Power Correction in Three phase

Power factor is the ratio between the Power (Watt) and the Apparent Power measured in VA(voltage-ampere)  drawn by an electrical load . It is written as PF= P/S.

Using the Power triangle:

P= VpIpcosθ

S= VpIp

Q=Ssinθ

S= VpIp*   w/ angle

We can sum up all the Power and  Apparent Power (S).

P = P1 + P2 + P3

S= S1 + S2 + S3

EXAMPLE

LINE VOLTAGE = 34.5     60Hz             Find C when the power factor is 0.94                                                                                                                                                                                                    leading

Solve first the P, S and Q using the 0.78 pf lagging.

Using the power triangle:

pf= arcos(0.78) = 38.74

The given is the Apparent Power equal to 24 MVA, we only need to solve the P and Q  to complete the triangle.

P = Scos38.74

P = 24cos38.74

P =  18.72 MV

Q= Ssin38.74

Q = 24sin38.74

Q = 15.01 MVAR

These values is for the old.

FOR THE NEW ONE

Using the 0.94 pf leading and the P= 18.72   ( note that the power will just the same to the old one)

θ= arcos(0.94)

θ= 19.94

solve the Q;

using the trigonometric function

Q=Ptanθ

Q= 18.72tan(19.94)

Q= 6.79MVAR

we can disregard the S, since we only need the  Q.

the power triangle for the New one is:

Since the Q in the new is in below, the sign is  negative

Qc= Qnew – Qold

Qc= -6.79 -15.01

Qc = -21.79

Solving for the Capacitor

C= Q/2πf V^2rms

C= -21.79×10^6/ 2π(60) (34.5×10^3)^2

C= + 48 microFarad   (disregard the sign)

 

THREE-PHASE POWER ANALYSIS

There are three kinds of power- real (P) , reactive (Q) and apparent(S).

REAL POWER :

P = VpIpcos(θv – θi)

it is measured in Watts(W)

where p is the phase voltage and phase current in the system.

if the system is in line voltage and line current:

P= (VoIocos(θv-θi))/sqr(3)   

 

REACTIVE POWER:

 

Q =  VpIpsinθ

 

or 

Q = VoIosinθ

 

APPARENT POWER:

 

S= VpIp

S = sqr(P^2 + Q^2)

or 

S = VoIo/sqr(3)

 

 

 

            POWER TRIANGLE

 

pf( power factor) = cosθ

 

 

Example:

 

Y-Y connected

 

z= 0.4 + j0.3 ohms / phase

Zy = 24 + j19 ohms / phase

Zo = 0.6 + j0.7 ohms / phase

pf= 0.8 lagging

Van = 120< 30° 

Find 

S ,P, Q

Solution:

 

Using ohm’s law

V= IZ

I=V/Z

I= (120< 30° )/(0.4 + j0.3 + 24 + j19 + 0.6 + j0.7)

 

I = 3.75<-8.66°  A

 

P= (120)(3.75)cos(30+8.66) 

P= 351.39 W

 

Q= Ptanθ

 

θ=arcos(pf)

θ= arccos(0.8)

θ= 36.87

 

Q=  351.39tan36.87

Q – 263.54 VAR

 

S= sqr( 263.54 ^2 + 351.39^2)

S= 439.24 VA

 

 

THREE-PHASE CIRCUITS

Before we discuss about this circuit, the first thing is to discuss about the power system.

Power System

The power system used three phase circuits , this means that the alternating current(AC) have three different phases.(see below)

But why three phases?

There are two reasons for this:

1.  Three-phase generators can be driven by constant force or torque.

2. Industrial applications, such as high-power motors, welding equipments, have constant power output if they are three-phase systems

How ac generator work?

If you know about the electromagnetism theory, that there is  an electric field present in a wire. This had been proven by the famous scientist Micheal Faraday. That there is a current present in a wire when there is a magnetic force acting on it.

One-phase voltage sources

Three-phase circuits

CONFIGURATION OF THREE PHASE CIRCUIT SYSTEM

Wye (left) and Delta (right) connected voltage sources

Va = Vb = Vc    same magnitude but the angle is different by a factor of 120 degree.

The figure below is the circuit for a three phase circuit ( Y-Y)

This is a sample of Y-Y circuit. Load is combination of the load and line impedance  ( Zy). And in the end of it is the neutral.

To solve this kind of problem, we should only get one of those three phase circuit because the value used is just equal.  To solve for the line current ( that is going to the line impedance)  use the ohm’s law I = V/Zy. Used this formula but have different angle.

If Load is Delta,   The Zy divide it  by 3.

If the Voltage source, Vp/sqr(3)   ( θ old – 30)

SEQUENCES

abc (positive) sequence b is lag a by 120 degree.

acb (negative) sequence b is lead a by 120 degree.

ENERGY CONSUMPTION COST

Electrical consumption cost is the amount of electricity consumed by the household appliances. It is measured in kWh.

HOW DOES IT MEASURED

The electric company used a device to measured the amount of electric of each houses in the area. This is called the meter.

Digital meter

Every end of the month, they read the meter and using their device they instantly give the electrical statement.

All appliances have the same electrical consumption (Watt). Look at the picture below:

The refrigerator is the 24/7 on in all appliances and it consumed 727W per hour, think about it.  The television (flat screen) consumed only 133 W/h not like the tube TV consumed almost 330W/h.

HOW TO CALCULATE THE MONTHLY USED OF ELECTRICITY

The formula to compute the monthly electricity consumtion is:

Wattage of appliance x average hrs used in a day x no. of days in a month that it is used x the electrical cost

For example the flat screen TV consumed 133W and  you used it 8 hrs every weekends but on weekdays you used it 3 hrs only. The electrical cost is 7 pesos per KWh  To solved this problem, used the table to know the wattage of this appliance and then compute how many days in weekends and weekdays in a month. Then solved it using the formula.

For weekends

133W x 8hrs = 1064Wh or 1.064KWh

1.064 x 8 days = 8.512KWh in a month

For weekdays

133W x 3hrs = 399 W

399W x 22 days =8.778KWh in a month

Sum it up

8.512 + 8.778 =17.29

17.29 x 7 = 121. 03 pesos

The cost is 121.03 pesos. This is only in your TV not included the refrigerator iron and air conditioner.

TIPS TO LESSEN THE AMOUNT OF COST

1. Unplug the appliance is not in used. like the tv if it is in standby mode it also consumed electricity.

2. Using the washing machine and iron, create a schedule to used it for example once a week.

3. Before going to bed, check all the appliances if it is turn off /unplug and also turn down the thermostat of the refrigerator to lessen the cool.

4. Used only the rice cooker if it is needed, like you are busy for something and you can used this but if not you can used the stove or the coal to cook rice.

5. Used only the CFL to save more energy.

6. In case of brownout, Unplug all the appliance for safety and do not plug it immediately if the electricity came back.

7.  If buying new appliances, look the Energy efficiency first before buying it.

8. Used only the computer or laptop in important things.

MAXIMUM POWER TRANSFER

MAXIMUM AVERAGE POWER TRANSFER

To obtain the  maximum average power transferred from a source to a load, the load impedance should be chosen equal to the conjugate of the Thevenin equivalent impedance representing the reminder of the network. This means that the load impedance must be solved using the thevenin’s theorem

The formulas to solve for ZL:

STEPS IN SOLVING THE ZL AND POWER TRANSFER

1. Solve the load impedance (ZL) using the Thevenin’s theorem. Short the voltage source and open the current source.

2. Solve the Vth where Zl parallel to it, this means that the ZL is not concluded to solve the Vth.

3. The ZL must be in rectangular form . Used only the real part ( Rth +/- jXth)

4. The RL is equal to the conjugate of Zth. (see in formula)

5. Solve the power using the  formula below:

where: Vth=Voc

(Click the image to enlarge)

      The maximum average power transfer is a method to identify the maximum impedance load that is needed in the given ac circuit. This can be solve using thevenin’s theorem.

EFFECTIVE OR RMS VALUE

   The effective or root mean square(RMS) value of a periodic current is the DC value that delivers the same average power to a resistor as the periodic current.

IN SINUSOID

The Root Mean Square (RMS) value of a sinusoidal voltage or current is equal to the maximum value divided by square root of 2.

APPARENT AND COMPLEX POWER

• Real power (P) [Unit: W]
• Reactive power (Q) [Unit: VAR]
• Complex power (S) [Unint :VA]
• Apparent Power (|S|)

Real power  (P) is equal to the Voltage magnitude times Current magnitude time cos(angle of v -angle o i).

P=VrmsIrmscos(Av-Ai)

Reactive power (Q)= to Ssin(angle v-angle i)

Apparent power is the power that is “appears” to flow to the load. (S= |Vrms||Irms|)

Complex power (solve in polar form) = Vrms(Irms)*.    note * is the conjugate or the angle will be in opposite sign.

or S = P + jQ  (if converted in rectangular)

where P is the real power and Q is the reactive power.

power factor(pf) = P/S = cos(angle v-angle i)

Example (Click to enlarge)

USING THE POWER TRIANGLE

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

Example: Find the complex power of the three loads

Solution:

Using the power triangle, it is more easily to solve the given problems and if you know how to used trigonometric functions then the problem is a piece of cake to you. But first things to do in solving the problem is to understand it, what is asked and what method is easily to solve the given problem

AC POWER ANALYSIS

INSTANTANEOUS AND AVERAGE POWER

Instantaneous Power is a power at any instant of time. It is measured as watts.

p(t)=v(t)i(t) or

p(t)= VmImcos(ωt + θv)cos(ωt + θi)

Vm and Im is the amplitudes (or peak values) and the θv, θi are the phase

Using trigonometric Identity

p(t)= VmImcos(ωt + θv)cos(ωt + θv)= 1/2(VmImcos(θv-θi)+ VmIm(2ωt +θv+θi)

The instantaneous power entering the circuit

The instantaneous power is changing with respect to time so it is difficult to measure the power. The other method to solve the power is used the AVERAGE POWER formula, in fact the instrument used to measured the power is the wattmeter the used average power.

P = 1/2 VmImcos(θv-θi)

Thevenin’s and Norton Theorem

Any combination of batteries and resistances with two terminals can be replaced by a single voltage source e and a single series resistor r. The value of e is the open circuit voltage at the terminals, and the value of r is e divided by the current with the terminals short circuited.

 

STEPS

First is to solve for the Zth, thevenin’s impedance. The things to do is to turn off the sources ( short the voltage and open the current).

Second is to solve the Vth, thevenin’s voltage.  In the sample given above, Just solve the voltage on the R3, the R2 is cannot be used because it is open circuit, not connected to the circuit. The voltage across R3 is equal to the Vth.

 

Source Transformation AC ANALYSIS

Since ac circuit are linear, the Source Transformation theorem applies to ac circuits the same way it applies on the dc circuit. Source transformation is to transform the circuit into a simpler form. The formula used to transform the circuit is the Ohm’s law. 

STEPS:

First is to analyse the circuit how to attack it. 

To transform voltage to current source, the source must be series on the impedance and used ohm’s law. If it is current to voltage source the current source must be parallel to the impedance. (see below)  

 

 

After transforming, there is a time that the impedance can combine to the others (parallel or series) or the sources can be combine as one.   

Repeating the steps above until the circuit can be solved easily using (current or voltage division)

Just like below:

 

Circuit A the original circuit, Is is parallel to R so it transform the voltage source. After that i can be solved easily the V1 and V2 using voltage division formula.

 

 

          I learned that it is not all sources (voltage and current) be converted in the circuit. The first thing to do is to know what is the unknown or to understand the problem. Second is how to make the circuit simplify and where to used the source transformation. If the unknown is the current or voltage drop, do not touch that part or there is a dependent variable. 

 

 

 

Superposition

A superposition a  method to solve a unknown by killing the source( voltage or current). 

STEPS 

* First is to know what is the goal or what we find on the problem.

* Second  is to kill a source. only remain one source to used.

             Killing the source means that we remake the circuit( if it is voltage source, short it, but if it is current source, open it)

* Solve the unknown by using any methods ( ohms law, voltage or  current division)

 * Repeat it to the next source.

* Add it all. 

 

Sinusoidal Steady-State Analysis (Mesh Analysis)

To solve a sample circuit using mesh analysis:

1. Create a loop on the circuit and label it (e.g. I1, I2, I3)  Just like below:

2. Create an equation on each loop using KVL method.

@ I1

Vs -I1R1 – (I1-I2) R4 – (I1-I3)R5 = 0

@ I2

-I2R2  – I2R3 – (I2 – I3)R6 -(I2 – I1)R4  = 0

@ I3

there is a current source so that will be the -I3.

3. Solve the given equations using any methods. I prefer to use cramer’s rule.

4.Note that the current solved here is not really current pass through each impedance.

there is a formula to solve in a given branch ( if there is a two mesh current passes through. used a formula..)

I= Iccw – Icw

5. Used the currents to solve the parameters or the unknown.

 

 

The mesh analysis is  a method to solve a parameters  for a given circuit. This is more useful if there are many current source in a given circuit. The method used here is KVL as an equations. The number of loop is also the number of equations to create. I learned that you should not always to create a loop immediately. The first thing to do is to analyse first the problem if it can be simplify before attacking it.