Answer / kang chuen tat (malaysia - pen

Answer 17 : Let V / t = 80 L / s = 0.08 cubic metres / s. V / t = Av : (a) pai = 3.142 A = (pai)(radius square). (3.142)(0.16)(0.16)v = 0.08, v = 0.995 m / s. (b) (3.142)(0.08)(0.08)v = 0.08, v = 3.98 m / s. (c) (18000) + (1000)(9.81)(0) + 0.5(1000)(0.995)(0.995) – (1000)(9.81)(2) – 0.5(1000)(3.98) = 158885 Pa at point 2. 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.

QUANTUM COMPUTING - EXAMPLE 32.4 : A system of linear congruences consists of 3 equations : X ≡ 1 (mod 2), X ≡ 3 (mod 3), X ≡ 4 (mod 5). X has positive values. (a)(i) List the values of these equations from 1 to approximately 40. (ii) Find the first smallest value and second smallest value of X. (iii) Guess the third smallest value of X. (b) Let X ≡ Aa (mod Ma), X ≡ Ab (mod Mb), X ≡ Ac (mod Mc). According to Chinese remainder theorem, X ≡ (Aa x Ya x Md + Ab x Yb x Me + Ac x Yc x Mf) [ mod (Ma x Mb x Mc) ]. (i) Show that Ma, Mb and Mc have the greatest common divisor of Ma x Mb x Mc. (ii) Find the values of Md, Me and Mf if Md = Mb x Mc, Me = Ma x Mc and Mf = Ma x Mb. (iii) Find the values of Ya, Yb and Yc if Ya = Remainder of (Md / Ma), Yb = Remainder of (Me / Mb) and Yc = Remainder of (Mf / Mc). (iv) Use Chinese remainder theorem to find X.

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What are the greenhouse gases in earth's atmosphere?

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Question 48 - Let a ^ 2 = a x a and a ^ 3 = a x a x a where ^ is power function. Niobium is a metal with a body-centered cubic structure. The length of the unit cell structure is b = 0.3349 nm. (a) Find the volume for a unit cell structure for niobium. (b) There are 2 atoms per unit cell structure of niobium. The metal has a molar mass of 92.9 g / mol. One mole of the metal consists of 6.02 x 10 ^ 23 atoms. Find the mass of niobium per unit cell and the density of niobium.

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Is there any way to slow coke formation in ethylene furnaces?

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sir i am appearing in railway exam of section engineer and my branch is chemical engineering i want previous question paper of section engineer of chemical branch please tell me as soon as possible my email id is surendrasaroj18@gmail.com THANKYOU

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CHEMICAL ENERGY BALANCE - EXAMPLE 11.4 : Calculate the bubble temperature T at P = 85-kPa for a binary liquid with x(1) = 0.4. The liquid solution is ideal. The saturation pressures are Psat(1) = exp [ 14.3 - 2945 / (T + 224) ], Psat(2) = exp [ 14.2 - 2943 / (T + 209) ] where T is in degree Celsius. Please take note that x(1) + x(2) = 1. Please take note that y(1) + y(2) = 1, y(1) = [ x(1) * Psat(1) ] / P, y(2) = [ x(2) * Psat(2) ] / P, * is multiplication. P is in kPa.

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Chemical Engineering Material Balance - Three hundred gallons of a mixture containing 75.0 wt % ethanol and 25 wt % water (mixture specific gravity = 0.877) and a quantity of a 40.0 wt % ethanol - 60 wt % water mixture (specific gravity = 0.952) are blended to produce a mixture containing 60.0 wt % ethanol. The specific gravity of a substance is the ratio of density of a substance compared to the density of water. The symbol of weight percent is wt %. (a) Estimate the specific gravity of the 60 % mixture by assuming that y = mx c where y is wt % ethanol, x is mixture specific gravity. Values for m and c are constants. (b) Determine the required volume of the 40 % mixture.

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viscosty of crude use for petroliun?

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