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.
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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from which book shud i will prepare for the hpcl and various company exma i m fresher and my basic is not strong could u suggest from where to start
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PROCESS CONTROL - EXAMPLE 6.3 : The differential equation is 3 dy / dt + 2y = 1 with y(0) = 1. (a) The Laplace transformation, L for given terms are : L (dy / dt) = sY(s) - y(0), L(y) = Y(s), L(1) = 1 / s. Use such transformation to find Y(s). (b) The initial value theorem states that : When t approaches 0 for a function of y(t), it is equal to a function of sY(s) when s approaches infinity. Use the initial value theorem as a check to the answer found in part (a).
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