APSC 150 Case 2 Assignment 3&4 Solution

APSC 150 Case 2 Assignment 3&4 Solution - APSC 150...

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APSC 150 Biological Engineering Tutorial Solutions APSC 150: Engineering Case Studies Assignment #3 Solution 3-1. Define the system = Atmosphere of Earth (an open system) Specify the components: Component C CO 2 N u m b e r 1 2 Molar Mass 12 44 Initial amount of CO 2 = n = PV/RT (ideal gas law) = (101.3 kPa)(4E18 m 3 )(0.028E-2)/ ((8.314 kPa.m 3 /kmol.K)(273 K)) = 50.0E12 kmol Mole balance on CO 2 ACC = IN – OUT + GEN – CON IN = fossil fuel + (volcano + biomass decomposition) = (220E12 kg C)/(12 kgC/kmol C) +440E12 kg CO 2 /(44 kgCO2/kmol CO 2 ) = 28.3E12 kmol CO 2 OUT = photosynthesis +absorption = 350E12 kg CO 2 /(44 kg/kmol CO 2 ) + 370E12 kg CO 2 /(44 kg/kmol CO 2 ) = 16.4E12 kmol CO 2 GEN = 0 CON = 0 ACC = 28.3 E12 – 16.4 E12 +0 – 0 = 11.9 E12 kmol CO 2 Final CO 2 =Initial CO 2 + ACC CO 2 = 50.0E12 + 11.9E12 = 61.9 E12 kmol CO 2 Concentration of CO 2 in the atmosphere in 1990 = 0.028% (61.9E12 kmol CO 2 / 50.0E12 kmol CO 2 ) = 0.0347 % (vol) = 347 ppm (vol) c) It fluctuates because of seasonal biomass growth (plants take up CO 2 ). 3-2. a) C 8 H 18 density = 0.71 kg/L
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APSC 150 Biological Engineering Tutorial Solutions 22 US gal x 3.785 L/gal = 83.27 L 83.27 L x 0.71 kg/L = 59.12 kg of Octane 1 mole = 96 +18 = 114 g 1 kmole = 114 kg Since there are about 59.12 kg/tank, then 114 kg represents 1.9 tankfulls or about two tanks. b) C 8 H 18 + 25/2 O 2 8CO 2 +9H 2 O c) 1 mole of C 8 H 18 8 moles CO 2 1 kmole produces 8 kmoles CO 2 8 x 44 kg/kmole = 352 kg CO 2 d) person kgCO CO
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APSC 150 Case 2 Assignment 3&4 Solution - APSC 150...

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