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Oct 18th 2013, 00:34, by pseudo0405

作者pseudo0405 (Kasim)

看板Examination

標題[考題] 中油97年新進職員單元操作第3題

時間Fri Oct 18 00:34:51 2013

抱歉，考題我就不列上來了，我只寫出我的解答 磚壁排列為 (l800 oF) Fireclay (T1) insulating (T2) building (100 oF) 因為250 Btu/hr-ft2 > qt/A = (2000-100)/R 又Rt = R1 + R2 + R3 = (L1/0.90) + (L2/0.12) + (L3/0.40) 由於串聯所以 qt/A = q1/A = q2/A = q3/A 因為insulating磚材耐熱只有1800 oF 250 = (2000-1800) / (L1/0.90) 所以若Fireclay和insulating接觸面要低於1800 oF的話 則L1至少要0.72*12 = 8.64 inch，故Fireclay磚材必須要兩塊 也就是實際上的L1 = 2*4.5 = 9 inch = 0.75 ft (這裡如果要算界面溫度 250 = (2000-T1) / (9/12/0.90) 則T1 = 1791 oF，基本上這裡 可以不用特別去算，因為0.72和0.75差不多，所以T1和1800 oF不會差太多) 因為building磚材耐熱只有300 oF 250 = (1800-300) / (L2/0.12) 所以若insulating和building接觸面要低於300 oF的話 則L2至少要0.72*12 = 8.64 inch，故insulating必須要3塊 也就是實際上L2 = 3*3 = 9 inch = 0.75 ft (這裡同樣因為0.72和0.75差不多，所以T2和300 oF不會差太多) 最後building 250 = (300-100) / (L3/0.40) 可得到L3 = 0.32*12 = 3.84 inch 故building必須要為1塊，也就是實際上L3 = 4*1 = 4 inch = 1/3 ft 所以總和厚度L = L1 + L2 + L3 = 9 + 9 + 4 = 22 inch 實際熱傳通量qt/A = (2000-100) / [(0.75/0.90) + (0.75/0.12) + (4/12/0.40)] = 240 Btu/hr-ft2 我的計算應該沒有問題吧，還望各位多多指教，感謝~ -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 1.163.139.84

→ sweetycool:怎感覺好像是熱傳學的題目 10/18 00:52

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