chapter12-part2-solutions

chapter12-part2-solutions - PROBLEM 12.106 KNOWN: Solar...

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Unformatted text preview: PROBLEM 12.106 KNOWN: Solar irradiation of 1100 \‘Wm2 incident on a flat roof surface of prescribed solar absorptivity and emissivity: air temperature and convection heat transfer coefficient. FIND: (a) Roof surface temperature, (b) Effect of absorptivity. emissivity and convection coefficient on temperature. SCHENIATIC: '* a , I ch(Ts) u i ‘ G q eonv ‘ l as 8 fl i ’_\ / F {as j; Gs = 1100 mm? W/ 7 °C —~ r .r T3, 8 = 0.10.5 = 0.6 2 .0c ' I h = 25 WJmZ-K Metal plate, roof surface , Insulation ASSEMPTIONS; (l) Steady-state conditions, (2) Back-side of plate is perfectly insulated. (3) Negllgible 11rad1atronto plate by aunospheric (sky) enusslon. ANALYSIS: (a) Performing a surface energy balance 011 the exposed side of the plate, aSGS — qgom. — eEb (T5) = 0 aSGS — Htrs — L, ) — sari = 0 Substituting nutrierical values and using absolute temperatm‘es, \ I. 0.6x l 100—— 25 m2 1112 -K (t, — 300m — 0.26.6? x10_8 “7,.-'f1n2-K4)T:1 = 0 Regrouping . 8160 : ESTS +1.134D ><10_8T54 . and performing a trial-and-elror solution. r,:321.5K:4s.5°c. < (b) Using the IHT F first Law Model for a plane wall. the following results were obtained. 12D 100 so " 60 40 . , __ ___k_ P ate temperature. I [C] .‘ IF; 20 —- -- ' D 0 20 40 SD 80 100 Convection eoeffi crent, h(Wr'm fl2. K) + eps = 0.2, alphaS= 0.5 —.|.— eps = 0.3, alphaS= 0.5 —l— eps = D 8, alphas: 02 Irrespective of the value of 11. T decreases with increasing 8 (due to increased emission) and decreasing org (due to reduced absorption of solar energy). For moderate to large on; and-"or small a (net radiation transfer to the surface) T decreases with increasing h due to enhanced cooling by convection. However. for small as and large 3, emission exceeds absolption, dictating convection heat transfer to the surface and hence T < Tm. With increasing h. T a Tm. irrespective of the values of org and 8. COMMENTS: To minimize the roof temperatm‘e. the value of a“ch should be maximized. PROBLEM 12.115 KNOWN: Glass sheet. used on greenhouse roof. is subjected to solar flux. G3. atmospheric emission. Gatm. and interior surface emission. G. as well as to convection processes. FIND: (a) Appropriate energy balance for a unit area of the glass. (b) Temperature of the greenhouse ambient air. Tm. for prescribed conditions. SCHEMATIC: C and: H 0115 .- 7;=27°c 1;,0=24°C h;=10W/mz'K hozsswfmz-K 65:1100W/m2 GmsBSOW/MZ G,- = 44 0 Mmz ASSUMPTIONS: (1) Glass is at a uniform temperature. Tg. (2) Steady-state conditions. PROPERTIES: Glass: 11 = 1 for 2*. g 1pm: 1;” = 0 and oh = 1 for 3*. 1 inn. ANALYSIS: (a) Performing an energy balance on the glass sheet with Em —Eout : 0 and considering two convection processes. emission and three absorbed irradiation terms. find where as : solar absorptivity for absorption of Gm ~- EM, (In. 5800K) 01mm : ai : absorptivity of long wavelength irradiation 0. lab 1 pm) a 1 a = {1}” for It. .'>:» 1 pm. emissivity for long wavelength emission 2 1 (b) For the prescribed conditions. Tm can be evaluated from Eq. (1). As noted above. 013m] = on = 1 and a = 1. The solar absorptivity of the glass follows from Eq. 12.45 where GL3- ~ EM, (3c. 5800K). as = GL3 d/UGS 2133611 EU) (' -,5800K)d/1.='Eb (5800K) as = a1F(O_>1;£1n} + a2 [1—F(0_,1#m)] = OK 0720 +1.0[1— 0.720] = 0.28. Note that from Table 12.1 for ).T = 1 pm >< 5800K = 5800 um-K. Fm _ a = 0.720. Substituting numerical values into Eq. (1). 0.28 >< 1 100VW1112 +1 X 25.0 w ran + 5.5 W 1113 - K [24 — 27) K +1>< 440 W; 1113 + 10w 1 1112 «(Tm 727m 72><I>< 5.67x10’8 wln2 -K[27 + 273:)4 K4 = 0 find that Tm‘i = 35.5%.: < PROBLEM 12.127 KNOWN: Spectra] distribution of coating on satellite surface. Irradiation from earth and sun. FIND: (a) Steady-state temperature of satellite on dark side of earth. (b) Steady-state temperature on bright side. SCHEKIATIC': RIF. ix ‘— ASSUMPTIONS: (1) Steady-state conditions. (2) Opaque. diffuse-gray surface behavior. (3) Spectral distributions of earth and solar emission may be approxinmted as those of blackbodies at 280K and 5800K. respectively. (4) Satellite temperature is less than 500K. ANALYSIS: Perforating an energy balance on the satellite. Ein — Eout = 0 tag GE {ED2 54‘] +a5 (33 (3D2 4)—60'T:' (:rD2 = 0 \14 (IE GE +0363 \ 480' J (I TS : From Table 12.1. with 98% of radiation below 3 pm for LT = 17.400uni-K. as '2' 0.6. With 98% of radiation above 311111 for ?.T = Sum >< 500K = 1500 tun-K. g =3 0.3 dB 2 0.3. (a)011 dark side. I f ' T _’ aE GE ‘1'4 _ 0.3 x340 W.-"'1n2 S _ _ \ 480 i \4x0.3><5.67><10_8 Wf'm2 K4 TB :19? K. < (b) On bright side. , 1..»‘4 r \1:""4 T _ aE GE +513 63 l _ 0.3 x340 w..--'1113 +0.6x1353 w; 1113 S 7 — 7 , t. 480 i .\ 4x0.3><5.67><10_8 w..-"mz -K4 T.3 2 340K. < PROBLEM 12.137 i KNOW.\': Solar panel mounted on a spacecraft of area 1 111 haying a solar-to-electrical power conversion efficiency of 12% with specified radiative properties. FIND: (a) Steady-state temperature of the solar panel and electrical power produced with solar irradiation of 1500 W..-"'m2. (b) Steady-state temperature if the panel were a thin plate (110 solar cells) with the same radiative properties and for the same prescribed conditions. and (c) Temperature of the solar panel 1500 s after the spacecraft is eclipsed by the earth; thermal capacity of the panel per unit area is 9000 J/lllz-K. SCHEMATIC': Solar panel, T, Ap = 1 m2, g g 5 G3 = 1500 Wlm2 thermal capacity, 9000 JImZ-K Array, ca = 0.8, “8,3 = 0-8 conversion efficiency. 9 = 12% Backside, ch = 0.7 ASSUNIPTIOXS: (1) Solar panel and thin plate are isothermal. (2) Solar 11radiation is nonnal to the panel upper surface. and (3) Panel has unobstructed View of deep space at 0 K. ANALYSIS: (a) The energy balance on the solar panel is represented in the schematic below and has the form Ein — Eout = 0 “SGS'Ap_(3a+3blEb (Ts )‘Ap_Pelec :0 (1) where E, (T) = 6T4. o = 5.6? x 10'8 Wx'1112-K4. and the electrical power produced is Pelec : E‘GS 'Ap (2) Pclec : 0.12 x 1500 W1“ In2 x 1 1112 : 180 W < Substituting numerical values into Eq. (1 J. find 0.8 x1500 w 1112 x 1 1112 — (0.8 + 0.7)crrgflJ X 1 1112 — 180 w = 0 TSp 2330.9 K=57.9°C < (a) Solar panel EAp (b) Thin plate (b) The energy balance for the thin plate shown in the schematic above follows from m (1) with PmC = 0 yielding 0.8 x1500 w 1112 >< 11113 — (0.8+ 0.7)ch1 X 1 1113 = 0 (3) T132344] K=71.70C < Continued PROBLEIVI 12.137 (Cont) (c) Using the lumped capacitance method. the energy balance on the solar panel as illustrated in the schematic below has the form Ein _ Eout : Est dt where the thelmal capacity per unit area is TC ” 2 (Mc r” AP) = 9000 .T r" in2 -K. —[ga+ab)ar;}).Ap=rC"-Ap (4) Eq. 5.18 provides the solution to this differential equation in terms of t = t (Ti. TSP). Alternatively. use Eq. (4) in the 1H Tworkspace (see Comment 4 below) to find T5p(1500 s) = 242.6 K : _30_40C < Tspn). Ti = 57.900 (Mei/nip) = 9000 Jim2 -K EAP / (c) Transient energy balance COMMEXTS: (1) For part (a). the energy balance could be written as Ein _ Eout + Eg = 0 where the energy generation tenn represents the comers-ton process-float thermal energy to elecn'iml energy. That is. E = —e - G . -A e S p [2) The steady-state temperattu‘e for the thin plate. part (b). is higher than for the solar panel. part (a). This is to be expected since. for the solar panel. some of the absorbed solar irradiation (thermal energy) is converted to electrical power. (3) To justify use of the lumped capacitance method for the transient analysis. we need to know the effective thermal conductivity or internal thermal resistance of the solar panel. (4) Selected portions of the [HT code using the Models- Lin-npeni | Capacitance tool to perfomt the transient analysis based upon E-q. (4) are shown below. it Energyr balance, Modell Lumped Capacitance * Conservation of energy requirement on the control volume. CV ‘ .-" Edotin — Edotout : Edotst Edotin = 0 Edotout = Ap “ (+q“rad) Edostat : rhovolcp “ Ap * Der[T:t) rhovolcp = rho * vol ’ cp thermal capacitance per unit area. Jim’Q-K Radiation exchange between Cs and large surroundings q"rad : (eps_a + eps_b) " sigma ’ (TM a Tsuer sigma = 5.6'r'e—8 Stefan—Boltzmann constant Wr'm"2-K"4 ll Initial condition Ti = 57.93 + 2?3 = 330 9 From part {a}: steady-state condition T_C : T — 2??) ...
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This note was uploaded on 06/03/2008 for the course MEM 345 taught by Professor Cho during the Spring '07 term at Drexel.

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chapter12-part2-solutions - PROBLEM 12.106 KNOWN: Solar...

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