# project power electrical

hello the project is mostly done though the professor marked it with yellow as wrong I need someone to correct them and compelete 7th harmonic from q 10 and Q11
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Question 1
Single line diagram
Question 2
According to the impedance table, a transformer with 1500kVA has
ð??
= 6.15, R= 0.83%, X=5.1% and Z= 5.18%.
ð??
Transformers inductance (L)
X= (0.4802/1.5) x5.75% = 0.008832ohms
R= 0.008832/6
X= XL
XL = 2ð?ð??ð?? ð??ð??
0.008832
L= 2â??3.142â??60
L= 2.343 â?? 10â??5 ð??ð»
1.50
MVA 5.18%
Z (ohms)
60
Hz
0.001472
R (ohms)
5.75% Z
0.008832
X (ohms)
â??5
6
X/R
2.343 â?? 10 ð??ð» L (mH)
Question 3
1,500ð??ð??ð´â??1000
1.732â??480
= 1,804.27ð´
1500â??100
ISC= 1.732â??0.480â??5.75% = 31,377.7ð´
Available short circuit current is therefore: 31,778A
1500
60
0.48
5.75%
kVA
Hz
kVZ
Z
1,257A
31,778A
I (rated)
I Sc
Question 4
Combined fundamental frequency power factor
373
kVA = 0.95 = 392.6ð??ð??ð´
522
kVA= 0.80 = 652.5ð??ð??ð´
Total kVA = 1045.1kVA
Total kW = 895kVA
895
PF = 1045.1 = 0.856
ð¾ð?£ð´2 = ð??ð??ð´ð?? 2 ð??ð?? 2
ð??ð??ð´ð?? = â??ð??ð??ð´2 â?? ð??ð?? 2
kVAR =
VFD
Motors
Total
kW
373
522
895
PF
0.95
0.80
0.856
kVA
392.6
652.5
1045.1
Total kVAR required for â?¥0.95PF = 122.5
Round up to standard rating = 125
Estimate about 2/3 for 5th harmonic filter = 100.9
Estimate about 1/3 for 7th harmonic filter =18.88
Question 5
Effective impedance of transformer
FLA ratings = 1257A
VFD base
472.2
Transformer Base
1500
Transformer impedance
5.75
Zeff = %Zxfmr x VFDbase/XFMRbase
= 5.75% * 472.2/1500
Effective Z=1.81%
kVA
kVA
%
kVAR
122.5
391.5
514
Arms
472.2
784.8
1257
Question 6
Harmonic current distortion
Reactor Size = 5% TOTAL %Z effective XFMR Line reactor = _____%Z
I=472.2A
I(1)
I(5)
I(7)
I(11)
I(13)
(17)
I(19)
Total
%
100%
32.02% 12.38% 5.76% 3.90% 2.22% 1.74% 158.02%
Arms
472.2
151.4
58.45
27.20
18.42
10.39
8.22
500.56
Question 7
% of total =
392.6
652.5
â?? 100% ð??ð??ð??
â?? 100% = 37.56% ð??ð??ð?? 62.44%
1045.1
1045.1
VFD
Motor
Total
kVA
392.6
652.5
1045.1
% of Total
37.56%
62.44%
100%
%THD-i
35.17%
0%
35.17%
Question 8
Voltage Distortion
IEEE-519,2014 Table 1
Bus Voltage at Individual Voltage Distortion (%)
PCC
5.0
ð?? â?¤ 1000ð??
Current Distortion- 120V thru 69kV Systems:
IEEE-519,2014 Table 2
I sec/I Load < 11 11 ð?? 17 ð?? â?¤ ð??â??, 17 â?¤ ð??â??, 23 <20 4.0 2.0 1.5 20<50 7.0 3.5 2.5 50<100 10.0 4.5 4.0 100<1000 12.0 5.5 5.0 1000 15.0 7.0 6.0 ISC: 31,377.7A Demand Load (IL): 1257 Arms ISC /IL = 24.96 % THD-I contribution 13.209% 0% 100% 0% 13.2% = 13.2% Total Voltage Distortion THD (%) 8.0 23 ð?? â?¤ ð??â??, 35 0.6 1.0 1.5 2.0 2.5 35 ð?? â?¤ ð??â?? 0.3 0.5 0.7 1.0 1.4 TDD 5.0% 8.0% 12.0% 15.0% 20.0% IEE limit for %TDD = 8.0% EEE limit for THD-v = 8% Question 9 Capacitor rating: 125 ð??ð? = 1/2Ï?fC C= ð??ð??ð´ð??/2ð?ð??ð??ð??ð?? 2 =1.43912 â?? 10â??6 125 kVAR 1.43912*10-6 uF (wye) Xc= 1,843.2 Ohms Ic= 1257 Arms Base on kVAR Inductor Rating: 2.343 â?? 10â??5 ð??ð» Required: 1.013 â?? 10â??5 Harmonic Frequency -Xc (ohms) XL (ohms) -Xc 1 60 -1,843.2 XFMR (XL) XFMR XL (XC) VFD Harmonic Current from 6 Current at PCC 2.81 â?? 10â??3 -1843.9 472.2 5 300 -368.64 3.82 â?? 10^ 0.01909 â??3 -1843.196 -368.62 1257 Question 10 a. Capacitor rating: 87.5 ð??ð? = 1/2Ï?fC = 1/ 2Ï? C= ð??ð??ð´ð??/2ð?ð??ð??ð??ð?? 2 =1.43912 â?? 10â??6 b. Capacitor rating: 41.67 7 420 -263.31 11 550 -201.08 0.02672 0.0350 -263.28 13 780 -141.78 0.0496 17 1020 108.42 0.0649 19 1140 -97.01 0.0726 -141.73 201.045 -96.94 108.36 0.014058 0.01968 0.0257 0.0366 0.0477 0.0534 368.61 -263.26 -201.02 141.6934 108.31 -96.88 151.4 58.45 27.20 18.42 10.39 8.22 ð??ð? = 2Ï?fC C= ð??ð??ð´ð??/2ð?ð??ð??ð??ð?? =1.43912 â?? 10â??6 2 Question 11 I(1) I(5) current 1264A 82.1 Current I(1) I(5) harmonic Nearest 2 45.61 52%THDi Irms/If1.31 1257 I(7) I(11) I(13) (17) I(19) Total 41.166 I(7) 16.38 I(11) 10.314 I(13) 6.444 (17) 5.076 I(19) 500.56 Total 22.87 9.10 5.73 3.58 2.82 ... Purchase answer to see full attachment

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