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There is absolutely no intention to mislead ...
if any of these calculations seem to present a significant error
please let us know .
Gravitational Settling
Aim: to evaluate distances of travel for dust particles.
As a starting point need to evalute the 'still-air' characteristics .
Particles in still air have two forces acting on them; (1) a
gravitational force downward and (2) the air resistance (or drag) force
upward. When particles begin to fall, they quickly reach a terminal
settling velocity, which represents the constant velocity of a falling
particle when the gravitational force downward is balanced by the air
resistance (or drag) force upward. The terminal settling velocity can
usually be expressed using Equation 3.
Note: Equation 3 is applicable for particles less than 80
micrometers in size (aerodynamic diameter) and having a Reynolds
number less than 2.0 and a low velocity.
Example Problem 1.
Calculating the Terminal
Settling Velocity
To calculate the terminal settling velocity for the following particles in
still air having a gas temperature of 20°C. Use an air viscosity of
0.000182 gm/cm sec and the (Cc) provided. We can assume in all cases that
the Reynolds numbers are less than 2.0.
- 80-micrometer particle with a density of 1.0
gm/cm3,
Cc = 1.0 (Calculate the velocity in
terms of ft/min.)
- 8.0-micrometer particle with a density of 1.0
gm/cm3,
Cc = 1.0 (Calculate the velocity in
terms of cm/sec.)
- 0.8-micrometer particle with a density of 1.0
gm/cm3,
Cc = 1.05 (Calculate the velocity in
terms of cm/sec.)
Solution:
Since the particles are in still air, the particle diameters are not
greater than 80 micrometers, and the NRe(p) < 2, the
following equation can be applied. Also, because the particles have a
density of 1gm/cm3, the aerodynamic diameter is equivalent to
the Stokes diameter.
Part i
For an 80-micrometer particle,
Part ii
For an 8.0-micrometer particle, vt = 0.191 cm/sec since the
particle diameter has changed by a factor of 10 and the diameter is raised
to the power of 2.
Part iii
For a 0.8-micrometer particle, the equation will be used again since
the Cunningham slip correction factor has changed.
This problem illustrates the large difference in settling rates for
particles in this common size range. Particles of 0.8 micrometer and less
have essentially negligible settling rates. The terminal settling
velocities calculated indicate that even 80-micrometer particles settle at
the relatively slow velocity of approximately 37.7 ft/min. A
0.8-micrometer particle settles at the extremely slow rate of 0.002
cm/sec. thus indicating a PM ref distribution distance of 860 metres from a quarry-plateau height of 20 metres at a typical applied windspeed of 8 knots. At higher windspeeds this distance will increase, and the channeling effect of the valley sides will be a major factor.
A full section will be dedicated to a clearer explanation of this aspect when the graphics are completed.
Aim : To determine the number of 26tonne lorry sightings that these figures indicate.
From Surrey CC PMZ71 pdf document ( also on this website ), estimations are that zone contains a net mineral reserve of approximately
2 900 000 tonnes of soft sand and 500 000 tonnes of Bargate stone.
VEHICLES If taken to be 3-axle 26 tonne Rigid / Tippers.
Using the MB Axor-R 2628K vehicle specification as a reference this has a UK-Authorised GVW
( Gross Vehicle Weight ) of 26 tonnes.
The unladen vehicle has a kerb weight of 7350 kg ( taken to be 7 tonnes ).
Thus the load - carrying capability can be taken to be 26 minus 7 = 19 tonnes per quarry visit.
Combining the rock and sand totals given, ( i.e. 2 900 000 plus 500 00 tonnes ) gives a total projected extraction of 3 400 000 tonnes.
If every vehicle was full to capacity ... certainly not always the case ...
then this derives a figure of ( 3 400 000 divided by 19 = 178 948 loads ).
But we need to ascertain the number of vehicle routings ,
i.e. unladen lorries approaching the site , and laden lorries leaving the site , often on a different route.
The total number of vehicle routes would therefore be approximately 357 896 over the whole period of extraction.
Presuming five-and-a-half-days per working week, taken over 50 weeks per year as a nod towards public holidays ...
this gives 50 times 5.5 working days per year = 275 days per year to generate 357 896 vehicle sightings.
If the extraction takes ten years this gives 357 896 divide by ten = 35 789 per year .
With a working year of 275 days this gives 130 lorry routes per day.
If the extraction takes five years this gives 357 896 divide by five = 71 578 per year .
With a working year of 275 days this gives 260 lorry routes per day.
For a fair 'minutes-based' perspective take a functional day to be 9 hours ,
this is 9 times 60 = 540 minutes.
Now, the figures that are derived in the next line are averages ,
i.e. matters would be even worse early in the day as the first 'bunched' arrivals arrive at similar times ...
morning rush-hour.
So, for a ten year plan we have 540 divided by 130, giving 4.15 ...i.e. 4 minutes.
And for a five year plan we have 540 divided by 260, giving 2.07 ...i.e. 2 minutes.
THIS GIVES A TIME PATTERN OF ONE LORRY ROUTING EVERY TWO MINUTES FOR THE FIRST FIVE YEARS
PLUS THE EFFECT OF SMALLER OWNER_DRIVER VANS ON LOCAL COLLECTIONS...
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Aim : To place the quantities into some sort of visual perspective .
From Surrey CC PMZ71 pdf document ( also on this website ), estimations are that zone contains a net mineral reserve of approximately
2 900 000 tonnes of soft sand and 500 000 tonnes of Bargate stone.
Source : Road Haulage Association Bulk Load Factors also used.
SAND predicted mass is 2 900 000 tonnes
Density of Sand in it's compressed 'in-ground' state is 1 400 kg per cubic metre
which is 1.4 tonne per cubic metre
Inverting ... this is 0.7143 cubic metres per tonne
As there is 2 900 000 tonnes of Sand projected ...
This is 2 900 000 * 0.7143
= 2 071 470 ' volume in-place ' cubic metres
but a 'bulking factor ( volume increase during excavation ) of 1.2 applies ...
which means 2 071 470 * 1.2
= 2 485 764 cubic metres once extracted.
So visualise as a conical pile of sand 220 metres across the base...
( i.e. a radius of 110 metres ) this pile having a height of 196 metres.
To check this assertion, enter these figures into the geometrical formula
for a cone which is ...
( pi * radius * radius * height ) divided by three.
= ( 3.1416 * 110 * 110 * 196 ) / 3
= 2 483 539 Correct, as just slightly less than the 2 485 764 target figure.
STONE predicted mass is 500 000 tonnes
Density of Sandstone in it's compressed 'in-ground' state is 1 600 kg per cubic metre
which is 1.6 tonne per cubic metre
Inverting ... this is 0.625 cubic metres per tonne
As there is 500 000 tonnes of Sandstone projected ...
This is 500 000 * 0.625
= 312 500 ' volume in-place ' cubic metres
but a 'bulking factor ( volume increase during excavation ) of 1.6 applies ...
which means 312 500 * 1.6
= 500 000 cubic metres once extracted.
So visualise as an ideal cube of tightly packed rock 79 metres square at the base...
this would have a height of 79 metres.
To check this assertion,
enter these figures into the geometrical formula
for an ideal cube which is ...
( width * depth * height )
= ( 79 * 79 * 79 )
= 493 039 Correct, as just slightly less than the 500 000 target figure.
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From Surrey CC PMZ71 pdf document ( also on this website ), estimations are that zone contains a net mineral reserve of approximately
2 900 000 tonnes of soft sand and 500 000 tonnes of Bargate stone.
Aim : To estimate the depth of excavation.
SAND predicted mass is 2 900 000 tonnes
Density of Sand in it's compressed 'in-ground' state is 1 400 kg per cubic metre
which is 1.4 tonne per cubic metre
Inverting ... this is 0.7143 cubic metres per tonne
As there is 2 900 000 tonnes of Sand projected ...
This is 2 900 000 * 0.7143
= 2 071 470 ' volume in-place ' cubic metres
STONE predicted mass is 500 000 tonnes
Density of Sandstone in it's compressed 'in-ground' state is 1 600 kg per cubic metre
which is 1.6 tonne per cubic metre
Inverting ... this is 0.625 cubic metres per tonne
As there is 500 000 tonnes of Sandstone projected ...
This is 500 000 * 0.625
= 312 500 ' volume in-place ' cubic metres
This gives a total 'volume in place' of 2 383 970 ' volume in-place ' cubic metres
as the declared surface area is 278 000 square metres and this is estimated to be 86% applicable,
the depth of the excavation should average 2 383 970 / 278 000 = 8.575 metres.
Allowing for access ramping and taper of the perimeters , this hints at a peak excavation depth of about 10 metres?
However the slope of the virgin surface of the location as it now stands, means that
the Nort-West corner is at an altitude amsl which is perhaps 19 metres higher than the river-facing ridge at East.
Does this mean a graduated face at west of 16 to 18 metres ( 6 storeys high )
Notes
What is the legislative framework to prevent further depths being excavated ?
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To calculate the volume of rainwater 'processed' by the potential quarry site each year .
There are no major considerations re outside areas affecting the calculations
as there is a natural ridge along the 1000m long side of the site
which borders the A3 highway. This forms a natural 'watershed'.
The area of the site which is available for excavation to a depth of at least two metres
would form a right-angle triangle with two sides of approx 400m and 800metres
Using 'half width times height' derive an area of 160,000 square metres.
as rainfall is taken to be 100mm ( 0.1 metre ) p.a. this gives ...
( from Met Office website ... climate figures 20 year average ... South England region.)
as rainfall is taken to be 700mm ( 0.7 metre ) p.a. this gives ...
( ... climate figures 20 year average ... South England region.)
a volume of rainwater processed by the potential area of excavation of
112,000 cubic metres...
As 48x48x48=110,592
this can be visualised as a cube of sides all 48 metres in length.
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