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# HW2 - PROBLEM 1.71 KNOWN Surface-mount transistor with...

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PROBLEM 1.71 KNOWN: Surface-mount transistor with prescribed dissipation and convection cooling conditions. FIND: (a) Case temperature for mounting arrangement with air-gap and conductive paste between case and circuit board, (b) Consider options for increasing g E ± , subject to the constraint that c T = 40 ° C. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Transistor case is isothermal, (3) Upper surface experiences convection; negligible losses from edges, (4) Leads provide conduction path between case and board, (5) Negligible radiation, (6) Negligible energy generation in leads due to current flow, (7) Negligible convection from surface of leads. PROPERTIES: (Given): Air, g,a k = 0.0263 W/m K; Paste, g,p k = 0.12 W/m K; Metal leads, k = 25 W/m K. ANALYSIS: (a) Define the transistor as the system and identify modes of heat transfer. in out g st E E E E 0 - + = ∆ = ± ± ± ± conv cond,gap lead g q q 3q E 0 - - - + = ± ( 29 c b c b s c g s c g T T T T hA T T k A 3k A E 0 t L - - - - - - + = ± where s 1 2 A L L = × = 4 × 8 mm 2 = 32 × 10 -6 m 2 and c A = t × w = 0.25 × 1 mm 2 = 25 × 10 -8 m 2 . Rearranging and solving for c T , ( 29 { } ( 29 c s g s c b g s g s c T hA T k A /t 3 k A /L T E / hA k A /t 3 k A /L = + + + + + ± Substituting numerical values, with the air-gap condition ( g,a k = 0.0263 W/m K) { ( 29 2 6 2 6 2 3 c T 50W/m K 32 10 m 20 C 0.0263W/m K 32 10 m /0.2 10 m - - - = × × × + × × × ( 29 } 8 2 3 3 3 3 3 25 W/m K 25 10 m /4 10 m 35 C / 1.600 10 4.208 10 4.688 10 W/K - - - - - + × × × × + × + × c T 47.0 C = . < Continued.….

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PROBLEM 1.71 (Cont.) With the paste condition ( g,p k = 0.12 W/m K), c T = 39.9 ° C. As expected, the effect of the conductive paste is to improve the coupling between the circuit board and the case. Hence, c T decreases. (b) Using the keyboard to enter model equations into the workspace, IHT has been used to perform the desired calculations. For values of k = 200 and 400 W/m K and convection coefficients in the range from 50 to 250 W/m 2 K, the energy balance equation may be used to compute the power dissipation for a maximum allowable case temperature of 40 ° C. 50 100 150 200 250 Convection coefficient, h(W/m^2.K) 0.3 0.4 0.5 0.6 0.7 Power dissipation, Edotg(W) kl = 400 W/m.K kl = 200 W/m.K As indicated by the energy balance, the power dissipation increases linearly with increasing h, as well as with increasing k . For h = 250 W/m 2 K (enhanced air cooling) and k = 400 W/m K (copper leads), the transistor may dissipate up to 0.63 W. COMMENTS: Additional benefits may be derived by increasing heat transfer across the gap separating the case from the board, perhaps by inserting a highly conductive material in the gap.
PROBLEM 2.2 KNOWN: Hot water pipe covered with thick layer of insulation. FIND: Sketch temperature distribution and give brief explanation to justify shape. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional (radial) conduction, (3) No internal heat generation, (4) Insulation has uniform properties independent of temperature and position.

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