Question

An indoor air quality specialist records following information for a room:

Total number of persons including smokers = 12

Number of smokers                                       =   6

Smoking rate                                                  =   5 cigarettes/hr

Room dimensions (length, width and height) = 12m, 10m, and 4m

Mixing factor                                                  = 0.3

Filter efficiency                                               = 0.7

Recirculated indoor air                                   = 2000 m³/hr

Installed fan capacity                                     = 300 m³/hr

For calculations make the following assumptions:

Assumptions :
Mixing Factor, k			0.3
Makeup air , Q0     =       ACH*V         =			300.00	m3/hr
Filter efficiency for makeup air, F0			0
Recirculated air, Q1			2000	m3/hr
Filter efficiency for recirculated air, F1			0.7
Infiltration from outdoors, Q2			0
Deposition or reaction term , R			0
At beginning of day, assume Cs = C0
where, Cs is steady state concentration
Exfiltration (Q3)	0
Infiltration (Q2)	0

Outside air concentrations (C_o) in g/m³ for respirable particles (RP), carbon monoxide (CO) and carbon dioxide (CO₂) are 0.00070, 0.0035, and 0.650 respectively. Emission factor (mg/cig) for RP, CO, and CO₂ are 25.625, 37.5, and 318.75 respectively. The emission factor of CO₂ due to respiration is 35000 mg/hr/person

The room is being used for the following time periods: 6 A.M to 9 A.M; 1 P.M to 5 P.M   Assume at the beginning of the day, the steady state concentration is equal to outdoor air concentration i.e. C_s = C_o

1. Total emission rate of CO₂ (in mg/hr) (approximate value)

318

9560

420000

429560

2. The concentration (g/m³) of RP at 10 A.M is

0.00052

0.00059

0.00063

0.00070

3. The concentration (g/m³) of CO at 1 P.M is

0.00656

0.00818

0.00888

0.00951

4. The source strength (g/m³) of CO₂ at 2 P.M is

1.819

2.435

2.703

2.946

5. The concentration (g/m³) of CO₂ at 5 P.M is

1.466

2.352

4.011

5.056

Answer 1

1)

Total cigarettes consumed per hour = 30

total emission of CO₂ = 30*318.75 mg/cig = 9562.5 mg/hr

emission factor of CO₂ due to resipiration is 35000 mg/hr/ person, so total emission rate due to resipiration is 35000*12 = 420000

total emission rate = 420000+9562.5 = 429562.5

= 429560(Appr.) (Ans.)

2)

Emission factor (mg/cig) for RP = 25.625

Number of smokers =   6

Smoking rate =   5 cigarettes/hr

6Am to 9 Am = 3 hr

total Cigarettes consumed in 3 hr = 15

Total emission in 3 hr = 15*25.625 = 384.375

for 6 smokers, 387.375*6= 2306.25

Recirculated indoor air = 2000 m3/hr

so At 10 am means after 1hr the weight of RP = 2306.25 - 2000 = 306.25

total volume of the room = 12mL* 10m W* 4m H = 480 m³

concentration of RP at 10 am = 306.25 mg /480m³ = 0.63 mg/m³ = 0.00063 g/m3 (Ans.)

similarly others can be solved