QUANTUM COMPUTING - EXAMPLE 32.9 : In quantum computing, find the equations of S = (a | 0 > + b | 1 >) (g | 0 > + d | 1 >) in term of | 00 >, | 01 >, | 10 > and / or | 11 > when ad = 0.
QUANTUM COMPUTING - ANSWER 32.9 : If ad = 0, then a = 0 but d ≠ 0, or d = 0 but a ≠ 0, or a = d = 0. When a = 0 but d ≠ 0, S = (b | 1 >) (g | 0 > + d | 1 >) = bg | 10 > + bd | 11 >. When d = 0 but a ≠ 0, S = (a | 0 > + b | 1 >) (g | 0 >) = ag | 00 > + bg | 10 >. When a = d = 0, S = (b | 1 >) (g | 0 >) = bg | 10 >. 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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In Galvonised iron and stainsteel pipes, in which pipe friction losses will more and why ?
CHEMICAL ENERGY BALANCE - EXAMPLE 11.5 : According to Margules Equation, P = x(1) p(1) g(1) + x(2) p(2) g(2) for a two-component mixture where P is bubble pressure, x is mole fraction, p is saturation pressure, g is constant given by ln g(1) = x(2) A x(2). Find the value of A as a constant when P = 1.08 bar, p(1) = 0.82 bar, p(2) = 1.93 bar in a 50 : 50 mole fraction mixture. Estimate the pressure required to completely liquefy the 30 : 70 mixture using the same equation, by proving P = 1.39 bar. Take note that ln g(2) = x(1) A x(1), ln g(1) = x(2) A x(2).
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Question 8 - 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.
CHEMICAL FLUID MECHANIC - EXAMPLE 3.2 : The terminal velocity of a falling object, v is given by v = sqrt [ 4g (R - r) D / (3Cr) ] where sqrt is the square root of, g = 9.81, D = 0.000208, R = 1800, r = 994.6, m = 0.000893. The Reynold number, L is given by L = rD (v) / m. The C for various conditions are : C = 24 / L for L < 0.1; C = 24 (1 + 0.14 L^0.7) / L for 0.1 <= L <= 1000; C = 0.44 for 1000 < L <= 350000; C = 0.19 - 80000 / L for 350000 < L. Find the value of v for the situation above by trial and error, ^ is power, <= is less than or equal to.
PROCESS DESIGN - EXAMPLE 21.1 : According to rules of thumb in chemical process design, consider the use of an expander for reducing the pressure of a gas when more than 20 horsepowers can be recovered. The theoretical adiabatic horsepower (THp) for expanding a gas could be estimated from the equation : THp = Q [ Ti / (8130a) ] [ 1 - (Po / Pi) ^ a ] where 3 ^ 3 is 3 power 3 or 27, Q is volumetric flowrate in standard cubic feet per minute, Ti is inlet temperature in degree Rankine, a = (k - 1) / k where k = Cp / Cv, Po and Pi are reference and systemic pressures respectively. (a) Assume Cp / Cv = 1.4, Po = 14.7 psia, (temperature in degree Rankine) = [ (temperature in degree Celsius) + 273.15 ] (9 / 5), nitrogen gas at Pi = 90 psia and 25 degree Celsius flowing at Q = 230 standard cubic feet per minute is to be vented to the atmosphere. According to rules of thumb, should an expander or a valve be used? (b) Find the outlet temperature To by using the equation To = Ti (Po / Pi) ^ a.
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HOW WOULD YOU CALIBRATE A ROTAMETER
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The water is superheated steam at 440 degree Celsius and 17.32 megapascals. Estimate the enthalpy of the steam above. From the steam table for water at 440 degree Celsius, enthalpy of steam, h at 18 megapascals is 3103.7 kilojoules per kilogram and at 16 megapascal is 3062.8 kilojoules per kilogram. Assume that h = mP c where P is pressure; m and c are constants at fixed temperature with small differences in P.
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QUANTUM CHEMISTRY AND CHEMICAL ENGINEERING - EXAMPLE 31.4 : In a rigid rotor model in quantum chemistry, the moment of inertia I is given by an Equation E as I = Ma x La x La + Mc x Lc x Lc = m x L x L, where m = (Ma x Mc) / (Ma + Mc) and L = La + Lc, m is the reduced mass, Ma is the mass of a, Mc is the mass of c, La is the radius of a from point O, Lc is the radius of c from point O. Prove by simplest method that Equation E is wrong.