An indoor air quality specialist records following information for a room: Total
ID: 1063982 • Letter: A
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 (Co) 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. Cs = Co
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
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
Explanation / Answer
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
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