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Unformatted text preview: \ I . ' PROBLEM92 chl §ollxlﬁm KNOWN: Object with speciﬁed characteristic length and temperature diﬁ'erence
above ambient ﬂuid. ‘ '
FIND: Grashof number for air, hydrogen, water, ethylene glycol for a pressure of 1
atm. . ' Quiescen+ fluid a
7;,157‘721 ASSUWTIONS: (1) Thermophysical properties evaluated at Tf = 350K, (2) For
gases, ,5 =‘1/Tf PROPERTIES: Evaluate at 1 atm, Table 114, Air: U=20.92X10—6 m2 /s; Hydrogen: u= 143x10*6 In2 /s
Table A6, Water (Sat, liquid): u=pf vf =37.5><1o6 m2 /s, a =0.624><10'3 K‘1
_ Table A5, Ethylene glycol: U=3.17><10_6 rnz/s, ﬂ=0.65x10‘3 K‘l. Tf 4— 350K: ANALYSIS: The Grashof number is
Gm = stirsTog)? _
V2
Substituting numerical values for the ﬂuid air with ,6 =' l/Tf, ﬁnd GrL _ 9.8m/s‘2 x (1/350K) (25K) (0.25m)3
’ ' '— 3" (20;92x10 8rug/s)2 given by Eq. 9.12, Gm“, = 2.50x107 . <1
Performing similar calculations for the other ﬂuids, ﬁnd
Gmyd = 5.35x105 . .<]
Grimm, = 1.70m),6 <l
— ’ ,9 . . . <1
COMNIENTS: Higher values of GrL impltr increased free convection ﬂows. However, other properties affect the value of the heat trans fer coeﬂicients. Note that for the
gases, ﬂ = 1 /Tf, assuming perfect gas behavior. ' m W . rafﬂe/L T/fo €39 ' ‘ _ p ' PROBLEM 9.92 KNOWN: Cross ﬂow over a cylinder with'prescribed surface temperature and free stream
conditions. '  FIND: .Whether free convection will be signiﬁcant if the ﬂuid is water or air. SCHEMATIC: D=50mml Cylinder V=0.05m/s —+i> L20 C 7;:55‘73 ASSUMPTIONS": (1) Uniform properties, (2) Combined free and. forced heat transfer
situation.  . a . ' PROPERTIES: Table A6, Water. (T§=(T.,+Ts)/2=300K): V v=ttvf=855><10‘6
Ns/m2x1.003x103 m3/lcg=8.576x10‘7.m2/s, =276.1><10',6 K‘l; TableA‘4, Air (30032.1
arm): v=15.89x10‘5 mZ/s,a=1/r;=3.333x10 Kl. ~ . ANALYSIS: Following the discussion of Section 9.9, the general criterion for delineating the
a relative signiﬁcance of free and forced convection depends upon the value of Gr/Re3. If free
convection is signiﬁcant, _ ' ' _ » . ' ‘
‘GrD/Re2D21.'  .  3 ' ' «(1)
where t ‘ _ .
.ch = g‘B (T..,—T..,)D3/v2  and I RC]; = Vii/v. I  (2,3)
(a) When the surIOunding ﬂuid is water, ﬁnd ‘ '
GrD = 9.8 m/s2 x 276.1x10‘5 K:1 (35—20)K (0.05m)3/(8.576x107 1312/5)2 = 68., 980 to, 'ReD=0.OS m/sxo.oSm/8.576x107 mzls = 2915 “C
Q S7 CLSZOHO , » ' {Big
CED/Reg = sense/29152 ==9£08T2f.(9, g I  b/ WM 7% mad ‘ 4 We conclude that since GrD'lReIZJ << 1, free convection is w‘signiﬁcant. It is apparent that
forced convection dominates the heat transfer process. ' (b) When the surrounding ﬂuid is air, ﬁnd' ‘ ,
GID = 9.8 m/s2 S< 3.333x10’3 K‘l _(35—20)K(0.058803/(15.89x1o6 mgA/s)2 = 242.558
RED = 0.05 m/s x O.O5m/15.89x10‘6 mz/s = 157 GrD/Rc% = 242,558/1572 = 9.8 . ,  ' i <1 We conclude that, since GrD/Reﬁ >> 1, free convection dominates the heat transfer process. COMMENTS: Note also that for the air ﬂow situation, surface radiation exchange is likely to be signiﬁcant. r  y . A? [1407 M90 Calm/(an [(A4 [9.07% am it ComﬁMQU ' ‘ g PROBLEM 11.4 ' KNOWN: Dimensions of heat exchanger tube with or without ﬁns. Cold and hot side
convection coefﬁcients. . . : V FIND: Cold side overall heat transfer coefﬁcient without and with ﬁns. SCHEMATIC: [esow ‘FiTl 00W/,,,a./< +.=2mm§, ‘N=16 ASSUMPTIONS: (1) Negligible fouling, (2) Negligible contact resistance between ﬁns and
tube wall, (3) hh is not affected by ﬁns, (4) Onedimensional conduction in ﬁns, (5) Adiabatic .
ﬁn tip. . a _ ANALYSIS: Froni Equation 11.1, L = 1 + DimCDn/Di) +  Ac (Home 2k ("PhAhl "
Without ﬁns: TIM =1‘\'(,‘}1 =1. ' . . .
a _1_ 1 + (0.02 m)ln(26/20) + a 1 20 ' U. = 8000 W/mzK 100 W/mK . 200 W/mzK 36— ‘ ‘
1/Uc = (1.25x10‘4 + 5415x10s Jr 3.85x10‘3) mZK/w =.4.oz><103 mzK/W Uc=249WIm2K. '. _ . " _ v .. V . 4
‘ Withﬁns: 110,; = 1,110... = 1  (Af/Ax1  hf) Per unit length along the tube axis, A;=N(2L;+t)=16(30+2)mm=51in1m ’ ‘ ' '11,, = Af+ (11D. — 160 = (512+ 81.7  32)mm = 561.7 m
I I With m = (21mm = (400 W/mzK/SO whim1:20.002 m)”2 = 53.3 m1
t (63.3 m1)(0.01s m) = 0.95 i ‘ '
and Eq. 11.4 yields ' x ' " ,
'ﬂr é tanh(er)/mLr = 0.739/0.95 =, 0.7713.  The overall surface efﬁciency. is then ' ‘ ‘11., = l — (As/Ath — 115) = 1 — (512/561.7)(1 f 0.778) = 0.798. , .. 1 ‘ 1: ySéL
é;l(/LJI0,VLS F 1 .4 . ‘ .5  I 2 ‘ V 4
.._. = 2 + . __.___.__.  — . .
Hence. UP. (1. 5x10 5 25x10 + 0.798(200)56L7 m m g73;.<104 m 1 UC _ Vilmzi'K. ' 1 ‘ . f I ‘  < ‘ COMMENTS: The ﬁns provide signiﬁcant heat transfer enhancement. ' v ' ‘ ' , PROBLEMIIJO KNOWN: A shell and tube Ileer (two shells, four tube passes) heats 10,000 kg/h of
pressurized water from 35 ° C to 120 ° C with 5,000 kg/h Water entering at 300 C. FIND: Required heat transfer area, A4,. _ scant/mirror ‘ ,
777150012 Th  Ll=1500w/m2/<
‘ rp'zaﬂzofc. Lf{>7i2f5000kg//1 A72 mc =10,000kg//1
l a . ASSUMPTIONS: (1) Negligible heat lossto surroundings, (2) Negligible kinetic and
potential energy changes, (3) Constant properties. ' PROPERTIES: TabliAB, Wat'e1"'('.[‘c =350K):‘ cp =419'5 J/kgK; Table A6, Water
(Assume TM, {3150 ° C, Th 8500K): cp =4660 J/kg‘K." p A ' ‘ ANALYSIS: The rate equation, Eq. 11.14, can be written in the form  _
‘ As = Q/U ATrm ‘ i i ' 3 (1)
andf’rom Eq. 11.18, ' ' ' i. H
AT1.— AT2 ATgm = F ATgm'GF Where I ATngCF = m . (2.3.) From an energy balance on the cold ﬂuid, the heat rate is _ ‘ 'q= 1h, cp,c_(Tc,c,4Tc,i) = W x 4195'ﬁ(120—35)K = 9.905x105 W . From an energy balance on the hot ﬂuid, the outlet temperatureis 5000 kg' J ‘ .
———— 4660—— = 147 C .
. . 3600 s x, kgK 
From Fig. 11.11, determine F from values 'of P and R Where'P =(120—35)°C
/(300—35)°C =0.32, R=(300—147)‘°C /(120—35)°C =1;8 giving F~0.97. The
log mean temperature based upon OF arrangement follows from Eq. (3); ﬁnd Thin .= Th), _ q/Ihh cm}, = 300" C — 9.905x105 W/ 300—120) '
AT = .300—120 — 147—35 K 4? (———— = 3.3 .
A,’= 9.90.5x105W/1500W/m2‘K><0.97><143.3K =4.7.5m2 . _ . 0. <1 QOMMENTS:  (1) Check Thz500K used in: property ' _ determination;
Th =(300+147) ° 0/2 =497K. (2) Using the NTUE method, determine ﬁrst the capacity
rate ratio, Cm], /Cmax = 0.56 . Then ' , E: q = Cmax(Tc,o—Tc,i) = 1 X (120—35)°C =057
qmax Cmin (Th,i_Tc,i) 0'56 (300—35) a C ' I From Fig. 11.17, ﬁnd that NTU = AU/Cmi,1 $31.1 giving As =4.7 m2. é _ ' PROBLEM 11.24 KNOWN: ConCentric tube heat exchanger operating in parallel flow (PF) conditions with a thinwalled '
separator tube of lOOmm diameter; fluid conditions as specified in Problem 11.23. FIND: (a) Required length for the exchanger; (b) Convection coefficient for water flow, assumed to be
fully developed; (0) Compute and plot the heat transfer rate, q, and ﬂuid inlet temperatures, Tm, and Tm,
as a function of the tube length for 60 s L S 400 m with the PF arrangement and overall coefficient (U = ZOOW/m2 K), inlet temperatures (Th, = 225°C and Tc, = 30°C), and ﬂuid flow rates from Problem 11.23; (d) Reduction in required length relative to the value found in part (a) if the exchanger were i
' 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 m for the CF arrangement of part (c). SCHEMATIC: '
u = 200 W/mzK
TM: 225 "C _ a ‘
T TN, 100 c
a, /—
\D‘ ,
A7,,
“’ —»
To): 30 °C Tm, = 50 °C (D ' ®
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 properties are those of atmospheric air. PROPERTIES: Table A4, Hot ﬂuid, Air (1 atm, T: (225 +100)°C /2 = 436 K): cP = 1019 J/kgK; Table A6, Cold fluid, Water T = (30 + sow: /2 = 328 K): p = 1/v, = 935.4 kg/mﬁ, 'c, = 4183 J/kgK, k = '
0.648 W/mK, u = 505 x 10'6 N‘s/ml, Pr ; 3.58. ' ANALYSIS: (a) From therate equation, Eq. 11.14, with A ; 1rDL‘, the length of the exchanger is
L=q/U.nDA'Tg,.pp  . . " . (I)
The heat rate follows from an energy balance oni'the cold fluid, using Eq, 11.7, find
q = rhccc(Tm ~T,,) = 3'1cg/s >< 41831/kg K(80— 30)K = 6275 x103W.
Using an energy balance on the hot ﬂuid, find r'nh for later use. I
En, = q/ch (Tm — T“) = 627.5 >‘< 103 W/1019J/kgi K(225'—100)K = 4.93 kg/s (2)
For parallel ﬂow, Eqs. 11.15 and 11.16, ' AT — AT (225— 30)°c —(1oo—so)°c,
AT =——1——“"‘—=———————=763°CI
“m” tnAT, /AT2 €n(225— 30)/(1oo— so)
Substitutingnumerical values into Eq. (1), find ‘ I V
L=6275x1o‘w/200W/m2 K(itx0.1m)76.8K=130rn. ' ' ‘ < . Continued... PROBLEM 11.24 (Cont) (b) Considering the water flow within the separator tube, from Eq. 8.6,
ReD = 4th/1tDtL = 4 x 3kg/s/(1t X 0.1m x 505 X 10—5 N/s  m2) = 75,638. I
Since Rep > 2300, the flow is turbulent and since flow is assumed to be fully developed, use the Dittus ' 4 Boelter correlation .With n ; 0.4 for heating; ' ' ' '
NuD ; 0.023 Reg8 Pr“ = 0023054533)"8 (358)"4 = 306.4 h=NuDg=3O6.4><O.648W/mK/(0.1m)=1935w/m2,Kl ' ' < (c) Using the [Wm Concentric Tube.'Parczllel Flaw, Eﬁ‘eciiveness relation, and'the
Properties Tool for Water and Air, a model was developed for the PF arrangement. With U = 200 W/rnle and prescribed inlet temperatures, TM = 225°C and Tel = 30°C, the outlet temperatures, TI”, and _'
Ten. and heat rate, q, werepcornputed as ‘a function of tube length L. Parallel llnw arrang'amanl . o aw P0300“ 115
90 65 Temperature (C) or Heal BIB (W‘10A4) 4o . . V .
BO ' 120 130 240 300 350 Tube length. L (m) ', N Cold outlet lemperalure.Tuo (C)
9— Hot outlet tern erature. The (C)
— Heat rate, q'1 (W) As the tube le‘ngth'increases, the outlet temperatures approach one another and eventually reach Tm = ‘ I H
Tc‘g = SSIEOC~ ' > ‘ . ‘ (d) If the exchanger as for part (a) is operated in T counterflow (rather than parallel flow), the log . mean temperature difference is , ' _ . p ' Th,/= 225 °C ‘ ' ‘ _ ATw, _  3;}:
AT =——‘———=— , , TI =80 °o
£m.CF enATl /4T2 . t , ca (223—80)—(100'— 30) ATgmﬂF = “1(225 _.80)/100  3O = 103.0”C . 'Using Eq. (1). the required length is ' .
' L=627.5x103W/200W/m3Kx1tx0.1mx103.OK=97m. The reduction in required length of relative to PF operation'is ,
 ' ‘ ' Continued... PROBLEM 11.24. (Cont) I? 0 \‘30. v  ‘
ﬂea/er =(l195f97l/‘m=5£‘% ' < (e) Using the [H T Heat Exchanger Tool, Concentric Tube. Counterﬂow, Eﬁectiveness relation, and the '
Properties Tool for Water and Air, a 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. ' ‘ J“ i :140
120 100 Countadlow arrangement Issesiiae...iiiuu so iliiﬂlllllllllll \ e . _ to IIIIIIIIEE!!!IIII
~o)\l\‘ . 2° IIIIIIIIII
' 50 120 130 240 300 >\ I I . Tubs length. le) —>+— Gold outlet temperature. Ten to)
9— Hal outlet temperature. The (C)
— Ettacllvanese.eps'1uu BO Tho, Tm (C) or eps‘mﬂ Note that as the length increases, the effectiveness tends toward unity, and the hot fluid outlet temperature tends toward Tu = 30°C. Remember the heat rate for an infinitely long CF heat exchanger is
qm and the minimum ﬂuid (hot in our case) experiences the temperature change, TM ‘Tc_i. COMMENTS: (1) As anticipated, the required length for CF operations was less than for PF operation. (2) Note that U is. substantially less than hi implying that the‘gasside coefficient must be the controlling
thermal resistance. r ~ ' ' ' t,  ’ ...
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This note was uploaded on 06/06/2008 for the course ME 114 taught by Professor Okamoto during the Spring '08 term at San Jose State University .
 Spring '08
 Okamoto
 Heat Transfer

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