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Unformatted text preview: PROBLEM 12.29 KNOWN: Spectral emissivity. dimensions and initial temperature of a tungsten filament. FIND: (a) Total hemispherical emissivity. 8. when ﬁlament temperature is T. = 2900 K; (b) Initial rate
of cooling. defdt. assuming the surroundings are at Tm = 300 K when the current is switched off: (c) Compute and plot E as a function of T. for the range 1300 S '[‘s E 2900 K; and (d) Time required for
the filament to cool from 2900 to 1300 K. SCHEMATIC:
qrad
Evacuated Tungsten
045 mm ﬁlament.
‘ D = 0.8 mm.
5A L = 20 mm,
T8 = 2900 K.
0.10 Tf= 1300 K
7t
0 2 4 (m) ASSUMPTIONS: (l) Filament temperature is uniform at any time (lumped capacitance). (2) Negligible
heat loss by conduction through the support posts. (3} Surroundings large compared to the ﬁlament, {4) Spectral emissivity. density and speciﬁc heat constant over the temperature range, (5) Negligible
convection. PROPERTIES: ’I‘ableAJ. Tungsten (2900 K}; p = 19, 300 lag/m3. er, a 135 J/kg~K. ANALYSIS: (a) The total emissivity at Ts = 2900 K follows from Eq. 12.38 using Table 12.1 for the
band emission factors. 5 : IO EAELbUs )d’l' : ElﬁOHEttm) T 52“ "FOHme) (1) e = 0.45 x 0.72 + 0,1t1~ 0.72) : 0.352 <
where Fwazlum) = 0.72 at RT = ZumXZQOOK = 5800 MIH‘K. (b) Perform an energy balance on the filament at the instant of time at which the current is switched off.
;_  CITS
hill _Eout : Mcp dl AS (ethur —E): A.) (GUT: —£oTS4) : Mep de/dt and find the change in temperature with time where A. = rtDL. M : p0, and ‘v’ : (TIDE/4H“ cl’l‘S a‘DLcrteT: #a’lﬁlr) 7 40 t'
C“ p (2119/4)ch Dept) (5T: 0T3” ) 7 t' '
0T. 7 4x507x10 Swat"mgK4(0.352x29004—0.l><3004)K4 dt 19. 300 kg; m2 x 185 ig'kg  K X00008m (c) Using the [HT Tool, Radiation, Band Emission Factor. and liq. (l), a model was developed to
calculate and plot 8 as a function of T5. See plot below. = 1977 K/s < PROBLEM 12.29 (Cont.) (d) Using the [HT Lumde Capacitance Model along with the IHT workspace for part (C) to determine E
as a function of 1;, a model was developed to pl'edlCt 11 as a function ol'cooling time. The results are
shown below for the variable emissivity case (2 vs. 'l‘S as per the plot below left) and the ease where the
emissivity is ﬁxed at 50900 K): 0352. For the variable and fixed emissivity eases. the times to reach
T‘. = 1300 K are Iv\'.r = S [lth i S <
0.4
0.3 7 E
a 3
CD
0.2  I _
0 2 4 6 8 10
01 I r .   a Elapsed time, 1(5)
1000 1500 2000 2500 3000 ‘ ‘ I I
. — Variable emisswity
Filament temperature, Ts (K) _+ Fixed emissivity, eps = 0.35 COMMENTS: (l)From the E vs. Tq plot, note that E increases as T5 increases. Could you have
surmised as much by looking at the spectral emissivity distribution, at vs. It? (2) How do you explain the result that t“.Ir > tax? PROBLEM 12.52 KNOWN: Pewer dissipation temperature and distribution of spectral emissivity for a tungsten
filament. Distribution of spectral absorptivity for glass bulb. Temperature of ambient air and
sun'oundings. Bulb diameter. FIND: Bulb temperature.
SCHEMATIC: Glass bulb, T5 iris. 1
D = 75 mm. at, I ‘11 ‘13 Tungsten ﬁlament 0.7L
T = 3000 K, sf
052
if} ./ 0"
Tm = 25°C Clitionv Pelee = 75 W D 04 20 101m) ASSUMPTIONS: (l) Steady—state. (2) Uniform glass temperature TS‘ and uniform irradiation of
inner surface. (3] Surface of glass is diffuse, (4) Negligible absorption of radiation by filament due to
emission from inner surface of bulb, (5) Net radiation trtmsferfrem outer surface of bulb is due to
exchange with large surroundings, (6) Bulb temperature is sufficiently low to provide negligible
emission at it < 2am. (7) Ambient air is quiescent. PROPERTIES: Table M. air (assume rf : 323 K): v = 1.82 x 10'5 mils. o: = 2.59 x 105 mzlth =
0028 WhirK. a: 0.0031 K‘l, Pr = 0.704. ANALYSIS: From an energy balance on the glass buib‘ a If a 1 r 4 _ 71
qrad.i : Qradm +Clconv : JL'b‘3"(r.s ’Tsur)+ h _ loo) (.1) where Eb : £l>2wn 2 a;_>2lum : l and h is obtained from Eq. {9.35) _ 0.589 Ralj‘l ED
NU Z 2 + : (2)
D 9/10 4’9 k
[1+ (0.416911%) ]
with Ran : g ('1‘s —'l1,¢, )D3 Iva. Radiation absorption at the inner surface of the bulb may be
expressed as
I 2
Clrad.i : CCG : O5 (Pelee HID ) [3) where. from Eq. (12.46)‘ or :aIJ‘gA (Gil IONA +052J‘2'0 0'4 (G;_/G)dl+a3j: (GA/omit The irradiation is due to emission from the filament, in which ease {GA/G} ~ (E;_ll£)f: (Er,aLEg‘b/EfEb).
Hence. a = (“i “'1‘ lily45:12. (EQUb’JEb )d/L +012 lFr )LiffrJ. (ELIHEI; )‘M '’ ((13 l 1‘1 (Fit: " Eta)” (4) where. from the spectral distribution ofProblem 12.25. 8ij E 81 = 0.45 for 2. < 2pm and EM 2 £3 =
0.10 for l “2 211m. From Eq. (12.38) Er =l:5r,1 (E/lJﬂﬁbliUL = 51 F(0—>2,um) +52 (1 —F(n—>2pm))
With AT,‘ : 2ymx3000 K 26000 hmK. F(0..)2#m) = 0.738 from Table 12.1. Hence.
ff 2 0.45x0.738 + 0.1(1 —0.738) = 0.358
Equation (4) may now be expressed as
(I :(alLEI)ElF(0—>0.411I11)+(02}£I')El(WU—Emu)_F(0—>0.4um))+(a315f)52(1_F(0—)3t1m)) where. with AT = 041th 3000 K = 1200 me. F(0_,0_4#m) : 0.0021. Hence. a (1m.358)0.45 X00021 + (0.110.358)0.45><(0.738 — 0.0021) + (110.358)0.1 (i 70.738) 0.168 Substituting Eqs. (2) and {3) into liq. (l). as well as values of £1, : l and 06 : 0.168, an iterative
solution yields TS 2348.1K < COMMEN'l‘S: For the prescribed conditions, (lgildyi = 713 mez. qf—adﬂ = 385.5 \Vi’mz and qgom, = 327.5 Win12. ...
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This note was uploaded on 08/08/2008 for the course ME 364 taught by Professor Rothamer during the Fall '08 term at Wisconsin.
 Fall '08
 Rothamer
 Heat Transfer

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