Question

Power Calculations 4 of 4 Review Learning Goal In this trial, you will practice the callation of power in circuts cortaring loads, including the effects of non-ideal transmission Ines, and practice how to improve the power factor for an entire system. Many appliancese.. hair dryers, coffee makers, and refrigerators) and indoads are powered by AC sources The folly to be able to calculate the power needs is important ma number of towns. Before completing the Moral make sure you are all with the relationships between average roactive, and apparent power and the power factor Part A. Power and power factor for parallel loads Consider the two loads in the circut in Figure 1). Lond 1 (L1) absorbs - 50 KVA at a leading power factor of ph-0.66. Load 12 (12) absorbs S, -56 KVA sta laging power factor of ph0.08. Ut normation to calculate the real (P) and reactive (Qpower provided by the V, -425 V-rms 60H source and its power factor Express your answers numerically to three significant figures separated by commas. View Available in VALD ? POW W, VAR, Submit Figure 102 > - Part B. Power for a parallel load with a transmission in impedance Consider the court in Figure 2). The load (1.1) absorbs P. W and Q - 1 VAR of power. Use this information to calculato come power provided by the source of the voltage across the food is V., V-mature and the impedance of the transion ine is 0.055.000 Express your answer in rectangular complex form to live significant figures. View Available PO LI VAXDC ? VA Submi

Answer 1

. 다 = (051.66 And P2 = (05!66 = 48-7 °E part Given- L; Sya56 KVA at.66 PB -1564-48.7") kumleading P = (56kVA) X Pf = Active power of Loadt = (568.66) kw P = 37kw Power factor angle le (^,) 48.7° Reactive power of Loada 8 = Sex sind, --564 Sin 48 7) KUARA Q; =-42 KUAR Lq: Sg. = 56kVA at .66 Pf lagging =(56<48.7") kva .: P2 = Szt Pf = (56X Q2 = S₂x Sin 2 f 66 ) kw = 37kW = (56x Sin 48-7) KUHR = = 42KVAR total active power supplied by source (P) P = PitP2 =137+37) kw = 74kw total reactive power supplied by source () S =' B1+8, = (-42+42) KUAR - OKUNR 8 = 0 KVAR

.! power factor = cos Quietor = cos (tan'la.) = cos (turbo) EI Part B). Given {line Load L Pi=s6kw 8,= IVAR My ve Vs=4250 . I Zline -7.055 +3.005322 How apparent power of Lord = S, = P, +j8 Si= 56 til = Vs I* I' 56+ j 425 I * = .131 (1.02 (-13121.02)* =:1315-1.02 Now active power taken by line = Siine 12 1.131<-1-02) oss tj.oos Sline = .308 -1.039 Siine = Pune + j Oine so, complex pamer supplied by source Sit seine 56+j1 +.308 3.039 → 56.308 +3.961 Seine Zine ; voltage

Parta for making power factor unity we have to add Capacitor in parallel with Loads u given - 4 S=56VA at .61 lagging ㅗ P=568.6) = 34.16 W 8, = 56* sin(605!61) = 56 x.792 = 44.37 VAR دا P2 = 56w 82 IVAR for making unity Power factor cirunit the reactive power taken by load must be supplied by capacitor Reactive power supplied by capacltor () Qc = 8,+d2 = (44.37 +1JVAR = $5-37 VAR And dc = (load voltage )? (Xe) 45.37=(425)2 Xc * * 2 4 S3 + = .25 ma (42512 so, empedence added to circunt = .25mar be