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  #2580ÀÇ 2Â÷ Áú¹®
  ±Û¾´ÀÌ : kipa   °íÀ¯ID : kipa     ³¯Â¥ : 09-03-24 23:27     Á¶È¸ : 4232    
#2580ÀÇ #4½Ä
 
k/x*A*(Ta-T)dt = M*Cp*dT
k/x*A*dt = M*Cp*dT/(Ta-T) ----------(5)
5¹ø ó·³ ÀûºÐÇÒ °æ¿ì, Áº¯Àº ¿­Àü´Þ·ü Q'(kcal/h), ¿ìº¯Àº ÃÑ¿­·® Q(kcal)À¸·Î
´ÜÀ§°¡ ¼­·Î ´Ù¸£°Ô µÇ´Âµ¥, ¾î¶»°Ô ÇØ¾ß µÇ³ª¿ä?
 
ÀûºÐ °á°ú ¾Æ·¡¿Í °°Àº °á°ú°¡ ³ª¿Ô°í
k/x*A*t = M*Cp*ln[Ta-T]+C ----------(6)
»ó¼ö C´Â ½Ã°£ t=0, ¿Âµµ T=10 À¸·Î Çؼ­ °á°ú¸¦ ã¾ÒÀ¾´Ï´Ù.

½ºÅ×Æijë Stefano   09-03-24 23:44
(1) Á ¿ìº¯ ´ÜÀ§°¡ ¿Ö kcal/h,  kcal¿Í °°ÀÌ ´Þ¸® ³ª¿À´Â Áö k, x, A, µî ¸ðµç º¯¼öÀÇ ´ÜÀ§¸¦ »ç¿ëÇؼ­ ´Ù½Ã È®ÀÎÇØ º¸¼¼¿ä.

(2) k/x*A*(Ta-T) dt = M*Cp*dT    ÀÇ ÀûºÐ

dT/(Ta-T)=(Ak/x)/(MCp) dt
ln(Ta-T)=-(Ak/x)/(MCp) t + C 

(Ta-T) = e^(-¥á t + C) = (e^C) * e^(-¥át) = Co e^(-¥át)......(7)  Co, C, ÀûºÐ»ó¼ö
¥á = (Ak/x)/(MCp)..........................................................(8)
T=10, Ta=25, at t=0  µû¶ó¼­  (25-10) = Co =15..............(9)  À̸¦ (7)½Ä¿¡ ´ëÀÔ

T = 25 - 15 e^(-¥át) .....................................................(10)
À­½Ä °Ë»ê:  t =0ÀÏ ¶§ T=10,  t=¡ÄÀ϶§  T =25  O.K.

(3) Ta =25ÀÌ°í t=1 ÀÏ ¶§ÀÇ ¿Âµµ T °è»ê
À§ÀÇ (10)½Ä¿¡ (8)½Ä, t=1, Ta=25 ´ëÀÔÇؼ­ T¸¦ ±¸ÇÔ

T = 25 - 15*(e^-¥á).......................................................(11) ¿©±â¿¡ ¥á °ª ´ëÀÔÇϸé 1h ÈÄÀÇ T°ª°è»êµÊ.
kipa    09-03-28 12:11
Á¦°¡ °è»êÀ» À߸ø Ç߳׿ä.

¿­Àü´Þ°è¼ö U¸¦, k/x·Î ÇÏÁö ¾Ê°í, 5kcal/hm2c (ÀÚ¿¬´ë·ù)·Î ÀÔ·Â °è»êÇÏ¿´´Âµ¥
½ÇÇè°ª°ú ºñ½ÁÇÏ°Ô ³ª¿ÔÀ¾´Ï´Ù. (½ÇÇèµµ ÀÚ¿¬´ë·ù·Î)
°¨»çÇÕ´Ï´Ù.
   

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