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  À¯Ã¼¿ªÇÐ °ü·Ã Áú¹®ÀÔ´Ï´Ù..
  ±Û¾´ÀÌ : ³ì»öµ¹ÀÌ   °íÀ¯ID : jsi1985     ³¯Â¥ : 09-05-28 15:42     Á¶È¸ : 4205    
   ²Ù¹Ì±â_13113.tif (507.8K), Down : 5, 2009-05-28 15:42:28
ºÎÇÇ°¡ ÀÏÁ¤ÇÑ ÅÊÅ© ( 70ft^3 ) ¿¡ Áø°ø ÆßÇÁ·Î °ø±â¸¦ »©³À´Ï´Ù. (¿Âµµ´Â 70'F ·Î ÀÏÁ¤)
ºüÁ® ³ª¿À´Â ºÎÇÇ À¯·®Àº ¾Ð·Â¿¡ °ü°è¾øÀÌ 1.0 ft^3/min À» »Ì¾Æ ³»°í
µé¾î°¡´Â ºÎÇÇ À¯·®Àº 10^-4 lbm/min atm ( P_outside - P_tank)
¶ó°í ÇÕ´Ï´Ù.
ÅÊÅ© ¾ÈÀÇ ¾Ð·ÂÀÌ 1atm ¿¡¼­ 0.01  atmÀ¸·Î ÁÙ¾îµå´Â ½Ã°£À» ±¸ÇÏ°í
ÀÌ ÅÊÅ©ÀÇ Á¤»ó»óÅ ¾Ð·ÂÀ» ±¸Ç϶ó´Â°Ô ¹®Á¦ÀÔ´Ï´Ù.
 
´Ù½Ã Á¤¸®Çϸé
 
V(tank) = 70 ft^3
output Q = 1.0 ft^3/min
input  m(dot) = 10^-4 lbm * (P_outside - P_tank) / (min atm)
 
Question 1 : ÅÊÅ© ¾Ð·ÂÀÌ 1 atm -> 0.01atm À¸·Î ÁÙ¾îµå´Â ½Ã°£Àº?
Question 2 : ÀÌ ÅÊÅ©ÀÇ Á¤»ó»óÅ ¾Ð·ÂÀº? 
(Á¤»ó»óÅ´ input = output À϶§Àε¥.. ÀÌ ¹®Á¦¿¡¼­ ¾î¶»°Ô Àû¿ëÀÌ µÇ´Â°É±î¿ä?? ;; )
 

Stefano    09-05-28 17:39
"³ì»öµ¹ÀÌ" À̸§ °íÄ¡¼¼¿ä..
Àü»óÀÏ jsi1985   09-05-28 19:49
¾Æ..±ÛÀ» »èÁ¦ÇÏ·Á Çߴµ¥ »èÁ¦°¡ ¾ÈµÇ³×¿ä..
À̸§ ¹Ù²Ù°í »õ·Î ¿Ã·È´Âµ¥.. °Ô½ÃÆǸ¸ ¾îÁö·´Çû³×¿ä.. ¤¸¤µ..
   

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