Question

The South Coast Air Basin (Los Angeles, Orange, and San Bernardino counties) does not meet the air quality standards for carbon monoxide (CO). Computer modeling indicates that meeting the standard in 2020 will require reducing total annual CO emissions to the atmosphere to 30% of the 1990 value. If the automobile CO emissions factor in 1990 was 0.9 g CO per mile, what must it be in 2020 to meet the air quality goal? Assume the miles driven per person per year remains constant at 8700 mi/yr.person. The combined populations of Los Angeles, Orange, and San Bernardino counties were 10.3 million in 1980 and 12.5 million in 1990. Assume exponential growth. (Hint: Calculate the allowable mass of CO based on 0.9 g/mi. Use the populations given to project a 2020 population and a new allowable g/mi CO.) Answer: 0.15 gC

Answer 1

Answer:

0.2225 gr/mile

Step-by-step explanation:

Let's work out first the amount of CO sent out in 1990.

The population we estimated in 12.5 million people with a total of driven miles per year of about  

12.5 million*8,700 = 108,750 million miles.

With an CO emission factor of 0.9 g per mile, we would have a total of CO emitted rounding 0.9*180,750 = 97,875 million grams

Now, we must estimate the population for 2020.

Since we are assuming an exponential growth, the population in year t is given by a function

$\bf P(t)= Ce^{kt}$

where C and k are constants to be determined.

We can take 1980 as year 0. This way calculations are lighter. 1990 is year 10 and 2020 is year 20.

So P(0) = 10.3 and C=10.3

So far we have

$\bf P(t)= 10.3e^{kt}$

Given that P(10)=12.5

$\bf 10.3e^{k*10}=12.5\rightarrow e^{10k}=\frac{12.5}{10.3}=1.21359\rightarrow \\10k=log(1.21359)\rightarrow k=0.01936$

And the function that models the population growth is

$\bf P(t)= 10.3e^{0.01936t}$

We need P(20)

$\bf P(20)= 10.3e^{0.01936*20}=10.3e^{0.38717}=15.17\;million$

If the miles driven per person per year remains constant at 8700 mi/yr.person, then we have a total miles driven of

15.17*8,700=131,979 million miles, so the CO emitted would be 0.9*131,979=118,781.1 million grams.

The 30% of the CO sent out in 1990 is 0.3*97,875=29,362.5 million grams.

We must reduce 118,781.1 down to 29,362.5

Hence the new CO emission factor would be

29,362.5/131,979 = 0.2225 gr/mile