Presentation is loading. Please wait.

Presentation is loading. Please wait.

Lesson 37 AC Generators II

Similar presentations

Presentation on theme: "Lesson 37 AC Generators II"— Presentation transcript:

1 Lesson 37 AC Generators II

2 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.

3 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 3

4 DC Power Conversion Diagram
Review DC Power Conversion Diagram Electrical Mechanical

5 AC Generator power conversion diagram
Mechanical Electrical PIN = Trotor=746*hp POUT Pelectr loss Pmech loss NOTE: ω is the speed of the rotor, not the angular velocity of the AC current.

6 Example Problem 1 Consider a 3-phase, 4 pole, 60Hz, 450 V synchronous generator rated to supply kVA to a ship distribution system requiring a 0.8 lagging power factor. If this machine was operating at rated conditions, what would the real (P) and reactive (Q) power and the current being supplied? If the generator has an efficiency () of 95%, what torque does the prime mover provide? What is the speed of the rotor (rpm)?

7 Single-Phase Equivalent Circuit
Just like 3-phase loads, it is useful to look at just a single phase of the generator. Einduced + EAN - XS RS N A Ia  Single-phase equivalent 3-Phase Generator

8 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. Einduced + EAN - XS RS N A Ia

9 AC Generator Power Balance
Mechanical Input Power can be calculated: Electrical (Armature) Losses can be calculated (notice 3 sets of armature windings, so must multiply by 3) Electrical output power can be calculated The total overall power balance:

10 Solution steps Determine the rms value of IL
Determine phase angle of IL from the given power factor FP (using phase voltage as the reference)

11 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. Determine PIN Determine torque supplied to the generator if needed

12 Example Problem 2 A submarine has a 3-phase, Y-connected, 2-pole, 60Hz synchronous generator rated to deliver kVA at a FP = 0.8 lagging with a line voltage of 450-V. The machine stator resistance RS = 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. 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)

Download ppt "Lesson 37 AC Generators II"

Similar presentations

Ads by Google