Upload
mrsplooge
View
107
Download
3
Tags:
Embed Size (px)
Citation preview
Induction Machines 1
Induction Motor Stator-2 pole
concentrated winding
Induction Machines 2
Induction motor stator- Distributed
4 pole lap winding
Ref: Electrical Machine Design by A.K. Sawhney
& A. Chakrabarti
Induction Machines 3 1 Cycle
Amp
time t0
t1 t2 t3 t4
t01 t12 Currents in different phases of AC Machine
Induction Machines 4
Axis of phase a
aβ aβ
-90 -40 10 60 110 160 210 260 -1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Fa
Space angle (Σ¨) in degrees
t0
t01
t12
t2
a
MMF Due to βaβ phase current
Induction Machines 5
a Fc
-93 10 113 216 -1.5
-1
-0.5
0
0.5
1
1.5
aβ
cβ bβ
b c
a
aβ
cβ bβ
b c
a
aβ
cβ bβ
b c
a
aβ
cβ bβ
b c
Fb
Fa F
Fb Fc
F
Fa
F
Fb
Fc Fc Fb
F
Space angle () in degrees
F Fa Fc
Fb
t = t0= t4
t = t1 t = t2 t = t3
t = t0= t4
RMF(Rotating Magnetic Field)
Induction Machines 6
Mathematical expression for RMF
;sec/2
);(#
2
3
;)2
cos(),(
radinvoltageappliedoffrequencyf
bookthewithconsistentbetospoleofP
INF
tP
FtF
or
msm
m
3
2
2cos
3
2cos
3
2
2cos
3
2cos)
2cos()cos(),(
PtIN
PtIN
PtINtF
ms
msms
Speed of RMF (synchronous speed)
The RMF will become constant if
02
tP
That can happen if the rate of change of
.sec/.2
1 radPdt
d
is
This is called the synchronous speed.
rpmP
f
P
f
Pdt
dn
120
2*
60*2*2
2*
60*2
2
601
In rpm,
7 Induction Machines
Induction Machines 8
Mathematical expression and speed of
RMF with space and time harmonics
nP
h
hn
thnP
FtF
nh
mnh
2by given is speed RMF thegeneralIn
6. slidein described procedure thefollowingsign out the findcan Onenegative are they etc. 5,11,17For
positive arethey etc.1,7,13,For.harmonicsondependentis RMF theofrotationofspeedThe
...,5,7,6,11.harmonics1timeoforder....1,5,7,6,11 harmonicsspaceoforder
;)2
cos(),(
Induction Machines 9
RMF- How it looks for different poles
Induction Machines 10
# of poles βversus synchronous speed in
rpm for a machine excited with 60 Hz
n1
Induction Machines 11
Induction Motor
β’Most popular motor today in the low and medium horsepower range
β’Very robust in construction
β’Speed easily controllable using V/f or Field Oriented Controllers
β’Have replaced DC Motors in areas where traditional DC Motors
cannot be used such as mining or explosive environments
β’Of two types depending on motor construction: Squirrel Cage
or Slip Ring
β’Only Disadvantage: Most of them run with a lagging power factor
Induction Machines 12
Squirrel Cage Rotor
Induction Machines 13
Slip Ring Rotor
β’The rotor contains windings similar to stator.
β’The connections from rotor are brought out using slip rings that
are rotating with the rotor and carbon brushes that are static.
Induction Machines 14
Torque Production in an Induction Motor
β’In a conventional DC machine field is stationary and the
current carrying conductors rotate.
β’We can obtain similar results if we make field structure
rotating and current carrying conductor stationary.
β’In an induction motor the conventional 3-phase winding
sets up the rotating magnetic field(RMF) and the rotor
carries the current carrying conductors.
β’An EMF and hence current is induced in the rotor due to
the speed difference between the RMF and the rotor,
similar to that in a DC motor.
β’This current produces a torque such that the speed
difference between the RMF and rotor is reduced.
Induction Machines 15
Slip in Induction Motor
β’However, this speed difference cannot become zero because that
would stop generation of the torque producing current itself.
β’The parameter slip βsβ is a measure of this relative speed difference
1
1
1
1
mm
n
nns
where n1,1 are the speeds of the RMF in RPM and rad/sec
respectively
nm,m are the speeds of the motor in RPM and rad/sec respectively
β’The angular slip frequency and the slip frequency at which voltage
is induced in the rotor is given by
sff,s slipslip
Induction Machines 16
Induction Motor Example 1
A 100 hp, 8 pole, 60 Hz, 3 phase induction motor runs at 891 rpm
under full load. Determine the i) synchronous speed in rpm, ii) slip, iii)
slip frequency at full load. Also, iv) estimate speed if load torque becomes
half of full load torque, given the fact that torque is proportional to
slip in the region between breakdown torque or zero torque.
Solution:
i) Synchronous speed= π1 = 120 π
π=
120 . 60
8 = 900
rpm.
ii) Slip = π =π1βππ
π1=
900β891
900=
9
900= 0.01
iii) Slip frequency = π . π = 0.01 .60 = 0.6 Hz.
Example I: Solution
Induction Machines 17
T
S, nm
Tstart
Tmax
}Toruqe is proportional to
slip between maximum
(break down) torque and zero torque
Tfull load
iv) Following the figure above
ππΉπΏ = πΎπ πΉπΏ and ππΉπΏ/2 = πΎπ πΉπΏ/2 where βFLβ implies full load and βKβ is the constant of proportionality.
β΄ π πΉπΏ/2 =ππΉπΏ/2
ππΉπΏπ πΉπΏ =
π πΉπΏ2
= 0.005
β΄ speed at half load torque= 1 β π πΉπΏ2
ππ = 1 βπ πΉπΏ
2 ππ = 1 β 0.005 900 = 895.5 rpm.
Induction Machines 18
Rotor Equivalent Circuit
R2
sE2
jsX2
I2
E2
jX2
R2/sI2
The equation for the circuit on the left is
sE2=I2(R2+jsX2)
Dividing both sides by s we can obtain
E2=I2(R2/s+jX2)
Induction Machines 19
Induction Motor Equivalent Circuit
Note: Prot includes the iron losses along with the rotational losses. The sum of the
iron and rotational losses and hence Prot is constant over the normal operating range at rated
frequency of operation. This is because as the speed increases the rotational losses increase, but
the rotor iron losses decrease as the slip frequency is lower. The converse is true when speed
decreases.
V1
I1
R1
jX1 jX2'
R2β
((1-s)/s)R2βjXm
I2β
Im
Pin=3|V1||I1|cosf Pag
P1=3|I1|2R1
Stator copper
loss
Pdev =
3|I2'|2{(1-s)/s}R2
'Pout
P2=3|I2'|2R2
'
Rotor copper
loss
Prot
Rotational loss
Power Relationships
ssPPP
ssPP
PPP
devag
dev
devag
1::1::
1::
2
2
2
20 Induction Machines
Induction Motor Example 2 A, 3 phase, 15 hp, 460V, 4 pole, 60 Hz,1728 rpm IM delivers full
output power to a load connected to its shaft. Windage and friction
losses amount to 750W. Determine i) gross output(mechanical)
power, ii) Air-gap power and iii) Rotor copper loss.
Solution:
(i) Full load shaft power, πππ’π‘ = 15 hp = 15 * 746 =11190 W.
Gross ouput power = ππππ£ = Shaft power + windage and friction loss = 11190+ 750 = 11940 W.
(ii) Synchronous speed= π1 = 120 π
π=
120β 60
4 = 1800 rpm.
Slip = π =π1βππ
π1=
1800β1728
1800=
72
1800= 0.04
πππ =ππππ£1 β π
=11940
1 β 0.04= 12437.5 W
(iii) Rotor copper loss = π2 = π πππ = 0.04 β 12437.5 = 497.5 W.
OR
π2 = πππ β ππππ£ = 12437.5 β11940 W =497.5 W.
21 Induction Machines
Induction Motor Example 3
Induction Machines 22
A 30 hp, 4 pole, 440V,60 Hz delta connected induction motor
has a rotational loss of 900W at1764 rpm. Find the pf, line
current, output power, air-gap power, copper loss, output torque,
efficiency at 1764 rpm. R1 =1.2 ohms,
R2'=0.6 ohms X1=2 ohms, X2'=0.8 ohms , Xm=50 ohms.
Solutions:
V1 =440<00V
I1
R1=1.2 Ξ©
jX1=j2 Ξ© jX2β=j0.8 Ξ©
R2β=0.6 Ξ©
((1-s)/s)R2β =29.4Ξ©jXm= j50 Ξ©
I2β
Im
Induction Motor Example 3 (contd..)
Induction Machines 23
Synchronous speed= ππ = 120 π
π=
120 . 60
4 = 1800 rpm.
Slip = π =ππ βππ
ππ =
1800β1764
1800=
36
1800= 0.02.
β΄1βπ
π π 2β² =
1β0.02
0.02 . 0.6 = 29.4 Ξ©.
From the above figure
π1 = π 1 + ππ1 + πππ . 1 β π
π π 2β² + π 2
β² + ππ2β² / πππ +
1 β π
π π 2β² + π 2
β² + ππ2β²
=22.75+j 15.51 Ξ© =27.53β 34.290 Ξ©
β΄ πΌ1 = π1
π1= 440β 00
27.53β 34.290 = 15.98β β 34.290A
β΄ ππ = πππ π = cos 34.29 = 0.83 lag.
Line current = 3 πΌ1 = 27.68 A.
Induction Motor Example 3 (contd..)
Induction Machines 24
Input power = πππ = 3 π1 πΌ1 πππ π = 3. 440. 15.98 . 0.83 = 17.43 kW.
πΌ2β² =
π1βπΌ1(π 1+ππ1)
π 2β²
π +ππ2
β²=
440β 00β15.98β β34.290 .(1.2+π2)
(30+π0.8)= 13.55β β 3.730 A
Output power =πππ’π‘ = ππππ£ β π2 = 3 πΌ2β² 2 1βπ
π π 2β² β 900 = 16193.74 β 900 = 15293.74 W
Copper loss = π1 + π2 = 3 πΌ12π 1 + 3 πΌ2
β² 2π 2β² = 3 β 15.982 β 1.2 + 3 β 13.552 β 0.6 = 919.30 +
330.48 = 1249.78 W
Output Torque = πππ’π‘ =πππ’π‘
ππ=
15293 .74
2πβ 1764
60
= 82.79 π βπ.
% Efficiency =π =πππ’π‘
πππ. 100 =
15293 .74
17430. 100 = 87.74 %
πππ = πππ β π2=17430-3 β 15.982 β 1.2 =17430-919.30=16510.7 W.
Three Phase Table
Induction Machines 25
Relationships of balanced three phase circuits
Star Delta
Voltage (V) ππΏβπΏ = 3ππβ ππΏβπΏ = ππβ
Current (A) πΌπΏ = πΌπβ πΌπΏ = 3πΌπβ
Active Power (P) (W) 3ππΏβπΏπΌπΏ cosπ
=3ππβ πΌπβ cosπ 3ππΏβπΏπΌπΏ cosπ
=3ππβ πΌπβ cosπ
Reactive Power (Q)
(VAR) 3ππΏβπΏπΌπΏ sinπ
=3ππβ πΌπβ sinπ
3ππΏβπΏπΌπΏ sinπ
=3ππβ πΌπβ sinπ
Apparent Power(VA) 3ππΏβπΏπΌπΏ=3ππβπΌπβ 3ππΏβπΏπΌπΏ=3ππβπΌπβ
Note: ππΏβπΏ= Line-Line Voltage, πΌπΏ= Line Current, ππβ= Phase
voltage, πΌπβ= Phase Current, cosπ = Power Factor, π = Power Factor
Angle.