ELECTRICAL POWER AlternAting Current (AC)
45
Electrical Power Summary
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ELECTRICAL POWER AlternAting Current (AC)
Electrical Power Summary
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= ½Emax2[– Sin 2] from 0 to 2
= ½Emax2(2– 0 – 0 + 0) = Emax2 Therefore average value of E2 = ½Emax2
Take the square root to obtain the root mean square or rms value Erms= Emax /√ 2 = 0.707Emax
The power of an electrical circuit is proportional to E2 therefore use theErmsvalue for power.
Unless stated otherwise, the value given for an AC voltage or current is the rms value.
The power in a DC circuit is equivalent to the power in an AC circuit with the same rms values.
Peak factor and Form factor
Peak factor = (peak value) / ( rms value)
Form factor = (rms value) / (av value for ½ cycle) Sine wave Form factor = (0.707) / ( 0.636Emax) = 1.11 Triangular wave formE=Emaxθ/( π/2) between 0 and π /2
Eav= 0.5Emax
Erms=Emax √[ ∫{θ2/(π/2)2}dθ /(π/2)] with the integral from 0 to π/2 Erms= [Emax /(π/2)3/2] √[ ∫ θ2dθ]
Erms= [Emax /(π/2)3/2] √[ θ3/3 ] from 0 to π/2
Electrical Power Summary
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= ½Emax2[– Sin 2] from 0 to 2
= ½Emax2(2– 0 – 0 + 0) = Emax2 Therefore average value of E2 = ½Emax2
Take the square root to obtain the root mean square or rms value Erms= Emax /√ 2 = 0.707Emax
The power of an electrical circuit is proportional to E2 therefore use theErmsvalue for power.
Unless stated otherwise, the value given for an AC voltage or current is the rms value.
The power in a DC circuit is equivalent to the power in an AC circuit with the same rms values.
Peak factor and Form factor
Peak factor = (peak value) / ( rms value)
Form factor = (rms value) / (av value for ½ cycle) Sine wave Form factor = (0.707) / ( 0.636Emax) = 1.11 Triangular wave formE=Emaxθ/( π/2) between 0 and π /2
Eav= 0.5Emax
Erms=Emax √[ ∫{θ2/(π/2)2}dθ /(π/2)] with the integral from 0 to π/2 Erms= [Emax /(π/2)3/2] √[ ∫ θ2dθ]
Erms= [Emax /(π/2)3/2] √[ θ3/3 ] from 0 to π/2
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ELECTRICAL POWER AlternAting Current (AC)
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Electrical Power Summary
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Erms= [Emax /(π/2)3/2] √[ (π/2)3/3] =Emax /√3 = 0.577Emax Triangular wave Form factor = 0.577Emax/0.5Emax = 1.15
Square wave Eav=Emax and Erms=Emax Square wave form factor =Emax/Emax = 1 Summarising
Square wave form factor = 1 peak factor = 1
Sine wave form factor = 1.11 peak factor = √2 = 1.41 Triangular wave form factor = 1.15 peak factor = √3 = 1.73 Thus the form factor and peak factor are a measure of how peaky the wave is.
Example
Find the average value for half cycle, the rms value and the form factor of the wave shown where 3T = π
Figure 51;Wave form example
Average value for ½ cycleEav= 2200/3 = 133 volts
Erms = √[{2 ∫2002(t/T)2dt+ 2002T}/π] where the integral is from 0 to T = 200 √[{2t3/( 3T2) +T}/π] where the integral is from 0 to T = 200 √[{2T3/( 3T2) +T}/π] = 200 √[(5T/3)/π ] where T = π/3 = [200 √(5π/9)/π] = 200 √[5/9]
= 149 volts
Form factor =Erms/Eav= 149 / 133 = 1.12 Frequency
The number of complete cycles per second is the frequency in Hertz (Hz) Mains electricity is at a frequency of 50 Hz in Europe and 60 Hz in America.
E=EmaxSint so the time taken forNcycles is given by T = 2πN Therefore T = 2πN/ But the time taken for N cycles = N/f Therefore N f = 2πN/
Therefore = 2πf
AC VoltageE=EmaxSin (2f t) Vector representation of AC
Figure 52;Vector representation of AC
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ELECTRICAL POWERElectrical Power AlternAting Current (AC)Summary
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Let the vectorVprotate anti-clockwise at a speed ofω radians/sec.
The projection ofVpon a vertical is sinusoidal with respect to time.
Let another vectorIpat an angle of withVp also rotate anti-clockwise at the same speed of ω radians/sec.
The projection ofIpon the vertical is sinusoidal displaced behindVp.
Figure 53;Vector representation of voltage and current The vectorsVpand Ip represent the two sinusoidal quantities.
Thus alternating currents and voltages can be represented by vectors.
Addition of two AC voltages or two AC currents
LetV1andV2be two AC voltages at the same frequency but of different magnitude and phase angle.
Figure 54;Addition of two AC currents or voltages V1=Vp1Sin (ωt+θ1) andV2=Vp2Sin (ωt+θ2)
V1+V2=Vp1[SinωtCosθ1+ CosωtSinθ1] +Vp2[SinωtCosθ2+ CosωtSinθ2] V1+V2 = Sinωt[Vp1Cosθ1+Vp2Cosθ2] + Cosωt[Vp1Sinθ1+Vp2Sinθ2] Put V1+V2=Vp3[Sin (ωt+θ3)]
= Sinωt(Vp3Cosθ3) + Cosωt(Vp3Sinθ3) Thus [Vp1Cosθ1+Vp2Cosθ2] =Vp3Cosθ3
And [Vp1Sinθ1+Vp2Sinθ2] =Vp3Sinθ3
Inspection of the vector diagram of the two voltages shows that this result could have been obtained directly from the vector diagram.
tanθ3=Vp3Sinθ3 / (Vp3Cosθ3)
= [Vp1Sinθ1+Vp2Sinθ2] / [Vp1Cosθ1+Vp2Cosθ2] andVp3 = √[(Vp3Sinθ3)2 + (Vp3Cosθ3)2]
= √[(Vp1Sinθ1+Vp2Sinθ2)2 + (Vp1Cosθ1+Vp2Cosθ2)2]
If two AC voltagesV1andV2(at the same frequency but different phase) are added together, the result is another AC voltage whose magnitude is the vector additionV1+V2.
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Similarly if two AC currentsI1andI2(at the same frequency but different phase) are added together, the result is another AC current whose magnitude is the vector addition I1+I2. The vectors have been shown as the peak value of the vector. However the rms value of a sine wave is always 1/√2 times the peak value. Thus the vector diagram of the rms values is exactly the same to a different scale as the vector diagram for the peak values.
The vector diagrams of voltage and current are the rms values unless stated otherwise.
Power in a single phase AC circuit
The power in an AC circuit is the product of Volts and Amps.
Let the phase angle between voltage and current be.
LetV=VpSinωt and I =Ip Sin (ωt+) W=Vp IpSinωtSin (ωt+)
=Vp IpSinωt(SinωtCos+ CosωtSin)
=Vp Ip[Sin2ωtCos+ (1/2) Sin 2ωtSin]
The mean value of Sin 2ωtover a complete cycle is zero, W=Vp IpCosSin2ωt
Putωt=x
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Similarly if two AC currentsI1andI2(at the same frequency but different phase) are added together, the result is another AC current whose magnitude is the vector addition I1+I2. The vectors have been shown as the peak value of the vector. However the rms value of a sine wave is always 1/√2 times the peak value. Thus the vector diagram of the rms values is exactly the same to a different scale as the vector diagram for the peak values.
The vector diagrams of voltage and current are the rms values unless stated otherwise.
Power in a single phase AC circuit
The power in an AC circuit is the product of Volts and Amps.
Let the phase angle between voltage and current be.
LetV=VpSinωt and I =Ip Sin (ωt+) W=Vp IpSinωtSin (ωt+)
=Vp IpSinωt(SinωtCos+ CosωtSin)
=Vp Ip[Sin2ωtCos+ (1/2) Sin 2ωtSin]
The mean value of Sin 2ωtover a complete cycle is zero, W=Vp IpCosSin2ωt
Putωt=x
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ELECTRICAL POWER AlternAting Current (AC)
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The mean value of Sin2x= (1/2π ) ∫Sin2xdx from 0 to 2π
= (1/2π ) ∫[(1 – Cos 2x)/2] dx from 0 to 2π = (1/4π ) [x– Sin 2x] from 0 to 2π
= (1/4π ) [2π – 0 – 0 + 0] = ½ ButVp Ip = 2Vrms Irmsr
ThereforeW=Vrms IrmsCos
In a single phase circuit
W=V ICos whereVandIare the rms values
Figure 55;Power in a single phase AC crcuit
This can be written by the vector equationW=V● I Cos is called the power factor (or pf).
Three phase system
If the generator has three coils at 1200spacing, three voltages will be produced with a phase angle of 1200between them.
Figure 56;Three phase system
If each voltage is connected to a circuit with the same impedance and the three currents return along the same conductor, then the vector sum of the three return currents is zero. Thus instead of three full sized return cables, only one of smaller size is needed. If none of the load is single phase, then the neutral is not needed at all. High voltage supplies are nearly always three phase without a neutral conductor. There is a great economy in distribution costs if the
electricity can be supplied in three phases.
The vector diagram shows the common return point, called the Neutral point, at N and a three phase supply with voltagesVA,VBandVC.
These are called the phase voltages orVph.
The voltagesV1,V2andV3are called the line voltages orVline.
It can be seen by the 300and 600 triangles that the magnitude of the line voltage is √3 times the magnitude of the phase voltage.
The voltage of a three phase supply is defined by the line voltage.
Thus a 415 volt three phase supply can provide three separate 240 volt single phase domestic supplies .
If all three phases have currents of the same magnitude and power factor, then
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Total powerW= 3Vph I Cos = √3 Vline I Cos W = √3 Vline I Cos
Power to a balanced three phase system is constant Ea Ia = √2ESin (ωt) √2ISin (ωt –)
Eb Ib = √2ESin (ωt + 1200) √2ISin (ωt + 1200–) Ec Ic = √2ESin (ωt – 1200) √2ISin (ωt – 1200–) whereEandIare rms values
Sin A Sin B = ½[Cos(A – B) – Cos(A + B)]
Ea Ia=E I[Cos() – Cos(2ωt –)]
Eb Ib=E I[Cos() – Cos(2ωt + 2400–)]
Ec Ic=E I[Cos() – Cos(2ωt – 2400–)]
W=Ea Ia+Eb Ib+Ec Ic
W=E I[3 Cos() – Cos(2ωt –) – {Cos(2ωt + 2400–) + Cos(2ωt – 2400–)}]
CosA + Cos B = 2 Cos{(A + B)/2} Cos{(A – B)/2}
W=E I[3 Cos() – Cos(2ωt –) – 2{Cos(4ωt – 2)/2 Cos(4800)/2}]
W=E I[3 Cos() – Cos(2ωt –) – 2{Cos(ωt –) Cos(2400)}]
Cos 240 = – ½
W=E I [3 Cos() – Cos(2ωt –) + Cos(ωt –)]
W=E I[3 Cos]
W= 3E ICoswhereEis the phase voltage W = √3 VICoswhereVis the line voltage
This does not includet, therefore it is constant for all values oft.
Measurement of power
The power in a three phase, three wire, system can be measured by two single phase wattmeters.
Figure 57;Power measured by two wattmeters W1 reads the vector dot product (Va–Vb)●Ia W1 = (Va–Vb)●Ia
W2 = (Vc–Vb)●Ic
W1 +W2 =Va●Ia–Vb●(Ia+Ic) +Vc●Ic For a three wire system,Ia+Ib+Ic= 0
W1 +W2 =Va●Ia+Vb●Ib+Vc●Ic= total power in the three phases
The sum of two wattmeter readings gives the power in a three phase three wire system. The phases do not need to carry the same current or have the same power factor.
Figure 58;Power measured by two wattmeters
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Power Factor measurement by two wattmeters.
If the load is an electric motor, the voltages are usually balanced three phase system with the currents and power factors the same on each phase. In this case, the power factor can be obtained from the two wattmeter readings.
W1=Va IaCos–Vb IaCos(1200–)
=V ICos–VI{Cos(1200)Cos() + Sin(1200)Sin()}
W2=Vc IcCos–Vb IcCos(1200+)
=V ICos–VI{Cos(1200)Cos() – Sin(1200)Sin()}
W2–W1= 2V ISin(1200)Sin() = 2V I (√3/2)Sin() =V I (√3) Sin() ButW1+W2= 3V ICos()
(W2–W1)/(W1+W2) = (1/√3) Tan() Tan() = √3 (W2–W1)/(W1+W2)
Power factor = Cos() = 1 / √{1 + Tan2()}
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ELECTRICAL POWER resistAnCe, induCtAnCe And CApACitAnCe on AC
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