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3 years ago

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5. Before vehicle emissions were well-regulated CO emissions were 66 g/mile. Assume this emission rate applies for an airshed. T

he airshed has dimensions of 9 km by 18 km with wind speeds of 1.5 m/s parallel to the longer dimension. There is a temperature inversion at 100 m. The average vehicle miles traveled is 800,000 miles each hour between 5 AM and 8 PM, and the concentration of CO before the morning rush hour and the incoming wind concentration are both 200 ppb. Temperature is 0 °C and pressure is 0.9 atmospheres. a. Find C0 and Cin in mg/m3 . b. What is the pollutant residence time in the box? c. If morning rush hour starts at 5 AM. What is the CO concentration seven hours later?

1 answer:

Aleks [24]3 years ago

4 0

Answer:CALINE4 Model and Geographical Information System were used for

the present study to predict the CO concentrations and prepare thematic

maps for the study area.

CALINE4 is a latest model that predicts the concentration of carbon

monoxide impacts near the roadways. The California Department of

Transformation (CALTRANS) developed the model and its purpose is to help

planners protect public health from the adverse effects of carbon monoxide.

The model predictions along with the aid of GIS based model help to arrive at

air shed levels of carbon monoxide. CALINE4 is a simple line source

Gaussian plume dispersion model. The user defines the proposed roadway

geometry, worst-case meteorological parameters, anticipated traffic volumes

and receptor positions to predict the concentrations of pollutant.

CALINE4 is graphical with windows based user interface, designed to

ease data entry and increase the online help capabilities. The model was

developed for predicting the concentrations of relatively inert pollutants such

as carbon monoxide and it is now used for several other pollutants like NO2

and SPM. The model is based on fine tuned Gaussian diffusion equation and

it employs mixing zone concept to characterize the pollutant dispersion over

the roadways. Given the exhaust emission concentrations, meteorology and

site geometry, the model can predict the concentration of pollutants for any experimentation

Explanation:

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3241004551 [841]

Answer:

V2 = 600ml

Explanation:

dilution principle formula

M1V1 = M2V2

C1V1 = C2V2

3 X 20 = 0.1 x V2

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3 years ago

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What is the coefficient for CO2?

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3 years ago

An iron pipe has a density of 7.86 g/mL and a volume of 324.4 mL. What is the mass of the iron pipe?

wlad13 [49]

Answer:

$Mass=2549.784 \ g$

Explanation:

The equation for mass is:

$Mass = Density * Volume$

We plug in the given values into the equation:

$Mass = 7.86 \ g/mL * 324.4 \ mL$

$Mass=2549.784 \ g$

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If you were givin 2.50 moles of.sodium bromide, how many grams of salt do you have

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3 years ago

How much ice (in grams) would have to melt to lower the temperature of 353 mL of water from 26 ∘C to 6 ∘C? (Assume the density o

Rashid [163]

Answer:

The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg

Explanation:

Heat gain by ice = Heat lost by water

Thus,

Heat of fusion + $m_{ice}\times C_{ice}\times (T_f-T_i)=-m_{water}\times C_{water}\times (T_f-T_i)$

Where, negative sign signifies heat loss

Or,

Heat of fusion + $m_{ice}\times C_{ice}\times (T_f-T_i)=m_{water}\times C_{water}\times (T_i-T_f)$

Heat of fusion = 334 J/g

Heat of fusion of ice with mass x = 334x J/g

For ice:

Mass = x g

Initial temperature = 0 °C

Final temperature = 6 °C

Specific heat of ice = 1.996 J/g°C

For water:

Volume = 353 mL

$Density (\rho)=\frac{Mass(m)}{Volume(V)}$

Density of water = 1.0 g/mL

So, mass of water = 353 g

Initial temperature = 26 °C

Final temperature = 6 °C

Specific heat of water = 4.186 J/g°C

So,

$334x+x\times 1.996\times (6-0)=353\times 4.186\times (26-6)$

$334x+x\times 11.976=29553.16$

345.976x = 29553.16

x = 85.4197 kg

Thus,

<u>The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg</u>

7 0

3 years ago