Question

Example #1 A chicken pieces (a=1.4x10m²/s) and k=5.5(W/mk) of diameter of 2cm is initially at a uniform temperature of 7°C, is dropped suddenly in a boiling water at 100°C and the heat transfer coefficient is 150 (W/m²K), the chicken pieces is considered to be cooked and under consumer's taste if it's center temperature is 80 °C, how long will it take the center of the pieces to be cooked?

Answer 1

Given, a 2=1.4x10-7 (m/s) K=5.5 (wink) d=2am = 0.02m h=150 (W/m²K) Ti= 7°C To= 100°c T=80°c Biot Number is Bi= hlc le= Lc= Length Characteustic Volune Area.

Le = 4 T 43 4512 le= r ad 36 : Bi - 150 (0.02/6) 5.5. Bi = 0.091 Since Bicool therefore lumped Capacitance analysis is applicable Therefore its formula is - a exp (that) To-To fvce ? And a=k And, Lea v - Sce And ( L x K 80-100 es 7-100 & exp/ -15ox t. 10.02 x 5.5 6 1:4810-7

0.21505 = esp (-1.14545x10°3t). los (0.21505) = -1-14545×10 -3 t. -1.53688= -10 14545x10 m3 xt. t = 1.53688 1.14545x10-3 t = 1341.726 (seconds) t=1341726 (5) t = 1341.726 minutes) t=22.362 (mis) It will take 22.362 (min)
Using lumped capacitance analysis to get the result.