SISTEM SATU FASA EET103 TEKNOLOGI ELEKTRIK. SISTEM SATU FASA Pengenalan dan ciri-ciri sistem satu...
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Transcript of SISTEM SATU FASA EET103 TEKNOLOGI ELEKTRIK. SISTEM SATU FASA Pengenalan dan ciri-ciri sistem satu...
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