ENGINEERING MATHEMATICS - EXAMPLE 8.1 : A local utility burns coal having the following composition on a dry basis : Carbon (C) 83.05 %, hydrogen (H) 4.45 %, oxygen (O) 3.36 %, nitrogen (N) 1.08 %, sulfur (S) 0.7 % and ash 7.36 %. Calculate the ash free composition of the coal with reference to C, H, O, N and S.
ENGINEERING MATHEMATICS - ANSWER 8.1 : Total percentage without ash is 83.05 + 4.45 + 3.36 + 1.08 + 0.7 = 92.64 %. For ash free composition percentage calculations, C = 83.05 / 92.64 x 100 % = 89.65 %. H = 4.45 / 92.64 x 100 % = 4.80 %. O = 3.36 / 92.64 x 100 % = 3.63 %. N = 1.08 / 92.64 x 100 % = 1.17 %. S = 0.7 / 92.64 x 100 % = 0.76 %. 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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ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.20 : Assume an engineer buys a $1 bond in period T while the nominal interest rate is R. The inflation rate at T + 1 is anticipated to be I. (a) If the bond is redeemed in period T + 1, how much money will the buyer engineer receive, in term of R, which is not affected by inflation? (b) Find the present value, PV of the proceeds from the bond, in term of R and I. (c) If the bond is redeemed in period T + 1, calculate the real growth or real value of the money that the buyer engineer will receive, in term of r = real interest rate, which is affected by inflation. (d) From the answers in (b) and (c), find the values of x, y and z in the following Fisher equation : (1 + x) = (1 + y) / (1 + z), in term of r, R and I.
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ENGINEERING PHYSICS - EXAMPLE 30.3 : (a) The quantum number m is given by m = -s, -s + 1. If s = 0.5, find the values of m. (b) | T > = (cos T) | V > + (sin T) | H >. The V and H states form a basis for all polarizations. Let cos T = 0.8. (i) If (sin T)(sin T) + (cos T)(cos T) = 1, find the value of sin T. (ii) For | T > = a | V > + b | H >, where a x a represents the probability of | V > and b x b represents the probability of | H >. Which one is more abundant, | V > or | H >? (iii) Find the value of T without using any mathematical tools.
Question 51 - A batch reactor is designed for the system of the irreversible, elementary liquid-phase hydration of butylene oxide that produces butylene glycol. At the reaction temperature T = 323 K, the reaction rate constant is k = 0.00083 L / (mol - min). The initial concentration of butylene oxide is 0.25 mol / L = Ca. The reaction is conducted using water as the solvent, so that water is in large excess. (a) Let the molecular weight of water is 18 g / mol and the mass of 1 kg in 1 L of water, calculate the molar density of water, Cb in the unit of mol / L. (b) Determine the final conversion, X of butylene oxide in the batch reactor after t = 45 min of reaction time. Use the formula X = 1 - 1 / exp [ kt (Cb) ] derived from material balance. (c) Find the equation of t as a function of X.
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ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.29 : An engineering company that produces small biochemical tools applies Installment Sales Method in its accounting. Let A = Installment Sales, B = Cost of Installment Sales, C = Gross Profit, D = Gross Profit Ratio. In year 20X8, let A = $400, B = $250, C = $150. In year 20X9, let A = $450, B = $315, C = $135. If the value of D = 37.5 % in year 20X8 : (a) find the value of D for year 20X9; (b) calculate the realized gross profit = ED / 100, for year 20X8 when the cash collected from sales is E = $100.
Question - Chemical Engineering Material - In crystal material, hexagonal crystal system could form 4-digit index in certain direction of solid. For [1(-1)0] direction in the hexagonal crystal systems of particular catalyst applied in fume removal of incinerator, what is the four-digit index for this direction? Hint : The transformation equations between the 3-digit [h’k’l’] and the 4-digit [hkil] indices are : h = (1/3) (2h’ – k’); i = - (h + k); k = (1/3) (2k’ – h’); l = l’. A. [(-1)100] B. [1(-1)00] C. [(-1)000] D. [00(-1)(-1)] E. [(-1)0(-1)0]
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