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Example- The 1 MVA, 2.3 kVrms, 3-phase, Y-connected, 6-pole, 60 Hz,
synchronous generator of the previous example is supplied the
rated field current of 12 A and is turned by a steam turbine at
1200 RPM. If the core losses are 15 kW and the friction and
windage losses are 20 kW, determine the following:
a) What are the terminal voltage, phase voltage and induced voltage if the
load draws the rated current at a power factor of 0.866 lagging?
Assume we are looking at phase A with an assumed terminal voltage
phase angle of zero (other phases will be at ±120°).
IA
+
+
EA -
From (A-31),
I rated
Srated
106
=
=
= 251.022 A rms
3 Vrated
3 (2300)
The phase/armature current angle is θ = cos −1 ( pf ) = cos −1 (0.866) = 30° .
Then, the armature current is I A = 251.022∠ − 30° A rms .
From the previous example, VT = 2470.8 Vrms from the OCC at a field
current of If = 12 A. For a Y-connection, then
Vφ ,OC =
2470.833
= 1426.5363 Vrms .
3
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a) cont.
By KVL,
E A = I A ( RA + jX S ) + Vφ ∠0
which yields
1426.5∠δ = ( 251.022∠ − 30° )( 0.3 + j1.57 ) + Vφ ∠0
1426.5 e jδ = 1426.5cosδ + j 1426.5sinδ = 262.285 + j 303.6379 + Vφ
Equate imaginary parts to solve for δ-
j 1426.5sinδ = j303.6379
sinδ =
303.6379
⎛ 303.6379 ⎞
⇒ δ = sin −1 ⎜
⎟ = 12.2897°
1426.5
1426.5
⎝
⎠
Equate real parts to solve for phase voltage magnitude-
1426.5cosδ = 262.285 + Vφ
Vφ =1426.5cos (12.2897° ) − 262.285
Vφ = 1131.525 Vrms
Since, we assume the phase voltage is at zero degreesVφ = 1131.525∠0° Vrms ,
and the line-to-line terminal voltage for the Y-connected generator
VT =
( 3 )1131.525 = 1959.86 V
rms .
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b) What are the voltage regulation VR, input power Pin, output power Pout,
armature/stator copper losses PSCL, and efficiency η?
VR =
VNL − VFL
2470.8 − 1959.86
∗ 100% =
∗ 100%
VFL
1959.86
VR = 26.1%
PSCL = 3 I A RA = 3 ( 251.022 ) 0.3
2
2
PSCL = 56,710.8 W = 56.7 kW
Pout = Sout cosθ = 106 (0.866)
Pout = 866,000 W = 866 kW
Pin = Pout + Plosses = Pout + ( Pfriction/windage + Pcore + PSCL )
Pin = 866 + 20 + 15 + 56.7 = 957.7 kW
η=
Pout
866
*100% =
*100% = 90.4%
957.7
Pin
c) What are the applied torque τapp (from the prime mover), converted
power Pconv, and the induced torque (acts against prime mover) τind?
⎛π ⎞
⎛π ⎞
Now Pin = τ app ωm , where ωm = nm ⎜ ⎟ = 1200 ⎜ ⎟ = 125.6637 rad/s .
⎝ 30 ⎠
⎝ 30 ⎠
So,
τ app =
Pin
ωm
=
957,711
= 7621.2 Nim .
125.6637
Pconv = Pin − Pfriction/windage − Pcore
Pconv = 957.7 − 20 − 15 = 922.7 kW
τ ind =
Pconv
ωm
=
922,711
= 7342.7 Nim
125.6637