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.