4th Homework Solution

4th Homework Solution - E 1 14 Solo/$0M it 3-116 A...

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Unformatted text preview: E 1 14 Solo/$0M it 3-116 A commercially available heat sink is to be selected to keep the case temperature of a transistor below 55°C in an environment at 18°C. Assumptions 1 Steady operating conditions exist. 2 The transistor case is isothermal at 55 ° C. 3 The contact resistance between the transistor and the heat sink is negligible. Analysis The thermal resistance between the transistor attached to the sink and the ambient air is determined to be AT ‘ T . —T... (55—18)°C a '__—) Reuse-ambient = —'uln2itg___ = W = 1.5 C/W Q: Reese—ambient The thermal resistance of the heat sink must be below 1.5 horizontal and vertical positions, HS6071 in vertical position, positions can be selected. Problem 2-71 Tb = 100°C Ta = 20°C h = 20 W/m2 °c °C/W. Table 3-6 reveals that HSSOSO in both and H86115 in both horizontal and vertical d = 2 mm L = 50 mm ~ Assumptions: 1. Steady state. 2. Constant properties. 3. One-dimensional temperature distribution. /fl@ P=rd=nx0.002=6.283x10'3m . 2 _ 2 ‘ A=i=n9;%°—2—=3.1416x105m 2/ x From TableAl, for carbon steel, k = 60.5 W/mK, h = h = 20 W/m2 K C . = 100 - 20 = 80 °c ' ab V hP hrtd 4h 20 m = / = / = /_ = ____.—— = 25.71/meter kAx 1“Edi/4 kd 60. 0.002 mL = 25.71 x 0205 = 1.286 Urb >4 >4 (c) If the tip is insulated (he = 0) q = /h P k Ax 9b ta'nh mL = /20 x 6.283 x 10'3 x 60.5 x 3.1416 x 10'6 x 80 x tanh(1.286) q=0.335W $510: C170,E> _ b _ 80 _ o . . __ 0 611:0 — cosh mL — cosh 1.286 — 41'1 C " Tx=o — ex=o + Tea = 41'1 + 20 = 61'1 C 3-79 3-123 A hot plate is to be cooled by attaching aluminum pin fins on one side. The rate of heat transfer fiom the 1 m by 1 m section of the plate and the effectiveness of the fins are to be determined. Assumptions 1 Steady operating conditions exist. 2 The temperature along the fins varies in one direction only (normal to the plate). 3 Heat transfer fi'om the fin tips is negligible. 4 The heat transfer coefficient is constant and uniform over the entire fin surface. 5 The thermal properties of the fins are constant. 6 The heat transfer coefficient accounts for the effect of radiation from the fins. ‘ I Properties The thermal conductivity of the aluminum plate and fins is given to be k = 237 W/m-°C. Analysis Noting that the cross-sectional areas of the fins are constant, the efficiency of the circular fins can be determined to be 2 a m: h_p= hzrD ___ Eli: 4(35W/m .C) =15'37m4 kAc [c7702 /4 kD (237 W/m.°C)(0.0025 In) - -1 Ufin = tanh mL = tanh(_13.37 m x0.03 m) = 0935 mL 15.37 m" x 0.03 m The number of fins, finned and unfinned surface areas, and heat transfer rates fiom those areas are 2 "» D=0.25 cm 1m n = -——————— = 27,777 (0.006 m)(0.006 m) 2 2 Ah = 27777[7rDL +537] = 27777[7r(0.0025)(0.03) +—_”(°'O:25) ] =6.68m2 2 2 Aunfimd =1— 27777[”%) = 1—27777[”—(‘%2_5)—] = 035 m2 . Qfinned = 77mefin,max = UfiuhAfin (Tb ‘77:») = 0.935(35 W/m2 .°C)(6.68 m2 )(100 — 30)°c = 15,300 W 9‘“de = hAunfimd (Tb — T1,) = (35 W/m2 .°C)(0.86 m2 )(100—30)°C = 2107 W Then the total heat transfer from the finned plate becomes Q'mmn = Q'finned + 9%.. = 15,300 +2107 =1.74><104 W = 17.4kW The rate of heat transfer if there were no fin attached to the plate would be Anofin = (1 m)(1m) =1m2 Q’nofln = hAnofin (Tb — 7;, ) = (35 W/m2 .°C)(1m2)(100—30)°C = 2450 W Then the fin effectiveness becomes Q'fin _ 17,400 g = _ _ fin Qnufin = 7.10 O R=2.Sm d=7m Tl=-180C IV v 0W,” T2 = k = DC Kgfi/I 1b Assumptions: 1. Steady state 2. Cc‘mstant prope ' . 0.7/5 V 1905’ 3. Conduction shape factor applicable Solution: Entry 8, Table aw tion shape factor for (amephere in an infin' medium «1’92; Wt our law lc} q = heat transfer rate to nitrogen = Slc (T2 - T1) = 38.25 x 0.17 [10 - (-180)] = 1235 W Mass rate of vaporization of liquid nitrogen q 1235 -3 — E — m — 6219 x ‘0 kg/S Total mass of LN 2 = pV = 808.6 X[§]Tt x 2.53 = 5.292 x 10“ kg -3 _', LN2 Vapofized/hour = x 100 = 0_042% per hour 5.292 x10 kg V Ti = 20°C Tf = 95°C ' - Air Ten Tm = 100°C h = 20 W/m2 °C 1 l ‘l, l d = 1 cm L = 3 cm For copper, from Table A1, k = 401 W/m K T cp = 385 J/kg °c p = 8933 kg/m3 Bi = fl = W = 2.49x10'4 << 1 KW k 401 L Lumped analysis is appropriate. : lg {n 0W” E30?“ ' Tr - Tm From Eq. 2.8.2 Ti _ TN = exp - » 2 a. 2 : As=ndL +2x%=n[0.015(0.03+2x0401]=1.1x10'3m2 f 0.012 ‘ m = pV = 8933 x n: x x 0.03 = 2.105 x 10'2 kg 4 . _ mp Tr‘Tm _ 2.105 x 10% 335 95-100 “‘WlnT-T “ -3 1“ 20—100=10218 s i co 20x1.lx10 My Ti1 = 30. °C ‘11 = 60 s hI = 40 W/m2 °c Tal = 95°C Tfi = temp. at the end of ,one minute in the first stream h2 = 20 W/m2°c - T 2 = 20 °c 0 T = S . ' = \ ‘2 Ti2 Tn ' ~ ' W KWO V909? ‘ Tfl - Tml “:1 ’ 61111:l As From Eq.2.8.2, = exp [ w ] = exp pdc ] E— 11 cml H 13 From Table A1, for 316'Stainless steel, p = 8238 kg/m3 cp = 468 Mg K T - T 6h 1 . T - 9s ' r1 cox H H __ -6x40x60 _ o T_i"1"-_T:; ‘ exp pdcp ] 30 - 95 ' “p [8238 x 0.004 x 468}: TH ‘ 6945 C For the second bath Ti2 = Tfl = 69.45 °'c T1322 = exp Ti2 - Twz pdcp “‘22 _> Trz' 20 _ex - 6 3020 x 60 " 697215" - 2'0” ‘ P O '82‘3"8——x 0._00‘_‘4 x 46—8] sz ‘ 51 C ...
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4th Homework Solution - E 1 14 Solo/$0M it 3-116 A...

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