Unformatted text preview: Chapter 12: ThreePhase Circuits
Exercises
Ex. 12.31
VC = 120∠ − 240° so VA = 120∠0° and VB = 120∠ − 120° Vbc = 3 (120 ) ∠ − 90° Ex. 12.41
Fourwire YtoY Circuit Mathcad analysis (12v4_1.mcd):
Describe the threephase source:
j⋅ Va := Vp⋅ e π
180 Vp := 120 ⋅0 j⋅ Vb := Va⋅ e Describe the threephase 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) = 134.036 121 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. 12.42
Fourwire YtoY Circuit Mathcad analysis (12x4_2.mcd):
Describe the threephase source:
j⋅ Va := Vp⋅ e π
180 Vp := 120 ⋅0 j⋅ Vb := Va⋅ e Describe the threephase 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) = 83.13 122 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. 12.43
Threewire unbalanced YtoY Circuit with line impedances Mathcad analysis (12x4_3.mcd):
Describe the threephase source:
j⋅ Va := Vp⋅ e π
180 Vp := 120 ⋅0 j⋅ Vb := Va⋅ e Describe the threephase 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) = −150.475 123 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. 12.44
Threewire balanced YtoY Circuit with line impedances Mathcad analysis (12x4_4.mcd):
Describe the threephase source:
j⋅ Va := Vp⋅ e π
180 Vp := 120 ⋅0 Describe the threephase 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 124 − 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. 12.61
Balanced delta
load: (See Table 12.51)
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 125 Ex. 12.71
Threewire YtoDelta Circuit with line impedances Mathcad analysis (12x4_4.mcd):
Describe the threephase 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 threephase 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 126 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 Yconnected 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 linetoline 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) = 87.553 127 Ex. 12.81
Continuing Ex. 12.81:
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. 12.91
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. 12.92
Consider Fig. 12.91 with P = 60 kW P2 = 40 kW .
1
(a.) P = P + P2 = 100 kW
1
(b.) use equation 12.97 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 128 Problems
Section 123: Three Phase Voltages
P12.31
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°
P12.32
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 P12.33
Vab = Va × 3∠30° ⇒ Va = Vab
3∠30° In our case, the linetoline voltage is
So the phase voltage is Vab = 1500 ∠30° V
1500 ∠30°
Va =
= 866∠0° V
3∠30° 129 Section 124: The YtoY Circuit
P12.41
Balanced, threewire, YY circuit: where Z A = Z B = Z C = 12∠30 = 10.4 + j 6
MathCAD analysis (12p4_1.mcd):
Describe the threephase source: j⋅ Va := Vp⋅ e π
180 Vp := 208
3 ⋅0 j⋅ Vb := Va⋅ e Describe the balanced threephase 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 1210 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 . P12.42
Balanced, threewire, YY 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): 1211 Describe the threephase 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 threephase load: ZA := 10 + j⋅ 37.7 ZB := ZA ZC := ZB Describe the threephase 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 linetoline 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 linetoline 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
1212 P12.43
Balanced, threewire, YY 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 threephase source: j⋅ Va := Vp⋅ e π
180 Vp := 10
2 ⋅0 j⋅ Vb := Va⋅ e Describe the balanced threephase 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 1213 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 P12.44
Unbalanced, threewire, YY 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 threephase 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 3phase source: ω := 377
Describe the threephase load: ZA := 20 + j⋅ ω⋅ 0.06 Describe the threephase line: ZaA := 10 + j⋅ ω⋅ 0.005 ZbB := ZaA ZB := 40 + j⋅ ω⋅ 0.04 ZC := 60 + j⋅ ω⋅ 0.02
ZcC := ZaA 1214 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 P12.45
Balanced, threewire, YY 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 1215 Mathcad Analysis (12p4_5.mcd):
Describe the threephase 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 3phase source: ω := 377
Describe the threephase load: ZA := 20 + j⋅ ω⋅ 0.06 ZB := ZA ZC := ZA Describe the threephase 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 1216 P12.46
Unbalanced, threewire, YY 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 threephase 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 3phase source: ω := 4
Describe the threephase 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 1217 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 P12.47
Unbalanced, threewire, YY 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 threephase 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 3phase source: ω := 4
Describe the threephase 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 1218 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 126: The ∆ Connected Source and Load
P12.51
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 12.64
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 1219 P12.52 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 126: The Y to ∆ Circuit
P12.61
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 1220 P12.62
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 P12.63
Vab = Va × 3∠30° ⇒ Va = Vab
3∠30° In our case, the given linetoline 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 1221 P12.64
Vab = Va × 3∠30° ⇒ Va = Vab
3∠30° In our case, the given linetoline 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 1222 1223 Section 127: Balanced ThreePhase Circuits
P12.71
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 P12.72
Convert the delta load to an equivalent Y connected load:
ˆ
Z ∆ ⇒ Z Y = 50 Ω
3
To get the perphase 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 1224 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 P12.73 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 1225 Section 128: Power in a Balanced Load
P12.81
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 P12.82
Assuming a lagging power factor:
cos θ = pf = 0.8 ⇒ θ = 36.9° The power supplied by the threephase 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 1226 P12.83
(a) For a ∆connected load, Eqn 12.85 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 12.84 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 P12.84 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 1227 P12.85
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 P12.86
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 * 1228 Let I2 be the line current required by the first Yconnected 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 Yconnected 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 P12.87
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 * 1229 Let I2 be the line current required by the first Yconnected 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 1230 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 1231 Section 129: TwoWattmeter Power Measurement
P12.91
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 P12.92
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 1232 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 P12.94
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 1233 P12.95 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 = P12.96 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 1234 PSpice Problems
SP 121 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 8.436E01 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)
2.045E14 FREQ
6.000E+01 IM(V_PRINT2)IP(V_PRINT2)IR(V_PRINT2)II(V_PRINT2)
3.142E+00
7.557E+01
7.829E01
3.043E+00 VP(N01496)
2.211E+01 VR(N01496)
1.895E14 VI(N01496)
7.698E15 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 1235 SP 122 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 9.023E01
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 1236 Verification Problems
VP 121
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 linetoline voltage was
The report is not correct. (Notice that
3
mistakenly used in place of the phase voltage.)
IA = VP 122
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 121
P = 400 W per phase,
0.94 = pf = cos θ ⇒ θ = cos1 ( 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 1237 DP 122 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 123
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 11.67 to calculate the value of capacitance needed to correct the power factor.
0.75 = pf = cos θ ⇒ θ = cos1 ( 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) 1238 DP 124
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 = 1239 ...
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This note was uploaded on 06/10/2011 for the course PHYSICS physics taught by Professor Physics during the Spring '11 term at HKU.
 Spring '11
 physics
 Physics

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