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Unformatted text preview: PROBLEM 104 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 res3mm Tm {a} = Primary Cnnsrant Temperature Him = Constant Mean Velocity ;, m Omanirm Tum wall (1;: heat ﬂux from 11'); In 2r:Fr side at r = r1 _'. Energy Equation. pe m = — '1? ” + q‘” + BTm+ It! SEW5mm. Pop iiiVT =  vq" (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 = Earndz 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 Lil94 {156 Nu = [1.023  11:03 P93 = {111123 [?.454x105)ﬁ'8 31.94103 _ = 11215.1 h _ m _ USE11126.1 = 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} ovariﬂe l‘.u1:Jv: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
clam1* {Tm  Tm} = Q. whate 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 "rmr“ ﬁﬂﬂ} 2T6 +311 3111.: (2 Tuna = Tm'. * 23.2 = 2,7911%: Page 4 0f 4 pages ...
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 Fall '08
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 Physics, Trigraph, TM, Wail temperature, Omanirm Tum wall, Tm E Primary, pep Vm dTm

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