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1.1 SINGLEPHASECIRCUIT

IR

VmSin t

IR V

Figure 1.Single-phase pure resistance circuit and its time response and phasorrepresentation.

Both Current and voltage are in phase. Instead of representing in time domain i.e. as sinusoids

we could represent voltage and current in phasor form. A phasor is a complex number that

represents the amplitude and phase angle of a sinusoid signal. ost of the time drawing vector

diagram we assume voltage is used as reference! but it is not necessar".

I#

VmSin t

Figure 2.Single-phase pure inductive circuit and its time response

In pure inductive circuit current lags the voltage b" $%%. &e can also sa" this as voltage leads the

current b" $%%. 'igure ( shows the phasor representation

V

I#

Figure 3. Inductive circuit phasor representation.

In pure Capacitive circuit the current leads voltage b" $%%

)

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VmSin t

IC

VFigure 4. Capacitive circuit time domain & phasor represebtation.

In power s"stems and other power area we alwa"s use rms *root mean s+uare, value of voltage

and current. If we write V - 1)% %% this is rms value. n the other hand if we write,*1)%,* += tSintV then! 1)% is maximum value of the voltage V.

R-L-C Circuit

Impedance! / - * ,L CR ! !+ &here!1

))

L C! "L and !

"C

= =

EXAMPLE1:

R#C series circuit

Figure# $. RLC

Series Circuit,* CL !!R% += - )0% 2*)30.%3 4 1567.3, - )0% 4 210)(.( - 103(.5 87%.67

(

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.67.7%1%5(.5567.7%103(.5

%1)% %( &%

VI =

==

About 57 mA is flowing through the circuit and the current is leading the voltage b" 7%.67

degrees. 9his is a highl" capacitive load.

1.2 POWER& REACTIVEPOWER

Real :ower! : can be calculated in several wa"s;

: - I)R watts is an eas" wa" to calculate power in series circuit

: -)

RV

Rwatts is an eas" wa" to calculate power in parallel circuit. Rem. VRis

the voltage across the e+uivalent resistance.

9he most general formula to calculate power is;

: - V I Cos wattswhere! Cos is the power factor and is the angle difference between the voltage and current.9he power factor is called lagging if the current is lagging the voltage and power factor is called

leading if the current is leading the voltage.

Reactive :ower! < can be similarl" calculated;

< - I)=# and < -

)

L!

L

V

!and < - V I Sin

9he unit for reactive power is var.

9he complex power! S - V I>- : ? 2< and the unit

for S is VA. @S is an important in power anal"sisbecause it contains all the information pertaining to

the real reactive power absorbed b" a given load.

capacitive.

:ower 'actor! Cos -''

VI S=

: - S Cos And < - S Sin

Figure (. 'o)er *riangle.

3

Power Triangle for Inductive Load

Power Triangle for Capacitive Load

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Example 2:

'or a circuit shown the parallel impedances are /1- )% (%% and /)- 13.13 830% . 9he input voltage is 1%% 6%. CalculateS1! S)and S9 for the s"stem.

SOLUTION: 1%% 6%

I1-1

1%% 6%0 (%

)% (%

V

%

= =

I1is lagging V b" (%

%.

I)-)

1%% 6%5.%5 1%0

13.13 30

V&

%

= =

Figure +. ,ample .

S1- V I1> - 1%% 6% 0 8(% - 0%% (% - 3(( 2)0% VA9herefore :1- 3(( &atts and -1%% 6% 5.%5 81%0 - 5%5 830 - 0%% 4 20%%:) - 0%% &atts and

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*c, A capacitor connected in parallel with the loads. 'ind GVar and the capacitance re+uired

to improve the overall power factor to unit". Also calculate the new line current.

SOLUTION:

1

1

)%% %I 0 (0%.7 0.6

V % = = = + A

>

1 1 )%% % *0 (0, 1%%% 5%%%S V I = = + = + VA

)

)

)%% %I 0 1%

7 16

V

%

= = = +

>

) ) )%% % *0 1%, 1%%% )%%%S V I = = = VA

S(- 0 GVA at %.7 pf lag - 0%%% (6.75 - 3%%% 2(%%%

S9- S1 S) S(- 1%%% 2 5%%% 1%%% 4 2)%%% 3%%% 2 (%%% - 6%%% 27%%%- 1% 0(.1( GVA

S9is the power drawn from the suppl".

I9-

>1%%%% 0(.1(

0% 0(.1()%% %

*S

V

= = verall :ower factor is Cos 0(.1( - %.6 #agging

9o maGe the overall power factor unit"! the capacitance must cancel the inductance of the

combined load. 9herefore;

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9hree8phase s"stems are used for normal transmission and distribution s"stems. 9his is because

for the same amount of power! the three8phase s"stem is more economical than the single8phase.

Another advantage of three8phase s"stem is that the instantaneous power in a three8phase isconstant *not pulsating,! which results in better performance for machines.

A balanced s"stem is one in which the ( sinusoidal voltages have the same magnitude andfre+uenc"! and each is 1)%L out8of8phase with the other two.

Figure 2. *hree-phase sstem.&e will for our purpose assume that source is alwa"s balanced. Also assume that source is

alwa"s &"e even though a source can be w"e or delta. It is easier to solve problems b" this

assumption. &e do have to worr" if the load is connected in w"e or delat and also if the load is

5

( )

( )

( ) ( )+==

=

=

1)%cos)3%cos,*

1)%cos,*

cos,*

tVtVtv

tVtv

tVtv

44cn

4bn

4an

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balanced or unbalanced. A balanced three-phase circuitis one in which the loads are such that

the currents produced b" the voltages are also balanced. 'or a balanced s"stem! IA IB IC- %.

Figure 56. *hree-phase 7-connection )ith neutral

9he line voltages and phase voltages are shown in the diagram above. #ine voltages can be

calculated as;

VAB- VA4 VB VBC- VB4 VC VCA - VC4 VA

'or Balanced M8 #oad

IN-V

%

! V#IJ-

%( (%LI8,

V and I I + =

:ower! : - (L L

V I Cos - (* ,V I Cos

and Reactive :ower! < - ( L LV I Sin

:ower 'actor - Cos where is the angle between thephase voltageandphase current.9his is a ver" important to remember.

7

( )

( )

( )=

=

=

)3%cos,*

1)%cos,*

cos,*

tIti

tIti

tIti

4c

4b

4a

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Figure 55. *hree-phase 7-load )ith vector diagram

9his diagram is drawn assuming a8b8c se+uence. Also assuming that current lags the voltage.

!or "elta Loa#

:hase currents

IAB-&0

V

%

IBC-0C

V

%

ICA-C&

V

%

#ine currents

IA- IAB4 ICAIB- IBC4 IABIC- ICA8 IBC

'or a balanced three8phase delta load s"stem;

I#IJ- ( IN 8(% and V:EASJ- V #IJ

9he e+uation for power and reactive power is exactl"

same for delta load as the &"e load. 9he complex power!

S ->(* , (

' ' L LV I ' 1 V I = + =

Figure 5. *hree-phase delta load and vector

diagram.

$

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1$% ANAL&SISTEC'NI(UE)*ALANCE"LOA"+

1. Replace (8phase circuit with its single8phase e+uivalent circuit.

). nl" one current or voltage needs to be calculated. Rest of the +uantities can be found

easil".(. It is easier *but not re+uired, to use VAas the reference! if no reference is given.

3. 'or balanced load /M- /OK(. 9his is a useful formula for man" parallel balanced loads.

0. 'or M8load; I# - IN and V#- 3 VN (%6. 'or O8load; V#- VN and I#- IN 8(%

Example %: A balanced (8phase deltaconnected load is supplied b" a )%7 V 6% EF

suppl". Jach phase impedance is / - 1) 2

$ D per phase. Calculate the line current!

power and the complex power supplied b"the source.

SOLUTION:

)%7 % )%7 %I 1(.75 (6.75

1) $ 10 (6.75

ab

ab

ab

V

%

= = = =

+

NOTE: Vabis assumed reference

Ibc- 1(.75 8(6.75 8 1)% - 1(.75 8106.75 AIca- 1(.75 8(6.75 1)% - 1(.75 7(.1( A

IA- ( Iab 8(%- ( 1(.75 (6.75 (% )3.%) 66.75 & =

IAcan also be calculated b" the difference of two

phase currents or IA- Iab8 Ica

IB- )3.%) 8 66.75 4 1)% - )3.%) 8176.75 AIC- )3.%) 0(.1( A

: - ( V#I#Cos

- ( )%7 )3.%) cos(6.75 6$)).39att =

: -) )( 1) ( 1(.75 1) 6$)).6

'I 9att = =

>( ( 1)% 1(.75 (6.75 6$)).6 01$1.7' '

S V I Var = = = +

1%

ZZ

Z

Z = 12 +j 9

IAA

!

B

Iab

Ib"

I"a

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Example ,: A (8phase delta load is supplied b"

a (8phase source through a line whose impedance

is %.%6 2 %.1) . 9he voltage of the source is)%7 V and the load impedance is 1) 2$ each

phase. Hetermine the *a, line current! I#! *b, loadcurrent in each phase! IN ! *c, power delivered to

the load and *d, power supplied b" the source.

SOLUTION

Convert the delta load to &"e and draw a single8phase

e+uivalent. /M- /OK( - 3 2(

I#-

)%7

1)%.%$ %( (

3 ( %.%6 %.1) 3.%6 (.1)

)(.30 (5.03

L

* 7 LI8,

V

V

% % %

= = =+ + + + +

= Eere we assumed VN phase as our reference. ow we calculate the phase voltage of the loadVN#AH- I# /M- )(.30 8(5.03 *3 2(, - 115.)6 81.(5 V.

V#*#oad, - ( VN#AH- )%(.1 )7.6( V

IN - V#*#oad,K /O-)%(.1 )7.6(

1) $

+ - 1(.03 87.)3 A.

NOTE: &e can not calculate IN b" dividing the I#b" ( and shifting (%%.

:#AH- ( V#*#oad, I#Cos - ( )%(.1 )(.30 Cos *)7.6( 7.)3, Note: - VN 8 IN

- 60$$.3 &atts

:#AHCan be calculated various wa"s; (*RO IN), - (*1) 1(.03), - 60$$.$ &

:SPRCJ- ( V#*Source, I#Cos - ( )%7 )(.30 Cos*% 4 (5.03, - 66$7.7 &

Approximate #ine loss - :SPRCJ4 :#AH- 66$7.7 8 60$$.3 - $$.3 &

9his can also be calculated b"; (*R#IJ I#), - (*%.%6 )(.30), - $$

It is also important to verif" our results if and when possible. In real life there is no answer sheetavailable to checG our results

11

I#

IN

I#

ZY = 4 + jZ

!" !# + j!" 12

Z Z

Z

!" !# + j!" 12

!" !# + j!" 12

!" !# + j!" 12

Z = 12 + j

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Example : 9hree loads are connected in parallel across a 1).35 GV three8phase suppl".#oad 1; Inductive! 6% G& and 66% Gvar

#oad ); Capacitive! )3% G& at %.7 power factor.

#oad (; Resistive! 6% G&.

*a, 'ind the total complex power! power factor! and the suppl" current

*b, A M8connected capacitor banG is connected in parallel with the three loads. 'ind the totalGvar and capacitance per phase in #' to improve the overall power factor to %.7 lagging.&hat is the new line currentQ

SOLUTION:

a, #oad1; S1- 6% 2 66% G

#oad); )3% G& and %.7 pf capacitive.

))3% (%% (%% (6.75%.7

' :S SCos= = = =

r S)- )3% 8 217% G

#oad(; S(- 6% G& 2%

9otal Complex :ower! S9- S1 S) S(- *6% 2 66%G )3% 8 2 17% 6%, G - *(6% 2 37%, G

S9- 6%% 0(.1( GVA

I#-

> >6%% 0(.1(

)5.57 0(.1(( ( 1).35

S :

V :

= =

b, S9to be %.7 pf lagging(6%

30% 30% (6.75%.7

*

*8 *8

' :S : S :

Cos= = = =

S9- *(6% 2)5%, GVA

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Example .: *9aGen from Chapman, 'igureshows a (8phase s"stem with two loads.

9he generator is producing 37% and the line

impedance is %.%$ 2 %.16 . #oad 1 is Mconnected! with impedance of ).0 (6.75 and load ) is delta connected with aphase impedance of 0 8)% .

*a, &hat is the line voltage of the two

loadsQ

*b, &hat is the voltage drop on thelineQ

*c, 'ind the real reactive powers

supplied to each load.*d, 'ind the real reactive power supplied b" the source

Vab- 37% %

Van-37% %

(

= )55 8(%

/1- ).0 (6.75 D

/)-0 8)%

- 1.665 8)%(

%Converting&to M

/J1, - ( *)0(.(6 1%1.(3, (6.75

1(

4)0

v= 2'' (

!"!9 + j!"1# .

I*

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- 55%)6 (6.75 - 61!6)1 236!)16 VA

S)* Complex :ower of #oad ), - ( *V#AH I>), - ( *)0(.(6 10), 8)%- 110!033.( 8)% - 1%7!056 8 2($!017.0 VA

S9* Complex :ower of Source, - ( *V#AH I>A, - ( *)0(.(6 ))3, $.0

- 176!)0$.6

8$.0 - 17(!5%0 2(%!53) VA

9ransmission #oss - S94 *S1 S),

- 17(!5%0 2 (%!53) 4 *61!6)1 236!)16 1%7!056 4 2($017,- 1(!0%7 2)3!%33 VA

9his problem can be solved in other wa"s too. 'or example;

9ransmission #oss - ( * IA) R# 2 IA

) =#, -- (*))3) %.%$ ))3) %.16,

- 1(!035 2)3!%73 VA

9he slight difference is due to rounding error.

13

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1$, PO/ER!ACTORCORRECTION:

:ower factor is the ratio between the & and the VA drawn b" an electrical load where the

& is the actual power and the VA is the apparent power. Similar to power current also has

real and reactive components. All current will causes losses in the distribution s"stem. A loadwith a power factor of 1.% results in the most efficient loading of the suppl" and a load with a

power factor of %.0 will result in much higher losses in the suppl" s"stem.

:oor power factor is generall" the result of an inductive load such as an induction motor! power

transformer! lighting ballasts! etc. #ow power factor can also be due to a high harmonic content

or distortedKdiscontinuous current waveform resulting from power electronics devices. A poorpower factor due to an inductive load can be improved b" the addition capacitors! but! a poor

power factor due to a distorted current waveform re+uires a change in e+uipment design or

expensive harmonic filters to gain an appreciable improvement. &e will onl" discuss addition of

capacitors to correct the power factor. :ower factor correction is achieved b" the addition of

capacitors in parallel with the connected motor circuits and can be applied at the starter! orapplied at the switchboard or distribution panel.

Some of the benefits of improving power factor are as follows;

Ptilit" bill will be smaller. #ow power factor re+uires an increase in the electric utilit"sgeneration and transmission capacit" to handle the reactive power component caused b"

inductive loads. Ptilities usuall" charge a penalt" fee to industrial customers with power

factors less than %.$0. ne can avoid this additional fee b" increasing "our power factor.

Jlectrical s"stems branch capacit" will increase. Pncorrected power factor will cause

power losses in distribution s"stem. It will increase voltage drops as well as power losses.

Jxcessive voltage drops can cause overheating and premature failure of motors and otherinductive e+uipment.

Capacitors can be connected at each motor or at the distribution panel as shown below.

10

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&hen power factor is corrected! the real

power! : does not change! but reactivepower < and complex power S changes

because of the addition of capacitor.

In the diagram;

#H- ld pf angleJ&- ew pf angle or desired pf angle: - S#H*Cos #H,

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EXAMPLE: A large (8phase industrial plant is

supplied power at 1).35 GV and has a demand of 3& at a power factor of %.5 lag. Calculate the

amount of capacitanceKphase and reactance value

re+uired to improve the power factor to %.$5lagging. Assume the capacitors banG is connected

in delta. Calculate the new and old current.

SOLUTION:

ld power factor - %.5 lag #H- 30.05%

Hesired power factor - %.$5 lag J&- 13.%5%

S#H-3%%%

0513.(%.5

;L- : ? 2(* , ('hase 'hase L LV I ' 1 V I = + =

:ower! : - ( L LV I Cos - (* ,'hase 'haseV I Cos

and Reactive :ower! < - ( L LV I Sin

:- (*Single8phase :ower,

U7alace# S94tem

9hree8phase formulas are not valid for unbalances"stems.

Solve each phase separatel". Hraw vector diagram for currents voltages :9- :A :B :C

T5o /att-meter

&A- VAB IACos *angle between VABand IA,&C- VCB ICCos *angle between VCBand IA,

Power Triangle for Inductive Load

Power Triangle for Capacitive Load

)5

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