Answer / kangchuentat

Answer 79 - (a) Substitute API gravity = 141.5 / SG - 131.5 into V = (API gravity + 131.5) / (141.5 x 0.159) will produce V = (141.5 / SG - 131.5 + 131.5) / (141.5 x 0.159) = 1 / (0.159 SG). Then SG = 1 / (0.159 V). (b) Let V = 3.142 x d x h x d / 4 = 3.142 x 50 x 50 x 50 / 4 = 98187.5 cubic centimetres = 98.1875 litres = 98.1875 litres x (1 oil barrel / 159 litres) = 0.6175 oil barrel approximately. (c) Oil density and temperature. 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.

what is mean by chemical engineering and its scope in industry

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ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.16 : An engineer would like to invest his money in a home, business and bond. The implicit interest payment frequency is monthly for home loans, quarterly for business loans; semi annually for bonds. A generalized mathematical formula to calculate I = interest rate equivalence is I = (1 + i / N) ^ N - 1 where i = annual interest rate, N = number of payment per year. (a) Calculate the value of N for : (i) home loans; (ii) business loans; (iii) bonds. (b) For i = 0.08, find the value of I for : (i) home loans; (ii) business loans; (iii) bonds.

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Question 44 - In a non-dilute absorber, graphical method is used to represent the process. In an X - Y coordinate system, X-axis represents mole of carbon dioxide / mole of water and Y axis represents mole of carbon dioxide / mole of nitrogen. The inlet gas stream consists of 8 mol % of carbon dioxide in nitrogen. (a) Find the S / G minimum as a slope that goes through the point (0, 0.0304) and (0.0000488, 0.086957). (b) Find the actual slope of operating line when it is 1.5 times the S / G minimum! (c) Find the value of x for inlet gas stream when y = 1640 x, y is mole fraction of carbon dioxide in nitrogen.

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ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.24 : (a) The net present value of an engineering project is given by V. Let Q = summation of [ T / (1 + I) ^ N ] from N = A to N = B, where P = project cost, T = tax inflow, N = duration. If V = P + Q where P = -$1000, T = $275, I = 0.12 (means 12 % , capital cost of project per year), A = 1 year, B = 5 years, calculate the value of V. (b) The present value tax shields are W = $ (11.9725, 9.1486, 6.5579, 4.1811, 2.0005) in 5 beginning balances of an engineering project. Find the adjusted present value of the engineering project X, when X = V + (summation of the values of W).

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