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Nataly [62]
2 years ago
8

Consider an area-source box model for air pollution above a peninsula of land. The length of the box is 15 km, its width is 80 k

m, and a radiation inversion restricts mixing to 15 m. Wind is blowing clean air into the long dimension of the box at 0.5 m/s. On average, there are 250,000 vehicles on the road, each being driven 40 km in 2 hours and each emitting 4 g/km of CO.
Required:
a. Estimate the steady-state concentration of CO in the air. Should the city be designated as "nonattainment" (i.e., steady-state concentration is over the NAAQS standard)?
b. Find the average rate of CO emissions during this two-hour period.
c. If the windspeed is zero, use the formula to derive relationship between CO and time and use it to find the CO over the peninsula at 6pm
Engineering
1 answer:
In-s [12.5K]2 years ago
3 0

Consider an area-source box model for air pollution above a peninsula of land. The length of the box is 15 km, its width is 80 km, and a radiation inversion restricts mixing to 15 m. Wind is blowing clean air into the long dimension of the box at 0.5 m/s. On average, there are 250,000 vehicles on the road, each being driven 40 km in 2 hours and each emitting 4 g/km of CO.

Required:

a. Estimate the steady-state concentration of CO in the air. Should the city be designated as "nonattainment" (i.e., steady-state concentration is over the NAAQS standard)?

b. Find the average rate of CO emissions during this two-hour period.

c. If the windspeed is zero, use the formula to derive relationship between CO and time and use it to find the CO over the peninsula at 6pmConsider an area-source box model for air pollution above a peninsula of land. The length of the box is 15 km, its width is 80 km, and a radiation inversion restricts mixing to 15 m. Wind is blowing clean air into the long dimension of the box at 0.5 m/s. On average, there are 250,000 vehicles on the road, each being driven 40 km in 2 hours and each emitting 4 g/km of CO.

Required:

a. Estimate the steady-state concentration of CO in the air. Should the city be designated as "nonattainment" (i.e., steady-state concentration is over the NAAQS standard)?

b. Find the average rate of CO emissions during this two-hour period.

c. If the windspeed is zero, use the formula to derive relationship between CO and time and use it to find the CO over the peninsula at 6pmConsider an area-source box model for air pollution above a peninsula of land. The length of the box is 15 km, its width is 80 km, and a radiation inversion restricts mixing to 15 m. Wind is blowing clean air into the long dimension of the box at 0.5 m/s. On average, there are 250,000 vehicles on the road, each being driven 40 km in 2 hours and each emitting 4 g/km of CO.

Required:

a. Estimate the steady-state concentration of CO in the air. Should the city be designated as "nonattainment" (i.e., steady-state concentration is over the NAAQS standard)?

b. Find the average rate of CO emissions during this two-hour period.

c. If the windspeed is zero, use the formula to derive relationship between CO and time and use it to find the CO over the peninsula at 6pm

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if the mine winch drum diameter is 6 m, calculate how far the cage will drop for each single rotation of the drum
vesna_86 [32]

Answer:

The depth to which the cage will drop for each single rotation of the drum is approximately 18.85 m.

Explanation:

The mine winch drum consists of a drum around around which the haulage rope is attached

Therefore, given that the diameter of the winch drum = 6 m

The length of rope unwound by each rotation = How far the cage will drop for each single rotation of the drum

The length of the rope unwound by each rotation = The rope that goes round the circumference of the winch drum, once

∴ Since the rope that goes round the circumference of the winch drum, once = The circumference of the winch drum, we have;

The length of the rope unwound by each rotation = The circumference of the winch drum = π × The diameter of the winch drum

The length of the rope unwound by each rotation = π × 6 m = 6·π m

The length of rope unwound by each rotation = How far the cage will drop for each single rotation of the drum = 6·π m

How far the cage will drop for each single rotation of the drum = 6·π m ≈ 18.85 m

The depth to which the cage will drop for each single rotation of the drum ≈ 18.85 m.

3 0
2 years ago
Suppose that a class CalendarDate has been defined for storing a calendar date with month, day and year components. (In our sect
laiz [17]

Answer:

note:

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2 years ago
While at a concert you notice five people in the crowd headed in the same direction. Your tendency to group them is due to? *
Vlada [557]

Answer:

common fate

Explanation:

The gestalt effect may be defined as the ability of our brain to generate the whole forms from the groupings of lines, points, curves and shapes. Gestalt theory lays emphasis on the fact that whole of anything is much greater than the parts.

Some of the principles of Gestalt theory are proximity, similarity, closure, symmetry & order, figure or ground and common fate.

Common fate : According to this principle, people will tend to group things together which are pointed towards or moving in a same direction. It is the perception of the people that objects moving together belongs together.

7 0
2 years ago
The rate of flow through an ideal clarifier is 8000m3 /d, the detention time is 1h and the depth is 3m. If a full-length movable
Fittoniya [83]

Answer:

a) 35%

b) yes it can be improved by moving the tray near the top

   Tray should be located ( 1 to 2 meters below surface )

   max removal efficiency ≈ 70%

c) The maximum removal will drop as the particle settling velocity = 0.5 m/h

Explanation:

Given data:

flow rate = 8000 m^3/d

Detention time = 1h

depth = 3m

Full length movable horizontal tray :  1m below surface

<u>a) Determine percent removal of particles having a settling velocity of 1m/h</u>

velocity of critical sized particle to be removed = Depth / Detention time

= 3 / 1 = 3m/h

The percent removal of particles having a settling velocity of 1m/h ≈ 35%

<u>b) Determine if  the removal efficiency of the clarifier can be improved by moving the tray, the location of the tray  and the maximum removal efficiency</u>

The tray should be located near the top of the tray ( i.e. 1 to 2 meters below surface ) because here the removal efficiency above the tray will be 100% but since the tank is quite small hence the

Total Maximum removal efficiency

=  percent removal_{above} + percent removal_{below}

= ( d_{a},v_{p} ) . \frac{d_{a} }{depth}  + ( d_{a},v_{p} ) . \frac{depth - d_{a} }{depth}  = 100

hence max removal efficiency ≈ 70%

<u>c) what is the effect of moving the tray would be if the particle settling velocity were equal to 0.5m/h?</u>

The maximum removal will drop as the particle settling velocity = 0.5 m/h

7 0
2 years ago
Determine F12 and F21 for the following configurations: (a) A long semicircular duct with diameter of 0.1 meters: (b) A hemisphe
uysha [10]

Answer:

long duct: 1.0 and 0.424

Hemisphere 1.0 ; 0.125; 0.5

Explanation:

For a long duct:

By inspection, F_{12} = 1.0

Calculating by reciprocity, F_{21} = \frac{A_{1} }{A_{2}F_{12}  }  = \frac{2RL}{\frac{3}{4}*2\pi RL  }* 1.0\\                                                         = 0.424

Hemisphere:

By reciprocity gives = 0.125

using the summation rule: F_{21} + F_{22} + F_{23} = 1

However, because this is a hemisphere, the value will be= 0.5 * 1

                                                                                               = 0.5

6 0
2 years ago
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