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  ±Û¾´ÀÌ : now   °íÀ¯ID : naomily     ³¯Â¥ : 11-05-09 12:40     Á¶È¸ : 3359    

In the dehydration of diced potatoes, the following weights were recorded at various times in the process when dehydration rate was the slowest.

t= 6 hours from start of drying: weight= 2350 g

t= 8 hours from start of drying: weight is 2275 g; moisture content= 15.6%.

In this range of moisture content, the drying rate is proportional to the moisture content, expressed in mathematical form as follows:

dW/dt = - kW

W is the moisture content in g water/g dry matter, and k is a constant.

 

(a) Derive an equation for the moisture content, W, as a function of time, t, which satisfis the experimental conditions given above.

(b) How long would it take for a product to be dehydrated from 22.5% to 12.5% moisture?

 

µµ¿ÍÁÖ¼¼¿ä.


½ºÅ×Æijë Stefano   11-05-09 23:54
(1) °ÇÁ¶µÇ´Â ¼Óµµ´Â ½Ã°£¿¡ µû¶ó ¼öºÐÇÕ·® WÀÌ °¨¼ÒµÇ´Â ºñÀ²Àε¥ °ÇÁ¶¼Óµµ°¡ ¹Ù·Î  Moisture Content W ¿¡ ºñ·ÊÇÑ´Ù°í ÇϹǷΠ   Áï -dW/dt = kW ..........................(1)

(2) À­½ÄÀ» ºÎÁ¤ÀûºÐÇÏ¸é  -¡òdW/W =¡òkdt ----->  lnW = - kt + C'........(2) 
    W = exp(-kt+C) = (exp C')*(exp(-kt)  or W=C*e^(-kt)  ...........(2a, 2b) 

(3) À§ÀÇ (2a) ȤÀº (2b)½Ä¿¡ Boundary Condition(BC)À» µÎ°³ ´ëÀÔÇÏ¿©  C¿Í k¸¦ ±¸ÇÏ¸é µË´Ï´Ù.   
BC 1:  at t=6,  Dry Matter, D + Moisture, M6 =2350 g,  M6/D = ........            (M6: t=6hÀÏ ¶§ÀÇ ½ÀºÐ Áú·®)
BC 2:  at t=8,  Dry Matter, D + Moisture, M8 =2275 g,  M8/D = 0.156.............(M8: t=8hÀÏ ¶§ÀÇ ½ÀºÐ Áú·®)

(4) (ÈùÆ®) BC 2 ¿¡¼­ D°ªÀ» ¾Ë¾Æ³¾ ÇÊ¿ä°¡ ÀÖ½À´Ï´Ù.
D + M8 = 2275,  M8/D = 0.156  ------>  D=1968 g  M8=307g ==> M6= 2350-1968=382 g, 

µû¶ó¼­ BC 1, 2¸¦ ´Ù½Ã Á¤¸®ÇÏ¸é  W=0.156 at t=8.  W=382/1968=0.1941 at t=6 <---À̸¦ ÀÌ¿ëÇؼ­ (2a, 2b)¸¦ Ç®¾î »ó¼ö¸¦ ±¸ÇÏ°í ¶Ç (b)¹ø ¹®Á¦µµ Ç®¾î¼­ ´ä±ÛÀ» ¿Ã·Áº¸¼¼¿ä.
     
now naomily   11-05-10 07:11
´äº¯ Á¤¸» °¨»çµå¸³´Ï´Ù. Ç®¾îºÁ¼­ ´ä±Û ¿Ã¸®°Ú½À´Ï´Ù. Ç®¾îº¸·Á°í ÇßÁö¸¸ Á¦°¡ ÀûºÐÀ» ¾È ¹è¿ö¼­ ³Ê¹« ¾î·Á¿ö¿ä.
   

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