Circuits

Power Triangle

Complex Power and Reactive Power

Apparent Power and Power Factor

Power in a Balanced System

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





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







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







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 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

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.





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 (Ia, Ib, and Ic). This definition does not change with connection type.

Line Voltages

the voltages between any two of the lines (Vab, Vbc, and Vca). 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 (Va, Vb, and Vc). 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 (Ia, Ib, and Ic).

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

_________________________________________________________________________



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°




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





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 threerequirements 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”
All 3 variables have the same amplitude
All 3 variables have the same frequency
All 3 variables are 120 degrees in phase





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°





Thus,

Vb = Va (1 ∠ -120o) , and Vc = Va (1∠ +120o)

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.



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 (Vab, etc.)Phase currents = currents from line to line (Iab, 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:



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

_________________________________________________________________________


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 theimpedance 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 )





Vab = Va – Vb = Vm ∠Φ – Vm ∠Φ – 120o

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

thus, Va = Vm∠ 0o, and Vb = Vm∠ -120o, so

Vab = Vm ∠ 0o – Vm ∠– 120o = Vm (1 – 1 ∠– 120o ) = Vm (1 – (cos 120o – j sin 120o))

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

Converting to polar form,

Vab = Vm √[ (3/2)2 + (√3 / 2)2 ] tan-1 {(√3 / 2) / (3/2) }

= Vm √[ 9/4 + 3/4 ] tan-1 {1/√3 }

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

= Vm √3 tan-1 {tan 30o } = Vm ∠ 30o

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

Vab = Va √3 ∠ 30o

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

Ia = IAB – ICA

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

Ia = IAB √3 ∠ -30o

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 (Va and -Vb ) we obtain the resulting Vab shown in the figure.

Since Vab is longer, we know . . . . |Vab| = √3 |Va| ,

and since Vab is ahead of Va, we know that, . . . . (the angle of Vab) = (the angle of Va) + 30o

Graphical Voltage Relationship

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

Since Ia is longer, we know

|Ia| =√3 |Iab| ,

and since Ia is behind Iab, we know that,

(the angle of Ia) = (the angle of Iab) – 30o



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 Y- Δ Connection:






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




_________________________________________________________________________


A balanced Δ-Δ system





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




_________________________________________________________________________


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°




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