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Engr_Lec_07RM

# Engr_Lec_07RM - SUSTAINABLE DEVELOPMENT AND ENVIRONMENTAL...

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SUSTAINABLE DEVELOPMENT AND ENVIRONMENTAL STEWARDSHIP ENGR 202 /4R Lecture 07 March 04 /2010 FACULTY OF ENGINEERING AND COMPUTER SCIENCE Instructor: Dr. Saleh Kaoser Winter session

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Agenda 1. Group selection and presentation schedule 2. Solution to Assignment 1 3. Solution to Midterm 4. Water Pollution Continues (if time permits)
Solution to Assignment 1 Solutions:1 1. (a) Montreal present population = 2. 5 million Total car as per given data = 1.5 cars/capita x 2.5 million people = 3. 75 million cars. Projected population of Montreal after 5 years with population increase rate of 0.7% will be: P 5 = P 0 e rt = 2.5 . e 0.007 x 5 = 2.5 x 1.0356 = 2.589 million people The number of cars in Montreal will be : 2.589 (people) x 2 cars/capita = 5.178 million cars The number of increased cars in Montréal therefore will be: 5.178 ( million) - 3.75 (million) = 1.428 million 1. (b) Carbon (C) emission for additional cars will be: = 1.428 million cars x 853 g = 1218.08g x10 6 = 1218.08 tons C. The amount of CO 2 therefore, will be : (44/12 ) x 1218.08 Tons = 4466.308 Tons of CO 2 / day.

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Solution 2 We consider the Bar act as CSTR (Continuous Stirred Tank Reactor). Out put rate (QC) = Reaction rate While Q = 1000 m3/ hr , k= 0.40 /hr and V = 500 m3 ; C = ? First calculate the Generation rate: Generation rate of HCHO: G = 50 (smokers) x 2 (cigs/hrs) x 1.4 (mg/hr) = 140 mg/hr. Since Reaction rate consists of Generation rate (G) and Decay rate (-VkC), we can write: QC = G – kCV, or QC + kCV = G Or, C (Q + kV) = G; C = G (Q +kV) = 140 (mg/m 3 ) 1000 (m3/h) +0.4/h x 500 m 3 = 0.116 mg/m 3 ppm = 24.45 x 0.116 30 = 0.09 ppm> 0.05 ppm
Solution 3. (a) Net energy out put = 500 MW = 5 00x 10 3 kW = 5 x 10 5 x (24 hr/day) kW-hr /day = 1.20 x 10 7 kW-hr/day = 1.20 x 10 7 x 3600 kJ /day = 4.32 x 10 10 kJ/day Running at full capacity means with 33 efficiency with heating value (HV) of 30, 940 KJ/Kg Efficiency (η) = E out E in Or, E in = E out η (E in = Mass coal x HV coal ) = E out η = 4.32 x 10 10 kJ/day 0.33 x 30,940 = 4231 Tons/day (b) Since, ash represents 12% of the coal. Therefore total ash is : 4231 x 0.12 = 507.72 Tons. A total of 75% is entrained as flue gas. Therefore, 507.72 x 0.75 =380.8 tons of ash will be emitted as flue gas. The total S produced is 2 % of the coal, which is 4231 x 0.02 = 84.62 tons of S. A total of 97% of such S will produce SO 2 . That is : 0.97 x 84.62 = 82.08 tons of S will produce: 82.08 x 2 = 164 Tons of SO 2 (c)

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(d) Thermal energy input to plant = Mass coal x (HV) = 4231 x10 3 (kg) x 30,940 kJ/kg = 130907140 x 10 3 kJ/day = 1.309 x 10 11 kJ/day Mass Nox = (1.309 x 10 11 kJ/day) x(300 ng/J) x (10 3 J/kJ) = 39.27 x 10 15 ng/day = 39.27 tons/day (Since 1 ton =10 6 g and 1 g = 10 9 ng; 1015 ng )
MIDTERM SOLUTION

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