Heat Transfer Homework

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ASSUMPTIONS: (1) Negligible radiation effects, (2) Constant properties. 5.10 to a sphere (Lc = ro/3), Hence, the temperature of the steel remains approximately uniform during the cooling process, and the lumped capacitance method may be used. 5.4 and 5.5, COMMENTS: Due to the large value of Ti, radiation effects are likely to be significant during the early portion of the transient. Problem 19.2 (WWWR) Air Water Benzene Mercury Glycerin υ 1.7× 10 − 5 0.474× 10 − 5 0.473× 10 − 5 1.06× 10 − 6 0.138× 10 − 2 Cρ 1.008× 103 1.0 0.45 0.033 0. Re 2.3× 105 1.02× 107 1.02× 107 4.57× 107 37800 Pr 0.699 2.72 5.21 0.021 1305 Nu 348 15.4 77.3 1.17 35. Fw-300 #ya-qn-sort h2 /* Breadcrumb */ #ya-question-breadcrumb #ya-question-breadcrumb i #ya-question-breadcrumb a #bc .ya-q-full-text, .ya-q-text #ya-question-detail h1 html[lang="zh-Hant-TW"] .ya-q-full-text, html[lang="zh-Hant-TW"] .ya-q-text, html[lang="zh-Hant-HK"] .ya-q-full-text, html[lang="zh-Hant-HK"] .ya-q-text html[lang="zh-Hant-TW"] #ya-question-detail h1, html[lang="zh-Hant-HK"] #ya-question-detail h1 /* Trending Now */ /* Center Rail */ #ya-center-rail .profile-banner-default .ya-ba-title #Stencil . Bgc-lgr .tupwrap .comment-text /* Right Rail */ #Stencil . Fw-300 .qstn-title #ya-trending-questions-show-more, #ya-related-questions-show-more #ya-trending-questions-more, #ya-related-questions-more /* DMROS */ .ANALYSIS: (a) On the basis of a unit axial length, the circuit, thermal resistances, and heat rates are as shown in the schematic.COMMENTS: Contact resistances between the heater and materials A and B could be important.St 2.1× 10 − 3 5.55× 10 − 7 1.45× 10 − 6 1.22× 10 − 6 7.21× 10 − 7 Problem 19.8 (WWWR) 2 sin 900 2500sin 1.in / , in qx x ab AL q Wm x m A ⎛⎞ππ⎛ ⎞ = ⎜⎟= ⎜ ⎟ ⎝⎠ ⎝ ⎠ 77.4 83.5 1 cos 1.D  The cross section of a storm window is shown in the sketch.How much heat will be lost through a window measuring 1.83m by 3.66 m on a cold day when the inside and outside air temperatures are, respectively, 295K and 250K?Problem 18.18 (WWWR) 22 0 1 0.556 n= 90(0.01) i h V W m K D L B k A W m K DL TT Y TT k m hx π π ∞ ∞ i 3 22 1 tt x t x α − == =× 25 8 3 From Figs. x tt s − (i) 2 2 0 exp 1 22 TT xhxht htx erf erf TT ttk k k αα αα ∞ ∞ 0 0 at the surface x= 0 71 1/ 2 1 / 2 0 erf (0) 0 1- (0) 1 2 x erf t ht ttz k α α − Eq (i) reduces to 2 0.484 1 (1 ) z =− −eerfz Trial & error 1/ 2 ⇒≅ =zt0.73 0.⇒=ts 551 KNOWN: Diameter and initial temperature of steel balls cooling in air.

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