Homework 7
Chapter 6 Synchronous Motors
1. A 2300 V, 1000 hp. 0.8 PF leading, 60 Hz, two poles, Y-connected synchronous
motor has a synchronous reactance of 2.8 Ω and an armature resistance of 0.4 Ω. At
60 Hz, its friction and windage losses are 24 kW, and it’s core losses are 18 kW. The
filed circuit has a dc voltage of 200 V, and the maximum I F is 10 A. The open circuit
characteristic of this motor is shown in below Figure.
Answer the following questions about the motor, assuming that it is being supplied by
an infinite bus.
(a) How much field current would be required to make this machine operate at unity
power factor when supplying full load?
(b) What is the motor’s efficiency at full load and unity power factor?
(c) If the filed current were increased by 5 percent, what would the new value of the
armature current be? What would the new power factor be? How much reactive
power is being consumed or supplied by the motor?
(d) What is the maximum torque this machine is theoretically capable of supplying at
unity power factor? At 0.8 PF leading?
Solution
(a) At full load, the input power to the motor is
PIN = POUT + Pmech + Pcore + PCU
We can’t know the copper losses until the armature current is known, so we will find the
input power and armature current ignoring that term, and then correct the input power
after we know it.

W
PIN = (1000 hp )  746
 + 24 kW + 18 kW = 788 kW
hp 

Therefore, the line and phase current at unity power factor is
788 kW
= 198 A
3 (VT )( PF )
3 ( 2300 V )(1.0 )
The copper losses due to a current of 198 A are
2
PCU = 3I A2 RA = 3 (198 A ) ( 0.4 ) = 47.0 kW
IA = IL =
P
=
Therefore, a better estimate of the input power at full load is

W
PIN = (1000 hp )  746
 + 24 kW + 18 kW + 47 Kw = 835 kW
hp 

and a better estimate of the line and phase current at unity power factor is
835 kW
= 210 A
3 (VT )( PF )
3 ( 2300 V )(1.0 )
The phasor diagram of this motor operating a unity power factor is shown below:
IA = IL =
P
The phase voltage of this motor is
voltage is
=
2300
= 1328 V . The required internal generated
3
E A = Vφ − RA I A − jX S I A
E A = 1328∠0o − ( 0.4 Ω ) ( 210∠0o A ) − j ( 2.8 Ω ) ( 210∠0o A )
E A = 1328 − 84 − j 588 = 1244 − j 588 = 1376∠ − 25.3o V
This internal generated voltage corresponds to a terminal voltage of
3 (1376 ) = 2383 V .This voltage would require a field current of 4.6 A.
(b) The motor’s efficiency at full load and unity power factor is
P
746 kW
η = OUT × 100% =
× 100% = 89.3%
PIN
835 kW
(c) To solve this problem, we will temporarily ignore the effects of the armature
resistance RA . If RA is ignored, then E A sin δ is directly proportional to the power
supplied by the motor. Since the power supplied by the motor does not change when I F
is changed, this quantity will be a constant.
If the field current is increased by 5%, then the new field current will be 4.83 A, and
the new value of the open-circuit terminal voltage will be 2450 V. The new value of E A
2450
will be
= 1415 V . Therefore, the new torque angle δ will be
3
E
 1376 V 
o
o
δ 2 = sin −1 A1 sin δ1 = sin1 
 sin ( −25.3 ) = −24.6
EA2
1415
V


Therefore, the new armature current will be
Vφ − E A
1328∠0o V − 1415∠ − 25.3o V
IA =
=
= 214.5∠3.5o A
RA + jX S
0.4 + j 2.8 Ω
The new current is about the same as before, but the phase angle has become positive.
The new power factor is cos 3.5 = 0.998 leading, and the reactive power supplied by the
motor is
Q = 3 (VT )( I L ) sin θ = 3 ( 2300 V )( 214.5 A ) sin 3.5 = 52.2 kVAR
(d) The maximum torque possible at unity power factor (ignoring the effects of RA ) is:
3V E
3 (1328 V )(1376 V )
P
τ ind ,max = conv = φ A =
= 5193 N ⋅ m
r  1 min  2π rad 
ωm ωm X S 
 3600


 ( 2.8 Ω )
min  60 s  1 r 

so I A = 247∠36.87° A . The internal generated voltage at 0.8 PF leading (ignoring copper
losses) is
E A = Vφ − RA I A − jX S I A
E A = 1328∠0o − j ( 2.8 Ω ) ( 247∠36.87o A )
E A = 1328 − j 691.6∠36.87o
E A = 1328 − j 553.23 + 414.96 = 1742.96 − j 553.23 = 1829∠ − 17.6o V
Therefore, the maximum torque at a power factor of 0.8 leading is
3V E
3 (1328 V )(1829 V )
P
τ ind ,max = conv = φ A =
= 6093 N ⋅ m
r  1 min  2π rad 
ωm ωm X S 
 3600


 ( 2.8 Ω )
min  60 s  1 r 

2. A 480 V, 100 kW, 50 Hz, four pole, Y connected synchronous motor has a rated power
factor of 0.85 leading. At full load, the efficiency is 91 percent. The armature resistance
is 0.25 Ω, and the synchronous reactance is 4.4 Ω. Find the following quantities for this
machine when it is operating at full load:
(a) Output torque
(b) Input power
(c) nm
(d) E A
(e) I A
(f) Pconv
(g) Pmech + Pcore + Pstray
Solution
(a) Since this machine has 4 poles, it rotates at a speed of
120 f e 120 ( 50 Hz )
nm =
=
= 1500 r/min
P
4
if the output power is 100 kW, the output torque is
P
100, 000 W
τ load = out =
= 637 N ⋅ m
r  2π rad  1 min 
ωm 
1500



min  1 r  60 s 

(b) The input power is
P
100 kW
PIN = OUT =
= 110 kW
η
0.91
(c) The mechanical speed is nm = 1500 r/min
(d) The armature current is
P
110 kW
IA = IL =
=
= 156 A
3 VT PF
3 ( 480 V )( 0.85 )
I A = 156∠ cos −1 ( 0.85 ) = 156∠31.8o A
Therefore, E A is
E A = Vφ - RA I A - jX S I A
E A = ( 277∠0o V ) - ( 0.25 Ω ) (156∠31.8o A ) - j ( 4.4 Ω ) (156∠31.8o A )
E A = 277∠0o − 39∠31.8o − j 686.4∠31.8o
E A = 277∠0o − ( 33.15 + j 20.55 ) − j ( 583.37 + j 361.70 )
E A = 277 − 33.15 − j 20.55 − j 583.37 + 361.70
E A = ( 277 − 33.15 + 361.70 ) − j ( 20.55 + 583.37 )
E A = 605.55 − j 603.92 = 855.23∠ − 44.9o V
(e) The magnitude of the armature current is 156 A.
(f) The power converted from electrical to mechanical form is given by the
equation Pconv = Pin - Pcu
Pcu = 3I A2 RA = 3 (156 A ) ( 0.25 ) = 18.25 kW
2
Pconv = Pin − Pcu = 110 kW - 18.25 kW = 91.75 kW
(g) The mechanical, core, and stray losses are given by the equation
Pmech + Pcore + Pstray = Pconv − POUT = 91.75 kW - 100 kW = - 8.25 kW