IAQ Numericals



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 = f x C

= 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 = [pic]= 84

Amount of fresh air required per person = 15 l

Total amount of fresh air required for 98 people = [pic]= 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) |Post Mitigation (pCi/L) |

|1 |80 |2 |

|2 |10 |1.5 |

|3 |9.2 |0.5 |

|4 |11 |2 |

|5 |20 |2.5 |

Find the Average removal efficiency and the Standard Deviation of the System.

Removal Efficiencies:

|Pre Mitigation (pCi/L) |Post Mitigation (pCi/L)|Removal Efficiency= |

| | |(Pre-Post/Pre)*100% |

|80 |2 |97.50 |

|10 |1.5 |85.00 |

|9.2 |0.5 |94.57 |

|11 |2 |81.82 |

|20 |2.5 |87.50 |

Average Removal Efficiency = (97.50+85.00+94.57+81.82+87.50)/5 = 89.28%

Standard deviation

[pic]

For each value xi calculate the difference [pic]between xi and the average value [pic].

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) |0.17  |

|Dinner (avg. of 4/family) |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 air-cleaner 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 (PERSONS / |SMOKING |NON-SMOKING |

| |1000 ft2) | | |

|CONFERENCE ROOM |50 |35  |7 |

|(cfm/person) | | | |

|OFFICE SPACE |7  |20 |5 |

|(cfm/person) | | | |

|MEETINGS AND WAITING ROOMS |60 |35 |7 |

|(cfm/person) | | | |

| | | | |

|  | | | |

|CORRIDORS AND UTILITY ROOMS |- |0.02 |0.02 |

|(cfm/ft2) | | | |

|PUBLIC RESTROOMS |100 |75 |- |

|(cfm/stall) | | | |

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.

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