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Homework3

# Homework3 - MAE118A Introduction to Energy Systems Homework...

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MAE118A: Introduction to Energy Systems Homework 3 Kowsik Bodi November 16, 2009 Problem 5.1 Power: P = 1000 MW Efficiency: η = 0 . 35 Fuel Heating Value: q = 30 MJ/kg 1 kg of fuel : 0 . 02 kg of S + 0 . 01 kg of N 2 + 0 . 10 kg of Ash + 0 . 87 kg of CH a. Electricity Output: Q = P T = 1000 · 10 6 · 365 × 24 = 8 . 76 · 10 9 kWh/y b. Coal Usage: M = Q ηq = P T ηq = 10 9 · 365 · 24 · 3600 0 . 35 · 30 · 10 6 = 3 · 10 6 ton/y c. 1 kg of S 2 kg of SO 2 M SO 2 = 2 M S = 2 · 0 . 02 M = 2 · 0 . 02 · 3 · 10 6 ton/y = 1 . 2 · 10 5 ton/y d. 1 kg of N 3.28 kg of NO 2 from fuel alone 2 · 3 . 28 kg of NO 2 in total M NO 2 = 2 · 3 . 28 M S = 2 · 3 . 28 · 0 . 01 M = 2 · 3 . 28 · 0 . 01 · 3 · 10 6 ton/y = 2 · 10 5 ton/y e. 1 kg of fuel 0.10 kg of fly ash M ash = 0 . 10 M = 0 . 10 · 3 · 10 6 ton/y = 3 · 10 5 ton/y 1

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Problem 5.2 1 kg 2 . 205 lbs, 1 J = 9 . 4782 · 10 - 4 Btu. M SO 2 Q = 1 Q · 2 · 0 . 02 · Q ηq = 3 . 81 · 10 - 9 kg/J = 8 . 86 lb/MBtu M NO 2 Q = 1 Q · 2 · 3 . 28 · 0 . 01 · Q ηq = 6 . 25 · 10 - 9 kg/J = 14 . 5 lb/MBtu Compared to the 1970 values, SO 2 emissions have increased 7 times, and NO 2 emissions have increased 20 times. Problem 5.3 Power: P = 1000 MW Efficiency: η = 0 . 35 Capacity factor: η c = 0 . 90 Fuel Heating Value: q = 30 MJ/kg Coal Consumption rate: ˙ M = P η c ηq CH + 5 4 O 2 -→ CO 2 + 1 2 H 2 O From the above reaction, 13 kg of CH requires 40 kg of O 2 . = 40 × 4 . 319 kg of air. Since Coal is being burnt in 20% excess air, air mass is 1 . 2 × 4 . 319 × 40 / 13 kg ( 16 kg) per 1 kg of Coal. Flue Gas is the term for the exhaust gases from the power plant. So, the mass of flue gas is the sum of masses of fuel and air.
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Homework3 - MAE118A Introduction to Energy Systems Homework...

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