Indoor Air Quality Problems
(Concentration Design Problems Not Included)
Q. SO2 concentration is given as 700 g / m3 at 90o C and 6 atm. What is the
concentration in ppm?
(Molecular weight of SO2 = 64 g/mol, Universal Gas Constant = 0.08206 (atm.liter)
/ (g mol* K)
Solution:




Concentration of SO2 is 700 ug/m3
AT STP conditions (25°C and 1 atm), one mole of gas occupies 24.5 L
(V=nRT/P); Therefore, at 90 °C and 6 atm, V = (1*0.08206*363)/6 = 4.964 L
Molecular Weight of SO2 is 64g/mol
Concentration in decimal form
= (concentration g/L x volume L/mol) / (molecular weight (g/mol)
The concentration expressed in decimal form is unit-less. To find the concentration
of in ppm, multiply the answer from equation above with 1,000,000.
For example, a concentration of Compound A of 0.000100 (decimal form from
equation above) would be 100 ppm, while a concentration of 0.010000 would be
10,000 ppm or 1%.
1 g = 106 ug; 1 m3 = 1000 L
Concentration in decimal form
= (700 ug/m3) *(10-6 g/ug )*(10-3 m3/L) x 4.964 L/mol) / (64 (g/mol)
= 54.29 * 10-9
Concentration in ppm = 54.29 * 10-9 *106
= 0.054 ppm
Q. What is the density of CO at 1 atm absolute and 80o F?
(Use R =0.08206 (atm.liter) / (g mol* K), Molecular weight of CO =28 g/mol)
Solution:
oF
= (1.8 * 0C +32)
oC
= [(F-32)/1.8]
oK
= 273 + oC
Density, D = P * M / (R * T)
Given, pressure P = 1.5 atm
Molecular weight (M) CO = 28 g/g-mol
R = 0.08206 L atm/g-mol 0K (standard)
T = 80 F = (80-32)/1.8 = 26.67 oC = 273+26.67 K = 299.67 K
Hence, density = 1.5 * 28 / (0.08206 * 299.67) = 1.71 g/l
(1 m3 =1000 liters
Answer in g/m3 = 1710 g/m3)
Q. A scientist exposed 75 male rats to 822 μg/cum Nickel sulfide by inhalation for
90 weeks (5days/week, 5hr/day). Convert animal dose to lifetime dose.
Solution:
Exposure time of the rodents = 90 x 5 x 5 = 2250 hr
Concentration to which the rodents were exposed = 822 μg/cum
Lifetime dose = concentration x time
= 822 x 2250
= 1849500 μg-hr/cum
= 1.84 g-hr/cum
Q. A family was continuously exposed to 0.010mg/cum of an indoor carcinogen.
After 10 years they moved to another house. Calculate the risk due to the
chemical if the potency for inhalation is 10-6kg-day/mg
Solution:
Assume the breathing rate (B) to be 15 cum/day
Assume weight (W) of each person to be 70 kg.
Indoor concentration (C) = 0.010mg/cum
Potency of inhalation = 10-6 kg-day/mg
Frequency of exposure (f) = (B * C) / W
= (15 * 0.010) / 70 = 0.0021 mg/kg-day
Consequence (c) = a x Kah x I
= 10 –6 x 1 x 1
=10-6 kg-day/mg
Risk
=fxC
= 0.0021 x 10-6
Risk
= 2.1 x 10–9 (harm/ unit time)
The above value of risk implies that an individual has a 0.0021 chance in a million of
getting cancer over his or her lifetime because of the daily exposure to the carcinogen
at a concentration level of 0.01 mg/m3 of air. If an individual has a life span of 70 years,
then chance of getting cancer in any one year is 0.0021/70 = 3*10-5 in a million
Q. What is the discharge velocity if rate of flow is 150 m3/ min through a 0.24 ft
diameter vent?
Solution:
1 m = 3.28 ft
1 minute = 60 seconds
Q = 150 m3/ min = 150 * (3.28)3 ft3/min = 5293.13 ft3/min = 88.22 ft3/sec
Area = (π d2/4) = (3.14*(0.24)2/4) = 0.045 ft2
Discharge velocity = Q / A
= 88.22/0.045
= 1960.4 ft / sec.
Q. What is the gage pressure if the absolute pressure is 130 kN/m2 and the
atmospheric pressure is 101.3 kPa?
Solution:
Gage pressure = Absolute pressure - Atmospheric pressure = 130 – 101.3
= 28.7 kN/m2 or 28.7 kPa
Q. According to the ASHRAE standard 62-1989, estimated maximum occupancy
for dining rooms is 70/100 sqm and the fresh air requirements is 15 l/person.
Calculate the amount of fresh air required for a 120-sqm dining room of a Mexican
restaurant in Toledo, OH.
Solution:
For 100 sqm, maximum occupancy = 70
Therefore, for 140 sqm, maximum occupancy =
70
 120 = 84
100
Amount of fresh air required per person = 15 l
Total amount of fresh air required for 98 people = 8415 = 1260 l
Q. The results of a radon mitigation program that used the Sub-Slab
Depressurization system are as follows:
No.
Pre Mitigation
(pCi/L)
1
2
Post Mitigation
(pCi/L)
80
10
2
1.5
3
4
5
9.2
11
20
0.5
2
2.5
Find the Average removal efficiency and the Standard Deviation of the System.
Removal Efficiencies:
Pre Mitigation
(pCi/L)
80
10
9.2
11
20
Post
Mitigation
(pCi/L)
2
1.5
0.5
2
2.5
Removal
Efficiency= (PrePost/Pre)*100%
97.50
85.00
94.57
81.82
87.50
Average Removal Efficiency = (97.50+85.00+94.57+81.82+87.50)/5 = 89.28%
Standard deviation
For each value xi calculate the difference
between xi and the average value
.
Calculate the squares of these differences.
Find the average of the squared differences. This quantity is the variance σ2.
Take the square root of the variance.
Standard Deviation
= Square root (1/5 * [(97.50-89.28)2 + (85-89.28)2 +(94.57-89.28)2 +(81.82-89.28)2
+(87.50-89.28)2]) = 5.87
Note: This is the standard deviation for a population. If the given dataset were a sample
set drawn from a larger population, and the question at hand was the standard deviation
of the population, then we would replace the N in the equation with N−1.
Q. Estimate the amount of moisture added to your house over a one-day period
due to the following:

Three occupants

Three showers

No breakfast.

Lunch and dinner cooked in a gas stove

12 house plants

One load of laundry with the electric drier vented outdoors.
Solution
Three occupants:
Respiration or Perspiration (avg. of 4/family) 0.21 /hr
Therefore, Moisture due to 3 occupants = Amount of Moisture added per person
* Occupants * Time
Moisture added = 0.21 * 3 * 24 / 4 = 3.78 kg
Three showers:

Shower
0.25
Therefore, Moisture due to 3 showers = 3 * 0.25 = 0.75 kg
Food:
Cooking



Breakfast (avg. of 4/family)
Lunch (avg. of 4/family)
Dinner (avg. of 4/family)
0.17
0.25
0.58
No breakfast: No moisture is added when no breakfast is cooked.
Lunch and dinner cooked in a gas stove:
Lunch = 0.25 kg
Dinner = 0.58 kg
Total from food = 3/4 *(0.25 + 0.58) = 0.62 kg
12 houseplants:
House plants 0.4 to 0.45 /day
Therefore, 12 houseplants will add: 12 * (0.4 to 0.45) = 4.8 to 5.4 kg of moisture
Assume 5.1 kg of moisture
One load of laundry with the electric drier vented outdoors:
Washing clothes
Negligible
Drying clothes


Vented outdoors
Not vented outdoors
Negligible
2.21 to 2.92
Negligible moisture will be added from washing and drying the clothes with venting
outdoors.
Total moisture added = 3.78 + 0.75 +0.62 + 5.1 = 10.25 kg
Q. The outdoor air concentration of SO2was measured as 200 μg/cum around the
building. The building has no sources of SO2. Calculate the efficiency of aircleaner for a 100% outdoor air ventilation system such that the indoor air
concentration is not exceeded by 100 μg/cum.
Solution:
Efficiency of Ventilation system = [(200 – 100) / 200] *100%
= 50%
Q. Calculate the ventilation requirements for the building having following
spaces:
Office space
500 sqm
Corridor
60 sqm
Reception area
75 sqm
Conference room
85 sqm
Rest rooms (16 WCs) 15 sqm
Use ASHRAE standards.
Solution:
Ventilation Requirements as given by ASHRAE Standards:
ROOM TYPE
ESTIMATED
OCCUPANCY
SMOKING
(PERSONS / 1000
ft2)
NON-SMOKING
CONFERENCE ROOM
(cfm/person)
50
35
7
OFFICE SPACE
(cfm/person)
7
20
5
60
35
7
CORRIDORS AND UTILITY
ROOMS
(cfm/ft2)
-
0.02
0.02
PUBLIC RESTROOMS
(cfm/stall)
100
75
-
MEETINGS AND WAITING
ROOMS
(cfm/person)
Office space:
Area = 500 m2 = 500 * 3.28 2 = 5379.2 ft 2
Estimated occupancy = 5379.2 * 7 / 1000 = 37.65 = 38 people
Cfm required per person = 20 (smoking), 5 (non-smoking)
For smoking area, cfm required = 38 * 20 = 760 cfm
For non-smoking area, cfm required = 38 * 5 = 190 cfm
Corridor:
Area = 60 m2 = 60 * 3.28 2 = 645.504 ft 2
Cfm required per ft 2 = 0.02 (smoking and non smoking)
For both smoking and non-smoking areas, cfm required = 645.504 * 0.02 = 12.91 cfm
Reception area (Taken as a waiting room):
Area = 75 m2 = 75 * 3.28 2 = 806.88 ft 2
Estimated occupancy = 806.88 * 60 / 1000 = 48.41 = 49 people
Cfm required per person = 35 (smoking), 7 (non-smoking)
For smoking area, cfm required = 49 * 35 = 1715 cfm
For non-smoking area, cfm required = 49 * 7 = 343 cfm
Conference room:
Area = 85 m2 = 85 * 3.28 2 = 914.464 ft 2
Estimated occupancy = 914.464 * 50 / 1000 = 45.72 = 46 people
Cfm required per person = 35 (smoking), 7 (non-smoking)
For smoking area, cfm required = 46 * 35 = 1610 cfm
For non-smoking area, cfm required = 46 * 7 = 322 cfm
Restrooms:
Number of Stalls = Number of WCs = 16
Cfm required per stall = 75 (smoking)
Total cfm required = 75 * 16 = 1200 cfm
Q. The concentration levels of nicotine in an office were found to be 2.6 µg/cum.
Compute the mass of nicotine collected on the filter paper of a personal sampling
pump. The sampler was run for 2 hours at a flow rate of 2 liters/min.
Solution:
Time of operation of the sampler = 3 hours = 2 * 60 = 120 min
The total volume of air flowing = 120 x 2
= 240 liters
= 0.240 m3
Nicotine collected = 2.6 µg/ m3
Therefore, in 3 hours, Nicotine collected = 0.240 x 2.6
= 0.624 µg
Q. Calculate the time required for saturation of 140 kg of adsorbent. The retentivity
fraction is 0.5 and the efficiency for the adsorbent is 0.90. The average molecular
weight of adsorbed vapor is 198 g/mole. The airflow rate is 160 cfm and the pollutant
concentration is 100 ppm in the air stream.
ts = (2.41 * 107 * Rt * W)/( η * q * M * C)
Where:
ts = time of adsorbent service to saturation (hr) = ?
Rt = retentivity (fraction) = 0.5
W = weight of adsorbent (kg) = 140
η = adsorption efficiency (fraction) = 0.90
M = average molecular weight of adsorbed vapor (g/mole) = 198
q = air flow rate (m3/hr) = 160 cfm
C = entering pollutant concentration (ppm) = 100 ppm
Air flow rate, q = (140 ft3/min)*(60min/hr)*(0.02832 m3/ft3) = 271.9 m3/hr
Time required for saturation,
ts = [(2.41)*(10^7)*(0.5)*(140)] / [(0.9)*(271.9)*(100)*(198)
= 348.17 hours or 14.5 days
Q. The following observations for RP concentration were obtained in houses located in
the Midwest:
The right hand column of the above table includes information on predicted
concentrations using a model. Compute correlation coefficient between observed and
predicted values.