{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Engr_Lec_07RM - SUSTAINABLE DEVELOPMENT AND ENVIRONMENTAL...

Info icon This preview shows pages 1–8. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Agenda 1. Group selection and presentation schedule 2. Solution to Assignment 1 3. Solution to Midterm 4. Water Pollution Continues (if time permits)
Image of page 2
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.
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Image of page 4
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)
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
(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 )
Image of page 6
MIDTERM SOLUTION
Image of page 7

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 8
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern