HW 9 Solutions

HW 9 Solutions - 53:050 Natural Environmental Systems...

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53:050 Natural Environmental Systems Solutions to Homework No. 9 – Spring 2011 1. Ecological footprint results will be summarized and announced in class. Below is my printout. I did a bit worse this year than I did last year (5.69 earths and 221 acres) because I did a bit more traveling by plane (a biggie) and a bit more driving (although I do drive a Prius, which gets over 40 mpg). Flying seems to make a huge difference! You should know that there are several other sites that estimate your personal carbon or ecological footprint. 1
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2. Problem 8-3 in Davis and Masten. I accessed and navigated the site on March 8, 2011. The pH recorded for 14:00 (EST) on 3/8/11 was 6.6 . The Tioga River near Mansfield, PA, is probably not impacted by acid mine drainage since the pH is not less than 6. For the period of March 1 through March 8, it appears the low reading was about 6.25 and the high reading was perhaps 6.85. 3. Problem 8-7 in Davis and Masten Power required for 305 x 10 6 100 W bulbs for one year = 305 x 10 6 (100) = 3.05 x 10 10 W Coal required to provide this power, assuming 33% efficiency and given that 1 W = 1 J/s: Coal required = ) J/MJ)(0.33 MJ/kg)(10 (28.5 J/s 10 x 3.05 6 10 = 3.24 x 10 3 kg/s Coal required for a year = 3.24 x 10 3 kg/s(86,400 s/d)(365 d/yr) = 1.02 x 10 11 kg/yr Since the CFL bulb only uses 25% of the power of an incandescent bulb, Coal Savings = 0.75 x 1.02 x 10 11 kg/yr = 7.65 x 10 10 kg/yr = 76.5 Tg/yr 4. Problem 8-8 in Davis and Masten. Given: 32 machines, 400 W per machines, 100% of demand given off as heat, 8 hr at night Energy loss per day = 32 machines(400 W/machine)(8 hr/d) = 102,400 W-hr/d Convert to J for one year: 102,400 W-hr/d(10 -3 kW/W)(3.6 x 10 6 J/kW-hr)(365 d/yr) = 1.35 x 10 11 J = 135 GJ 5. A plethora of information on this subject (
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HW 9 Solutions - 53:050 Natural Environmental Systems...

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