Exam 1 fa - Name E Fell ECE 393 Exam#1 Tuesday October 8...

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Unformatted text preview: Name: E Fell ECE 393 Exam #1 Tuesday, October 8, 2009 Closed book, closed notes One note sheet allowed Several potentially useful constants are given on the last page l. /15 2. /l6 3. /20 4. /20 5. /l4 6. /15 Total / 100 1. (15 points total) Use the Load Duration Curve shown below to answer the following questions. 9’ .5"? A 9mm on LDC-‘- ags W’s/9r wou- loacl‘m greater 6000 Mom or eq/Uai t0 “ti/1W“ Vague" 5500 A )3, B .C i g t/ "0 ................................................................ -. - *0 , a 0 2000 ; 0 < ' \jxeow 5 100 O O 0 1000 HRS 6000 8760 Find the number of hours per year that the load is less than 1000 MW. Find the number of hours per year that the load is between 2000 and 5500 MW. If nuclear plants with a combined rated output of 2000 MW are base loaded, what is their capacity factor? What is the average power from the plants in part 0.? o hrs/yr _ \oobwiiws/tif ~Load§s Qwemmn $500M”: 1.; 5000 m} { teams Min/ear} (0000 NS lyre mold is gromwaynacn 20wa ’ mob 3‘ 5500 MW C? : gghnAghegj WWW t inoompoagiwégj5914300” moo + J5: logo energy @ mlecl 00090015 "765on x 87600 : 03,9000” W“ - Mm l’lBJ. xlonh “Fm” ABQUS .powovr av: E0936? (1?: \ uaf {00% a go» 0‘6 power: {flieclvflwfl X 5": 01802.04? MW W (or, actual met/37 / 8 7&3“) e ,0, M ff: MS” (“A 351% WW 2. (16 points total) \0 (3% fig )1: 60 Hz, single phase motor with a power factor of 0.6 lagging draws 4.2 kW of power at LLOW Qgpfiom a 10 kVA transformer. Suppose a second motor is needed and it too draws LN ?. 4.2 kW with the same power factor. With both motors on-line, the transformer will be overloaded. a. What is the minimum power factor is needed to continue to use the 10 kVA transformer (with out it being overloaded)? b. How much kvar of capacitance is needed to provide this level of power factor correction? c. Now assumed it is desired to correct the two motor load to a power factor of 0.95 leading. How much total capacitance in kvar is needed? d. If the orginal motors (that is without any power factor correction) are supplied through a transmission line with an impedance of l + 2j Q, what are the real power losses in the transmission line? 0 Com; mm was Wendi/e) W/Eflfgfl (Walt NQWO‘Y‘+“> ti . ' ‘ werieeds. =QWU6—Qdes\\-—€6 g» t W 33 gnaw/w} “30‘ ,3 it 3 ' A; W G Q \oVr QM ._ 5.1125KV24V‘ (MW 0% Cf Wed. ()7 8H KN P: 6H Kw @1W 5 (\nggrcht‘br ts “Q J26 WAN a< > a I © 6: figoflsyzawq 9=<amaw tome : Q :7Q= 2.7(06‘5 war 3 mt“ (“Q3 P K/Gldesireglz 23 (pl \LVAT‘ (Ngafiw Q» €03 Ame” alpqddanm muted g5 Hill/AIM ‘r inlet KW? '2 8.959%??? it ® W} sail" * H' 23L (aw'llifi ‘ Aesxm a --«~4-~, [ m w we; aa - .. iiifiwis‘is \(Ll‘O-OEEELl ) (3' UL ltl‘ .0351.” j “ “lg-6““ 9 met: a 2.93! W - PA: = l2 bazw lay 3 0 \0 . , has .7. i W} w M5 4mg; more \osses to 3. (20 points total) H 0 (354% Suppose a 6-inch diameter penstock delivers 900 gpm (gallons per minute) though an elevation change of 150 feet. The pressure in the pipe is 40 psi when it reaches the power house. a. What fraction of the available head is lost in the pipe? b. What amount of power (in kW) is available for the turbine? c. If the turbine/ generator has a 60% efficiency and a 50% capacity factor, how much yearly energy (in kWh) does this system deliver? (1. Assuming that the head loss can be approximated as sz (where Q is the flow rate), what is the optimal flow rate in gpm? @- ’Vt‘lheoml > 2L2 @HLtH yH : 5 _ 31:; .Si 2 h We? (3 N W (gate N:Q-M&WQ M _,———~ ' E Co" 3~ (9 vanity» gewéfi A “t <T’7131lr63 :xso—Lblfi5wa2aw23 1 meg—3&5 .- HM: @7585?» “’6‘ We : \orzxzsfaua‘; Lul4‘18-l3lr ZAngZ‘FEIISL I“ /’—-—-- ® V'emehew ‘3: , r1 (game %b)f"t? 69 P: one (9&3 z 5%) iJ-w ‘0? V 2 8v , 3 lgflLtBKW ‘ Wfifiwfig 612%?6 @ Eta” «We : ‘LSQG KW XQB'IbOh/erOFb) '3 LlilBCHQMN KWh/jr Wk prevail/ho) KQQ QWWM flow tale: we Mow oatoiswtwn HLz 43 HC, : 59% ““ Claw r y or M: 5 H UM" 0W a” . 3 ‘7 EMWL‘} [El—iii?th St» was l< flwlfi?u 0‘ mam L (2:) Mam LN 4 (15 points total). Assume that Z=ZG : 29. IA (9 _.+ «swig-VMWM * ~. 1 two; tSD‘ s55“ 507” \ N ‘01 H5 A V > ‘W @oiséatrw’s 0A 33 gm) ZG Ni. 9 (Av-{cg s qflui'l ” L “3‘ 3393‘s; «Egrflwscasfiof; A7 b/Uégoic- i103) sCGfiQWH: ‘3tg’O =(0 “t5 {s%?$:o;zooj “‘3 Io = £59 {2' L03 {soc 9:; a .,_, ,_ ...c,.__,._.r,w,,,\ 2’? :: ~ . Ca“ (Q3: 71 O 9513(in 9, e SM; a {sec , Lé =5 ; \ “mg Sx'tw": 4.213") ‘ ‘ (X3 5»! A v 9 , QM: am) t 3, by, Cosfls’ofé; t2!) Wwifi 2x soc} ~; cosmwtj extra, 2 (,05460‘é3“ rm defiant-63v mfimfiwéj a. Suppose the currents in the figure are given by a? «0C9: z'A = 200J§ cos(a)t+0°) o. l \3 Z A at WW“ W, ‘ i3 =200J§ cos(a)t—120°) lMMa“ s; suntsi’ffi‘" W 9“ z‘C = 200J§ cos(a)t +120°) (i) Find the RMS current of each phase QSMLA (ii) Find the real power consumed by ZG O N (iii) Justify your answer to (ii) and explain whyhit is an important result ' o co Above, %s m “imam 1 o W wwtfw‘ d . ‘W‘I . l} ‘ Vfg‘f‘eswtu 3:36 5),“?me GUN-— mm‘fiwbl 99¢ Luféwmfl Na 604‘ o Paperworse“ elbows a v’ belowgmamm witc- b. Now suppose the currents are glven by “A?” J “0 lofi‘sfi m % 4:" 5 = zoos/E cos(a)t + 0°) + 50¢? cos 3(a)t + 0°) + 50J§ cos 5(601‘ + 0°) + SOs/E cos 9((02‘ + 0°) i3 = 200$ cos(a)t — 120°) + 50J§ cos 3(wt —120°) + 50J§ cos 5(a): —120°) + 50J§ cos 9(a2t —120°) iC = 200J§ cos(a)t+120°)+ 50\/§ cos 3m .+ 120°) + 505 cos 5(a)t +120°) + 50¢? cos 9(wt +120°) (1) Find the RMS current of each phase 1 W gig/A (ii) Find the Total Harmonic Distortion (THD)"5?“each current 0&3 (iii) Find ID in the time domain and its RMS value kg): 635—2 Lb) @WQwéié’wg (iv) Find the real power consumed by ZG 3: MS 2 2. PA 317°; 5 (14 points total) (True/false) Two points each. Circle T if statement is true, F if statement is False. As of September 2009 natural gas prices in the US are at an all time high. The world’s largest operational hydro plant is Grand Inga on the Congo River in the Democratic Republic of the Congo. Fossil fuels currently supply at least 80% of the energy used in the US. Flourescent light bulbs are a source of power system harmonics Combined cycle power plants usually have efficiencies of no more than 40%. Modern computer technology now allows most power to be “routed” through the grid from a seller to a buyer. As a direct consequence of Kirchoff 5 current law the locational marginal price (LMP) at a bus in the power grid can never be less than zero since this would require negative resistive losses. 6 (15 points total) (Short answer, five points each) 1. As discussed in class, explain in two or three sentances howthe area control error (ACE) and automatic generation control (AGC) are used to operate the power grid. Amt coan mo“ 13 we the between We wedded power «Plow mm one arm» to anal/hag and 444:: aCf-UM 2. In two or three sentences explain how concentrated solar works. Also as discussed in class explain what role (if any) water and/or carbon dioxide play in cOncentrated solar power. Gomerfirottea &0\ar POW use; mirmrsfofimb 1], met! Wop a Peéer’VOW‘ Oxc WOYK‘YB‘HUFA C‘h/‘ofwliltj W01 VJ.‘7‘4’)£ W4“ “F‘Wd mafia Iner an m.» aw“: ldbfi I (argyle or 9mm durbw, Typpwuygm Calm wagging emswam Cool waterma‘i EC warm Mner Macy van chl he a prome in 444$ elem. Warm Nata/r“ Can also be used +9 ewe] waste hear}. unless coma! M whvenhmgu generalth 3. Explain the advantages and disadvantages of distributed electric generation with respect to traditional central station generation. Fax/arrays: F€wer Wmmmrbrt W125, mgr enA‘ user, meow,qu can general-la 0W, PWJ [0055mm To 9&1! 1 WWW Wick“ +0 h’3)usgwasi€ him J 3130:4660? Com do“? as reannes mine (mpg/5} DBad-Whlafiéaf NQQA rfo M yolk, 3+ me W 943 (Lewd (9&4???ka Wed to control 50mm: méngwF—grttmoh Small anxth 010W Useful Constants H One cubic foot 7.4805 gallons 0.02832 m3 One foot per second 0.6818 mph 0.3048 m/s Water density 1000 kg/m3 1 kW = 737.56 ft—lb/s 1000 N—m/s Gravity = 32.2 ft/secz 9.81 m/s,2 One BTU = 777.9 ft—lb 1055 J 1 PSI = 2.307 feet of water. One Meter = 39.37 inch K = °C+273.15 You may assume the atomic weight of carbon is 12, and oxygen is 16. ...
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Exam 1 fa - Name E Fell ECE 393 Exam#1 Tuesday October 8...

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