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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PROCESS DESIGN - EXAMPLE 21.2 : The names of the flow streams could be represented by : H1 for first hot stream, H2 for second hot stream, C1 for first cold stream, C2 for second cold stream. Data of supply temperature Ts in degree Celsius : 150 for H1, 170 for H2, 30 for C1, 30 for C2. Data of target temperature Tt in degree Celsius : 50 for H1, 169 for H2, 150 for C1, 40 for C2. Data of heat capacity Cp in kW / degree Celsius : 3 for H1, 360 for H2, 3 for C1, 30 for C2. (a) Find the enthalpy changes, dH for all streams of flow H1, H2, C1 and C2 in the unit of kW. Take note of the formula dH = (Cp) (Tt - Ts). (b) Match the hot streams H1 and H2 with the suitable cold streams C1 and C2 to achieve the maximum energy efficiency.
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