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Unformatted text preview: PROBLEM 5.41 if? $ KNOWN: Hot dog with prescribed thermophysical properties, initially at 6°C, is immersed in
boiling water. I ' FIND: Time required to bring centerline temperature to 80°C.
SCHEMATIC: k=0.52W/mK
p =580k9/m5
c =3550J/k9K ——‘> h=100W/mZ'K ASSUMPTIONS: (1) Hot dog can be treated as inﬁnite cylinder, (2) Constant properties. ANALYSIS: The Biot number, based upon Eq. 5.10, is Bi: th _ bro/2 _ 100 W/mZK(10x103m/2) _096 J
‘ 'k ‘ k ‘ 0.52 W/mK ': ' Since Bi > 0.1, lumped capacitance analysis is not appropriate. Using the Heisler chart, Figure C.4with , aka g 60 7A1 ‘6‘ i hr0 = lOOW/mZKx10x10'3m
k . 0.52 W/mK =1.92 or Bi“=0.52 and 9* _ E _ T(0,t)—Ts _ (80—100)°C 2021  ° 0 .Ti—Ts (6—100>°c ' [Ml (TM/W 0“ 7,
ﬁnd 0/ answers W Vth 0» '
at ' r3 (10x103m)2 ' A’ ‘
gFo = t* = —2 =0.8 t= —Fo= x08 :453.5s =7.6 min to ' on 1.764x10‘7m2/s where on = k/pc = 052 W/m'K / 880 kg/m3x3350 J/kgm3 = 1.764x10‘7m2/s . COMMENTS: (1) Nete that LG =To/2~ when evaluating the Biot number for the lumped
._ capacitance analysis; however, in the Heisler charts, Bi 5 hro/k (2) The su ace erature of the hot dog follows from use of .
BF] = 0.52; ﬁnd 0(1,t)/0o ~ . . From Eq. (1), n o = 0.21 0i giving 9(1,t) = T(ro,t)—T,., = 0.459 = 0.45x0.21[6—100]°C = 8.9°C (3) Since F0 2 0.2, the approximate solution for 0*, Eq. Wis valid. From Tablﬂ’ with
Bi: 1.92, ﬁnd that £1 = 113245 rad and C1 = 172334? Rearrangng Eq. 5.49 and substituting With r/ro = 1 and values, 1444?  r32 5?
___1_ _ 1 0.213 _3 gzlg’
Fo— ;% 1n (ask/C1) — (1.3245 rad)2 In [12334  mg 0‘ This result leads to a value oft = 9.5 min or 20% higher than that of the graphical method. a» ‘ KNOWN: Properties and thickness L of ceramic coating on rocket nozzle wall. Convection conditions.
‘Initial temperature and maximum allowable wall temperature. PROBLEM 5.34 FIND: (a) Maximum allowable engine operating time, tum, for L = 10 mm, (b) Coating inner and outer
surface temperature histories for L = 10 and 40 mm. ' SCHEMATIC:
L Metal wall, . r , . Ceramic ‘ a = 6x10'6 m2/s
Tmax=1500K Ti=3ooK k=10W/mK
Combustion gases
T T T Too= 2300 K h = 5000 W/m2K
ASSUMPTIONS: (1) Onedimensional conduction in a plane wall, (2) Constant properties, (3) Negligible thermal capacitance of metal wall and heat loss through back surface, (4) Negligible contact
resistance at wall/ceramic interface, (5) Negligible radiation. 1 ANALYSIS: (a) Subject to assumptions (3) and (4), the maximum wall temperature corresponds to the
ceramic temperature at x = 0. Hence, for the ceramic, we wish to determine the time trmlx at which T(O,t)
= To(t) = 1500 K. With Bi = hL/k = 5000 W/m2~K(0.01 m)/ 10 W/mK = 5, the lumped capacitance method cannot be used. Assuming Fo > 0.2, obtaining C1 = 1.3138 and C1 = 1.2402 from Table and
evaluating 0; = (To — Tug/(Ti  Ta) = 0.4, Equation 5.41 yields _ 111(6); /C,) _ _ ln(0.4/1.2402)
q; ' (1.3138)2 conﬁrming the assumption of F0 > 0.2. Hence, F0 = = 0.656 t _ Fo(LZ) _ 0.656(001m)2
_ _ 6x 10'6 mz/s =10.9s max a. A (b) Using the IHT Lumped Capacitance Model for a Plane Wall, the inner and outer" surface temperature
histories were computed and are as follows: = ’ 2300 I Z .
2:52:3— 1900 1500 1100 Temperature. T(K) _ 700 ll!E
llz‘IIIII 300 O 30 60 90 120 150
Tlme.t(s) 4+ L=0.01.X=L + L=0.01,X=O + L=0.04.X=L _’— L=0.04.x=0 ' Continued... PROBLEM 5.34' (Cont) The increase in the inner (x = 0) surface temperature lags that of the outer surface, but within t= 458 both
temperatures are within a few degrees of the gas temperature for L = 0.01 m. For L = 0.04 m, the
increased thermal capacitance of the ceramic slows the approach to steady—state conditions. The thermal
response of the inner surface significantly lags that of the outer surface, and it is not until I = 137s that the inner surface reaches 1500 K. At this time there is still a significant temperature difference across
the ceramic, with T(L,tm) = 2240 K. COMMENTS: The allowable engine operating time increases with increasing thermal capacitance of
the ceramic and hence with increasing L. E 2:: om _‘ o: oo _\ Om om ov ON E Ednx .E 3.0“; aux .E :35 _, E Edux .E SOL Ia:
aux .E 5.05 IT mE_._. .m> 2323:.me mcwmoo. com 002‘ come ooom comm ()1) aJnmJadwa; (“3
PROBLEM 5.68 KNOWN: Thick oak wall, initially at a uniform temperature of 25°C, is suddenly exposed to
combustion products at 800°C with a convection coefficient of 20 W/m'zK. FIND: (a) Time of exposure required for the surface to reach an ignition temperature of 400°C, (b)
Temperature distribution at time t = 325s. ‘  SCHEMATIC:
l Too = 800 °C T(0,t) = 400 PC Combustion
products T(x,t) T,= 25 °C
ff T“, = 800 0c L
h = 20 W/m2'K ' ‘ ’
‘ . L» x ,
ASSUMPTIONS: (1) Oak wall can be treated as semiinfinite solid, (2) Onedimensional conduction,
(3) Constant properties, (4) Negligible radiation. PROPERTIES: Table A3, Oak, cross grain (300 K): p = 545 kg/m'l, c = 2385 J/kgK, k = 0.17 W/mK,
a = k/pc = 0.17 W/mK/545 kg/nﬁ x 2385 J/kgK : 1.31 x 10'7 mz/s. ‘ ’ ANALYSIS: (a) This situation corresponds to Case 3 of Figure 5.7. The temperature distribution is? given by E . . . or by Figureé’ﬁ’.v Using the ' ure with _ ' ‘
, t 4 ., 2, . A V ww .
, 4/ = = o 43 ‘ and _x_ = o .
4/4 T... — Ti ‘ 800 — 25 ' ‘ 2(oa)”2 .  0‘ we obtain h(oct)”2 which case t = (0.75k/h0c1/2)2. Hence, Temperature, T(c) Q01 0.02 Distance [rum the suﬂaca. x(m) “k X The temperature'decay would become more pronounced with decreasing 0t (decreasing k, increasingpcp) [\0 {p and in this Case the penetration depth of the heating process corresponds to x = 0.025 m at 3253.
LQSWYN COMMENTS: 'The result of part (a) indicates that, after approximately 5 minutes, the surface of the wall will ignite and‘combustion will ensue.' Once combustion has started; the present model is no longer
appropriate. ' PROBLEM 5.73 ' 3:? A KNOWN: Steel (plain carbon) billet of square crosssection initially at a uniform temperature
of 30°C is placed in a soaking .oven and subjected to a convection heating process with
prescribed temperature and convection coefﬁcient. FIND: Time required forbillet center temperature to reach 600°C. SCHEMATIC: 770,0,r)=600°C ASSUMPTIONS: (l) Twodimensional conduction mm and x2 directions, (2) Constant
properties, (3) Heat transfer to billet is by convection process only. * PROPERTIES: Table A1, Steel, plain carbon (T = (30+bOO)°C/2 ﬁ 588K = 600K): '
p = 7854 kg/m3, cp = 559 J/kgK, k = 48.0 W/mK, a = k/pcp = 1.093x105m2/s. ANALYSIS: The billet corresponds to Case (e), Figure 5.11, inﬁnite rectangular bar, for which
the temperature distribution is of the form 9*(x1,x2,t>=P<x1,t>><P(xz,t> where P(x,t) denotes the distribution corresponding to the plane wall. Because ‘of symmetry in
x1 and x2 direction, the P functions are identical. Hence, ' ' ' e (0 t) 2 e = T4“ I
I i i Plane wall 90 = T(0, t) — Tm Substituting numerical values, ﬁnd 1/; I  1/2
90 (O, t) T(0,0,t)—Tm (600.750)°C 
ei ' Ti—Tw ‘ (30—750)°c ‘ 0'46 ' O bur. L @ [4
Consider now the Heisler chart for the plane wall, Figure C.l. Forth “CU” f
* ‘ 90 k ‘ K) ' 
e =——=o.46 B1‘1=——= ' :32
° 91 hL 100 W/m K><O. 15m ﬁnd answers W16le WM 4 932142 =M=65875=L83h 6 3,4 '8’ fr? <1
7.093x10'5m2/s ' 0.. I
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\ 0' a: mm a < “*5” 1"“
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N . J 1%“ 4—1an7 1; MM?
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This homework help was uploaded on 04/15/2008 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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