ees3040HW9solns_fa10

ees3040HW9solns_fa10 - Fall 2010 1 EES 3040 Homework 9...

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Unformatted text preview: Fall 2010 1 EES 3040 Homework 9 Solutions Reading Assignment: D&M1 Ch 1 (pp. 18-21); Ch 11 (pp. 436-458, 471-484) or D&M2 Ch 12, pp. 524-547, 571-582; and internet Problem 1. (3 pts.) Convert the following indoor air quality standard from ppm to mg/m 3 (at 25 o C and 1 atm), or vice versa. Hint: see Chapter 2 and HW1. a.) Naphthalene (C 10 H 8 ), 10 ppm MW = 10(12)+8(1) = 128 Conversion equation: mg/m 3 = (ppm)(MW)/24.465 for 25 o C and 1 atm mg/m 3 = (10 ppm)(128) / 24.465 = 52.3 mg/m 3 (2 pts.) b.) Nicotine (C 10 H 14 N 2 ), 1.0 mg/m 3 MW = 10(12)+14(1)+2(14)=162 ppm = (1.0 mg/m 3 )(24.465) / 162 = 0.15 ppm (2 pts.) Problem 2. ( 4 pts.) Indoor air quality Given: I = 0. 35 /hr, Area A = 2 50 m 2 , Volume V = 7 2 5 m 3 , Radon emission rate =0. 6 pCi/m 2-s. Decay rate constant k = 7.6 x 10-3 /hr. Steady-state concentration, C ∞ = ? Assume Ambient (outdoor) conc Ca = 0 Emission rate E = (0. 6 pCi/m 2-s)(2 5 0 m 2 ) = 1 5 0 pCi/s (0.5 pts.) Q = IV = (0. 35 /hr)(7 2 5 m 3 ) = 25 3.7 5 m 3 /hr (0.5 pts.) C ∞ = (QC a + E) / (Q + kV) (Eqn 11-23 in text) = [0 + 1 5 0 pCi/s)(3600s/hr)] / [ 25 3.7 5 m 3 /hr + (7.6 x 10-3 /hr)(7 2 5 m 3 )] = 2083 pCi/m 3 (3 pts.) Problem 3. (3 pts.) Same as Problem 2, but now a 2-story house with A = 12 5 m 2 per floor New emission rate E = (0. 6 pCi/m 2-s)(12 5 m 2 ) = 75 pCi/s since radon enters house through the ground C ∞ = (QCa + E) / (Q + kV)...
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This note was uploaded on 10/29/2011 for the course EES 3040 taught by Professor Chan-hilton during the Fall '10 term at FSU.

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ees3040HW9solns_fa10 - Fall 2010 1 EES 3040 Homework 9...

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