EE 204 Lecture 25 Mre Examples n Pwer Factr and the Reactve Pwer The pwer factr has been defned n the prevus lecture wth an example n pwer factr calculatn. We present tw mre examples n ths lecture. Example : lad absrbs 5 KW f pwer at a leadng p.f. f 0.8 when a 60 Hz snusdal vltage surce, wth a peak vltage f 40 V s appled acrss t. Calculate: a) ssumng the lad vltage V has a zer phase angle, calculate the lad current b) The lad mpedance c) s the lad mpedance capactve, nductve r purely resstve? V = 40 0 Fgure Slutn: The 5KW pwer absrbed by the lad s actually the average pwer absrbed by the lad. t s cmmn practce n pwer engneerng t refer t the average pwer, smply as pwer. Usng P = 0.5V csθ n the abve equatn, we knw: P = 5 KW = 5000W p.f. = 0.8 = csθ V = 40V 5000 = 0.5 (40) 0.8 5000 = = 89.286 0.5 40 0.8 = 89.286 s the ampltude f the lad current
[Nte that = θ = 89.286 θ we need t knw θ t cmpletely fnd ] Snce p.f. = 0.8 = csθ cs 0.8 =± 36.870 s θ =+ 36.870 r θ = 36.870? Because the p.f. s leadng θ < 0 θ = 36.870 θ 0 v = (the vltage acrss the lad has a zer phase angle) 36.870 θ = = θv θ 36.870 = 0 θ θ 36.870 = = θ = 89.286 36.870 = 89.29 36.87 V = 40 0 Fgure 2 b) the lad mpedance L V 40 0 = = =.568 36.870 Ω 89.286 36.870 [Nte that: the lad angle = p.f. angle = 36.870 = θ ] V = 40 0 = 89.29 36.87 L =.568 36.87 Ω Fgure 3 c) The lad mpedance s capactve. Ths can be cncluded easly frm ether: ) θ = 36.870 < 0 2) the pwer factr s leadng
= 89.29 36.87 V = 40 0 L =.568 36.87 Ω Fgure 3 Example 2: Lad and lad 2 n the crcut are cnnected n parallel t the vltage surce V = 220 0 V. Lad absrbs 0 KW at a leadng p.f. f 0.6 and lad 2 absrbs 6 KW at a laggng p.f. f 0.90. Calculate: ) the mpedance f the cmbned lad. 2) the p.f. f the cmbned lad. Slutn: V S = 220 0 Lad 2 Lad 6 KW p.f.=0.9 laggng Fgure 4 0 KW p.f.=0.6 leadng Fr lad P = 0.5 V (cs θ ) 0000 = 0.5 220 0.6 = 5.55 θ cs 0.6 53.30 = = (because p.f. f lad s leadng) θ = θ θ 53.30 = 0 θ = 53.30 v θ = θ = 5.55 53.30
repeatng a smlar prcedure fr lad 2: V S = 220 0 Lad 2 Lad 6 KW p.f.=0.9 laggng Fgure 5 0 KW p.f.=0.6 leadng P2 = 0.5 2V(cs θ ) 2 6000 = 0.52 220 0.9 2 = 60.606 θ 2 cs 0.9 25.842 =+ = (because p.f. f lad 2 s laggng) θ = θv θ2 25.842 = 0 2 θ 2 = 25.842 θ 2 = 2 θ 2 = 60.606 25.842 2 V S = 220 0 = 5.52 53.3 Lad 2 Lad 6 KW p.f.=0.9 laggng Fgure 6 0 KW p.f.=0.6 leadng KCL = + 2 ( s the current thrugh the cmbne lad) = 5.55 53.30 + 60.606 25.842 = (90.909 + 2.22 j) + (54.545 j26.48) = 45.454 + 94.794 = 73.67 33.093
V S = 220 0 = 5.52 53.3 Lad 2 Lad 6 KW p.f.=0.9 laggng The lad mpedance f the cmbne lad s: 2 = 60.6 25.84 Fgure 7 0 KW p.f.=0.6 leadng V 220 0 = = =.267 33.093 Ω 73.67 33.093 = 73.67 33.093 + V S = 220 0 (Cmbned Lad) 220 0 Fgure 8 b) The mpedance angle f the cmbned lad s θ = 33.093 p.f. = cs( 33.093 ) = 0.838 (leadng) = 73.67 33.093 V S = 220 0 =.267 33.093 Ω 220 0 Fgure 9 The Reactve Pwer:
n the prevus lecture we have seen that the energy strage elements (nductrs and capactrs) d nt absrb any average pwer. These elements actually absrb electrc pwer (nstantaneus pwer) durng part f the snusdal cycle. The absrbed pwer s stred by these elements. Electrc energy absrbed and stred Crcut Energy Strage Element Fgure 0 Durng anther part f the cycle, thse elements delver all the prevusly stred pwer back t the crcut. Ths s why n average, the pwer absrbed by these elements equals zer. prevusly stred electrc energy s delvered back t the crcut Crcut Energy Strage Element Fgure The pwer stred and then delvered back t the crcut s called the reactve pwer Q. The reactve pwer absrbed by an electrc element s gven by: Q= 0.5Vsnθ Q has the unt f VR [Vlt-mpere-Reactve] = θ V = V θ v θ θ θ = v Fgure 2
The Reactve Pwer f the Resstr, nductr and Capactr: Fr the resstr, the phase dfference θ = 0 Q = 0.5Vsnθ = 0.5V sn 0 = 0 R The resstr has n reactve pwer, because t des nt stre electrc pwer V = V θ v Fr the nductr, the phase dfference Q = 0.5V sn 90 = 0.5V 0 L = θ θ = θ R θ = v 0 θ = 90 The nductr has a pstve reactve pwer Fgure 3 V = V θ v Fr the capactr, the phase dfference Q = 0.5Vsn( 90 ) = 0.5V 0 c = θ jwl θ = θ v -θ = 90 Fgure 4 θ = 90 The capactr has a negatve reactve pwer + V = V θv = θ jwc θ = θ v -θ = 90 Fgure 5 Example 3:
Fr the cmbne lad f the prevus example, calculate: a) the reactve pwer absrbed b) usng the result f part b), s the cmbned lad capactve, nductve r purely resstve? = 73.67 33.093 V S = 220 0 =.267 33.093 Ω 220 0 Fgure 9 Slutn: a) Fr the cmbned lad = 73.67 33.093 & V = 220 0 V θ = 33.093 Q= 0.5Vsnθ = 0.5 73.67 220 ( 0.546) = 0427.4 VR c) Q= 0.5Vsnθ the sgns f Q and θ are always the same Fr the cmbned lad: Q = 0427.4 < 0 θ < 0 the cmbned lad s a capactve mpedance = 73.67 33.093 V S = 220 0 =.267 33.093 Ω 220 0 Fgure 9