three-phase circuitsfaculty.citadel.edu/potisuk/elec202/notes/3phase2.pdf · three-phase circuits...
TRANSCRIPT
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THREE-PHASE CIRCUITS PART II
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POWER IN A BALANCED SYSTEM
• For a Y-connected load with θ∠= ZZY , the phase voltages and currents are:
,cos2 tVv pAN ω= ),120cos(2 o−= tVv pBN ω )120cos(2 o+= tVv pCN ω ),cos(2 θω −= tIi pAN ),120cos(2 o−−= θωtIi pBN )120cos(2 o+−= θωtIi pCN where pV and pI are the rms voltage and current, respectively.
• The total instantaneous power absorbed by the load is
)()()()( tptptptp CBA ++= CNCNBNBNANAN iviviv ++= )120cos()120cos()cos([cos2 oo −−−+−= θωωθωω ttttIV pp
)]120cos()120cos( oo +−++ θωω tt
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• Applying )]cos()[cos(21coscos BABABA −++= gives
)240cos(coscos3[)( o−++= ααθppIVtp )]240cos( o++ α
where θωα −= t2
• Applying BSBABA sinsincoscos)cos( −=+ yields
oo 240sinsin240coscoscoscos3[)( αααθ +++= ppIVtp ]240sinsin240coscos oo αα −+
]cos)21(2coscos3[ ααθ −++= ppIV
θcos3 ppV I=
→ The total instantaneous power in a balanced three-phase system is constant regardless of whether the load is Y- or Δ-connected while that of each phase is still time varying.
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→ The average power consumed by each phase of the load is one- ).( third of tp
θcosppp IVP =
→ The reactive power consumed by each phase of the load is
θsinppp IVQ = → The apparent power consumed by each phase of the load is
ppp IVS = → The complex power consumed by each phase of the load is
∗=+= ppppp IVjQPS
Note that pV is the phasor rms voltage whose magnitude is pV pI is the phasor rms current whose magnitude is pI
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• The total complex power is
*
22 3
333p
pppppp
ZV
ZIIVSS ==== ∗
where θ∠= Yp ZZ or θ∠ΔZ (the load impedance per phase)
• Alternatively, since pLpL VVII 3, == for a Y-connected load or pLpL VVII == ,3 for a Δ-connected load,
θθ cos3cos3 LLpp IVIVP == θθ sin3sin3 LLpp IVIVQ ==
θθ ∠=∠=+= LLpp IVIVjQPS 33
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Ex. Practice problem 12.6
For the Y-Y circuit of Practice Prob.12.2, calculate the complex power at the source and at the load.
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Ex. Practice problem 12.7
Calculate the line current (magnitude only) required for a 30-kW three-phase motor having a power factor of 0.85 lagging if it is connected to a balanced source with a line voltage of 440 V.
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UNBALANCED THREE-PHASE SYSTEMS
• An unbalanced system is due to unbalanced voltage sources or an unbalanced load.
• Solved by direct application of mesh and nodal analysis or ohm’s law
• For a four-wire Y-Y system, the neutral line current is not zero
)( cban IIII ++−=
• Each phase of the load consumes unequal power. Thus, the total power is the sum of the powers in the three phases.
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Ex. Practice problem 12.9
The unbalanced Δ-load shown is supplied by balanced line-to-line voltages of 240 V in the positive sequence. Find the line currents using Vab as reference.
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Ex. Practice problem 12.9
For the unbalanced circuit shown, find: (a) the line currents, (b) the total complex power absorbed by the load, and (c) the total complex power absorbed by the source.
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RESIDENTIAL WIRING
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SINGLE-PHASE THREE-WIRE RESIDENTIAL WIRING
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GROUNDING & SAFETY
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GROUND FAULT CIRCUIT INTERRUPTER (GFCI)
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GENERATION & DISTRIBUTION OF AC POWER
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