Download AC Generators Part 2

Lesson 37
AC Generators II
Learning Objectives

Use the power conversion diagram to describe
power flow for a three phase generator.

Find line voltages and current for a Y-connected
three phase generator.
Large AC generator
•Unlike our generator model with a fixed magnetic field and
rotating armature, it is more practical to fix the armature
windings and rotate the magnetic field on large generators.
•Brushes and slip rings pass EXCITATION voltage to the field
windings on the rotor to create the magnetic field
•Minimizes current flow through brushes to rotor windings
Review
DC Power Conversion Diagram
Electrical
Mechanical
AC Generator power conversion diagram
Mechanical
Electrical
PIN = Trotor=746*hp
POUT
Pmech loss
Pelectr loss
 2 
NOTE: ω is the speed of the rotor, not 

rotor

 2 f
the angular velocity of the AC current.
Poles


Example Problem 1
Consider a 3-phase, 4 pole, 60Hz, 450 V synchronous
generator rated to supply 1687.5 kVA to a ship distribution
system requiring a 0.8 lagging power factor.
a. If this machine was operating at rated conditions, what
would the real (P) and reactive (Q) power and the current
being supplied?
b. If the generator has an efficiency () of 95%, what
torque does the prime mover provide?
c. What is the speed of the rotor (rpm)?
Single-Phase Equivalent Circuit

Just like 3-phase loads, it is useful to look at just
a single phase of the generator.
XS
RS
Ia 
Einduced
A
+
EAN
N
Single-phase equivalent
3-Phase Generator
Single-Phase Equivalent Circuit




EAN is the phase voltage of the a-phase Ia is the
line current
Einduced is the induced armature voltage.
RS is the resistance of the generator’s stator coil.
XS is the synchronous reactance of the stator
coil.
XS
Einduced
RS
Ia
A
+
EAN
-
N
AC Generator Power Balance

Mechanical Input Power can be calculated:
PIN  T rotor  746* hp

Electrical (Armature) Losses can be calculated (notice 3
sets of armature windings, so must multiply by 3)
PELEC  LOSS  3I Rarmature
2
L

Electrical output power can be calculated
POUT  3I LVL cos

The total overall power balance:
PIN  POUT  PLOSS  MECH  PELEC  LOSS
Solution steps

Determine the rms value of IL
PL  3VL I L cos 

 IL 
PL
3VL FP
(since FP  cos  )
Determine phase angle of IL from the given
power factor FP (using phase voltage as the
reference)
  cos1 FP
 I  V  
Solution steps

Determine Electrical losses (zero for a “negligible
stator resistance”). This is PER-PHASE, so must
multiply by 3 when adding to other power.
PELEC  LOSS  3I L 2 Rarmature

Determine PIN
PIN  POUT  PLOSS  MECH  PELEC  LOSS

Determine torque supplied to the generator if needed

2

rotor  
 2 f
 Poles 
TIN 
PIN
rotor
Example Problem 2
A submarine has a 3-phase, Y-connected, 2-pole, 60Hz synchronous
generator rated to deliver 1687.5 kVA at a FP = 0.8 lagging with a line
voltage of 450-V. The machine stator resistance RS = 0.004 Ω. The
synchronous reactance is Xs=0.08 Ω. The actual system load on the
machine draws 900 kW at FP = 0.6 lagging. Assume that a voltage
regulator has automatically adjusted the field current so that the terminal
voltage is its rated value. Mechanical losses are 100 kW.
a.
b.
c.
Determine the reactive and apparent power delivered by the generator.
Find the current delivered by the generator
What is the overall efficiency?
The rated voltage (here, 450 V) is always a line voltage (this is the voltage we can
measure between any two cables in a 3-phase system)