7.1) Convert the following (8-hr) indoor air quality standards established by the U.S. Occupational Safety and Health Administration (OSHA) from ppm to mg/m3 (at 25°C and 1 atm), or vice versa:

a. Carbon dioxide (CO2), 5,000 ppmb. Formaldehyde (HCHO), 3.6 mg/m3

c. Nitric oxide (NO), 25 ppm

Solution:a. Using 24.465x10-3 m3/mol

M = 12 + 2(16) = 44 g/mol

5,000 ppm( 1m3CO2

106 m3 air )( 44 g /mol )

24.465×10−3 m3 /mol=8,992 mg CO2/m

3

b. M = 2(1) + 12 + 16 = 30 g/mol

3.6 mg HCHO /m3( 1g

103 mg ) (24.465×10−3

m3/mol )

30g /mol ( 106 m3 air1 m3 HCHO )=2.936 ppm

c. M = 14 + 16 = 30 g/mol

25ppm( 1m3NO106 m3 air )(30g /mol )

24.465×10−3 m3/mol=30.656 mg NO /m3

7.13) Suppose the bonfire emits CO at the rate of 20 g/s on a clear night when the wind is blowing at 2 m/s. if the effective stack height of the fire is 6 m, (a) what would you expect the ground-level CO concentration to be at 400 m downwind? (b) Estimate the maximum ground-level concentration.

Solution:a. From Table 7.7 (Masters):

Stability Class = F (stable)

From Table 7.9 (Masters):σ y=15 m and σz=7m

C ( x ,0 )= QπuH σ y σ z

exp (−H2

2σ z2 )= 20×106 μg/s

π (2 m /s ) (15 m ) (7 m )exp [−(6 m )2

2 (7 m )2 ]C (0.4,0 )=20,995.4 μg /s

b. From Fig. 7.52 (Masters):x=0.2 km

(CuHQ )max

≅ 4.0×10−3 m−2

(C )max=20×106 μg/s

2 m /s( 4.0×10−3 m−2 )

(C )max=40,000μg /m3

7.14) A coal-fired power plant with effective stack height of 100 m emits 1.2 g/s of SO2 per megawatt of power delivered. If winds are assumed to be 4 m/s at that height and just over 3 m/s at 10 m, how big could the plant be (MW) without having the ground-level SO2 exceeds 365 μg/m3? (First decide which stability classification leads to the worst conditions.)

Solution:The stability classification that would lead to the worst condition is F. From Fig. 7.52:

xmax = 11 km

Using Table 7.8:σ y=a x

0.894=34 (11 )0.894=290.057 m

σ z=c xd+ f=62.6 (11)0.18+(−48.6 )=47.788 m

Q=Cmax π uHσ y σz

exp(−H2

2σ z2 )

=365 μg/ m3π (4 ) (290.057 m ) (47.788 m )

exp[ − (100 m )2

2 ( 47.778 m )2 ]Q=567,760,264.1 μg /s

Plant Size =567,760,264.1μg /s

1.2×106 μg/s perr MW=473 MW

7.16) A stack emitting 80 g/s of NO has an effective stack height of 100m. the wind speed is 4 m/s at 10 m, and it is a clear summer day with the sun nearly overhead. Estimate the ground-level NO concentration:

a. Directly downwind at a distance of 2 km.b. At the point downwind where NO is a maximumc. At a point located 2 km downwind and 0.1 km off the downwind axis.

Solution:Stability Classification B

a. x = 2 km

From Table 7.6 (Masters):P = 0.15

uH=ua(Hza )p

=(4 m /s )( 100 m10 m )

0.15

=5.65m /s

From Table 7.9 (Masters):σ y=290 m and σ z=234 m

C (2,0 )= Qπ uH σ y σ z

exp(−H2

2σ z2 )= 80×106 μg/s

π (5.65 m /s ) (290 m ) (234 m )exp [−(100 m )2

2 (234 m )2 ]C=60.62 μg/s

b. From Fig. 7.52 (Masters):

(CuHQ )max

=1.5×10−5 m−2

Cmax=QuH

(CuHQ )max

=80×106 μg /s5.65 m /s

(1.5×10−5 m−2 )=¿

Cmax=212.39 μg /s

c. y = 0.1 km

C ( x , y )= Qπ uH σ y σ z

exp(−H2

2σz2 )exp(− y

2

2σ y2 )= 80×106 μg/s

π (5.65m /s ) (290 m ) (234 m )exp [−(100 m )2

2 (234 m )2 ]exp[−(100 m )2

2 (290 m )2 ]C (2,0.1 )=57.12μg /m3

7.35) Consider a “tight” 3300 m3 home with 0.2 ach infiltration rate. The only source of CO in the home is the gas range and the ambient concentration of CO is always zero. Suppose there is no CO in the home at 6 pm, but then the oven and two burner are on for 1 hr. Assume the air is well-mixed in the house and estimate the CO concentration in the home at 7 pm and again at 10 pm.

Solution:From Table 7.13 (Masters):

Soven = 1900 mg/hrSburner = 1840 mg/hr

Stotal=Soven+2 Sburner=1900 mg /hr+2 (1840 mg / hr )=5580 mg /hr

@ 7 pm:

C ( t )=( SnV ) (1−e−nt )=[ 5580 mg / hr

(0.2 ach ) ( 3300 m3 ) ] (1−e−(0.2ach )( 1 hr ) )

C (1 )=1.53mg /m3

@ 10 pm:

C ( t )=( SnV ) (1−e−nt )=[ 5580 mg / hr

(0.2 ach ) ( 3300 m3 ) ] (1−e−(0.2ach )( 4 hr ) )

C (t )=4.66 mg /m3