ELG3311: Assignment 3

Problem 6-12:

The Y-connected synchronous motor whose nameplate is shown in Figure 6-21 has a per-
unit 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
VT 6600
Vφ ,base = = = 3810V
3 3
I A,base = I L ,base = 1404 A
S base = Prated = 3VT I L PF = 3 (6600)(1404)(1.0) = 16.05MW

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(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

XSP (0.90)(0.623)
δ = sin −1 = sin −1 = 24.9°
3Vφ E A (1.0)(1.33)

At rated voltage and 10 MW of power supplied, the armature current will be

Vφ − E A 1∠0° − 1.33∠ − 24.9°

IA = = = 0.663∠20.2° pu
R A + jX S j 0.90

In actual amps, this current is

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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

Pout (21000hp)(746W / hp)

η= × 100% = × 100% = 97.6%
Pin 16.05MW

(f) The output torque in SI and English units is

Pout (21000hp)(746W / hp )
τ load = = = 124700 N ⋅ m
ωm (1200r / min)(1min/ 60s )(2πrad / 1r )
5252 Pout (5252)(21000hp)
τ load = = = 91910lb ⋅ ft
nm 1200r / min

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?

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Solution:

(a) The input power to the motor at rated conditions is

Pout (100hp )(746W / hp)

Pin = = = 83.8kW
η 0.89

(b) The line current to the motor at rated conditions is

P 83.8k
IL = = = 137 A , θ = cos −1 PF = cos −1 0.8 = 37.86°
3VT PF 3 (440)(0.8)

The phase current to the motor at rated conditions is

IL 137
Iφ = = = 79.4 A
3 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

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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

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  E A1 sin δ 1   (603) sin( −19.5°) 
δ 2 = sin −1   = sin −1   = −21.8°
 E A2   543 

Therefore

Vφ − E A 440∠0° − 543∠ − 21.8°

IA = = = 70.5∠17.7° A
jX S 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
R1 = 0.200Ω R2 = 0.120Ω X M = 15.0Ω
X 1 = 0.410Ω X 2 = 0.410Ω
Pmech = 250W Pmisc ≈ 0 Pcore = 180W
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:

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(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

1 1
ZF = = = 2.220 + j 0.745 = 2.34∠18.5°Ω
1 1 1 1
+ +
jX M Z 2 j15 2.40 + j 0.41

The phase voltage is

Vφ = VT / 3 = 208 / 3 = 120V

So the line current IL is

Vφ 120∠0°
IL = IA = = = 44.8∠ − 25.5° A
R1 + jX 1 + RF + jX F 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

R22 2
(c) The air gas power is PAG = 3I 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

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(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

PAG 13.4k
τ ind = = = 35.5 N ⋅ m
ω sync (
120(60)
)(1min/ 60s )(2πrad / 1r )
2

(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

Pout 12.3k
τ load = = = 34.3 N ⋅ m
ωm (3420r / min)(1min/ 60 s )(2πrad / 1r )

(g) The overall efficiency is

Pout Pout 12.3k

η= × 100% = × 100% = × 100% = 84.5%
Pin 3Vφ I A cosθ 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

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R1 = 0.075Ω R2 = 0.065Ω X M = 7.2Ω
X 1 = 0.17Ω X 2 = 0.17Ω
PF &W = 1.0kW Pmisc = 150W Pcore = 1.1kW
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:

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The equivalent impedance of the rotor circuit in parallel with jXM is

1 1
ZF = = = 1.539 + j 0.364 = 1.58∠13.2°Ω
1 1 1 1
+ +
jX M Z 2 j 7.2 1.625 + j 0.17

The phase voltage is

Vφ = VT / 3 = 440 / 3 = 254V

So the line current IL is

Vφ 254∠0°
IL = IA = = = 149.4∠ − 18.3° A
R1 + jX 1 + RF + jX F 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

X2 0.17
θ R = tan −1 = 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 2
(e) The air gas power is PAG = 3I 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.)

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 2 R2 2
PAG = 3I 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

PAG 103k
τ ind = = = 327.9 N ⋅ m
ω sync (
120(50)
)(1 min/ 60 s)(2πrad / 1r )
2

(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

Pout 98.9k
τ load = = = 327.6 N ⋅ m
ωm (2880r / min)(1min/ 60s )(2πrad / 1r )

(i) The overall efficiency is

Pout Pout 96.6k

η= × 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

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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 per-
phase 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

jX M ( R1 + jX 1 ) ( j 30)(0.33 + j 0.42)
Z TH = = = 0.321 + j 0.418 = 0.527∠52.5°Ω
R1 + j ( X 1 + X M ) 0.33 + j (0.42 + 30)
jX M j 30
VTH = Vφ = (265.6∠0°) = 262∠0.6°V
R1 + j ( X 1 + X M ) 0.33 + j (0.42 + 30)

(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 )

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The induced torque is given by the equation

2
3VTH R2 / s
τ ind =
ω sync [( RTH + R2 / s ) 2 + ( X TH + X 2 ) 2 ]

Substituting known values and solving for R2/s yields

3(262) 2 R2 / s
205.7 =
(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

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 R2 0.172
s max = = = 0.192
2
RTH + ( X TH + X 2 ) 2
(0.321) + (0.418 + 0.420) 2
2

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

2
3VTH 3(262) 2
τ max = =
2ω sync [( RTH + RTH + ( X TH + X 2 ) 2 ]
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

2
3VTH R2 / s 3(262) 2 (0.172)
τ ind = =
ω sync [( RTH + 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

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 1 1
Z F , start = = = 0.167 + j 0.415 = 0.448∠68.1°Ω
1 1 1 1
+ +
jX M Z2 j 30 0.172 + j 0.42

The phase voltage is

Vφ = VT / 3 = 460 / 3 = 266V

So the line current IL is

Vφ 266∠0°
I L ,start = I A = = = 274∠ − 59.2° A
R1 + jX 1 + RF + jX F 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

218
kVA / hp = = 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:

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(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.

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