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½ºÅ×Æijë Stefano   22-01-18 01:00
(1) ÀÌ ¹®Á¦´Â È­°ø°ú ÀÔÇÐÈÄ ÈçÈ÷ Ç®¾îº¸°ÔµÇ´Â °ÍÀε¥µµ ÃâÁ¦µÇ´Â ¸ð¾çÀÔ´Ï´Ù.  ÀÌ°÷¿¡¼­ ¹®Á¦¸¦ Ç®¾îÁÖÁö´Â ¾Ê½À´Ï´Ù.  ´Ù¸¸ ÈùÆ®¸¸ ¿Ã·Á³õ°Ú½À´Ï´Ù.

A¹°Áú ¹°Áú¼öÁö:    (ÅõÀÔ·®)+(»ý¼º·®)-(¹èÃâ·®)=(´©Àû·®)

(ÅõÀÔ·®) = (ÅõÀÔ³óµµ*À¯·®)= Cf*v;  (»ý¼º·®) = - (¼Ò¸ð·®) = -(r_A*V); (¹èÃâ·®)= (¹ÝÀÀ±â³» ³óµµ*À¯·®)=C_A*v;  (¹ÝÀÀ±â³» ´©Àû·®) = (¹ÝÀÀ±â ºÎÇÇ* AÀÇ ³óµµÁõ°¡¼Óµµ) = V * dC_A/dt    .......................(1)

Cf=10.0 mol A/L, v=1.15 L/s, r_A=0.007*C_A, V=10L C_A = ¹ÝÀÀ±â³» AÀÇ ³óµµ(½Ã°£ tÀÇ ÇÔ¼ö).

À§¸¦ ±âÁØÀ¸·Î ¹°Áú¼öÁö¸¦ ³ªÅ¸³»´Â ¼ö½ÄÀ» ¸¸µé¾î¼­  ¹ÌºÐ¹æÁ¤½ÄÀ» Ç®¸é µÇ´Âµ¥  C_A¸¦ »ó¼ö(Cf, v, V)·Î ³ªÅ¸³»¸é µÇ°í, ÀûºÐ»ó¼ö¿¡ »ç¿ëÇϴ  ¹ÌºÐ¹æÁ¤½ÄÀÇ °æ°èÄ¡(B.C)´Â  Ãʱâ³óµµ (t=0À϶§ÀÇ C_A ³óµµ)´Â 2.0 mol A/L¸¦ ´ëÀÔÇÏ¸é µË´Ï´Ù.

(2) ÀûºÐ¼ö½Ä Çü½Ä :  t = ¡òdt = ¡ò {dC_A /(a - b*C_A)}  =-(1/b) ln (a-b*C_A) + C  ..............(2) 
a, b´Â »ó¼ö(Cf, v, V)·Î Ç¥½ÃµÈ °ª.  ÀÌ ¼ö½Ä¿¡ B.C °ª (C_A=2.0 at t =0)À» ´ëÀÔÇÏ¿© C¸¦ ±¸ÇÑ´ÙÀ½  C_A¸¦ tÀÇ ÇÕ¼ö·Î ³ªÅ¸³»¸é µË´Ï´Ù.
     
±èÀÎ¹è    22-01-18 05:40
ÇØ°áÇß½À´Ï´Ù! Á¤¸» °¨»çµå¸³´Ï´Ù
   

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