chapter 12 - Chapter 12: Three-Phase Circuits Exercises Ex....

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Unformatted text preview: Chapter 12: Three-Phase Circuits Exercises Ex. 12.3-1 VC = 120∠ − 240° so VA = 120∠0° and VB = 120∠ − 120° Vbc = 3 (120 ) ∠ − 90° Ex. 12.4-1 Four-wire Y-to-Y Circuit Mathcad analysis (12v4_1.mcd): Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 Vp := 120 ⋅0 j⋅ Vb := Va⋅ e Describe the three-phase load: Calculate the line currents: IaA = 1.079 − 0.674i IaA = 1.272 180 π ⋅ arg( IaA ) = −32.005 IaA := π 180 ⋅ − 120 Vc := Va⋅ e ZA := 80 + j⋅ 50 Va IbB := ZA j⋅ Vb ZB IbB = 1.061 π 180 ⋅ 120 ZB := 80 + j⋅ 80 IbB = −1.025 − 0.275i 180 π ⋅ arg( IbB) = −165 IcC := ZC := 100 − j⋅ 25 Vc ZC IcC = −0.809 + 0.837i IcC = 1.164 180 π ⋅ arg( IcC) = 134.036 12-1 Calculate the current in the neutral wire: INn := IaA + IbB + IcC INn = −0.755 − 0.112i Calculate the power delivered to the load: SA := IaA ⋅ IaA ⋅ ZA SB := IbB⋅ IbB⋅ ZB SA = 129.438 + 80.899i SC := IcC⋅ IcC⋅ ZC SB = 90 + 90i SC = 135.529 − 33.882i Total power delivered to the load: SA + SB + SC = 354.968 + 137.017i Calculate the power supplied by the source: Sa := IaA ⋅ Va Sb := IbB⋅ Vb Sc := IcC⋅ Vc Sa = 129.438 + 80.899i Sb = 90 + 90i Sc = 135.529 − 33.882i Total power delivered by the source: Sa + Sb + Sc = 354.968 + 137.017i Ex. 12.4-2 Four-wire Y-to-Y Circuit Mathcad analysis (12x4_2.mcd): Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 Vp := 120 ⋅0 j⋅ Vb := Va⋅ e Describe the three-phase load: Calculate the line currents: IaA = 1.92 − 1.44i IaA = 2.4 180 π ⋅ arg( IaA ) = −36.87 IaA := π 180 ⋅ − 120 Vc := Va⋅ e ZA := 40 + j⋅ 30 Va IbB := ZA j⋅ Vb ZB IbB = 2.4 π ⋅ 120 ZB := ZA IbB = −2.207 − 0.943i 180 π 180 ⋅ arg( IbB) = −156.87 IcC := ZC := ZA Vc ZC IcC = 0.287 + 2.383i IcC = 2.4 180 π ⋅ arg( IcC) = 83.13 12-2 Calculate the current in the neutral wire: INn := IaA + IbB + IcC INn = 0 Calculate the power delivered to the load: SA := IaA ⋅ IaA ⋅ ZA SB := IbB⋅ IbB⋅ ZB SA = 230.4 + 172.8i SC := IcC⋅ IcC⋅ ZC SB = 230.4 + 172.8i SC = 230.4 + 172.8i Total power delivered to the load: SA + SB + SC = 691.2 + 518.4i Calculate the power supplied by the source: Sa := IaA ⋅ Va Sb := IbB⋅ Vb Sc := IcC⋅ Vc Sa = 230.4 + 172.8i Sb = 230.4 + 172.8i Sc = 230.4 + 172.8i Total power delivered by the source: Sa + Sb + Sc = 691.2 + 518.4i Ex. 12.4-3 Three-wire unbalanced Y-to-Y Circuit with line impedances Mathcad analysis (12x4_3.mcd): Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 Vp := 120 ⋅0 j⋅ Vb := Va⋅ e Describe the three-phase load: π 180 ⋅ − 120 ZA := 80 + j⋅ 50 j⋅ Vc := Va⋅ e π 180 ⋅ 120 ZB := 80 + j⋅ 80 ZC := 100 − j⋅ 25 Calculate the voltage at the neutral of the load with respect to the neutral of the source: VnN := ZA ⋅ ZC⋅ e 4 j⋅ ⋅ π 3 + ZA ⋅ ZB⋅ e 2 j⋅ ⋅ π 3 + ZB⋅ ZC ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC VnN = −25.137 − 14.236i VnN = 28.888 ⋅ Vp 180 π ⋅ arg( VnN) = −150.475 12-3 Calculate the line currents: IaA := Va − VnN ZA IaA = 1.385 − 0.687i π Check: Vb − VnN IcC := ZB IbB = −0.778 − 0.343i IaA = 1.546 180 IbB := 180 π ZC IcC = −0.606 + 1.03i IbB = 0.851 ⋅ arg( IaA ) = −26.403 Vc − VnN IcC = 1.195 180 ⋅ arg( IbB) = −156.242 π ⋅ arg( IcC) = 120.475 IaA + IbB + IcC = 0 Calculate the power delivered to the load: SA := IaA ⋅ IaA ⋅ ZA SB := IbB⋅ IbB⋅ ZB SA = 191.168 + 119.48i SC := IcC⋅ IcC⋅ ZC SB = 57.87 + 57.87i Total power delivered to the load: SC = 142.843 − 35.711i SA + SB + SC = 391.88 + 141.639i Ex. 12.4-4 Three-wire balanced Y-to-Y Circuit with line impedances Mathcad analysis (12x4_4.mcd): Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 Vp := 120 ⋅0 Describe the three-phase load: j⋅ Vb := Va⋅ e π 180 ⋅ − 120 ZA := 40 + j⋅ 30 j⋅ Vc := Va⋅ e ZB := ZA π 180 ⋅ 120 ZC := ZA Calculate the voltage at the neutral of the load with respect to the neutral of the source: VnN := ZA ⋅ ZC⋅ e 4 j⋅ ⋅ π 3 + ZA ⋅ ZB⋅ e 2 j⋅ ⋅ π 3 + ZB⋅ ZC ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC ⋅ Vp 12-4 − 14 VnN = −1.31 × 10 − 14 Calculate the line currents: IaA := VnN = 2.301 × 10 Va − VnN ZA IaA = 1.92 − 1.44i π ⋅ arg( VnN) = 124.695 Vb − VnN IcC := ZB 180 π − 15 IaA + IbB + IcC = 1.055 × 10 180 ⋅ arg( IbB) = −156.87 π − 15 ⋅ arg( IcC) = 83.13 − 2.22i × 10 SB = 230.4 + 172.8i Total power delivered to the load: ZC IcC = 2.4 Calculate the power delivered to the load: SA := IaA ⋅ IaA ⋅ ZA SB := IbB⋅ IbB⋅ ZB SA = 230.4 + 172.8i Vc − VnN IcC = 0.287 + 2.383i IbB = 2.4 ⋅ arg( IaA ) = −36.87 Check: IbB := π IbB = −2.207 − 0.943i IaA = 2.4 180 180 − 14 + 1.892i× 10 SC := IcC⋅ IcC⋅ ZC SC = 230.4 + 172.8i SA + SB + SC = 691.2 + 518.4i Ex. 12.6-1 Balanced delta load: (See Table 12.5-1) Z ∆ = 180∠ − 45° VAB 360∠0° = = 2∠45° A Z ∆ 180∠− 45 I BC = phase currents: I AB = VBC 360∠−120° = = 2∠− 75° A ° Z ∆ 18∠− 45 I CA line currents: VCA 360∠120° = = = 2∠165° A ° Z ∆ 180∠− 45 I A = I AB − I CA = 2∠45° − 2∠165° = 2 3∠15° A I B = 2 3∠ −105° A I C = 2 3∠135° A 12-5 Ex. 12.7-1 Three-wire Y-to-Delta Circuit with line impedances Mathcad analysis (12x4_4.mcd): Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 Vp := 110 ⋅0 j⋅ Vb := Va⋅ e Describe the delta connected load: π 180 ⋅ − 120 Z1 := 150 + j⋅ 270 j⋅ Vc := Va⋅ e π 180 Z2 := Z1 ⋅ 120 Z3 := Z1 Convert the delta connected load to the equivalent Y connected load: ZA := Z1⋅ Z3 Z1 + Z2 + Z3 ZA = 50 + 90i Describe the three-phase line: ZB := Z2⋅ Z3 Z1 + Z2 + Z3 ZB = 50 + 90i ZaA := 10 + j⋅ 25 ZC := Z1⋅ Z2 Z1 + Z2 + Z3 ZC = 50 + 90i ZbB := ZaA ZcC := ZaA 12-6 Calculate the voltage at the neutral of the load with respect to the neutral of the source: VnN := ( ZaA + ZA ) ⋅ ( ZcC + ZC) ⋅ e − 14 − 14 IaA := IaA = 0.392 − 0.752i IaA = 0.848 Check: − 14 + 1.784i× 10 Calculate the line currents: π + ( ZaA + ZA ) ⋅ ( ZbB + ZB) ⋅ e 2 j⋅ ⋅ π 3 + ( ZbB + ZB) ⋅ ( ZcC + ZC) ( ZaA + ZA ) ⋅ ( ZcC + ZC) + ( ZaA + ZA ) ⋅ ( ZbB + ZB) + ( ZbB + ZB) ⋅ ( ZcC + ZC) VnN = −1.172 × 10 180 4 j⋅ ⋅ π 3 ⋅ arg( IaA ) = −62.447 VnN = 2.135 × 10 Va − VnN IbB := ZA + ZaA ⋅ arg( VnN) = 123.304 IcC := ZB + ZbB IbB = 0.848 π π Vb − VnN IbB = −0.847 + 0.036i 180 180 ⋅ arg( IbB) = 177.553 ⋅ Vp Vc − VnN ZC + ZcC IcC = 0.455 + 0.716i IcC = 0.848 180 π ⋅ arg( IcC) = 57.553 IaA + IbB + IcC = 0 Calculate the phase voltages of the Y-connected load: VAN := IaA ⋅ ZA VAN = 87.311 180 π ⋅ arg( VAN) = −1.502 VBN := IbB⋅ ZB VBN = 87.311 180 π ⋅ arg( VBN) = −121.502 VCN := IcC⋅ ZC VCN = 87.311 180 π ⋅ arg( VCN) = 118.498 Calculate the line-to-line voltages at the load: VAB := VAN − VBN VAB = 151.227 180 π ⋅ arg( VAB) = 28.498 VBC:= VBN − VCN VBC = 151.227 180 π ⋅ arg( VBC = −91.502 ) VCA := VCN − VAN VCA = 151.227 180 π ⋅ arg( VCA) = 148.498 Calculate the phase currents of the ∆ -connected load: IAB := VAB Z3 IAB = 0.49 180 π ⋅ arg( IAB) = −32.447 IBC := VBC Z1 IBC = 0.49 180 π ⋅ arg( IBC) = −152.447 ICA := VCA Z2 ICA = 0.49 180 π ⋅ arg( ICA) = 87.553 12-7 Ex. 12.8-1 Continuing Ex. 12.8-1: Calculate the power delivered to the load: SA := IaA ⋅ IaA ⋅ ZA SB := IbB⋅ IbB⋅ ZB SA = 35.958 + 64.725i SC := IcC⋅ IcC⋅ ZC SB = 35.958 + 64.725i Total power delivered to the load: SC = 35.958 + 64.725i SA + SB + SC = 107.875 + 194.175i Ex. 12.9-1 P1 = VAB I A cos(θ +30° ) + VCB I C cos(θ − 30° )= P + P2 1 pf = .4 lagging ⇒ θ = 61.97 ° So P T = 450(24) cos 91.97° + cos 31.97° = 8791 W ∴ P 1 = − 371 W P2 = 9162 W Ex. 12.9-2 Consider Fig. 12.9-1 with P = 60 kW P2 = 40 kW . 1 (a.) P = P + P2 = 100 kW 1 (b.) use equation 12.9-7 to get tan θ = 3 P2 − P 40− 60 1 =3 = − .346 ⇒ θ = − 19.11° PL + P2 100 then pf = cos ( − 19.110°) = 0.945 leading 12-8 Problems Section 12-3: Three Phase Voltages P12.3-1 Given VC = 277 ∠45° and an abc phase sequence: VA = 277 ∠ ( 45−120 ) ° = 277 ∠ − 75° VB = 277 ∠( 45° +120 )° = 277 ∠165° VAB = VA − VB =( 277 ∠ − 75° )−( 277 ∠165° ) =( 71.69 − j 267.56 ) −( −267.56+ j 71.69 ) =339.25− j 339.25 = 479.77 ∠ − 45° 480 ∠ − 45° Similarly: VBC = 480 ∠ − 165° and VCA = 480 ∠75° P12.3-2 VAB 3∠30° = 12470 ∠145° V VAB = VA × 3∠30° ⇒ VA = VAB = −VBA = − (12470 ∠−35° ) In our case: VA = So 12470 ∠145° = 7200∠115° 3∠30° Then, for an abc phase sequence: VC = 7200 ∠ (115 + 120 ) ° = 7200 ∠235° = 7200 ∠ − 125° VB = 7200 ∠ (115 − 120 ) ° = 7200 ∠ − 5° V P12.3-3 Vab = Va × 3∠30° ⇒ Va = Vab 3∠30° In our case, the line-to-line voltage is So the phase voltage is Vab = 1500 ∠30° V 1500 ∠30° Va = = 866∠0° V 3∠30° 12-9 Section 12-4: The Y-to-Y Circuit P12.4-1 Balanced, three-wire, Y-Y circuit: where Z A = Z B = Z C = 12∠30 = 10.4 + j 6 MathCAD analysis (12p4_1.mcd): Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 Vp := 208 3 ⋅0 j⋅ Vb := Va⋅ e Describe the balanced three-phase load: π 180 ⋅ − 120 j⋅ Vc := Va⋅ e ZA := 10.4 + j⋅ 6 π 180 ⋅ 120 ZB := ZA ZC := ZB Check: The voltage at the neutral of the load with respect to the neutral of the source should be zero: VnN := ZA ⋅ ZC⋅ e 4 j⋅ ⋅ π 3 + ZA ⋅ ZB⋅ e IaA := Va − VnN ZA IaA = 8.663 − 4.998i Check: − 14 ⋅ Vp IbB := VnN = 2.762 × 10 Vb − VnN IbB = −8.66 − 5.004i IaA = 10.002 π + ZB⋅ ZC ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC Calculate the line currents: 180 2 j⋅ ⋅ π 3 IbB = 10.002 180 ⋅ arg( IaA ) = −29.982 π − 15 IaA + IbB + IcC = 4.696 × 10 IcC := ZB Vc − VnN ZC −3 IcC = −3.205 × 10 + 10.002i IcC = 10.002 ⋅ arg( IbB) = −149.982 180 π ⋅ arg( IcC) = 90.018 − 14 − 1.066i× 10 12-10 Calculate the power delivered to the load: SA := IaA ⋅ IaA ⋅ ZA SB := IbB⋅ IbB⋅ ZB 3 SA = 1.04 × 10 + 600.222i Total power delivered to the load: SC := IcC⋅ IcC⋅ ZC 3 SB = 1.04 × 10 + 600.222i 3 3 SC = 1.04 × 10 + 600.222i 3 SA + SB + SC = 3.121 × 10 + 1.801i× 10 Consequently: (a) The phase voltages are Va = 208 ∠0° = 120∠0° V rms, Vb = 120∠ − 120° V rms and Vc = 120∠120° V rms 3 (b) The currents are equal the line currents (c) I a = I aA = 10∠ − 30° A rms, I b = I bB = 10∠ − 150° A rms and I c = I cC = 10∠90° A rms (d) The power delivered to the load is S = 3.121 + j1.801 kVA . P12.4-2 Balanced, three-wire, Y-Y circuit: where Va = 120∠0° Vrms, Vb = 120∠ − 120° Vrms and Vc = 120∠120° Vrms Z A = Z B = Z C = 10 + j ( 2 × π × 60 ) (100 × 10−3 ) = 10 + j 37.7 Ω and Z aA = Z bB = Z cC = 2 Ω Mathcad Analysis (12p4_2.mcd): 12-11 Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 Vp := 120 ⋅0 j⋅ Vb := Va⋅ e π 180 ⋅ − 120 π j⋅ 180 Vc := Va⋅ e ⋅ 120 Describe the three-phase load: ZA := 10 + j⋅ 37.7 ZB := ZA ZC := ZB Describe the three-phase line: ZaA := 2 ZbB := ZaA ZcC := ZaA Calculate the voltage at the neutral of the load with respect to the neutral of the source: ( ZaA + ZA ) ⋅ ( ZcC + ZC) ⋅ e VnN := 4 j⋅ ⋅ π 3 + ( ZaA + ZA ) ⋅ ( ZbB + ZB) ⋅ e 2 j⋅ ⋅ π 3 + ( ZbB + ZB) ⋅ ( ZcC + ZC) ( ZaA + ZA ) ⋅ ( ZcC + ZC) + ( ZaA + ZA ) ⋅ ( ZbB + ZB) + ( ZbB + ZB) ⋅ ( ZcC + ZC) − 15 VnN = −8.693 × 10 − 14 Calculate the line currents: IaA = 0.92 − 2.89i IaA := VnN = 2.396 × 10 Va − VnN ZA + ZaA π 180 π − 15 Calculate the phase voltages at the load: VA = 118.301 180 π π Vc − VnN ZC + ZcC IcC = 2.043 + 2.242i 180 π ⋅ arg( IcC) = 47.656 − 15 − 3.109i× 10 VA := ZA ⋅ IaA 180 IcC := IcC = 3.033 VB = 118.301 ⋅ arg( VA) = 2.801 ⋅ arg( VnN) = 111.277 ZB + ZbB ⋅ arg( IbB) = 167.656 IaA + IbB + IcC = −1.332 × 10 π Vb − VnN IbB = 3.033 ⋅ arg( IaA ) = −72.344 Check: IbB := IbB = −2.963 + 0.648i IaA = 3.033 180 180 − 14 + 2.232i× 10 ⋅ Vp ⋅ arg( VB) = −117.199 VB := ZB⋅ IbB VC := ZC⋅ IcC VC = 118.301 180 π ⋅ arg( VC) = 122.801 Consequently, the line-to-line voltages at the source are: Vab = Va × 3∠30° = 120∠0° × 3∠30° = 208∠30° Vrms, Vbc = 208∠ − 120° Vrms and Vca = 208∠120° Vrms The line-to-line voltages at the load are: VAB = VA × 3∠30° = 118.3∠3° × 3∠30° = 205∠33° Vrms, Vbc = 205∠ − 117° Vrms and Vca = 205∠123° Vrms and the phase currents are I a = I aA = 10∠ − 72° A rms, I b = I bB = 3∠168° A rms and I c = I cC = 3∠48° A rms 12-12 P12.4-3 Balanced, three-wire, Y-Y circuit: where Va = 10∠0° V = 7.07∠0° V rms, Vb = 7.07∠ − 120° V rms and Vc = 7.07∠120° V rms and Z A = Z B = Z C = 12 + j (16 ) (1) = 12 + j16 Ω MathCAD analysis (12p4_3.mcd): Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 Vp := 10 2 ⋅0 j⋅ Vb := Va⋅ e Describe the balanced three-phase load: π 180 ⋅ − 120 j⋅ Vc := Va⋅ e ZA := 12 + j⋅ 16 π 180 ⋅ 120 ZB := ZA ZC := ZB Check: The voltage at the neutral of the load with respect to the neutral of the source should be zero: VnN := ZA ⋅ ZC⋅ e 4 j⋅ ⋅ π 3 + ZA ⋅ ZB⋅ e IaA := IaA = 0.212 − 0.283i IaA = 0.354 π + ZB⋅ ZC ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC Calculate the line currents: 180 2 j⋅ ⋅ π 3 ⋅ arg( IaA ) = −53.13 Va − VnN ZA − 15 ⋅ Vp IbB := VnN = 1.675 × 10 Vb − VnN IbB = −0.351 − 0.042i IbB = 0.354 180 π ⋅ arg( IbB) = −173.13 Calculate the power delivered to the load: SB := IbB⋅ IbB⋅ ZB SA := IaA ⋅ IaA ⋅ ZA SA = 1.5 + 2i Total power delivered to the load: SB = 1.5 + 2i IcC := ZB Vc − VnN ZC IcC = 0.139 + 0.325i IcC = 0.354 180 π ⋅ arg( IcC) = 66.87 SC := IcC⋅ IcC⋅ ZC SC = 1.5 + 2i SA + SB + SC = 4.5 + 6i 12-13 Consequently (a) The rms value of ia(t) is 0.354 A rms. (b) The average power delivered to the load is P = Re {S} = Re {4.5 + j 6} = 4.5 W P12.4-4 Unbalanced, three-wire, Y-Y circuit: where Va = 100∠0° V = 70.7∠0° V rms, Vb = 70.7∠ − 120° V rms and Vc = 7.07∠120° V rms Z A = 20 + j ( 377 ) ( 60 ×10−3 ) = 20 + j 22.6 Ω, Z B = 40 + j ( 377 ) ( 40 × 10−3 ) = 40 + j 15.1 Ω Z C = 60 + j ( 377 ) ( 20 ×10−3 ) = 60 + j 7.54 Ω and Z aA = Z bB = Z cC = 10 + j ( 377 ) ( 5 × 10−3 ) = 10 + j 1.89 Ω Mathcad Analysis (12p4_4.mcd): Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 Vp := 100 ⋅0 j⋅ Vb := Va⋅ e π 180 ⋅ 120 j⋅ Vc := Va⋅ e π 180 ⋅ − 120 Enter the frequency of the 3-phase source: ω := 377 Describe the three-phase load: ZA := 20 + j⋅ ω⋅ 0.06 Describe the three-phase line: ZaA := 10 + j⋅ ω⋅ 0.005 ZbB := ZaA ZB := 40 + j⋅ ω⋅ 0.04 ZC := 60 + j⋅ ω⋅ 0.02 ZcC := ZaA 12-14 Calculate the voltage at the neutral of the load with respect to the neutral of the source: VnN := ( ZaA + ZA ) ⋅ ( ZcC + ZC) ⋅ e 4 j⋅ ⋅ π 3 + ( ZaA + ZA ) ⋅ ( ZbB + ZB) ⋅ e 2 j⋅ ⋅ π 3 + ( ZbB + ZB) ⋅ ( ZcC + ZC) ( ZaA + ZA ) ⋅ ( ZcC + ZC) + ( ZaA + ZA ) ⋅ ( ZbB + ZB) + ( ZbB + ZB) ⋅ ( ZcC + ZC) VnN = 12.209 − 24.552i Calculate the line currents: IaA := Va − VnN ZA + ZaA IaA = 2.156 − 0.943i IaA = 2.353 180 π 180 VnN = 27.42 IbB := π IcC := ZB + ZbB 180 π ⋅ arg( IbB) = 100.492 ( SA = 55.382 + 62.637i Total power delivered to the load: ZC + ZcC IcC = 1.244 Calculate the power delivered to the load: IbB⋅ IbB IaA ⋅ IaA ⋅ ZA SB := ⋅ ZB SA := 2 2 ) Vc − VnN IcC = −0.99 − 0.753i IbB = 2.412 ⋅ arg( IaA ) = −23.619 ( ⋅ arg( VnN) = −63.561 Vb − VnN IbB = −0.439 + 2.372i ⋅ Vp ) SB = 116.402 + 43.884i 180 π ⋅ arg( IcC) = −142.741 SC := (IcC⋅ IcC) 2 ⋅ ZC SC = 46.425 + 5.834i SA + SB + SC = 218.209 + 112.355i The average power delivered to the load is P = Re {S} = Re {218.2 + j112.4} = 218.2 W P12.4-5 Balanced, three-wire, Y-Y circuit: where Va = 100∠0° V = 70.7∠0° V rms, Vb = 70.7∠ − 120° V rms and Vc = 7.07∠120° V rms Z A = Z B = Z C = 20 + j ( 377 ) ( 60 ×10−3 ) = 20 + j 22.6 Ω and Z aA = Z bB = Z cC = 10 + j ( 377 ) ( 5 × 10−3 ) = 10 + j 1.89 Ω 12-15 Mathcad Analysis (12p4_5.mcd): Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 Vp := 100 ⋅0 j⋅ Vb := Va⋅ e π 180 ⋅ 120 π j⋅ 180 Vc := Va⋅ e ⋅ − 120 Enter the frequency of the 3-phase source: ω := 377 Describe the three-phase load: ZA := 20 + j⋅ ω⋅ 0.06 ZB := ZA ZC := ZA Describe the three-phase line: ZaA := 10 + j⋅ ω⋅ 0.005 ZbB := ZaA ZcC := ZaA Calculate the voltage at the neutral of the load with respect to the neutral of the source: VnN := ( ZaA + ZA ) ⋅ ( ZcC + ZC) ⋅ e 4 j⋅ ⋅ π 3 + ( ZaA + ZA ) ⋅ ( ZbB + ZB) ⋅ e 2 j⋅ ⋅ π 3 + ( ZbB + ZB) ⋅ ( ZcC + ZC) ( ZaA + ZA ) ⋅ ( ZcC + ZC) + ( ZaA + ZA ) ⋅ ( ZbB + ZB) + ( ZbB + ZB) ⋅ ( ZcC + ZC) − 15 VnN = −8.982 × 10 − 14 Calculate the line currents: IaA := IaA = 1.999 − 1.633i IaA = 2.582 180 π − 14 + 1.879i× 10 ⋅ arg( IaA ) = −39.243 VnN = 2.083 × 10 Va − VnN ZA + ZaA IbB := 180 π ⋅ arg( VnN) = 115.55 Vb − VnN IcC := ZB + ZbB IbB = 0.415 + 2.548i π ⋅ arg( IbB) = 80.757 Vc − VnN ZC + ZcC IcC = −2.414 − 0.915i IbB = 2.582 180 ⋅ Vp IcC = 2.582 180 π ⋅ arg( IcC) = −159.243 Calculate the power delivered to the load: SA := (IaA⋅ IaA) ⋅ ZA 2 SA = 66.645 + 75.375i Total power delivered to the load: SB := (IbB⋅ IbB) ⋅ ZB 2 SB = 66.645 + 75.375i SC := (IcC⋅ IcC) ⋅ ZC 2 SC = 66.645 + 75.375i SA + SB + SC = 199.934 + 226.125i The average power delivered to the load is P = Re {S} = Re {200 + j 226} = 200 W 12-16 P12.4-6 Unbalanced, three-wire, Y-Y circuit: where Va = 10∠ − 90° V = 7.07∠ − 90° V rms, Vb = 7.07∠150° V rms and Vc = 7.07∠30° V rms and Z A = 4 + j ( 4 ) (1) = 4 + j 4 Ω, Z B = 2 + j ( 4 ) ( 2 ) = 2 + j 8 Ω and Z C = 4 + j ( 4 ) ( 2 ) = 4 + j 8 Ω Mathcad Analysis (12p4_6.mcd): Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 Vp := 10 ⋅ − 90 j⋅ Vb := Vp⋅ e π 180 ⋅ 150 j⋅ Vc := Vp⋅ e π 180 ⋅ 30 Enter the frequency of the 3-phase source: ω := 4 Describe the three-phase load: ZA := 4 + j⋅ ω⋅ 1 ZB := 2 + j⋅ ω⋅ 2 ZC := 4 + j⋅ ω⋅ 2 Calculate the voltage at the neutral of the load with respect to the neutral of the source: VnN := ZA ⋅ ZC⋅ Vb + ZA ⋅ ZB⋅ Vc + ZB⋅ ZC⋅ Va ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC VnN = 1.528 − 0.863i Calculate the line currents: IaA := IaA = −1.333 − 0.951i IaA = 1.638 180 π 180 VnN = 1.755 ⋅ arg( IaA ) = −144.495 Va − VnN ZA IbB := π Vb − VnN IbB = 0.39 + 1.371i IbB = 1.426 180 π ⋅ arg( VnN) = −29.466 ⋅ arg( IbB) = 74.116 IcC := ZB Vc − VnN ZC IcC = 0.943 − 0.42i IcC = 1.032 180 π ⋅ arg( IcC) = −24.011 12-17 Calculate the power delivered to the load: IaA ⋅ IaA IbB⋅ IbB SA := ⋅ ZA SB := ⋅ ZB 2 2 ( ) ( SA = 5.363 + 5.363i ) SC := SB = 2.032 + 8.128i Total power delivered to the load: (IcC⋅ IcC) 2 ⋅ ZC SC = 2.131 + 4.262i SA + SB + SC = 9.527 + 17.754i The average power delivered to the load is P = Re {S} = Re {9.527 + j17.754} = 9.527 W P12.4-7 Unbalanced, three-wire, Y-Y circuit: where Va = 10∠ − 90° V = 7.07∠ − 90° V rms, Vb = 7.07∠150° V rms and Vc = 7.07∠30° V rms and ZA = ZB = ZC = 4 + j ( 4 ) ( 2 ) = 4 + j 8 Ω Mathcad Analysis (12p4_7.mcd): Describe the three-phase source: j⋅ Va := Vp⋅ e π 180 Vp := 10 ⋅ − 90 j⋅ Vb := Vp⋅ e π 180 ⋅ 150 j⋅ Vc := Vp⋅ e π 180 ⋅ 30 Enter the frequency of the 3-phase source: ω := 4 Describe the three-phase load: ZA := 4 + j⋅ ω⋅ 2 ZB := ZA ZC := ZA The voltage at the neutral of the load with respect to the neutral of the source should be zero: VnN := ZA ⋅ ZC⋅ Vb + ZA ⋅ ZB⋅ Vc + ZB⋅ ZC⋅ Va ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC − 15 VnN = 1.517 × 10 12-18 Calculate the line currents: IaA := IaA = −1 − 0.5i π IbB := ZA Vb − VnN 180 π ⋅ arg( IbB) = 86.565 SA = 2.5 + 5i Total power delivered to the load: ( ZC IcC = 1.118 Calculate the power delivered to the load: IaA ⋅ IaA IbB⋅ IbB SA := ⋅ ZA SB := ⋅ ZB 2 2 ) Vc − VnN IcC = 0.933 − 0.616i IbB = 1.118 ⋅ arg( IaA ) = −153.435 ( IcC := ZB IbB = 0.067 + 1.116i IaA = 1.118 180 Va − VnN ) SB = 2.5 + 5i 180 π ⋅ arg( IcC) = −33.435 SC := (IcC⋅ IcC) 2 ⋅ ZC SC = 2.5 + 5i SA + SB + SC = 7.5 + 15i The average power delivered to the load is P = Re {S} = Re {7.5 + j15} = 7.5 W Section 12-6: The ∆- Connected Source and Load P12.5-1 Given I B = 50∠ − 40° A rms and assuming the abc phase sequence we have I A = 50∠80° A rms and I C = 50∠200° A rms From Eqn 12.6-4 I A = I AB × 3∠ − 30° ⇒ I AB = so IA 3∠ − 30° 50∠80° = 28.9∠110° A rms 3∠−30° = 28.9∠ − 10° A rms and ICA = 28.9∠ − 130° A rms I AB = I BC 12-19 P12.5-2 The two delta loads connected in parallel are equivalent to a single delta load with Z ∆ = 5 || 20 = 4 Ω The magnitude of phase current is 480 Ip = = 120 A rms 4 The magnitude of line current is I L = 3 I p = 208 A rms Section 12-6: The Y- to ∆- Circuit P12.6-1 We have a delta load with Z = 12∠30° . One phase current is I AB 208 208 ∠−30° − ∠−150° V V −V 3 3 = 208∠0° = 17.31∠ − 30° A rms = AB = A A = Z Z 12∠30° 12∠30° The other phase currents are I BC = 17.31∠ − 150° A rms and I CA = 17.31∠90° A rms One line currents is I A = I AB × 3∠ − 30° = (17.31∠ − 30° ) × ( ) 3∠ − 30° = 30∠0° A rms The other line currents are I B = 30∠ − 120° A rms and I C = 30∠120° A rms The power delivered to the load is P = 3( 208 ) (30) cos ( 0 − 30° ) = 9360 W 3 12-20 P12.6-2 The balanced delta load with Z ∆ = 39∠− 40° Ω is equivalent to a balanced Y load with ZY = Z∆ = 13∠ − 40° = 9.96 − j 8.36 Ω 3 Z T = Z Y + 4 = 13.96 − j 8.36 = 16.3∠ − 30.9 Ω 480 ∠−30° 3 then I A = = 17∠0.9° A rms ° 16.3 ∠−30.9 P12.6-3 Vab = Va × 3∠30° ⇒ Va = Vab 3∠30° In our case, the given line-to-line voltage is Vab = 380 ∠30° V rms 380 ∠30° So one phase voltage is Va = = 200∠0° V rms 3∠30° So VAB = 380∠30° V rms VA = 220∠0° V rms VBC = 380∠-90° V rms VB = 220∠−120° V rms VCA = 380∠150° V rms VC = 220∠120° V rms One phase current is IA = VA 220∠0° = 44∠ − 53.1° A rms Z 3+ j4 The other phase currents are I B = 44∠ −173.1° A rms amd I C = 44∠66.9° A rms 12-21 P12.6-4 Vab = Va × 3∠30° ⇒ Va = Vab 3∠30° In our case, the given line-to-line voltage is Vab = 380 ∠0° V rms Va = So one phase voltage is So 380 ∠0° = 200∠ − 30° V rms 3∠30° Vab = 380∠0° V rms Va = 220∠ − 30° V rms Vbc = 380∠-120° V rms Vb = 220∠ −150° V rms Vca = 380∠120° V rms Vc = 220∠90° V rms One phase current is IA = Va 220∠−30° = = 14.67∠ − 83.1° A rms Z 9 + j1 2 The other phase currents are I B = 14.67∠ − 203.1° A rms and I C = 14.67∠36.9° A rms 12-22 12-23 Section 12-7: Balanced Three-Phase Circuits P12.7-1 Va = IA 25 ×103 ∠0° Vrms 3 25 ×103 ∠0° Va = =3 = 96∠ − 25° A rms Z 150 ∠25° 25 ×103 96 cos(0 − 25°) = 3.77 mW P = 3 Va I A cos (θ v -θ I ) = 3 3 P12.7-2 Convert the delta load to an equivalent Y connected load: ˆ Z ∆ ⇒ Z Y = 50 Ω 3 To get the per-phase equivalent circuit shown to the right: The phase voltage of the source is Z ∆ = 50 Ω Va = 45×103 ∠0° = 26∠0° kV rms 3 The equivalent impedance of the load together with the line is 50 3 + 2 = 12 + j 5 = 13∠22.6° Ω Z eq = 50 10 + j 20 + 3 (10 + j 20 ) The line current is Ι aA = Va 26 × 103 ∠0° = = 2000∠ − 22.6° A rms Z eq 13∠22.6° The power delivered to the parallel loads (per phase) is 50 (10 + j 20 ) 3 2 PLoads = I aA × Re = 4 ×106 × 10 = 40 MW 50 10 + j 20 + 3 The power lost in the line (per phase) is 12-24 PLine = I aA × Re {Z Line } = 4 × 106 × 2 = 8 MW 2 The percentage of the total power lost in the line is PLine 8 × 100% = × 100% = 16.7% PLoad + PLine 40 +8 P12.7-3 Ia = Va 5∠30° = = 0.5∠ − 23° A ∴ I a = 0.5 A Z T 6 + j8 2 PLoad I = 3 a Re {Z Load } = 3 × 0.125 × 4 = 1.5 W 2 also (but not required) : PSource = 3 (5) (0.5) cos(−30 − 23) = 2.25 W 2 2 Pline I = 3 a Re{Z Line } = 3×0.125× 2 = 0.75 W 2 12-25 Section 12-8: Power in a Balanced Load P12.8-1 Assuming the abc phase sequence: VCB = 208∠15° V rms ⇒ VBC = 208∠195° V rms ⇒ VAB = 208∠315° V rms Then VA = VAB 208∠315° 208 = = ∠285° V rms 3∠30° 3∠30° 3 also I B = 3∠110° A rms ⇒ I A = 3∠230° A rms Finally P = 3 VAB I A cos (θ V − θ I ) = 3 ( 208 ) (3) cos(285° − 230°) = 620 W 3 P12.8-2 Assuming a lagging power factor: cos θ = pf = 0.8 ⇒ θ = 36.9° The power supplied by the three-phase source is given by Pin = Pout η = Pin = 3 I A VA pf 20 ( 745.7 ) = 17.55 kW where 1 hp = 745.7 W 0.85 ⇒ Pin 17.55 ×103 IA = = = 26.4 A rms 3 VA pf 480 3 ( 0.8 ) 3 480 ° I A = 26.4∠ − 36.9° A rms when VA = ∠0 V rms 3 12-26 P12.8-3 (a) For a ∆-connected load, Eqn 12.8-5 gives PT 1500 = = 4.92 A rms 3 VP I L pf 3( 220 )(.8) 3 The phase current in the ∆-connected load is given by PT = 3 VP I L pf ⇒ I L = I IL 4.92 ⇒ IP = L = = 2.84 A rms 3 3 3 The phase impedance is determined as: IP = Z= V 220 VL VL = ∠ (θ V − θ I ) = L ∠ cos −1 pf = ∠ cos −1 0.8 = 77.44∠36.9° Ω IP IP IP 2.84 (b) For a ∆-connected load, Eqn 12.8-4 gives PT = 3 VP I L pf ⇒ I L = PT 1500 = = 4.92 A rms 3 VP I L pf 3( 220 )(.8) 3 The phase impedance is determined as: 220 VP VP V Z= P = ∠ (θ V − θ I ) = ∠ cos −1 pf = 3 ∠ cos −1 0.8 = 25.8∠36.9° Ω IP IP IP 4.92 P12.8-4 Parallel ∆ loads Z1Z 2 (40∠30° ) (50∠−60° ) Z∆ = = = 31.2 ∠−8.7° Ω ° ° Ζ1 + Ζ 2 40∠30 + 50∠− 60 VL = VP , Ι P = VP 600 = = 19.2 A rms, Z∆ 31.2 IL = 3 Ι P = 33.3 A rms So P = 3 VL I L pf = 3 (600) (33.3) cos ( − 8.7° ) = 34.2 kW 12-27 P12.8-5 We will use In our case: S = S ∠θ = S cosθ + S sin θ = S pf + S sin ( cos −1 pf ) S1 = 39 (0.7) + j 39 sin ( cos −1 ( 0.7 ) ) = 27.3 + j 27.85 kVA 15 sin ( cos −1 ( 0.21) ) = 15 − j 69.84 kVA 0.21 S S3φ = S1 + S 2 = 42.3 − j 42.0 kVA ⇒ Sφ = 3φ = 14.1− j 14.0 kVA 3 The line current is S 2 = 15 + * S (14100+ j 14000) S = Vp I L ⇒ I L = = = 117.5 + j 116.7 A rms = 167 ∠45° A rms V 208 p 3 208 ∠0° = 120∠0° V rms. The source must The phase voltage at the load is required to be 3 provide this voltage plus the voltage dropped across the line, therefore * VSφ = 120∠0° + (0.038 + j 0.072)(117.5 + j 116.7) = 115.9 + j 12.9 = 116.6 ∠6.4° V rms Finally VSφ = 116.6 V rms P12.8-6 The required phase voltage at the load is VP = 4.16 ∠0° = 2.402∠0° kVrms . 3 Let I1 be the line current required by the ∆-connected load. The apparent power per phase 500 kVA required by the ∆-connected load is S1 = = 167 kVA . Then 3 S1 = S1 ∠θ = S1 ∠ cos −1 ( pf ) = 167 ∠ cos−1 ( 0.85) = 167∠31.8° kVA and * 3 S1 (167 ×10 ) ∠31.8° = 69.6∠ − 31.8° = 59 − j36.56 A rms ⇒ I1 = = 3 VP ( 2.402 ×10 ) ∠0° * S1 = VP I1 * 12-28 Let I2 be the line current required by the first Y-connected load. The apparent power per phase 75 kVA required by this load is S 2 = = 25 kVA . Then, noticing the leading power factor, 3 S 2 = S 2 ∠θ = S 2 ∠ cos −1 ( pf ) = 25 ∠ cos −1 ( 0 ) = 25∠ − 90° kVA and * 3 S 2 ( 25 ×10 ) ∠ − 90° = 10.4∠90° = j10.4 A rms ⇒ I2 = = VP ( 2.402 ×103 ) ∠0° * S 2 = VP I 2 * Let I3 be the line current required by the other Y-connected load. Use Ohm’s law to determine I3 to be 2402∠0° 2402∠0° I3 = + = 16 − j 10.7 A rms 150 j 225 The line current is I L = I1 + I 2 + I 3 = 75− j 36.8 A rms 4.16 ∠0° = 2.402∠0° kVrms .The source 3 must provide this voltage plus the voltage dropped across the line, therefore The phase voltage at the load is required to be VP = VSφ = 2402∠0° + (8.45 + j 3.9) (75 − j 36.8) = 3179 ∠−0.3° Vrms Finally VSL = 3 (3179) = 5506 Vrms P12.8-7 The required phase voltage at the load is VP = 4.16 ∠0° = 2.402∠0° kVrms . 3 Let I1 be the line current required by the ∆-connected load. The apparent power per phase 1.5 MVA required by the ∆-connected load is S1 = = 0.5 MVA . Then 3 S1 = S1 ∠θ = S1 ∠ cos −1 ( pf ) = 0.5 ∠ cos −1 ( 0.75) = 0.5∠41.4° MVA and * 6 S1 ( 0.5 ×10 ) ∠41.4° = 2081.6∠ − 41.4° = 1561.4 − j1376.6 A rms ⇒ I1 = = VP ( 2.402 ×103 ) ∠0° * S1 = VP I1 * 12-29 Let I2 be the line current required by the first Y-connected load. The complex power, per phase, is 0.67 S 2 = 0.67 + sin ( cos −1 ( 0.8 ) ) = 0.67 + j 0.5 MVA 0.8 * 6 S 2 ( 0.67 + j 0.5 ) ×106 ( 0.833 ×10 ) ∠ − 36.9° = I2 = = 3 3 VP ( 2.402 ×10 ) ∠0° ( 2.402 ×10 ) ∠0° = 346.9∠ − 36.9° = 277.4 − j 208.3 A rms The line current is I L = I1 + I 2 = 433.7 − j 345.9 = 554.7∠ − 38.6 A rms * * 4.16 ∠0° = 2.402∠0° kVrms .The source 3 must provide this voltage plus the voltage dropped across the line, therefore The phase voltage at the load is required to be VP = VSφ = 2402∠0° + (0.4 + j 0.8) (433.7 − j 345.9) = 2859.6 ∠ − 38.6° Vrms Finally VSL = 3 (2859.6) = 4953 Vrms The power supplied by the source is PS = 3 (4953) (554.7) cos (4.2° + 38.6° ) = 3.49 MW The power lost in the line is PLine = 3 × ( 554.7 2 ) × Re {0.4+ j 0.8} = 0.369 MW The percentage of the power consumed by the loads is 3.49 − 0.369 ×100% = 89.4% 3.49 12-30 P12.8–8 The required phase voltage at the load is VP = 600 ∠0° = 346.4∠0° Vrms . 3 θ = cos −1 (0.8) = 37 ° Let I be the line current required by the load. The complex power, per phase, is S = 160 + 160 sin ( cos −1 ( 0.8 ) ) = 160 + j 120 kVA 0.8 The line current is S (160 + j 120 ) × 103 I= = = 461.9 − j 346.4 A rms 346.4∠0° VP * * 600 ∠0° = 346.4∠0° Vrms .The source 3 must provide this voltage plus the voltage dropped across the line, therefore The phase voltage at the load is required to be VP = VSφ = 346.4∠0° + (0.005 + j 0.025) (461.9 − j 346.4) = 357.5 ∠1.6° Vrms Finally VSL = 3 (357.5) = 619.2 Vrms The power factor of the source is pf = cos (θ V − θ I ) = cos (1.6 − ( − 37)) = 0.78 12-31 Section 12-9: Two-Wattmeter Power Measurement P12.9-1 W = 14920 W hp P 14920 Pin = out = = 20 kW 0.746 η Pout = 20 hp × 746 Pin = 3 VL I L cos θ Pin 20 × 103 = = 0.50 3 VL I L 3 (440) (52.5) ⇒ cos θ = ⇒ θ cos -1 ( 0.5 ) = 60° The powers read by the two wattmeters are P = VL I L cos (θ + 30° ) = (440) (52.5)cos ( 60° + 30° ) = 0 1 and P2 = VL I L cos (θ − 30° ) = (440) (52.5)cos ( 60° − 30° ) = 20 kW P12.9-2 VP = VL = 4000 V rms IP = VP 4000 = = 80 A rms Z∆ 50 Z ∆ = 40 + j 30 = 50 ∠36.9° Ι L = 3 I P = 138.6 A rms pf = cos θ = cos (36.9° ) = 0.80 P1 = VL I L cos (θ + 30° ) = 4000 (138.6) cos 66.9° = 217.5 kW P2 = VL I L cos (θ −30° ) = 4000 (138.6) cos 6.9° = 550.4 kW PT = P1 + P2 = 767.9 kW Check : PT = 3 Ι L VL cos θ = 3 (4000) (138.6) cos 36.9° = 768 kW which checks 12-32 P12.9–3 Vp = Vp = 200 = 115.47 Vrms 3 VA =115.47∠0° V rms, VB = 115.47∠ −120° V rms and VC = 115.47∠120° V rms IA = VA 115.47∠0° = = 1.633∠ − 45° A rms Z 70.7∠45° I B = 1.633 ∠ − 165° A rms and I C = 1.633 ∠75° A rms PT = 3 VL I L cos θ = 3 (200) (1.633) cos 45° = 400 W PB = VAC I A cos θ1 = 200 (1.633) cos (45° − 30° ) = 315.47 W PC = VBC I B cos θ 2 = 200 (1.633) cos (45° + 30° ) = 84.53 W P12.9-4 ZY = 10∠ − 30° Ω and Z ∆ = 15∠30° Ω Convert Z ∆ to Z Y → Z Y = ˆ ˆ then Zeq = Z∆ = 5∠30° Ω 3 (10∠−30° ) ( 5∠30° ) = 10∠−30°+5∠30° 208 Vp = Vp = = 120 V rms 3 VA = 120∠0° V rms ⇒ I A = 50∠0° = 3.78∠10.9° Ω 13.228 ∠−10.9° 120∠0° = 31.75 ∠−10.9° 3.78 ∠10.9° I B = 31.75∠−130.9° I C = 31.75∠109.1° PT = 3VL I L cos θ = 3 ( 208 ) ( 31.75 ) cos (10.9 ) =11.23 kW W1 = VL I L cos (θ −30°) = 6.24 kW W2 = VL I L cos (θ + 30°) = 4.99 kW 12-33 P12.9-5 PT = PA + PC = 920 + 460 = 1380 W tan θ = 3 ( −460 ) = −0.577 ⇒ θ = −30° PA − PC =3 1380 PA + PC PT = 3 VL I L cos θ so I L = IP = P12.9-6 IL = 4.43 A rms 3 ∴ Z∆ = Z = 0.868 + j 4.924 = 5∠80° VL = 380 V rms, VP = I L = I P and I P = 1380 PT = =7.67 A rms 3 VL cos θ 2 ×120×cos( −30 ) ⇒ 120 = 27.1 Ω ο r Z ∆ = 27.1 ∠−30° 4.43 θ = 80° 380 = 219.4 V rms 3 VP = 43.9 A rms Z P = ( 380 ) ( 43.9 ) cos (θ −30° ) = 10,723 W 1 P2 = ( 380 ) ( 43.9 ) cos (θ + 30° ) = −5706 W PT = P + P2 = 5017 W 1 12-34 PSpice Problems SP 12-1 FREQ 6.000E+01 IM(V_PRINT3)IP(V_PRINT3)IR(V_PRINT3)II(V_PRINT3) 3.142E+00 -1.644E+02 -3.027E+00 -8.436E-01 FREQ 6.000E+01 IM(V_PRINT1)IP(V_PRINT1)IR(V_PRINT1)II(V_PRINT1) 3.142E+00 -4.443E+01 2.244E+00 -2.200E+00 FREQ 6.000E+01 VM(N01496) 2.045E-14 FREQ 6.000E+01 IM(V_PRINT2)IP(V_PRINT2)IR(V_PRINT2)II(V_PRINT2) 3.142E+00 7.557E+01 7.829E-01 3.043E+00 VP(N01496) 2.211E+01 VR(N01496) 1.895E-14 VI(N01496) 7.698E-15 3.1422 20 = 98.7 W 2 3.1422 I B = 3.142∠75.57° A and RB = 20 Ω ⇒ PB = 20 = 98.7 W 2 3.1422 I C = 3.142∠ − 164.4° A and RC = 20 Ω ⇒ PC = 20 = 98.7 W 2 I A = 3.142∠ − 43.43° A and RA = 20 Ω ⇒ PA = P = 3 ( 98.7 ) = 696.1 W 12-35 SP 12-2 FREQ 6.000E+01 IM(V_PRINT3)IP(V_PRINT3)IR(V_PRINT3)II(V_PRINT3) 1.612E+00 -1.336E+02 -1.111E+00 -1.168E+00 FREQ 6.000E+01 IM(V_PRINT1)IP(V_PRINT1)IR(V_PRINT1)II(V_PRINT1) 2.537E+00 -3.748E+01 2.013E+00 -1.544E+00 FREQ 6.000E+01 VM(N01496) VP(N01496) 1.215E+01 -1.439E+01 FREQ 6.000E+01 IM(V_PRINT2)IP(V_PRINT2)IR(V_PRINT2)II(V_PRINT2) 2.858E+00 1.084E+02 -9.023E-01 2.712E+00 VR(N01496) VI(N01496) 1.177E+01 -3.018E+00 2.537 2 20 = 64.4 W 2 2.8582 I B = 2.858∠108.4° A and RB = 30 Ω ⇒ PB = 30 = 122.5 W 2 1.6122 I C = 1.612∠ − 133.6° A and RC = 600 Ω ⇒ PC = 60 = 78 W 2 I A = 2.537∠ − 37.48° A and RA = 20 Ω ⇒ PA = P = 64.4 + 122.5 + 78 = 264.7 V 12-36 Verification Problems VP 12-1 416 = 240 V = VA 3 Z = 10 + j 4 = 10.77 ∠21.8° Ω VA = VA 240 = = 22.28 A rms ≠ 38.63 A rms Z 10.77 38.63 = 22.3 . It appears that the line-to-line voltage was The report is not correct. (Notice that 3 mistakenly used in place of the phase voltage.) IA = VP 12-2 VL = VP = 240∠0° Vrms Z = 40 + j 30 = 50 ∠36.9° Ω IP = VP 240∠0° = = 4.8 ∠−36.9° A rms Z 50∠36.9° The result is correct. Design Problems DP 12-1 P = 400 W per phase, 0.94 = pf = cos θ ⇒ θ = cos-1 ( 0.94 ) =20° 208 I L 0.94 ⇒ I L = 3.5 A rms 3 I I ∆ = L = 2.04 A rms 3 V 208 = 101.8 Ω Z= L = 2.04 I∆ 400 = Z = 101.8 ∠20° Ω 12-37 DP 12-2 VL = 240 V rms PA = VL I L cos (30° + θ ) = 1440 W PC = VL I L cos (30° − θ ) = 0 W ⇒ 30−θ = 90° or θ = −60° then 1440 = 240 I L cos (−30° ) I L = 6.93 A rms IL = IP = VP Z ⇒ ⇒ 240 V Z = P = 3 = 20 Ω IP 6.93 Finally, Z = 20 ∠ − 60° Ω DP 12-3 Pin = Pout η 100 hp × (746 = W ) hp 0.8 = 93.2 kW, P = φ Pin = 31.07 kW 3 VL = 480 V rms, pfc = 0.9 and pf = 0.75. We need the impedance of the load so that we can use Eqn 11.6-7 to calculate the value of capacitance needed to correct the power factor. 0.75 = pf = cos θ ⇒ θ = cos-1 ( 0.75) = 41.4° 480 I P 0.75 ⇒ I P = 149.5 A rms 3 480 VP 3 = 1.85 Ω Z= = IP 149.5 31070 = Z = 1.85 ∠41.4° Ω = 1.388 + j1.223 Ω The capacitance required to correct the power factor is given by tan (cos −1 0.75) − tan (cos −1 0.9) 1.365 = 434 µ F C= × 2 2 1.365 +1.204 377 (Checked using LNAPAC 6/12/03) 12-38 DP 12-4 VL = 4∠0° kV rms n2 25 VL = 4000∠0° = 100∠0° kVrms n1 1 Try n2 = 25 then V2 = VL 4×103∠0° = 3∠0° kA rms = IL = 4 ZL 3 3000∠0° The line current in 2.5 Ω is I = = 120∠0° A rms 25 Thus V1 = ( R + j X ) I + V2 = (2.5 + j 40) (120∠0°) + 100×103 = 100.4 ∠2.7° kV Step need : n1 = Ploss = I 2 100.4 kV = 5.02 ≅ 5 20 kV R = 120 2 (2.5) = 36 kW, P = (4×103 ) (3× 103 ) = 12 MW 12 − .036 × 100% = 99.7 % of the power supplied by the source 12 is delivered to the load. ∴η = 12-39 ...
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