Question

Wayne Industries has developed a revolutionary computer processor which dissipates 0.25 Watts of thermal energy through a heat sink by convection and radiation (black body) into the surroundings (20 oC). You are tasked with determining the steady-state surface temperature of the heat sink if the convective heat transfer coefficient is 5 ? /?2.? , and the heat sink has an effective area of 0.001 m2 ?

Answer 1

Given that Heat dissipates =0.25 Watts heat transfer coefficient = 5W/m2.K. Heat dissipated by convexstion & radiation, To=20° ho 5w/mek, A-01001 H is At Steady state. At Steady state- → Heat dissipated = Heat last due to convectionę Radiations. =) 0,25=5X0.001 (T5–209)+Tblack body x A {ts4_(204273)4] -) 0125 = 0.005 (75-293)+51677108x0.00) (T54–2934). 5) 250 = 5(15-293) +5167X168 (T54_2934). 5) 250 5575 -1465 +5167x108x584-417.8818. ) 2132.88 = T5 (5+5167X108Ts 3).

let Ts=x 2132-88=215+5167 X10-9x})| solve, it by Newton-Raphson-methods f(a) = 567X10 8x4 +58 - 2132188 0 Let the initial guess be Tg = 3ook. 5) 50,24 = x - Flas) = f(300) = +316). f'(21) &#42)=(475167 X10883+5) So, f'(300)=4x5167X1038x (300)375 -11,1236 80,82 = 300-(417354) 2315,607357 k. 44= 315607357K 29 3X, -f(0) =) $(20) 2716678.V. HICHO #112) 512.1299. )- 31516073 - 716678 = 314.975K 12.1299 e8808 between X & Xe are very Less Hence, at Steady state Tsurface = 314,975. =41.975°C