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Unformatted text preview: mm ‘ BaumLu 7.1 Air at a pressure of 6 kam2 and a temperulure 01' JUNE ﬁlms with a velocin of
1G W5 over :1 Hal plate {1.5 m long. Estimate the mmling rate per unit width of the
plan: needcd tn maintain it at a Surface 1emperature n’r‘ 21°C. ﬁrm I'I‘II'IN Known. Airﬁow over an isnthumal ﬁat plant.
Find: lelng raic per unit width of the plain q' [Wini]. Srhmlmﬂ'r: ‘ T_ I SUDJC —II a“ =10 mfs —l*
p,=5 kNa‘rnz—JI 1: 2m; i
 L—L ELLEirI'I —H
i L” Assumptions:
1. Steadystate conditions.
2. Negligible radialidn effects. Properties: Table n.4, air ti"I = 43? K. p = I mm}: P = 30.84 x it)"I
it: = 36.4 x in", Wi'm ' K, Pr = 0.531 Properties such as It. Pr, and p. Ina}r
assumed to be independent of pressure to an excellent approximation. However.
a gas, the kinematic viscosity P = mp will vary with pressure through its
denee on density. From the ideal gas law. ,0 = MT. it follows that the ratio at '
matte viscosities fur a gas at the same temperature but at different pressures, p.
p2. is intuit m (Mpg. Hence the kinematic viscosity of air at 43? It. and p, = ti
10“ Him: is 1 2
LUIJJ X 112* Mm = 5m x "r; it
IS x 10’ Him? m 5 v = 30.34 x it)"tn:ts x Analysis: Fer a plate of unit width. it follows from Newtﬂn's law of coating
the rate of convection heat transfer to the plate is q' = Eur.” m T3 T0 determine the appropriate convection comlatter: for computing 3, the Reyn
number must ﬁrst be determined Hut'3 iﬂmts x {1.5 m
R = —=—=959'i‘
6' P 5.21 x10—4mtis Hence the ﬁﬂw is laminar over the entire plate, and the appropriate correlation 'a
ﬁiven ’0? Equation 73$. N—HL = damsel” Fri” m omotsssti'ﬂtoosn'” = 514 The average convection coefﬁcient is then _ —lt ' 1
PM? =514EW=4szimnK {15 m
and the required cooling rate per unit width of plate is q’=4.13WIm2'K.XD.5ml3m2'?}"C=5?DWIm <1 Ennuunnts: The results of Table A»! apply u: gases at ammopheric pressure
Except for the kinematic viscosity. mass density. and thermal diffusivity. the}! may
generally be used at tailtor pressures without commotion. The kinematic viscosityad
thermal dil‘l'usivltjlr for pressures m! then ] stm may be obtained by dividing the
tabulated value by the pressure (atm}. Emma 12 A ﬂat Flatt: uf widﬂ‘: w = l m is maintajmd at & unifurm surface temperaturc,
'1'} = 230°C. by using independentty cuntrulled. electrical strip hcamrs. each at which
is. 51} mm long. If annmphedc air at 25°C ﬂows out: the plat: at a vtlatity of 60 1111's.
1!! what heater is the elecutcal input a maximum? What is the value oFthjs input? Snwrmn Known: Airﬂow uvtr a ﬂat plate wilh aegmtnted heaters. F ind : Maximum heater pnwur rcquirement. Sthnmir: assumptions:
1. Steady—state conditions.
I. Negligible radiation effects.
3. Bottom Surface of plate adiabatic. Properties: Table n.4, air ii"Jr = 44]} K. p = 1 atm}: I' = 215.4] X lﬂ'ﬁ mzi'a.
it = iliﬂlli Wim  K. Pr = 0.6911 Analysis: The location of the heater requiring the maximum electrical power
may be determined by ﬁrst ﬁnding the point of boundary layer transition. The
Reynolds number based on the length L, of the ﬁrst heater is "all 2 fit] mis X ﬂﬂﬁ m =1_iai<idS
" zeal x lﬂ'ﬁmzfs Rm = If the transition Reynolds number is assumed to he Relic = 5 X ICP. it follows that
transition will occur on the ﬁfth heater. or more speciﬁcally at _ tr _ 26.4l X lﬂ'anfa r _
,Irc — Hm Re”. — wmm 5 K II] — {1.22 TI': The heater reoairing the maximum eiemrical power is that for which the average
convection coefﬁcient is largest Knowing hoar the local convection coefﬁcient Titan 1' .4 Constanta of Equation 1.53 for the circular
cylinder in cross ﬂow [lift] If” C m
Ito tits on
dentin cat on
103—2 x it? one on 2 x ire—toﬂ care at a cylinder in cross ﬂow. it is simply a matter of rcplac'tng ED by ED and Fr by 5:.
mass transfer problems, boundary layer property variations are typically small.
when using the mass transfer analog of Equation 153, the property ratio.
accounts for nonconstant property effects. may be neglected.  EWPLE 1' .4 Experimean have been conducted on a metallic cylinder 12.? mm in diameter
94 mm long. The cylinder is heated internally by an electrical heater and
subjected to a cross flow of air in a towspeed wind tunnel. Under a speciﬁc set
operating conditions for which the upstream air velocity and temperature at _
maintained at V = II} mfs and 262°C. respectively. the heater power dissi ‘
was measured to be P = as W. while the average cylinder surface temperature it
determined to be i"1 = HEAT. It is estimated that IS% of the power dissipatan lost through the cumulative effect of surface radiation and conduction throuﬁi
endpieces. Thermoc on pie for I‘I‘IEHHI'II'IE
airstream temperature " Heated F Insulated
cylinder endptece P'rtot tube for _
determining velocity Thmbﬂﬂh tends Potter leads
to electrical heater 1. Determine the convection heat transfer coefficient from the experimenul
observations. 2. Compare the experimental result with the convection coefﬁcient computed 'rl‘ﬁl'l'l n.“ enamelsliens ﬁhfl‘allﬁl;hﬂ SOLUTION Known. Operating conditions for a heated Cylinder. Find:. Convection coefﬁcient associated with the operating conditions. 2. Convection coefﬁcient from an appropriate conelation. Schematic:
T, = 262°C T,=128.4°c 
q = 391w
‘—
——I
P = we w I
D = 12.? mm
alaarurrptr'ons: l. Steadystate conditions. 2. Uniform cylinder surface temperature. Properties: Table n.4, air {Ta = 262°C 2 300 K): v: [5.39 X It)"El mitts,
t = 215.3 X It)" mm  It. Pr = 0.701 Table n.4, air (T! a: 350 K): p = zoo: x
ID" mas. I: = 30 x It)" th K. Pr = moo. Table M. air it; 2 were 2
am K]: Pr = 0.590. Analysis. I. The c0nuection heat transfer coefﬁcient ma}r be determined from the data by
ming Newton‘s law of cooling. Thai. is. — _ l?
h ‘ Ate  m
With a = 0.35.” and A = 'ITDL, itfollows that 9.35 X 46 W —— = 2 , K. H: 2. Working with the Zultauakas relation, Equation 153‘ _ Pr IM
NHHECRE’HFI‘" P—r, all properties. except Pr“ are evaluated at T3. Accordingly. VD lﬂ‘ﬂ'lfa X 0.012? m
= — = _.. = H 15.39 x it)" m'fa Hence, from Table 14, C = DIE and m == '16. Mao. since Pr <1' II}. a =
It follows that Rep _ on at... = d.20t1992)”tﬂ.?01t“ “( ﬁ) = 50.5
~_— 5: UﬂEﬁJWIm'K: =_
aauﬂﬂ ace—ﬂﬂmm lﬂﬁWIrn x CorrineLittle:
1. Using the lChurchill relation. Equation 154. 0.52M,“ Pr'” [I +( Re” )1“ F = 0.3 + —
"D [t + {BMW13”?“ With all properties evaluated at Tr Pr = {Hi} and ” 1’ 20.92:: lﬂ'ﬁmlfs Hence the Nuaselt number and the convection coefﬁcient are 0.62tﬁW1}'”t0.?0}"’‘ I + m“ are as = 4M
[1 + (omnjof’irt‘ 202.000 ‘ m... = 0.3 + _—i= ﬂﬂﬁﬂ‘t‘th'mKE 1,
H—Nanﬂ ace—W 95.0mm a Alternatively. from the Hilpcrt correlation. Equation 152.
Hi?“ = C Raﬁ Pr'” Witt: all properties evaluated at the ﬁlm temperature. Rep = fulﬁl and e
010. Hence. from Tahie T2. (3' = {LIE} and m I: 0.613. The Nasselt nu and the convection coefﬁcient are then
m, = ammo? I Macedon3" = 37.3 —_— lt_ D.ﬂJﬂWIm'K_ 2
ft  Nltpﬁ  313W — 33 WIRE ' K
I. Uncertainties emaciated with measuring the air velocity. estimating the
loss from cylinder ends. and averaging the cylinder surface tent
which varies: axially and circumferentially. render the experimental
accurate It} titt better than Iii}. Accordingly. calculations based on each of
three correlations are within the experimental uncertainty of the result. 3. Recogniee the importance of using the proper temperature when eval
ﬂuid properties. EXAMPLE Tuf Thc dwomtiw: plastic ﬁlm on a. copper sphere of lﬂ—mm diam:ch is cured in
oven at HT. Upon removal From me man. the alarm: is subij [1} an '
at I am and 23°C having a velocity of IE! mfs. Esrjmate haw long it will take: cool the splat to 35°C. SOLUTION Known: Sphere cooling in an aimrtam.
Find. Time I required to can] from T. = “35°C m T“) = 35°C. Schemmic: 1rI . ?5°C.. m}: 35H: Ass tempt Erma: l. Negligible thermal resistance and capacitance fer the plastic ﬁlm. 2. Spatially isothermal sphere.
3. Negligible radiation effects. Pauperﬂeee Table A. I. copper {T == 323 K}: p = 3933 kgl'm". .l't = 39"} “Elm ' K.
CF = 333 Mtg  K. Table M, an tn. = 295 Ki: a = lﬂlb x In" N — 5th.
:u = 5.53 X Ill—'5' mzals1 k = [1.025] me ' K113!" = DEBS. Table A31. air {T5 =3 328 K]: pi = 1913 x It)" N'ei'mz. Atlﬂlyeis: The time required te complete the ceniing process ma}.r he chtainetl
from results for a lumped capacitance. in particular, from Equations 54 and 5.5 pvcp T:_Tm
f=":'_—l1
ha, TTm
unwith v = when“, = “:qu.
t=pr£inE—TH
ea TTw Ftttm Equaticn 15E: Nun = 2 + [ﬂAReH'z + {Jﬂtjﬂegﬁprﬂ‘ ( in. H4
a) when: Hence the Nusselt number and the cunvccliun cuefﬁcicni are if?” = 2 + {Miaum“ + ﬁ.ﬂ6(ﬁ44ﬂ}“liﬂ.?ﬂsl°" 1r _ z m
x(32.ﬁx m N arm) :41]
IQTEX IU'TN'sfmz J: E=EDD=47IIW : 2
ﬂlmm HEme K The lime [equirad fur cooling is then ,_§Wln(?ﬁﬂj=v:ss
ﬁxnswimJK 35’23 ' Cum menu: I. The validity of the lumped Capacitance math may be determined by caiculal
ing the Bict number. From Equation 5.1!] .‘ ._hirﬁl_nawmIKxunme ...
B“k,‘ k, ' 399WIm'K “135‘”: and the crilerimi is satisﬁed. 2. Atﬂiough their deﬁnitions are similar. the Noosett number is deﬁned in
the thermal conductivity of the ﬂuid. whereas the Biot number is de
terms of the theme! muttivily of the solid. 3. Options for enhancing production rates include accelerating the cooling
by increasing the ﬂuid veiocity andtor using a different ﬂuid. Applyi
foregoing procedures. the cooling time is computed and plotted for air
helinn't over the range of velocities. 5 E v 5 25 into. 125
1m
El 50 its} 25 5 ID 15 ED 25
'la' tn'u’sl' Although Reynolds numbers for He are much smaller than those for air.
thermal conductivity is much larger and, as shown below, convection
transfer is enhanced. Pressurized water is often available at elevated temperatures and may be used for
space heatin g or industrial process applications. in such cases it is customer},r to use
a tube bundle in which the water is passed through the tubes. while air is passed in
cross ﬂow over the tubes. Consider a staggered arrangement for which the tube out
side diameter is lost mm and the longitudinal and transverse pitches are 5}, = 34.3
mm and ST = 3.3 mm. There are seven rows of tubes in the airﬂow direction and
eight tubes per row. Under typical operating conditions the cylindm surface tern perﬁ
atore is at NFC while the air upstream temperature and velocity ate 15°C and
t5 nits, respectively. Determine the airside convection coefﬁcient and the rate ol'
heat transfer for the tube bundle. What is the airside pressure drop? SoLLr'rton Known: Goometry and operating conditions of a tube bank. Find:
1. Airside convection coefﬁcient and beat rate. 2. Pressure drop. Schema! Etc: Assumptions:. Steadyvstate conditions.
2. Negligible radiatien effects. 3. Negligible effect at change in air temperature across tube bank en ‘
properties. Properties: Table a4. air it"... = 15°C}: a = 1.21? kgrmt, a, = let)? tag
a = 14.82 x “3'6 mzr's, a = eases wrm ~ K. Pr = tutti. Table as. air it: =
P: = mm. Table AA, air {a} = are}: r = 114 a: [ti—‘5 are, a = sum wrait—
Pr = ates. Analysis:. Frﬂm Equations T154 and 1155, the airside Husselt number is _ Ira
Nun = Ca: aremam arm (g) Sinee SD = [SE + {Efﬂﬂm = 31? mm is greater than {ST + mm. the matti
mum velocity occurs on the transverse plane+ A... of Figure 1 ]. Henee item
Equatien 162 _ Sr _ 31.3 mm _
pm“ ' sT— D 1L“Hula  16.4]mm 5W5_12"5"”5
1With
tr” D
Rap M = “f = —‘2‘5 "“5 X Uﬂlﬁfm = 13.943
‘ 14.32 X It] m'r's
and
3r _ 31.3 mm _
3—;  34.3 mm _ '19] a: 2
it follows from Tables "3.? and "LB that
_ “5
C = U.35( = 0.34. m =ﬁ.ﬁt}. and C3 = 0.95
L
Hence _ ‘ e25
Nun = are x 0.34t13.943}“'wtﬂ.T110‘( %) = era and 2 a.” X 0.0253 wan  a .i.’ = :'I ,
5 Dﬂlmm 135.15 me K all From Equation 15? 50545 )
PVNrS'rfp mo.o154m}554135.5wrm1K1 )
.21?kgfm3[ﬁmfs}8(ﬂ.ll1313m}IDDTJIICg K T. F Tﬁ= IE?"I — T,} exp(— 1"”‘ H T” = [55"C) exp(
TJ  TE. 2 445°C
Hence from Equations 166 and 163 = (T, — T.) — or.  m = .155 —44.55*C on,“ = 45.55::
I T‘ _ l “ T, — 5:, " 44.5
and
q’ = 4454055.; = 5547 5: I355 wxm= + K x 0.5554 n1 5: 49.5%:
q" = 15.4 kWIm <5 2. The plﬂﬁﬁlll't.‘ drop may ho ubtainod from Equation 16'}. IVEax With Ream = 13,943. Pr= {STE}; = I3]. and {PTIPL} = 0.9!. it follows
from Figure 114 that; == 1.534 and f = 1135. Hence with NL = T _ 5 3 _ 1
121 kym24125ws}]mj sip = 245 Wm2 = 146' X 113—3 bars 5:] op=TX1.ﬂ4[ 5
. Utﬂﬂlﬂ'ﬂ H .T I. With properties evaluated at Tr. Rea m = 1 1.376. With STID 2 2 and SJ!) 2 2, it
follow: from Tables 15 and 115 that C] = 11432, H: = 11556. and C? = 13.91 From Equations loft and TEN. the Husselt number is then mp = 36.7". and h I '
th: ' K. Values of h obtained from Equations Inf[l and 164 therefore '
within 'ttita, which is well within their uncertainties. . Had tit?? E I} — T, been used in lieu of tilTI.” in Equlation 7.63, the would have been overpredicted by  “it. . Since the air temperature is predicted to increase by only I'll.5”C. evﬂu ‘_
the air properties at T. = 1513 is a reasonable approximation. Heat improved accuracy is desired. the calculations could be repeated Wili't the etties reevaluated at {21 + Tram = NEST. An exception pertains to the
sity p in the esponentia] term of Equation "not. As it appears iii
denominator of this term, p is matched 1tvith the inlet velocity to provide a
M {pm that is linked to the "less ﬂow rate of air entering the lube =!
Hence. in this term. ,9 should be evaluated at T,. I . The air outlet temperature and heat rate may be increased by increasing.
number of tube rows. and for a ﬁtted number of rows, they may be van
adjuszing the air velocity. Fen5 E t'tt'I '5 25 and ir’ = El mils. parametric c #
tions based on Equations 1154 through 7.63 yield the following results: TH cc] The air uutlct temperature would asymptotically approach the surface
mm with increasing m. at which paint thc hear ran: appmachcs a cansram
and there is no advantage tn adding more rubc rows. Nutc that ﬁp incramcs
early with increasing Ni. For N,“ = 25 and I s V s 21) mix. we obtain ...
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 Fall '08
 Any
 Heat Transfer

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