𝐼𝑛𝑙
2
2
𝑄𝑛𝑙 = √𝑆𝑛𝑙
− 𝑃𝑛𝑙
𝐸2
𝐸2
𝑅𝑚 = 𝑃𝑛𝑙
𝑋𝑚 = 𝑄𝑛𝑙
𝑛𝑙
𝑛𝑙
Lock Rotor Test (Block Rotor Test)
-
Under rated line voltage, when the rotor is lock or blocked, stator current (𝐼𝑃 ) is almost
six times its rated value. Furthermore, the slip is equal to 1 which means that
𝑅𝑅
𝑆
is still 𝑅2
where 𝑅2 is the resistance of the rotor reflected into the stator and since 𝐼𝑃 is much
greater than compared to 𝐼1 thus the magnetizing branch can be neglected.
2
2
𝑄𝐿𝑅 = √𝑆𝐿𝑅
− 𝑃𝐿𝑅
𝑆𝐿𝑅 = √3 ⋅ 𝐸𝐿𝑅 ⋅ 𝐼𝐿𝑅
Assuming Wye Connected (per phase)
𝑅𝑒𝑞 =
𝑃𝐿𝑅
2
3𝐼𝐿𝑅
𝑋𝑒𝑞 =
𝑄𝐿𝑅
2
3𝐼𝐿𝑅
𝑍𝑒𝑞 =
𝐸𝐿𝑅
√3𝐼𝐵𝑅
DC Stator Resistance Test
𝑟1 =
𝑅
3
𝑟1 = 𝑅
2
2
𝑃𝐿𝑅 = 𝑊1 + 𝑊2
Note: voltage applied when doing lock rotor lost should be 20-30% of the rated voltage. The value
of rotor resistance and rotor reactance whether SCRIM or WRIM can be determined by
performing the so-called Lock Rotor Test
Sample Problem
1. A 50𝐻𝑝1 , 60Hz, 115V, 8-Pole 3𝜙 IM was tested and the following data were obtained:
𝐸𝑛𝑙 = 115𝑉
𝐸𝐿 = 115𝑉
𝐼𝑛𝑙 = 10𝐴
𝐼𝐿 = 27.3𝐴
𝑃1 = 725𝑊
𝑃1 = 3,140𝑊
𝑃2 = −425𝑊
𝑃2 = 1,570𝑊
𝑁𝑅 = 810 𝑟𝑝𝑚
DC resistance between terminals equal to 0.128 Ω. Calculate (a) hp output, (b) torque, (c)
%efficiency, (d) power factor of the motor for the given load value. Assume 1.25 times the DC
resistance.
Solution;
𝑅𝑒𝑞 =
0.128
(1.25)
2
= 0.08Ω
2
⋅ 𝑅𝑒𝑞 = 3(10)2 (0.08) = 24𝑊
𝑃𝑛𝑙 = 𝑃1 + 𝑃2 = 725 − 425 = 300𝑊
𝑃𝑓𝑤 + 𝑃𝑤 = 300 − 24 = 276𝑊
Load copper loss at the stator = 3𝐼𝐿2 ⋅ 𝑅𝑒𝑞 = (3)(0.08)(27.3)2 = 179𝑊
Total loss of stator under load = 𝑃𝑓𝑤 + 𝑃𝑤 + 𝑃𝑐𝑢𝑠𝐿 = 276 + 179 = 455𝑊
𝑅𝑃𝐼𝐿 = 𝑃𝑖𝑛 − 𝑃𝑙𝑜𝑠𝑠 = (3140 + 1570) − 455 = 4,255𝑊
𝑆=1−
𝑁𝑅 𝑃
120𝑓
(850)(8)
= 1 − (120)(60) = 0.10
RCL = S⋅RPI = (0.1) (4,255) = 425.5W
RPD = RPI – RCL = 4,125 – 425.5 = 3829.5W
a.
Hp = 3829𝑊
ℎ𝑝
760𝑊
𝑅𝑃𝐼
= 5.13ℎ𝑝
T = 7.04 𝑁 = 7.04
b.
𝑆
𝑃𝑙𝑜𝑠𝑠
c. %eff = (1 − 𝑃
𝑙𝑜𝑠𝑠 +𝑃𝑜𝑢𝑡
%eff = (1 −
(4255)
(900)
= 33.28 𝑙𝑏. 𝑓𝑡
) × 100%
455+425.5
)×
4710
100%
%eff = 81.30%
d. P.f =
𝑃𝑡𝑛
√3⋅𝐸𝐿 ⋅𝐼𝐿
=
4710
√3(110)(27.3)
P.f = 0.866 lagging
2. A block rotor test was performed in the rotor of the last example and the following data were
obtained: 𝐼𝐵𝑅 = 32𝐴; 𝑃1 = 1420𝑊 ; 𝑃2 = −860𝑊 ; 𝐸𝐵𝑅 = 26𝑉. Calculate; (a) 𝑅𝑒𝑞 , 𝑋𝑒𝑞 , 𝑍𝑒𝑞
(b) 𝑅𝑅 , 𝑋𝐵𝑅 (c) speed at which the maximum torque will occur.
Solution;
a. 𝑅𝑒𝑞 =
𝑍𝑒𝑞 =
𝑃1 +𝑃2
2
3𝐼𝐵𝑅
𝐸𝐿𝑅
√3𝐼𝐵𝑅
=
1420−860
3(32)2
=
26
√3(32)
=
=
0.1833 Ω
0.4691 Ω
2 − 𝑅2
𝑋𝑒𝑞 = √𝑍𝑒𝑞
𝑒𝑞
𝑋𝑒𝑞 √(0.4691)2 − (0.1833)2 =
0.4318 Ω
b. 𝑅𝑅 = 𝑅𝑒𝑞 − 𝑅𝑠𝑡𝑎𝑡𝑜𝑟 = 0.1833 − 0.08 =
𝑋𝐵𝑅 =
𝑋𝑒𝑞
2
=
0.13
2
=
0.2159 Ω
0.1033 Ω
c. @ max torque
𝑅
0.1033
𝑆 = 𝑋 𝑅 = 0.2159 = 0.4785
𝐵𝑅
𝑁𝑅𝑚𝑎𝑥𝑇 =
120𝑓
𝑃
(1 − 𝑠) =
(120)(60)
8
(1 − 0.4785)
𝑁𝑅𝑚𝑎𝑥𝑇 = 469.35 𝑟𝑝𝑚
Starting Method of Induction Motor
 Three methods are Generally Used:
1. Full-Voltage Starting
2. Reduced-Voltage Starting
3. Part Winding Starting
1. Full-Voltage Starting [Direct-On-Line (DOL)]
-
Direct applying the full voltage rating full voltage starting depends upon the following
factors:
a. Size and design of motors
b. Kind of application
c. Location of the motor in the distribution system
d. Capacity of the power system
2. Reduced-Voltage Starting
1. Compensator method – uses autotransformer to reduce voltage
2. Line resistance method
3. Wye-Delta method
1. Cheaper and more
2. Slightly higher efficiency and power factor
3. Are explosive group since the risk sparking is eliminated by the absence of slip ring process
1. Have a much highly starting torque
2. Have a much lower starting current
3. Have a means of varying speed by the use of wound rotor
Motor Operation
Before starting, inspect the following
 Foundation and base bolt are tight
 Insulation resistance
 Rating plate voltage & frequency agree with the power supply
 Correct phase sequence for desired rotation
 Each bearing is properly connected
 Motor is free of axial thrust
 Brushes contact to slip ring is perfectly with correct spring tension
 Rotors turns freely
 Any protective devices connected and operating properly
When motor is running, inspect the following:
 Vibration
 Unusual noise
 Unusual rise temperature
 Lubrication is sufficient
 Brushes contact (whether it is producing spark)
 Proper ventilation
Starting of Induction Motor
-
At the instant the motor is started, the rotor is not running or turning, therefore, the
following friction and windage loss is zero. However, the iron loss at starting is greater
than the normal operating value because the rotor frequency 𝑓𝑅 is the same. Usually,
starting current is slots times the full-load current.
Sample Problem:
3. Calculate the starting torque of the induction motor in the previous problem.
Solution;
Iron loss = 276W
𝐸𝐵𝑅1
𝐸𝐵𝑅2
By ratio and proportion:
𝐸
𝐵𝑅2
115
26
𝐼𝐵𝑅2 = 𝐼𝐵𝑅1 𝐸𝐵𝑅2 = 32
𝐵𝑅1
@ 115V w/ rotor block =
𝐼
= 𝐼𝐵𝑅1
2
𝑉𝑟𝑎𝑡𝑒𝑑
𝑉𝐵𝑅
= 142𝐴
⋅ 𝑃𝐵𝑅 =
(115)2
(570)
20
= 11,160𝑊
Stator copper loss
2
𝑃𝑐𝑢𝑠 = 3𝐼𝐵𝑅
⋅ 𝑅𝑒𝑞 = 3(142)2 (0.08) = 4830𝑊
𝐸𝐵𝑅1 𝑅𝑃𝐼 = 𝑃𝑖𝑛 − (𝑃𝑖𝑟𝑜𝑛 + 𝑃𝑐𝑢𝑠 ) = 11,160 − (4830 + 276) = 6,054𝑊
Solving for starting torque
𝑇𝑠𝑡 = 7.04
4.
𝑅𝑃𝐼
𝑁𝑠
= 7.04
6054
900
=
47.3 lb.ft
The following information is given in connection with a 50hp, 440V, 1,160rpm induction motor,
full-load torque and current at 44Ov are 227 lb.ft and 63A respectively. Starting torque and
current @440V are 306lb.ft and 362A. (a) calculate the starting current and torque at 254V. (b)
what percentage or rated values are calculated in (a)?
Solution;
a.
𝐸254
𝐸440
𝐼
= 𝐼254
440
𝐼254 =
b. %𝐼 =
𝑇𝑟𝑎𝑡𝑒𝑑
𝑇254
(363)(254)
440
209
×
63
=(
=
100% =
209 A
332 %
𝑉𝑟𝑎𝑡𝑒𝑑 2
)
𝑉𝐿
254 2
𝑇254 = 306 [440] =
102
%𝑇 = 227 × 100% =
102lb.ft
45 %
Induction Motor Characteristics Under Various Load Conditions
Formula:
𝑆=
𝐾𝑇𝑅
𝐸2
Where;
K- constant;
T- torque ;
𝑇
𝑅
𝐸
𝑆𝑥 = 𝑆𝑛 [𝑇𝑥 ] [𝑅𝑥 ] [𝐸𝑛 ]
𝑛
𝑛
R – rotor resistance ;
S – slip
2
𝑥
n subscript refers to the initial or given load condition (the given condition may correspond to the nominal
rating of the rotor)
x subscript refers to the new load condition
Note: in applying the formula, the only restriction is that the new torque 𝑇𝑥 must not be greater than
𝐸
𝑇𝑛 [𝐸𝑛 ]
2
𝑥
European Standard: (380V. 50Hz- 3𝜙)
Sample Problem:
1. A 3𝜙 208V induction motor having a synchronous speed 𝑁𝑠 of 1,200 rpm runs @ 140 rpm when
connected to a 215V line. Calculate the speed if the voltage increases to 240V.
Solution:
𝑆=
𝑁𝑠 −𝑁
𝑁𝑠
=
1200−1140
1200
= 0.05
When the voltage rises to 240V, the load torque and rotor resistance remains the same.
𝐸
2
215 2
𝑆𝑥 = 𝑆𝑛 (𝐸𝑛 ) = 0.05 (240) = 0.04
𝑥
The slip speed is 0.04(1200)=48 rpm
The new speed @ 240V
𝑁𝑥 = 1200 − 48
𝑁𝑥 = 1152 𝑟𝑝𝑚
Another solution:
𝑆𝑥 =
𝑁𝑠 −𝑁𝑥
𝑁𝑠
→ 𝑁𝑥 = 𝑁𝑠 (1 − 𝑠) = 1200(1 − 0.04)
𝑁𝑥 = 1152 𝑟𝑝𝑚
2. A 3-phase IM driving a compressor runs at 873 rpm immediately after it is connected to a fixed
460V, 60Hz line. The initial cold rotor temperature is 23 degrees Celsius. The speed drops to 864
rpm after the machine has run for several hours. Calculate (a) the hot rotor resistance in terms of
the cold resistance (b) the approximate hot temperature of the rotor bars knowing they are made
of copper.
Solution:
𝑁𝑠 = 900𝑟𝑝𝑚
a.) Initial and Final slips are;
𝑆𝑛 =
900−873
900
= 0.03
𝑆𝑥 =
900−864
900
= 0.04
The voltage and torque are fixed. Consequently, the speed change is entirely due to the
change of the rotor resistance.
𝑅
𝑅
𝑆𝑥 = 𝑆𝑛 (𝑅𝑥 ) ; 0.04=0.03[𝑅𝑥 ]
𝑛
𝑛
𝑅𝑥 = 1.33𝑅𝑛
∴ the hot rotor resistance is 33% greater than the cold resistance
b.) The hot rotor temperature is:
𝑅2
𝑅1
=
𝑇+𝑡2
𝑇+𝑡1
Where; 𝑡1 = 23°𝐶
𝑅
𝑡2 = [𝑅2 (𝑇 + 𝑡1 )] − 𝑇
1
Note: 𝑇𝑐𝑢 = 234℃ (referred to 1001 EE p.4 & Elec. Machines p.121)
𝑡2 = [
1.33𝑅1
(234
𝑅1
+ 23)] − 234
𝑡2 = 107.81℃
Remember:
Assume the synchronous speed of the motor is nearest to its operating speed
(rotor speed).
3. A 3-phase wound rotor IM has a rating of 110kW, 1760rpm. 2.3kV 60Hz. Three external resistance
of 2 Ω are connected in wye, across the rotor slip ring. Under this condition, the motor develops
a torque of 300N.m at the speed of 1000rpm. Calculate (a) the speed for a torque of 400N.m (b)
the value of the external resistance so that the motor develops 10kW at 200rpm.
Illustration
Solution:
𝑇𝑛 = 300𝑁. 𝑚
𝑆𝑛 =
𝑁𝑠 −𝑁𝑟
𝑁𝑠
=
1800−1760
1800
𝑆𝑛 = 0.444
Since the voltage and resistance are fixed;
𝑇
400
𝑆𝑥 = 𝑆𝑛 (𝑇𝑥 ) = 0.444 [300] = 0.592
𝑛
∴ the speed is;
𝑁𝑥 = 𝑁𝑠 (1 − 𝑠) = 1800(1 − 0.592)
𝑁𝑥 = 733 𝑟𝑝𝑚
The torque corresponds to 10Kw at 200rpm is,
𝑇𝑥 = 9.55
𝑃𝑜𝑢𝑡
𝑁𝑟
10,000
)
200
= 9.55 (
= 478𝑁. 𝑚
The rated torque of the motor is;
𝑇𝑥 = 9.55
𝑃𝑜𝑢𝑡
𝑁𝑟
110,000
)
1760
= 9.55 (
The slip is;
𝑆𝑥 =
𝑁𝑠 −𝑁𝑥
𝑁𝑠
=
1800−200
1800
= 0.89
= 597𝑁. 𝑚
𝑆𝑥
𝑆𝑛
𝑇
𝑅
0.89
478
𝑅
= [𝑇𝑥 ] [𝑅𝑥 ]; 0.444 = [300] { 2𝑥 }
𝑛
𝑛
𝑅𝑥 = 2.5Ω
4. A 3-phase, 4 pole, 1740 rpm, 440V, 60Hz induction motor is used in a 3-phase, 380V 50Hz supply.
Determine the speed if the torque is the same.
Solution:
𝑁𝑠 =
𝑆𝑥 =
𝑆𝑥 =
120𝑓
120(60)
=
= 1800 𝑟𝑝𝑚
𝑃
4
𝑁𝑠 −𝑁𝑟
1800−1740
=
= 0.033
𝑁𝑠
1800
𝐸 2
440 2
𝑆𝑛 (𝐸𝑛 ) = 0.033 (380) = 0.04424
𝑥
@ 50Hz supply;
𝑁𝑠 =
120𝑓
𝑃
120(50)
= 4 = 1500 𝑟𝑝𝑚
∴ 𝑁𝑟 = 𝑁𝑠 (1 − 𝑠) = 1500(1 − 0.04424)
𝑁𝑟 = 1433.64 𝑟𝑝𝑚
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