ELG3311: Assignment 3
Problem 6-12:
The Y-connected synchronous motor whose nameplate is shown in Figure 6-21 has a perunit synchronous reactance of 0.9 and a per-unit resistance of 0.02.
(a) What is the rated input power of this motor?
(b) What is the magnitude of EA at rated conditions?
(c) If the input power of this motor is 10 MW, what is the maximum reactive power the
motor can simultaneously supply? Is it the armature current or the field current that limits
the reactive power output?
(d) How much power does the field circuit consume at the rated conditions?
(e) What is the efficiency of this motor at full load?
(f) What is the output torque of the motor at the rated conditions? Express the answer
both in newton-meters and in pound-feet.
Solution:
The base quantities for this motor are:
VT ,base = 6600V
Vφ ,base =
VT
3
=
I A,base = I L ,base
6600
= 3810V
3
= 1404 A
S base = Prated = 3VT I L PF = 3 (6600)(1404)(1.0) = 16.05MW
1
(a) The rated input power of this motor is
Pin =
3V T I L PF =
3 (6600)(1404)(1.0) = 16.05MW
(b) At rated conditions, Vφ , pu = 1.0∠0° pu and I φ , pu = 1.0∠0° pu , so EA is given in per-
unit quantities as
E A, pu = Vφ , pu − R A I φ , pu − jX S I φ , pu = (1∠0°) − (0.02)(1.0∠0°) − j (0.90)(1∠0°)
E A, pu = 1.33∠ − 42.6° pu
And the EA is
E A = E A, puVφ ,base = (1.33∠ − 42.6°)(3810) = 5067∠ − 42.6°V
(c) From the capability diagram, we know that there are two possible constraints on the
maximum reactive power – the maximum stator current and the maximum rotor current.
We will have to check each one separately, and limit the reactive power to the lesser of
the two limits.
The stator apparent power limit defines a maximum safe stator current. This limit is the
same as the rated input power for this motor, since the motor is rated at unity power
factor. Therefore, the stator apparent power limit is 16.05 MVA. If the input power is
10MW, then the maximum power that still protects the stator current is
Q = S 2 − P 2 = (16.05) 2 − (10) 2 = 12.6MVAR
Now we must determine the rotor current limit. The per-unit power supplied to the motor
is 10 MW / 16.05 MW = 0.623. The maximum EA is 5067 V or 1.33 pu, so with EA set to
maximum and the motor consuming 10 MW, the torque angle (ignoring armature
resistance) is
δ = sin −1
XSP
(0.90)(0.623)
= sin −1
= 24.9°
(1.0)(1.33)
3Vφ E A
At rated voltage and 10 MW of power supplied, the armature current will be
IA =
Vφ − E A
R A + jX S
=
1∠0° − 1.33∠ − 24.9°
= 0.663∠20.2° pu
j 0.90
In actual amps, this current is
2
I A = I A,base I A, pu = (1404)(0.663∠20.2°) = 931∠20.2° A
The reactive power supplied at the conditions of maximum EA and 10 MW power is
Q = 3Vφ I A sin θ = 3(3810)(931)(sin 20.2°) = 3.68MVAR
Therefore, the field current limit occurs before the stator current limit for these
conditions, and the maximum reactive power that the motor can supply is 3.68 MVAR
under these conditions.
(d) At the rated conditions, the field circuit consumes
Pfield = VF I F = (125)(5.2) = 650W
(e) The efficiency of this motor at full load is
η=
Pout
(21000hp)(746W / hp)
× 100% =
× 100% = 97.6%
Pin
16.05MW
(f) The output torque in SI and English units is
τ load =
Pout
τ load =
5252 Pout (5252)(21000hp)
=
= 91910lb ⋅ ft
nm
1200r / min
ωm
=
(21000hp)(746W / hp )
= 124700 N ⋅ m
(1200r / min)(1min/ 60s )(2πrad / 1r )
Problem 6-15:
A 100-hp, 440-V, 0.8-PF-leading, ∆-connected synchronous motor has an armature
resistance of 0.22 Ω and a synchronous reactance of 3.0 Ω. Its efficiency at full load is 89
percent.
(a) What is the input power to the motor at rated conditions?
(b) What is the line current of the motor at rated conditions? What is the phase current of
the motor at rated conditions?
(c) What is the reactive power consumed by or supplied by the motor at rated conditions?
(d) What is the internal generated voltage EA of this motor at rated conditions?
(e) What are the stator copper losses in the motor at rated conditions?
(f) What is Pconv at rated conditions?
(g) If EA is decreased by 10 percent, how much reactive power will be consumed by or
supplied by the motor?
3
Solution:
(a) The input power to the motor at rated conditions is
Pin =
Pout
=
η
(100hp )(746W / hp)
= 83.8kW
0.89
(b) The line current to the motor at rated conditions is
IL =
P
83.8k
=
= 137 A ,
3VT PF
3 (440)(0.8)
θ = cos −1 PF = cos −1 0.8 = 37.86°
The phase current to the motor at rated conditions is
Iφ =
IL
3
=
137
= 79.4 A
3
(c) The reactive power supplied by this motor to the power system at rated conditions is
Qrated = 3Vφ I A sin θ = 3(440)(79.4)(sin 36.87°) = 62.9kVAR
(d) The internal generated voltage at rated conditions is
E A = Vφ − R A I A − jX S I A = 440∠0° − (0.22)(79.4∠36.87°) − j (3.0)(79.4∠36.87°)
E A = 603∠ − 19.5°V
(e) The stator copper losses at the rated conditions are
2
Pcu = 3I A R A = 3(79.4) 2 (0.22) = 4.16kW
(f) Pconv at rated conditions is
Pconv = Pin − Pcu = 83.8k − 4.16k = 79.6kW
(g) If EA is decreased by 10%, the new value of EA is
E A 2 = 90% E A1 = (0.9)(603) = 543V
Then the quantity EAsinδ is constant as EA changes, so
4
 E A1 sin δ 1 
 (603) sin( −19.5°) 
 = sin −1 
 = −21.8°
543


 E A2 
δ 2 = sin −1 
Therefore
IA =
Vφ − E A
jX S
=
440∠0° − 543∠ − 21.8°
= 70.5∠17.7° A
j 3.0
The reactive power supplied by the motor to the power system will be
Q = 3Vφ I A sin θ = 3(440)(70.5)(sin 17.7°) = 28.3kVAR
Problem 7-7:
A 208-V, two-pole, 60-Hz, Y-connected wound-rotor induction motor is rated at 15-hp.
Its equivalent circuit components are
R2 = 0.120Ω
X M = 15.0Ω
R1 = 0.200Ω
X 1 = 0.410Ω
X 2 = 0.410Ω
Pcore = 180W
Pmisc ≈ 0
Pmech = 250W
For a slip of 0.05, find
(a) The line current IL
(b) The stator copper losses PSCL
(c) The air-gas power PAG
(d) The power converted from electrical to mechanical form Pconv
(e) The induced torque τind
(f) The load torque τload
(g) The overall machine efficiency
(h) The motor speed in revolutions per minute and radians per second
Solution:
The equivalent circuit of this induction motor is shown below:
5
(a) The easiest way to find the line current (or armature current) is to get the equivalent
impedance ZF of the rotor circuit in parallel with jXM, and then calculate the current as
the phase voltage divided by the sum of the series impedances, as shown below:
The equivalent impedance of the rotor circuit in parallel with jXM is
ZF =
1
1
1
+
jX M Z 2
=
1
1
1
+
j15 2.40 + j 0.41
= 2.220 + j 0.745 = 2.34∠18.5°Ω
The phase voltage is
Vφ = VT / 3 = 208 / 3 = 120V
So the line current IL is
IL = IA =
Vφ
R1 + jX 1 + RF + jX F
=
120∠0°
= 44.8∠ − 25.5° A
0.20 + j 0.41 + 2.22 + j 0.745
(b) The stator copper losses are
2
PSCL = 3I A R1 = 3(44.8) 2 (0.20) = 1205W
R2
2
= 3I A RF
s
2
2 R
(Note that 3I A RF is equal to 3I 2 2 , since the only resistance in the original rotor
s
circuit was R2/s, and the resistance in the Thevenin equivalent circuit is RF. The power
consumed by the Thevenin equivalent circuit must be the same as the power consumed
by the original circuit.)
2 R
2
PAG = 3I 2 2 = 3I A RF = 3(44.8) 2 (2.220) = 13.4kW
s
(c) The air gas power is PAG = 3I 2
2
6
(d) The power converted from electrical to mechanical form is
Pconv = (1 − s ) PAG = (1 − 0.05)(13.4k ) = 12.73kW
(e) The induced torque in the motor is
τ ind =
PAG
ω sync
=
13.4k
120(60)
(
)(1min/ 60s )(2πrad / 1r )
2
= 35.5 N ⋅ m
(f) The output power of this motor is
Pout = Pconv − Pmech − Pcore − Pmisc = 12.73k − 250 − 180 − 0 = 12.3kW
The output speed is
120(60)
nm = (1 − s )nsync = (1 − 0.05)(
) = 3420r / min
2
Therefore the load torque is
τ load =
Pout
ωm
=
12.3k
= 34.3 N ⋅ m
(3420r / min)(1min/ 60 s )(2πrad / 1r )
(g) The overall efficiency is
η=
Pout
Pout
12.3k
× 100% =
× 100% =
× 100% = 84.5%
3Vφ I A cosθ
Pin
3(120)(44.8)(cos 25.5°)
(h) The motor speed in revolutions per minute is 3420 r/min. The rotor speed in radians
per second is
ω m = (3420r / min)(2πrad / 1r )(1 min/ 60s ) = 358rad / s
Problem 7-14:
A 440-V, 50-Hz, two-pole, Y-connected induction motor is rated at 75-kW. The
equivalent circuit parameters are
7
R1 = 0.075Ω
R2 = 0.065Ω
X 1 = 0.17Ω
X 2 = 0.17Ω
X M = 7.2Ω
Pcore = 1.1kW
PF &W = 1.0kW
Pmisc = 150W
For a slip of 0.04, find
(a) The line current IL
(b) The stator power factor
(c) The rotor power factor
(d) The stator copper losses PSCL
(e) The air-gas power PAG
(f) The power converted from electrical to mechanical form Pconv
(g) The induced torque τind
(h) The load torque τload
(i) The overall machine efficiency η
(j) The motor speed in revolutions per minute and radians per second
Solution:
The equivalent circuit of this induction motor is shown below:
(a) The easiest way to find the line current (or armature current) is to get the equivalent
impedance ZF of the rotor circuit in parallel with jXM, and then calculate the current as
the phase voltage divided by the sum of the series impedances, as shown below:
8
The equivalent impedance of the rotor circuit in parallel with jXM is
ZF =
1
1
1
+
jX M Z 2
=
1
1
1
+
j 7.2 1.625 + j 0.17
= 1.539 + j 0.364 = 1.58∠13.2°Ω
The phase voltage is
Vφ = VT / 3 = 440 / 3 = 254V
So the line current IL is
IL = IA =
Vφ
R1 + jX 1 + RF + jX F
=
254∠0°
= 149.4∠ − 18.3° A
0.075 + j 0.17 + 1.539 + j 0.364
(b) The stator power factor is
PFS = cos(18.3°) = 0.949lagging
(c) To find the rotor power factor, we must find the impedance angle of the rotor
θ R = tan −1
X2
0.17
= tan −1
= 5.97°
R2 / s
1.625
Therefore the rotor power factor is
PFR = cos(5.97°) = 0.995lagging
(d) The stator copper losses are
2
PSCL = 3I A R1 = 3(149.4) 2 (0.075) = 1675W
R2
2
= 3I A RF
s
R
2
2
(Note that 3I A RF is equal to 3I 2 2 , since the only resistance in the original rotor
s
circuit was R2/s, and the resistance in the Thevenin equivalent circuit is RF. The power
consumed by the Thevenin equivalent circuit must be the same as the power consumed
by the original circuit.)
(e) The air gas power is PAG = 3I 2
2
9
PAG = 3I 2
2
R2
2
= 3I A RF = 3(149.4) 2 (1.539) = 103kW
s
(f) The power converted from electrical to mechanical form is
Pconv = (1 − s) PAG = (1 − 0.04)(103k ) = 98.9kW
(g) The induced torque in the motor is
τ ind =
PAG
ω sync
=
103k
120(50)
(
)(1 min/ 60 s)(2πrad / 1r )
2
= 327.9 N ⋅ m
(h) The output power of this motor is
Pout = Pconv − Pmech − Pcore − Pmisc = 98.9k − 1.0k − 1.1k − 150 = 96.6kW
The output speed is
120(50)
nm = (1 − s)nsync = (1 − 0.05)(
) = 2880r / min
2
Therefore the load torque is
τ load =
Pout
ωm
=
98.9k
= 327.6 N ⋅ m
(2880r / min)(1min/ 60s )(2πrad / 1r )
(i) The overall efficiency is
η=
Pout
Pout
96.6k
× 100% =
× 100% =
× 100% = 89.4%
Pin
3Vφ I A cos θ
3(254)(149.4)(cos18.3°)
(j) The motor speed in revolutions per minute is 2880 r/min. The rotor speed in radians
per second is
ω m = (2880r / min)(2πrad / 1r )(1min/ 60s) = 301.6rad / s
10
Problem 7-19:
A 460-V, four-pole, 50-hp, 60-Hz, Y-connected, three-phase induction motor develops its
full-load induced torque at 3.8 percent slip when operating at 60 Hz and 460 V. The perphase circuit model impedance of the motor are
R1 = 0.33Ω
X M = 30Ω
X 1 = 0.42Ω
X 2 = 0.42Ω
Mechanical, core, and stray losses may be neglected in this problem.
(a) Find the value of the rotor resistance R2.
(b) Find τmax, smax and the rotor speed at maximum torque for this motor.
(c) Find the starting torque of this motor.
(d) What code letter factor should be assigned to this motor?
Solution:
The equivalent circuit for this motor is
The Thevenin equivalent of the input circuit is
Z TH =
jX M ( R1 + jX 1 ) ( j 30)(0.33 + j 0.42)
=
= 0.321 + j 0.418 = 0.527∠52.5°Ω
R1 + j ( X 1 + X M ) 0.33 + j (0.42 + 30)
VTH =
jX M
j 30
Vφ =
(265.6∠0°) = 262∠0.6°V
0.33 + j (0.42 + 30)
R1 + j ( X 1 + X M )
(a) If losses are neglected, the induced torque in a motor is equal to its load torque. At full
load, the output power of this motor is 50 hp and its slip is 3.8%, so the induced torque is
120(60)
nm = (1 − s )nsync = (1 − 0.038)(
) = 1732r / min
4
(50hp)(746W / hp)
τ ind = τ load =
= 205.7 N ⋅ m
(1732r / min)(1min/ 60 s )(2πrad / 1r )
11
The induced torque is given by the equation
τ ind =
2
ω sync [( RTH
3VTH R2 / s
+ R2 / s ) 2 + ( X TH + X 2 ) 2 ]
Substituting known values and solving for R2/s yields
205.7 =
3(262) 2 R2 / s
(188.5)[(0.321 + R2 / s ) 2 + (0.418 + 0.42) 2 ]
R2 / s = 0.156,4.513
R2 = 0.0059Ω,0.172Ω
These two solutions represent two situations in which the torque-speed curve would go
through this specific torque-speed point. The two curves are plotted below. As you can
see, only the 0.172Ω solution is realistic, since the 0.0059Ω solution passes through this
torque-speed point at an unstable location on the back side of the torque-speed curve.
(b) The slip at pullout torque can be found by calculating the Thevenin equivalent of the
input circuit from the rotor back to the power supply, and then using that with the rotor
circuit model. The Thevenin equivalent of the input circuit was calculated in part (a). The
slip at pullout torque is
12
s max =
R2
2
RTH + ( X TH + X 2 )
2
=
0.172
(0.321) + (0.418 + 0.420) 2
2
= 0.192
The rotor speed at maximum torque is
120(60)
n pullout = (1 − s max )nsync = (1 − 0.192)(
) = 1454r / min
4
And the pullout torque of the motor is
τ max =
3VTH
2
2ω sync [( RTH + RTH + ( X TH + X 2 ) 2 ]
2
=
3(262) 2
2(188.5)[(0.321 + (0.321) 2 + (0.418 + 0.420) 2 ]
τ max = 448 N ⋅ m
(c) The starting torque of this motor is the torque at slip s = 1. It is
τ ind =
2
ω sync [( RTH
3VTH R2 / s
3(262) 2 (0.172)
=
+ R2 / s ) 2 + ( X TH + X 2 ) 2 ] (188.5)[(0.321 + 0.172) 2 + (0.418 + 0.420) 2 ]
τ ind = 199 N ⋅ m
(d) To determine the starting code letter, we must find the locker-rotor kVA per
horsepower, which is equivalent to finding the starting kVA per horsepower. The easiest
way to find the line current (or armature current) at starting is to get the equivalent
impedance ZF of the rotor circuit in parallel with jXM at starting conditions, and then
calculate the starting current as the phase voltage divided by the sum of the series
impedances, as shown below.
The equivalent impedance of the rotor circuit in parallel with jXM at starting conditions
(s=1) is
13
1
Z F , start =
1
jX M
+
=
1
Z2
1
1
j 30
+
1
= 0.167 + j 0.415 = 0.448∠68.1°Ω
0.172 + j 0.42
The phase voltage is
Vφ = VT / 3 = 460 / 3 = 266V
So the line current IL is
I L ,start = I A =
Vφ
R1 + jX 1 + RF + jX F
=
266∠0°
= 274∠ − 59.2° A
0.33 + j 0.42 + 0.167 + j 0.415
Therefore, the locker-rotor kVA of this motor is
S = 3VT I L ,start = 3 (460)(274) = 218kVA
And the kVA per horsepower is
kVA / hp =
218
= 4.36kVA / hp
50
This motor would have starting code letter D, since letter D covers the range 4.00-5.00.
Problem 7-23:
A would-rotor induction motor is operating at rated voltage and frequency with its slip
rings shorted and with a load of about 25 percent of the rated value for the machine. If the
rotor resistance of this machine is doubled by inserting external resistors into the rotor
circuit, explain what happens to the following:
(a) Slip s
(b) Motor speed nm
(c) The induced voltage in the rotor
(d) The rotor current
(e) τind
(f) Pout
(g) PRCL
(h) Overall efficiency η
Solution:
14
(a) The slip s will increase.
(b) The motor speed nm will decrease.
(c) The induced voltage in the motor will increase.
(d) The rotor current will increase.
(e) The induced torque will adjust to supply the load’s torque requirements at the new
speed. This will depend on the shape of the load’s torque-speed characteristic. For most
loads, the induced torque will decrease.
(f) The output power will generally decrease: Pout = τ ind ↓ ω m ↓ .
(g) The rotor copper losses (including the external resistor) will increase.
15