HW4Soln - 18.1(0.5 ft)2 L Btu 4 15 2(0.5 ft)L V hrft F h A...

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18.1 () Charts Heilser use hrftF Btu L ft L ft F hrft Btu k A V h Bi 151 . 0 4 . 12 5 . 0 5 . 0 4 15 2 2 = = = π 381 . 0 2300 200 2300 1500 = = = F F F F T T T T Y o 0 25 . 0 0 = = ft ft n 3 . 3 25 . 0 15 4 . 12 2 1 = = = ft F hrft Btu hrftF Btu hx k m From Figure F.5: X ~ 1.7 7 . 1 ~ 25 . 0 15 . 0 2 2 2 1 t ft hr ft x t X = = α t = 0.71hrs s ft ft hr ft hr ft time Length V 008 . 0 min 47 . 0 2 . 28 71 . 0 20 = = = = = 18.2 Check Biot number 004 . 0 10 5 . 0 488 3 3 2 3 2 = = hrftF Btu ft ft lb lb F hrft Btu Bi 1 st law energy balance onvection toc due heat loss of iron P V C p - ower t T = ρ amb p T T hA g mC = - t T & , define θ = T-T amb θ = lbF Btu lb ft hrFft Btu lbF Btu lb W hr Btu W 11 . 0 3 5 . 0 3 - 11 . 0 3 2931 . 0 1 500 t 2 2 = hr 1 .54 4 - hr F 169 5 t Initial condition is T=80F, or θ = 0 Integrating this expression: = 54 . 4 5169 5169 ln 54 . 4 1 t For T = 240F, or θ = 160F Time = 0.03hrs = 2 minutes
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This note was uploaded on 09/07/2009 for the course CBME Kinetics & taught by Professor Lobban during the Spring '09 term at The University of Oklahoma.

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HW4Soln - 18.1(0.5 ft)2 L Btu 4 15 2(0.5 ft)L V hrft F h A...

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