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EEE463_Homework5_Solution_F07

EEE463_Homework5_Solution_F07 - :umw.YEAEASE" 53001an...

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Unformatted text preview: :umw .YEAEASE" 53001an 13782 42 381 42382 10 MfifflNaIianal @Brand I ! l A l—GWe power plant with a heat rate of 8800 Btu/kWh operates at full power for a period of 1 day. Find: (a) the mass of coal, in tons per day, and (b) the volume of oil, in barrels per day, required for operation at full power. Assume that the energy content (heating value) of the coal is 27,800 kJ/kg, and that of the oil is 141,000 kJ/gal. There are 42 gallons of oil per barrel. 20% = 5:: x9 (me): (o kWQfiiSii“ (227 900 kT/kj‘D 34/2 flaw,» . M [+0 =<8. OléK log ija;/ :2 if; (W?) = 8/ 8 20 flnJ/‘Iay flag :68 axloc’ gfuflar%_______4____ // /<J‘ secY 2w,» @) V51“ f‘f—l coo kT/fia 34/2 Efl/Arl/ gig) 9’47 M) =6 562x10 Wax/{eyX 2237:: zg 37/ éBO éorralsf/Aa/l flack ”is of o Jud: A Fuel Aas 74¢ “pol/0w} '75 elemernlq/ aflolygas : W40 C <52. 1 S o, 2 H1 17, 7 :1: 714,5 QC] ,2; burngg! in .2079 excess aim] CZertLermffle 71‘Ae perCErfiL: $7 vo/ume op geek coms+f+uen+ in +176. p/ue 30$- \ 2.5+! + + 2;”) (01+ 3.76 Ml) ~:> C2; #2119] 321+: + (Zi+ Z; (col)+ Z—EJ— Afio+ 23130.1) + (z; + 23—” + 24‘”) 3.76 /\/',1 .8 / Carbon: 2": (W72 2 00684 Mo/e A O./'77; M/clroaarfi 24:31.] : J % (no/c : 0.1 77 VHS/C»: O.CDOQ§: Su/‘pu/ : 24“,; 2 3.2 g/me/e : 0.000063 man/a i i i E E E E E % E E § E E i i Am 1 2; + ’27?" + 232 :- 0.06844» 02:37 +©.voo©63 = 0.1127me/e: E i § § E E i Z i E E § i E E E E <9me (Oz) -‘ (Am) [0. a): (0.1427>(o-2) = 0.02254 nae/Q Nawaen (Iva) i(A;n)(a.7é>(1,2) :(o.11;27)(3.7é)(1,.23 = 0.5085 rho/e. COQ : 2' = 9.0584” male 2‘ I__ 0.127 _ #010 C :4: — 9., ~ 0.0885 mole 80,1 : 21,31: 0.000063 nae/fir 4% er; Slvflce one (no/e o‘pan); ideal gas occu/ofcs +146 some. vo/un’ia/ 'l—Aa nae/P, Qac-fil‘on is Cgur‘ualen‘lé *9 194a Va/ume. predict). 72+?) Moles = cams-4 + asoaSJr aoee4~+ 0.0885 + 0.000063 3 O. é88 ”Io/as 0.02254— 0; -‘ ”672%? = 0.037 .: 0.5065 2/ A/ = mama = 0.739 73.9% /t/ /2.7% o; ; 1 I 51 _ 0.0684- __ 4 a f C01. 762%? - 0.6999 5 C7,? /0 CC) . BBS ' Hgo= Wiiae = 0.1201 151.9% #40; 0.000063 5 50;: Wage = 090001 ; 0.01% sag/I 100.41 0/6 ”WWWWWW 515544-53 g \ A SVLQQM F/an?“ produces 600 MW 910 e/ecnl-r‘fca/ pow/en (xv/31% or; overs?” 7‘7/flerma/ amok/b7 0p 35 79. 7745 acme/m4 a: fiua/ Aas Q Ewen Ago-Hana Vo/ofi 910' RS; 000 kT/Ag ancf a sat/QM confinvL do 11.3 %. We 5+4c/< 34$ scruljéer WGVES 957a 910 +Aa SO: pmazcz o/ and disc/mzraes a s/uflry 0-10 69 30+ W/u‘cLz (‘5 45°43 w/o CQS‘Q ‘ ‘ W/va vo/ume $10 3 u/‘ry is produced Pen year I-p‘b/ve. In/grm‘ oFena¥7es 0+0 7S 0/0 Ca/oacfi? pacifier 7 Ba. ._ $39.11“: IO/anJ— TAern/Zq/ fQG7L'1'n; 7' [04+] : flfig, w O. 35 = 1429/ MW i C6: i ; E 3 E i /4“;2°7x103kW \(3 l x Calif f‘efi‘U/remernL: 07L FUN /D<:We,v-:’ Sig/GOO /<T//< :W'%c> E == 571 Ay/sfi = 205,735 kg/Ap E E i E E é % E CoaI actual/7 usec/ : (205/ 700 %%.7S><%:fi;yi%§éy> \ 2.3 11%»3 \ kg .\ Sulfur LurneJ =((OO eyed/j LBSXIOQ V7} 2 3.1.ix [06 $2 MCQSCZF : 40 + 3 2 + (flag) : 13$ kg/mo/é é i3é é C2523? ———___._.§_.__~._ M453 m0 030.; (3.1,ixl?é ‘fl>(32 k? s >(o-9s> iQQK /O k? CQSO+/Yn H II SCQSQ+ = '2’ gé gy/GM 3 Ca/C/Yum Su/pa7le’ _ m__ .LQQK/Oe’k f' [000 /:-:z 3 VOI, CQSO¢ - 5 W(—H§ (loom ; 42/570 mg/yn Mm op Ago =Q§§££Mmewoé 1:; Cam} 4,”) = 154- x 105 I? Ago/W val. Alp: W?§O:ffio "(%:i§)(ré~‘”az)3 = 154/000 m3/y, 72M vo/we = vol, Cosofvr VGA #10 = (4:2 570%) +(Isfiooo 7; W“— 14-1 A 27 mi/h wind at 14.65 psia and 70°F enters a turbine wheel that has a IOOO-ft2 cross-sectional area. 5 Calculate (a) the power of the incoming wind, (b) the theoretical maximum attainable turbine power, (c) L a reasonably attainable turbine power, all in horsepower and kilowatts, (d) the torque, and (e) the axial thrust, in pound force, if the turbine wheel rotates at 30 r/min. r, E 5! a: E § § ~E( / l ,/@.) Using Eq, [/4f~2‘f> g“ x \ at: +33 (I )u a «(a 2} >1 81‘ A‘ S M “l ,\l 00 N E ”Q m i 1‘ Elm k/ (2;) ASSUME/l? n: Lil-0% (as fr) Fara/Ac) examp/e /Z+‘I_> (Muff-la E7” (/7530) P = n PM = (0.4+)é756kw3 = ml (Lav-w 2%) (631) {15,715 E7, 54w) 4w; ' __ .13.” = 5:2 v .. ::j . ,. _i- _ amax‘a fAV; _ jc - a: [ (gs/s Mi / 444' mer mm ’ lira—«>530 aerYié—gxa) [1 25:0 {:07‘ TAR? Ffablem f3 gimp/ar- +29 EKamF/é’ l4‘l 13482 42-381 42—382 MAI National <”Brand l ‘1 I z I 14-4 A wind turbine with a diameter of 100 ft operates during a 24-h period in which the wind velocity may be appmximated by half a sine wave as V(6) = V,,' sin (170/24), whene 0 is time in hours and V... = ‘05 M. The wind density is 0.076 lb../ft3. Calculate (a) the theovetical maximum turbine energy during that 24-h pen'od, in foot-poundgforce and kilowatt hours, and (b) the wind mean energy velocity, in feet PEI hour. («a 5’ A v; (42) +46 VHOKI‘V‘WHM wi‘nJ pom/aw 13‘ Ioflmx = 3756 7776. i’VIQ/‘ZI'I’HLz/V) £M¢rgy [:5 (#3 fn‘liggrq/ eve» ”lime 2+ .2 85’ - 02 EMX: SID/NQK it : a R7 (Z Z“ \/»{3 4f _ wao“ 4% ‘i " 375:“ {5 VM3 si‘n3 if?) 45" FWD/n47 4able a‘p I‘nvLeaP—ojd‘: 3c, coszaxMZn (ma +2] CQSKGX’) (32+x/o /A;£+>§[‘3+lj*: “5”]? *" L140 39x K)? /ég'£+ Q. égst‘fl: £453 I65 a) 4527 jet/VGA» _ Q?) .7736 50556 e7uc4-i’er) +0 finci 4/26, QUE/‘55; vo/ae ['5 V §3+V(e’3 (JG: EL/(Q‘va Si?) 77,46) [email protected] qu 1:???) COS \(f: :)7:: 2+ qr— .— '—— 3:? \/ a3; ééQ welt/Ar A?“ Me aver e value is I’IO‘IL WAQ‘P +46 Proé/em has aséed 2g 1". 6446.) ' 42— Nanonalmamnd 42. a I ”A?” I ‘ (b) ’15 ”lea”! 9‘43” y Velma-47 f5 «Cram 0 mad/‘Qcm‘r‘orr or L9 (i+ 26 _ . ‘74 , - “\7 = (,2 VLUVJ" : {2 veg»)? 4&)'/3 E V7 0 2+ Fea/ y an average of V3 M[ 524%43 777;?) (1(9ng :53“: 1?“: 9322537 §3 =— 7711‘s )5 consis‘vten4 LIVE/414 4/15 vLex+ em pajgfiéflf’; w/n‘c; 54:04:: 4-5268} .1— \/&v5, < V1:- Agra Q 3: 4&2 < 75; /SO / ...
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