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Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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calculate the convective heat transfer coefficient → calculate the convective heat transfer rate (Qoon) sh . Qemvh Aes (Ts-Tin) 85-57 (x(0.076232) (600-300) Qconv 2 Qconv 2 1256-534 I 1000 Apply the energy balance equation to Gnd Tout- ui. сре (Tout -Tin) = Qcom (0.05) 1005) (Tout-300) = 1256.53. Tout-300 Tout = = 2-5 302-5K Tre-sic channel is not assumed smooth!- wh we will use the sieder Tate correlation to adjust it for the channel. Assume The non-smooth turbulent flow. sieder Tate correlation equation 11'1 去 2 hemoth + (0.79033 fos Re°·s) 3. 0 +
Transcribed Image Text:calculate the convective heat transfer coefficient → calculate the convective heat transfer rate (Qoon) sh . Qemvh Aes (Ts-Tin) 85-57 (x(0.076232) (600-300) Qconv 2 Qconv 2 1256-534 I 1000 Apply the energy balance equation to Gnd Tout- ui. сре (Tout -Tin) = Qcom (0.05) 1005) (Tout-300) = 1256.53. Tout-300 Tout = = 2-5 302-5K Tre-sic channel is not assumed smooth!- wh we will use the sieder Tate correlation to adjust it for the channel. Assume The non-smooth turbulent flow. sieder Tate correlation equation 11'1 去 2 hemoth + (0.79033 fos Re°·s) 3. 0 +
Assume, f= 0.02 0 = 85-57 + 805 Re= 1 0-0762 (0.79-10-708) (48802-4) 00257 605) (0.0762) 10.07623x1.9996x105 kg/m.s ==== Re 48802.42 Au 11= 1.996×105 at 300k. d = 0.02 602 +13.32 h h = 40074. +3.33602 2 0.075 wlwk Using the energy mcp fout-Ting 0 balance equation:- = Qunv. &A Acs = width x height = (0.02) (0.5) =0.00532 0.05 x (1005) (Tout - 200k) = Acs ( 600-300) 0.05 005) (rout -300k) = 0.075X0.005 (300) 1 Sout = 300-6K)
Transcribed Image Text:Assume, f= 0.02 0 = 85-57 + 805 Re= 1 0-0762 (0.79-10-708) (48802-4) 00257 605) (0.0762) 10.07623x1.9996x105 kg/m.s ==== Re 48802.42 Au 11= 1.996×105 at 300k. d = 0.02 602 +13.32 h h = 40074. +3.33602 2 0.075 wlwk Using the energy mcp fout-Ting 0 balance equation:- = Qunv. &A Acs = width x height = (0.02) (0.5) =0.00532 0.05 x (1005) (Tout - 200k) = Acs ( 600-300) 0.05 005) (rout -300k) = 0.075X0.005 (300) 1 Sout = 300-6K)
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