Assignment_3 Solution

Assignment_3 Solution - ELECTRICAL ENGINEERJNG DEPARTMENT-...

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Unformatted text preview: ELECTRICAL ENGINEERJNG DEPARTMENT- CALIFORNIA POLYTECHNIC STATE UNIVERSITY BB 255 Assignment #3 Problem 1: A 3 KVA, 480:240-V, 60-Hz single-phase ideal transformer is supplying .a load impedance of 15 + j20 Q at 240 V. Determine the load current, source current and the impedance seen by the sourCe. Problem 2: ‘ A 2200/220—V, 60-Hz, single-phase ideal transformer is supplying a load through a line impedance of 9 + j12 Q. The load is a heating load with a resistance of 180 Q. The line is in series with the load and is connected to the 220 V side. Determine the power supplied to the load and the power supplied by the source. What is the efficiency of the system? Problem3: A 115/5 V, 60 Hz, single-phase ideal transformer is delivering a power of 15 W to a pure resistive load at the 5 V side. Determine the resistance of the load, secondary current, the primary current, and the power supplied by the source. Problem4: ‘ An ideal transformer is delivering power to a 6 + j8 9 load impedance. The power supplied to the load is measured to be 150 W. Determine the primary and secondary voltages, primary and secondary currents if the primary current is half the secondary current. Problem 5: - - A 140-KVA, ll,000:2,200—V, 60-Hz single-phase transformer has the following circuit parameters: - Primary (HVS) Secondary (LVS) R1=6.IQ R2=0.3§2 X1=31.29 ' X2=1.2§2 Xm=124,000 9 R6 = 57,300 g The transformer is supplying at the 2200—V the rated load at a power factor of 0.7 lagging. Using the exact equivalent circuit, determine: ‘ (a) The voltage regulation of the transformer, ' (b) The efficiency of the transformer. Problem A single-phase, SO-KVA, 400/2000-V transformer has the following parameters: R1= 002 Q R2 = 0.5 Q X1=0.06Q X2=l.5§2 Xm = 166.7 Q Xm is a Low Voltage Side parameter. The transformer supplies a load of 40 KVA at 2000 V and 0.8 pf lagging. Calculate the efficiency and the voltage regulation of the transformer for the given load. Use the approximate circuit with the excitation circuit across the supply. Problem 7: A 25-KVA, 4200/480-V, (SO-Hz single—phase transformer has the following circuit parameters: ' - Primary (HVS) Secondary (LVS) R1 = 6 £2 ' R2 = 0. 19) X1 = 459 X2 = 0.69 RC = 44 K9 Xm=14 KS2 The transformer is supplying at 480-V full load at a power factor of 0.8 leading. Determine: ' . . (c) The voltage regulation of the transformer, (d) The efficiency of the transformer. (e) The all-day efficiency for the following load cycle: 10 hours at full load and 0.8 power factor leading 14 hours at no-load. Problem 8: A single-phase, 60-KVA, 2200/460-V transformer has the following parameters: R1=2£2 R2=O.OSQ X1=8£2 X2=0.32§2 Xm = 4.2 KQ RC = 6 Kg ’ Km and Re are High Voltage Side parameters. The transformer Supplies a rated load at 460 V and unity pf. Calculate the efficiency and the voltage regulation of the transformer for the given load. Use the exact equivalent circuit. - ' Problem 9: _ A 7.16-KV generator supplies power to a load through a transmission line. The load impedance is Z, d 22894363212; and the transmission line impedance is 00 Z a 34.64160o Q. line 1. If the generator is directly connected to the load, what is the ratio of the load voltage to the generator voltage? What are the transmission losses of the system? 2. If a 1:10 step-up transformer is placed at the generator side of the transmission line and a 10:1 step-down transformer is place at the load side of the transmission line, what is the ratio of the load voltage to the generator voltage? What are the transmission losses of the system? (Assume ideal transformers.) No. 937 8115 A Esmenmsw Engineer‘s Computalion Pad ’ . Engineer's Computation Pad . No. 937 811E "E a m N0. 937 811 E I Eflginear‘s Computation Pad $5MEDTLER° Engineer‘s Computation Pad No. 937 811E 435542131153" , 2.5 e V; :JZ:°({¢-J€’) ail—55”“ 7/? ;/&a15_~§i3 l/ I m H...“- .. - " ‘ ‘ 1' .i Q55!” mmmt 4%— 3 I (j - = 140K149 (21:61.13, 22 70.3.2. . mam/€202") 1/ x, :31- 2x2 X2 :Azfi- «50 2’1;- &: 573mg. ‘ l gé L.an ; Ill-O [(qu 9 2200 V g . _' ' 0 7 PF (“66% E a.) Kw W2. . ,I E). : )Lmofiio +/2.73é:fl5;0(7,5-+Jr30 w::— }‘/33?r;j {(161.2 V i -_-.. 1!,34L3Z/‘i/f y. I - i i l .355: a 0 mg '— u ., . 0 10 s. Izwzfia/ 1 pl Hg NM [4’ 53% 573570 “I ,— gmw F823? 74‘ - lZQoaoLE _ F.»/ I: :nyI—0 rel—vim a ‘ c? 1:. IZ7‘3L—"L‘g'57 + aeg‘PE’ g] +0»O?/5 ’3?" 7 I 435mm : 3'71}; 1f ?V‘9?04 y— p-Zg? «bio-CD94; :— ?,;agj_.j?./§2/ b I2-934 (—1523 H! II t .- lbw/aw 1.}2,?34Z:_€Z'239(6.!+}3L2% {- .-.- ulz'ifi‘v'fi‘t: +j [971? + 3%. .U‘ 223.2 I i g :- Ir,5‘5i.é I+J£f234 7.:- 19637413: 1/ I E _ “6874—41000 Mm H 090 .1 5.27%- 33 . P +~ :15 b '7' L cm fig: ) Q L aw ; 2w 2 E: 8m .2: M25}, 3' M“ 341'; a ' Ra. 573w E ‘ a 2.9.45 KW {13 Pm a (1;) 2,” + (2-9-12, g'észfla) (7.7) +(12.734)‘?(41) LQZE'j-J-Oi —Q-935 KW 80} 7/40K0.7— s“ ?3 KW- Pa = 93 +2245+2235 .7: [02-44; :22 (Mara :- ! J g l i i 1 gay/fa? x“ 444.732.. 1.4.9045- 402’1’5 Zml/ 0”? H; , ' EngineeI's Computafion Pad ;. i g. No.93? a11E ‘5: ’ 00 ‘2- E X15715<§w€>~7006J24 (E V; -= 400 90V a 3 1: ' i z (40):!» {.3187 _____ MO («zip/$7]? #00 22 stanza—0:0? :aaéL—SL. tfiuag JI— X2: "'5 VI -'=‘ 4001—2 .1- 10§V3é87(0,04f+j&./2j { twp-#flfiz get/0.sz W .3 watfé—téwfldo :24; f h; I l Engineer‘s Computation Pad NO.‘ 93? 811E 435MEUnLER“ ELI 1- 1'- CD h. C9 03 d ‘2‘ 9 4m :0 = E5“, =* 6- Egg”, .. 5,373 Mr .707KH/H' s .. P11 " I n a VII! -_:_ 2547,2239: 277 _ , WOO“? flé3.‘75m ' ‘ [+1.99 ‘ 0/0 ' (Z M00 =3 [22 _ 6 3.75 r‘s Computation Pad N0. 937 511E Enginee asmenmsfl“ _ ‘_ a lime ... 34.64 ('90 .51— I. Use V‘s/751.52. decJer .... Educ Qt. :G4 ,2: L—‘J 1‘- Z" +Zunn a VL a :71!!!“ 2‘3" by? Lx’ 2€7§£57+ :51 1449-0 1, ‘ :asa' '27-léér affiz'fl M .. 6%”;(45. J: 75’ Z )vé) 7.16 fmgl '-"" g Lit-(+5.5 “2.345 Ixzaao 287L195? U a KW l E I __ 2. L; E PI,qu :- -.'.."' l E :Hgfl Kw ' I iflm V) _-: #696 = r E b I19-7+€6 , J m. _ _ . E I \ 2 ' A ' O . Ci? 2 ZLHHB '3 7;) Ztma =' 9.54-{4' {é J1" AG E if, .._. MAO: N I 5.3 45415+28fifiéfl° 2% 173+) 173-9: = 24.7% 4474” _ E vm; = in. a. 28315557 =- 2/5411” tjx/ \/ “a ' - ' ° . xlw = 'ZZ'E‘fz/ea { yd, 7.5! I i I W z fimfflx‘ffloq L? {E 6’5 .. I l =- N73 w w. l ...
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This homework help was uploaded on 04/07/2008 for the course EE 255 taught by Professor Shaban during the Fall '05 term at Cal Poly.

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Assignment_3 Solution - ELECTRICAL ENGINEERJNG DEPARTMENT-...

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