Answer / kangchuentat

PETROLEUM ENGINEERING - ANSWER 25.3 : (a) 1 cubic metre of liquid octane has 703 kilograms or 703 kilograms x mole / (0.11423 kilograms) = 6154.25 mole. Energy needed = [ 255.68 J / (mol K) ] x (6154.25 mol) K = 1573518.64 J where K is the unit symbol for Kelvin and mol is the unit symbol for mole. (b) Density of liquid octane = 703 kilograms per cubic metre = 0.703 grams per cubic centimetre = 703 g / L. There is 0.007 mg = 0.000007 g or 0.000007 g x (1 L / 703 g) = 0.0000000099573257 L of liquid octane in 1 L of water. With negligible amount of liquid octane in 1 L of water, octane is considered insoluble in water. 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.

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.6 : In a random walk, a stochastic process starts off with a score of A. This is a score for a chemical engineering test. At each discrete event, there is probability p chance you will increase your score by B and a (1 - p) chance you will decrease your score by C. The event happens T times. Let D = expected value of score. (a) Form an equation of D as a function of A, B, C, T and p. (b) Find the value of D if A = 60, B = 1, C = -1, T = 50, p = 0.5.

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