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Unformatted text preview: 529M7an 79W ﬂay/1mm? 3 530.334 Heat Transfer ASSIGNMENT 8
Assignment due: Friday, May 5, 2695 by 4PM o Read chapters 7 and 8 in your textbook
0 Study solved problems in these chapter Problem 1. Ethyiene glycol ﬂows at (L02 1ch 5 through a 4— mm diameter, thin—waned tube, The tube is coiled and submerged in a well—stirred water bath maintained at 25"C ,.
If the ﬂuid enters the tube at 85"C, what heat rate and tube length are required for the
fluid to leave at 30"C '2 Neglect heat transfer enhancement associated with the coiling. Problem 2. To cool a summer home without using a vapormompression refrigeration
cycle, air is routed through a piastic pipe (k = (1.15 W/mnK , D, #0.! m, D" =On13 m) that is submerged in an adjoining body of water: The water temperature is nominally at
Tm MTG , and a convection coefﬁcient of 11" wiSOO W/mzK is maintained at the outer surface of the pipe. Ifair from the home enters the pipe at a temperature of TIN m 30?? and a volumetric ﬂow rate of ‘v’,=0.025 1713/5 , What pipe length L is needed to provide a discharge temperature of 77W, m18°C ‘2 What is the fan power required to move the air through this
iength of'pipe if its inner sulfaee is smooth? 60AM??? '. Problem 3. Water is to be heated from 15“C to 65 “C as it ﬂows through a 3*(337 inner
diameter 5—»: long tube» The tube is equipped with an electric resistance heater that
provides uniform heating throughout the surface of the tube. The outer surface of the
heater is well insulated, so that in steady operation all the heat generated in the heater is
transferred to the water in the tube. If the system is to provide hot water at a rate of 0.01 1123 (min, determine the power rating of the resistance heater Also, estimate the
inner surface temperature of the pipe at the exit. . gs = constant ;. 7:, at A
'1? gig? in ‘ were”. .a Problem 4. The surface of a 35111 long ﬂat plate is maintained at 40”C, and water at a
temperature of 4 “C and a velocity of 0.5 m/ s flows over the surface.
(a) Using the ﬁlm temperature T! for evaluation of the properties, calculate the heat transfer rate per unit width of the plate, (1' (W / m) . (1:) Calculate the error in {1' that would be incurred in part (a) if the thermophysical properties of the water were evaluated at the free stream temperature and the same
empirical correlation were used. (c) In part (a), if a wire were placed near the leading edge of the plate to induce
turbulence over its entire length, what would be the heat transfer rate? Problem 5. In an industrial facility, air is to be preheated before entering a furnace by
geothermal water at 120°C flowing through the tubes of a tube bank located in a duct“ Air enters the duct at 20°C and 1 atm with a mean velocity of 4.5 m/ s, and ﬂows over
the tubes in normal direction, The outer diameter of the tubes is 1.5 cm , and the tubes are
arranged in—line with longitudinal and transverse pitches of SL =5} : S cm . There are 6 rows in the ﬂow direction with 10 tubes in each row, as shown in the ﬁgure. Determine the rate of heat transfer per unit length of the tubes, and the pressure drop across the tube
bank. Office hours:
C. Herman, Latrobe 102 after class MW and by appointment PROBLEM 8.2.6 KNOWN: Ethylene glycol ﬂowing through a coiled, thin walled tube submerged in a well—stirred
water bath maintained at a constant temperature FIND: Heat rate and required tube length for prescribed conditions SCHEMATIC: 30
mfg? Tm: °c
nga’td“ "" " ' engine 8‘ z ‘73,,=25°c ' ASSUMPTIONS: (i) Steadystate conditions, (.2) Tube well thermal resistance negligible, (3)
Convection coefﬁcient on water side inﬁnite; cooling process approximates constant wall surface
temperature distribution, (4) KB, PE and flow work changes negligible, (5) Constant properties, (6)
Negligible heat transfer enhancement associated with the coiling B 1
Ba 5; «:0 gﬁafk
PROPERTIES: Table A5, Ethylene glycol (Tm = (85 +E°Cl2 m 0°C m 333 K): cp m 2562 7 .
I/kgK, tt== 07522 X 10 u N 5/1112, lc F» 07260 W/mK, Pr = 51.3 ANALYSIS: From an overall energy balance on the tube,
ﬂ oz "3818 2— 30
gem = ran up (Tm  itm) wmlcg/sx 2562 eagles," 85 )° C emu! (1) <
For the constant surface temperature condition, from the rate equation, As m cloonv I HATlini I m b (2)
3905 ' EEG2? {g . .22.! c AT  25
AT 2 AT HAT [tinMimi: ~25 ° C 85—n25 °C:}/ (In =2?.9°C7 3
Em ( o I) ATE (35‘ ) i ) 85_25 ()
Find the Reynolds number: to determine flow conditions,
4th 4X0, 1 I
RED 3 m m 03. cg s 2 2 1217 (4)
7r DH 2‘: x0301}, 1n><0 522 x10" N s/m Hence. the ﬂow is laminar and, assuming the flow is hilly developed, the appropriate con‘elation is —« 2.3?»
NuD “4 112=3..66, ii: Nu—IE‘: 3 66x0 260ﬂm/0iooimmm7W/mz K (5)
k D m K. From Eq (2), the required area, As, and tube length, L, are
zat3~> 25747 2 324 0.936 2
AS 51%} WISH W/m KineWC = serene n1 2 42.55
L =AS fit D :0, 1448111 I71: (0.003m) mm" < COIMMENT S: Note that for ﬁtlly developed laminar ﬂow conditions, the requirement is satisﬁed:
C12 2 (LID) 1’ RED Pr = (15 3/0003) 1' (813 X 51.3)= (1122 > G ()5. Note also the sign of the heat
rate qcmw when using Eqs. ( l) and (2) Pmlalm 2 : 6W
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a?“ = 0‘0263‘Wwe, Wr=07t>7 . . “1.1% L ‘ v I ‘
540% ’1 \le 7b.; ~ I? o .q many; I. . by , D
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(a) eurfme, Maci J911.3% [‘5 antimm
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I Ell H .33 33E “cams: Friction factor f and correction
factor x for {ulna banks (Imm Zuknuskus. Ref 16. 1985) FIGURE 7w27 iii?!" l
m
a
C
R
wk
Illwannwusum
I‘m“?! m
r. m
H c
m . w
n
m. m
d
m m
Nb
u..... m.
.m S
W .\. dude 53th “We dln'a bank. 3 f heat trarisfer per y. g'Air kw ﬁﬁﬁthérféémﬂér ii! #“Wﬁé '
5 [law direction with ID tubes Erie the rate 0 thmngh thé tubes 65 a tpbe'b'anlv. luca'ie
d th'e_ psassare drap miss the gube ; air is tq bé preheated henna eﬁteringa furna'ce by 3&0
duct A]! enters thg du‘ct at 20°cf'and 1 atm with is mean oajnc'uy cf 45 mls. _§he_tuba's hernia! diréclion. The b‘u’ger diam'éter'of thé 111ng Is
and thé tubes are arranged tilline wlth lunglﬁudlnal and transveme
N Alf Is haated by geothermal water tn 3 tube bank“ The rate. of heat d e .m m e I. e d E b m m l 3
Wm h mD a 50% .m m4 m 8:7. .m n Wu .El.m B
.ﬂ‘, 5.. “Nu r:
fa W. .Wu W
73.. D. .F._.l.... M.
...H..K.. H .m n.a F
tummy". C ELI. mm.“ D.
..P,..B‘.. .‘ .ﬂ. my .5 be .m
. .. "WE =.Dw t
31 Thu d fium 85“ .H 7! .mr 9. m
L mm w Lam m
E ma 0 WW.“ 0
n... dw Wpurmw U”
D: .mqm DMSHH “e
M I ECG I:
"m :lChBl. HUS nn 6... 65th.“ lﬂ
VA E H.“ n gm
pr. Inmrhm 3.19m“. cull
EEEEE‘ La wamm—rgeeegze’ ’.ﬁﬁiﬁfﬂ‘aﬁgrﬁzew _u..~.. ....  a u—nﬂé '  HEATTRANSEEWW» " V: 4 5 m 7;. 120%: m Assumpiions 1 Steady operating cenditiens exist 2 The surface temperature a;
T] u: we Cf the tubes is equai to the temperaluje of geothermal water "‘" O O O O G Pmperﬂes The exit temperature of air, and thus [he mean temperature, is nut
'. knowm We evaluate the air prepedies at the assumed mean temperature a: 59?? {wili be checked Eater) and 1 atm are Tabieaﬁ: M. (3 (3 Q (3 Q g k '2 002893 Wlm ‘ K. p m 1&6 icgi'm3 ~— (3 G 0 0 (“.9 (3 gammy/eye. ‘Pr=07202 M G) o (a) O O G) p, new; X._1E_i‘5§cglea~s 1.1.3:,I=Pr437,= 07073 H" 0 G O 0 O O ' éiee, the density pfaire‘ttﬁe inlettemperatme of 20%] (for use in the massiiew W O (D {19 G (9 C3 [aﬁe'eeiee‘iaijori atthe inlet) is pl 2" 102.01} kglma a a... .. a Q G {a a May's}; {it is given the; D "9: 9.0.15 m'. 5;. m Sr? 0 05 m, and “V = 4 5 :11]; Mags) 0 a O ‘ Then the maximuguieieciiy age! the Reyneiﬁfs number based on the maximum
SL=ST=5cm Dm15cm :ST', . are». (4:5 W = 543 "1’s wmgp _; (3.06 kgfm=xe43 ml5)(0.015 m)
u ﬂ X"ig"§kglms 7 FIGURE 7—28
Schematic for Example 7?? = 5093. emigrated ageing praper reieiiee [mm Table
53    =  N ' t 5:92." .'  . ~le dew!er
“ I 292197.073?!” = 5312.4th I 3' ;; "7'; "I . q 5 ' ' ‘ T032 137
{we Nueeeit aerate: is eppiieebie gq tube banks with N; > 18 in our case. the
 nun:qu 'rdweis NL @715. egg the: corresponding coneciien faster from 'i'abie
3}? i: 0,945.:Then the aqegage‘Nusseit number arid heat transfer mam" f0 .éii met as in the'teige Eankrbeceme .. :3I33'lH . . m. ' ' Res. J: '49.3{u.ozaoswrm = 922. Win13 i°C length“: m}. the heaut treesifer s'uﬁetge area and the mass ﬂow rate oi air
' (evaluateg ettneiniet} are 5 ‘ ' ‘ ~u '.  ﬁne {oi—Lil! number ef'tuhee [SW = ML X N; = 6 x 10 = 50 For n ueii lube : A, ewe We: mm me) :—~ 2.827 m?
§. ‘1. : ‘ '.. '.:'::...:  .1
ﬁr =.. rix: := ewes») . #05394, kgméites genomes mm m) = 2 709 kgls Theethe fluid exit termzxelraimeI the tag mean temperature difference, and 813
I reie pf heat transier beeeme ' HT“? (T—T}ex I: .r s I P may (2 709 kglsiiiﬂm mfg " um) w 29(1) C * 120  (120 ~ 20)cxp(— at, (fl9ED _ tT,— f.) w (I; I.) m (320  29.11) ~ (120  20) m a
M“ " lam".  Tana",  To] mitten « as miner)  '20)} 954 C
Q = MAT!" = (seawrm? “CW. 327 m3)l95 4°C) = 249 x 1&4 w The rate of heat transfer can atso be determined in a simpler way "Iran's Q = neon” 2 racprr, — r.)
= (2 799 against)? iflrg “ones 1: ~ zone a 249 >< :04 w For this square Errline tube bank, the irintion coefﬁcient corresponding to
HEg = 5088 and SLID =1 511 5 m 3.33 is, from Fig 7475. f = 0 16 Also,
x =1 I for the square arrangements Then the pressure drop across the tube hank becomes d J3 d X
vim . a «Pin 9"
APsNrixch 5K1? > ' sum {39% 1H“
(I 06 trglrn3)(6.43 info)J ( m a S“) Hill!) 2 1):: 21 Pa lirg‘rnls Discussion The arithmetic mean fluid temperature is (7} + Talia = (20 +
1109)]? =1 65 4‘0, which is fairly close to the assumed value of STD There»
tare, there Is no need to repeat calculations by reevaluatlng the pronertles at
65 riff} (It can be shatvrr that doing so would change the results by less than 1 percent. which is much less than the uncertainty in the equations and
the charts used). ,,__m.m.__.W——H.W . “'"tfn‘winﬁttl'ugﬁgtﬁvTH f.
assassinateviii  , em m...rmge.t:n‘.:Jr if? omit? Thermal insulations are materials or combinations of matcn'aIs that are
used primarily to provide resistance to heat ﬂow (Fig. 7&9) You are prob»
“my familiar with sovorai kinds of insulation available in the market Most
insulations are heterogeneous materials made of low thermal conductivity
materials, and they involve air packets This is not surprising since air has
Gm: of the lowest thertnai conductivities and is readily ovoilainte The Sty
mfaam commoniy used as a packaging material for TVs. VCRs, oomput~
(its. and just about anything because of its tight weight is aiso an excelient
Insulator , Ternpemtrrre diﬁismttce is the driving force for heat flow, and the greater
“it: 1t‘vlrtporatorra difference. the larger the rate of heat transfer. We can slow
"“two the heat ﬂow between two mediums at different temperatures by
smiling "barriers" on the path of heat ﬂow Thelma] institutions serve as
such barriers. and they piay a major role in the design and manufacture of
“‘5 enorgyefﬁoien: devices or systems, and they are usually the cornerstone
0f energy conservation projects A 1993 Drexel University study of the
a“""EiMntensive {LS industries revealed that insulation saves the U S W "(his section can be stunned without a loss in continuity Lg' FIGURE 729 Thermal insulation retards heat
transfer by acting as a barrier in the
path of heat flow ...
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 Spring '08
 HERMAN
 Heat Transfer

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