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# prob.x-4 - PROBLEM 10-4 SOLUTION 1— Determining the...

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Unformatted text preview: PROBLEM 10-4 SOLUTION -_ 1—- Determining the Temperature of the "' ' Primary Side Ceelant of a Steam Generator TM] 2 Seedndaij.r side wall Temperature = 36°C Tm E Primary side Wail temperature — in be determined to = 9.5 mm 5 =12 mm res-3mm Tm {a} = Primary Cnnsrant Temperature Him = Constant Mean Velocity -;, m Omani-rm Tum wall (1;: heat ﬂux from 11'); In 2r:Fr side at r = r1 -_'. Energy Equation. pe m = — '1?- ” + q‘” + BTm+ It! SEW-5mm. Pop iii-VT = - v-q" (r) - I q" ds, as E Surface of as! = Z'Iridz {19L} as HEB, Wish 3 advantage of die azimuthal sFl'ﬂmﬂﬂ'F and aﬁﬁuming fun!" dawlﬂped HDW’ :1, dv = Zardrda, ds = Earn-dz and ii" = ‘1; TI WIT-“- - 5T H 1'3: we have: I pep v2 — litrdrdz = a qw litrfdz £313} 2' ' 0 3'2 Heir Mwemhar 2]. 1993 Page i did- pages Tl I va Ettrdr U [I I vzlttrdr o on equation yields m]? _ pep Vm dTm [z]: - Ettri q:rdz {2C} Eq. (2C) could t1qu boon formulated from an anorgy oajaooo across a toot: sum: of 33.912- 2 I} mutt}:- Jm—dz {3} pop Vm r1 ton, q; = h {Tm :2) mm} (4} hoot conduction oquation in tho wall, wo obtain title tompcratoro distribution in the: wall. Twlrj=C1ﬂm + c: ' q q r: BIC. atr=r- —"'=-rl_}cl=_ w 1' 3: k4: kc 1. El r: q '- Tw {I} = - "E ri in {org} 4- Two kc [I Ti TW1=Tw {rilszo + 1” ﬂnﬁﬁhi] {5} k; Pogo 2 of 4 pagos ﬁﬂgﬁq (5} inm Eq. {4}, we get after snme rearranging, "' = h {Ln {3} ' TWEI} {5) w [I + inﬁnfrﬂl L1 'iEq. {6} intu Eq. {3], we get __ m Tm (:1 — Twu pup um ri [1 + en ram] I ' m = i where ﬁrr : med Pﬁmary mass flew = 51:10 kgfﬁ "-' a A ETuheﬂuwma=E£t=1L§£ L1 5 number nf tubes = STUU p = 125 kgrnﬁ um :__5_HJ£J_ T26><%Xlﬁ.ﬁ1xlﬂ‘ﬁx5mﬂ = 5.69 misec: _ meDE _ Tzﬁxsegxmﬁxmﬁ “ 92x iﬂ‘ﬁ Re = 1454x105 —ﬁ 3 Pr=p_l-:E=92xm {5.7.5410 Lil-94 {156 Nu = [1.023 - 11:03 P93 = {111123 [?.454x105)ﬁ'8 31.94103 _ = 11215.1 h _ m _ USE-11126.1 = 3393mm“ xvme “K D [5.5.x 1 {r3 Page 3 efd pages: the given values and h into Eq. [’F}, we get ﬁrm _ 2 (3.?939x1ﬂ4} dz 3 {a} "2} * Turn a 3.?939x104 [3.3x1ﬂ‘3) ) - 5.45.9 . -3 —-—- 126(5Jx1ﬂ ] [3 3x1!) } 1+ 26 £n(—1ilﬁ_ﬁ) ff: =—U.14sz .1 I. ‘g Eq. {3} ova-riﬂe l‘.u1:J-v:11;1ngllr1+ we obtain T“ M I J T T ~ '11 z =£ (M]=-D.14TXL=H.14T><115:2.352 J; TD: [11' Two n Tmi ' Two Tmn = Tum + {Tmi ' Two} €235: {9} {11¢ Energy Balance we get number aquacan as fulluws clam-1* {Tm- - Tm} = Q. what-e 2 total htat transfar rate = 820 MW ﬂ EEleuﬁ Tm'l" rmg =—=23.2UC 5.?xlﬂ3{51llﬂ} Eqs. {9) and {1U}, we obtain . = J31“: : *1 Tml "rm-r“ ﬁﬂﬂ} 2T6 +311 3111.: (2 Tuna = Tm'. * 23.2 = 2,7911%: Page 4 0f 4 pages ...
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