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Unformatted text preview: PROBLEM 8.70 KNOW N: Inner and outer radii and thermal conductivity of a teﬂon tube. I‘lowratc and temperature
of conﬁned water. Heat ﬂux at outer surface and temperature and convection coefﬁcient of ambient
air. FIND: Fraction of heat transfer to water 11
. . . . . 911
and temperature of tube outer surface. T
m
SCHEMATIC: 010103"
_9.p E0
EHé'Q/rf)
ZﬁLk
r0 =13mm
(hiAil‘I
Teﬂon, 9:! 7;”
k=0.55WlmK m ’ ASSUMPTIONS: (I) Steady—state conditions. (2) Fully~devclopcd flow, (3) One—dimensional
conduction, [4) Negligible tape contact and conduction resistances. —(
PROPERTIES: Table A43, Water (Tm = 290K}: it I 1080 x 10 ) kgrlsm, k = 0.598 WhitK. Pr = 7.56.
ANALYSIS: The outer surface temperature follows from a surface energy balance
T 4‘... r.  r
(2?! equ”=L_l+#
(1102111101.) (111(10 Mimi: Lk]+(11’21r 11m.)
Tso * Tin ”:1 '1‘. 40., —_
q 10( 5‘0 )+(r0fk]ln(rbfri)+(1‘01’r'1]1"'hi With ReD =4 rr11"[71' Dy) :4[0.2kgfs)f[rr[0.02in)1080xl(}_6 kgr's m] :1 1.789
the ﬂow is turbulent and liq. 8.60 yields r1] : (loDi )0023ReD’S P104 = (0.598 W."m K1002 m)(0.023 )(I 1,?09)4"5(756)U'4 : 2m: wrm2 K. llcnce
T SI 0 —290 K 2000 Via«1113 = 25 was1112 K (TM, 300K )+ 7
(00131111035 wrm K)111 (1.3}+(1.3).‘[2792 \\'.'1111K) and solving for T510. Tsﬁ. = 308.3 K. <
The heat flux to the air is L113: h11lTs.11— Too] 2 25 vii/1112 K (3003—300) K : 207.5 w..'1113. Hence. q"; 111’: (2000 207.5) w1‘11121'2000 won? 2 0.00. < COMMENTS: The resistance to heat transfer by convection to the air substantially exceeds that due
to conduction in the tellon and convection in the water. Hence. most or the heat is transferred to the
water. PROBLEM 3.76 KNOWN: Fluid enters a thinrwalled tube of SAmm diameter and 27m length with a ﬂow rate of 0.04
kgfs and temperature of Tm‘i = 85°C.; tube surface temperature is maintained at T8 = 25°C; and, for
this base operating condition, the outlet temperature is Tm0 = 311°C. FIND: The outlet temperature it'the flow rate is doubled? SCHEMATIC:
T5 = 25°C
Tmib = 31.300
' ”Tr”? 7
Tmli = 85°C M)” " '
m1, = 0.04 kgls Thinwalled tube
_ or _ _ —————"i D = 5 mm
rnn = kaJ ie———————_ 1 ‘ 2 m ASSUMPTIONS: [1) Flow is fully developed and turbulent, (2) Fluid properties are independent of
temperature, and (3) Constant surface temperature cooling conditions. ANALYSIS: For the broe operating condition [b]. the rate equation. Eq. 8.42b. with C = rhcp. the capacity rate, is TS _Tm._.0)b = exp _ PLEb
Ts ’de Cb (1) Substituting numerical values, with P = TED, ﬁnd the ratio, hb J'Cb,
25 v 31.1 —
m _ exp[n' x0005 rn><2 m(hb rcb )] Eb rcb = 72.77 m_2 For the new operating condition (n), the flow rate is doubled, C” = 2Cb. and the convection coefﬁcient
scales; according to the Dittus—Boelter relation. Eq. 8.60, hi Rc%8l rho8 En 2 203m, and (En Jen): 208 r2(Eb/Cb) (2)
Using the rate equation for the new Operating condition. ﬁnd
mzexp[_PLHn ]:exp[PLXO.87l(hb/Cb)] (3)
15 ‘TmJ C11
25 —Tm!0) —” : cxp[—jr><0.005 m><2 m><0.87l X7277 III—2]
25—85 PROBLEM 11.33 KNOWN: Concentric tube heat exchanger operating in paratlel flow (PF) conditions with a thinwalled
separator tube of lOOntm diameter; ﬂuid conditions as speciﬁed. FIND: (a) Required length for the exchanger; (b) Convection coefﬁcient for water flow. assumed to be
fully developed; to) Compute and plot the heat transfer rate. q, and fluid inlet temperatures, TM and '1"
as a function of the tube length for 60 S L S 400 m with the PF arrangement and overall coefﬁcient [1.1: 200w/m2  K). inlet temperatures (rm = ’22an and 'rm = 30°C). and fluid flow rates from Problem 1123; ((1) Reduction in required length relative to the value found in part (a) ifthe exchanger were
operated in the counterﬂow (CF) arrangement; and (e) Compute and plot the effectiveness and fluid
outlet temperatures as a function of tube length for 60 S L S 400 tn for the CF arrangement of part (c). (2.0! SCHEMATIC: o E u=2oowrm2K
r :225 0
Q) T "" hrmzmot’c \Ll ;.(s;;;;p ASSUMPTIONS: (1) No losses to surroundings, (2) Negligible kinetic and potential energy changes.
(3) Separation tube has negligible thermal resistance, (4) Water ﬂow is fully developed, (5) Constant
properties, (6) Exhaust gas prepetties are those of atmospheric air. PROPERTIES: Table A4, Hot fluid. Air(l attn. T = (325 +l[}0)°C I2 = 436 K): on 2 1019 J/kg‘K: Table A6. Cold fluid. Water T: (30 + 80)°C l2 2: 328 K): p = l/vr: 9854 kg/m". cl, : 4183 Jfkg‘K‘ k =
0.648 W/mK. u = 505 x 10"” N‘s/ml. Pr = 3.58. ANALYSIS: (a) From the rate equation. Eq. 11.14, with A = TIDL. the length of the exchanger is
l. =q/Uerarrn‘PF. (l) The heat rate follows from an energy balance on the cold fluid. using Eq. 11.7. find
q = mccc (Tm —'1‘L.,i ) : 3kg/sx41831/kgK(80—30)K = 627.5xtn3w.
Using an energyr balance on the hot fluid. find n1“ for later use. mh = q/Ch ("on ﬁrm ) = 627.5x103 w 10191;"ikg . K (225 —100)K : 4.93. kg/s (2)
For parallel How. Eqs. 11.15 and 11.10. an — er; _ (3:5 — 30)“ C — (loo —: 80)° t'.‘ A'nm H. — . _ — _ J _ 768°C.
' mathat: 111(225 — 30),..(100—30)
Substituting numerical values into Eq. ( l ). find
.= 027.5 x103 xiv/”gooair”m2 K(rr><().1m)7o.8K : 130m. < Continued... PROBLEM 11.33 (Cont.) (b) Considering the water flow within the separator tube, from Eq‘ 8.6,
ReD = 4zn/7rDtt : 4x3kgf’s/(rr XOmeSOleO—6 N/s n12 ) : 75.638. Since Rm > 2300. the flow is turbulent and since ﬂow is assumed to be fully developed. use the Dittus
Boelter correlation with n = 0.4 for heating, NuD = 0023113958 Pr0‘4 = 0.023(75.638)0'8 (358)0'4 = 306.4 k .
h : Nun — = 306.4x0.648W/tn<K/(0.lm) =1985W/m2 ‘K. <
D
(c) Using the [HT Heat Exchanger loaf, Concentric Tube. Parallel Flow, Effa‘tivcncss relation, and the
Properties Tool for Water and Air. :1 model was developed for the PF arrangement. With U = 200
\Wln“K and prescribed inlet temperatures. '1".,_, = 225°C and Tc.i = 30°C, the outlet temperatures. Th0 and
Tm and heat rate, q. were computed as a function oftuhe length I... Parallel llow arrangement i; 140 1
a: ' l a  = 
.— . . 1....___I.__+. ..
a l l l .l !
3 Its : i ; . . .
E l I l
a; l : E . . —"'————"'."—'
E i l i l
6 90  _ .“i
Q i ' i
a: i
g 65
H)
D.
E 
.3  . 4t) . I ———¢; ' . 50 120 130 240 300 350 Tune length, L {m} ‘’*— Cold outlet temperature. Tco (CJ
“9— Hot outlet temperature The {C}
Heat rate. q‘tD‘Nd (W) As the tube length increases, the outlet temperatures approach one another and eventually reach Tm, :
TN, = 816°C. (d) lithe exchanger as for part (a) is operated in T
Counterflow (rather than parallel flow). the log
mean temperature difference is Tm: 225 DC
AT — A'l‘
ATl“ m C1: '— l ‘7'";' Too = 80 DC
‘ {HATI \12 r—\
Ix.)
I
 2 80)—(00—30) .
—.—= 103.0 (7.
m [225 — 801.5100 — 30 [_Jsing Lit]. (I). the required length is L:(:27.5><103WfEOOW/n12K xn x0.lm><103.0l< 2 97 m.
The reduction in required length 01' CF relative to FF operation is PROBLEM 11.33 (Cont.) AL = (LH; —L(_;I.)/LPF 2(103797)/103= 5.8% < (C) Using the IHT Heat hitclronger Tool, Concentric Tube. Cotmterﬂow, hﬂec'tiveness relation. and the
Properties Tool for Water and Air, 21 model was developed for the CF arrangement. For the same
conditions as part (c). but CF rather than PF. the effectiveness and fluid outlet temperatures were
computed as a function of tube length L. Counterﬂow arrangement 140 120 100 80 50 Tho, Too (0] or eps'too 40 20 . . .
60 120 180 240 300 360 Tube length. L (m) #7 Cold oullet temperature. Too [8}
—9— Hot outlel temperature. The (C!
— Ettectrveness, eps‘t 00 Note that as the length increases. the effectiveness tends toward unity, and the hot ﬂuid outlet
temperature tends toward TW = 300C. Remember the heat rate for an infinitely long CF heat exchanger is
(1...... and the minimum Fluid {hot in our case) experiences the temperature change. Tm  '1}... COMMENTS: (1) As anticipated. the required length for CF operations was less than for PE operation. (2) Note that U is substantially less than hl implying that the gas—side coefficient must be the controlling
thermal resistance. PROBLEM 11.69 KNOWN: Inlet temperatures and flow rates of water (c) and ethylene glycol (h) in a shell—and—tube heat
exchanger (one shell pass and two tube passes) of prescribed area and overall heat transfer coefficient. FIND: (:1) Heat transfer rate and fluid outlet temperatures and (b) Compute and plot the effectiveness, E
and fluid outlet temperatures. Tm and Tim as a function ofthe ﬂow rate of ethylene glycol. tin1 . for the range 0.5 S iiih S 5 kg/s. s SCHEMATIC: l 1;”,th = 2 kgts ' r
One Shell Too
pass and
two tube
passes Tam n'qc : 5 kgls Ethylene glycol ‘— Tw = 10 ”C. water
l Tim A =15 m2. U = 300 wrmZK ASSUMPTIONS: (l) Negligible heat loss to surroundings. (2)I Negligible kinetic and potential energy
changes, (3) Constant properties, and (4) Overall coefﬁcient remains unchanged. PROPERTIES: Table 45, Ethylene glycol {Tm = 40°C.): c. = 2474 J/kgK; Tabché, Water (Tm =
15°C): c. = 41 so J/kgK.
ANALYSIS: (3) Using the erNTU method we first obtain
Ch = (mhcp’h ) = (2kg/sx2474J/kg  K) : 4948 w/K
cc =(meow):(5kg/sx41361/kg K): 20,930w/K.
Hence with Cmm = Ch : 4948 W/K and Cr = Cum/Cm. : 0.236,
UA (sonwr’in2 K)15m2 N'ru : =———=2.43.
Cmin 4948 W] K
From Fig. [1.16. 8 = 0.81 and from Eq. 11.23
q 2 ecmin (TM :13. ) = (181(4948 W/K)(60—10]K : 2x105 W. <
From qu. 11.6 and 11.7, energy balances on the fluids,
5
0 2x10 W O
Th.,:’1*hi—i=60 01,:1943 c <
' ' ch 494swn<
S
., 2x10 w o
rc(,=rci+i=10C++=19.ec. <
1 ‘ cC 20,930w,ri<; (b) Using the IHT Heal? Exchanger Tool, Shell and Tube. and the Properties Tool for Water and .8: 100
Ethylene Glycol. Tm. T1,... and E as a function of g so 4
m}. were computed and plotted. g :3
O
r:
At very low Cmm. (low l‘hh ) note that E ——> 1 while E 2: _
Tim—Sr '1‘“... As mh increases. both fluid outlet F
temperatures increase and the effectiveness Hot fluid llow rate. rndoln (kgis) decreases. + TeolC}, max ﬂuid
+ Tho (C), min lltlld —~— Elfectlveness.eps‘100 ...
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 Fall '08
 Rothamer
 Heat Transfer

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