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practice - PROBLEM 1.65 KNOWN: Conditions associated with...

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PROBLEM 1.65 KNOWN: Conditions associated with surface cooling of plate glass which is initially at 600 ° C. Maximum allowable temperature gradient in the glass. FIND: Lowest allowable air temperature, T SCHEMATIC: ASSUMPTIONS: (1) Surface of glass exchanges radiation with large surroundings at T sur = T , (2) One-dimensional conduction in the x-direction. ANALYSIS: The maximum temperature gradient will exist at the surface of the glass and at the instant that cooling is initiated. From the surface energy balance, Eq. 1.12, and the rate equations, Eqs. 1.1, 1.3a and 1.7, it follows that () ( ) 44 ss s u r dT -k h T T T T 0 dx εσ −− = or, with (dT/dx) max = -15 ° C/mm = -15,000 ° C/m and T sur = T , C 2 WW 1.4 15,000 5 873 T K mK m  =   84 4 4 24 W 0.8 5.67 10 873 T K . × T may be obtained from a trial-and-error solution, from which it follows that, for T = 618K, 21 000 1275 19 730 ,, . W m W m W m 22 2 ≈+ Hence the lowest allowable air temperature is T K = 345 C. 618 < COMMENTS: (1) Initially, cooling is determined primarily by radiation effects. (2) For fixed T , the surface temperature gradient would decrease with increasing time into the cooling process. Accordingly, T could be decreasing with increasing time and still keep within the maximum allowable temperature gradient.
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PROBLEM 3.27 KNOWN: Operating conditions for a board mounted chip. FIND: (a) Equivalent thermal circuit, (b) Chip temperature, (c) Maximum allowable heat dissipation for dielectric liquid (h o = 1000 W/m 2 K) and air (h o = 100 W/m 2 K). Effect of changes in circuit board temperature and contact resistance. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Negligible chip thermal resistance, (4) Negligible radiation, (5) Constant properties. PROPERTIES: Table A-3 , Aluminum oxide (polycrystalline, 358 K): k b = 32.4 W/m K. ANALYSIS: (a) (b) Applying conservation of energy to a control surface about the chip () in out EE 0 −= , cio qqq 0 ′′ −− = c, i o c it , c o b TT q 1h Lk R ∞∞ −− =+ ++ With TT ` qW m c 31 0 42 , h o = 1000 W/m 2 K, k b = 1 W/m K and t,c R1 0 m K W =⋅ , ( ) cc 2 T2 0 C 0 C 0Wm 1 1000 m K W 1 40 0.005 1 10 m K W ×= + 2 3 10 W m 33.2T 664 1000T 20,000 W m K + 1003T c = 50,664 T c = 49 ° C. < (c) For T c = 85 ° C and h o = 1000 W/m 2 K, the foregoing energy balance yields 2 c q 67,160 W m = < with o q = 65,000 W/m 2 and i q = 2160 W/m 2 . Replacing the dielectric with air (h o = 100 W/m 2 K), the following results are obtained for different combinations of k b and R .
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This note was uploaded on 08/08/2008 for the course ME 364 taught by Professor Rothamer during the Spring '08 term at Wisconsin.

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practice - PROBLEM 1.65 KNOWN: Conditions associated with...

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