Practice midterm I spr 04 solution

# Practice midterm I spr 04 solution - CHE/ME 109 Midterm...

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Unformatted text preview: CHE/ME 109 Midterm Exam I Section 01 Name 50 Uﬂ O M Instructions. PLEASE read each question carefully so that you ansWer the question that is being asked. Please CIRCLE YOUR ANSWERS on problems HI'end' Oi?“”ﬂ Wig} l‘l Gig Mai-15 1» Calculate [:1 szmiﬂd 5 pts. Problem I. On the following solid cylinder, lumped system analysis does not apply. Draw only two temperature proﬁles and label them (1) and (2): 1. the temperature proﬁle at t = o . 2. the temperature proﬁle at a short time later than t=0 Too ((1)0 air T J t7 initially, T= Ti to Ti 4 5 pts. Problem II. Explain why most materials have a thermal conductivity that varies with temperature. M49. COHAWC+.|\)\ ALflexKXS 0V) Vibrzh‘onqk ’ J Moi/ions a‘F tint. meﬁeqt s as Mu «S W \ Van cad-2L ins, (:V‘QQ Q'Q’QCWHS. Sir/764 +k1 clf/ZH7W q ' m; MA VhG/QlCLxU‘S in ‘fLQ, Lot‘HR (X and rho/QLQMRW Mesh)” QK aHggW [pal Wflzf’oc‘l‘ml/C/ Condko‘m‘qu . t, K. lk§ UCLW .nj ((9? rtaJ/ on W Tw=37°C h = 80 W/m2°C Small cylindrical candies (p = 2100 kg/m3, k = .57 W/mOC, Cp = 1000 J /kg°C and on: 3.91 x 10-6 m2/s) with a diameter of 1.0 cm and a length of 3.0 cm have a chemical at the center (r2 = 0.0005 m) which gives off ﬂuorescent light when the center (r2 = 0.0005 m) of the candies reach 35°C. The temperature inside the mouth is 37°C and the'initial temperature of the candies is 24°C. The heat transfer coefﬁcient inside the mouth is 80 W/mZOC. 10 pts. IIIa. Calculate the Biot number for the heating of the balls. Can Lumped System Analysis be used? / 1L 2 — . : : 000 J 0 = 3.91200 33’— (J-atoo tag/ms k 0.51 CF 1 #5 c. >4 5 D:l.ocm =I0|m Lc:\ﬁ':‘ ' ' 15 pts. IIIb. How long must the child suck the candy until the light will appear? (Assume the physical properties (p = 2100 kg/m3, k = .57 W/mOC, CP = 1000 J /kg°C and a: 3.91 x 10'6 mQ/s) are the same in the center where the light-emitting chemical is.) (Hint: Determine the time it takes for the candies at r=r2 to reach 35°C.) 8;:- hL : ‘éov’009M: 3407; 7(=[,08’?3 ' K ‘ ' '5} AF (.1633 @ “(Z lira : = .10 2B l ‘ m .._ _ ’ IT 9691:) 5 =A‘e/ 2‘1 Jo(\%> . t { >.io§2?3 30%} 3345 * ~ TQM“): a9 ‘2 _ S DC W / ‘ _ttog}3 I LSA 59. 3} : Mgme ( [399 241-3} 41.13211“ 7 .6364» 1 6‘, - , _ -150124g : 4487,21" T: 43-02 .. n ' O z 4!, L1 1 n2 [GLOZ \$152- “)6 1M Jg;|©.?\$7 SQC /\$ Problem IV. A 0.4 cm thick, 10 cm high and 15 cm long circuit board houses electronic components on one side that dissipate a total of 5 W of heat uniformly at the surface of the board. The board is made of a special polymer with very high thermal conductivity (k) of 10 W/mOC. All the heat generated in the components is conducted across the circuit board and is dissipated from the back side of the board into a space that is maintained at 15°C. 10 pts. IVa. Neglecting heat transfer from the sides of the board, and any heat transfer off the top of the components, show a coordinate axis with x=o at the interface of the components with the board, and x=L at the back edge of the board. Draw an arrow clearly showing the direction of the heat transfer. T‘ 6 Q Too 0 L. M) , 10 pts. Write the appropriate form of the differential equation describing the one- dimensional heat conduction in the board and the bonndary conditions for the heat ﬂux through the board to the space behind the board. There is NO HEAT GENERATION in the board. The convection heat transfer coefﬁcient at the back of the board opposite the components is 40 W/m2 oC. AZT , W L s I : ~ 2. ' @iiO ‘kir: ' o q? O'IM'IlSCW 3 33.3 LL) , 20 pts. Solve the equation in part (b) and write an expression for the temperature distribution, T(x) in the board from x=o to x=L in terms of qo , k, x, L, h and T00. —l/?l]\:‘ V I“ Eli-q \dx i" ‘ iﬁ Problem V. An alluminum heat sink is commercially available for cooling of electronic equipment. 4—— .076m ———> A/ 10 pts. Va. Calculate Aﬁn and Aunﬁn, assuming the ﬁns are 1 mm thick. 49m: :0707‘024 X4 1' E5}le +,0242<4](’Oob == . 007K50 1- ,00029 :: ,001—95 “42’ ‘ . Ame“; , 03(0 x.o‘+o : , 00?}4 ML 15 pts. Vb. The k for alluminum is 237 W/mOC. If the ﬁns are 95% efﬁcient, calculate the ﬁn effectiveness and the overall ﬁn effectiveness. 0033x,o?é ~— / (2C; ire/.15" (hp 2‘ “003?? V75 ‘ 44.101 “‘ A b V‘ .o 1030M ...
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## This note was uploaded on 11/08/2010 for the course ME 109 taught by Professor Staff during the Spring '08 term at San Jose State.

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Practice midterm I spr 04 solution - CHE/ME 109 Midterm...

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