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Unformatted text preview: IJS AmmbﬂwmmﬂmmﬁmmmnﬁMIﬂWH.
Apipentmeﬁacal line receives 450 IMﬂMItmpiilandu lOD'FaMujuu 130 psia. Calming
the pipeaxi canditim. Axum cameraItch 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' 1r'laf/am‘a Std" "HIarcs are 55’; fuss; so
6'2: (aquﬁsrjkw) = (43.3 kw Q“ r5651”; ‘ imam) :9ng upon» sIIanm Joén'u; 9%: [ALJ we: :3 can«Framed;
naf saPura'hd' LiftNJ; :39 H: can"! hm‘pcrH'urc Sfﬁfd'
i‘rafaw alumni 37I°Fj +1523 éar'fr'nj arcEr rte'4' occur, 62 '=' n? C1, 537" Q _ f‘fEﬁ. .w 442. A—
AT= ”a cf ‘ ‘f'S'OJénfﬁr IEh/fbm‘ﬁj
= ID?D.3 BR Hence! bﬂllff‘r‘j drag: cactiIr ;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 1311 ALEmimdrnllWd'uﬂknﬁanDImhwludlj x In"
WIHﬂumeWdﬂ.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” aufFyJ waufg’ 3:“. (5?2.6%X‘F¢.f8¢m2¥%)1 = 2.6F8 W rTeLfﬂ 1"3‘“'Il {Marita1‘15; 1‘1"}; ,}1 1‘A: mucfen‘j'p‘é
ran ‘— op O.7—CL?/#m 45::4 *Hg mnvdﬂabﬂ
aﬁcfency r}. 6.73; J's: @.éf5 w)(o.73) = 1.91 ML} 1313 mummﬁammmmhmrnWmmm
mmmalmlcmya.euh5hnondnaide.Ame0.95pckinzﬁwﬁunafdwoelh. I:
[Him 32:: " JT’
EurHa C
ICEW
~ . _ M =
Elfinaenc? " Pan' Carnivalf £5,9an é Paaéiﬂﬁ 99:4"!!! " "7—,,qu Areq _ O'?S __ (5km): ExJra'farrtsJﬁfal' Safer Isahan‘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 stir11:13:. 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. kWehr = 4.1153140” kWehr First, calculate the solar electricity potential based. on the incident energy and the solar cell
efﬁciency {a} Elam = .9 Emma]: = (1112332 x111” kwhrryr)
= 2.6%: 10“ kWhrfyr 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
264x 111” kWhrfyr = (4.053 x112]12 kWhrlyr}[1+ —Yr] 1 periodfyr 5514 = (1.11253T
Now take the 10311} beoth sides
loglﬂ{651_4] =1ugm[(1_1125)TJ
281383 = 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'DD 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 kWhrfmZday for a
tracking system, and only 1.35 kWhrfmzday for a horizontal ﬁxed system. Solution
Daytime load = (l (10 We) (06) (24 hrsfday) = 1440 MWhrfday Nighttime load = (100 MWe} {0.40) (24 hrsr'day} = 960 MWhrfday Total energy needed to ﬁrlﬁll the nighttime load, including stored energy loss
= (96D MWhrfday) f (0.9) = lﬂﬁ? MWhrfday Total daily electric energy needs = Daytime load + Nighttime load
=l41G +1D6T = 250? MWhrfday Required thermal load = (250? MWhrfday} f (025} = mess marham {a} For a tracking system using reflectors:
Energy collected = 6 kWhn’nfday 10323 103 kwhrma
Area =# = (3} ma“ mZJ[l huntm m}2
(a kWhrfdayIDAﬁ) =3.:r'luu= (b) For a ﬁxed horizontal system:
Energy collected = 1.35 kWhrr’mEday _ 11:111st 1s3 kWhrr’day
{125 kwhsdaymo) = 12 km3 Area = (12>: m“ m2 )(1 manna m)2 ...
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 Spring '08
 HOLBERT

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