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hw10_sol_printable

hw10_sol_printable - IJ-S AmmbflwmmflmmfimmmnfiMIflWH...

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Unformatted text preview: IJ-S AmmbflwmmflmmfimmmnfiMIflWH. Apipentmefiacal line receives 450 IMflMItmpiilandu lOD'FaMujuu 130 psia. Calming the pipeaxi canditim. Axum camera-Itch Rustin Mandfipement. Lav—are: {Iniflf‘ fe:——-———saa’ | m Ifa'f- QHr‘f' I"; = #50 {gm/fir =200 {It} = (BC) run}. was r}:— F 5'; 905' 1-r'laf/a-m‘a Std" "HI-arcs are 55’; fuss; so 6'2: (aqufisrjkw) = (43.3 kw Q“ r5651”; ‘ imam) :9ng upon» sII-anm J-oén'u; 9%: [AL-J- we: :3 can-«Framed; na-f- sa-Pura'hd' Lift-NJ; :39 H: can"! hm‘pcr-H'urc Sffifd' i‘rafaw alumni 37I°Fj +1523 éar'fr'nj arc-Er rte-'4' occur, 62 '-=' n? C1, 537" Q _ f‘f-Efi. .w 442. A— AT= ”a cf ‘ ‘f'S'OJé-nffir IEh/fbm‘fij = ID?D.3 BR Hence! bflllff‘r‘j drag: cacti-Ir ;4¢:erflpnrt, av mfg? 57;,” + AA] €43.3kW1/3WRJ %J=Gso%?)f%wfi 7’ —ch 7:) + A 1"] A In == 616?. *+ astu/fém 14+ PM PM,» g% = am; '2 - 0.3 (seas—M3,?) : 35:13 7%: 2=~§Efig flosréli 13-11 ALE-mimdrnll-Wd'uflknfianDImhwludlj x In" WI-HflumeWdfl.mm.thuwmmfln «thin “tripe-”gunmenandtbnhcmuimummnllpowcnhlmbepmduwdbymeflll. inwalts. | I . Gib = 15:. {GD—L‘M/rm M 515:1”: 2': 0.83%»? I E : I 7.5cm Q) Curricula-#3 +3.59. flflffif Fer/4704a» E 2 be: = 5,:3255xIa—34JT; 2.??7725'xf08m r I“ ‘2 0.53Bxlfl'ém = 2.37KIIO' q T roar- fga‘finr: '77: 59;?” EflérJ; film: r3 4: EF = (2.5.: to ”Jewfim :o*’j;;§;)(£?§m)’(#§j £32.; W] G") Ci” Area = Wag“ = «(3%? '1‘ = 4+4. Iscm= I 1:? +5: Sofa!" one.” was IDO‘J’Q e.'quc.5au¢fi 1%.,” | Hie. at” au-f-Fy-J- waufg’ 3:“.- (5?2.6%X‘F¢.f8¢m2¥%)1 = 2.6F8 W rTeLffl 1"3‘-“'Il {Marita-1‘15; 1‘1"};- ,}-1 1‘A: mucfen‘j'p‘é ran ‘— o-p O.7—-CL?/#m 45::4- *H-g mnvdflabfl aficfency r}.- 6.73; J's:- @.éf5 w)(o.73) = 1.91 ML} 13-13 mummfiammmmhmrnWmmm mmmalmlcmya.euh5hnondnaide.Ame-0.95pckinzfiwfiunafdwoelh. I: [Him 32:: " JT’ Eur-Ha C ICE-W ~ . _ M = Elfinaenc? " Pan' Carnival-f £5,9an é Paaéiflfi 99:4"!!! " "7—,,qu Are-q _ O'?S __ (5km): ExJ-ra'farrtsJ-fifal' Safer Isa-han‘riy 5 S: L- 353 kW/MJ According to El—Wakil (p. 541), the terrestrial radiation received by the continental US. is about 2.2x 111115 stir-11:13:. Ifa typical solar cell has a conversion efficiency of 12%, determine how long it would be before the entire surface of the US. is covered with solar panels if all of the electricity was produced this way. Utilize the electricity use datum from Ell-[1'6 (see Homework #1} as the baseline, and assume that the electrical consumption rate grows at a constant rate of 2.5% per year. Solution From Homework #1, the total electricity use in the [1.3. in 21306 was 4,052,961345 thousand. kWe-hr = 4.1153140” kWe-hr First, calculate the solar electricity potential based. on the incident energy and the solar cell efficiency {a} Elam = .9 Emma]: = (11123-32 x111” kw-hrryr) = 2.6%: 10“ kW-hrfyr We can use the financial type equations for the future value due to the growth ofan initial principal ammmt of money (in this case energy} in order to find the number of years until the electric energy use equals the possible production (at today’s solar cell efi'iciency) f 111' F =P[1+—] H 11 0251 MM 2-64x 111” kW-hrfyr = (4.053 x112]12 kW-hrlyr}[1+ —Yr] 1 periodfyr 5514 = (1.11253T Now take the 10311} be-oth sides loglfl{651_4] =1ugm[(1_1125)TJ 2-81383 = r logm (1.025) 2.31333 = — = 151.4 years lugmu .1125) A solar thermal plant with energy storage capability is to be constructed at about a 40" latitude to provide the electric energy to a system whose 24—hr average load is I'D-D We during the winter. Assume that the solar thermal power plant will have a 25% energy conversion efficiency; that the nighttime load is 440% of the total load; and that 10% of the energy that must he stored is lost {a} Determine the total land area (in 1:11:12) required for the reflectors. {b} For comparison also determine the area (in 1:11:12) required for a fixed horizontal system. In both cases, it is believed that a parabolic reflector system can be set up so that the reflectors cover 45% of the land area on which they are placed The reflected energy collected at 4G” in winter is 6 kW-hrfmZ-day for a tracking system, and only 1.35 kW-hrfmz-day for a horizontal fixed system. Solution Daytime load = (l (10 We) (0-6) (24 hrsfday) = 1440 MW-hrfday Nighttime load = (100 MWe} {0.40) (24 hrsr'day} = 960 MW-hrfday Total energy needed to firlfill the nighttime load, including stored energy loss = (96D MW-hrfday) f (0.9) = lflfi? MW-hrfday Total daily electric energy needs = Daytime load + Nighttime load =l4-1-G +1D6T = 250? MW-hrfday Required thermal load = (250? MW-hrfday} f (025} = mess mar-ham {a} For a tracking system using reflectors: Energy collected = 6 kW-hn’nf-day 10-323 103 kw-hrma Area =# = (3} ma“ mZJ-[l hunt-m m}2 (a kW-hrfdayIDAfi) =3.:r'luu= (b) For a fixed horizontal system: Energy collected = 1.35 kW-hrr’mE-day _ 11:111st 1s3 kW-hrr’day {1-25 kw-hsdaymo) = 12 km3 Area = (12>: m“ m2 )(1 manna m)2 ...
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