EEE463_Homework11_Solution_F07

# EEE463_Homework11_Solution_F07 - A reservoir has a water...

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Unformatted text preview: A reservoir has a water surface of 1200 acres and an available storage of 4000 million gal. The watershed contributing to and including the reservoir covers 35 mi2. The dam of the reservoir contains a hydroelectric turbine with an available head of 120 ft and a mechanical efﬁciency of 83%. If the runoff of rainwater is 36 in/year and the evaporation from the surface is 8 in/year, compute the average electrical output of the generator, in kW, assuming that all the overﬂow goes through the turbine. wwwwwww w .c 1.13%.” >{éé org-1:) :71... we : 2.92‘72x/O? 41/43 [oer y/x 9{' 0“; EvaFowalnl-r‘am= (/5200 acresécigl/T )‘% #3 560.”: acre. 42- Nalianal ®Eirand ”‘2‘ ~— - ‘ ' ‘ 3.1? a]: 6“,: aver-«Flavor =(VO/ 6‘; )QL)ﬁO#>‘[\/el c'p Evﬁfcran/Ori) Q 927QQ/c9 — 3 (+ng 2'07 «£43 .52., EQQx /O‘; #3 per 7% N055 cf Ob’e/‘nva : <\/¢f> (ﬁg/:57? ‘:‘—' L/ f = (Q. wax/calm? é-JZ. tr /érm/++3) '= / EOSXIC‘” /é[email protected] Fe,” 7/“ ll 1! «.24 para“: 4 (PE): (0°63>(’2 4“? APX! {air/17p.) = l773 kWe/ Lu 3: < 3 0 w h’) 0: Lu .1 « /\SE‘3 5 SQUARE - ASE?a 5 SQUARE ASE“ 5 SQUARE mmmmm 200 RECYCLED WHITE 5SQUARE 500SHEETS Fl 50 SHEETS e 100 SHEETS 389 200 SHEETS EYE- 392 100 RECYCLED WHKE 5 SQUARE _ 16-2 A power grid has a load pattern during one 24—h period that averages 600 MW during 18 h and 1200 5 i a . ' . l l i MW during 6 h. A pumped-hydro energy storage system with an elevation of 100 ft is considered. Calculate (a) the pOWer output of a powerplant, in megawatts, that would meet the load demand with and without storage, and (b) the volume, in cubic feet, of water that must be pumped to meet storage demand. The MM” “’1 ' efﬁciency is 0.8. Water density is 62.4 [hm/{(3. . [kin/y LOQo/ paﬁézérn Gt) W/‘1L'LOW‘; 5+9 nasal, ,l—Ae p i r‘QOOMW loam/er“ Plat/:4" Wgulcf Aim/c3 36’"; ‘*-‘ -- m f: — — -- - - via: be, gar/gaggle? émLq meg-Hr]?- ‘ ' ‘ Fiﬁ- v41: emir‘retlﬁoa @Iaécﬂ Wi‘i'lﬂ g‘fio ”1235/ #4 Fey/var /a/4‘/:“' mu5‘% 1’70‘7’: an/y ”49565,; ¢Aee gas}? ﬁred cpép 500 MIA/j bu‘i’ nL mus‘f Giro Frau/2]: Paw‘em '—/-o Pam/a Wd‘IL‘EI’" [n+0 “the garage :5"’/3+eme 7718 alufnm‘ouvzc/ (cf‘ngCFenCy 7.3 Jaﬂ‘hec/ a: E :_ +o‘7ig/ energy oval-31071,: O 8 +o4a/ aria-K317 fn’mm/v We camera); released claw/VIC] [beak 4/4/75 Musﬁc écf fgpuo/ +0 .14).»: tarmac/rm!“ 54ored' Chart-(my- +53 aman?ﬂk/ /es=;: any [he {bitty/Cy, w M é: (AgnouM-I- cap ene/jy .s‘fl’ored‘) random/r; 52¢ ewe/S27 USEKC/ﬂ) l8 ‘ U’Yk’lj aw—pcuk ’iﬂiedv. dbl/”("17 [Dave/C erfoc/ f {‘0 (jo‘en ‘— I‘D/Cad) 5% C #5 (JD/cad hpgen> 4% (O, 91>(p en ‘ éOOMi/V ()8 Ales) : ﬁlOO/WW‘ILELJ>(6 Arts:> QGeZ-F foe,” i“: /58‘f0 / are, ‘: 7755Mw'] 635:1; ,. 18. , , , (b) Essie/e " £69 an “Dieeﬂoa :("77é931600 MWJQ’S Agﬁaooew/mg = ‘30/"77Xi06 Aw‘me Fran/1 E7 (‘Ié"/> ;' 105:. M H z 23: f V H v2 £5141; _ .f " A” L {gas—52x IO ewe”, 3* f H i; \/='i 1‘. 35x /0 7 1&3 o 6‘ I0. #5 24 £59; ....\ 15-1 Ocean water, just below the topmost layer (where reﬂection back to the sky and a complex mechanism of heat transfer and evaporative mass transfer occurs) receives 65 W/m1 of solar energy in one locality. The extinction coefﬁcient 15 0. 4m ‘ .Calculate (a) the volumetric heat-generation rate 4” , in watts per cubic meter, near the surface and l m below the surface, and (b) the distance below the surface, in meters, where 99.99 percent of solar energy has been absorbed. 32:63 w/ma ﬂ=004nr1 (a) Accom/V , 740 Ea}, (Bu/D ”ii 9 * IVI“ A/zfar Saﬁpac: 6/3,: E/LIC I(0 44')(5 W) :lQéz’l'“ W/ 7—4:: reduced in¢en§ﬂ61 45L .1 M below 71-45 Sur-ﬁoce /5 also found Urfj/z E7 (/S’~I> I(y)“: IO 87“)“ :(IM) «2 (es 2%) e‘(0°+/’”>(i’"3 = 43.57 (AZ/74" UV 4722:, =/aI/1m)-= {a Lew/4357:— .47 M For 999? 0/0 aéscirs/o/vam 2’ €402)— :- i” (9&9979: (20067.1 £61 : 090001 : e‘ (a'ﬁ/AV k “LO V: 23:6’14/7 15-2 An OTEC powerplant of the Claude type has an ideal turbine operating between 25 and 15°C. For a turbine output of 1 MW, calculate (a) the mass ﬂow rate of steam, in kilograms per second, (b) the mass ﬂow rate of warm water, in kilograms per second, if the evaporator temperature drop is 2°C, (c) the mass ﬂow rate of cold water, in kilograms per second, if the condenser temperature drop is 2°C, and (d) the cold water pipe diameter, in meters, if the water velocity in it is not to exceed 1 m/s. ﬂi/CTE: I I 211;; Frog/am 1.3:. 25‘3‘cem‘r‘r‘o/1} 744351097442 (2.5 awn/01’} fir-L, Vacuum PWT‘P Low-pressure steam S+e¢2WI £16241? I") 7—be lghlu Powerplanr Surface water 17° C Direct- contact \ . , t Varm Wdlkf condenser discharge Tmupuramrc mam 03KB: ((<( 333 000 mini/3 mm“) m gun/1 £355. ”55 U3 PM we Ina (”tum 011 ON mm m N m m 'U a e: E K2“ ﬂag“ Cold-water Cold deep discharge water 389 200 SHEETS EYE-EASE" 5 SQUARE 392 100 RECYCLED WHlTE SSOUAHE 399 200 RECYCLED WHlTE SSQUARE 15°C 15°C Deep water ll” C Eniropy.1 Figure 15.3 Flow diagram and schematic of a Claude (open-cycle) OTEC powerplanl. 53“" 15" T" “”8”“ “Dmsvondins to Fisr 15-3. @ \$53: 523 = 5"}é2525' C>= 515570 kT/(ZﬁKJ .2 ._ 3 »~ 35- ___ 85976: 0. 224%, , "7‘3“ 7 W 8 “-7‘Eioe‘ o 2245‘ = 607739 1755‘ 717% + he =(0373‘?>/2%6§+é2.97= 2%:522‘ if: F’ , p ., rﬂ __ i533 kW JOB/«7%? W‘s-4&0”, = Ah ’= 5;— AJS. ‘ 254722493 é ﬁg: 3 #K/kf 31.2 kg/efj “ ﬁliimm,_ _""3"3 _ __ ‘L? ”3‘2— 16"‘1‘8 _ . .. Cb) 27mm — n2, '~ 79 ' Water mm ~ 0-003414 .» __ 77151-5 __~ #2 ka/s‘ec __ " ”wt/warm - R’:M _ 0uOOB‘I-LI— _ 3‘7‘88 kg/g'cic Ca) ’fﬁ’sieom 7155+ moguld 56 = (/éstﬁwmﬁ" ﬂ2¢6q>57 ._, 555 59M 2’+é&: — é2f/‘7 maid _ mﬁeom In 41\$ ‘7 2(25 ézz <77— 5‘4 60 = i 3‘1‘1‘ 2 kg/S‘cg’ 9. Av mac-“v __ 2t 0"); > Vle : f A V L: xv— = ”U— ““ t—f- ”U"- Where ’U" IS ‘H’Ia \$€e6(:€‘t Vo/ume 0‘7" /U"5_» (29 137C) {)7 z m: “.7, ‘7‘ VVE(Q‘3 ‘C> /Q:42__LLY_.§;J WW2— 4:20.ch ”77' (/m/mc 15-9 A wave-energy generator system is to produce 1 MW in the ocean where the waves have a steady 4- ft amplitude and 227.1-ft wavelength. The ﬂoats measure 12 X 12 X 1 ft. Calculate the number of ﬂoats {mm necessary if the overall efﬁciency of the system is 9.6 percent. Sea water density is 64 lb,,./ft3. F : / M W Q2L7L+5+ 63:9,é‘é/o 2 =' 2:27" I +“+' 5‘” Q: Li- / [y m / 41+ 3 F’W+ Area =QQ>O2> = #47042 From 5:” (I939 >7: 5/2 7.2 , Lag, _/ 2%? 15+ Fla-fried ’ZI'IZ 5' i5? —éa é é Sac: *- , ._.L./)’ ._. :5. “ ~ éJééSec i- O‘ISO [1+2 LLEH 5 SQUARE ? I E E g: E 3:: §= e§§ ...
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