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AME331-S08-FinalExamStudyGuide

# AME331-S08-FinalExamStudyGuide - AME 331 Final Exam Study...

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AME 331, Final Exam Study Guide May 13, 2008 Format of the exam The final exam will be similar to the midterms - unlimited open book. You may use any reference materials you want. Material covered The exam may cover any material covered during the course, which includes book chapters 1 - 8 and 10. The exam will place about 50% of the emphasis on radiation and heat exchangers since these subjects have not been covered on an exam yet, with about 25% devoted to conduction and 25% to convection. The material on heat exchangers includes: Types of heat exchangers – concentric tube, shell-and-tube, cross-flow Overall heat transfer coefficient Log mean temperature difference & modifications for non-parallel flow Effectiveness-NTU method The material on radiation includes: Planck's law Black and gray bodies Absorption, reflection, transmission, emission Radiation from surfaces Radiation shape factor Reciprocity and other relations between shape factors Radiation to/from non-black surfaces (those radiative resistance networks)

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Last year’s final exam Problem #1 (radiation) (35 points total) A paint-drying oven for cars is 4 m tall and 8 m wide. It is very deep in the 3 rd dimension (into the page), so the radiative transfer in the 3 rd dimension is negligible. It has a radiant heater on top with emissivity 0.8, a temperature of 400K and transfers 661 Watts/m 2 to the interior of the oven. The side walls are insulated with emissivity 0.5. The bottom (floor) has an emissivity of 1. If there is no car in the oven : (a) (4 points) Draw a radiative network for this system (b) (4 points) What are the radiation shape factors F top-bottom and F top-side ? (F top-side meaning 1 side, not both sides). Figure 8-12: X = , Y = 8 m, D = 4 m: X/D = , Y/D = 2: F ji = F top-bottom = F bottom-top = 0.60 Note that since the side walls are symmetrical we can treat them as identical. Summation: F top-top + F top-side + F top-side + F top-bottom = 1, F top-top = 0, F bottom-top = 0.60, Thus F top-side = F bottom-side = 0.20. Also F top-sides = F bottom-sides = 2(0.20) = 0.4 (both sides here) (c) (6 points) What is the radiosity (J) of the top surface? q top = E b , top " J top 1 " # top # top A top \$ J top = % T top 4 " q top A top 1 " # top # top = (5.67 & 10 " 8 W / m 2 K 4 )(400 K ) 4 " (661 W / m 2 ) (1 " .8)/.8 [ ] = 1286 W / m 2 (d) (10 points) What is the temperature of the bottom?
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