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1Power in Single Phase AC Circuits• Real Power• Reactive Power• Apparent Power• Power Factor• Complex Power• ExamplesReal Power PReal Power: The same as the average power and is sometimes called the active power. It has units of watts.• P = Pavg= VrmsIrmscos(Φv - Φi)= VrmsIrmspfPower factor = pf = cos(Φv - Φi)Power Factor Angle Φpf= Φv - ΦiLagging PF: For inductive loads, the current lags the voltage so that 0 < Φpf< 180. Leading PF: For capacitive, the current leads the voltage so that -180 < Φpf< 0.ELI the ICE manP(t) for general RLC Load• Recall that:P(t) = v(t)i(t) = V cos(ωt +Φv) I cos (ωt +ΦI)P(t) = ½ V I {cos(Φv - ΦI) + cos(2 ωt +Φv + ΦI)}• Can you show thatP(t) = VrmsIrmscos Φpf{1 + cos[2(ωt +Φv )]} +VrmsIrmssin Φpfsin[2(ωt +Φv )] ?Real, Reactive, and Apparent Power•Real Power (units = watts)= P = VrmsIrmscos Φpf• Reactive Power (units = VAR)Q = VrmsIrmssin Φpf•Apparent Power (units = VA)|S| = VrmsIrmsComplex Power• Given the general RLC case let =< Φand != < Φ"• Then the complex power is defined#!=  !∗= (< Φ)( < −Φ")=  < (Φ−Φ").• In rectangular formS =  &'((Φ−Φ") + j  ()*(Φ−Φ").S = P + j Q2Power TriangleS = P + j QS = complex powerP = real powerQ = reactive powerThe power triangleImQ sRePPassive sign conventionWith + conventional current entering the + voltage terminal of a device, then P > 0 : real power absorbed by deviceP < 0 : real power delivered by deviceQ > 0 : reactive power absorbed by deviceQ < 0 : reactive power delivered by deviceMeasuring PF• In the lab we can find PF without measuring voltage and current phase angles since;PF = +|-|=./ 012.00.34 012.=244 .4. .5"361/4 .4. 7 .4.Ex 1 … Derive P, Q and S for the source 89="3:;=120?@( < 0°10Ω= 12C ?@( < 0° D9="3:EFD=120?@( < 0°E5Ω= 24C ?@( < −90° 414:= 89+ D9= 26.8C ?@( < −63.4°• EX 1 cont..and the current into the ‘+’ side of Vin is "3= − 414= 26.8 C ?@( < +116.6°# = "3: "3∗O= (120 < 0°)(26.8C < −116.6°)S = 3220 VA < -116.6°S = -1440 w – j 2880 var• P = 1440 w delivered• Q= 2880 var deliveredPower Factor Correction• What is the big deal about the power factor?Since |S| = P+ QP=   then for a fixed supply voltage and a given real power P, the larger the Q the larger the required . This mean all the distribution equipment (lines, cables, transformers, circuit breakers, etc.) must be sized for the larger . Additionally, larger P; losses will occur in transmission lines.Thus we want to keep Q small so thatR = P+QP~13Example 2Calculate the power to be delivered by a capacitorconnected in parallel with the previous load in order to increase the source power factor to 0.95 lagging. Ex 2…• Recall: 1/5 = 1440 T, Q1/5= 2880 C;And #1/5= 3220 C• We require VW = &'(∅3.2= 0.95 ⇒ ∅3.2= 18.2°• Using the power triangleZ[*∅3.2=\]^_+`abQ S Q3.2= 1/5Z[*∅3.2Q3.2= 1440 tan (18.2°)= 473 VARP Qc0= Q1/5− Q3.2= 2407 VARExample 3Compare currents circuits of Ex 1 and Ex 2• Circuit 1 hasRd=+-=deefgPPf=0.44 d= 26.8C• Circuit 2 hasRP= 0.95 P=-6=deefhiejghdPf=dkdldPf= 12.6C• Clearly the higher PF leads to smaller required current.Adding Complex Powers• The total power is the sum of the component powers regardless of their interconnection. • #!= !*={: !1*+ !2*}• =(P1+jQ1) +(P2+jQ2)• =(P1+P2) +j(Q1+Q2)• = PTOT + j QTOTEX 4 Find S1and S2 !=nd+ nP=120 < 02 + E4 + (3 − E2)= 22.3C < −21.81:= nd != 4.47 < 63.4 22.3 < −21.8 = 99.6V < 41.62:= nP != 3.61 < −33.7 22.3 < −21.8 = 80.3V <-55.5EX 4 cont.#d= 1: 1:∗= 2220 VA <63.4 = 993 W + j 1986 VAR#P= 2: 2:∗= 1790 VA <-33.7 = 1489 W - j 993 VAR#pqp=  9∗= (120V <0)(22.3A<21.8) = 2673VA <21.8#pqp= 2482.3W + j 993 VARAnd therefore#pqp= pqp+ j Qpqp= #d+#P= d+ P+ E(Qd+


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The Citadel ELEC 316 - Power in Single Phase AC Circuits

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