COMPUTER PROGRAMMING FOR ENGINEERS - EXAMPLE 17.3 : (a) The byte is the basic building block of computer data used in chemical engineering process simulation where 16 bits make a word, 4 bits make a nibble, 32 bits make a quad word and 8 bits make a byte. Then how many nibbles are there in a megabytes? (b) In computer data items, let : 1 bit - counts from 0 to 1, 8 bits - counts from 0 to 255, 16 bits - counts from 0 to A. What is the value of A? (c) In a binary system of 4 bits, if 1100 = 12, 1101 = 13, 1110 = 14, 1111 = 15, B = 16, then guess the value of B. (d) By using any form of tools, find the exact value of 2 power 64 or 2^64.
COMPUTER PROGRAMMING FOR ENGINEERS - ANSWER 17.3 : (a) 4 bits as a nibble, then 8 bits as 2 nibbles and as a byte. Finally a megabytes make 2 meganibbles. (b) By induction, final number of count = 2 power (number of bit) - 1. For 16 bits, A = 2 power 16 - 1 = 65536 - 1 = 65535. (c) Undetermined and no answer for B since 1111 = 15 is maximum in a binary system of 4 bits. (d) By using web calculator, 2^64 = 18446744073709551616. 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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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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BIOCHEMICAL ENGINEERING INSTRUMENTATION - EXAMPLE 29.4 : The resolution of separation, Rs for chromatography is given by the formula Rs = (difference in retention time) / (average width at the base). In a chromatogram, 3 peaks a, b and c are found. Average widths W at the bases of the solutes are : Wa = 20 s, Wb = 40 s, Wc = 30 s. Resolutions of separation, Rs for solutes b and c in comparison to a are 2 and 4 respectively. The differences in retention times T for b and c in comparison to a are (Tb - Ta) and (Tc - Ta), Ta = Tc - Tb : (a) Form 2 equations involving Rs as a function of Wa, Wb, Wc, Ta, Tb and Tc. (b) Find the values of Ta, Tb and Tc.
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