Chapter 6 HW solutions

Chapter 6 HW solutions - PROBLEM 6.6 KNOWN: Expression for...

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Unformatted text preview: PROBLEM 6.6 KNOWN: Expression for the local heat transfer coefficient of a circular. hot gas jet at T35 directed normal to a circular plate at TS of radius r0. FIND: Heat transfer rate to the plate by convection. SCEiEllAITCE IE!" Too 1 hfr] = a +br" Plath!J 7; a?" 7;: dgcanr ASSUMPTIDXS: (1) Steady-state conditions. (3) Flow is axisynnnetric about the plate. (3) For htr). a and b are constants and n e -2. ANALYSIS: The convective heat transfer rate to the plate follows from Newton‘s law of cooling 1. Cicom; =J [kiconr =1. hiri'dA'iT-r ‘Ts A A The local heat transfer coefficient is known to have the form. 11(1'] = a + hrn and the differential area on the plate surface is dA 2 Ear dr. Hence. the heat rate is $0 qmm: =JO (a +brni-23r r Cir-(T.Clo —TS] a 3 b 11+? 0 (icon? = 2'7iT-so _T9] —1' + _ 1' 2 11 +2 0 a 2 b n+2— fCiconv=hr[jro +n_21"o {Tm—T5]. <2 COMMENTS: Note the importance of the requirement. 11 e -2. Typically. the radius of the jet is much smaller than that of the plate. PROBLEl‘t-‘I 6.10 KNO WEN}: Expression for face-averaged Nusselt numbers on a cylinder of rectangular cross section. Dimensions of the cylinder. FIND: Average heat transfer coefficient over the entire cylinder. Plausible explanation for variations in the face—averaged heat transfer coefficients. SCHEJIATIC: "— C 2 40' mm ‘5‘ v=10mrs —-» dz3umm Tw=300K —p ASSUMPTIONS: [1} Steady—state conditions. (2) Constant properties. PROFERTIES: Table A4. air (300 K): k = 0.0263 RR-’.-’111-K._t-'= 1.589 x 10': 1113.953. Pr = 0.?01 ANALYSIS: For the square cylinder. 0-0 = 40 min-’30 min = 1.33 Yd 10111:; x 30 r- 10'3 n1 Red : — : — 5 1' 218.880 V 1.589 >< 10' 111'.-'s Therefore. for the front face C = 0.61%. n1 = 1/3. For the sides. C = 0107". 111 = 2.53 while for the back C = 0.153.111 = 2:3. Front face: 311“ = 0.514 x 18.8301"3:-- 0.10?“3 = 02.44 — k): 0.0263 141-" -K >< 82.44 « hf = “d = m = 13.2? mm‘ -K d 30 x 10'3 in Side faces: and.) = 0.101 >< 13.3302’3r 0.10?” = 6.3.36 kNu 0 0253 w.-' -K >< 5? 36 ‘1‘ = + = 59.05 11.3fm} K d 30 >1 10'3 n1 HS: Back face: mm = 0.153 x 13.3302’3x 0.101” = 95.43 _ kNu 0.025310! -K><96.43 . 1 ab = “J = + = 34.54 w..-'m-' .14 d 30 N 10'3 n1 Continued... PROBLE M 6. 17" KNOTWN: Pressure dependence of the dynamic Viscosity. thermal conductivity and specific heat. FIND: (a) Variation of the kinematic viscosity and thermal diffilsis'ity with pressure for an incompressible liquid and an ideal gas. (13} Value of the thermal diffusivity of air at 350 K for pressures of l. 5 and 10 atni. (0) Location where transition occurs for air flow over a flat plate with T,3 = 350 K. p = 1. 5 and 10 atm. and ux = 3 iii-"s. ASSUMPTIONS: (1} Steady-state conditions. (2) Constant properties. (3} Transition at Rex.c = 5 x 105. (4) Ideal gas behavior. PROFERTIES: Table A4. air (3:30 K): n = 208.2 x 10J Its-"ml. k = 0.030 W.:’1n-K. cl: = 1009 l-"kg-K. p = 0.995 kg-51113'. ANALYSIS: (a) For an ideal gas 10 = pRT or p = p.-"RT (I) while for an incompressible liquid. p = constant (2) The kinematic viscosity is i' = pip (3) Therefore. for an ideal gas i'= itRTI'p or v 00 p':' (4) ‘1: and for an incompressible liquid i' = ILL-"p or v is independent of pressure. ‘1: The thermal diffusivity is 01'. = l: pc Therefore. for an ideal gas. [I = kRTI'pC or 0 3C p'l‘ (6} < For an incompressible liquid at = lt.-"pc or or is independent of pressure <- (b) For T = 350 K. p = 1 atm. the thermal diffusivity of air is a _ = 29.9 x we "‘ 0.99s leg-"mg >< 1009 J.-"ltg — K Using Equation 5. at p = 5 atrn. Continued. . . PROBLEI‘t—‘I 6.1]r ( Cont.) o = 29.9 X 1'0"6 1112.-"'s.-"5 = 5.98 >< 10"5 1112.58 At p = 10 mm. a = 29.9 x 10‘5 mats-’10: 2.99 :-- 10-51858 [0) For transition ot-‘er a flat plate. XU -: Re = H‘==5><10- 1+. '.|L_C There fore it: = 5 X IUD-{t'r’ttm} For To: = 350 K. p = 1 atrn. v: pip = 208.2 x 10'? N .s..-"m3/0.995 kgs'm3 = 20.92 x 105 1113.95 Using Equation 4. at p = 5 attn V: 20.92 x 10‘6 Hilts/5 = 4.18 >+ 10"5 1:12.95 At 10 = 10 8011. v = 20.92 x 10'6 n13.-'s/10 = 2.09 x 10'5 1112.55. Therefore. atp = 1 Min xc = 5 x i05 >< 20.92 x 10‘5 m3.-'s.-"(2m.-'s) = 5.23 hi At p = :3 atm. xC = 5 x 105 x 4.18 >< 10-5 n11.-'s.-'(2ui-'s) = 1.05 In At 1;: =10 attn xc = 5 x 105 >< 2.09 >< i0"S m3..-"s.-'(2m-'s) = 0.523 m < COMMEIT: Note the strong dependence of the transition length upon the pressure for the gas (the transition length is independent of pressure for the incompressible liquid). PROB LEE-I 6.2T KNOWN: Expression for the local heat transfer coefficient of air at piescribed velocity and temperature flowing over electronic elements on a circuit board and heat dissipation rate for a 4 >: 4 mm chip located lZU mm from the leading edge. Atmospheric pressure in Mexico City. FIND: (a) Surface temperature of chip. (b) Air velocity required for chip temperature to be the same at sea level. SCHEJIATIC: Approp riaf‘e corr e t‘a flan : —. = 76.5 kPa P g: 4,, 112553115 E=25flc m Nux =a04‘REX T 52 Ch‘ 5| ____ -_¢l.‘a"v _|:u:% tails: 'P saw; W Wbmrd L3"! L=I20mm ‘ ASSL‘IEIPTIONS: (1) Steady—state conditions. (2) Power dissipated in chip is lost bey convection across the upper surface only. (3) Chip sulface is isothermal. (4} The average heat transfer coefficient for the chip stuface is equivalent to the local value at X = L. (5} Negligible radiation. (6) Ideal gas behavior. PROFERTIES: Table A4. air {p = 1 atm. assume T5 = 45 “C. If: (45 GC + 25 T}? = 35 “(7): ls = 0.0259 ‘W.='m-K. v = 16.69 x 10'5 1112-"s. Pr: 0.?05. ANALYSIS: (a) From an energy balance on the chip (see above]. Elcont = Eg = 301N- (1) Newton's law of cooling for the upper chip surface can be rwritten as Ts = To: + ILIcons; ’Ir E Achip (3) 2 where Achip = I." . Prom Assiunption 4. hem-p == hx where the local coefficient can be evaluated from the correlation provided in Problem 6.35. h X V}; [3'85 1:3 Nux = x = anal ] Pr -' (3) 1; The kinematic viscosity is 1' = i (4) P while for an ideal gas. P _ = _ 3 a RT ( ) Combining Equations 4 and 5 yields v at: p'1 (5} Continued... PROBLENI 6.2? (Cont) The Prandtl number is t' 1113c pc Pr: — = — = — .7 [1 pk k ( ) which is independent of pressure. Therefore. at sea level (13 = 1 atm) 1-: = 0.0269 twin-K. 1: = 16.69 :-- 10'6 1112.55. Pr = 0.?06 0.85 k UL .- 11x = 0.04 —[—] P1‘1'3 L 1' 0.35 11K = 0.04[ ] (0.?06}1"3 = 101 1.191112 -K a = 25°C + 3*] -‘ 1'34 W 4 1 =42s°c 10? VS".-"111_ -K >< (0004111)" In Mexico City (1) = 7’65. kPa) 0.0269 win—K];— 10 1n-"s >< 0.120111 0.120 in £16.69 >< 10‘6 1111:"s 4. 101.319. 4. 4-: 16.69 >< 10"i infra >< —q = 22.10 :-- 10'6 m‘.-'s 26.51cpa k = 0.0269 ‘W-"in- K. Pr = 0.706 0.0269 W.-"1n-K]|:10 IIL-"S x 01201117035 0.120 111 2210 r-- 10-6 1112.55} 11K = 0.04[ (02061“3 = 84.5 114.51112 .1: a =2s°c + 3'3 -‘" 10's W _ a. 4. =4?.2°C *1 04.5 w..-m- -K :-- (0.004 m}- (b) For the same chip temperature. it is required that 11,5 2 10? Wiring-K. Therefore _ -.:- 0.35 . 4 . 0.0269 19: ><I’ VX 0.120 .- 11K = 107 11r_..-m- -K = 0.04: m X 5 my ] (0.106)” ! 0.120 n1 _.'!_ 22.10 x 10' 111'5'5. From rwhich we may find V = 13.2 1113's "Z COMMENTS: (1} In Part (a). the chip surface temperature increased from 42.4 “C to 4?.2 “C. This is considered to be significant and the electronics packaging engineer needs to consider the effect of large changes in atmospheric pressure on the efficacy of the electronics cooling scheme. (2) Careful consideration needs to be given to the effect changes in the atmospheric pressure on the kinematic viscosity and. in turn. on changes in transition lengths which might dramatically affect local cons-‘ectix-‘e heat transfer coefficients. PROBLEJ'I 6.35 EIGHT: Air flow conditions and. drag force associated with a heater of prescribed surface temperature and area. FI‘ND: Required heater power. SCHENL—XTIC: A = 0. 25m 3‘- Um=15m{s, —D T =14!) °C unfit, lfip—P f9 5 p=1afm —3-FD=O.25N ASSUMPTIONS: (1) Steady-state conditions. (2) Reynolds analogy is applicable. (3) Bottom surface is adiabatic. PROPERTIES: TableA—4‘. Air (Tf= 350K. latinj: p = 0.995 leg-inf. cp = 1009 J.-"kg-K. Pr = 0.?00. ANALYSIS: The average shear stress and friction coefficient are _ FD _ 0.25. N r5=— ? A 0.25111— =1 Nfin2 — .7 1 }~l.--"m2 _ Cf = j — 28.93x10 3. p 11.32 0.995 l<g.--’rn3 {lfirnf's )2 2 From the Reynolds analogy. — l— T _'2 sr=—1 =TfPr -- 3 p uch Solving for h and substituting numerical values. find E = 0.995 kgf'1113{:15111-"s] 1009 ring -K [8.93x10'3 :2] [oat—3"? E = as tram2 Hence. the heat rate is q = E A {Is —rm] = BEER-31112 -K ("0.251113 [140—15]{c c q = 2.66 kW. < COB-'Il-IENTS: Due to bottom heat losses. which have been assumed negligible. the actual power requirement would exceed 2.66 kW. PROBLEM 6.34 KND‘WN: Drag force and air flow conditions associated with a flat plate. FIB'D: Rate of heat transfer from the plate. SCHEE‘IATIC: um=40mf5 _,, Tm:20°C —D' 0.2m H P =1 5+”? —p</:{:.;=.‘ ASSUMPTIONS: (1) Chilton-Colbuin analogy is applicable. PROPERTIES: Tabled—4, Air (TUDC‘J atnl): p = 1.018 kgr'1n3. c1) = 1009 _T_.-"kg-K. Pr = 0.70. '3 -6 L. v=2022 » 10 in Is. 5:120? ANALYSIS: The rate of heat transfer from the plate is q=2h LLB} {Ts _Tu:] ; where h may be obtained from the Chilton-C'olbuln analogy. 3H =C—f = st Ping-"'3 =—11 P133 2 p 11,: c1) ‘7 -_ 0.075 Nil r 0.2111' 3 _. ICf=1 3 =IJL l 2“ 2576,104. 2 2 p 11.30 .-"2 3 1.018 kg.-’rn" (40 in-’s_]‘ Hence. E _ Cf P 2.3 — T9 “or: ‘31: T _ 5-. _4 .- _ I 3 -. I _ II r _ H _2 11 = 5._.-sx1o (1.013Lg-m )40111-‘s (_1009.T.-kg-l~.) (0. r0} E = 30 wall2 The heat rate is q = 2(30 wrm2 {0.2m}‘ (120—20? c q = 240 w. < CTOl—‘Il—‘IEXTS: Although the flow is laminar over the entire surface {ReL = LLILr'v = 40 nr's _ — 9 . . . . x 0.2111f2022 x 10 6rn'.--'s = 4.0 x 105). the pressure gradient 1s zero and the Cillllton-Cfolburn analogy is applicable to average. as 1arell as local, surface conditions. Note that the only contribution to the drag force is made by the surface shear stress. PROBLEM 6.3 KNOWN: Boundary layer temperature distribution. FI‘NIII: Surface heat flux. SCHEl-IATIC: Air, 7;: 400K W 0mm £00032" —'> PF =07 [ PROPERTIES: Tabled—4.1Mr (T3 = 300K): k = H.0263 KKK-"111K. ANALYSIS: Applying Fourier's law at y = O. the heat flux is , s T . _ ' q; =_1; (a =—k{T:0 —T$] Pru—Elexp' —Pr "30} r- }? V: 1/ J '_ V . 3:0 r 11. q; = ‘kiT-ac ‘TsiPl'i v _ q; = —0.0253 W.-"m-K{100K_]O.T , 5000 1.-"rn. q; = —9205. 001112. CfOl-‘Il-‘IEXTS: (1) Negative flux implies convection heat transfer to the surface. (2) Note use of Is at T5 to evaluate q; from Fourier‘s law. ...
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This note was uploaded on 08/17/2008 for the course MEM 345 taught by Professor Cho during the Summer '07 term at Drexel.

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Chapter 6 HW solutions - PROBLEM 6.6 KNOWN: Expression for...

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