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Descripción: Solution file for youtube video "Server Error in '/' Application Solution"
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ECE330 – Spring 2014 2.1 v(t ) = 100 cos(377t + 10 0 ) V => V
2
(
100
∠55
0
0
V
A
100 1
)
= VI cos θ v − θ i =
Power factor: PF = cos(θ v
∠10
2
1
i (t ) = cos(377t + 55 0 ) A => I =
Average power: P
=
2
cos(10 0
2
) cos(− 45
− θ i =
0
− 55
0
) = 50 cos(− 45 ) = 35.36 W 0
) = 0.7071 leading
2.2 a) v(t ) = 100 cos(377t + 15 0 ) V => V
=
100
∠15
2
0
V 10
i (t ) = −10 sin (377t + 45 0 ) = −10 cos(377t − 45 0 ) = 10 cos(377t + 135 0 ) A => I =
Average real power: P
=
100 10 2
2
cos(15 0
b) v(t ) = 100 cos(377t − 75 0 ) V => V
=
− 135
100
0
=
100 10 2
2
cos(− 75 0
A
0
0
V
i (t ) = −10 sin (377t + 15 0 ) = −10 cos (377t − 75 0 ) = 10 cos(377t + 105 0 ) A => I =
Average real power: P
0
) = 500 cos(− 120 ) = −250 W
∠ − 75
2
2
∠135
− 105
0
10 2
∠105
0
A
) = 500 cos(− 180 ) = −500 W 0
c) v(t ) = 100 2 sin (377t − 30 0 ) = 100 2 cos(− 120 0 ) V => V
= 100∠ − 120
0
V
i (t ) = 10 2 cos(377t − 60 0 ) A => I = 10∠ − 60 A 0
Average real power: P
(
0
) = 1000 cos(− 60 ) = 500 W + 60 ) = cos(− 60 ) = 0.5 leading
= 100 × 10 cos − 120 + 60
Power factor: PF = cos(θ v
) cos(− 120 0
− θ i =
0
0
0
0
d) v(t ) = 100 sin (377t + 30 0 ) = 100 cos(377t − 60 0 ) V => V i (t ) = 10 sin (377t + 120 0 ) = 10 cos(377t + 30 0 ) A => I =
2.8 Total current: I = I 1 + I 2 = 50 − j 40 + 50 + j 60 = 100 + j 20 = 101.98∠11.310 A Complex power: S = V ⋅ I *
I1
I2
2 (100)∠45 0 × 101.98∠ − 11.310
=
= 14422∠33.69
Power factor: PF = cos(45 0
0
= 12000 + 0
− 11.31
j 8000 VA
) = 0.832 lagging
2.9 Load 1: S 1 = 5(0.8 + j sin (cos −1 (0.8))) = 4 + j 3 kVA Load 2: P −1 S 2 = P2 + j 2 sin (cos (PF 2 )) = 3 + j1.453 kVA PF 2
Load 1
Load 2
Total complex power: S = S 1 + S 2 = 4 + j 3 + 3 + j1.453 = 7 + j 4.453 kVA 0
S = 8.296∠32.46 kVA Total current: I
*
=
S
=
8296∠32.46
0
=
0
0
36.07∠32.46 A
V 230∠0 Hence, I = 36.07∠ − 32.46 0 A Power factor: PF = cos(0 − (− 32.46 0 )) = 0.844 lagging
2.10 Load 1: S 1
=
P1
+
j
P1 PF 1
sin (cos
−1
(PF 1 )) = 8 + j 6 kVA
Load 2: S 2 = 20(0.6 − j sin (cos −1 (0.6))) = 12 − j16 kVA Load 3: Z 3 = 2.5 + j 5 = 5.59∠63.430 Ω
is Feeder
I
L1
L2
L3
V * S 3 = V ⋅ I 3 = V ⋅ Z 3 Or S 3
=
*
=
V
2
* 3
=
Z
250 2 5.59∠ − 63.435
0
= 11180∠63.435
0
=
5000 + j10000 VA
5 + j10 kVA
Total complex power: 0 S = S 1 + S 2 + S 3 = 8 + j 6 + 12 − j16 + 5 + j10 = 25 + j 0 = 25∠0 kVA Total current for the load combination: S 25000∠0 0 * 0 I = = = 100∠0 A 0 V 250∠0 Therefore, 0 I = 100∠0 A Source current: i s (t ) =
2 (100) cos(2π × 60t ) =
2 (100) cos(377t ) A (is is the same as I)
Source voltage: 0 V s = V + (0.1 + j1.0) I = 250 + 10 + j100 = 278.57∠21.04 V v s (t ) =
2 (278.57) cos(377t + 21.04 0 ) V
2.11 Load 1: S 1 = 250(0.5 + j sin (cos −1 (0.5))) = 125 + j 216.5 kVA Load 2: P −1 S 2 = P2 − j 2 sin (cos (PF 2 )) = 180 − j135 kVA PF 2
I
L1
L2
L3
Load 3: S 3 = 283 + j100 kVA Total complex power: 0 S = S 1 + S 2 + S 3 = 125 + j 216.5 + 180 − j135 + 283 + j100 = 588 + j181.5 = 615.37∠17.154 kVA Overall power factor: PF = cos(17.154 0 ) = 0.955 lagging (since positive Q) New reactive power: P 588 −1 − Q' = − PF ( ) ( ) = − sin cos ' sin (cos 1 (0.8)) = −441 kVAR ' 0 .8 PF Additional kVAR from capacitor: Qcap = Q'−Q = −441 − 181.5 = −622.5 kVAR Special problem #1 Assuming load current phasor is I = 75∠0 0 A. The voltage drop on the wire impedance is: 0 V f = 2 × (0.05 + j 0.05) I = 7.5 + j 7.5 = 10.6∠45 V a) When the load power factor is unity, the load voltage can be expressed as V L = 120∠0 0 = 120 + j 0 V The required source voltage is then 0 V s = V f + V L = 7.5 + j 7.5 + 120 + j 0 = 127.5 + j 7.5 = 127.7∠3.366 V with the RMS value of 127.7 V. b) When the load power factor is 0.707 lagging, the load voltage can be expressed as
with the RMS value of 130.6 V. c) When the load power factor is 0 leading, the load voltage can be expressed as V L = 120∠ − 90 0 = 0 − j120 V Therefore, 0 V s = V f + V L = 7.5 + j 7.5 + 0 − j120 = 7.5 − j112.5 = 112.75∠ − 86.186 V with the RMS value of 112.75 V. Special problem #2 Load 1: S 1 = 6(0.8 + j sin (cos −1 (0.8))) = 4.8 + j 3.6 kVA Load 2: P 1 S 2 = P2 + j 2 sin (cos − (PF 2 )) = 4 + j1.937 kVA PF 2 Load 3: S 3 = 240 × 13 + j 0 = 3.12 + j 0 kVA
I
L1
L2
L3
a) Total complex power: 0 S = S 1 + S 2 + S 3 = 4.8 + j 3.6 + 4 + j1.937 + 3.12 + j 0 = 11.92 + j 5.537 = 13.143∠24.916 kVA b) Source current magnitude: S 13143 I = = = 54.76 A V 240 c) New reactive power: P 11.92 −1 − Q' = ( PF ) = ( ) sin cos ' sin (cos 1 (0.95)) = 3.918 kVAR ' 0.95 PF Additional kVAR from capacitor: Qcap = Q'−Q = 3.918 − 5.537 = −1.619 kVAR
