Daniel_Lougee_Wk6 Assignment
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Running head: ASSIGNMENT WEEK 6 (MATH)
1
Assignment Week 6 (Math)
Question A
NAAQS were established for six principal pollutants. Describe each and give examples of each applicable pollutant.
The six principal pollutants established under the National Ambient Air Quality Standards in accordance with 40 CFR part 50 are Carbon Monoxide (CO), Lead (Pb), Nitrogen Dioxide (NO
2
), Ozone (O
3
), Particle Pollution (PM), and Sulfur Dioxide (SO
2
) (US EPA, 2021.)
Carbon Monoxide air pollutants are produced by exhaust byproducts of transportation, farming, energy production, and other industries. CO is measured in parts per million (US EPA, 2021.)
Lead air pollutants derive from leaded fuel and the metal production industry. Lead is measured in one-millionth of a microgram per cubic meter of air (US EPA, 2021.)
Nitrogen Dioxide like Carbon Monoxide comes from multiple sources to include transportation, farming, energy production, and other industries. NO
2 is measured in parts
per billion (US EPA, 2021.)
Ozone originates from both manmade gasses and natural occurring production when UV light radiates Oxygen in the atmosphere. O3 is measured in parts per million (US EPA, 2021.)
ASSIGNMENT WEEK 6 (MATH)
2
Particle Pollution comes from any finite airborne particle that is either solid or liquid less than 10 micrometers in size. PM like lead is measured in one-millionth of a microgram per cubic meter of air (US EPA, 2021.)
A great example of Sulfur Dioxide air pollution is any diesel vehicle’s emissions. However, it also occurs naturally through volcanic activity. Sulfur Dioxide is measured in
parts per billion (US EPA, 2021.) Question B
In Hepburn’s Speed Model, the coefficients of vehicles are indicated for C and D. As
the chief of operations in your organization, you are responsible for presenting the yearly budget for the semi-trucks in your company’s inventory. Since your safety officer is insisting that each of your drivers must maintain an average speed of 55mph, what would be the vehicle operating cost of your company for each semi-
truck in cent per mile? To determine the companies’ vehicle operating cost for each semi-truck in cents per mile one must use Hepburn’s Speed Model. Sinha & Labi define the formula as vehicle operating cost (VOC) = constant (a
0
)
– constant (a
1
) *
vehicle speed (S) + constant
(a
2
)
*
speed (S)
2 (2007.) The example defines speed as 55 miles per hour. Given the vehicle is a large automobile on the Sinha & Labi’s chart, constant (a
0
) is 38.1, constant (a
1
) is 0.093, and constant (a
2
) is 0.00033. The formula calculates as VOC = 31.1 – (0.093
* 55) + (0.00033 * 55
2
). This simplifies to VOC = 38.1 – (5.115) + (0.99825) and VOC then equals 33.98325 and rounds up to 33.9833 cents per mile for each semi-truck.
VOC = 31.1 – (0.093 * 55) + (0.00033 * 55
2
)
VOC = 38.1 – (5.115) + (0.99825)
ASSIGNMENT WEEK 6 (MATH)
3
VOC = 33.98325 or rounded up to 33.9833
Question C
A taxi driver plans to pick you up at the airport and drop you off at the bank so you
can complete some financial transaction before you head home. He notes that the change in vehicle operating cost (VOC) is 42 cents per mile. Given that his fuel consumption per minute is 0.2, what is the approximate price of fuel for this given arrangement if you delayed the driver for 36 minutes at the bank? To determine the approximate price of fuel for the 36-minute delay at the bank
Sinha & Labi’s formula for change in fuel vehicle operating cost is the best method. In
the formula Change in Fuel Vehicle Operating Cost (VOC) = fuel consumption per
minute (g) * (change in delay d
0 - d
1
) * price of fuel (P) or VOC = g * (d
0
-d
1
) * p (2007.)
The problem defines vehicle operating cost as 42 cents per mile, VOC = 42 cents / mile.
Fuel consumption per minute equals 0.2, g = 0.2. Change in delay is 36 minutes, (d
0
-d
1
) =
36. Therefore, the formula is 42 = 0.2 * (36) * p which simplifies to p = 42 / 0.2 * (36).
This reduces to p = 42 / 7.2 where price of fuel for a 36-minute delay at the bank =
$5.8333 and rounds down to $5.83.
42 = 0.2 * (36) * p
p = 42 / 0.2 * (36)
p = 42 / 7.2
p = $5.8333 = $5.83 Question D
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ASSIGNMENT WEEK 6 (MATH)
4
The simple interest for buying a passenger transit rail is shown as the product of the
principal amount (P), time (in years), and annual rate (R). The City of Phoenix plans to buy five additional mass transit cars for $15 million, and pay off its loan in 30 years. What would the annual percentage rate be if the city plans to make an interest payment of $2 million? Simple interest formula is the way to find out the annual percentage rate if the city
plans to make an interest payment of two million dollars. Simple interest formula is annual interest rate (r) = interest paid (i) / (principal loan amount (p) * time (t)) or r = i / (p * t) (Sinha & Labi, 2007.) For this problem the following variables are defined, interest paid is $2,000,000, principal loan amount is $15,000,000, and time is 30 years. Therefore, in accordance with the formula r = 2,000,000 / (15,000,000 * 30) which yields
the annual percentage rate of 0.44% if the city plans to make an interest payment of two million dollars. r = 2,000,000 / (15,000,000 * 30)
r = 2,000,000 / (450,000,000)
r = 0.44%
Question E
In a box model, the maximum distance to transport particle across a city is calculated as follows: Approximate length of one side is 30 miles, width is 28 miles, and the mixing height is 2 miles. To the nearest mile, what will be this distance? To calculate this distance to the nearest mile Sinha and Labi’s box model formula represents the most accurate solution. This solution however postulates that where the problem does not define the precise disbursement of particles, they spread in a consistent
ASSIGNMENT WEEK 6 (MATH)
5
manner. With this assumption in place the box model formula is maximum distance (D) =
the square root of (length squared (a
2
) + width squared (b
2
) + mixing height squared (H
2
) (Sinha & Labi, 2007.) The problem defines length as 30 miles, width as 28 miles, and mixing height as 2 miles. Therefore, the formula populates as D = √ (30
2 + 28
2 + 2
2
). This simplifies as D = √ (900
+ 784
+ 4) and then D =√ 1688 which equals 41.085 miles or 41 miles distance to the nearest mile.
D = √ (30
2 + 28
2 + 2
2
)
D = √ (900
+ 784
+ 4)
D = √ 1688
D = 41.085 = 41 Miles
Question F
In a box model, the maximum distance to transport particle across a city is calculated as follows: Approximate length of one side is 30 miles, width is 28 miles, and the mixing height is 2 miles. For a pollutant particle emitted on one side of the town, what is the maximum time it will take to be transported across the city with wind velocity of 10 miles per hour (to the nearest hour)? To calculate the maximum transport time it will take to transport particle across a city with a wind velocity of 10 miles per hour, Sinha and Labi’s box model formula combined with a formula for maximum transport time represents the most accurate solution. This solution however postulates that where the problem does not define the precise disbursement of particles, they spread in a consistent manner. With this assumption in place the box model formula is maximum distance (D) = the square root of
(length squared (a
2
) + width squared (b
2
) + mixing height squared (H
2
) and the maximum
ASSIGNMENT WEEK 6 (MATH)
6
transport time formula is maximum transport time (tmax) = distance (D) / velocity (v) (Sinha & Labi, 2007.) The problem defines length as 30 miles, width as 28 miles, mixing height as 2 miles, and velocity at 10 miles per hour. Therefore, the formula populates as D = √ (30
2 + 28
2 + 2
2
). This simplifies as D = √ (900
+ 784
+ 4) and then D =√ 1688 which
equals 41.085 miles or 41 miles distance to the nearest mile. Maximum transport time then solves with tmax = 41 miles / 10 miles per hour which equals 4.1 or to the nearest hour, 4 hours maximum transport time.
D = √ (30
2 + 28
2 + 2
2
)
D = √ (900
+ 784
+ 4)
D = √ 1688
D = 41.085 = 41 Miles
tmax = 41 miles / 10 miles per hour
tmax = 4.1 hours = 4 hours Question G
In a box model, the maximum distance to transport particle across a city is calculated as follows: Approximate length of one side is 30 miles, width is 28 miles, and the mixing height is 2 miles. To the nearest hour what will be the average time it will take a particle to travel from one side to the other (to the nearest hour)?
Sinha and Labi’s box model formula combined with the formula for solving average transport time is the most accurate solution to calculate the average time to the nearest hour that it will take a particle to travel from one side to the other. This solution however postulates that where the problem does not define the precise disbursement of particles, they spread in a consistent manner as well as the velocity still holds true from
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7
the previous problem. With this assumption in place the box model formula is maximum distance (D) = the square root of (length squared (a
2
) + width squared (b
2
) + mixing height squared (H
2
) and the average transport time formula is time average (Tavg) = distance (D) / 2 * velocity (v) (Sinha & Labi, 2007.) The problem defines length as 30 miles, width as 28 miles, mixing height as 2 miles, and velocity is assumed from the previous problem at 10 miles per hour. Therefore, the formula populates as D = √ (30
2 + 28
2 + 2
2
). This simplifies as D = √ (900
+ 784
+ 4) and then D =√ 1688 which equals 41.085 miles or 41 miles distance to the nearest mile. Average transport time then solves with Tave = 41 miles / 2 * 10 miles per hour and reduces to 41 / 20 which equals 2.05 or to the nearest hour, 2 hours average transport time.
D = √ (30
2 + 28
2 + 2
2
)
D = √ (900
+ 784
+ 4)
D = √ 1688
D = 41.085 = 41 Miles
Tave = 41 miles / 2 *10 miles per hour
Tave = 41 miles / 20 miles per hour
tmax = 2.05 hours = 2 hours Question H
If you are the planning engineer for an apartment complex your organization plans to build close to the highway and you know that your potential tenants will complain of traffic noise hazards. You want to assure them that you have adequately carried out the noise impact analysis. Given that the sea level pressure is
ASSIGNMENT WEEK 6 (MATH)
8
0.00002N/m, to ensure that pressure level does not exceed 12 decibels, the sound pressure of concern must be (in N/m). Sinha and Labi’s sound pressure formula best determines the sound pressure of concern in n/m from this problem to ensure that the pressure level does not exceed 12 decibels. The sound pressure formula is SPL (dB) = 10 * log
10
* p
2
/ p
0
2
(Sinha & Labi, 2007.) The problem identifies the pressure threshold at 12 dB and places the sea level pressure as point of reference which is 0.00002N/m. Therefore, it calculates as 12 dB = 10 * log
10
* p
2
/ 0.00002
2
N/m. This reduces to p = 10 √ 10 √ 0.0000000004 which equals
a pressure point of concern of 0.00007 N/m.
12 dB = 10 * log
10
* p
2
/ 0.00002
2
N/m
p = 10 √ 10 √ 0.0000000004
p = 0.00007 N/m
Question I
In the Rocky Mountain region, there are endangered species like jumping mice and burrowing owls that are protected by the EPA. If you want to build a roadway to facilitate
traffic congestions in Denver, part of your plan will include using Simpson’s diversity index to calculate the number of organisms in two communities. If EPA warned you that there are 400 burrowing owls in proposed region Community A and 500 burrowing owls in proposed region Community B, determine Simpson’s diversity index for community A.
Determining Simpson’s diversity index for community A requires diversity index formula. Sinha and Labi define this formula as diversity (D) = total number (N) * (N - 1) / number in group one (n
1
) * (n
1
– 1) + (number in group two (n
2
) * (n
2
-1)) (2007.)
ASSIGNMENT WEEK 6 (MATH)
9
The problem defines the population of group one as n
1
= 400 and the population of group two as n
1
= 500 where the total number extrapolates as N = 900. Accordingly, D = 900 (900-1) / ((400 * (400-1)) + ((500 * (500-1)) which reduces to D = 900 * 899 / ((400 * 399)) + (500 * 499)). The problem further simplifies to D = 809,100 / (159,600 + 249,500) which is D = 809,100 / 409,100. It solves as 1.9777, but rounds up to a 1.98 diversity index for community A. D = 900 (900-1) / ((400 * (400-1)) + ((500 * (500-1))
D = 900 * 899 / ((400 * 399)) + (500 * 499))
D = 809,100 / (159,600 + 249,500)
D = 809,100 / 409,100
D = 1.9777 = 1.98
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References
Sinha, K. C., & Labi, S. (2007). Transportation decision making: Principles of project evaluation and programming
. Hoboken, NJ: John Wiley & Sons, Inc.
US EPA. (2021, February 10). NAAQS Table
. EPA. https://www.epa.gov/criteria-air-
pollutants/naaqs-table.