5th Homework Solution

5th Homework Solution - PROBLEM 5.41 if? $ KNOWN: Hot dog...

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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/m-K p =580k9/m5 c =3550J/k9-K ——‘> 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/mZ-K(10x10-3m/2) _096 J ‘ 'k ‘ k ‘ 0.52 W/m-K ': ' 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/mZ-Kx10x10'3m k . 0.52 W/m-K =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 (10x10-3m)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 = 0-52 W/m'K / 880 kg/m3x3350 J/kg-m3 = 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 = 1723-34? 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 [1-2334 - 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/m-K Combustion gases T T T Too= 2300 K h = 5000 W/m2-K ASSUMPTIONS: (1) One-dimensional 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/m-K = 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----- l-lz‘I-I-I-I-I 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'z-K. 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 semi-infinite solid, (2) One-dimensional conduction, (3) Constant properties, (4) Negligible radiation. PROPERTIES: Table A-3, Oak, cross grain (300 K): p = 545 kg/m'l, c = 2385 J/kg-K, k = 0.17 W/m-K, a = k/pc = 0.17 W/m-K/545 kg/nfi x 2385 J/kg-K : 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 cross-section 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) Two-dimensional conduction mm and x2 directions, (2) Constant properties, (3) Heat transfer to billet is by convection process only. * PROPERTIES: Table A-1, Steel, plain carbon (T = (30+bOO)°C/2 fi 588K = 600K): ' p = 7854 kg/m3, cp = 559 J/kg-K, k = 48.0 W/m-K, a = k/pcp = 1.093x10-5m2/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 93-2142 =M=65875=L83h 6 3,4 '8’ fr? <1 7.093x10'5m2/s ' 0.. I (sew book— (“0L5 Cl 4(flréféfilfil; (Va/(OW 7%fo< Corflaf mo&e_LU{#M c;nV€a#¢mq ‘ \ 0' a: mm a < “*5” 1"“ :2. x v hm. " + ' * N . J 1%“ 4—1an7 1; MM? .1 . I I a» we "2-: WW +KMT >53 .2 WLM'E ...
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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 .

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5th Homework Solution - PROBLEM 5.41 if? $ KNOWN: Hot dog...

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