Question { 2994 }

CHEMICAL MATERIAL BALANCE – EXAMPLE 2.2 : 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.

Answer

CHEMICAL MATERIAL BALANCE – ANSWER 2.2 : (a) Let X = specific gravity of the 60 % mixture. By interpolation for a straight line with constant m as gradient, then (75 - 40) / (0.877 - 0.952) = (60 - 40) / (X - 0.952). Then 35 / (-0.075) = -466.67 = 20 / (X - 0.952). X = -0.0429 + 0.952 = 0.9091. (b) Let V = the required volume of the 40 % mixture. Mass of ethanol in 75 % mixture + Mass of ethanol in 40 % mixture = Mass of ethanol in 60 % mixture. Then, 300 (0.877) (0.75) + V (0.952) (0.4) = (V + 300) (0.9091) (0.6). Then 197.325 + 0.3808 V = 0.5455 V + 163.638. V = 204.53 gallons.

Question { 1753 }

CHEMICAL MATERIAL BALANCE - EXAMPLE 2.1 : Two methanol-water mixtures are contained in separate tanks. The first mixture contains 40.0 wt % methanol and the second contains 70.0 wt % methanol. If 200 kg of the first mixture is combined with 150 kg of the second, what are the mass and composition of the product? The symbol of weight percent is wt %.

Answer

CHEMICAL MATERIAL BALANCE - ANSWER 2.1 : Total mass of product mixture = 200 kg + 150 kg = 350 kg. Total mass of methanol in product mixture = 0.4 x 200 + 0.7 x 150 = 80 kg + 105 kg = 185 kg. Mass fraction of methanol in product mixture = 185 / 350 = 0.5286 or 52.86 wt %. Mass fraction of water in product mixture = 1 - 0.5286 = 0.4714 or 47.14 wt %. 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.

Question { 1758 }

CHEMICAL MATERIAL BALANCE - EXAMPLE 2.3 : A 1.5 weight % aqueous salt solution is concentrated to 4 weight % in a single-effect evaporator. The feed rate to the evaporator is F = 7500 kg / h and the feed is at 85 degree Celsius. The evaporator operates at 1 bar. By assuming that only pure solvent of water exists in the form of vapor from the feed, calculate the flow rate of such vapor V.

Answer

CHEMICAL MATERIAL BALANCE - ANSWER 2.3 : Mass balance of salt : Mass of salt in feed = Mass of salt in concentrated liquid. (0.015) (7500 kg / h) = 0.04L where L = flowrate of concentrated liquid = (0.015) (7500) / (0.04) = 2812.5 kg / h. Let F = V + L as overall mass balance, then V = F - L = 7500 - 2812.5 = 4687.5 kg / h.

Question { 1697 }

CHEMICAL MATERIAL BALANCE - EXAMPLE 2.4 : A mixture consists of benzene (B), toluene (T) and xylene (X). At a temperature of 353 K, the data of vapor pressures : B : 754.12, T : 289.71, X : 91.19. Unit is mm Hg. The pressure P is 0.5 atm. The value of k for each substance is k = (vapor pressure) / P. (a) Calculate k for B, T and X. Let L / V = 0.65. (b) By using the equation V = F / [ (L / V) + 1 ], find the value of V when F = 100, then what is the value of L?

Answer

CHEMICAL MATERIAL BALANCE - ANSWER 2.4 : (a) 0.5 atm is 1 / 2 x 760 = 380 mm Hg. Values of k could be obtained : B : 754.12 / 380 = 1.985; T : 289.71 / 380 = 0.762; X : 91.19 / 380 = 0.240. (b) V = F / [ (L / V) + 1 ] = 100 / (0.65 + 1) = 100 / 1.65 = 60.61. L = 0.65 V = 39.39. 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.

Question { 1761 }

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.

Answer

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.

Question { 2280 }

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?

Answer

Answer 79 - (a) Substitute API gravity = 141.5 / SG - 131.5 into V = (API gravity + 131.5) / (141.5 x 0.159) will produce V = (141.5 / SG - 131.5 + 131.5) / (141.5 x 0.159) = 1 / (0.159 SG). Then SG = 1 / (0.159 V). (b) Let V = 3.142 x d x h x d / 4 = 3.142 x 50 x 50 x 50 / 4 = 98187.5 cubic centimetres = 98.1875 litres = 98.1875 litres x (1 oil barrel / 159 litres) = 0.6175 oil barrel approximately. (c) Oil density and temperature. 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.

Question { 1909 }

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.

Answer

CHEMICAL MATERIAL BALANCE – ANSWER 2.6 : (a) For component 1, ln (PSAT) = 15 - 3010 / (150 + 250) = 7.475, PSAT = 1763 kPa. For component 2, ln (PSAT) = 14 - 2700 / (150 + 205) = 6.394, PSAT = 598 kPa. Let x(1) PSAT(1) = y(1) P, x(2) PSAT(2) = y(2) P, then x(1) PSAT(1) + x(2) PSAT(2) = [ y(1) + y(2) ] P = P, x(1) (1763) + [ 01 - x(1) ] (598) = 1210, x(1) (1763 - 598) = 1210 - 598, x(1) = 0.525, x(2) = 1 - x(1) = 0.475. (b) y(1) = x(1) PSAT(1) / P = 0.525 (1763) / 1210 = 0.765, y(2) = 1 - y(1) = 0.235. Overall mass balance is assumed 1 mole for F = V + L where F is incoming mole, V is flashed vapor in mole and L is outgoing liquid in mole. For component 1, zF = xL + yV then 0.6(1) = 0.525 L + 0.765 (1 - L), then L = 0.6875, V = 1 - L = 0.3125. Let z to be 60 mole %. 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.

Question { 1757 }

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.

Answer

CHEMICAL FLUID MECHANIC - ANSWER 3.1 : 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) (180000) + (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.

Question { 2306 }

Question 80 - 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.

Answer

Answer 80 - (a) 1 cubic metre of liquid octane has 703 kilograms or 703 kilograms x mole / (0.11423 kilograms) = 6154.25 mole. Energy needed = [ 255.68 J / (mol K) ] x (6154.25 mol) K = 1573518.64 J where K is the unit symbol for Kelvin and mol is the unit symbol for mole. (b) Density of liquid octane = 703 kilograms per cubic metre = 0.703 grams per cubic centimetre = 703 g / L. There is 0.007 mg = 0.000007 g or 0.000007 g x (1 L / 703 g) = 0.0000000099573257 L of liquid octane in 1 L of water. With negligible amount of liquid octane in 1 L of water, octane is considered insoluble in water. 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.

Question { 2570 }

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.

Answer

CHEMICAL FLUID MECHANIC - ANSWER 3.2 : By trial and error, v = 0.016425. Excel function of program (Insert Function) in FORMULAS could be used to find a suitable value of C. Excel (Goal Seek : What-If-Analysis) in DATA could perform iteration on v. 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.

Question { 1756 }

CHEMICAL FLUID MECHANIC - EXAMPLE 3.3 : The drag coefficient Cd = 0.05 and lift coefficient Cl = 0.4 for a levelled flow aircraft are measured. The velocity of the aircraft is v = 150 ft / s with its weight W = 2677.5 pound-force. (a) Find the value of the lift of the aircraft, L, when it is also its weight. (b) The drag of the aircraft, D = Cd M, L = Cl M. Find the value of D. (c) The power required is P = Dv. If 1 pound-force x (ft / s) = 1.356 W, find the value of P in the unit of Watt or W.

Answer

CHEMICAL FLUID MECHANIC - ANSWER 3.3 : (a) L = W = 2677.5 pound-force. (b) D / L = Cd / Cl, then D = (Cd / Cl) L = (0.05 / 0.4) (2677.5) = 334.6875 pound-force. (c) P = Dv = 334.6875 (150) (1.356) = 68075.4375 W, where D has a unit of pound-force and v has a unit of ft / s. 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.

Question { 1703 }

Question 81 - (a) In natural gas pipe sizing, the length of the pipe from the gas source metre to the farthest appliances is 60 feet. The maximum capacities for typical metallic pipes of 60 feet in length are : 66 cubic feet per hour for pipe size of 0.5 inches; 138 cubic feet per hour for pipe size of 0.75 inches; 260 cubic feet per hour for pipe size of 1 inch. By using the longest run method : (i) Find the best pipe size needed for the capacity of 75 cubic feet per hour. (ii) Estimate the suitable range of capacities for the pipe size of 1 inch. (b) The maximum capacities for typical metallic pipes of 50 feet in length are : 73 cubic feet per hour for pipe size of 0.5 inches; 151 cubic feet per hour for pipe size of 0.75 inches; 285 cubic feet per hour for pipe size of 1 inch. By using the branch method find the best pipe size needed for the capacity of 75 cubic feet per hour when the length of the pipe from the gas source metre to the appliance is 52 feet.

Answer

Answer 81 - (a)(i) Best pipe size needed = 0.75 inches for the capacity of 75 cubic feet per hour. Data for a metallic pipe of 60 feet in length and maximum capacity of 138 cubic feet per hour instead of 66 cubic feet per hour are selected. (ii) Range is more than 138 but less than or equal to 260 cubic feet per hour for pipe size of 1 inch. (b) Best pipe size needed = 0.75 inches for the capacity of 75 cubic feet per hour. Data for a metallic pipe of 60 feet in length instead of 50 feet and maximum capacity of 138 cubic feet per hour instead of 66 cubic feet per hour are selected. 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.

Question { 1901 }

Question 82 - (a) The Hyperion sewage plant in Los Angeles burns 8 million cubic feet of natural gas per day to generate power in United States of America. If 1 metre = 3.28084 feet, then how many cubic metres of such gas is burnt per hour? (b) A reservoir of natural gas produces 50 mole % methane and 50 mole % ethane. At zero degree Celsius and one atmosphere, the density of methane gas is 0.716 g / L and the density of ethane gas is 1.3562 mg / (cubic cm). The molar mass of methane is 16.04 g / mol and molar mass of ethane is 30.07 g / mol. (i) Find the mass % of methane and ethane in the natural gas. (ii) Find the average density of the natural gas mixture in the reservoir at zero degree Celsius and one atmosphere, by assuming that the gases are ideal where final volume of the gas mixture is the sum of volume of the individual gases at constant temperature and pressure. (iii) Find the average density of the natural gas mixture in the reservoir at zero degree Celsius and one atmosphere, by assuming that the final mass of the gas mixture is the sum of mass of the individual gases. Assume the gases are ideal where mole % = volume % at constant pressure and temperature.

Answer

Answer 82 - (a) 1 foot = (1 / 3.28084) metre = 0.3048 metre, 1 cubic foot = 0.3048 x 0.3048 x 0.3048 cubic metre = 0.0283 cubic metre, 8 million cubic feet = 8 million cubic feet x (0.0283 cubic metre / cubic feet) = 0.2264 million cubic metres. If 0.2264 million cubic metres of gas is burnt per day, then 0.2264 million / 24 = 9433 cubic metres of gas is burnt per hour. (b)(i) Let 50 mole of methane and 50 mole of ethane in the gas. Mass (g) = Mole (mol) x Molar Mass (g / mol), then mass of methane = 50 x 16.04 = 802 and mass of ethane = 50 x 30.07 = 1503.5. Total mass of natural gas = 802 g + 1503.5 g = 2305.5 g, then mass % of methane = (802 / 2305.5) x 100 = 34.786 % and mass % of ethane = 100 - 34.786 = 65.214 %. (ii) Density of methane gas = 0.716 g / L, density of ethane gas = 1.3562 mg / (cubic cm) = 1.3562 mg / mL = 1.3562 g / L. Volume (V) = Mass (m) / Density (r). Then V = m / r = 0.34786 m / 0.716 + 0.65214 m / 1.3562 = 0.9667 m, average density, r = m / (0.9667 m) = 1.0344 g / L. (ii) Mass (m) = Volume (V) x Density (r). Then m = Vr = 0.5V x 0.716 + 0.5V x 1.3562 = 1.0361V, average density, r = 1.0361V / V = 1.0361 g / L. 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.

Question { 1718 }

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).

Answer

MASS TRANSFER - ANSWER 4.1 : Rearranging the given equation, then FW = (FO)(CO)(TIO – TOO) / [ (CW)(TOW – TIW) ] = (0.1)(2131)(100 – 55) / [ (4178) (42 – 35) ] = 0.327891 kg / s. 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.

Question { 1689 }

MASS TRANSFER - EXAMPLE 4.2 : In a non-dilute absorber, graphical method is used to represent the process. In an X - Y coordinate system, X-axis represents mole of carbon dioxide / mole of water and Y axis represents mole of carbon dioxide / mole of nitrogen. The inlet gas stream consists of 8 mol % of carbon dioxide in nitrogen. (a) Find the S / G minimum as a slope that goes through the point (0, 0.0304) and (0.0000488, 0.086957). (b) Find the actual slope of operating line when it is 1.5 times the S / G minimum! (c) Find the value of x for inlet gas stream when y = 1640 x, y is mole fraction of carbon dioxide in nitrogen.

Answer

MASS TRANSFER - ANSWER 4.2 : (a) S / G minimum = (0.086957 - 0.0304) / (0.0000488 - 0) = 1160. (b) S / G actual = 1.5 (1160) = 1740. (c) Let x = y / 1640 = 0.08 / 1640 = 0.0000487805. The value of y corresponds to 8 % or 0.08. 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.