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The road project




CALCULATIONS

The design:
Type of ground substrate - clay brief
The quality of the land for the road - pretty good
Easy transition " l " - 40 m
Arc radius " R " - 80 m
Design speed " v " - 30 km/h
For a clean dry surface " φ " - 0,7

CALCULATION OF OPTIMUM MIXES (Ferett triangle)

Where:
P - sand
PΠ - dusty sand
Pg - loamy sand
G - clay
Gp - clay sand
GΠ - dusty clay
Gpz - sandy clay brief
Gz - clay brief
GΠz - silty clay brief
Ip - sandy loam
I - loam
IΠ - dusty clay

X GROUND A GROUND B
The content of the sand fraction Fp 44% 18%
The content of clay fraction Fi 33% 14%
The content of silty fractions FΠ 23% 68%



X GROUND A GROUND B
Fp 44 • 0,67 = 29,48 18 • 0,33 = 5,94
Fi 33 • 0,67 = 22,11 14 • 0,33 = 4,62
FΠ 23 • 0,67 = 15,41 68 • 0,33 = 22,44
67 33


OPTIMAL MIX

Fp
44 • 0,67 + 18 • 0,33 =
35,42 %
Fi
33 • 0,67 + 14 • 0,33 =
26,73 %
FΠ
23 • 0,67 + 68 • 0,33 =
37,85 %

CALCULATION OF THE SURFACE THICKNESS

h = 3 • a • b1 + 15 • a • b2 • c • d1 + 10 • a • b3 • c • d2 • e + 5 • d4 • d2
where:
a - depends on the load road surface ( 1.0 - 1.2 )
b1 - depends on the type of bitumen used for road surface ( 1.2 )
b2 - depends on the type of upper layer fabrication ( 1.2 )
b3 - depends on the type of material used for the bottom layer of foundation (Sand fine = 1.2; coarse sand = 1,8 )
b4 - depends on the type of land to the cut-off layer (Sand = 1. )
c - depends on the maximum load per wheel of car (Tones) ( 1.12 )
d1 - depends on hydrological conditions of land
d2 - depends on hydrological conditions of land

h = 3 • 1,2 • 1,2 + 15 • 1,2 • 1,2 • 1,12 • 1,0 + 10 • 1,2 • 1,3 • 1,12 • 1,0 • 1,0 + 5 • 1,0 • 1,6
h = 54 cm


ARC ROAD WIDENING

e = 50 / R

e = 50 / 80

e = 0,63 m


Check the path of maximum tilt


CALCULATION OF HORIZONTAL ARC



CALCULATION OF VERTICAL ARC





The project of small family house




Pieniny Mountains - cross altitude



Excursion in the Pieniny (2008). Motivated me to do that in AutoCad programm draw a cross section of the route which went through on that day.


Small wastewater plant




CALCULATIONS

The calculation of the required degree of wastewater purification.
I. Calculation of the dilution of sewage waters of the receiver:

n = ( q + Q0 ) / q [-]

where:
q - amount of wastewater discharged into the receiver [m3*h-1] ⇔ q = 30 [dm3*s-1]= 108 [m3*h-1]
Q0 - minimum flow in the river [m3*h-1] ⇔ Q0 = 600 [dm3*s-1]= 2160 [m3*h-1]

n = ( 108 + 2160 ) / 108 [-]
n = 21 [-]


II. Calculation of complete mixing of the road from the point of admitting water to the receiver section to be analyzed:

L = 3,6 ∙ V0 ∙ tc [km]

where:
V0 - average flow velocity of water in the river [m*s-1] ⇔ V0 = 0,60 [m*s-1]
tc - total time of mixing dirt with water receiver [h] ⇔ tc = 5,5 [h] dla Q / q = 20


L = 3,6 ∙ 0,60 ∙ 5,5 = 11,88 [km]


III. Calculation of the mixing ratio of waste water from the waters of the receiver:

α = Lm / L [-]

where:
Lm - the distance between analyzed sections [km] ⇔ Lm = 9 [km]


α = 9 / 11,88 = 0,76[-]


IV. Time flow of water from the outlet of the collector to the test section of the watercourse:

t = 0,0116 ∙ Lm / V0 [day]
t = 0,0116 ∙ 9 / 0,60 = 0,17 [doby]


V. Calculation of BOD5 mixture of water and sewage directly below the drain outlet after days of starting the process of mineralization:

L5m = LT / 10-k1 ∙ t [mg*dm-3]
where:
LT - BOD5 water in the receiver for a given category ⇔ for III category Lt = 10 [mg*dm-3]
k1 - constant rate of biochemical oxygen consumption by the sewage is mixed with water ⇔ dla T = 5o k1 = 0,05
L5m = 10 / 10- 0,05 ∙ 0,13 = 10,15 [mg*dm-3]


VI. Calculation of the allowable effluent BOD5 entering the receiver:

Ldop5s = ( α ∙ Q0 / q ) ∙ ( L5m - Lrz ) + L5m [mg∙dm-3]
where:
Lrz - BOD5 river water directly above the collector ⇔ Lrz = 2,2 [mg∙dm-3]

Ldop5s = ( 0,76 ∙ 2160 / 108 ) ∙ (10,15 - 2,2 ) + 10,15 = 130,99 [mg*dm-3]


VII. The calculation of the required degree of wastewater purification:

P = [( L5s - Ldop5s ) / L5s] ∙ 100 [%]
where:
L5s - BOD5 raw sewage existentially-economic ⇔ L5s = 300 [mg∙dm-3]

P = [( 300 - 130,99 ) / 300] ∙ 100 = 56,34 [%]


The calculation of the balance of oxygen-day receiver.
a) Active page.
1. Calculation of the amount of oxygen dissolved in water after the first day:

O1 = O2rz - Z24 [mg*dm-3]
where:
O2rz - the actual amount of oxygen dissolved in water receiver [mg*dm-3] ⇔ O2rz = 8,3 [mg*dm-3]
Z24 - oxygen consumption within 24 hours [mg*dm-3] ⇔ Z24 = 0,72 [mg*dm-3]

O1 = 8,3 - 0,72 = 7,58 [mg*dm-3]


2. Calculation of the remaining amount of oxygen available to the receiver:

Q2 = O1 - Oc [m*dm-3]

where:
Oc - the minimum amount of oxygen required for a particular species of fish living [m*dm-3] ⇔ for carp Oc = 4 [m*dm-3]
O2 = 7,58 - 4 = 3,58 [m*dm-3]


3. Calculate the total amount of oxygen in the river water during the day:

Qd = O2 ∙ O0 ∙ 86400 [mg]

where:
O0 - minimum flow in the river [dm3*s-1] ⇔ O0 = 600 [dm3*s-1]
Od = 3,58 ∙ 600 ∙ 86400 = 185587200 [mg]


4. The calculation of oxygen saturation:

Pnas = Orz2 / Op ∙ 100 [%]

where:
Op - volume for the full saturation temperature of the receiver [mg*dm-3] ⇔ for T = 50C Op = 12,80 [mg*dm-3]
Pnas = 8,3 / 12,8 ∙ 100 = 64,84 [%]


5. Calculation of water surface:

W = B0 ∙ V0 ∙ 86400 [m2]

where:
B0 - width of the river [m] ⇔ B0 = 10 m
V0 - flow velocity in the river [m*s-1] ⇔ V0 = 0,60 [m*s-1]
W = 10 ∙ 0,60 ∙ 86400 = 518400 [m2]


6. Calculation of the amount of oxygen supplied to the waters of the receiver:

R = W ∙ A [mg]

where:
A - daily absorption by the water's surface at a given level of oxygen saturation for a given river ⇔ for large river A = 3,5
R = 518400 ∙ 5,45 = 2825280 [mg]


7. The amount of oxygen for the active:

OA = Od + R [mg]
OA = 185587200 + 2825280 = 188412480 [mg]


b) Passive page.
1. Calculation of oxygen to neutralize the waste water:

Ośc = 0,3 ∙ L5s ∙ q ∙ 86400 [mg]
Ośc = 0,3 ∙ 300 ∙ 30 ∙ 86400 = 233280000 [mg]


2. Calculation of the amount of oxygen required to neutralize the sediment:

Or = 0,01 ∙ e ∙ W ∙ G [g]

where:
G - oxygen demand to the process of reducing sediment [g*m-2] ⇔ G = 1,7 [g*m-2]
e - amount of oxygen in the bottom sediment [%] ⇔ e = 90 [%]
0r = 0,01 ∙ 75 ∙ 518400 ∙ 1,7 = 660960 [g]


3. Calculation of oxygen to the passive page:

Op = Ośc + Or[mg]
Op = 233280000 + 660960000 = 894240000 [mg] = 894,24 [kg]


CONCLUSIONS:
0A = 188,41 [kg] and Op = 894,24 [kg] and 0A < Op
SHOULD BE BUILT SEWAGE TREATMENT PLANTS

DESIGNED

MONOLITHIC SEPTIC TANK - a preliminary treatment of waste water (sedimentation). The total capacity is 90 m3.
It is composed of four chambers:
chamber VI = 51 m3
chamber VII, VIII, VIV = 17 m3 each
Each chamber has a diameter manhole Ø 60 cm.
The guide wire wastewater has a diameter of Ø 150 mm.
Designed sewage supply pipe is located 10 cm above the mirror treatment. The residence time of sewage in septic is 2 - 3 days.


FEEDER PLANT - its purpose is to periodically and systematically acquired dispensing water from the settling and transport them to the sand filter.
Periodic water supply is to improve the efficiency and effectiveness of wastewater treatment by artificial filters.
Feeder plant is well paved with a diameter Ø 160 cm.
Here the dish is mounted and the dosing cycle is dependent upon the speed of filling the vessel.


ARTIFICIAL GROUND FILTER - a second degree of treatment, immediately after the septic tank.
Placed at the bottom of the treated effluent drainage collector is made of perforated pipe with a diameter of 10 cm with 1% slope studded with
35 cm layer of washed and graded gravel.
On the surface this was placed another 15 cm layer of small pebbles washed with a diameter of 0.6 - 1 cm, in direct contact with fluid.
Ventilation of drains was performed using a vertical vent pipe connected to the highest point of the catheter.


IMHOFF SETTLER - is an alternative to a septic tank.





Drainage




Calculations


1. Inflow of water into the drain unit:

where:
r0 - radius of the drain [m] ⇔ r0 = 0,10 [m]
k - coefficient of filtration of the aquifer [m/d]
PΠ - 1,7[m/d]
Pd - 2,1[m/d]
h1 - thickness of the dusty sand ⇔ 2,00 [m]
h2 - thickness of fine sand ⇔ 2,50 [m]

k = 1,92 [m/d]
R - radius of the curve of depression [m]
S - reduction in depression [m] ⇔ 2,00 [m]

2. The range of the curve of depression:

R = 2 ∙ S ∙ √k ∙ √H [m]

where:
H - thickness of the aquifer [m] ⇔ H = 4,5[m]


R = 2 ∙ 2 ∙ √1,92 ∙ √4,5 = 11,76 [m]


3. Water flow in drain:

Q = q ∙ L [m3/s]

where:
L - length of drain [m]
q - inflow of water into the drain unit [m3/s/mb]



q = 0,885 [m3/s/mb]


4. The size of the groundwater table elevations inside the ring over the water level in drains:


where:
r0 - radius of the drain [m]
R0 - radius of the curve of depression + drainage spanning half the width [m]
a - half width of the drainage [m] ⇔ 13,00 [m]
φ1 - f(a/m) (reading of the graphs)
φ2 - f(Ra/m) (reading of the graphs)
φ = 5,2 - 3,8 = 1,4



ha = 0,39 [m]


5. Calculation of the offset from the foundation drains:

L = a + b/2 + (H - h)/tgφ

where:
a - plinth foundation [m] ⇔ 0,4 [m]
b - 0,5 [m]
H - distance from the foundation drain to the ground surface [m]
h - distance from the foundation of the building to the ground surface
Drain: 5.4 Position
L = 0,4 + 0,5/2 + 2,1/0,45 = 0,68 [m]
Drain: Position 1,2,3,6
L = 0,4 + 0,5/2 + 0,4/0,45 = 0,49 [m]


5. Calculation of the amount of filter protection:

V = F ∙ L [m3]

where:
L - pipe length [m] ⇔ 44 [m]
F - cross-sectional area ⇔ 0,84 [m2]
V = 0,84 ∙ 50 = 42 [m3]