SISTEM SATU FASA

EET103 TEKNOLOGI ELEKTRIK

SISTEM SATU FASA

• Pengenalan dan ciri-ciri sistem satu fasa

• Kebaikan dan keburukan sistem satu fasa

• Pengiraan voltan, arus dan kuasa

• ALTERNATING CURRENT: a current or voltage varies periodically in magnitude and direction

PENGENALAN DAN CIRI-CIRI SISTEM SATU FASA

PERBEZAAN AC DAN DC

Difference between DC and AC:

• DC: a direct flow of electrons through a conductor such as a metal wire. A battery or DC generator usually provides a source of electrons and the potential or voltage between the positive (+) and negative (-) terminals.

• AC (Alternating Current): a back-and-forth movement of electrons in a wire. When the force of a negative (-) charge is at one end of a wire and a positive (+) potential is at the other end, the electrons in the wire will move away from the (-) charge, just like in DC electricity. But if the charges at the ends of the wires are suddenly switched, the electrons will reverse their direction.

AC voltage (1 cycle):

Mathematics of AC voltages

• An AC voltage v(t) can be described mathematically as a function of time by the following equation:

whereA is the amplitude in volts (also called the peak voltage), ω is the angular frequency in radians per second t is the time in seconds

)cos()(

)sin()(

tAtv

tAtv

• Since angular frequency is of more interest to mathematicians than to engineers, this is commonly rewritten as:

Where f is the frequency in hertz (Hz).

)2sin( tfAv(t)

AC voltage:

• Amplitude is the maximum voltage reached by the signal. It is measured in volts, V.

• Peak voltage (Vp) is another name for amplitude.

• Peak-peak voltage (Vp-p) is twice the peak voltage (amplitude). When reading an oscilloscope trace it is usual to measure peak-peak voltage.

RMS (effective) value:• The root-mean-square (RMS) value of an alternating

current is the steady direct current which converts electrical energy to other forms of energy in a given resistance at the same rate as the alternating current (AC).

• In power distribution work the AC voltage is nearly always given in as a root-mean-square (rms) value, written Vrms. For a sinusoidal voltage:

A707.02

AVrms

Vrms = 0.707 × Vpeak   and   Vpeak = 1.414 × Vrms

Frequency

• The number of complete cycles of alternating current or voltage completed each second

• Frequency is always measured and expressed in hertz (Hz).

Period

• An individual cycle of any sine wave represents a definite amount of TIME.

• The time required to complete one cycle of a waveform is called the PERIOD of the wave.

• The relationship between Period (T) in seconds and frequency (f) in Hz:

Tf

fT

11

Frequency measurement:

Wavelength

• The time it takes for a sine wave to complete one cycle is defined as the period of the waveform. The distance traveled by the sine wave during this period is referred to as WAVELENGTH.

Wavelength measurement:

Alternating current values:

• PEAK AND PEAK-TO-PEAK VALUES:

Contoh (1)Rajah berikut menunjukkan satu kitar gelombang arus sinus. Berikan persamaan bagi arus tersebut sebagai fungsi masa.

t (ms)

i (mA)

170

-170

0

20

10

Penyelesaian:

•Persamaan umum bagi gelombang sinus ialah

i(t) = Im sint •Dari rajah gelombang

diketahui;Im = 170 mA dan

T = 20 ms = 0.02 s

• Maka, frekuensi dapat dikira seperti berikut:

f = 1/T = 1/0.02 = 50 Hz• Seterusnya persamaan arus

menjadi,i(t) = Imsint

= 170 sin(2ft) = 170 sin(100t) mA

Contoh (2)

Satu voltan ulang-alik bentuk sin mempunyai frekuensi 2500 Hz dan nilai puncak 15 V. Lukiskan bentuk satu kitar gelombang voltan tersebut.

Penyelesaian• Dari soalan, diketahui nilai-nilai berikut:

Vm = 15 V dan T = 0.4 ms

• Rajah umum gelombang sinus adalah seperti berikut:

t (ms)

v (V)

15

-15

0

0.4

0.2

Contoh (3)

Satu voltan ulang-alik sinus diberikan oleh persamaan:

v(t) = 156kos(800t) V

Lukiskan rajah satu kitar gelombang voltan tersebut.

Penyelesaian:

Diberi, v(t) = Vm kos(t) = 156 kos(800t) volt

Maka, nilai puncak voltan, Vm = 156 = 2f = 800

f = 400Tempoh bagi gelombang ini ialah;

T = 1/f = 1/400 = 2.5 ms

v (V)

0

156

-1560.625

1.25

2.5

t (ms)

1.875

Gambarajah gelombang voltan:

INSTANTANEOUS VALUE

• The INSTANTANEOUS value of an alternating voltage or current is the value of voltage or current at one particular instant

• There are actually an infinite number of instantaneous values between zero and the peak value.

AVERAGE VALUE

• The AVERAGE value of an alternating current or voltage is the average of ALL the INSTANTANEOUS values during ONE alternation.

pavg

pavg

II

VV

637.0

637.0

Effective and

average value

Phase angle (sudut fasa):

• Apabila sesuatu gelombang sinus tidak melepasi nilai kosong pada t=0, maka persamaan gelombang tersebut mempunyai sudut fasa yang perlu dipertimbangkan.

• Sudut fasa menunjukkan ANJAKAN sesuatu gelombang dari sifar.

• Gelombang sinus boleh bermula dari apa-apa nilai seperti berikut. Ia tidak semestinya bermula dari sifar (atau dari nilai puncak bagi fungsi kos).

Persamaan gelombang berikut: y = Ymsin(x + a)

x ()

90

180 270

360

0

Ym

-Ym

a

ao adalah sudut fasa

(phase angle)

Sine Waves In Phase (sefasa)

• When two sine waves are precisely in step with one another, they are said to be IN PHASE. To be in phase, the two sine waves must go through their maximum and minimum points at the same time and in the same direction.

Voltage and

current are in phase

Sine Waves Out Of Phase (tidak sefasa)

• Voltage wave E1 which is considered to start at 0° (time one). As voltage wave E1 reaches its positive peak, voltage wave E2 starts its rise (time two). Since these voltage waves do not go through their maximum and minimum points at the same instant of time, a PHASE DIFFERENCE exists between the two waves. The two waves are said to be OUT OF PHASE.

Phase difference is 90o

GELOMBANG TIDAK SEFASA• Dua gelombang sinus yang tidak sefasa

boleh diwakilkan dengan persamaan:

v(t) = Vm kost ; i(t) = Im kos(t + )• Arus i(t) MENDAHULU (leading) voltan v(t)

dengan sudut . • Dalam sebutan masa, arus mendahului

voltan dengan tempoh (T/360) saat.• Boleh juga disebut voltan MENGEKOR

(lagging) arus dengan sudut

Dua Gelombang Tidak Sefasa:

v, i

Vm

Im

0

-Vm

-Im

Gelombang V mencapai nilai puncak di t2

t

T

Gelombang i mencapai nilai puncak di t1

v i

Θ is phase

difference

CONTOH (4)

Lukiskan satu kitar gelombang arus sinus yang diberikan oleh persamaan

i(t) = 70sin(8000t + 0.943 rad) mA.

Tandakan nilai-nilai kritikal.

PENYELESAIAN

Bentuk am gelombang arus ialah:

i(t) = Im sin(t + ) = 70 sin(8000t + 0.943 rad)

Dari persamaan diatas, nilai-nilai kritikal ialah:

Im = 70;

= 2f = 8000;

f = 4000 Hz = f = 4000 Hz = 4 kHz4 kHz;;

Maka,Maka,

T = 1/f = 1/4000 = T = 1/f = 1/4000 = 0.25 0.25 msms;;

Dan,Dan, = 0.943 rad = = 0.943 rad = 5454

180rad1

Rajah gelombang sinus bagi arus:

0

i (mA)

t (ms)

70

-70

57

54

0.25

0.125

SISTEM SATU FASA

• Pengenalan dan ciri-ciri sistem satu fasa

• Kebaikan dan keburukan sistem satu fasa

• Pengiraan voltan, arus dan kuasa

KEBAIKAN DAN KEBURUKAN SISTEM SATU FASA• Direct current has several disadvantages

compared to alternating current. Direct current must be generated at the voltage level required by the load. Alternating current, however, can be generated at a high level and stepped down at the consumer end (through the use of a transformer) to whatever voltage level is required by the load.

• The major advantage that AC electricity has over DC is that AC voltages can be transformed to higher or lower voltages. This means that the high voltages used to send electricity over great distances from the power station could be reduced to a safer voltage for use in the house.

• This is done by the use of a transformer. This device uses properties of AC electromagnets to change the voltages.

• It is easy to convert AC to DC but expensive to convert DC to AC.

SISTEM SATU FASA

• Pengenalan dan ciri-ciri sistem satu fasa

• Kebaikan dan keburukan sistem satu fasa

• Pengiraan voltan, arus dan kuasa

1. POLAR FORM1. POLAR FORM

Z= Z

Z :magnitude of Z : angle of Z

COMPLEX NUMBER

2. RECTANGULAR FORM2. RECTANGULAR FORM

Z = R + jX

R: real value of ZX: imaginary value of Z j : operator valued √-1

3. EXPONENTIAL FORM3. EXPONENTIAL FORM

Z = rej

r : magnitude : angle

magnitud:XRZ22

sudut:R

Xtan 1

RECTANGULARRECTANGULARFORM TO FORM TO POLAR POLAR FORMFORM

ZZ

cL jXR@jXRZ

POLARPOLAR FORM TO FORM TO RECTANGULAR RECTANGULAR FORMFORM

valuereal :θcosZR

valueimaginary:sinZjX

ZZ

cL jXR@jXRZ

COMPLEX NUMBER ALGEBRA OPERATION

• To ADD two number:

–Transform to Rectangular

• To MULTIPLY two number:

–Transform to Polar

• To SUBTRACTSUBTRACT two number:

–Transform to Polar

SINUSOID-PHASOR TRANSFORMATION

)sin(

)cos(

)sin(

)cos(

tI

tI

tV

tV

m

m

m

m

90

90

m

m

m

m

I

I

V

V

Time domain representation

Frequency domain representation

Ohm’s Law in AC circuit

RESISTOR IN ACRESISTOR IN AC

IN FREQUENCY DOMAIN:

V

IV

im

im

R

V

IR

R

NO PHASE DIFFERENCE BETWEEN VOLTAGE AND CURRENT

v, iv

it

IN PHASE

V and I waveform for resistor :V and I waveform for resistor :

INDUCTOR

dt

diLvL

• ININ TIME DOMAIN: TIME DOMAIN:

• ININ FREQUENCY DOMAIN: FREQUENCY DOMAIN:

ILjeLIjV ijmL

GELOMBANG LITAR L

v, iv

i t

90º

ARUS MENGEKOR VOLTAN (ELI)BEZA FASA SEBANYAK 90º = (1/4)T = 1/4f

KAPASITOR

dtiC

1vC

• IN TIME DOMAIN:IN TIME DOMAIN:

I

VC

Cj

1

eICj

1ij

m

• IN FREQUENCY DOMAIN:IN FREQUENCY DOMAIN:

GELOMBANG LITAR C

Arus mendulu voltan sebanyak 90° (ICE)

v

i

t

v, i

GELOMBANG v-i BAGI R,L,Cv, i

v

i t

V,I SEFASA

v

it

v, iI MENDAHULU V SBYK 90º

v, iv

i t

90º

I MENGEKOR V SBYK 90º

Penentuan Mendulu/Mengekor

ARUS MENGEKOR VOLTAN

Jika gelombang voltan mpy sudut fasa positif

Terletak disebelah kiri gelombang arus

Gelombang voltan mendulu gelombang arus.

Dengan menganggap gelombang Arus sbg RUJUKAN:

Jika gelombang voltan mpy sudut fasa negatif

Terletak disebelah kanan gelombang arus

gelombang voltan mengekor gelombang arus.

ARUS MENDULU VOLTAN

AC CIRCUIT ELEMENT

REACTANCE AND IMPEDANCE

REACTANCE

All elements in AC circuit (resistor, inductor and capacitor) should have same unit before you do the analysis.

• Inductor value (in henry) and capasitor value (in farad) must be transform into ohms ().

• Inductance and capacitance value in ohms are known as reactance which is inductive reactance for inductor, and capacitive reactance for capacitor.

• Symbol for inductive reactance is XL

and for capacitive reactance is Xc..• Formula to get XL dan Xc:

fC2

1j

C

1j

Cj

1X

fL2jLjX

C

L

HAFAL!

CAPACITOR

• Series capacitors:

NT CCCC

N

eq

XXXX

CCC

C

..

1..

111

21

21

• Parallel capacitors:

N

T

CCC

C

Neq

XXX

X

CCCC

1..

111

..

21

21

INDUCTOR

• Series inductors:

NT LLLL

Neq

XXXX

LLLL

..

..

21

21

• Parallel inductors:

N

T

LLL

L

N

eq

XXX

X

LLL

L

1..

111

1..

111

21

21

IMPEDANCE

Impedance is a element connecting the resistance, inductive reactance and capacitive reactance in time domain.

• Impedance is represent by Z symbol:

(ohms)

I

V

I

VZ

• is angle between voltage and current• Impedance is complex number because of magnitude and angle

• Resistance impedance, ZR:

R

0I

VI

V

I

VZ

R

RRR

• Inductive impedance, ZL

L

L

LLL

jX

90I

V90I

0V

I

VZ

• Capacitive impedance, Zc

cc

c

ccC

X

1j

jX

1

90I

V90I

0V

I

VZ

AI imj

m IeI i

VIV imR RIR

V)90(LIIX imLLV

• Transform circuit into frequency domain using these equation.

• Use KVL and KCL…

V)90(IC

1IXV imCC

KIRCHHOFF LAW IN AC ANALYSIS

R-L CIRCUIT IN FREQUENCY DOMAIN

+

V

L

-

RjL

V = V

m

+ V

R

-

I

V s

R

L

+ V R -

+V L-

CONTOH (1) : LITAR RLGiven Vs = 40 cos (2000t - 60º). Find,

a) Phasor voltage across R;

b) Phasor voltage across L;

c) Current, i(t) expression

V s

R = 1 0 0 0

L= 0 .2 5 H

+ V R -+V L-

Step 1: Circuit in frequency domain

Vs=40-60

R=1000

jL=j1570.8

+ VR -+VL

-

Step 2: Find circuit impedance

5.571.1862

1000

8.1570tan8.15701000

8.1570j1000

LjR

122

Z

mA5.1175.21

A5.1170215.0

5.57601.1862

405.571.1862

6040

Z

VI

Step 3: Calculate current

Phasor current is;

mAe5.21

atau

mA5.1175.21

5.117j

I

I

(a) Phasor voltage across R,

V5.1175.21

mA5.1175.211000

RR

IV

(b) Phasor voltage across L,

V5.278.33

mA)5.1175.21()908.1570(

mA)5.1175.21()25.02000j(

LjL

IV

(c) Current, i(t) expression:

»» Transform I in time domain

mA)5.117t2000(c5.21i os

mA5.1175.21 I

CONTOH (2) : LITAR RCGiven VC = 60 cos(1000t - 33º). Find,

a) Phasor current, I

b) Phasor voltage across R

c) Voltage source, vs(t).

2200R

CVsV

RV

Step 1: Circuit in frequency domain

2200R

CV1447j

Cj

1

3360VC

sV

RV

Step 2: Circuit impedance (Z)

3.332200

1447tan

2633)1447(2200Z

:sudutdanmagnitud

1447j2200C

1jR

1

22

Z

Z

Step 3: Find I, VR,VS (a) Phasor current;

A

CjC

570415.0901447

33601

VI

(b) Phasor voltage across R;

V

A

RR

573.91

570415.02200

IV

(c) Source phasor voltage;

V

A

7.233.109

570415.03.332633

ZIV

Transform V into time domain;

V)7.23t10000cos(3.109v

IMPEDANCE TRIANGLE

IMPEDANCE TRIANGLE• We also can get impedance by using

impedance triangle:

Galangan,Z Reaktan, X

Perintang, R

• Dari segitiga galangan;

• Ini bermakna, dengan mengetahui Galangan (Z), kita boleh juga mengetahui nilai perintang (R) dan Reaktan (X).

2X 2RZ

HUBUNGAN Z, R DAN X• Hubungan Z dan R;

• Hubungan Z dan X;

Z

Rkosθ

Z

Xθ sin

AC Power Calculation

POWER IN AC CIRCUIT

Source

)t(v ZLoad

)t(i

INSTANTENAOUS POWER (KUASA SEKETIKA)

• Kuasa seketika yg diserap oleh setiap peranti elektrik adalah hasil darab Voltan seketika yg merintanginya dan Arus seketika yang melaluinya.

)t(i)t(v)t(p

• Semua nilai Voltan dan Arus dalam pengiraan kuasa adalah menggunakan nilai RMS (root mean square). Iaitu:

• Dimana, Vm dan Im adalah Nilai Puncak

(peak) bagi voltan dan arus.

2

VV m

rms 2

II m

rms

RMS VALUE

1. AVERAGE POWER (KUASA PURATA)

• Kuasa purata atau kuasa sebenar ditakrifkan sebagai:

• Unit bagi kuasa purata ialah Watt.

)Watt(cos2

IVP

cos2

I

2

VPPurataKuasa

mm

mm

• Kuasa purata pada Perintang boleh juga diwakili oleh persamaan berikut:

• Kita tahu V=IR, maka kuasa purata turut diwakili oleh:

rmsrmsmmmm IV

2

IVcos

2

IVP

RIR

VIVP 2

rmsrms

2rms

rms rms

• Kuasa purata ialah kuasa berguna yg. disebabkan oleh elemen Perintang.

• Dengan itu, kuasa purata bagi beban reaktif (L atau C) adalah Kosong.

ialah Sudut antara Voltan dan Arus, ATAU dikenali juga sebagai Sudut Galangan,Z.

ZZ

-ZI

VZ

Nilai diperolehi seperti berikut:

CONTOH (1)

Voltan sinus mempunyai amplitud maksimum 625 Volt dikenakan pada terminal yg mempunyai Perintang 50. Dapatkan kuasa purata yang dihantar kepada perintang tersebut.

• Nilai rms:

• Kuasa purata bagi perintang diperolehi:

PENYELESAIAN

VVrms 94.4412

625

W25.3906

50

94.441

R

VP

22

2. REACTIVE POWER (KUASA REAKTIF)

• Kuasa Reaktif ditakrifkan sebagai:

)Var(sinIVQ

sin2

IV

sin2

I

2

VQreaktifKuasa

rmsrms

mm

mm

• Kuasa reaktif juga merupakan kuasa yg disimpan oleh elemen reaktif iaitu L atau C.

• Maka, kuasa reaktif boleh juga dicari dengan:

C

2C

CC2

C

L

2L

LL2

L

XVIVXIQ

,Dan

XVIVXIQ

CONTOH (2)Diberi v= 100 kos (t + 15º) dan i= 4 sin (t - 15º), Pada terminal rangkaian, dapatkan:

(a)Kuasa purata(b)Kuasa reaktif

i

+v-

Rangkaian

PENYELESAIAN• Tukarkan persamaan arus, i dalam bentuk

kos: i= 4 kos (t - 105º)

• Dari persamaan Kuasa Purata:

• Maka,

(Watt)cosθ2

IVP mm

W100

W10515cos41002

1P

• Dari pers. Kuasa reaktif:

• Maka,

(Var)sinθ2

IVQ mm

Var21.173

W10515sin41002

1P

• Nilai kuasa purata, P= -100 W:

Bermaksud kuasa purata telah dihantar balik dari beban kepada terminal bekalan.

• Nilai kuasa reaktif, Q=173.21 Var: Bermaksud kuasa reaktif telah diserap oleh beban.

AC POWER FORMULA

• AC power calculation can be done by using these formula:

)Watt(cosIVPPurataKuasa rmsrms )Watt(cosIVPPurataKuasa rmsrms

)Var(sinIVQaktifReKuasa rmsrms )Var(sinIVQaktifReKuasa rmsrms