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.

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Question 79 - (a) The American Petroleum Institute gravity, or API gravity, is a measure of how heavy or light a petroleum liquid is compared to water. Let SG = specific gravity of petroleum liquid, and V = barrels of crude oil per metric ton. Given the formula for API gravity = 141.5 / SG - 131.5 and V = (API gravity + 131.5) / (141.5 x 0.159), find the relationship of SG as a function of V. (b) An oil barrel is about 159 litres. If a cylinder with diameter d = 50 cm and height h = 50 cm is used to contain the oil, find the volume V of the cylinder in the unit of oil barrel by using the formula V = 3.142 x d x h x d / 4. (c) First reference : 1 cubic metre = 6.2898 oil barrels. Second reference : 1 cubic metre = 6.37 oil barrels. What are the 2 factors that cause the difference in such reference data?

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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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CHEMICAL FLUID MECHANIC - EXAMPLE 3.1 : Water flows through a pipe with circular cross sectional area at the rate of V / t = 80 L / s where V is the volume and t is time. Let Av = 80 L / s where A is cross sectional area and v is velocity of fluid. For point 1, the radius of the pipe is 16 cm. For point 2, the radius of the pipe is 8 cm. Find (a) the velocity at point 1; (b) the velocity at point 2; (c) the pressure at point 2 by using Bernoulli's equation where P + Rgy + 0.5 RV = constant. P is the pressure, R = density of fluid, V = square of fluid's velocity, g = gravitational constant of 9.81 N / kg and y = 2 m = difference of height at 2 points. The pressure of point 1 is 180 kPa.

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