ENGINEERING MATERIAL - EXAMPLE 12.2 : At 150 degree Celsius, a mixture of 40 wt % Sn and 60 wt % Pb present, forming phases of alpha and beta. Chemical composition of Sn at each phase : CO (overall) : 40 %, CA (alpha) : 11 %, CB (beta) : 99 %. (a) State 2 reasons for the existences of alpha and beta phases for the mixture of Sn - Pb at 150 degree Celsius. (b) By using Lever Rule, calculate the weight fraction of each phase for alpha, WA = Q / (P + Q) and beta, WB = P / (P + Q) where Q = CB - CO and P = CO - CA.
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ENGINEERING MATERIAL - ANSWER 12.2 : (a) Reasons : (1) atomic radii difference; (2) different crystal structure. (b) P + Q = CO - CA + CB - CO = CB - CA, then WA = Q / (P + Q) = (CB - CO) / (CB - CA) = (99 - 40) / (99 - 11) = 0.6705. WB = P / (P + Q) = (CO - CA) / (CB - CA) = (40 - 11) / (99 - 11) = 0.3295. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.
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CHEMICAL ENERGY BALANCE - EXAMPLE 11.3 : For liquid benzene, the CP constants are : a = 129440, b = - 169.5, c = 0.64781. Reference temperature is 298 K. The temperature of benzene is 60 degree Celsius. Calculate the enthalpy of benzene by using the formula H = a (DT) + (b/2) (T^2 - TREF^2) + (c/3) (T^3 - TREF^3) where ^ is power, DT is temperature difference with TREF = 298 K. H is in J / kmol. DT = T - TREF.
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