Department of Technology |
ET 332b Ac Machines and Power Systems Homework Answers |
This page contains the answers to the homework problems in the course. Use this page to check your work before you submit it for grading. I you can not find your errors after checking your work against these answers contact the course instructor or teaching assistant for additional help. Full solutions are available after your work is submitted for grading. |
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Copyright © 2016 |
Mailing Address: 1230 Lincoln Drive Room D110 Carbondale, IL 62901 |
Instructor Contact Information |
Phone: 618-453-7839 E-mail: powerguy@siu.edu Skype Name: carl.spezia |
Homework 1 |
|
Problem Number |
Answer and Comments |
1 |
-60+j103.9 |
5 |
50 ∠ 53.1 degrees |
10 |
167-j12 |
13 |
12-j10 |
Homework 2 |
|
Problem Number |
Answer and Comments |
sglephhw.wp5 |
ZT=16.144 ∠ 38.62 ohms IT =14.87 ∠ -38.62 A Vab = 161.61 ∠ -10.78 degrees V |
Homework 3 |
|
Problem Number |
Answer and Comments |
spqhw.wp5 |
1. P=2352.4 W Q=524 Vars 2. Downward triangle with height Q=524, base P=2352.4 S=2410 3. Fp =0.9769 Leading 4. IT =12.05 ∠ 12.55 degrees A |
Homework 4 |
|
Problem Number |
Answer and Comments |
pfhw.wp5 |
a. Qc=7634 Vars C=11.6 uF b. Qc= 90,344 Var C=137.5 uF
|
Homework 5 |
|
Problem Number |
Answer and Comments |
hw3phs1.wp5 |
a. Vp = 120.2 V b. IL=Ip=1.97 A c. ST=708.7 VA d. Fp = 0.574 Leading
|
Homework 6 |
|
Problem Number |
Answer and Comments |
hw3phs3.wp5 |
a. 3984 V b. 30 kVA c. IL = 2.51 A |
hw3phs4.wp5 |
a) Vp = Van = 2402 V b) IL = 485.3 ∠ -81.9 degree A c) ST = 3.49 MVA d) Sp - 1.16 MVA e) Fp = 0.141 Lag |
Homework 7 |
|
Problem Number |
Answer and Comments |
hw3phs2.wp5 |
a) Vp =VLN =VL =13.2 kV b) Iab = 126.43 ∠ -16.7 A, Ibc = 126.43 ∠ -136.7 A Ica = 126.43 ∠ 103.3 Ia = 218.98 ∠ -46.7 A, Ib = 218.98 ∠-166.7 A, Ic = 218.98 ∠ 73.3 A c) 5.007 MVA d) 0.9578 lagging |
pfcorr.wp5 |
a) PT = 231.263 W, QT = 167392 VARs, ST=285487 VA b) QcT = 91379 VARs, Qcp =30,460 VARs, Vp =480 V C=350 uF c) IL = 292.8 A |
Homework 8 |
|
Problem Number |
Answer and Comments |
wyedhw.wp5 |
a) Vab = 480∠ 30 Vbc = 480∠-90 Vca = 480∠150 b) Iab = 17.326∠70 Ibc =17.326∠50 Ica =17.326∠ 190 c) Ia = 30∠100 Ib =30∠-20 Ic =30∠220 |
tqspd.wp5 |
a) 178 N-m b) 98 N-m c) 23.6 HP |
Homework 9 |
|
Problem Number |
Answer and Comments |
Meterhw.wp5 |
Reading 10-1-97=8612 Reading 10-30-97=0254
1) 1642 kWh 2) PTR=120 CTR=160 Power Ratio=19,200 3) 31,526,4 MWh or 31,526,400 kWh 4) 41.47 kW 5) 796,262 kW |
Homework 10 |
|
Problem Number |
Answer and Comments |
2-9/8 |
a) 6600 V b) 45.83∠-46 A c) 1375∠-46 A d) 0.16 ∠-46 ohms e) S = 302,478∠-46 VA P= 210,119 W Q = 217,584 VARs |
Homework 11 |
|
Problem Number |
Answer and Comments |
2-11/8 |
a) 124.8∠ 0 V b) 624 V c) 3.12∠ -32 A d) S = 1947∠ 32 VA P= 1651 W Q = 1032 VARs |
Homework 12 |
|
Problem Number |
Answer and Comments |
2-13/10 |
a) Zeq = 8.97∠ 61.63 ohms b) Zeq = 0.56∠ 61.63 ohms |
Homework 13 |
|
Problem Number |
Answer and Comments |
2-15/10 |
a) Req = 6.21 ohms Xeq = 12.125j Zeq = 13.623∠ 62.9 ohms b) Zin = 531∠ 41.95 ohms c) Ip = 13.89∠-41.41 A d) VT= 7376.4∠0.53 V e) Ioc = 0.427∠-75.5 A f) Zeq = 17280∠76 ohms |
Homework 14 |
|
Problem Number |
Answer and Comments |
1 |
Zpu=0.0043+0.0123j |
2 |
Zpu=0.0043+0.0123j Impedance is the same on both sides of the transformer when rated power and voltages are used to find Z base |
3 |
a) Zactp=7.2+14.4j ohms b) Zacts=0.0018+0.036j ohms |
4 |
a) Ibase = 8.333 A Zbase=14.4 ohms ICpu=0.69∠106.7 b) VLpu=0.958∠16.7 c) IC=5.747∠106.7 A VL=114.94∠16.7 V |
5 |
Zppu=0.04995∠56.6 Vspu=1.0272∠1 Vs=246.53∠1 |
Homework 15 |
|
Problem Number |
Answer and Comments |
2-23/12 |
a) 0.194 ohms b) 0.012 ohms |
2-25/12 |
a) Isc = 45.45 pu Isc = 94,696 A b) Zpu = 0.0347, Z%=3.472% |
Homework 16 |
|
Problem Number |
Answer and Comments |
2-17 |
a) Zeq = 0.0435∠83.5 ohms b) ELS = 495.28∠1.97 V c) %VR=3.184% |
Homework 17 |
|
Problem Number |
Answer and Comments |
2-39 |
a) RfeHS = 11,901 ohms XmHS = 2962 ohms ReqLS = 1.743 ohms XeqLS = 3.233 ohms b) %VR = 2.511% c) 97.37% |
Homework 18 |
|
Problem Number |
Answer and Comments |
3phtxhw.wp5 |
1) Schematic diagram in notes 2) a=45.01 3) S=75 kVA |
Homework 19 |
|
Problem Number |
Answer and Comments |
3-11 |
a) IL = 416.7∠-43.95 A b) IA = 234.1∠-44.14 A IB = 182.6∠-43.71 A |
Homework 20 |
|
Problem Number |
Answer and Comments |
3-1 |
a) IL= 225∠-10 A b) IHS =44.02∠-10 A Itr = 181∠-10 A ILS = 225∠ -10 A |
3-9 |
a=10.074 |
Homework 21 |
|
Problem Number |
Answer and Comments |
Ind-hw1.wp5 |
1800 rpm |
4-4 |
a) s=0.01389 b) Er = 1.444 V c) fr = 0.8334 Hz |
Homework 22 |
|
Problem Number |
Answer and Comments |
4-6 |
a) 450 rpm b) 16 poles c) 3.6 Hz |
Homework 23 |
|
Problem Number |
Answer and Comments |
4-10 |
a) 951 rpm b) 20,250 W c) 150 ft-lb d) 147.7 ft-lb e) 89.1 |
Homework 24 |
|
Problem Number |
Answer and Comments |
4-16 |
a) 90% b) 1667 rpm c) 110.3 ft-lb d) 0.783 lagging |
Homework 25 |
|
Problem Number |
Answer and Comments |
indhw1.wp5 |
a) Zin = 119.23 ∠25.8 ohms I1 = 10.65∠-25.8 A I2 = 9.51∠-5.77 A P = 33621 W TD = 201.1 ft-lb or 272.7 N-m b) 92% c) 0.90 lagging |
Homework 26 |
|
Problem Number |
Answer and Comments |
5-22 |
a) 93.93 A |
Homework 27 |
|
Problem Number |
Answer and Comments |
5-31 |
R1 =0.1915 ohms |
5-32 |
R1 =0.200 ohms |
Homework 28 |
|
Problem Number |
Answer and Comments |
5-35/18 |
P=186.9 kW, Q=262.1 kVAR |
Homework 29 |
|
Problem Number |
Answer and Comments |
8-1 |
150 rpm |
8-2 |
30 Hz |
8-3 |
80 poles |
Homework 30 |
|
Problem Number |
Answer and Comments |
8-5 |
a) 5835 lb-ft b) 161∠32.9 A c) 29.17 degrees Ef=13679∠-29.17 V d) 12,412 lb-ft |
Homework 31 |
|
Problem Number |
Answer and Comments |
8-7 |
a) -27.75 degrees Ef=496∠-27.75 V b) 1332.5 lb-ft c) 666.3 lb-ft |
Homework 32 |
|
Problem Number |
Answer and Comments |
8-17 |
a) -0.784 lagging b) 0.896 lagging
|
Homework 33 |
|
Problem Number |
Answer and Comments |
9-1 |
1500 rpm |
9-5 |
a) 499,858 W or 670 hp b) 859.2 V c) 580.9Ð-34.1 A d) P=499,876 W Q=338,442 VARs e) 0.828 lagging |