QUANTUM BIOLOGY - EXAMPLE 33.9 : (a) Let ^ be the symbol of power where 10 ^ 1 = 10, 10 ^ 2 = 100, 10 ^ 3 = 1000 etc. Total energy consumption of the brain is about 25 Watts, whereas a Blue Gene computer requires 1.5 Mega Watts. Blue Gene computer performs at 1 petaflop. In a human body, there are approximately 10 ^ 16 synapse operations per second i. e. at least 10 petaflops. Prove by calculations that a human brain is more energy efficient than a Blue Gene computer. (b) Quantum effects and quantum entanglement in the brain are identical to quantum gravity and string theory. If one is true, the other is true. What conclusion can be made if quantum effects in the brain and quantum gravity are true?
QUANTUM BIOLOGY - ANSWER 33.9 : (a) Comparison of energy consumption per rate of operation is made : - Human brain : 25 Watts / (10 petaflops) = 2.5 Watts / petaflop = A. Blue Gene computer : 1.5 x (10 ^ 6) Watts / petaflop = B. Then A < B (Proven). (b) Quantum entanglement in the brain and string theory are true. 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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ENGINEERING PHYSICS - EXAMPLE 30.5 : (a) Let | A > = (Aa Ab Ac), | B > = (Ba Bb Bc), | C > = (Ca Cb Cc). Find | A > + | C > - | B > in term of Aa, Ab, Ac, Ba, Bb, Bc, Ca, Cb and Cc. (b) Let d | E > = d (Ea Eb Ec) = (d Ea d Eb d Ec). If | E > = (6 7 8), find the value of 10 | E >.
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CHEMICAL ENERGY BALANCE - EXAMPLE 11.1 : Please match the term A - E with the stated definition i - v. Terms : A. Yield. B. Selectivity. C. Relative saturation. D. Molal saturation. E. Absolute saturation. Definitions : i. (moles of desired product formed) / (moles that would have been formed if there were no side reactions and the limiting reactant has reacted completely); ii. (moles of desired product formed) / (moles of undesired product formed); iii. (relative humidity 40 % means partial pressure of water vapour equals 4 / 10 of the vapour pressure of water at the system temperature); iv. (moles of vapour) / (moles of vapour dry gas); v. (mass of vapour) / (mass of dry gas).
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Question 67 – In order to predict the wastewater production, the population number has to be understood. The population data is : 72000 (for year 1961 or P-1961), 85000 (for year 1971 or P-1971), 110500 (for year 1981 or P-1981). (a) Find the average population increase, or [ (P-1981 - P-1971) + (P-1971 - P-1961) ] / 2. (b) Find the average percentage population increase, or [ (P-1981 - P-1971) / P-1971 + (P-1971 - P-1961) / P-1961 ] / (2) X 100. (c) Find the incremental increase or P-1981 - 2 (P-1971) + P-1961. (d) Let Po = P-1981. After 2 decades or n = 2, the population is P-2001. By using arithmetical increase method, find P-2001 = Po + n (Answer for a). (e) By using incremental increase method, find P-2001 = (Answer of d) + n (n + 1) (Answer of c) / 2. (f) By using geometrical increase method, find P-2001 = Po [ 1 + (Answer of b) / 100 ] ^ n where ^ is power sign, or 1 ^ 2 = 1 x 1 = 1. (g) If the actual P-2001 = 184000, which method of estimation is more accurate, based on your answer in (d), (e) and (f)?
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PETROLEUM ENGINEERING - QUESTION 25.3 : Liquid octane has a density of 703 kilograms per cubic metre and molar mass of 114.23 grams per mole. Its specific heat capacity is 255.68 J / (mol K). (a) Find the energy in J needed to increase the temperature of 1 cubic metre of octane for 1 Kelvin. (b) At 20 degree Celsius, the solubility of liquid octane in water is 0.007 mg / L as stated in a handbook. For a mixture of 1 L of liquid octane and 1 L of water, prove by calculations that liquid octane is almost insoluble in water.
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CHEMICAL MATERIAL BALANCE – EXAMPLE 2.6 : According to Raoult's law for ideal liquid, x (PSAT) = yP where x is mole fraction of component in liquid, y is mole fraction of component in vapor, P is overall pressure and PSAT is saturation pressure. A liquid with 60 mole % component 1 and 40 mole % component 2 is flashed to 1210 kPa. The saturation pressure for component 1 is ln (PSAT) = 15 - 3010 / (T + 250) and for component 2 is ln (PSAT) = 14 - 2700 / (T + 205) where PSAT is in kPa and T is in degree Celsius. By assuming the liquid is ideal, calculate (a) the fraction of the effluent that is liquid; (b) the compositions of the liquid and vapor phases. The outlet T is 150 degree Celsius.
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