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  ¹°ÁúÀü´Þ_È®»ê¿¡ ´ëÇÑ Áú¹®ÀÔ´Ï´Ù.(¾Æ·¡ Áú¹®¿¡ À̾îÁý´Ï´Ù)
  ±Û¾´ÀÌ : ź»§   °íÀ¯ID : euphoria86     ³¯Â¥ : 09-03-11 02:02     Á¶È¸ : 5700    

Problem 1-1: Water in the bottom of a narrow metal cylinder is held at a constant temperature of 293 K.  The total pressure of air (assued dry) is 1 atm and the temperature is 293 K.  Water evaporates and diffuses through the air in the tube and the diffusion pathe is 0.15 m long.  Calculate the rate of evaporation at steady state in kg mol/s m2.  The diffusivity of water vapor at 293 K and 1 atm is 0.25 x 10-4 m2/s.  Assume that the system is isthermal.

 

 

Problem 1-2: In problem 1-1, the water is 0.015 m deep, and the water level is 0.15 m from the top.  As the diffusion proceeds, the level drops slowly.  Calculate the time to evaporate all the water in the cylinder.

 

 

Problem 1-3: An insulated glass tube and condenser are mounted on a reboiler containing benzene and toluene.  The condenser returens liquid reflux so that it runs down the wall of the tube.  At one point in the tube, the temperature is 170 oF, the vapor contains 30 mol% toluene, and the liquid reflux contains 40 mol% toleune.  The effective thickness of the stagnant vapor filem is estimated is to 0.1 inch.  Assuming the molar latent heats of benzene and toluene are equal, calculate the rate at which toluene and benzene are being interchanged by equimolar counter current diffusion at this point in the tube.  (Data: Dtoleuen,benzene = 0.2 ft2/hr, P=1atm, Pv,toluene at 170 oF = 400 torr.

 

 

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1.1¿¡¼­ È®»êÇüÅ°¡  One-way diffusion À̹ǷÎ

´ÜÀ§Á¶ÀÛÃ¥ »óÀÇ È®»ê½ÄÀ» È°¿ëÇÏ¿©

rate°¡ 1.61*10^-7 kgmol/m^2*S·Î ±¸ÇØÁ³½À´Ï´Ù.

 

±×·¯¸é 1.2¿¡¼­ ¾Æ·¡ Áú¹®ÀÇ ´äº¯¿¡ È®»êµÇ´Â °Å¸®¸¦ °í·ÁÇØ ÁÖ¶ó°í Çϼ̴µ¥,

ÀÌ°æ¿ì¿¡ Fick'sÀÇ È®»ê¹ýÄ¢¿¡ ¶Ç´Ù½Ã one-way diffusionÀÏ °æ¿ìÀÇ ½ÄÀ»

°°´Ù°í ³õ°í ÀûºÐÇÏ¿© °É¸®´Â ½Ã°£À» ±¸ÇÏ¸é µÉ·±Áö¿ä?

 

1.3ÀÇ »óȲÀº ±â»ó¿¡¼­ÀÇ Åç·ç¿£°ú ¾×»ó¿¡¼­ÀÇ º¥Á¨ÀÇ È®»êµÇ´Â ¸ôºÐÀ²ÀÌ 

µ¿ÀÏÇϹǷΠÀÌÁ߰渷ÀÌ·ÐÀÌ ¾Æ´Ñ ´Ü¼øÇÑ °æ¸··ÐÀ¸·Î Çؼ®µÇ´Âµ¥¿ä,

ÀÌ·²¶§ ºÐÀ²À» ÀÌ¿ëÇÏ¿© ¾î¶»°Ô Ç®¾î¾ß ÇÒÁö ¸ð¸£°Ú½À´Ï´Ù.

Ȥ, ¹«Â÷¿øÀ» ÀÌ¿ëÇÏ¿© Ǫ´Â °ÍÀ̶ó¸é Á÷°æÀÌ ÁÖ¾îÁ®¾ß ÇÒµíÇѵ¥,

±×°Íµµ ¾Æ´Ï¾î¼­ ¾î¶»°Ô ¹æÇâÀ» Àâ¾Æ¾ß ÇÒÁö ¸ð¸£°Ú½À´Ï´Ù. 

°¨»çÇÕ´Ï´Ù.


Stefano    09-03-11 19:02
(1.2) ¾×ü°¡ ¸ðµÎ È®»êµÇ¾î ¹Ù´Ú³ª´Âµ¥ ¼Ò¿äµÇ´Â ½Ã°£
È®»ê¼Óµµ°¡ ´ÜÀ§½Ã°£´ç ´ÜÀ§¸éÀû´ç È®»êµÇ´Â ¹°ÁúÀÇ ¸ô¼ö, RÀ̶ó°í Çϸé 

dt ½Ã°£µ¿¾È È®»êµÇ´Â ·®, dW kg = M *R *A * dt..............(1)
dW=¹ÌºÐÁú·®, M=ºÐÀÚ·®, R=È®»ê¼Óµµ, A= ´Ü¸éÀû, dt=¹ÌºÐ½Ã°£

¾×ÀÌ ´ã±ä ¿ë±âÀÇ ´Ü¸éÀûÀÌ È®»êÇÏ´Â Åë·ÎÀÇ ´Ü¸éÀû°ú °°´Ù¸é, dW = ¥ñ * A * dL ÀÏ °ÍÀ̹ǷÎ
dt ½Ã°£µ¿¾È LevelÀÇ °¨¼Ò, dL = dW/(¥ñ * A).....................(2)

À§ÀÇ (1),(2)½ÄÀ¸·ÎºÎÅÍ L=0~L±îÁö µÉ¶§±îÁö ¼Ò¿äµÇ´Â ½Ã°£À» ±¸ÇÏ¸é µÉ °Í °°½À´Ï´Ù.

(1.3) Equimolar Counter Diffusion
¹®Á¦´Â Á¤ÇØÁø À§Ä¡¿¡¼­ ¾×, ±â»ó¿¡¼­ÀÇ ³óµµ´Â º¯ÇÏÁö ¾Ê°í ÀÏÁ¤ÇÑ ³óµµ·Î À¯ÁöµÇ°í ÀÖÀ» ¶§
¾×»óÀÇ º¥Á¨ÀÌ ±â»óÀ¸·Î ¿Å°Ü°¡¸é °°Àº ¾ç(¸ô¼ö)ÀÇ Åç·ç¿£ÀÌ ±â»ç¿¡¼­ ¾×»óÀ¸·Î À̵¿ÇØ °£´Ù°í Ç®ÀÌÇÏ¸é µÉ °ÍÀÔ´Ï´Ù.
Equimolar Counter DiffusionÀ» »ç¿ëÇÏ´Â ÀÌÀ¯´Â ¹®Á¦¸¦ Á»´õ °£´ÜÈ÷ Çϱâ À§ÇØ »ç¿ëÇÏ´Â ¹æ¹ýÀ̱⠶§¹®¿¡ À̺κÐÀ» ÀÚ¼¼È÷ Àоµµ·Ï Çϼ¼¿ä.
   

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