Date post: | 15-Jun-2015 |
Category: |
Documents |
Upload: | hafiz-azman |
View: | 355 times |
Download: | 0 times |
EEE3233POWER SYSTEM
CHAPTER 3:3-Phase System and
Power Concept
nur diyana kamarudin
Introduction The generation, transmission
and distribution of electric power is accomplished by means of three-phase circuit.
At the generating station, three sinusoidal voltage have equal magnitude and an equal 1200-phase difference between any two phase.
This is called a balanced source.
Positive phase sequence when the voltages reaches their peak value in the sequential ABC
Negative phase sequence when the phase is order is ACB.
ECn
EBn
EAn
EAn
ECn
EBn
Positive or ABC phase sequence
Negative or ACB phase sequence
Y-Connected Load
A Y-connecter generator supplying a Y-connected load.
Y-Connected Load (cont.)
Assume a positive sequence to find the relationship between the line voltage (line-to-line voltage) and phase voltage (line-to-neutral voltage).
Line –to-neutral voltage of the phase is chosen as a reference.
|Vp|= magnitude of phase voltage
Y-Connected Load (cont.)
The line voltages at the load terminals in terms of the phase voltage are found by application of Kirchhoff’s voltage law.
Comparison of these line-to-line voltages with the line-to-neutral voltages of leads to the following conclusion:
In balanced three-phase Y-connected system with positive-sequence source, the line-to-line voltage are √3 times the line-to-neutral voltages and lead by 30°. That is;
Vab = √3Van +30°
Y-Connected Load (cont.)
Vbc = √3Vbn +30°
Vca = √3Vcn +30°
The relationship between the line voltage and phase voltage is shown as below:
Y-Connected Load (cont.)
Phasor diagram showing phase and line voltages
Vab
Vbc
Vca
Van
Vbn
Vcn
Voltages Triangle
RMS value of any of the line voltages is denoted by VL, then one of the important characteristic of the Y-connected tree-phase load may be expressed as
Y-Connected Load (cont.)
Three-phase current also possess three-phase symmetry and are given by
IL = IP
Example
Calculate the line voltage and line current of a Y-Y connection
A balanced ∆-connected load (with equal phase impedances) is shown in below:
∆-Connected Loads
Line voltage are the same as phase voltages.
The phase current Iab is chosen as reference
∆-Connected Loads (cont.)
The relationship between phase and line currents can be obtained by applying Kirchhoff’s current law at the corner of ∆.
Ip = magnitude of phase current
The RMS of any of the line currents and phase is denoted by IL
∆-Connected Loads (cont.)
The magnitude of line current is √3 times the magnitude of the phase current, and the line currents lags of phase currents by 30°.
The relationship between the line currents and phase currents is showing in phase diagram.
∆-Connected Loads (cont.)
Three Phase example
Three Phase example(cont.)
∆-Y Transformation For analyzing network problems, it is
convenient to replace the ∆-connected circuit with an equivalent Y-connected circuit.
Consider the Y-connected circuit of ZY Ω/phase which is equivalent to a balanced ∆-connected circuit of Z ∆ Ω/phase.
∆-connection Y-connection
∆-Y Transformation(cont.) For ∆-connected circuit
(1)
From the phasor diagram, the relationship between balanced phase and line-to-line voltage; (2)
∆-Y Transformation(cont.)
Substituting eq.(2) into eq.(1), we get
(3)or
(4)
For Y-connected circuit, we have
(5)
Substituting eq.(4) into eq.(5), we find that
(6)
A balanced, positive-sequence, Y-connectedvoltage source with Eab = 480 0° volts is
applied to a balanced-∆ load with Z∆ = 30 0° Ω. The
line impedance between the source and load is ZL= 1 85°Ω for each phase. Calculate the
line current, the ∆-load currents, and the
voltages at the load terminals.
example
Power in single-phase AC circuits
Figure show a single-phase sinusoidal voltage supplying a load
Instantaneous voltage be,
and Instantaneous current be,
The instantaneous power p(t) delivered to the load is the product of voltage v(t) and i(t) given by
By using the trigonometric identity
Which results in
RMS value of v(t) is |V|=Vm/√2, RMS value of i(t) is |I|=Im/√2 . Let θ=(θv – θi). The above equation, in terms of the rms values, is reduce to
Instantaneous power has been decomposed into two components. The first component is
Since the average value of this sinusoidal function is zero, the average power delivered to the load is given by
Also called Active power or Real powerThe second component is
The amplitude of this pulsing power is called reactive power
Q = |V|I|sin θ
Θ = θv– θi > 0 ;Q=+ve
Θ = θv– θi < 0; Q=-ve
The Power Factor (pf) is the cosine of the phase difference between voltage and current. It is also the cosine of the angle of load impedance.
The power factor may also be regarded as the ratio of the real power dissipated to the apparent power of the load.
Apparent Power
cos
Power Factor
( )v i
Ppf
S pS
P f
Example
Calculate the power factor seen by the source and the average power supplied by the load
Purely resistive load (R)
θv– θi = 0, Pf = 1 P/S = 1, all power are consumed
Purely reactive load (L or C)
θv– θi = ±90o, pf = 0
P = 0, no real power consumption
Resistive and reactive load (R and L/C)
θv– θi > 0 ;Q=+veθv– θi < 0; Q=-ve
• Lagging - inductive load• Leading - capacitive loadP/S < 1, Part of the apparent
power is consumed
Characteristics of the instantaneous power
For circuits operating in sinusoidal-steady state, real and reactive power are conveniently calculated from complex power, defined below.
Complex power S is product of voltage and the conjugate of the current:
Complex Power
IVIVS iv
sincos IVjIV
Thus, the complex power designated by S is given by
Complex Power(cont.)
jQPVIS
The magnitude of S, 22 QPS
* its unit is volt-amperes and the larger units are kVA or MVA The reactive power ,Q :
positive : θ(impedance angle) between V and I is +ve . In this case, load impedances is inductive (I lag V).
negative : θ(impedance angle) between V and I is -ve . In this case, load impedances is capacitive (I leadsV).
The impedances of the complex power S is given by
S
VZ
2
Power Triangle
a) Power Triangle b) Impedance Triangle
Power Triangle
The COMPLEX Power is represented by the POWER TRIANGLE similar to IMPEDANCE TRIANGLE. Power triangle has four items: P, Q, S and θ.
0 Resistive Loads (Unity )
0 Capacitive Loads (Leading )
0 Inductive Loads (Lagging )
Q Pf
Q Pf
Q Pf
Complex Power(cont.) The Complex Power contains all the information pertaining to the power absorbed by a given load.
2 2
Complex Power
Apparent Power
Real Pow
1
er
Reactive
= ( )2
=
= Re cos( )
=Q Im s Power
Power Fa
in( )
= =cctor os( )
Rms Rms v i
Rms Rms
v i
v i
v i
P jQ V I
S V I P Q
P S
S
P
S
S VI
S
S
S
• Real Power is the actual power dissipated by the load.
• Reactive Power is a measure of the energy exchange between source and reactive part of the load.
Complex Power(cont.) Examples
Complex Power(cont.)Solution
First method
Second method
Solution complex power (cont.)
Third method
Solution complex power (cont.)
Current phasor diagram and the complex power vector representation.
Power Factor Correction It is necessary to get PF is close to the unity. If
the PF < 1, the larger current I is supplied and additional cost to the utility company.
In order to maintain the power factor close to the unity, power companies install banks of capacitor throughout the network as needed.
A method of connecting a capacitor in parallel with an inductive load is known as power factor correction.
The effect of the capacitor is to increase the power factor of the source that delivers power to the load. The source apparent power also decreases.
Power Factor Correction(cont.)
Original Inductive Load Inductive Load with improved power factor correction
Effect of capacitor on total current
Power triangle of power factor correction
Increasing the power factor without altering the voltage or current to the load is called Power Factor Correction
Power Factor Correction(cont.)
Qc = Q1 – Q2
= P (tan θ1 - tan θ2)
= ωCVrms2
P = S1 cos θ1 Q2 = P tan θ2
c 1 22 2rms rms
Q P (tan θ tan θ )C
ωV ω V
Q1 = S1 sin θ1 = P tan θ1
Increasing the power factor without altering the voltage or current to the load is called Power Factor Correction.
Power Factor Correction(Cont.)
Power Factor Correction(Cont.)
Power Factor Correction(cont.)
2 cos( ) 2 cos( 120 )
2 cos( 120 )
2 cos( ) 2 cos( 120 )
2 cos( 120 )
cos( ) cos( ) cos( 120 )cos( 120 )2
cos( 120 )co
AN p BN p
CN p
a p b p
c p
a b c AN a BN b CN c
p p
v V t v V t
v V t
i I t i I t
i I t
p p p p v i v i v i
t t t tp V I
t
Using the above identity and simplifying, =2 t- we obtain
s( 120 )
1cos cos [cos( ) cos( )]
2 that:
13cos
2
cos 2 oos 3c c sp pp p
t
A B A B A B
VV I Ip
3Ø
Consider a balanced three-phase source supplying a balanced Y-or ∆-connected load with the following instantaneous voltage and current .
Balanced Three-Phase Power(cont.)
Balanced Three-Phase Power(cont.)
Three-phase reactive power is
The complex three-phase power is
So the real power of the three-phase is
ivLLpp whereIVIVP ;cos3cos33
sin3sin33 LLpp IVIVQ
333 jQPs or
ppIVs 33
Balanced Three-Phase Power(cont.)ExampleA three-phase line has an impedance of 2 + j4Ω as shown in figure below
The line feeds two balanced three-phase loads that are connected in parallel. The first load is Y-connected and has an impedance of 30 + j40Ω per phase. The second load is ∆-connected and has impedance of 60-j45Ω. The lines is energized at the sending end from a three-phase balanced supply of line voltage 207.85V. Taking the phase voltage Va as reference, determine:
Balanced Three-Phase Power(cont.)
a) The current, real power, and reactive power drawn from the supply.
b) The line voltage at the combined loads.
c) The current per phase in each load.
d) The total real and reactive powers in each load and the line.
Solution
Solution
(cont.)
Unbalanced Three Phase Systems
An unbalanced system is due to unbalanced voltage sources or unbalanced load. In a unbalanced system the neutral current is NOT zero.
Unbalanced three phase Y connected load.
Line currents DO NOT add up to zero.
In= -(Ia+ Ib+ Ic) ≠ 0
THANK YOU FOR YOUR ATTENTION!!