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PS12soln_Sp10

# PS12soln_Sp10 - PROBLEM#2 Assume CMFR Croom = C MB on room...

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1 PROBLEM #2 Assume CMFR. C room = C MB on room : dCV dt = QC in QC + E Assume steady state. dCV dt = 0, C = C ss , and C in = 0 0 = -QC ss + E E = QC ss Looking at the graph, we see that C ss = 20 ppm (mole fraction). Y = C ss C air 10 6 20ppm = C ss C air 10 6 C ss C air = 20 × 10 6 mol CO 3 3 C ss = C air 20 × 10 6 mol CO 3 3 We can use the ideal gas law to find C air C ss = n air V 20 × 10 6 mol CO 3 3 C ss = P RT 20 × 10 6 mol CO 3 3 = 1atm 82.05 × 10 6 atm m 3 mol K ( ) 298K ( ) 20 × 10 6 mol CO 3 3 C ss = 8.18 × 10 4 mol CO m 3 28 g CO mol CO = 0.023 g CO m 3 Substitute C ss into equation that we derived for E. E = QC ss = 30 m 3 h 0.023 g CO m 3 = 0.69 g CO h Emission factor = E F = 0.69 g CO h 6000 kJ h 10 6 μ g CO 1 g CO = 115 μ g kJ

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2 PROBLEM #3 (Paving Mixtures) (Ready-mix manufacture) (Rolled Steel Shape) Asphalt Concrete Rebar SRC Conventional Air Pollutants (mt) SO2 0.704 0.672 0.068 0.74 CO 1.88 1.81 0.341 2.151 Nox 0.658 0.843 0.063 0.906 VOC 0.561 0.602 0.04 0.642 Lead 0 0 0 0 PM10 0.352 0.11 0.028 0.138 Greenhouse Gases (mt CO2 eq) GWP 280 216 33.3 249.3 CO2 231 206 29 235 CH4 45.7 8.18 2.9 11.08 N2O 1.2 1.04 0.181 1.221 CFCs 1.9 1.02 1.23 2.25 Energy (TJ)
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PS12soln_Sp10 - PROBLEM#2 Assume CMFR Croom = C MB on room...

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