Power Factor Correction

This topic will help you to know the methods of power factor correction.

First, let me define power factor.

 Power factor is the ratio between the KW (Kilo-Watts) and the KVA (Kilo-Volt Amperes) drawn by an electrical load, where the KW is the actual load power and the KVA is the apparent load power. It is written as PF= P/S.

It is a measure of how effectively the current is being converted into useful work output and more particularly is a good indicator of the effect of the load current on the efficiency of the supply system.

As what our teacher said, everytime he talks about this topic, all current flow will causes losses in the supply and distribution system. A load with a power factor of 1.0 results in the most efficient loading of the supply, and a load with a power factor of 0.5 will result in much higher losses in the supply system.

     A poor power factor can be the result of either a significant phase difference between the voltage and current at the load terminals, or it can be due to a high harmonic content or distorted/discontinuous current waveform.

     Poor load current phase angle is generally the result of an inductive load such as an induction motor and power transformer.

As what I have learned from the other subject (electronics), which is related on this topic; is that a distorted current waveform can be the result of a rectifier, switched mode power supply, discharge lighting or other electronic load.

     A poor power factor due to an inductive load can be improved by the addition of power factor correction, but a poor power factor due to a distorted current waveform requires a change in equipment design or expensive harmonic filters to gain an appreciable improvement.

     Many inverters are quoted as having a power factor of better than 0.95 when in reality, the true power factor is between 0.5 and 0.75. The figure of 0.95 is based on the cosine of the angle between the voltage and current, but does not take into account that the current waveform is discontinuous and therefore contributes to increased losses on the supply.

Our teacher gave us an example figure of the balanced three phase system, with  balanced 3Φ load, balanced load with the apparent power of 24MVA, 0.78 pf lagging, and wye-connected capacitors. The capacitors are said to be wye-connected because of their common point, which is the neutral line.

This is the figure presented to us by our teacher:

CLICK THE IMAGE TO ENLARGE:

It is similar to single phase case. It uses capacitors to increase the power factor.

Just keep in your mind:

• Keep clear about total/phase power, line/phase voltages.
• To use capacitors this value should be negative. 

For the solution, first we identify the given.

Line voltage = 34.5 kVrms

Frequency= 60Hz

Next, we solve for the necessary parameters, inorder to answer the question, and after we arrived to the answers, we then label the calculated values in the power triangle.

Then, we focus on the question, what value of capacitor increase the pf to 0.94 leading?

I will show you how;

Solving for angle Θ:

Θ= arccos ( 0.78 )

Θ= 38.74°

Since, the apparent power was already given, you can now use different ways on how to solve the remaining parameters.

S= 24MVA

Solving for real power:

P= ScosΘ

P= 18.72 MW

Now, to complete the power triangle, solve for the reactive power:

Q=SsinΘ

Q= 15.01 MVAR

Power Triangle:

CLICK THE IMAGE TO VIEW IT CLEARER:

OLD:

NEW:

Using the 0.94 power factor leading:

Solving for angle Θ:

Θ= arccos ( 0.94 )

Θ=19.94°

Since, the real power is the same with the real power at the power triangle, you can now use different ways on how to solve the remaining parameters.

P= 18.72MW

Solving for real power:

S=P/PF ,

S= 19.91 MVA

Now, to complete the power triangle, solve for the reactive power:

Q=SsinΘ

Q= 6.79 MVAR

Power Triangle:

CLICK THE IMAGE TO VIEW IT CLEARER:

 

It is inverted because the PF is leading.

SUPER-IMPOSED:

Qcapacitor= Qnew- Qold

Solving for the 3Φ  capacitor:

C= Q/2πf V^2rms

C= -21.79×10^6/ 2π(60) (34.5×10^3)^2

C= |-48| microFarad

C= + 48 microFarad

Solving for the per phase capacitor:

jQc= -jωCV^2rms

You will come up with the same answer, which is

C= + 48 microFarad

CLICK THE IMAGE TO ENLARGE:

LESSON LEARNED:

In the power factor correction, you should know how to solve the angles, apparent, real, reactive power, and the other parameters involved to solve the capacitor needed to correct the power factor. You also consider the Power triangle , to analyze the problem better. Most loads on an electrical distribution system fall into one of three categories; resistive, inductive or capacitive. In a maunfacturing plant, the most common is likely to be inductive. Typical examples of this include transformers, fluorescent lighting and AC induction motors. Most inductive loads use a conductive coil winding to produce an electromagnetic field, allowing the motor to function. All inductive loads require two kinds of power to operate: Active power (kwatts) – to produce the motive force Reactive power (kvar) – to energise the magnetic field. Almost all loads are inductive.In order to cancel the reactive component of power, we must add reactance of the opposite type.  This is called power factor correction.

Why is Power Factor Important?

• For the load with Fp = 0.6, the generator had to supply 133 more amperes in order to do the same work (P)!
• Larger current means larger equipment (wires, transformers, generators) which cost more.
• Larger current also means larger transmission losses (think I^2R).

Because of the wide variation in possible current requirements due to power factor, most large electrical equipment is rated using apparent power (S) in volt-amperes (VA) instead of real power (P) in watts (W).

For more details, just click on these links:

http://www.youtube.com/watch?v=R2l0qSZPLbM

http://www.youtube.com/watch?v=dPFKcUxbNuQ

http://www.youtube.com/watch?v=9pWOm77KDJM

http://www.youtube.com/watch?v=6Le9_zyWULE

Power in a Balanced System

First, let’s review some informations about the previous lessons we encountered to refresh your minds.

Complex Power

CLICK THE IMAGES TO ENLARGE:

We already know that from the previous lesson on this topic, there are formulas we need to familiarize. This include:

S= | V | | I | [ cos ( θv-θi ) + jsin ( θv-θi ) ],

                                    =P + jQ     =  V I* ( the asterisk here means conjugate ) 

The student studies the method to calculate complex power where the Vrms of a circuit is multiplied by the complex conjugate ( if the angle of the current is positive, make it negative, vise versa ) of the total circuit current. 

P= REAL POWER ( units are in W, kW, MW)

Q= REACTIVE POWER ( VAr, kVAr, MVAr)

   = MAGNITUDE OF POWER INTO AN ELECTRIC AND MAGNETIC FIELDS

S= COMPLEX POWER  ( VA, kVA, MVA )

Power Factor (pf) = cosΦ

Remember if current leads voltage, then pf is leading, has negative Q on complex power rectangular form. If current lags voltage, then pf is lagging.

Now, let’s go to the relationships between real, reactive and complex power.

P= |S| cos Φ

Q= |S| sin  Φ = ± |S| sqrt of 1-pf squared

I’ll give you such example:

A load draws 100 kW with a leading pf of 0. 85. What are the power factor angle , Q and |S|?

SOLUTION:

To get the power factor angle> Φ= -arc cos (0.85) = -31.8º

For the |S|> |S|= 100 x 10 raised to 3 divided by 0.85, which is P/pf

= 100kW/ 0.85 = 117.6 kVA

For the Q> Q= |S|sin Φ= 117.6sin(-31.8º)= -62 kVAr

Moving on to the Conservation of Power

• At every node (bus) in the system:
  –Sum of real power into node must equal zero,
  –Sum of reactive power into node must equal zero.

• This is a direct consequence of Kirchhoff’s current law, which states that the total current into each node must equal zero.
  –Conservation of real power and conservation of reactive power follows since S = VI*

Power Consumption in Devices

CLICK THE IMAGE TO ENLARGE:

Below, you can see the image of the Distribution System Capacitors;

And now, let’s go to the Balanced 3 Phase (Φ) Systems:

• A balanced 3 phase (Φ) system has:
  –three voltage sources with equal magnitude, but with an angle shift of 120°,
  –equal loads on each phase,
  –equal impedance on the lines connecting the generators to the loads.

• Bulk power systems are almost exclusively 3Φ.
• Single phase is used primarily only in low voltage, low power settings, such as residential and some commercial.

Balanced 3Φ — Zero Neutral Current

CLICK THE IMAGE TO ENLARGE:

There are some advantages of 3Φ Power:

• Can transmit more power for same amount of wire (twice as much as single phase).
• Total torque produced by 3Φ machines is constant, so less vibration.
• Three phase machines use less material for same power rating.
• Three phase machines start more easily than single phase machines

Three Phase – Wye Connection 

There are two ways to connect 3Φ systems:

–Wye (Y), and Delta (D).

Here, you can see a Y-connected source labelled as Van, Vbn, and Vcn.

The Line Current is equal to the phase current, written as IL=IP.

Wye Connection Line Voltages

(α = 0 in this case)

Line to line voltages are also balanced.

• We call the voltage across each element of a wye connected device the “phase” voltage.
• We call the current through each element of a wye connected device the “phase” current.
• Call the voltage across lines the “line-to-line” or just the “line” voltage.
• Call the current through lines the “line” current.

Delta Connection

For Delta connection, voltages across elements equals line voltages. Simply VP=VL

CLICK THE IMAGE TO VIEW IT CLEARLY:

Three Phase Example  :

Assume a Delta-connected load, with each leg ZΔ = 100 ∠20°W, is supplied from a 3Φ 13.8 kV (L-L) source.

First, solve for the phase voltages of a delta-connected source:

Vab= 13.8  ∠ 0° kV

Vbc= 13.8  ∠ -120° kV

Vca= 13.8  ∠ +120° kV

Then, solve for the phase currents,

IAB= Vab/ ZΔ = 13.8∠ 0°/ 100 ∠ 20° = 138 ∠ -20°A

IBC= IAB ∠ -120° = 138 ∠-140°A

ICA= IAB ∠ +120° = 138 ∠ 100°A

Next, solve for the line currents, 

Ia= IAB-ICA
= 138 ∠ -20° – 138 ∠ 100° = 239 ∠ -50° A

You may use another formula, which is Ia= IAB √3 ∠ -30°

Ia= 138 (√3) ∠ -30° = 239  ∠ -30° + (-20)

= 239  ∠ -50° A

For Ib and Ic:

Ib=  239  ∠ -120° + (-50°)

=  239  ∠ -170°A

Ic=  239  ∠ +120° + (-50)

=  239  ∠ 70° A

Now, we will able to solve for the COMPLEX POWER :

S=3 Vab Iab*

=3 ( 13.8  ∠ 0° ) (138 ∠ +20° )

S= 5.7 ∠ +20° kVA

S=5.356 + j1.95 kVA

Lastly, we solve for the power factor,

pf= cos (Θv-Θi)

= cos ( 0 – (-20))

= cos (+20)

pf = 0.94 lagging (lagging because the Q (reactive power) in the complex power is negative)

You may also used the angle of the load impedance:

pf= cos ( +20 )

pf= 0.94

Delta-Wye Transformation

To simplify analysis of balanced three phase systems:

1.) Δ-connected loads can be replaced by Y- connected loads with 

2.)  Δ-connected soures can be replaced by Y- connected sources with

 

Delta-Wye Transformation Proof

Suppose the two sides have identical terminal behavior. For the Δ side we get:

Ia= Vab/ ZΔ  – Vca/ ZΔ = (Vab – Vca)/  ZΔ

Hence,  ZΔ= (Vab- Vca) / Ica

For the Y side we get:

The two pictures below are Three Phase Transmission Line:

Additional example: 

Simple Problem:

Answer:

To fully understand this topic, let me summarize all the formulas in getting the power for a balanced three phase system: You are not advised to memorize the formulas, but to understand how they are derived:

Complex Power ( when in polar form, it has an angle ): unit: VA

S= 3Sp
= 3VpIp*
= √3 VL IL ∠ Θ
= 3 Isquaredp Zp –> 3I^2Zp    (Zp could be Zy or ZΔ)
= 3V^2p/ Zp*
= P+jQ
= Vs IL*

Average Power/ Apparent Power (magnitude): unit: VA

S= VpIp
= P/pf
= √3 VL IL
= squareroot of 3P^2 + Q^2

Real Power: P= ScosΘ = Scos( Θv-Θi)

                        = VpIp pf ( where pf is cos ( Θv-Θi) )

                         = VL IL cos ( Θv-Θi) / √3

Reactive Power: Q=  SsinΘ = Ssin( Θv-Θi)

For the Y connected Load:

IL = Ip = 3/ SVp ( where S= √3 VL IL )

VL= √3 Vp

For the Δ connected Load:

Vp= VL

IL= √3 Ip

Power Factor:

pf= cos ( Θv-Θi)

    = P/S

Application:

Three Phase Electricity

Three-phase voltage, frequency and number of wires.

Although single-phase power is more prevalent today, three phase is still chosen as the power of choice for many different types of applications. Generators at power stations supply three-phase electricity. This is a way of supplying three times as much electricity along three wires as can be supplied through two, without having to increase the thickness of the wires. It is usually used in industry to drive motors and other devices.

Three phase electricity is by its very nature a much smoother form of electricity than single-phase or two-phase power. It is this more consistent electrical power that allows machines to run more efficiently and last many years longer than their relative machines running on the other phases. Some applications are able to work with three-phase power in ways that would not work on single phase at all.

Three phase is a common method of electric power transmission. It is a type of polyphase system used to power motors and many other devices.

For more details, just click the links below:

http://www.youtube.com/watch?v=2Ndfnta4Y8g

http://www.youtube.com/watch?v=m_xishFiAyI

http://www.youtube.com/watch?v=VMNPoxft7MQ

http://www.youtube.com/watch?v=R8A-GsC105o

Three-Phase Circuits

Before our teacher started the lesson proper about  this topic, he asked us what is the difference between three-phase circuit  and single phase circuit. At first, me and my classmates just looked each other, because we don’t know what is the right answer for that question, but we guessed that in single phase circuit, it is an alternating-current using only one, sine wave type, current flow, while a three-phase circuit, it consists of three different sine wave current flows, different in phase by 120 degrees from each other. Our teacher also agree in our answers. But in a more practical definition:

Single phase: a circuit that consists of three wires – live, neutral, and ground (earth). The main breaker in a single phase system is a single pole breaker, resembling the others in the panel, only with a higher capacity.

Three phase: a circuit where the main breaker switches off three poles. For most home owners this is the equivalent of having 3 separate main breakers that are divided among the circuits of the home. There are 5 wires that normally constitute a three phase line, although in many homes, the three phases simply supply the main and sub panels, but continue throughout most of the home as single phase lines. In most homes there are not many devices that run on three phase electricity.

CLICK THE IMAGE TO ENLARGE:

Almost all electric power generation and most of the power transmission in the world is in the form of three-phase AC circuits.

A three-phase AC system consists of three-phase generators, transmission lines, and loads.

Our teacher showed us his powerpoint slides about this topic, and it was all about the three phase circuits.

To start with, let me define first the terms which are usually used on this topic, and the naming conventions.

Phase

describes or pertains to one element or device in a load, line, or source. It is simply a “branch” of the circuit and could look something like this .

Line

refers to the “transmission line” or wires that connect the source (supply) to the load. It may be modeled as a small impedance (actually 3 of them), or even by just a connecting line.

Neutral

the 4th wire in the 3-phase system. It’s where the phases of a Y connection come together.

Phase Voltages & Phase Currents

the voltages and currents across and through a single branch (phase) of the circuit. Note this definition depends on whether the connection is Wye or Delta!

Line Currents

the currents flowing in each of the lines (I_a, I_b, and I_c). This definition does not change with connection type.

Line Voltages

the voltages between any two of the lines (V_ab, V_bc, and V_ca). These may also be referred to as the line-to-line voltages. This definition does not change with connection type.

Line to Neutral Voltages

the voltages between any lines and the neutral point (V_a, V_b, and V_c). This definition does not change with connection type, but they may not be physically measureable in a Delta circuit.

Line to Neutral Currents

same as the line currents (I_a, I_b, and I_c).

Now, you are familiar with the terms which will be used in the entire lesson.So, let’s go on on the main topic.

This is a single phase system ( a two- wire type)

You can see here  in this system , one source is present, and it is connected to a single load, where the voltage source is in phasor form, Vp is the rms magnitude and f is the phase. It does not mean that when you have only one source, you are limited in only one load also. No! it does not, because the load is dependent on the source. Our teacher gave us an example to clearly understand what does it mean. He asked our one classmate, named Quicee, about the power supply we made in our other subject’s project, which is the amplifier. I forgot the exact value, but then, we already understand that the loads in the system are dependent on the source.

This is a single phase system ( a two- wire type)

The common is this kind of system, it is the single phase three wire system. By simply looking the figure above, you can observed that there are two voltage sources which are equal, ( our teacher corrected us that “equal” is different from the words ” the same”), so the proper word to describe the sources is equal or identical ( equal in magnitude and the same phase) which are obviously connected to two loads by the two outer wires and the wire at the center, which is the neutral. However, examples may include a three phase central air conditioner, a three phase oven, a 3 phase swimming pool pump, or a large 3 phase hot water boiler.

There is a great deal of difference between a single phase and three phase circuit.

Both provide alternating current with the voltage or current following a sine wave that is starting at zero rising to a maximum say 220 volts then decreasing to zero followed by the voltage rising again but in a negative direction or minus 220 volts.

In a single phase circuit this just keeps repeating based on the frequency say 60 hertz) that is 60 times a second.

In a three phase circuit this occurs with each phase (three) the voltage rising falling going negative and increasing in the opposite direction. At any instant there are in fact three different voltages impressed on the three wires.

To sum it up any device, say motor designed for single phase operation must be used on a single phase circuit. A three phase motor can only run on a three phase circuit.

There is so called a kind of systems/ circuits which the ac sources operate at the same frequency but different phases, and that is the POLYPHASE.

Polyphase power is particularly useful in AC motors, such as the induction motor, where it generates a rotating magnetic field.

The images below are the examples of polyphase circuits:

For additional details, click this link. http://www.allaboutcircuits.com/vol_2/chpt_10/2.html

For some information, click on this links:

http://www.youtube.com/watch?v=m_xishFiAyI

http://www.youtube.com/watch?v=Rv1un0fqLWc

http://www.youtube.com/watch?v=s_7U4sZfBkw

_________________________________________________________________________

Balanced Three-Phase Circuits

Balanced Three-Phase Voltages

Comprised of three sinusoidal voltages identical in amplitude and frequency but out of
phase from one another by 120°.

It is referred to as a-phase, b-phase and c-phase.

We have learned the:

Two Types of Phase Sequences :

• abc (positive) phase sequence – based from what we have learned, it is when the phase b lags a by 120° and c leads a by 120°

CLICK TO ENLARGE:

• acb (negative) phase sequence – another is this type of phase, which phase c lags a by 120° and b leads a by 120°

CLICK TO ENLARGE:

TO CLEARLY UNDERSTAND, LET’S LOOK THE IMAGE BELOW, JUST CLICK TO ENLARGE:

I also understand the important characteristics, which are Va + Vb + Vc = 0 and va + vb +vc + 0 .

Let me add some information about the balanced three phase circuit. Below are the three requirements of a balanced three phase circuit to be satisfied, in order for a set of 3 sinusoidal variables (usually voltages or currents) to be a “balanced 3-phase set”

1. All 3 variables have the same amplitude
2. All 3 variables have the same frequency
3. All 3 variables are 120 degrees in phase

CLICK THE IMAGE TO ENLARGE:

Our teacher shown us another powerpoint slides, which was made by his 4th year electrical engineering students. The slide which was presented to us, tells about of what the sample of AC generator compose. It is a generator with three separate windings distributed around its stator, each winding
comprising one phase. The rotor is an electromagnet driven at speed by a
prime mover. The rotation induces sinusoidal voltages of equal amplitude and frequency,
that are out of phase 120° from one another.

 The generated voltages are 120 degrees apart from each other.

The image above is the Balanced 3-Phase Variables in Time Domain

In terms of phasors, we write the same balanced set as follows. Note that the phasors are in rms, as will be assumed throughout this topic.

Van = Vp∠  0°

Vbn = Vp ∠  -120°

Vcn = Vp ∠  -240° =  Vp ∠  120°

JUST CLICK THE IMAGE TO ENLARGE:

Thus,

V_b = V_a (1 ∠ -120^o) , and V_c = V_a (1∠ +120^o)

I want to show you the image below which illustrates the balanced 3-phase phasors graphically.

Having a balanced circuit allows for simplified analysis of the 3-phase circuit. In fact, if the circuit is balanced, we can solve for the voltages, currents, and powers, etc. in one phase using circuit analysis. The values of the corresponding variables in the other two phases can be found using some basic equations. If the circuit is not balanced, all three phases should be analyzed in detail.

CLICK THE IMAGE TO ENLARGE:

This illustrates a balanced 3-phase circuit  :

Below are the three-phase voltage sources: a) wye-connected source, b) delta-connected source

Wyes and Deltas

A summary of the characteristics of the two types of 3-phase circuit connections are given below.

The Wye = Y = “Star” connection	____	The Delta = Δ connection
each phase is connected between a line and the neutral		each phase is connected between two lines
Y Circuit
Phase voltages = Line to neutral voltages (Va, etc.)Phase currents = Line currents (Ia, etc.) Neutral connects the three phases Δ Circuit

Phase voltages = Line voltages (V_ab, etc.)Phase currents = currents from line to line (I_ab, etc.).

Neutral is not present

 

Two possible three-phase load configurations: a)a wye-connected load, b) a
delta-connected load:

For a balanced wye-connected load:

Z1=Z2=Z3=ZY, where ZY is the load impedance per phase.

For a balanced delta- connected load:

Za=Zb=Zc=ZΔ, where ZΔ is the load impedance per phase. ZΔ= 3ZY or ZY= 1/3 ZΔ

Now, let’s have example problems for this:

JUST CLICK THESE IMAGES BELOW TO ENLARGE AND TO SEE THESE PROBLEMS CLEARER:

NO.1 problem

ANSWER

: 

NO.2 problem

ANSWER

:

FOR SOME DETAILS, click these links:

http://www.youtube.com/watch?v=7yxPM4dAuXE

http://www.youtube.com/watch?v=xjSQ9VI9A4Y

http://www.youtube.com/watch?v=m_xishFiAyI

_________________________________________________________________________

Balanced Wye-Wye Connection

A balanced Y-Y system, showing the source, line and load impedances

It is a three phase system with a balanced Y- connected source and load.

There in the figure, ZY is the total load impedance per phase, Zs is the source impedance, Zl is the line impedance. and ZL is the load impedance per phase. However, Zn is the impedance of the neutral line.

ZY= Zs + Zl + ZL

You can have ZY equal to ZL,for Zs and Zl are often very small compared with the load impedance per phase, which is the ZL.

I want to show you the phase voltages or line-to line neutral voltages ):

Van = Vp∠  0°

Vbn = Vp ∠  -120°

Vcn = Vp ∠  +120°

The line-to line voltages Vab, Vbc, and Vca are related to phase voltages. 

VL= √3Vp ( the line voltages VL is square root of three (√3 ) times the magnitude of the phase voltage Vp.

Where Does that √3 Come From?

Let us determine the relationships between the line and line to neutral voltages. By applying Kirchoff’s Voltage Law (KVL) to the top “loop” of the source section in the Figure shown in the top part of this topic ( balanced 3-phase circuit )

CLICK THE IMAGE TO ENLARGE:

V_ab = V_a – V_b = V_m ∠Φ – V_m ∠Φ  – 120^o

Now, without loss of generality, let Φ  = 0^o

thus, V_a = V_m∠ 0^o, and V_b = V_m∠ -120^o, so

V_ab = V_m ∠ 0^o – V_m  ∠– 120^o = V_m (1 – 1 ∠– 120^o ) = V_m (1 – (cos 120^o – j sin 120^o))

= V_m (1 – (-1/2) + j (√3 / 2 ) ) = V_m (3 / 2) + j ( √3/ 2 ))

Converting to polar form,

V_ab = V_m √[ (3/2)² + (√3 / 2)² ]  tan^-1 {(√3 / 2) / (3/2) }

= V_m √[ 9/4 + 3/4 ]  tan^-1 {1/√3 }

= V_m  √3 tan^-1 {(1 / 2) / ( /2) } = V_m   tan^-1 {(sin 30^o) / (cos 30^o) }

= V_m √3  tan^-1 {tan 30^o } = V_m ∠  30^o

Thus we have the general equation (for abc sequence anyway)

V_ab = V_{a √3}  ∠ 30^o

The relationships between the currents can be developed similarly. Summing currents at the “A” node in Figure 3 yields the starting equation,

I_a = I_AB – I_CA

This time choose I_a to be the phasor reference (at 0^o). The final result is:

I_a = I_AB √3 ∠ -30^o

These relationships can also be remembered graphically using Figures 4 and 5 below. Figure 4 illustrates the voltage relationship. By looking at the phasor equation as the sum of two vectors (V_a and -V_b ) we obtain the resulting V_ab shown in the figure.

Since V_ab is longer, we know . . . . |V_ab| = √3 |V_a| ,

and since V_ab is ahead of V_a, we know that, . . . . (the angle of V_ab) = (the angle of V_a) + 30^o

 Graphical Voltage Relationship

Figure 5 illustrates the current relationship. Now the phasor equation is the sum of two vectors (I_ab and -I_ca ) we obtain the resulting I_a shown in the figure.

Since I_a is longer, we know

|I_a| =√3  |I_ab| ,

and since I_a is behind I_ab, we know that,

(the angle of I_a) = (the angle of I_ab) – 30^o

    

 Graphical Current Relationship

FOR CURRENTS:

• Ia= Van/ ZY
• Ib= Ia ∠  -120°
• Ic= Ia∠  -240°

In= -(Ia + Ib + Ic)= 0, so VnN= ZnIn = 0

>>>Neutral current is ZERO in a BALANCED three-phase system

>>>Can eliminate the neutral wire

Below is a single phase equivalent circuit:

It yields the line current Ia,

 Ia= Van/ ZY

Let’s have an example: (click the images to enlarge)

  answer:

Now, for more information, just click these links:

http://www.youtube.com/watch?v=kOFcI5olDNY

http://www.youtube.com/watch?v=tL9ROPLh9k4

_________________________________________________________________________

Balanced Wye-Delta Connection

Balanced Y- Δ Connection:

CLICK THE IMAGE TO ENLARGE:

It consists of a balanced Y- connected source having a balanced Δ-connected load.

Do you want to learn how to convert three phase balanced wye source to balanced delta source? .. If you do, then I will teach you how.

A three phase Y source, has the three voltage phases tied to a common point (neutral). V line to line is √3 times V phase. I line is equal to I phase. A three phase delta connected source has V line to line equal to V phase, I line is equal to √3 times I phase. V phase and I phase are respectively the voltage and current of the 3 sources that are to be wired as Y or delta.
Remove the common connection. Then connect what had been B’s common connection (neutral) to phase A, connect what had been C’ common connection (neutral) to phase B, and finally connect phase C to A’s common connection (neutral). The sources are now connected in Delta. I suggest you draw this out on paper, it will be much easier to see. The line to line voltage is now just V phase, if you want identical voltage the sources have to be increased by √3. In delta the line current is √3 times I phase. The new source will have to generate √3 times more voltage but 1/√3 the current for the same load. The source generate the same power for the same load, whether they are connected in Y or delta.

Having a positive sequence, the phase voltages are the ff:

Van = Vp∠  0°

Vbn = Vp ∠  -120°

Vcn = Vp ∠  +120°

The line voltages are:

Vab = √3Vp ∠ 30° = VAB

Vbc = √3Vp ∠ -90° = VBC

Vca = √3Vp ∠ -150° = VCA

From these, we can obtain the phase currents:

IAB= VAB/ ZΔ 

IBC= VBC/ ZΔ 

ICA= VCA/ ZΔ 

The line current:

IL= √3Ip

where IL=| Ia | = | Ib |= | Ic |

and Ip=| IAB | = | IBC |= | ICA |

Below is a figure of a single-phase equivalent circuit of a balanced Y-D circuit:

Using the Δ-Y transformation formula in the last topic we had,  ZY= 1/3 ZΔ

Now, i will give you one example:

(Just click the image to enlarge)

Answer:

 For more information, just click the link below:

http://www.slideshare.net/pradeepa_m/3phase-circuits

_________________________________________________________________________

Balanced Delta-Delta Connection

A balanced Δ-Δ system

CLICK THE IMAGE TO ENLARGE:

It is one in which both the balanced source and balanced load are Δ- connected

The phase voltages for a Δ- connected source are the ff:

Vab = Vp∠  0°

Vbc = Vp ∠  -120°

Vca = Vp ∠  +120°

You assume that there’s no line impedances:

Vab= VAB       ,     Vbc=VBC      ,      Vca= VCA

The phase voltages of the delta- connected source are equal to the voltages across the impedances.

The phase currents, however are the ff:

IAB= VAB/ ZΔ= Vab/ZΔ

IBC= VBC/ ZΔ= Vbc/ZΔ

ICA= VCA/ ZΔ= Vca/ZΔ

The line currents are obtained from the phase currents by applying the KCL at nodes A, B, C, like what we did in the previous topic:

Ia= IAB- ICA        

Ib= IBC- IAB

Ic= ICA- IBC

 

The magnitude IL of the line current is √3 times the magnitude Ip of the phase current.

IL=√3Ip

Now, let’s have an example for this topic:

Just click the images to enlarge:

PROBLEM:

ANSWER:

For some details, just click the link below:

http://www.youtube.com/watch?v=p7bi1tAYUn0

http://www.youtube.com/watch?v=ViVMu5srJsM

 

_________________________________________________________________________

Balanced Delta-Wye Connection

A balanced Δ-Y system

CLICK THE IMAGE TO ENLARGE:

It is consists of a balanced Δ- connected source and balanced load that are Y- connected

Assuming the abc sequence, the phase voltages for a Δ- connected source are the ff:

Vab = Vp∠  0°

Vbc = Vp ∠  -120°

Vca = Vp ∠  +120°

We can get the line currents in many ways. It’s just like “there are many ways to kill the cat”. Right?

Anyway, we are to solve the line currents, one way is to apply KVL to loop aANBba in the figure above.

–Vab + ZY Ia – ZY Ib= 0 or  ZY ( Ia –  Ib ) =Vab =Vp∠  0°

Therefore,

 Ia –  Ib = Vp∠  0° / ZY  (1)

But Ib lags  Ia  by 120°, since we have assumed abc sequence;

Ib =  Ia ∠ -120°, Thus:

 Ia –  Ib =  Ia (1-1∠ -120°)

= Ia (1+ 1/2 + j√3/2) = Ia √3 ∠ 30°    (2)

Substituting (2) into (1),

 Ia=  Vp√3∠ -30° / ZY

CLICK THE IMAGE TO ENLARGE:

Transforming a Delta connected source to an equivalent Wye connection

The above image will help you analyze how these formulas below were obtained:

Van = Vp / √3∠  -30°

Vbn = Vp / √3∠  -150°

Vcn = Vp ∠  +90°

JUST CLICK THE IMAGE TO ENLARGE:

You have seen above the the Single phase equivalent of Delta Wye connection, and the source is equal to Vp∠  -30° / √3, resulting to the formula.

  Ia=  Vp√3∠ -30° / ZY

which is actually the same on the formula we have obtained just a minute ago.

We can convert wye- connected load to an equivalent delta- connected load. This may result in a Δ-Δ system,

VAN= IaZY = Vp√3∠ -30°

VBN= VAN ∠ -120°

VCN= VAN ∠ +120°

So, let’s have an example :

Just click the images to enlarge:

PROBLEM:

ANSWER:

For some details, just click the link below:

http://www.youtube.com/watch?v=ViVMu5srJsM

http://www.youtube.com/watch?v=3AXrCpQTZw0

http://www.youtube.com/watch?v=p7bi1tAYUn0

 

ELECTRICITY CONSUMPTION COST

First, let me define what cost, consumption, and electricity mean:        

• Cost is an amount paid or required in payment for a purchase; a price.

• Consumption is the act or process of consuming ;the state of being consumed; an amount consumed.

• Electricity is the most common form of energy used in the home. Different characteristics of electricity are measured using different quantities and units,

Now, you already know their definitions, we must also consider the necessary terms about this topic, including:

VOLTAGE (symbol V, measured in Volts), which is a measure of the force of flow of electricity (and its “potential strength/ potential difference‟ when electricity is not actually flowing).

CURRENT (symbol I, measured in Amperes), which is a measure of the volume of flow of electricity.

However, neither voltage nor current is particularly
useful to the householder, because neither is
a measure of the rate of consumption of electricity,
which is what we really want to know…

POWER (symbol P, measured in Watts) is a
measure of the rate at which electricity flows
through an appliance, effectively the rate of
electricity consumption.  For example, use your imagination, just imagine the pipe
analogy, if you multiply the water pressure in a pipe
by its diameter, you will determine how much water
is flowing through that pipe every second. So, to
calculate the amount of power used by an appliance,
you simply multiply the force of the flow (in volts)
by the volume of flow (in amps) to get the rate at
which electricity is flowing through it (in watts):
P= V x I
If you look at an appliance’s compliance plate or
specifications, just look for “W‟ or “kW‟!
A 600W hairdryer uses 10 times more power
than a 60W lightbulb. Similarly, a 6kW
(6000W) air-conditioning unit uses 10 times
more power than a 600W hairdryer.

Our teacher gave us a photocopy, which is an electric bill (statement of account) in the year 2012 and 2014, that we need to analyze. The said electric bill is categorized in four groups, generation and transmission, second is distribution, the third one includes S.C discount subsidy and lifeline rate (others), and the fourth one is the government. Here in our place, electricity is solely provided by SOCOTECO II or South Cotabato II Electric Cooperative, Inc. It covers the SOCCSKSARGEN areas of South Cotabato, Cotabato, Sultan Kudarat, Sarangani and General Santos City

CLICK TO ENLARGE THE IMAGE:

SOCOTECO II provides Electricity to all the barangays of General Santos City and ensures that every household and business establishment enjoy the full benefits of a reliable electrical power, with as little power interruptions as possible.

The cooperative has likewise been instrumental in keeping the city abreast with the latest technology in communications by providing efficient Electricity to power communications infrastructure in the ICT field.

Our teacher let us calculate the amount to be paid relating the current charges total amount and the KWH used.

Based on the electric bill given to us students, we calculated this ff. values:

2012:

KWH used: 237

Current bill: P 1620.62

Calculation: P 1620. 62 divided by 237

P 1620.62 / 237 = P 6.838

2014:

KWH used: 293

Current bill: P 2218.03

Calculation: P 2218.03   divided by 293

P 2218.03 / 293 = P 7.570

Upon getting these values, our teacher gave us another activity about this topic. That was a triad activity, wherein we listed different appliances/ other things we used at our homes relating to electricity consumption, and we enjoyed it. In computing, I ended up to a P 2000 plus value of electricity bill in our own households. The value I had calculated in the classroom is almost the same in our actual bill at home, which is 2000 plus also.

Example:   

The electricity consumption cost per household depends on family size, living habits, number and age of electrical appliances and hours of usage. Customer can calculate the estimate electricity cost for different appliances if he knows the following:

1. Power rating of the electrical appliance and its efficiency
2. Number of hours appliances being used
3. The domestic tariff rate per kilowatt – hour (kWh)

Electricity consumption usually increases due to the following reasons:

• Additional electrical appliances as the family member grow
• Electrical loading or size of the appliances
• Modern life style leads to using more electrical appliances
• Longer usage of appliances
• Capacity of appliances which can be adjusted at maximum, result on high load factor, e.g air condition, fan, water heater, etc.
• Replacement of smaller appliances to bigger capacity
• Faulty appliance will result in appliance operating longer hour and wasting electricity, e.g. refrigerator with faulty thermostat, shortage of refrigerant, or defective door gasket

To calculate: Electricity consumption (kWh) = Power (watts) x Hour of Operation x 30 days ÷ 1000. ( depending on how many days in a month you used the appliance/s)

CLICK THE IMAGES TO ENLARGE:

  1000W x 5hr= 5 000 Wh

5 000 Wh x 15 days= 75 000 Wh

75 000/ 1000= 75 kWh

  750W x 7 hr= 5 250 Wh

5 250 Wh x 30 days= 157 500 Wh

157 500/ 1000= 157.5 Kwh

  150 W x 5 hr= 750 Wh

750 Wh x 30 days= 22 500 Wh

22 500/ 1000= 22.5 Kwh

75W x 7 hr= 525 Wh

525 Wh x 30 days= 15 750Wh

15 750/ 1000= 15.75 Kwh

36 W x 7 hr= 252 Wh

252 Wh x 30 days=  7 560 Wh

7 560/ 1000= 7.56 Kwh

850 W x .5 hr= 425 Wh

425 Wh x 20 days= 8 500 Wh

8 500/ 1000= 8.5 Kwh

850 W x .5 hr= 425 Wh

425 Wh x 15 days= 6 375 Wh

6 375/ 1000= 6. 375 Kwh

730 W x .75 hr= 547.5 Wh

547.5 Wh x 30 days= 16 425 Wh

16 425/ 1000= 16.425 Kwh

We make a table with Quantity, Equipment, Hours/day, Wattage(W), Wattage Hour (WHr), Estimated kWH monthly.

Putting all the necessary information and calculated values in the table we made. Then, to complete the table, we summed up all the electricity consumption (kWh): 

Electricity consumption (kWh)= 75 kWh + 157.5 Kwh +22.5 Kwh  + 15.75 Kwh + 7.56 Kwh + 8.5 Kwh + 6. 375 Kwh + 16.425 Kwh

Electricity consumption (kWh)=  309.61 kWh

After we had solved the total Kwh, we calculate the amount to be paid based on the values we obtained upon solving it. Different values in the year 2012 and 2014, which are P 6.838 and P 7.570  respectively.

2012: 309.61 kWh *P 6.838 = P 2 117.113   

2014:  309.61 kWh *P 7.570 =P 2343.7477

Thus, in the year 2014, the payment increases to 226. 6347 pesos..

The image above relates to the percentage of home energy usage.

Electricity is more than numbers on a utility bill; it is the
foundation of everything we do. All of us use electricity every
day—for cooking, heating and cooling rooms,
manufacturing, lighting, and entertainment. We rely on electricity
to make our lives comfortable, productive, and enjoyable.
There are many things we can do to use less electricity and use
it more wisely. These things involve electricity conservation and
electricity efficiency. Electricity conservation is any behavior that
results in the use of less electricity. Electricity efficiency is the use
of technology that requires less electricity to perform the same
function.

TIPS TO SAVE ELECTRICITY TODAY:

Easy low-cost and no-cost ways to save energy

• Set a proper schedule in ironing all your clothes.
• Use less hot water ,for example shower instead of bath, and take shorter showers. Only fill the kettle as much as you need it. Wash a full load of dishes, rather than one dish at a time. Use cold water where possible for laundry washing.
• Reduce/ lower the refrigerator’s thermostat to lower utility bills and manage your heating and cooling systems efficiently.
• Air dry dishes instead of using your dishwasher’s drying cycle.
• Turn things off when you are not in the room such as lights, TVs, entertainment systems, and your computer and monitor.
• Switch off equipment when not in use, turn appliances off at the wall plug, rather than leaving them on standby as this can still draw about 20% or more of normal electricity use. (Examples are TVs, music systems, computers, phone chargers etc.)
• Use efficient lighting, Compact Fluorescent Lamps (CFLs) use 75% less power than old incandescent bulbs, and last much longer.
• Air dry clothes.
• Check to see that windows and doors are closed when heating or cooling your home.

The activity gave us more additional details about electricity consumption and how the electricity conserved at home.

For more information, just click the links below:

http://www.youtube.com/watch?v=ZxN5Hx64jmk

http://www.youtube.com/watch?v=mTYwsSjWowU

http://www.youtube.com/watch?v=9qB9_z6AKLY

Lessons I’ve learned:

Electricity is beneficial because it is clean, cheap, safe and a convenient source of energy. Most of technology is based or needs electricity.   One advantage of electricity is that we have a more convenient way of life – light, power to our house, it can give you lighting, heat, and computers wouldn’t work without it, etc. Electricity is called an energy carrier because it is a safe way to move energy from one place to another, and it can be used for so many tasks. As we use more technology, the demand for electricity grows. Learning how to conserve energy and use it efficiently are important goals for everyone. Some important tips I want to advice to save energy for lesser amount/ payment are:

• Measure and monitor your home electricity consumption and costs.
• Educate everyone in the home, including children and domestic helpers.
• Remember that saving requires both behaviour and equipment. Example, it’s no use installing an efficient shower head, if you then shower for twice as long.

In calculating the total wattage:Total wattage = wattage of equipment x number of equipment. For Wattage hour, it is equal to wattage of equipment x hour/day and for Estimated kWh use monthly = (wattage hour x days use in a month)/1000 = kwH.

Therefore, it is better to have a right use of electricity inorder to save money. It’s hard to imagine life without electricity, but you must be responsible enough not to waste electricity, but to use it properly. There, you can minimize your bill. Make sure you’re paying a low price per kilowatt. Use as few kilowatts as possible. With this in mind, we have developed a simple way to help you lessen your bill.

Power Triangle

Once, you have solved the apparent power (S), the power(P), the reactive power (Q), and the angle PF, you can now make a representation of those parameters. This is what we call the power triangle.

The relationship between real power, reactive power and apparent power. which can be expressed by representing the quantities as vectors. Real power is represented as a horizontal vector and reactive power is represented as a vertical vector. The apparent power vector is the hypotenuse of a right triangle formed by connecting the real and reactive power vectors. This representation is often called the power triangle. Using the Pythagorean Theorem, the relationship among real, reactive and apparent power is:

(apparent power)^2 = (real power)^2 + (reactive power)^2

Real and reactive powers can also be calculated directly from the apparent power, when the current and voltage are both sinusoids with a known phase angle between them

They are related, due to the fact that everything above deals with electrical units (watts ) and as an electrical engineering student, I am also responsible for all electrical aspects of a project including any solutions and all calculations.

For some details, click this link:

http://www.wisc-online.com/Objects/ViewObject.aspx?ID=ACE7903

In AC circuits, current and voltage are normally out of phase and, as a result, not all the power produced by the generator can be used to accomplish work. By the same token, power cannot be calculated in AC circuits in the same manner as in DC circuits. The power triangle, shown in Figure 1, equates AC power to DC power by showing the relationship between generator output (apparent power – S) in volt-amperes (VA), usable power (true power – P) in watts, and wasted or stored power (reactive power – Q) in volt-amperes-reactive (VAR). The phase angle (θ) represents the inefficiency of the AC circuit and corresponds to the total reactive impedance (Z) to the current flow in the circuit.

 CLICK THE IMAGE TO ENLARGE:

The power triangle represents comparable values that can be used directly to find the efficiency level of generated power to usable power, which is expressed as the power factor (discussed later). Apparent power, reactive power, and true power can be calculated by using the DC equivalent (RMS value) of the AC voltage and current components along with the power factor.

 See more at:

http://openticle.com/2012/09/24/power-triangle/#sthash.nmWb3w4A.dpuf

 http://www.youtube.com/watch?v=0CwQ6nNGnFM

Complex Power and Reactive Power

Complex power: S.(written in bold letter S) 

Complex power is the product of the complex effective voltage and the complex effective conjugate current. In our notation here, the conjugate is indicated by an asterisk (*).Complex power can also be computed using the peak values of the complex voltage and current, but then the result must be divided by 2. Note that complex power is applicable only to circuits with sinusoidal excitation because complex effective or peak values exist and are defined only for sinusoidal signals. The unit for complex power is VA. 

It is the product of the phasor voltage and complex conjugate of the phasor current.

V=Vm angle of Θv

I= Im angle of Θi

S= 1/2 VI*= Vrms Irms angle (Θv-Θi)

example: I= 5 angle 35 ° = 4.096 + j2.86

I= 5 angle -35 ° = 4.096 – j2.86

Calculating True Power in AC Circuits

As mentioned before, the true power of a circuit is the power actually used in the circuit. This power, measured in watts, is the power associated with the total resistance in the circuit. To calculate true power, the voltage and current associated with the resistance must be used. Since the voltage drop across the resistance is equal to the resistance multiplied by the current through the resistance, true power can be calculated by the formula:

Reactive Power, Q

The units of reactive power are Volt Amperes Reactive or VAR. Most users have an interest in real power and power factor, so reactive power is not calculated automatically.

The Power Analysis software is useful in analyzing line power. It simplifies the determination of real power, apparent power, and power factor by eliminating the need to set up math traces and parameter math. It is even more convenient to use than dedicated line power analyzers. The scope is already on your bench and the answers are only a button push away.

Calculating Reactive Power in AC Circuits 
The reactive power is the power returned to the source by the reactive components of the circuit. This type of power is measured in Volt-Amperes-Reactive, abbreviated var.

Reactive power is calculated by using the voltage and current associated with the circuit reactance.

Since the voltage of the reactance is equal to the reactance multiplied by the reactive current, reactive power can be calculated by the formula:

Another way to calculate reactive power is to calculate the inductive power and capacitive power and subtract the smaller from the larger.

Either one of these formulas will work. The formula you use depends upon the values you are given in a circuit. For example, find the reactive power of the circuit:

Since this is a series circuit, current (I) is the same in all parts of the circuit.

MY LEARNINGS: 

P is the average power in watts delivered to a load and it is the only useful power. Q is the reactive power exchange between source and the reactive part of the load. It is measured in VAR. Q= 0 is a resistive load (unity pf), Q<0 is a capacitive load (leading pf), and Q>0 is an inductive load (lagging pf).  Active power is the same as real power. Apparent power is symbolize as letter S in plain, while complex power is written in bold letter S.

For further information, please click on these links:

http://www.youtube.com/watch?v=g0S-XV-BiUA

http://www.youtube.com/watch?v=iqX8VI6sH_w

http://www.youtube.com/watch?v=BIxAjjm9BjQ

http://www.youtube.com/watch?v=UbKr7aYKSPU

Apparent Power and Power Factor

Apparent power is automatically computed and displayed as the parameter (S). For resistive loads, the apparent and average powers are equal.

The ratio of average to apparent power is the power factor. In the sinusoidal case, the power factor is equal to the cosine of the phase angle between the current and voltage waveforms. It is more generally computed as the ratio of real to apparent power. In our example the power factor is also computed automatically and displayed using the parameter pf.

It is the product of the applied voltage and current in an ac circuit. Apparent power, or volt-amps, is not the true power of the circuit since power factor is not considered in the calculation.

Apparent power is the product of rms values of voltage and current. It is measured in volt amperes or VA to distinguish it from the average or real power which is measured in watts.

When we discussed about this topic, lots of funny moments happened. Like playing the word “apparent”.

Seriously, apparent means easy to see, open to view, and easy to identify.

APPLICATION:

Apparent Power is the Total Power Flowing

When electrical power in an AC system flows to a load (a computer, motor, lighting, cooling, etc.) all of the power is not necessarily absorbed and used to perform useful work – a portion of the power may be reflected back to the power source (power company).

The total power flowing is known as the “apparent power” and is measured as the product of the voltage and current (V * I).. For example, if 208 volts and 5 amps are measured – the apparent power is 1040VA (VA means volt-amps – the measurement unit of apparent power).

 Apparent Power in AC Circuits Calculations:

Apparent power is the power that appears to the source because of the circuit impedance. Since the impedance is the total opposition to ac, the apparent power is that power the voltage source . Apparent power is the combination of true power and reactive power. Apparent power is not found by simply adding true power and reactive power just as impedance is not found by adding resistance and reactance.

To calculate apparent power, you may use either of the following formulas:

For example, find the apparent power:

Given:

Z= 100Ω and I= 5A

SOLUTION:

Power Factor

The POWER FACTOR is a number (represented as a decimal or a percentage) that represents the portion of the apparent power dissipated in a circuit.

If you are familiar with trigonometry, the easiest way to find the power factor is to find the cosine of the phase angle Θ The cosine of the phase angle is equal to the power factor.

You do not need to use trigonometry to find the power factor. Since the power dissipated in a circuit is true power, then:

If true power and apparent power are known you can use the formula shown above.

Going one step further, another formula for power factor can be developed. By substituting the equations for true power and apparent power in the formula for power factor, you get:

Since current in a series circuit is the same in all parts of the circuit, I_R equals I_Z. Therefore, in a series circuit,

For example, to compute the power factor for the series circuit  any of the above methods may be used.

 Another method:

 If you are familiar with trigonometry you can use it to solve for angleӨ and the power factor by referring to the tables in appendices V and VI.

NOTE: As stated earlier the power factor can be expressed as a decimal or percentage. In this example the decimal number .6 could also be expressed as 60%.

 Power Factor Correction

The apparent power in an ac circuit has been described as the power the source. As far as the source is concerned the apparent power is the power that must be provided to the circuit. You also know that the true power is the power actually used in the circuit. The difference between apparent power and true power is wasted because, in reality, only true power is consumed. The ideal situation would be for apparent power and true power to be equal. If this were the case the power factor would be 1 (unity) or 100 percent. There are two ways in which this condition can exist. (1) If the circuit is purely resistive or (2) if the circuit “appears” purely resistive to the source. To make the circuit appear purely resistive there must be no reactance. To have no reactance in the circuit, the inductive reactance (X_L) and capacitive reactance (X_C) must be equal.

 

 The expression “correcting the power factor” refers to reducing the reactance in a circuit.

The ideal situation is to have no reactance in the circuit. This is accomplished by adding capacitive reactance to a circuit which is inductive and inductive reactance to a circuit which is capacitive. For example, the circuit shown in figure 4-10 has a total reactance of 80 ohms capacitive and the power factor was .6 or 60 percent. If 80 ohms of inductive reactance were added to this circuit (by adding another inductor) the circuit would have a total reactance of zero ohms and a power factor of 1 or 100 percent. The apparent and true power of this circuit would then be equal.

power factor is defined as PF= P/S

• In sinusoidal circuits, PF is the cosine of the phase difference between the voltage and current. It is also the cosine of the angle of the load impedance.

For further information, click these links.

http://www.youtube.com/watch?v=k0mdwQvn2iw

http://www.youtube.com/watch?v=6Le9_zyWULE

Effective or RMS Value

As our teacher showed us his power point slides about this topic, I had learned  that Effective value or RMS value of an ac waveform is an equivalent dc value. It tells how many volts or amps of dc that an ac waveform supplies in terms of its ability to produce the same average power. RMS value is a constant itself that depends on the function’s shape i(t). Effective values tell us about a waveform’s ability to do work. An effective value is an equivalent dc value. It tells you how many volts or amps of dc that a time-varying waveform is equal to, in terms of its ability to produce average. They are “Root Mean Squared” (RMS) values.

RMS stands for root mean square as this describes the operation we perform to find the effective value. The terms RMS and effective are synonymous.

In this lesson, the word “effective” is the value of voltage in the current that gives the same amount of power in watts. The effective value of periodic current is equal to the current in dc which delivers the same average power to a resistor as the periodic current.

The total power dissipated by R is given by:

Hence, I eff ( effective current) is equal to:

CLICK THE IMAGE TO ENLARGE:

Do you know what the word effective means in this lesson?

>>> Our teacher said that effective means the value of voltage in the current which delivers the same amount of watts.

The effective value of a periodic current is equal to the dc current that delivers the same average power to a resistor as the periodic current.

The average (or mean) and effective (or RMS) values, are common used terms to indicate the magnitude of a periodic signal. This can be a voltage, current, power or another quantity. This article lists the equations for the average and effective values for a number of different waveforms.

The RMS value of a sinusoid i(t)= Im cos(ωt) is given by:

The average power can be written in terms of the rms values:

I remember what my teacher said, ” Note, if you express amplitude of a phasor source(s) in rms, then all the answer as a result of this power source(s) must be in rms value also.

Power is not anymore P=IV but power is P= 1/2 Vm Im cos(Θv-Θi).

Example is that the transformer has a 225 V. It is not maximum, but rms.

Example: As you can see, the sine wave shown below has a peak voltage of 6 Vp. Also, its peak-to-peak voltage is 12 Vpp. And, as we’ll see below, it’s effective voltage is 4.24 Vrms. So you can’t just say something like “The sine wave had a voltage of 6 V.” You’ve got to be careful to say whether you’re talking about peak voltage, peak-to-peak voltage, or effective voltage.

• This may seem confusing, but you have to be able to deal with it. It’s similar to the situation that we have with temperatures or distances: when you give the distance between two cities, you can give it either in miles or in kilometers. It’s the same distance, but expressed with two different units.
• These distinctions apply only to ac, not to dc.

MY LEARNINGS:

The root-mean-square (rms) value or effective value of an ac waveform is a measure of how effective the waveform is in producing heat in a resistance.Example: If you connect a 5 V_rms source across a resistor, it will produce the same amount of heat as you would get if you connected a 5 V dc source across that same resistor. On the other hand, if you connect a 5 V peak source or a 5 V peak-to-peak source across that resistor, it will not produce the same amount of heat as a 5 V dc source. That’s why rms (or effective) values are useful: they give us a way to compare ac voltages to dc voltages.To show that a voltage or current is an rms value, we write rms after the unit: for example,V_rms = 25 V rms.

For further information, just click these links:

http://www.youtube.com/watch?v=9ogLqX5uFfM

http://www.youtube.com/watch?v=E6NhhQpPEjk

 

Instantaneous Power and Average Power

         Instantaneous power is the quantity of power moving at single instant in time. In general, it is defined as follows: p(t) = v(t) x i(t) , this is the equation 1.

CLICK THE IMAGE TO ENLARGE:

That is, the value of power P at time t is equal to the voltage at time t times the current at time t. For a sinusoidal signal, note that the voltage swings through zero volts twice a cycle, so the instantaneous power is zero at least twice a cycle. If the load has both resistance and reactance, then the current, which also swings through zero twice a cycle, is offset from when the voltage swings through zero, and thus the instantaneous power will swing through zero 4 times a cycle, namely when:
p(t) x 0V x i(t) = 0V, and
p(t) x v(t) x 0V = 0V

Also note that instantaneous power can be negative for part of the cycle and positive for part of the cycle, if there is both resistance and reactance in the circuit. The sign indicates which direction the power flows.

CLICK THE IMAGE TO ENLARGE:

p(t) >0 : power is absorbed by the current

p(t) <0 : power is absorbed by the source

Your first formula doesn’t show the time variable t, so it is not accurate maybe in solving the instantaneous power. Normally the instantaneous voltage v(t) would be given something as:
v(t)= Vm cos(wt+ theta v), where w is the omega

The current i(t) would be given as a sinusoidal of similar format but different phase and amplitude. i(t)= Im cos(wt+ theta i), where w is the omega.

           Then use equation 1 to calculate the instantaneous power, or if you know the voltage and current at one point in time, just multiply them together to calculate the instantaneous power.

What’s usually of more interest is to know how much power a circuit delivers over a period of time. To find that, one averages the instantaneous power over the period of interest. Since most applications deliver constant power, you only need to average the power over one cycle to find the average power. The averaging is done via integration:

Pavg = 1 over T, Integral of p(t) dt from 0 to  T

             When dealing with sinusoids, it turns out that it is useful to determine the RMS value (root mean squared value) of the voltage v(t), and separately of the current i(t). (Or measure them.) I usually hear the letters “RMS”, and I really don’t know what does it mean. My other classmates don’t know what RMS means too. We only heard the meaning of it in our electronics teacher, and in our electric circuits teacher.

        Going on, as far as I remember, some says ,that if you hear of power being expressed in “RMS watts”, but what really is meant is that the power being expressed is the “average power” as defined above. That’s in contrast with manufacturers that like to express things in “peak watts”, which is the peak instantaneous wattage (much higher than average power). In order for you not to be confuse on it, it so called RMS watts.

The other thing that is useful about average power, Vrms and Irms is that those values are equivalents to the DC values necessary to transmit the same amount of power using DC.

         

  For further explanation, I want you to go on this link: http://www.powershow.com/view/11e3d5-NTFhN/CHAPTER_3_AC_POWER_ANALYSIS_powerpoint_ppt_presentation

http://www.youtube.com/watch?v=HsryuUsEh28

EXAMPLE:

Calculate the instantaneous power absorbed by a passive linear circuit if:

My solution:

CLICK THE IMAGE TO ENLARGE:

 

>>>sine to cosine> just  subtract 90 degrees to the angle… ( 60+(-90)), afterwards substitute the given values to the formula given in getting the power.

Average Power

The average power P is the average of the instantaneous power over one period.

CLICK THE IMAGE TO ENLARGE:

• P is not time dependent
• when theta v= theta i, it is a purely resistive load, P= 1/2 VmIm
• when theta v-theta i= positive or negative 90, it is a purely resistive load, P=0

Our teacher introduced these two important figures. Just remember ICE and ELI.

EXAMPLE:

A current I= 10<30 flows through an impedance Z= 20<-22. find the average power delivered to the impedance.

My solution:

First, I solved for the voltage using ohm’s law, since current and impedance are already given.

Next, just substitute the values given to the formula to get the answer. The answer must be equal to P=927.18W

For further explanation, just click the links:

http://www.youtube.com/watch?v=lcXWii1ffuI

http://www.youtube.com/watch?v=qdmUtsXk6Y8

AC POWER ANALYSIS

Introduction:

We are studying electrical circuits subject, and we must know, of course, that every electrical device has a power rating, which indicates how much power the equipment or material requires. If we exceed the power rating, obviously it can cause permanent damage in such device, machine, or equipment. You must be careful.

Power is the most important quantity in electric utilities, and other electronic systems. Electric power is the rate of energy consumption in an electrical circuit. The electric power is measured in units of watts.

The electric power P is equal to the energy consumption E divided by the consumption time t: P= E/t

• P is the electric power in watt (W).
• E is the energy consumption in joule (J).
• t is the time in seconds (s).

As we solve for the electric power, we used these formulas depending on its given parameters:

P = V . I  or

P = I^2 . R ( read as I squared times R )  or

P = V^2 / R  ( read as V squared over R )

• P is the electric power in watt (W).
• V is the voltage in volts (V).
• I is the current in amps (A).
• R is the resistance in ohms (Ω).

             The choice of power delivery in 50 or 60 Hz AC form is due to the allowed high voltage power transformation.

             We’ve seen the formula for determining the power in an electric circuit: by multiplying the voltage in “volts” by the current in “amps” we arrive at an answer in “watts.”

            But, in the next topics, we will not use the same formula. Let’s take a look in my next posts on this blog about “Instantaneous Power and Average Power”, and ” Maximum Average Power Transfer”,  for you to know what i mean.