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Unformatted text preview: PROBLEM 5.24 KNOWN: Diameter and thermophysical properties of alumina particles. Convection conditions
associated with a two—step heating process. FIND: (a) 'l‘inte—in—ﬂight (tn) required for complete melting. (b) Validity ofassuming negligible
radiation. SCHEMATIC: Al203 sphere, Dp : 50 um, 7}: 300 K, kp=ill5 thK. p = 39m kgrm3, cp = 1560 thgK, = 2318 K, hsf= 3577 thkg mp n = 3x104 wrmZK "\
rm: 10,000 K qbonv ASSUMPTIONS: {1) Particle behaves as a lumped capacitance, (2) Negligible radiation, (3) Constant
properties. ANALYSIS: (a) The twostep process involves (i) the time t1 to heat the particle to its melting point and
(ii) the time 1: required to achieve complete melting. Hence, ti_f = t. + t;, where from Eq. {5.5). :ppvcp 91 pprCp 1n TiToo t1 In — = _
hAS 9 6h TIn —TDC,
3970 kg/m3 (50x10_6m)156OJ/kg  K (300_10 000)
t1=————2— n—’—= 4X10_4S
6(30,000w/m K) (231340000)
Performng an energy balance for the second step. we obtain
t1+t2
Ll (Iconvdt = AEst where qcom = hA,(T...,  Tmp) and ABS, 2 ppVhsf. Hence, ,, 3 —6
£22,0pr hsf :3970kg/m (50x10 m)x 3'577X106J/kg =5X104S
6h (Toe—Tmp) 6(30.000w/m3K) (10,000*2318)K
Hence Ii_f :9X10_4s==lms < (b) Contrasting the smallest value of the convection heat flux, quw‘mm = 11 (TM —Tmp ) : 2.3x108 W/m3 to the largest radiation ﬂux. qgﬂd’max = 50' ('lﬁm —T:,r) = 6.5 x 105 W/mz. we conclude that radiation is. in fact, negligible.
COMMENTS: [1) Since Bi = [hrp/3)fk : 0.05, the lumped capacitance assumption is good. (2) In an actual application. the droplet should impact the substrate in a superheated condition (T ) T,,,,,), which
would require a slightly larger tHc. PROBLEM 5.71 KN OWN: Asphalt pavement, initially at 50°C, is suddenly exposed to a rainstorm reducing the
surface temperature to 20D C. . 2 . .
FIND: Total amount 01 energy removed (Jan ) from the pavement for a 30 minute period. SC HEM AT] C: ASSUMPTIONS: ( I) Asphalt pavement can be treated as a semiinfinite solid, (2) Effect of rainstorm is to suddenly reduce the surface temperature to 20°C and is maintained at that level for
the period of interest. PROPERTIES: Table 213, Asphalt (300K): [3 = 2115 kg/m3, e = 920 J/kgK, k = 0.062
W/mK. ANALYSIS: This solution corresponds to Case 1, Figure 5.7, and the surface heat flux is given by
Eq. 5.58 as I
a lf2
QS(t):k(Ts_Ti)’x(ﬂal) (I) The energy into the pavement over a period of time is the integral of the surface heat flux expressed
as Q”: l; q; [t)dt. (2) Note that q: (t) is into the solid and, hence, Q represents energy into the solid, Substituting Eq. (I) for q; (t) into Eq. (2) and integrating find ,. 1/2‘4/2 k "
Q=k(Ts—Ti]/(:im) lot dt*—(—“§—><2 (3) Substituting numerical values into Eq. (3) with a_i_ 0.062 W/mK .0 c 2115 kgrm3 x920 J/kg K ﬁnd that for the 30 minute period,
A, 0.062 W/mK [20—50)K
Q _ 1:2 :3.isx10‘3m2/s x2(30x605)“2 2&4.99Xl05 erz. < (ax3.18X10'8m2/s] COMMENTS: Note that the sign for Q” is negative implying that energy is removed from the
solid. PROBLEM 6.23 KNOWN: Shaft of diameter 100 mm rotating at 9000 rpm in a journal bearing of 70 mm length.
Uniform gap of 1 mm separates the shaft and bearing filled with lubricant. Outer surface of bearing is water—cooled and maintained at Twe = 30°C. FIND: (a) Viscous dissipation in the lubricant. u¢(W/m'l}, (b) Heat transfer rate from the lubricant,
assuming no heat lost through the shaft, and (c) Temperatures of the bearing and shaft, Tb and "1",. SCHEMATIC:
Wmm) Bearing Bearing. Rb = 45 WImK l= ?0 mm . Lubricant, u = 0.03 Nsrm2
‘ k = 0.15 meK
ﬁbﬂ‘m mm Watercooled surface
: DC hf: 9000 rpm ﬂ [)0 = 200 mm i.— Two 30 // ASSUMPTIONS: (l) Steadystate conditions, (2) Fully developed Couette flow. (3) lneompressible
fluid with constant properties. and (4) Negligible heat lost through the shaft. ANALYSIS: (a) The viscous dissipation, uCD, Eq. 6.40, for Couette flow from Example 6.4, is 2 2 2
4 . 1‘
maze 93 =u E 20.03Ns/m2 13E :6.656><107 w/m3 <
dy L 0.00m where the velocity distribution is linear and the tangential velocity of the shaft is
U : JIDN = r: (0.100m)x9000rpm><(min/60s) = 47.1m/s. (b) The heat transfer rate from the lubricant volume V through the bearing is q zncpv = gownLt): 6.65x107 W/m3 (Hx0.100m><0.001mx0.070m): 1462W < where E: 70 mm is the length of the bearing normal to the page. PROBLEM 6.23 (Cont) (c) From Fourier‘s law, the heat rate through the bearing material of inner and outer diameters, D, and DD,
and thermal conductivity kl, is, from Eq. (3.27), _ Mikb (Tb rTwc) qr [n (Do/Di)
1 D “D
szTwc+Qr “( o/ l)
Zﬂfkb
1462Wl 200100
Tb=30°C+ n( / ) *81.2°C < 2rr><0.070m><45 W/m'K To determine the temperature of the shaft. T(0) = T5, first the temperature distribution must be found
beginning with the general solution, Example 6.4. U 2
in 2 T =—ﬁ— C +c (y) 2k[L]y+3y 4 The boundary conditions are. at y = 0. the surface is adiabatic —] =0 C3=0
dy 3:0 and at y = L, the temperature is that of the bearing, Th ,
IuU"3 M 2
TL=T =——— L+0+C C=T+—U
H b 2K{L] 4 4 b 2k Hence. the temperature distribution is and the temperature at the shaft‘ y = 0, is 0.03Ns/m2 ~———(47.lm/s)2 =303°c <
2x0.15W/mK TS =T(0):Tb Jrﬁuz :81.3°C+ PROBLEM 6.32 KNOWN: Variation of hx with x for flow over a ﬂat plate. FIND: Ratio of average Nusselt number for the entire plate to the local Nusselt number at x :
L. SCIlEMA’l‘IC: l/’—7719r'ma/ boundary Iayer; bx: Cx [a where
—’x L. C is a consf'anf'
ANALYSIS: The expressions for the local and average Nusselt numbers are l/2
hLL _ (CL )L _ CLU2 NuL=———
EkL k k
m =_t
where
, L L
h] =1] hx dx=EJ x‘l’2dngLl’222 (31:19.
‘ L 0 L 0 L
Hence,
_ 2CL‘U2 (L) 2CL“2
NUL :f: k
and
NuL _2 <
NUL — . COMMENTS: Note the manner in which EL is deﬁned in terms of FL. Also note that Nu l L T
L :EJU 1\uX dx. PROBLEM 7.23 KNOWN: Prevailing wind with prescribed speed blows past ten window panels. each of 17m length, on
a penthouse tower. FIND: (a) Average convection coefficient for the ﬁrst, third and tenth window panels when the wind
speed is 5 Ms; evaluate thennophysical properties at 300 K, but determine suitability when ambient air
temperature is in the range 15 5 Ta, 3 38°C; (b) Compute and plot the average coefﬁcients for the same panels with wind speeds for the range 5 S um S 100 km/h; explain features and relative magnitudes. SCHEMATIC: Window panel
WEI/f 1 m length in xdireetion
ASSUMPTIONS: (1) Steadystate conditions, (2) Constant properties. (3) Wind over panels approximates parallel flow over a smooth ﬂat plate, and (4) Transition Reynolds number is Re“ = 5 x
105. PROPERTIES: Table 11.4, Air m: 300 K. 1 alm): v = 15.89 x105 nil/s, k = 20.3 x 10'3 W/m~K, Pr =
0.707. ANALYSIS: (a) The average convection coefﬁcients for the ﬁrst, third and tenth panels are — q E X —E X3  E X H X
hl h2_3 :_ﬂ_2_~ hEHO =ﬂ_10__9_9 (1.2.3)
X3—X2 Klei—K9 where h2 = hz (212), etc. If ReM = 5 X 105, with properties evaluated at '1}: 300 K, transition occurs at —6 2 5. ,c :LRBX C =MX5X105 =1,ng
um ‘ Sm/s The ﬂow over the first panel is laminar, and hi can be estimated using Eq. (7.31). 313,1 = Ll)” : 0.664 Re“2 PrU3
k X  11'2
111=(0.0203w/mKx0.ss4/1m)(5mli’sx1m/1ss9x10‘6nil/s) (0.707)U3=8.73W/m2‘l( < The flow over the third and tenth panels is mixed, and hz , h3, h9 and hm can be estimated using Eq.
(7.41). For the third panel with x3 = 3 m and x3 : 2 m, N—uﬂ : “3"3 = 0.037 Rem—s71 PrU3
k X H3 = (0.0263W/m K/sm)
4/5
x 0.037(Sm/sxsm/issgx10—6mz/s) —871](0.707)“3=10.6w/m21< PROBLEM 7.23 (Cont.)
ﬁg 2 (0.0263 W/mK/zm) 4/5
x 0.037 5m"'s><2m 15.89><10‘6 mzfs e871 0.7071’32368w m2.K
.4 J" From Eq. (2).
2 o
— 10.6iw/m Kx3m—s.68w/m~Kx2m
h2—3 = —"_*—r———(3 2)
— m
Following the same procedure for the tenth panel. find hm = 11.64 mez‘K and hg = 11_71 W/mzK
and :14.5W/m2K < h9_10:ll.lW/m2K <
Assuming that the window pane] temperature will always be close to room temperature. TS = 23°C = 296 K. If Tm ranges from —15 to 38°C. the ﬁlm temperature, '1}: {TS + TW)/2, will vary from 275 to 310 K.
We'll explore the effect of 'I‘t subsequently. (b) Using the_IHT Tool. Correlations. External Flow, Flat Plate, results were obtained for the average
coefficients h . Using Eqs. (2) and (3), average coefﬁcients for the panels as a function of wind speed
were computed and plotted. Average coeﬂicie nl. hbar {WIIW‘ZKJ Wind speed. uinl (kmth) *7 First panel
—K— Third panel
—ﬁ Tenth panel COMMENTS: (1) The behavior of the panel average coefficients as a function of wind speed can be explained from the behavior of the local coefﬁcient as a function of distance for difference velocities as
plotted below. Local coeﬂiciem. hx [Wim’QM) Distance 1mm leading edge. x [m] "‘— uinl : 5 Math + Uinf = 15 km‘h
+ uinl . 25 km‘h
+ ulnl 7. 50 krwh Continued... PROBLEM 7.23 (Cont.) For low wind speeds, transition occurs near the mid—panel, making El and h9_10 nearly equal and very
high because of leadingedge and turbulence effects. respectively. As the wind speed increases,
transition occurs closer to the leading edge. Notice how h2_3 increases rather abruptly, subsequently becoming greater than h9_10 . The abrupt increase in h, around 30 km/h is a consequence oftransition occurring with x < lm. (2) Using the IHT code developed for the foregoing analysis with new : 5 m/s. the effect of T. is tabulated
below MK) 275 300 310
El (“z/mam 8.72 8.73 8.70
h2_3 (W/mZK) 15.1 14.5 143
H1340 (W/mEK) 11.6 11.1 10.8 The overall effect of '1} on estimates for the average panel coefﬁcient is slight. less than 5%. PROBLEM 7.54 KNOWN: Diameter and length of a copper rod, with ﬁxed end temperatures, inserted in an airstream
of prescribed veloctty and temperature. FIND: (a) Midplane temperature of rod. (b) Rate of heat transfer from the rod.
SCHEMATIC: 7;: 90°C ASSUMPTIONS: (1) Stead state conditions, (2) One—dimensional conduction in rod, (3) Negligible contact resrstance, (4) Negllgi le radiatlon, (5) Constant properties. PROPERTIES: Table A—!. Copper (T = 80C = 353 K): k = 398 meK; Tobie A4, Air (Toc =
6 2 25°C : 300K, 1 atm): v = 15.8 x 10 m /s, k = 0.0263 W/mK, Pr = 0.707; Table A4, Air (T5 : 80°C z 350K, 1 atm): Prs = 0.700. coshm(Lix) _ TiTw ANALYSIS: (a) For case B of Table 3.4. i = ﬂ
6}, cosh(rnL) Tb — Tm where HQ in : (FPr'kAC) : (rtE’kD)1 “i2. Using the Zhukauskas correlation with n = 0.37, NuD =C Reg,l Prn (131713101!4 VD ZSm/s 0.01m
ReD=—:~——#=15,823 V 15.8x10‘6 mz/s
and C = 026,111 2 0.6 from Table 7—4. Hence MD : 0.26(15,823)ﬂ'6 (0.707)“37 (0707:0700)‘ "'4 = 75.8
_ k — _ 0.0263 wrm.K D D 0.01m 1
4x199W/m2K J 75.8) =199 wrm2 1< :' 2 — =14.2 mi].
398 Wr'm‘ KXUD] m T L —T ‘ l 0
LL31: L03“ ) =_li=0.79 Hence, —
Tb ‘Tco cosh(l4.2 m'I X005 m) 1'26 T(t.)=25°C+0.79(90—25) =76.6°C. <
{b} From liq. 3.?6, q = qu = 2M tanh rnL, 112 M: HPkA
( c) m,K mK 4 M =20? w q =2(28.7 W) tanh (142111" x0.05m)=35 w. < If?
0b=[l99 2’" (“001111) [398 w F4001me 65°C COMMENTS: Note adiabatic condition associated with symmetry about midplane. ...
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 Fall '08
 Rothamer
 Thermodynamics, Heat, Heat Transfer, Orders of magnitude, heat transfer rate

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