CHEMICAL MATERIAL BALANCE - EXAMPLE 2.5 : In a non-dilute absorber, the inlet gas stream consists of 8 mol % carbon dioxide in nitrogen. By contact with room temperature water at atmospheric pressure, 65 % of the carbon dioxide from a gas stream has been removed. (a) Find the mole ratio of carbon dioxide and nitrogen gases at inlet and outlet gas streams. (b) The Henry's Law provides y = 1640 x for carbon dioxide in water. Find the mole ratio when x = 0.0000427. Mole ratio is y / (1 - y) for y.
CHEMICAL MATERIAL BALANCE - ANSWER 2.5 : Basis = 100 mol / h. For inlet, mole % of carbon dioxide = 8 % and mole % of nitrogen = 100 - 8 = 92 % then mole ratio = 8 / 92 = 0.086957. For outlet, mole % of nitrogen = 92 %, mole % of carbon dioxide = 8 % (1 - 0.65) = 2.8 %, then mole ratio = 2.8 / 92 = 0.030435. (b) When x = 0.0000427, y = 1640 x = 1640 (0.0000427) = 0.070028, then mole ratio = y / (1 - y) = 0.0753. 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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QUANTUM BIOLOGY - EXAMPLE 33.2 : (a) In a biomolecule, ATP hydrolysis will produce 3 kcal / mol to 60 kcal / mol of energy, according to energy scale. Find the range of such energy generation if 2 moles of molecules involved. (b) A formula is given by ln P = ln a + b ln W where P = metabolic rate, W = body size, a = dependent of taxa, b = scaling exponent. If b is approximately 1 for plant, and b is approximately 0.75 for animal, find the relationship of P as a function of a and W in plant.
MASS TRANSFER - EXAMPLE 4.1 : A concentric, counter-current heat exchanger is used to cool lubricating oil. Water is used as the coolant. The mass flow rate of oil into the heat exchanger is 0.1 kg / s = FO. For oil, the inlet temperature TIO = 100 degree Celsius and the outlet temperature TOO = 55 degree Celsius. For water, the inlet temperature TIW = 35 degree Celsius and the outlet temperature TOW = 42 degree Celsius. What is the mass flow rate of water in kg / s, FW needed to maintain these operating conditions? Constant for heat capacity of oil is CO = 2131 J /(kg K) and for water is CW = 4178 J /(kg K). Use the equation (FO)(CO)(TIO ?TOO) = (FW)(CW)(TOW ?TIW).
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