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Lecture 91 EEE 302 Electrical Networks II Dr. Keith E. Holbert Summer 2001
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Lecture 92 Three-Phase Circuits In three-phase circuits the 3 voltages sources are 120° apart Polyphase generation and transmission of electricity is more advantageous and economical (1) three-phase instantaneous power is constant over time (2) single-phase line losses are 50% greater than three-phase losses (for the same load power, voltage, pf), i.e., P Single =3/2×P Three
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Lecture 93 Balanced System A balanced system is one in which the 3 sinusoidal voltages have the same magnitude and frequency, and each is 120° out-of-phase with the other two
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Lecture 94 Three-Phase Voltages V an +–+– +–+– V bn +–+– V cn n c b a Balanced If: V an =V rms 0°V bn =V rms -120°V cn =V rms -240°
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Lecture 95 MATLAB Exercise Let’s create and plot balanced 3-phase voltages EDU» t=0:0.0005:0.04; EDU» va=120*sqrt(2)*cos(377*t); EDU» vb=120*sqrt(2)*cos(377*t-2*pi/3); EDU» vc=120*sqrt(2)*cos(377*t-4*pi/3); EDU» plot(t,va,'b:',t,vb,'g--',t,vc,'r') EDU» legend(’va',’vb',’vc') Don’t exit MATLAB, we shall return
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Lecture 96 Balanced Circuit A balanced three-phase circuit is one in which the loads are such that the currents produced by the voltages are also balanced
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Lecture 97 MATLAB Exercise Let’s now create and plot some corresponding balanced 3-phase currents EDU» figure; EDU» ia=10*sqrt(2)*cos(377*t-pi/8); EDU» ib=10*sqrt(2)*cos(377*t-pi/8-2*pi/3); EDU» ic=10*sqrt(2)*cos(377*t-pi/8-4*pi/3); EDU» plot(t,ia,'c:',t,ib,'k--',t,ic,'m')
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Lecture 98 Three-Phase Instantaneous Power Recall that the instantaneous power is for one phase is p(t) = i(t) v(t) The total instantaneous power for three phases is –the instantaneous power is a constant over time!
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Lecture 99 MATLAB Exercise Let’s now compute the individual and total instantaneous powers, and plot them EDU» figure; EDU» pa=va.*ia; EDU» pb=vb.*ib; EDU» pc=vc.*ic; EDU» pt=pa+pb+pc; EDU» plot(t,pa,'b:',t,pb,'g--',t,pc,'r',t,pt,'k-.')
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Lecture 910 Three-Phase Connections Standard notation specifies the phase sequence as abc (called positive phase sequence) where V an =V p 0°V bn =V p -120°V cn =V p -240° V bn =V an -120° V cn =V an -240° –V bn lags V an by 120° –where V p is the phase voltage which is the magnitude of the phasor voltage from the neutral to any line The balanced voltage set means that V an + V bn + V cn = 0 [Book Error]
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Lecture 911 Wye (Y) Connected Load ZYZY a ZYZY b ZYZY c Load ZYZY ZYZY ZYZY a b c n n
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Lecture 912 Delta ( ) Connected Load ZZ ZZ ZZ a c b Load ZZ ZZ ZZ a b c
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