Answer / kangchuentat

NATURAL GAS ENGINEERING - ANSWER 26.2 : (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.

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