BLOG 24 - FUNDAMENTAL CIRCUIT

Three-phase Y and Delta 

Initially we explored the idea of three-phase power systems by connecting three voltage sources together in what is commonly known as the “Y” (or “star”) configuration. This configuration of voltage sources is characterized by a common connection point joining one side of each source. (Figure below)

Three-phase “Y” connection has three voltage sources connected to a common point.

If we draw a circuit showing each voltage source to be a coil of wire (alternator or transformer winding) and do some slight rearranging, the “Y” configuration becomes more obvious in Figure below.

Three-phase, four-wire “Y” connection uses a "common" fourth wire.

The three conductors leading away from the voltage sources (windings) toward a load are typically called lines, while the windings themselves are typically called phases. In a Y-connected system, there may or may not (Figure below) be a neutral wire attached at the junction point in the middle, although it certainly helps alleviate potential problems should one element of a three-phase load fail open, as discussed earlier.

Three-phase, three-wire “Y” connection does not use the neutral wire.

When we measure voltage and current in three-phase systems, we need to be specific as to where we're measuring. Line voltage refers to the amount of voltage measured between any two line conductors in a balanced three-phase system. With the above circuit, the line voltage is roughly 208 volts. Phase voltage refers to the voltage measured across any one component (source winding or load impedance) in a balanced three-phase source or load. For the circuit shown above, the phase voltage is 120 volts. The terms line current and phase current follow the same logic: the former referring to current through any one line conductor, and the latter to current through any one component.

Y-connected sources and loads always have line voltages greater than phase voltages, and line currents equal to phase currents. If the Y-connected source or load is balanced, the line voltage will be equal to the phase voltage times the square root of 3:

However, the “Y” configuration is not the only valid one for connecting three-phase voltage source or load elements together. Another configuration is known as the “Delta,” for its geometric resemblance to the Greek letter of the same name (Δ). Take close notice of the polarity for each winding in Figure below.

Three-phase, three-wire Δ connection has no common.

At first glance it seems as though three voltage sources like this would create a short-circuit, electrons flowing around the triangle with nothing but the internal impedance of the windings to hold them back. Due to the phase angles of these three voltage sources, however, this is not the case.

One quick check of this is to use Kirchhoff's Voltage Law to see if the three voltages around the loop add up to zero. If they do, then there will be no voltage available to push current around and around that loop, and consequently there will be no circulating current. Starting with the top winding and progressing counter-clockwise, our KVL expression looks something like this:

Indeed, if we add these three vector quantities together, they do add up to zero. Another way to verify the fact that these three voltage sources can be connected together in a loop without resulting in circulating currents is to open up the loop at one junction point and calculate voltage across the break: (Figure below)

Voltage across open Δ should be zero.

Starting with the right winding (120 V ∠ 120o) and progressing counter-clockwise, our KVL equation looks like this:

Sure enough, there will be zero voltage across the break, telling us that no current will circulate within the triangular loop of windings when that connection is made complete.

Having established that a Δ-connected three-phase voltage source will not burn itself to a crisp due to circulating currents, we turn to its practical use as a source of power in three-phase circuits. Because each pair of line conductors is connected directly across a single winding in a Δ circuit, the line voltage will be equal to the phase voltage. Conversely, because each line conductor attaches at a node between two windings, the line current will be the vector sum of the two joining phase currents. Not surprisingly, the resulting equations for a Δ configuration are as follows:

Let's see how this works in an example circuit: (Figure below)

The load on the Δ source is wired in a Δ.

With each load resistance receiving 120 volts from its respective phase winding at the source, the current in each phase of this circuit will be 83.33 amps:

So each line current in this three-phase power system is equal to 144.34 amps, which is substantially more than the line currents in the Y-connected system we looked at earlier. One might wonder if we've lost all the advantages of three-phase power here, given the fact that we have such greater conductor currents, necessitating thicker, more costly wire. The answer is no. Although this circuit would require three number 1 gage copper conductors (at 1000 feet of distance between source and load this equates to a little over 750 pounds of copper for the whole system), it is still less than the 1000+ pounds of copper required for a single-phase system delivering the same power (30 kW) at the same voltage (120 volts conductor-to-conductor).

One distinct advantage of a Δ-connected system is its lack of a neutral wire. With a Y-connected system, a neutral wire was needed in case one of the phase loads were to fail open (or be turned off), in order to keep the phase voltages at the load from changing. This is not necessary (or even possible!) in a Δ-connected circuit. With each load phase element directly connected across a respective source phase winding, the phase voltage will be constant regardless of open failures in the load elements.

Perhaps the greatest advantage of the Δ-connected source is its fault tolerance. It is possible for one of the windings in a Δ-connected three-phase source to fail open (Figure below) without affecting load voltage or current!

Even with a source winding failure, the line voltage is still 120 V, and load phase voltage is still 120 V. The only difference is extra current in the remaining functional source windings.

The only consequence of a source winding failing open for a Δ-connected source is increased phase current in the remaining windings. Compare this fault tolerance with a Y-connected system suffering an open source winding in Figure below.

Open “Y” source winding halves the voltage on two loads of a Δ connected load.

With a Δ-connected load, two of the resistances suffer reduced voltage while one remains at the original line voltage, 208. A Y-connected load suffers an even worse fate (Figure below) with the same winding failure in a Y-connected source

Open source winding of a "Y-Y" system halves the voltage on two loads, and looses one load entirely.

In this case, two load resistances suffer reduced voltage while the third loses supply voltage completely! For this reason, Δ-connected sources are preferred for reliability. However, if dual voltages are needed (e.g. 120/208) or preferred for lower line currents, Y-connected systems are the configuration of choice.

LEARNING

I this topic the Three Phase, i learn that the conductors connected to the three points of a three-phase source or load are called lines. The three components comprising a three-phase source or load are called phases. Line voltage is the voltage measured between any two lines in a three-phase circuit. Phase voltage is the voltage measured across a single component in a three-phase source or load. Line current is the current through any one line between a three-phase source and load. Phase current is the current through any one component comprising a three-phase source or load. In balanced “Y” circuits, line voltage is equal to phase voltage times the square root of 3, while line current is equal to phase current. 

BLOG 23 - FUNDAMENTAL CIRCUIT

Complex Power

In power system analysis the concept of Complex Power is frequently used to calculate the real and reactive power.

This is a very simple and important representation of real and reactive power when voltage and currentphasors are known. Complex Power is defined as the product of Voltage phasor and conjugate of current phasor. See Fig-A

Let voltage across a load is represented by phasor V  and  current through the load is I.

If S is the complex power then,

                 S = V . I^*

V is the phasor representation of voltage and I* is the conjugate of current phasor. 

So if  V is the reference phasor then V can be written as |V| ∠0.

(Usually one phasor is taken reference which makes zero degrees with real axis. It eliminates the necessity of introducing a non zero phase angle for voltage)

Let current lags voltage by an angle φ, so  I = | I | ∠-φ

(current phasor makes -φ degrees with real axis)

                            I^*=  | I | ∠φ

So,

                           S = |V|  | I | ∠(0+φ) =  |V|  | I | ∠φ

(For multiplication of phasors we have considered polar form to facilitate calculation)

Writting the above formula for S in rectangular form we get

                           S =  |V|  | I | cos φ  +  j  |V|  | I | sin φ 

             The real part of complex power S is |V| | I | cos φ which is the real power or average power and the imaginary part  |V| | I | sin φ is the reactive power. 

             So,              S = P + j Q

             Where        P = |V| | I | cos φ    and    Q = |V| | I | sin φ

It should be noted that S is considered here as a complex number. The real part P is average power which is the average value, where as imaginary part is reactive power which is a maximum value. So I do not want to discuss further and call S as phasor. If you like more trouble I also advise you to read my article about phasor or some other articles on phasor and complex numbers.

Returning to our main point, from the above formula it is sure that P is always more than zero. Q is positive when φ is positive or current lags voltage by φ degrees. This is the case of inductive load. We previously said that inductance and capacitance do not consume power. The power system engineers often say about reactive power consumption and generation. It is said that inductive loads consume reactive power and capacitors produce reactive power. This incorrect terminology creates confusion.

The fact is that most of the loads are inductive and they unnecessarily draw more current from source. Although in each cycle both inductance and capacitance draw power from the source and return same amount of power to the source but the behavior of inductance and capacitance are opposing to each other. When capacitors are connected in parallel to inductive load the power requirement of inductive load is supplied by capacitor in half cycle and in next half cycle the reverse happens. Depending upon the values of capacitor this power requirement of inductance in the load may be fully or partially satisfied. If partially satisfied the rest will be drawn from the distant source. By properly selecting the capacitance the maximum value of reactive power (Q) drawn from the distant source (or returned to the distant source) is reduced. This reduction in reactive power results in reduction of line current so the reduction of losses in transmission line and improvement in voltage at load end.

Power Triangle

Returning to the complex power formula, P, Q and S are represented in a power triangle as shown in figure below.

S is the hypotenuse of the triangle, known as Apparent Power. The value of apparent power is |V|| I |

or    |S| = |V|| I |

 It is measured in VoltAmp or VA.

P is measured in watt and Q is measured in VoltAmp-Reactive or VAR. In power systems instead of these smaller units larger units like Megawatt, MVAR and MVA is used.

The ratio of real power and apparent power is the power factor of the load.

power factor = Cos φ = |P| / |S|

                      = |P| / √(P ²+Q ²)

The reactive power Q and apparent power S are also important in power system analysis. As just shown above the control of reactive power is important to maintain the voltage within the allowed limits. Apparent power is important for rating the electrical equipment or machines.

Total Power of Parallel Circuits

In real world the loads are usually connected in parallel. Here we will show the total power consumed by parallel branches. See Figure-C. It has two branches.

First we have to draw the individual power triangles for each branch. Next the power triangles are arranged back to back keeping real power in positive x direction as shown. The total power consumed is obtained by connecting starting point O to the tip of last triangle. This is actually the result of addition of complex numbers.

If   S₁ = P₁ +j Q₁

     S₂ = P₂ +j Q₂

       

     Then,  S = S₁+  S₂  

   or    S = (P₁+  P₂  ) + j  (Q₁+  Q₂ )

         P = P₁+  P₂ 

        Q = Q₁+  Q₂ 

_{In the above diagram} S₁  P_{1 ,}   Q_{1 and} φ ₁ correspond to branch1 and S₂  P_{2 ,}   Q_{2 and} φ ₂  correspond to branch2.

S, P, Q and φ correspond to total power consumption as seen by the generator

LEARNING:

I learn that the 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.The power triangle graphically shows the relationship between real (P), reactive (Q) and apparent power (S).

BLOG 22 - FUNDAMENTAL CIRCUIT

 Apparent Power in AC Circuits

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 "sees." 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 for the circuit shown in figure (22)
 

Recall that current in a series circuit is the same in all parts of the circuit.

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 Ө. 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 shown in figure (22), 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.

LEARNING

I learn about the apparent power and power factor. The apparent power is a combination of both reactive power and true power.Power factor is the ratio between the kW and the kVA drawn by an electrical load where the kW is the actual load power and the kVA is the apparent load power and it is a measure of how efficiently the load 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.

BLOG 21 - FUNDAMENTAL CIRCUIT

RMS VALUE or Effective

WHAT IS RMS VALUE?

When dealing with Alternating Voltages (or currents) we are faced with the problem of how we represent the signal magnitude. One easy way is to use the peak values for the waveform. Another common method is to use the effective value which is also known by its more common expression of Root Mean Square or simply the RMS value.

Note that the RMS value is not the same as the average of all the instantaneous values. The ratio of the RMS value of voltage to the maximum value of voltage is the same as the ratio of the RMS value of current to the maximum value of current. Most multi-meters, either voltmeters or ammeters, measure RMS value assuming a pure sinusoidal waveform. For finding the RMS value of non-sinusoidal waveform a “True RMS Multimeter” is required.

Having now determined the RMS value of an alternating voltage (or current) waveform, in the next tutorial we will look at calculating the “Average” value VAV of an alternating voltage and finally compare the two.

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• THERE TWO TYPE BASIC METHOD IN RMS VALUE

RMS Voltage Graphical Method

Whilst the method of calculation is the same for both halves of an AC waveform, for this example we will consider only the positive half cycle. The effective or RMS value of a waveform can be found with a reasonable amount of accuracy by taking equally spaced instantaneous values along the waveform.

The positive half of the waveform is divided up into any number of “n” equal portions or mid-ordinates and the more mid-ordinates that are drawn along the waveform, the more accurate will be the final result. The width of each mid-ordinate will therefore be no degrees and the height of each mid-ordinate will be equal to the instantaneous value of the waveform at that time along the x-axis of the waveform.

Graphical Method

 

Each mid-ordinate value of a waveform (the voltage waveform in this case) is multiplied by itself (squared) and added to the next. This method gives us the “square” or Squared part of the RMS voltage expression. Next this squared value is divided by the number of mid-ordinates used to give us the Mean part of the RMS voltage expression, and in our simple example above the number of mid-ordinates used was twelve (12). Finally, the square root of the previous result is found to give us the Root part of the RMS voltage.

Then we can define the term used to describe an RMS voltage (VRMS) as being “the square root of themean of the square of the mid-ordinates of the voltage waveform” and this is given as:

 

and for our simple example above, the RMS voltage will be calculated as:

 

So lets assume that an alternating voltage has a peak voltage (Vpk) of 20 volts and by taking 10 mid-ordinate values is found to vary over one half cycle as follows:

 

Voltage	6.2V	11.8V	16.2V	19.0V	20.0V	19.0V	16.2V	11.8V	6.2V	0V
Angle	18o	36o	54o	72o	90o	108o	126o	144o	162o	180o

The RMS voltage is therefore calculated as:

 

Then the RMS Voltage value using the graphical method is given as: 14.14 Volts.

RMS Voltage Analytical Method

The graphical method above is a very good way of finding the effective or RMS voltage, (or current) of an alternating waveform that is not symmetrical or sinusoidal in nature. In other words the waveform shape resembles that of a complex waveform. However, when dealing with pure sinusoidal waveforms we can make life a little bit easier for ourselves by using an analytical or mathematical way of finding the RMS value.

A periodic sinusoidal voltage is constant and can be defined as V(t) = Vm.cos(ωt) with a period of T. Then we can calculate the root-mean-square (rms) value of a sinusoidal voltage (V(t)) as:

 

Integrating through with limits taken from 0 to 360o or “T”, the period gives:

 

Dividing through further as ω = 2π/T, the complex equation above eventually reduces down too:

RMS Voltage Equation

Then the RMS voltage (VRMS) of a sinusoidal waveform is determined by multiplying the peak voltage value by 0.7071, which is the same as one divided by the square root of two ( 1/√2 ). The RMS voltage, which can also be referred to as the effective value, depends on the magnitude of the waveform and is not a function of either the waveforms frequency nor its phase angle.

From the graphical example above, the peak voltage (Vpk) of the waveform was given as 20 Volts. By using the analytical method just defined we can calculate the RMS voltage as being:

VRMS = Vpk x 0.7071 = 20 x 0.7071 = 14.14V

 

Note that this value of 14.14 volts is the same value as for the previous graphical method. Then we can use either the graphical method of mid-ordinates, or the analytical method of calculation to find the RMS voltage or current values of a sinusoidal waveform. Note that multiplying the peak or maximum value by the constant 0.7071, ONLY applies to sinusoidal waveforms. For non-sinusoidal waveforms the graphical method must be used.

LEARNING:

The r.m.s. value of an a.c. supply is the steady d.c. which would convert electrical energy to thermal energy at the same rate in a given resistance. In RMS  VALUE we need to use the formula of Irms = I/square root of 2 and Vrms = V/square root of 2. The RMS value have two type of basic method the graphical and analytical method.