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Electrotehnica si masini electrice
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AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 1/34
AC wave forms
AC = Alternating Currentpeak
peak-to-peak T
f = 1/T frequency [Hz]ω = 2πf angular frequency
[rad/s]
Instantaneous value: i(t) = Im sin(ωt + φ)
AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 2/34
Sinus wave
Average value: Iav = (1/T)∫i(t)dt = 0
Root Mean Squared value: I = √(1/T)∫i2(t)dt = Im/√2 = 0.707 Im
Instantaneous value: i(t) = √2 I sin(ωt + φ)
I = RMS value√2I = peak value (amplitude)f = frequencyT = 1/f = periodω = 2πf = angular frequencyφ = phase
325 V
230 V
0t
[ms]2010
u(t) = 325 sin(314t)
AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 3/34
Phase difference
ωt
φ=π/6
2π+φ
The two waveforms are π/6 (30o) “out-of phase”.The current lags the voltage by angle φ.
The voltage leads the current by angle φ.
The current cross the horizontal reference axis reaching its peak and zero values after the voltage waveform.
u(t) = Um sin(ωt)
i(t) = Im sin(ωt - φ)
φ = phase difference (phase shift)
AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 4/34
Complex quantities. Phasor diagram
the current phasor lags behind the voltage phasor
U = U·ej0 I =I·e-jπ/6
A Phasor Diagram can be used to represent two or more stationary sinusoidal quantities: voltage, current, or some other alternating quantity of the same frequency.
√2·A sin(ωt + φ) → A = A ejφ (j = √-1)
ejφ = cosφ + j sinφ
A = A (cosφ + j sinφ)
A = A cosφ + j A sinφ
reference axis
φ=- π/6
U
I
real axis
φ
A
Acosφ
Asi
nφ
imaginary axis
AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 5/34
Phasor addition
u1(t) = √2U1 sin(314t + φ1) u2(t) = √2U2 sin(314t + φ2)
u(t) = u1(t) + u2(t)
u (t) = √2U sin(314t + φ)
U = U1 + U2
U2 = U12+U2
2+2U1U2cos(φ1-φ2)
tan φ = (U1sinφ1+ U2sinφ2)/(U1cosφ1+ U2cosφ2)
u1(t) = √2·30 sin(314t + π/3)u2(t) = √2·20 sin(314t)u(t) = u1(t) + u2(t) = √2U sin(314t + φ)U = ?, φ = ?
U = 43.6 V, φ = 23.4 °
AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 6/34
Impedance, Phase angle, Resistance, Reactance
ImpedanceZ = U/I [Ω]
Phase angle φ = (φu - φi) [rad]
i(t) = √2(U/Z) sin(ωt + φu - φ)
u(t) = √2·100 sin(314t + π/3)i(t) = √2·2 sin(314t + π/6)Z = ?φ = ?
u(t)
i(t)
u(t) = √2U sin(ωt + φu)
i(t) = √2I sin(ωt + φi)
~Z, φ
AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 7/34
Impedance, Phase angle, Resistance, Reactance
ResistanceR = Z cosφ [Ω]
Reactance X = Z sinφ [Ω]
φ = (φu - φi) [rad]Z = √(R2 + X2)tan φ = X/R
u(t) = √2·100 sin(314t + π/3)i(t) = √2·2 sin(314t + π/6)R = ?X = ?
u(t)
i(t)
u(t) = √2U sin(ωt + φu)
i(t) = √2I sin(ωt + φi)
Z, φ
~ R, X
AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 8/34
Complex Impedance
Z = Uejφu/Iejφi Z = (U/I)ejφu-φi
Z = Zejφ
Z = Z(cos φ + j sin φ)Z = R + j X
Complex ImpedanceZ = U/I 1
j
R
X
Z
φ
u(t) = √2U sin(ωt + φu)
i(t) = √2I sin(ωt + φi)
Z, φ
R, X~Z
AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 9/34
Power in AC circuits
u(t)
i(t)
u(t) = √2U sin(ωt + φu)
i(t) = √2I sin(ωt + φi)
~
Instantaneous powerp(t) = u(t)·i(t)p(t) = 2UI sin(ωt + φu)· sin(ωt + φi)
AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 10/34
True power
P = (1/T)∫p(t)dt
P = UI cosφ [W] φ = φu - φi
Power in AC circuits
U = Z IP = Z I2
cosφP = R I2
Apparent power
S = U I [VA]
U = Z IS = Z I2
Power factor
kp = P/S = cos φ
Reactive power
Q = UI sinφ
[VAr]
U = Z IQ = Z I2
sinφQ = X I2
S = √(P2 + Q2)
S = U I* = Uejφu Ie-jφi = (UI)ej(φu-φi) = Sejφ = P + jQ
AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 11/34
Power in AC circuits. Applications
u(t) = √2·100 sin(314t + π/3)i(t) = √2·2 sin(314t + π/6)P = ?Q = ?S = ?
u(t)
i(t)
u(t) = √2U sin(ωt + φu)
i(t) = √2I sin(ωt + φi)
~
AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 12/34
AC Resistor with a sinusoidal supply
u(t)
i(t)
u(t) = √2U sin(ωt) (φu= 0)
i(t) = u(t)/R
i(t) = √2 (U/R) sin(ωt)
I = U/R → Z = U/I = Rφi = φu → φ = φu - φi = 0
R = Z cos φ = RX= Z sin φ = 0 Z = R + jX = R
AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 13/34
AC Resistor with a sinusoidal supply
P = RI2 > 0Q = XI2 = 0
S = ZI2 = RI2 = P (kP = 1)
AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 14/34
AC Inductor with a sinusoidal supply
u(t) = √2U sin(ωt + φu)
i(t) = √2I sin(ωt) ( φi = 0)u + e = Ri = 0 → u = - e
u(t) = dφ/dt = Ldi/dt
u(t) = √2ωLI cos(ωt)
u(t)
i(t)
uL(t)
L = ct u(t) = √2 ωLI sin(ωt + π/2)
U = ωLI → Z = U/I = ωL
φu = π/2 → φ = φu - φi = π/2
R = Z cos φ = 0X= Z sin φ = ωL > 0
Z = R + jX = jωL
AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 15/34
AC Inductor with a sinusoidal supply
U
I
P = RI2 = 0Q = XI2 = ωLI2 > 0
S = ZI2 = ωLI2 I=U/(2πfL)
XL=2πfL
AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 16/34
AC Capacitor with a sinusoidal supply
uC(t)
iC(t)
u(t)
u(t) = √2U sin(ωt) (φu = 0)
i(t) = √2I sin(ωt + φi)
i(t) = dq/dt = C du/dt
u(t) = (1/C)∫i(t)dt
u(t) = -√2I(1/ωC) cos(ωt + φi)
u(t) = √2I(1/ωC) sin(ωt + φi - π/2)
U = I/(ωC) Z = U/I = 1/(ωC)
φu = φi -π/2 = 0→ φi = π/2φ = φu - φi = -π/2, (kP = 0)
R = Z cos φ = 0X= Z sin φ = -1/(ωC) < 0
Z = R + jX = -j/(ωC)
AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 17/34
AC Capacitor with a sinusoidal supply
P = RI2 = 0Q = XI2 = -1/(ωC)I2 < 0
S = ZI2 = I2/(ωC) XC=1/(2πfC)
I=2πfCU
AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 18/34
Applications
ZC = 1/(2πfC) = 3.18 kΩ IC = 2πfCU = 15.7 mA
U = 50 Vf = 50 Hz 1 μFU = 50 Vf = 5 kHz
ZC = 1/(2πfC) = 31.8 Ω IC = 2πfCU = 1.57 A
U = 50 Vf = 5o Hz
ZL = 2πfL = 691.15 Ω IL = U/2πfL = 72.3 mA
U = 50 Vf = 5 kHz
ZL = 2πfL = 69.115 kΩ IL = U/2πfL = 0.723 mA
AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 19/34
Parallel RLC circuit
I
U
u(t) = √2U sin(ωt) (φu = 0)
i(t) = √2I sin(ωt + φi)
i(t) = iR(t) + iL(t) + iC(t)
I = IR + IL + IC
IR = U/ZR = U/R
IL = U/ZL = U/jωL = -jU/ωL
IC = U/ZC = U/(-j/ωC) =jωCU
I = U[1/R - j(1/ωL - ωC)]
1/Z = [1/R - j(1/ωL - ωC)]
1/Z = √[(1/R)2 + (1/ωL - ωC)2]
tanφ = (1/ωL - ωC)/(1/R)
AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 20/34
Parallel RLC circuit
realaxis
imaginaryaxis
UIR
IL
IC
I
φ
u(t) = √2·50 sin(628t)R = 50 ΩL = 20 mHC = 5 μFZ = ?, IR = ?, IL = ?, IC = ? I = ?, φ = ?
ZL = ωL = 628·0.02 = 12.6 ΩZC = 1/ωC = 1/(628·5·10-6)= 318.3 ΩZ = 1/√[(1/R)2 + (1/ωL - ωC)2] = 12.7 ΩIR = U/R = 1 A, IL = U/ZL = 3.9 A, IC = U/ZC = 0.16 A, I = √[IR
2 + (IL – IC)2] = 3.87 A, φ = arctg[(1/ωL - ωC)/(1/R)] = 75.3 °
AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 21/34
Parallel resonance circuit
I
U
Resonance : φ = 0
tanφ = (1/ωL - ωC)/(1/R)
1/ωL = ωC
ω0 = 2πf0 = 1/√LCf0 = 1/ (2π √LC)
1/Z0 = √(1/R)2
Z0 = R = Zmax
AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 22/34
Parallel resonance circuit
I0 = Imin = U/R
realaxis
imaginaryaxis
UIR = I
ILIC
φ = 0
3 PHASE AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 23/34
Three phase wave form
u1(t) = √2U sin(ωt)
u2(t) = √2U sin(ωt – 2π/3)
u3(t) = √2U sin(ωt – 4π/3) 120° 120° 120°
u2(t) = √2U sin(ωt + 4π/3)
u3(t) = √2U sin(ωt + 2π/3)
U1 = U ej0
U2 = U ej4π/3 = a2U1 a = ej2π/3
U3 = U ej2π/3 = aU1
U1
U2
U3
2π/32π/3
2π/3
THREE PHASE AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 24/34
Three phase voltage
U1
U2
U3
2π/32π/3
2π/3
U1 – U2 = U1(1 – a2)a2 = ej4π/3 = = cos(4π/3) + j sin(4π/3)(1 - a2) = 3/2 + j√3/2 = = √3(√3/2 + j1/2) == √3[cos(π/6) + j sin(π/6)]
U1 – U2 = √3 U1 ejπ/6
U1 –U2
π/6
THREE PHASE AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 25/34
Three phase connections
Uph = Phase VoltageUl = Line VoltageIph = Phase CurrentIl = Line Current
Il = Iph
Ul ≠ Uph Il ≠ Iph
STAR (Y) connection
Uph
Ul
Il Iph
neutral
line
line
Ul
Il
Iph
DELTA (∆) connection
THREE PHASE AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 26/34
Unbalanced 3-phase load, STAR connection
L1
L2
L3
N
Z1
Z2
Z3
Z0
I1
I2
I3
I0
U1 U2 U3U0
U’1
U’2
U’3
U1, U2 = a2U1, U3 = aU1
U’1= U1 – U0
U’2 = U2 – U0
U’3 = U3 – U0
I1 + I2 + I3 = I0
I1 = U’1/Z1 = (U1 – U0)/Z1
I2 = U’2/Z2 = (U2 – U0)/Z2
I3 = U’3/Z3 = (U3 – U0)/Z3
(U1 – U0)/Z1 + (U2 – U0)/Z2 + + (U3 – U0)/Z3 = U0/Z0
U0 = (U1/Z1 + U2/Z2 + U3/Z3)/(1/Z1 +
1/Z2+ 1/Z3+ 1/Z0)
THREE PHASE AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 27/34
Unbalanced 3-phase load, STAR connection
L1
L2
L3
N
Z1
Z2
Z3
Z0
I1
I2
I3
I0
U1 U2 U3U0
U’1
U’2
U’3
U’1 = U1 U2’ = U2 U’3 = U3
U0 = 0
U1
U2
U3
U’1
U’3
U’2
U0
N
N’
N’
THREE PHASE AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 28/34
Balanced 3-phase load, STAR connection
L1
L2
L3
N
Z
Z
Z
Z0
I1
I2
I3
I0
U1 U2 U3U0
U’1
U’2
U’3
N’
Z1 = Z2 = Z3 = Zejφ
U0 = [(U1 + U2 + U3)/Z]/(3/Z + 1/Z0) = 0
I1 = U1/Z = U1/(Zejφ) = (U1/Z)e-jφ
I2 = U2/Z = a2U1/(Zejφ) = a2I1
I3 = U3/Z = aU1/(Zejφ) = aI1
U1, U2 = a2U1, U3 = aU1
I1, I2 = a2I1 I3 = aI1
U1
U2
U3
I1
I2
I3
U12
U23
U31
U12
U23
U31
Ul =√3 Uph
THREE PHASE AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 29/34
Balanced 3-phase load, DELTA connection
Z12 = Z23 = Z31 = Zejφ
U12, U23 = a2U12, U31 = aU12
I12, I23 = a2I12 I31 = aI12
U12
U23
U31
I12I23
I31
I2
I3 I1
Il =√3 Iph
L1
L2
L3
U12
U23
U31
Z
Z
Z
I12
I23
I31
I1
I2
I3
I12 = U12/Z = U12/(Zejφ) = (U12/Z)e-jφ
I23 = U23/Z = a2U12/(Zejφ) = a2I12
I31 = U31/Z = aU12/(Zejφ) = aI12
THREE PHASE AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 30/34
Power in 3 - phase circuits
L1
L2
L3
N
Z
Z
Z
Z0
I1f
I2f
I3f
I0
U1 U2 U3U0 = 0
U1
U2
U3
U1 = Uf, U2 = a2Uf, U3 = aUf
I1f = U1/Z = Uf/(Zejφ) =
= (Uf/Z)e-jφ = If e-jφ
I*1f = If ejφ
I2f = U2/Z = a2Uf/(Zejφ) = a2If e-jφ
I*2f = aIf ejφ
I3f = U3/Z = aUf/(Zejφ) = aIf e-jφ
I*3f = a2If ejφ
Z1 = Z2 = Z3 = Zejφ S = U1f I*1f + U2f I*2f + U3f I*3f
THREE PHASE AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 31/34
Power in 3 - phase circuits
L1
L2
L3
N
Z
Z
Z
Z0
I1f
I2f
I3f
I0
U1f U2f U3fU0 = 0
U1
U2
U3
S = Uf If ejφ + a2Uf aIf ejφ +
+ aUf a2If ejφ = 3UfIf ejφ
(a3 = e2π = 1)
S = P + jQ
ejφ = cosφ + j sinφ
S = 3UfIf cosφ + j 3UfIf sinφ
P = 3UfIf cosφ
Q = 3UfIf sinφ
P = √3UlIl cosφ
Q = √3UlIl sinφ
“STAR” Uf = Ul/√3, If = Il
“DELTA”Uf = Ul, If = Il/√3
THREE PHASE AC CIRCUITS
Universitatea Tehnica din Cluj-Napoca, Facultatea de Constructii de Masini 32/34
Power in 3 - phase circuits. Application
L1
L2
L3
N
R
R
R
I1
I2
I3
U1 U2 U3
U1
U2
U3
Z1 = Z2 = Z3 = R = 100 Ω
PY = √3UlIl cosφ = = √3·400·(230/100)·cos0 = = 1600 W
L1
L2
L3
U12
U23
U31
R
R
R
I12
I23
I31
I1
I2
I3
Z1 = Z2 = Z3 = R = 100 Ω
P∆ = √3UlIl cosφ = = √3·400·(√3·400/100)·cos0 = = 4800 W
U1 = U2 = U3 = 230 V
U12 = U23 = U31 = 400 V