Power System Fundamentals EE-317 Lecture 3 06 October 2010

Aims  Finish Chapter 1 – Real and Reactive Power Real and Reactive Loads Power Triangle  Chapter 2 – Three Phase Circuits  Chapter 3 – Transformers

Sine Wave Basics (Review)  RMS – a method for computing the effective value of a time-varying e-m wave, equivalent to the energy under the area of the voltage waveform.

Real, Reactive and Apparent Power in AC Circuits  in DC circuits: P=VI but…= in AC circuits: average power supplied to the load will be affected by the phase angle  between the voltage and the current.  If load is inductive the phase angle (also called impedance angle) is positive; (i.e, phase angle of current will lag the phase angle of the voltage) and the load will consume both real and positive reactive power  If the load is capacitive the impedance angle will be negative (the phase angle of the current will lead the phase angle of the voltage) and the load will consume real power and supply reactive power.

Resistive and Reactive Loads

Impedance Angle, Current Angle & Power  Inductive loads  positive impedance angle current angle lags voltage angle  Capacitive loads  negative impedance angle current angle leads voltage angle  Both types of loads consume real power  One (inductive) consumes reactive as well while the other (capacitive) supplies reactive power

Useful Equations  First term is average or Real power (P)  Second term is power transferred back and forth between source and load (Reactive power- Q)

More equations  Real term averages to P = VI cos  (+)  Reactive term averages to Q = VI sin  (+/-)  Reactive power is the power that is first stored and then released in the magnetic field of an inductor or in the electric field of a capacitor  Apparent Power (S) is just = VI

Loads with Constant Impedance  V = IZ Substituting…  P = I 2 Z cos   Q = I 2 Z sin   S= I 2 Z  Since… Z = R + jX = Z cos  + jZ sin   P = I 2 R and Q = I 2 X

Complex Power and Key Relationship of Phase Angle to V&I  S = P + jQ  S = VI(complex conjugate operator)  If V = V  30 o and I = I  15 o THEN….. COMPLEX POWER SUPPLIED TO LOAD = S = (V  30 o )(I  -15 o ) = VI (  30 o -15 o ) = VI cos(15 o ) + jVI sin(15 o )  NOTE: Since Phase Angle  = V  - I  S = VI cos(  ) + jVI sin(  ) = P + jQ

Review V, I, Z  If load is inductive then the Phase Angle (Impedance Angle Z  ) is positive, If phase angle is positive, the phase angle of the current flowing through the load will lag the voltage phase angle across the load by the impedance angle Z .

The Power Triangle

Example  V = 240  0 o V  Z = 40  -30 o   Calculate current I, Power Factor (is it leading or lagging), real, reactive, apparent and complex power supplied to the load

Read Chapters 2 & 3  HW Assignment 2:  Problems 1-9, 1-15, 1-18, 1-19, 2-4

Example Problem  HW 1-19 (a)

Chapter 2  Three-Phase (3-  ) Circuits What are they? Benefits of 3-  Systems Generating 3-  Voltages and Currents Wye (Y) and delta (  ) connections Balanced systems One-Line Diagrams

What does Three-Phase mean?  A 3-  circuit is a 3-  AC-generation system serving a 3-  AC load   AC generators with equal voltage but phase angle differing from the others by 120 o

Multiple poles….

Benefits of 3-  circuits  GENERATION SIDE:  More power per kilogram  Constant power out (vs. pulsating sinusoidal)  LOAD SIDE:  Induction Motors (no starters required)

Common Neutral  A 3-  circuit can have the negative ends of the 3-  generators connected to the negative ends of the 3-  AC loads and one common neutral wire can complete the system  If the three loads are equal (or balanced) what will the return current be in the common neutral?

If loads are equal….  the return current can be calculated to be…  ZERO!  (see trig on p. 59 for more detail)  Neutral is actually unnecessary in a balanced three-phase system (but is provided since circumstances may change)

Wye (Y) and delta (  ) connection

Delta (  )

Y and   Y-connection I L = I  V LL =  3 V    -connection V LL = V  I L =  3 I 

Balanced systems

One-Line Diagrams  since all phases are same (except for phase angle) and loads are typically balanced only one of the phases is usually shown on an electrical diagram… it is called a one-line diagram  Typically include all major components of the system (generators, transformers, transmission lines, loads, other [regulators, swithes])

Chapter 3  Transformers Benefits of Transformers Types and Construction, The Ideal Transformer Transformer Efficiency and Voltage Regulation Transformer Taps Autotransformers 3-  Transformer connections –Y-Y, Y- ,  -Y,  - 

Benefits  Range of Power Systems  Power Levels  Seamless Converter of Power (Voltage)  Reduced Transmission Losses  Efficient Converter  Low Maintenance (min. moving parts)  Enables Utilization of Power at nearly all levels