Exam2_F10

Exam2_F10 - Name:_ Last First CIRCLE YOUR DIVISION: Div. 1...

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Name:__________________________________ Last First CIRCLE YOUR DIVISION: Div. 1 (9:30 am) Div. 2 (11:30 am) Div. 3 (2:30 pm) Prof . Ruan . Naik Mr . Singh School of Mechanical Engineering Purdue University ME315 Heat and Mass Transfer Exam #2 Wednesday, October 20, 2010 Instructions: Write your name on each page Closed-book exam – a list of equations is given Please write legibly and show all work for your own benefit. Write on one side of the page only. Keep all pages in order You are asked to write your assumptions and answers to sub-problems in designated areas. Only the work in its designated area will be graded. Performance 1 35 2 30 3 35 Total 100
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Name:__________________________________ Last First Problem 1 [35 pts] A sphere of diameter 18 mm has thermal conductivity of 10 W/m-K, density of 7800 kg/m 3 , and specific heat of 400 J/kg-K. Initially the sphere is at a uniform temperature of 27 C. The sphere is exposed to air at T = 2000 C with convective heat transfer coefficient h conv = 100 W/m 2 -K. The emissivity of the sphere surface (considered to be gray) is 0.8. The walls of the furnace are at T surr = 2000 C and can be considered large relative to the size of the sphere. (a) Neglecting radiation effects, calculate the time required (seconds) for the temperature of the sphere to reach 500 C. (b) Now consider that radiation from the furnace walls cannot be neglected. How much time (seconds) is required to reach 500 C at the center of the sphere? Comment on the time to reach 500 C in parts (a) and (b). In part (b), assume that radiation can be treated linearly and that the radiative heat transfer coefficient can be calculated based on initial temperature of the sphere, neglecting the variation of radiative heat transfer coefficient with changing surface temperature. List your assumptions here [3 pts]: One-dimensional radial conduction in the sphere Constant properties Uniform radiative heat transfer coefficient (remains constant) on the surface Start your answer to part (a) here [12 pts]: Biot number: 2 W3 100 m m -K 1000 0.03; W 6 10 m-K conv c c solid s hL VD Bi L kA     lumped capacitance approach is valid ( Bi < 0.1) Temperature distribution through the sphere:  exp i TT B iF o  500 2000 exp 0.03 27 2000 Fo  Solving we get: 2 9.14 c t Fo L Thermal diffusivity: 2 6 m 3.205 10 s p k C Therefore, the time required to reach 500 C: 25.7 seconds t
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This note was uploaded on 02/09/2012 for the course ME 352 taught by Professor Staff during the Fall '08 term at Purdue.

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Exam2_F10 - Name:_ Last First CIRCLE YOUR DIVISION: Div. 1...

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