题目
某厂吸收塔的填料层高度3m,用纯溶剂逆流等温吸-|||-收尾气中的有害组分。入塔气体中有害组分的含量为0.04(摩尔分-|||-数,下同),出塔气体中有害组分含量为0.008,出塔液体中有害组-|||-分含量为0.03,已知在操作范围内相平衡关系为 =0.8x, 试求:-|||-(1)填料塔的气相总传质单元高度Hoc为多少?-|||-(2)原塔操作液气比为最小液气比的多少倍?-|||-(3)因法定排放气有害组分含量为0.004,现增加塔高以使出口气体达标,若液气比不-|||-变,填料层总高应为多少?
题目解答
答案
解析
步骤 1：计算气相总传质单元数Nog
根据题意，已知入塔气体中有害组分的含量为0.04，出塔气体中有害组分含量为0.008，出塔液体中有害组分含量为0.03，相平衡关系为y=0.8x。首先，计算气相总传质单元数Nog。
\[ \frac{1}{A} = \frac{mG_G}{V_{UND} - y_m} = \frac{0.8 \times (0.03 - 0)}{0.04 - 0.008} = 0.75 \]
\[ N(x) = \frac{1}{1 - \frac{1}{A}} \ln \left[ (1 - \frac{1}{A}) \frac{y_m - mx}{y_m - mx} + \frac{1}{A} \right] \]
\[ = \frac{1}{1 - 0.75} \times \ln \left[ (1 - 0.75) \frac{0.04 - 0.8 \times 0}{0.008 - 0.8 \times 0} + 0.75 \right] \]
\[ = 2.773 \]
步骤 2：计算气相总传质单元高度Hog
根据步骤1计算出的气相总传质单元数Nog，计算气相总传质单元高度Hog。
\[ H_{OG} = \frac{H}{N_{OG}} = \frac{3}{2.773} = 1.082m \]
步骤 3：计算原塔操作液气比与最小液气比之比
根据题意，计算原塔操作液气比与最小液气比之比。
\[ \frac{L}{G} = \frac{y_{UND} - y_{UND}}{x_{UND} - x_{UND}} = \frac{0.04 - 0.008}{0.03 - 0} = 1.067 \]
\[ (\frac{L}{G})_{min} = \frac{y_{UND} - y_{UND}}{x_{UNDe} - x_{UND}} = \frac{0.04 - 0.008}{0.04/0.8 - 0} = 0.64 \]
\[ R = (\frac{L}{G}), (\frac{L}{G})_{min} = 1.67 \]
步骤 4：计算增加塔高后的填料层总高
根据题意，计算增加塔高后的填料层总高。
\[ y' = 0.004 \]
\[ \frac{1}{A} \] 不变，则：
\[ N'\omega = \frac{1}{1 - \frac{1}{4} \ln \left[ (1 - \frac{1}{A}) \frac{y_{总} - mx\erasure}{y_{m} - mx\erasure} + \frac{1}{A} \right] = 4.715 \]
\[ H = H_{OG} N_{OG} = 5.102m \]
根据题意，已知入塔气体中有害组分的含量为0.04，出塔气体中有害组分含量为0.008，出塔液体中有害组分含量为0.03，相平衡关系为y=0.8x。首先，计算气相总传质单元数Nog。
\[ \frac{1}{A} = \frac{mG_G}{V_{UND} - y_m} = \frac{0.8 \times (0.03 - 0)}{0.04 - 0.008} = 0.75 \]
\[ N(x) = \frac{1}{1 - \frac{1}{A}} \ln \left[ (1 - \frac{1}{A}) \frac{y_m - mx}{y_m - mx} + \frac{1}{A} \right] \]
\[ = \frac{1}{1 - 0.75} \times \ln \left[ (1 - 0.75) \frac{0.04 - 0.8 \times 0}{0.008 - 0.8 \times 0} + 0.75 \right] \]
\[ = 2.773 \]
步骤 2：计算气相总传质单元高度Hog
根据步骤1计算出的气相总传质单元数Nog，计算气相总传质单元高度Hog。
\[ H_{OG} = \frac{H}{N_{OG}} = \frac{3}{2.773} = 1.082m \]
步骤 3：计算原塔操作液气比与最小液气比之比
根据题意，计算原塔操作液气比与最小液气比之比。
\[ \frac{L}{G} = \frac{y_{UND} - y_{UND}}{x_{UND} - x_{UND}} = \frac{0.04 - 0.008}{0.03 - 0} = 1.067 \]
\[ (\frac{L}{G})_{min} = \frac{y_{UND} - y_{UND}}{x_{UNDe} - x_{UND}} = \frac{0.04 - 0.008}{0.04/0.8 - 0} = 0.64 \]
\[ R = (\frac{L}{G}), (\frac{L}{G})_{min} = 1.67 \]
步骤 4：计算增加塔高后的填料层总高
根据题意，计算增加塔高后的填料层总高。
\[ y' = 0.004 \]
\[ \frac{1}{A} \] 不变，则：
\[ N'\omega = \frac{1}{1 - \frac{1}{4} \ln \left[ (1 - \frac{1}{A}) \frac{y_{总} - mx\erasure}{y_{m} - mx\erasure} + \frac{1}{A} \right] = 4.715 \]
\[ H = H_{OG} N_{OG} = 5.102m \]