prob.xii-2 - gm (U PROBLEM 12-2 SOLUTION Tami availabie...

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Unformatted text preview: gm (U PROBLEM 12-2 SOLUTION Tami availabie powar = 3WD x 441'.) = 3.52 MW Let‘s put N E number of electrical rods Then,tnt:-1l hemad surface area = N 3: ID in: L = N k‘ :n: x CLGIS {:11} 3x: 2 (m) = {115? N m2 Thus, heai H Ln: = —LE— = Mme3 Li. 1 5 m I“ the critical heat flux in the puui boiling regime i5 Ubtained by Eq. [2-H]: Elfin: 13:5 .iahhl whnre: 4' r_ w m j: = crip. fig] P1: .J" and Cl '— 43.]3 {Ruhuemw-j: ur CL 16 {Kutattladza}, or 0.13 (Zuber). Thus, for C1: 0.13: [1.132356 31. - . . 1i“ x[ 35 mama £63wa- q-rru: {kgfmi} X X 16 3:. To run the heater at 30% oi'fllc critiual heat flux, :12: :13 x 3.63? wan: = liglmmfi N=IT 2: N=3 Analuguusly, far C1= U. | S and C; = £1.16, N=5_56 =5 N=Et H '— fifilfi = N = T , raspectivfilji. Fit the incipient pnint, q»;=L]"aq-: Ruv DECEth-l' 3.1993- Paga 1 0f 2 pages Pmblam 122 Solution where. q’fis iflcipicut lifiat flux q'fiurfi natural cunvecfiun heat flux Fur IJJE‘. pool boiling conditiun, the incipient heat flux and fiTm are relatedjay‘ mTFfli—m': K {Fri I) Cpl' iii-Tsar. = C I I qui __,U_. LLlrhiz EiPF'Psi' (Rotseuuw curreiaflcu fi'cm "Convective Boiling and Condensation" 2nd v01. by IGCnilier, pg 125): where, for water, Cir: 0.13132. 11 = 0.33, m & D. And, q"Nc-= 2.63 [flTml I" :- k‘wgmi' Nrm', putting qulzqfllfl in {Eq 1}, suiviug fur if“ and plugging in the pmperues ui' water at 3.35 Mpa sat. Mn: his 1“. 153 a: G {Haircpfll fiTs: = —>‘{U.Dl. — _ ' Err M‘iurhm iglprpgii LK : i flT-“lfifififl'i: lgfifi-lxfl-fllgmd’ 1'53 .Ln5[4'?fi9}cill9><m‘a : Lsz .. 4.?fi9 :ngx lfl-*K1"ffifi_1 9.3 {Blitz-MEI {1523 i, r an... = 1.45 "C Tharefurt, q”4=q"s._~= 2.53 H.431“: 4.19 kWEm: Page 2 (if 2 pagas ...
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prob.xii-2 - gm (U PROBLEM 12-2 SOLUTION Tami availabie...

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