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  ¹°¹æÀ» Áõ¹ß ¼Óµµ °è»ê Á» ÇØÁÖ¼¼¿ä
  ±Û¾´ÀÌ : ¸ð¸ð   °íÀ¯ID : sgip83     ³¯Â¥ : 07-11-23 20:11     Á¶È¸ : 9646    
30cm Á¤µµÀÇ ±Ý¼Ó ÆÇ À§¿¡ 10~25g Á¤µµÀÇ ¹°¹æ¿ïÀÌ Àִµ¥
 
±× ¹°¹æ¿ï¿¡ ´õ¿î ¹Ù¶÷À» °¡ÇØ Áõ¹ß½ÃÅ°´Â ¹®Á¦°¡ ÀÖ½À´Ï´Ù
 
3~5ºÐ Á¤µµ³»¿¡ ÆòÆòÇÑ ¹°À» ´Ù Áõ¹ß½ÃÅ°·Á¸é ¾î´ÀÁ¤µµÀÇ ¿­·®ÀÌ ÇÊ¿ä ÇÒ±î¿ä?
 

½ºÅ×Æijë Stefano   07-11-23 22:05
(1) Áõ¹ß¿¡ ÇÊ¿äÇÑ ¿­À» ±¸Çϸé

(Áõ¹ß¿¡ ÇÊ¿äÇÑ ¿­·®, Q) = (Áõ¹ß¿Âµµ±îÁö °¡¿­ÇÏ´Â ¿­·®, Q1) + (Áõ¹ßÀá¿­, Q2)..............(1)

Q1 =  (Mass, m)*(Specific Heat, Cp)*(Tb - To).....................(2a)
Q2 =  (Mass, m)*(Heat of Vaporization, ¥ë)............................(2b)

µû¶ó¼­  Q = m *{Cp*(Tb-T) + ¥ë} ............................................................................(3)
Tb = Boiling Point, 100¡É
To = Initial Temperature of Water Drops, 20¡É
m = Mass of Water Drops, 25 g
Cp = Specific Heat of Water Drops, 1 cal/g/¡É
¥ë = 539 cal/g

À­½Ä¿¡ ´ëÀÔÇϸé
Q = 25 * {1*(100-20)+539) = 25*619 = 15475 cal = 15.475 kcal

(2) Áõ¹ß¼Óµµ
À§ÀÇ (1)¿¡¼­ °è»êÇÑ Áõ¹ß¿­·®Àº Áõ¹ß¼Óµµ¿Í´Â °ü°è°¡ Àü¿¬¾ø´Â ¾çÀÌ¸ç ´Ü¼øÈ÷ ¹°¹æ¿ï Àüü¸¦ ¸ðµÎ Áõ¹ß½ÃÅ°´Âµ¥ ÇÊ¿äÇÑ ¿­·®¸¸ °è»êÇÑ °ÍÀÔ´Ï´Ù. 

±×·¯³ª ¸¸ÀÏ ÀÌ ¿­·®À» 3~5ºÐ »çÀÌ¿¡ Áõ¹ßÇϵµ·Ï ÇÏ·Á¸é Áõ¹ß¼Óµµ¿Í °°Àº °¡¿­¼Óµµ°¡ ÇÊ¿äÇÕ´Ï´Ù.

(°¡¿­¼Óµµ, q) = (Áõ¹ß¿¡ ÇÊ¿äÇÑ ¿­·®, Q)/(Áõ¹ß¼Ò¿ä½Ã°£, t)

q = Q/t
= 15.475 kcal/(3~5min) = 3.095 ~ 5.158 kcal/min = 185.7 ~ 309.5 kcal/h
   

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