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Spring09chbi302hw1 - CHBI 301 8630 Kl‘llel HOm ewomg.14 n...

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Unformatted text preview: CHBI 301 8630. Kl‘llel HOm ewomg .14 n SubsffiuI-e “3'" {4.1.49 my? (4.1-?) k r; T; q olr -— __ a — q b‘T DnL { r ;< + )dT r, dr‘ .3 1,1(1) : “(11-13.)+ b(T.-T1HT.+T1) 2 :. (TI—T1\( q + b (T___.1Tz)> 2 M .1 %(2>=LM(T,—Tj) km 4.2-3 k—. 15.23 vv/MK I xfiAPPfldn} AI t: 9'8 b+“/hr.f«l.F Al .- Q‘Wr, L a. D-Tt ( 0‘9?) 1. O_0653 {‘12 r‘._ 013.0ng [2.1 rzzopmbbsf-l- A; : arm-3L —. 1T\( 0'4 ) -. O_IOLOG {+2 01.3- Alm - ‘1 ‘A‘ —. 0063:? (4 2 L‘AJ/a| A a) -, k A!“ (ll) __ (3,9 (0.0838 (go-4°) (PL-PI) (o_o:66§—Olooc9\ /9= '-3‘Ib+u/.s -. 1393w 3 k.- q 1‘ bT+c T l ‘— — k i : ~ (n+bT-gcT3)°’_T A 4x 14x a ”1 £5. 3 3— fd" : -}(°‘+£3TJCT)c[T A X. T, .. ‘k‘ RA. ; AXA 0—5016, ‘ 1 Ta. 7’ kp J3 (0‘ ISI)(33) 7‘ RA=o.oo3l‘4~3</W A '5 C R _ Axg _ Axe, Frag curb. concred-g 3 ‘ "—_" ' *— «5 9’ (0.0u33)(33) 4* 9A!» 4x6 C RC: AXc - 0.0502 MA (oinmwfi) RC:— o,oo|70‘.’) K' VV ‘1 : T‘ ‘ I” ; "7‘8 _________2‘)'t‘ —_ 584, vv LRA ‘1ch 12¢ \ o_oo3l#3-¢R8+O.OOI10‘I RB = 0.0157flé ~_ A“ K/w (0‘0 #333 (3a) AX]; t, 043.11 m BA; RE». (1: kn - 031414 0" Ax Egg \AI/mK K830. 83° \Al/mi t 0.2% AKA - _________.. ) A ' (,30)(Io k9. X 0’93 _ A 3(10) A —(05‘+‘) k(5 '73. ‘l’. 1853 : ___,_.__.__ MMWWWQ 9'41 m AXB; I 30 V"/m.K - 19‘”! K I583 ‘- Tl -— T, 1. l3 AXA :1 635/000 :0.00€>33M Axe3 ; accessfls A = O.‘3|L¢xn.83=l.6?3 mi kn : kc : 0,86% W,/M'K kg = 0,015 V" /M.K AT ; 3543 K = 2.13? Eq‘n 4.3-3 RA : AXA- : 0-006557 = 0.00.4353 k“ A qaéq (L613) 0,006.35. RE 3 4‘6 - ___________ —_ owsqsq his, .A Q 0.026 ) (Lew) RC : AX: : 0.00639 1 0.004368 kc A ((1364)) 0.673) 7 .2 ,8 0‘ t A " 3 =7 0‘: waive 9.; 12,3 f g c o ,oou3£8+ OJHT918L4 4 Onau‘ssfl 505“““Hoo 43H 11:19.“ QC TAzggf‘Lec sirael L: 305m Lg : Ll? W/M.K kg, 2: 0.391 (ad/MK , // AS 2~. :e.. 13:55:) I m PP ‘ Q mm .05.: reams SghrokflA : . lmm 3b?“ 04.1? r;-r, 10.0039‘m Q J; = 0,0215% m D. __ :335‘3/“00a >O‘0523‘ M DA: 5150 ~r 9'. (3.9 D -— $0.3: mm 7.0.0603; M D3, = 60.39. + 1.(1S.H);|\\Jl (mm = O.HHlm A, ; TI'D‘L ,_ TI (0.05;?)(305) ; S‘O30m A; -_ TI D}L.: T‘ (O 7.. 06031) (30.?) = 5-790 m?- A?> = ’n DBL= ‘n (uninjggoj) =IO.6LI:I> 6mm iobles Ba interpoiohan 533“}... -. 6 9‘8 - 5.03 AA—Qyn5 AZ’A’ -:, 5 — 5396mL M53- 14» 5.?9 Al $.03 A ; AVA; : I0.6‘49‘-5.780 : 1%? m; BL“ his: A no.6q4 A1 5.13 ba'n (fins-TI») RA > 5"” .-_ 29211. = oncootwo K/UJ kF-AALM (qaflssqs) Re, ‘ r3 ‘0‘ = w -; 0.0!?‘3l3 K/uu kg Aalm (0.131)(q.%~+) a]: 1.1;; —_ Il|.( -26-1 2 53844 \A/ AA'HE‘E ODooolélo +0-OI?5'7 A.2 A ~.-. zafifiJ' hat/tap 5mm (glfifi.fx'OB-if'- {M%L>(36;o 5/3 m : 8.?! K/hr Solution The wall of a refrigerator is constructed of fiberglass insulation sandwiched between two layers of sheet metal. The minimum thickness of insulation that needs to be used in the wall in order to avoid condensation on the outer surfaces is to be determined. Assumptions 1 Heat transfer through the refrigerator walls is steady since the temperatures of the food compartment and the kitchen air remain constant at the specified values. 2 Heat transfer is one-dimensional. 3 Thermal conductivities are constant. 4 Heat transfer coefficients account for the radiation effects. Properties The thermal conductivities are given to be k = 15.1 W/m-°C for sheet metal and 0.035 W/m-°C for fiberglass insulation. Analysis The minimum thickness of insulation can be determined by assuming the outer surface temperature of the refrigerator to be 20°C. In steady operation, the rate of heat transfer through the refrigerator wall is constant, and thus heat transfer between the room and the refrigerated space is equal to the heat transfer between the room and the outer surface of the refrigerator. Considering a unit surface area, Q : ho A(Troom _ smut) ’V = (9 W/m2 -°C)(1 m2 )(25 — 20)°C = 45 w Using the thermal resistance network, heat transfer between the room and the refrigerated space can be expressed as insulation 1mm 1mm ' Tl'oom _ refi-ig Ri RI Rins R3 R0 9: R "— Tm Tm... total Q/ A = _L;_ff_g___ 1 (L) [L] ‘ — + 2 -— + — + __ h" k "19101 k insulation hi Substituting, (25 - 3)°C 45 W/ 2 = m 1 2 x 0.001 m L 1 + ——————-——— + —.___._____ + 9 W/m2 ~°C 15.1W/m2 ~°C 0.035 W/m2 .°c 4 Wm2 -°c Solving for L, the minimum thickness of insulation is determined to be L = 0.0045 m = 0.45 cm ...
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