Transcript AC Generators II

Lesson 34:
AC Generators II
1
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.
2
Large AC Generator
• Unlike the 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.
3
DC Generator
Power Conversion Diagram
Electrical
Mechanical
• Electrical power in yields a mechanical torque out.
4
AC Generator
Power Conversion Diagram
Mechanical
Electrical
PIN = (T)rotor=746*hp
POUT
Pmech-loss
Pelectr-loss
• Mechanical torque in yields a three-phase electrical output.
 2 
NOTE: ω is the speed of the rotor, not the 

rotor

 2 f
angular velocity of the AC current.
#
poles


5
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
6
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
7
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  PMECH  LOSS  PELEC  LOSS
8
Solution Steps
• Determine the rms value of IL
PL  3VL I L cos  ZT

IL 
PL
3VL FP
(since FP  cos  ZT )
• Determine phase angle of IL from the given power
factor FP (using phase voltage as the reference)
 Z  cos 1 FP
T

9
 I  V    Z
T
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  PMECH  LOSS  PELEC  LOSS
•
Determine torque supplied to the generator if needed

2

rotor  
 2 f
 Poles 
TIN 
PIN
rotor
10
AC Generator
Power Conversion Diagram
Mechanical
Electrical
POUTPOUT
 3I LVL cos 
PININ  Tinrotor
PMECH  LOSS  PIN  POUT PPmech-loss
ELEC  LOSS
2
PELEC
P

3
I
 LOSS
L Rarmature
electr
loss
• Mechanical torque in yields a three-phase electrical output.
 2 
NOTE: ω is the speed of the rotor, not the
rotor  
 2 f
angular velocity of the AC current.
 # poles 
11
Example Problem 1
Consider a 3-phase, 4 pole, 60Hz, 450V 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 is the real (P) power, reactive (Q)
power and the current (I) being supplied to the load?
b) When efficiency () is 95%, what torque does the prime mover provide?
c) What is the speed of the rotor (N) in rpm?
a)
ST  1687.5kVA => Remember, generators are rated in Apparent Power
PT  ST Fp  (1687.5kVA)(0.8)  1350kW  QT  S 2T  P 2T  (1687.5kVA) 2  (1350kW ) 2  1013kVAR
ST  3VI  I 
1687.5kVA
 2165 A
3(450V )
b)
 P  1350kW
Pin   out  
 1421kW

0.95


2
2
(2 f )  (2 60 Hz )  188.5 rad/s
Pin  (Tin )rotor  rotor 
# poles
Tin 
Pin
rotor

4
1421kW
 7.54kN  m
188.5rad / s
12
c)
N
120( f ) 120(60 Hz )

 1800 rpm
# poles
4
QUESTIONS?
13