STUDENT: TRINH DUC DUYEN/55XE/624455TUTOR: VUONG DO TUAN CUONGTHE PROJECT OF CONSTRUCTION ORGANIZATION

I.Introduction about building.1. Feature of building.- Architecture: the site is rectangular shape and the area is to build, BxL=29x70.4.This building have six stories that total height is 19.2m.- Height of each story. Height of story 1: H1=4.4m Rest of story height: H2=H3=H4=3.8m Height of roof story: Hm=3.4m

-Span and bay of column. Bay of the column: B= 4.0mEdge span: L1=7.0mMid-span: L2=7.0m-Foundation.NumberABC

b(m)1,41,41,4

a(m)2,42,42,4

t(cm)404040

-Cross-section of column.Cross-sectionOutside column bxh (C1)Inside column bxh (C2)

Story 125x4025x40

Story 2+325x3525x35

Story 4+525x3025x30

- Dimension of beam.+ Main beam D1: bxh= 25x70cm+ Auxiliary beam D2,D3: bxh= 25x35cm+ Roof beam Dm: bxh=20x70cm-Wall+ Outside wall : 220 (mm) ; inside wall : 110 (mm)+ Plastering 40% of outside wall area and 50% of inside wall area+ Coating 6% of outside wall area and 1% of inside wall area+ The area of door accounts for 60% of outside wall area and 10% inside wall area.-Electricity supply and water supply: 0.32h service/ 1m2.

- Slab.Thickness of slab: hs= 15 cmThickness of roof slab: hsm=15 cm- Size of foundation.

45+0.00ttt1001200Ground

2. Condition in construction.

A. Foundation I. Construction technologies analysis.a. Technical requirement. The methods of excavation must ensure thatthe natural structure of soil is not destroyed. If construction doesnt use the retaining wall during the excavation, we have to excavate according to step by step method. The size of foundation in site has tolarger than that in design 0.5m to make the formwork. If we use machines to excavate the foundation, we have to bring back the 10-20cm of soil in base of foundation. The soil is dug from the foundation that it has to transport far the hole to avoid the slide of rampart of foundation. To avoid the stagnant water in the hole of foundation we have to make a small cannel to drain the water.b. Method of excavation construction. To reduce the expense of construction we apply alternatively the mechanical method and manual method. According to the appendix 6-II we find out the slope of hole of foundation: m=0.67 Base of foundation: large than 0.5m.II. Works+ Main works: Excavation. Maintaining foundation ( equal to 5 percent volume of total soil) Pouring level concrete. Formwork installation ( foundation and bracing system ) Steel works for foundation and brace. Pouring foundation concrete and bracing system. Removing formwork and bracing system. Built wall. Fill hole of foundation.III. Calculation .2.1 Volume of soil The depth of foundation hole.Hd=0.1+hm=0.1+3x0.4=1.3m wSlide factor: choose m=0.67We see clear that the span of building(B=4m) is approximately the width of foundation hole.The Size of bottom foundation hole: am= a+ 2x0.5=2.4+1=3.4m bm= b+2x0.5=1.4+1=2.4mThe size of surface foundation: a=am+2h.m=3.4+2x0.67x1.35.1m b=bm+2h.m=2.4+2x0.67x1.34.1m

SECTION A-A

SECTION B-B

In which: + Bottom of foundation:Length A = 17.B + bf + 2x0.5 = 68 + 1.4 + 1 = 70.4 (m)Width B = am + 2x0.5 = 2.4 + 1 = 3.4 (m) + Surface of excavation pit: C = A + 2Hf.m = 70.4 + 2x1.2x0.67 = 72.008 (m)D = B + 2Hf.m = 3.4 + 2x1.2x0.67 = 5.008 (m) Total volume excavation soil :

Total volume of soil: 5V=1947.5(m3).In which: manual excavation and maintaining foundation make up five percent of total soil. Vtc= 0.05x1947.5=97.375(m3). Vm=1947.5-97.375=1850.125(m3).2.2 Volume of soil fill-up.- The first fill-up: up to surface of brace.- The second fill-up: up to ground. The first volume of soil:V1= Vd1-Vbt1 Vd1: The volume of soil is excavated from bottom to surface of brace. Vbt1: The volume of concrete is poured up to surface of brace.

In which: A=70.4m, B=3.4m C=A+2mHg=70.4+2x0.67x0.8=72.54m D=B+2mHg=3.4+ 2x0.67x0.8=4.47m

V1=1407.13-279.68=1127.45m3. The second volume of soil:V2=Vd2-Vbt2- Vd2 : The volume of soil from surface of brace to ground. Vd2=1947.5-1407.13=540.37m3

V2=540.37-67.34=473.03m3.

2.2 Choosing the excavator.- The volume of soil that is excavated is too big so we dig the foundation hole by machines and maintain the foundation by manual.- choosing the myogughch- Choosing back hole excavator HITACHI code UH015, with technical features: Volume of bucket : q = 0.45 (m3) Maximum radius for soil dumping : R = 7.82 (m) Rotation cycle : tck = 18.5 s Technical productivity :Nt = q x K / Kt x nck x ktgIn which : q = 0.45 (m3) volume of bucketK= 0.95 buckling filling factorKt = 1.15 unconsolidated factor of soilsktg= 0.8 time using factornck the number of excavation cycle in a hournck = 3600 / TckTck = tck x Kvt x KquayWith : Tck time of one cycle (s)tck =18.5s time of one cycle when rotation angle is 90o.Kvt = 1.1 soil pouring factorKquay = 1 factor depend on rotation angle (90o)=>nck = 3600/(18.5x1.1x1) = 176.9 Nt = 0.45 x 0.95 /1.15 x 176.9 x 0.8 = 52.6 (m3/h)

Productivity in 1 shift : N = 8 x Nt = 8 x 52.6 = 420.8 (m3/shift)

Number of shift for completing excavation :N = 1947.5/ 420.8 = 5 (shift)

2.3 Choosing the backfiller.The volume of soil is filled back the foundation hole.

Because the volume of soil that is filled back is highest therefore we use the bulldozer and maintain the foundation by manual.Choose machine: D2-53 with some technical parameters. Length: B=3.2m Height: h=1.2m Forward velocity: V1=10.1(km/h) Toward velocity: V2=5.3(km/h) Productivity: N=304(m3/shift) The number of shifts:(shift)2.4. Bracing works and wall works.- Choose the dimension of foundation brace: bxh=0.25x0.7m The volume of brace and lining concrete:V1=0.25x0.7x(70.4x5+29x19)=158.025m3V2=19x5x0.1x2.6x1.6=39.52m3- The wall is built from surface of brace to bottom concrete lining of slab.V=0.22x0.5(70.4x5+29x19)=99.33m32.5. Volume ofembanking sand, concrete lining and reinforcing concrete.- Volume of embanking sand: V1=0.8x29x70.4=816.64 m3- Volume of lining concrete(10cm): V2=0.1x29x70.4=204.16 m3- Volume of reinforcing concrete (10cm): V3=204.16 m3

- Volume of reinforcement(): V=0.015x204.16=3.1m3=> m=7850x3.1=24335(Kg)=24.34 (T)3. Reinforcement works, Formworks, and concrete of foundation.- After construction have finished the foundation holes work, we start doing workssuch as pouring concrete lining and brace, reinforcement, removing the formworks, fill the foundation hole- Some works such as concrete lining, concrete, reinforcement, formworks, build wall base on the geometry, size, and quantity, as shown in table below. Soil construction methods. Constructed by excavator combined with repairmen manual (UH015).

( EXCELL)IV. TECHNICAL METHOD.4.1. Tower crane.

Length of the building : 4.0x 17 = 68 m Calculate required height :H = Hct + Hat + Hck + HtWhere : Hct_ building height from the crane standing level, Hct = 0 mHat_ the safety distance, h1 = 1mHck_ maximum height of the structure components of craning ( the height of the concrete mortar barrel, Hck = 1.5 mHt_ height of the hanging, Ht = 0.75mThus, H = 0 + 1 + 1.5 + 0.75 = 3.25 m Calculate the required jib length : R = b + b1Where : b_ distance from the rotating base of crane to the building where is closest to the crane, b = 7.5 m b1_ building width (including the distance from the building to the scaffolding where is closest to the crane, b1 = 7.0x 4 = 28 mThus, R = 28 + 7.5 = 35.5 m Calculate loading weight : Q = qck + qtb (T)Choose the barrel which has 0.8 m3volumeWhere : qck_weight of the constructure components ( concrete ), qck = 2.5 x 0.8 = 2.0 tonqtb_ weight of equipment (barrel) : 0.1x 2.0 = 0.2 tonthus, Q = 2.0 + 0.2 = 2.2 ton Choose the tower crane : base on H, R, Q , we choose tower crane : GTMR 400, it has technical characteristics :Loading weight : Q = Qmin = 4,715 T >2.475TJib length : R = 36.5 > 35.5 mThe height : H = 32.6 m > 3.25 m Speed : + The lift and lower speed : 2.2 - 12 - 24 m/mins + the speed of car : 7.5 - 30 - 60 m/mins + the speed of crane : 12.5 - 25 m/mins + Rotating speed : 0.12 0.7- Determine the length of rail

Where

So, m ,

=>Determine productivity of crane: * Determine the cycle of delivering:

Formula: t1 : the time for hooking the load t1 = 0s

t2 : the time for lifting the load (s)

t3 : the time for rotating to the required location : (s)

t4 : the time for moving of tower crane : (s) We choose the distance from the center of tower crane to the concrete mixing station : l = 7(m)

t5 : the time for moving of car on the arm of the crane : (s) t6 : the time for removing the loads : t6 = 0

t7 : the time for lowering the load : (s)

, the cycle of delivering: (s) * Productivity of crane : Nshift = t.Q.K.Ktg.nLegand: t : the working time in a shift t = 8 (hours) Q =4.125 T K: weighting factor K = 0.8Ktg : factor considering time-used of crane Ktg = 0.85

n: the amount of cycle So, productivity of crane : Nshift = 8x2.2x0.8x0.85x12.2 =146.0 (T/shift)=> The weight of concrete in a shift:W=8 x 2.0 x 0.85 x 0.8 x 12.2 = 132.7 T=>The volume of concrete in a shift: V=132.7/2.5=53.1 (m3)

4.2.MixerThe concrete quantity in themaximum region is : We choose type mixer : free transportation, label SB-91AIt has technical parameters : The volume of concrete in 1 mixture : 750 l The volume of manufacturing tank : 500 l The velocity of rotary tank : 18.6 v/pht The engine capacity : Ndc = 5.1 kW The weight : m = 1.15 ton Time for 1 mixture : 80s The number of mixtures in 1 hour: Nck= 3600/ (20+ 20+ 80) = 30(mixtures )We have using productivity of concrete mixer :

Ns = = = 7.31 (m3/h)Thus, productivity of 1 machine shift : Nca = tcax Ns = 8x 7.31 = 58.48 (m3/ ca) V. The schedule of foundation construction.5.1. Division of construction joints. The number of areas must ensure continuous production line. The volume in the zone must equally and the difference does not exceed 25 percent The volume of concrete in a sector must ensure poured continuously in a work shift. We make sure concrete structure stability in period timeBase on the statement above We can divide the construction plan into some areas as shown in below.- The volume of lining concrete on base of foundation: V= 39.52(m3)- The volume of concrete of foundation: V=307.335(m3).=> The number of areas: 315.02/53.1=5.9 (areas)=> So, we can divide this plan into 6 areas.

Construction Schedule for Foundation

B. SUPERSTRUCTURE CONSTRUCTION.I. WORKS. Installing reinforcement of column. Formwork of column. Pouring column concrete. Removing column formwork Install beam and slab Formwork Install reinforcement of slab and beam. Pouring slab and beam concrete Removing slab and beam formwork Building wall Install power and water system Plastering inside wall Background work. Door installation. Lpthitbvsinh Painting inside wall Painting outside wallII. Table.All the data of superstructure work during construction is taken from the project of construction method 1.III. Technical method.1. Batch mixer.The volume of concrete for a maximum are.m3 So, We select automatic concrete mixer ( SB-91A).- Technical parameter:+ The volume of concrete for a batch of concrete:750 (l)+ The Volume of mixer drum:500 (l)+ The rotation speed:18,6 ( v/ph)+ The engine capacity: Ndc= 5,1 KW+ The weight : m = 1,15 (Ton)+ The mixing time a batch: 80 (s)t1 = 20s : The time for make material pour in bucket.t2 =80s : The mixing time a batcht3 = 20s: The time for batch mixer come back.=> the total time: Tc = t1+t2+t3= 20+80+20 = 120 s

=> The number of batch in 1 hours: Nck=3600/120=30.=>> Productivity in a machine shift:

=

2. Choosing the tower crane

In the design formwork, the bucket is chosen according to the load, which caused by pouring concrete. We choose the Hoa Phats bucket ,The volume of concrete 1,5 (m3), the height of cable hanger and hook is 0,75m. The bucket dont use hose lap and The distance from slab to bottom bucket which is 1(m).Because The building have many bay ( L>3B), the height of building is medium therefore we choose mobile tower crane, light balance-weight.- Determine the necessary height of tower crane.

The building height from the crane-standing level: 18,8m

The safety distance =1m

The maximum height of structure components of craning( the height of the concrete mortar barrel) =1,5m

The height of the hanging / tying equipment:=0,75mVy H = 22,05m- Duty of tower crane:According to the appendix 6.3.3, TCVN 4453:1995 The volume of concrete not exceed the 90-95% of volume of bucket.The weight of concrete: 2,51,2 = 3 Tons=> Qyc = 3.1,1=3,3Tons+) Determine the jib length: R = S + d * Distance from the rotating base of crane to the building (or the scaffolding) where is closest to the crane.

S = r/2+e+ldg r: The distance between two of rails .6m e: Safety distance: 2m ldg: The width of scaffolding+ Space for construction. 2,5m=> S = r/2+e+ldg =6/2+2+2,5=7,5m * Building width (including the distance from the building to the scaffolding where is closest to the craned = 7,02 + 7,02=28m=> R = 7,5 + 28 = 35,5m- Based on the some data about tower crane : Hyc=22,05mRyc=35,5mQyc=3,3 tnSo, We choose the tower crane GTMR 400, which is have some parameters :Loading weight : Q = Qmin = 4,715 T > 3,3TJib length : R = 36,5 > 35,5 mThe height : H = 32,6 m > 22,05 m Speed : + The lift and lower speed : 2,2 - 12 - 24 m/mins + the speed of car : 7,5 - 30 - 60 m/mins + the speed of crane : 12,5 - 25 m/mins + Rotating speed : 0,12 - 0,7- Determine the length of rail

Where

So, m ,

e. Determine productivity of crane: * Determine the cycle of delivering:

Formula: t1 : the time for hooking the load t1 = 0s

t2 : the time for lifting the load (s)

t3 : the time for rotating to the required location : (s)

t4 : the time for moving of tower crane : (s) We choose the distance from the center of tower crane to the concrete mixing station : l = 7(m)

t5 : the time for moving of car on the arm of the crane : (s) t6 : the time for removing the loads : t6 = 0

t7 : the time for lowering the load : (s)

, the cycle of delivering: (s) * Productivity of crane : Nshift = t.Q.K.Ktg.n Legand: t : the working time in a shift t = 8 (hours) Q =4,125 T K: weighting factor K = 0,8 Ktg : factor considering time-used of crane Ktg = 0,85

n: the amount of cycle So, productivity of crane : Nshift = 8.3,3.0,8.0,85.9,24 = 165,9 (T/shift)=> The volume of concrete in a shift:V=9,24.8.1,2.0,8.0,85=60,3 (m3)

C. ROOF WORKS.

- Waterproof concrete: Area : S= 28 x 68 = 1904 m2The thickness of this layer : 0,048mThe amount of waterproof concrete: V= 1904x 0,048= 91.39 m3- Heat resistant concrete:The thickness of this layer : 0,14mThe amount of heat resistant concrete: V= 1904x 0,14 = 266.56 m3- Two ceramic tile layer : Total area: S=1904 x 2 = 3808 m2

The productivity of the crane : 60,3 m3 The volume of concrete and mortar : V=91.39+266.56+3808.0,015=415.07m3=> The number of segments:

=> we divide the construction plan into 7 segment:The amount of materials for each segment showed below:

NOName of taskUnitQuantityNorm(h/m3)Working dayTime(day)Labour in an area

1Waterproof concretem391.3958.1519

2Heat resistant concretem3266.56523.8124

3Ceramic tile (2 layer)m238080.2517117

D. FINISHED WORKS.I. Main works.- build wall.- Plaster.-Tiling- Paint- Installing power and water system , installing window.II, Calculation of works.C. Volume of wall works. - The wall construction consists of two types that are 220mm and 110mm. - The total area of wall is calculated according to the percentage as below. The outside wall (220) follow center line having area of windows and doors accounting for 60 percent of total area. The inside wall (110) have area of window and door accounting for 10 percent total area. - The area of outside wall (220) in 1st story: Sng1= 0.4x3.6x2x(28+68)=276.48(m2) - The area of outside wall from 2nd to 4th story. Sng1=.. =Sng4 =0.4x3.8x2x(28+68)=291.84 (m2) - The area of outside wall (220) in 5th story: Sng5= 0.4x3.4x2x(28+68)=261.12 (m2)- The area of inside wall (110) in 1st story: Str1 = 0.9x3.6x(28x18+68x3)=2293.92 (m2)- The area of inside wall (110) from 2nd to 4th story: Str2 = .= Str6 = 0.9x3.8x(28x18+68x3)=2421.36 (m2)- The area of inside wall (110) in 5th story: Str5= 0.9x3.4x(28x18+68x3)=2166.48(m2). So, the volume of wall works: m3 m32. Volume of plastering wall work.

- The total area of outside wall must be plastered: Sng= m2- The total area of inside wall must be plastered:

Str= m2 The total area of wall must be plastered: S= Sng+ Str=7926.72m23. Volume of window and door installation works.- The area of window and door accounts for 60 percent of the total outside wall.Scua= 1413.12x0.6/0.4=2119.68 m2- The area of window and door accounts for 10 percent of the total inside wall.Scua=1290.68x0.1/(0.9x0.11)=1303.72 m2The total area of window and door in the building: S=3423.4 m24. Volume of painting works.- The painting for 220-wall ( 6%): S1=1413.12x0.06/0.4=211.95 m2- The painting for 110-wall(1%) : S2= 11724.48x0.01/0.9=139 m2=> S=350.95 m25. Volume of floor works. ( lat nen)The area of floor in building is tiled: 9112.71 m26. Volume of plastering ceiling works.Sceiling=Sfloor=9112.71 m27. Volume of tiling works.We can calculate this works about 5% of the area of plastering inside wall.So=0.05x13027.28. Volume of power and water system works.H=0.32x5x28x68=3046 (hours)

NOName of taskUnitQuantityNormWorking dayLabor in an area

1Building 110-wallm31290.680.662,43

2Building 220-wallm3310.895.124,45

3Installing electric and water systemh/m295200.328.59

4Plastering ceilingm29112.710.4711,8912

5Plastering inside wallsm26513.60.478.59

6Plastering outside wallsm21413.120.471.842

7Painting outside wallsm2211.9511.4277

8Painting inside wallsm213911.424.45

9Installing doors and windowsm23423.4 0.251711.739

10Paving slabsm29112.710.6616.717

PART II, CONSTRUCTION SITE LOGISTICS.I, Determine circulating factor of formwork :1, Slab and beam formwork.Using period of formwork : Tvk = T1 + T2 + T3 + T4 + T5Where: T1 _ installation time of formwork in 1 area, T1 = 2 dayT2 _ installation time for reinforce in 1 area, T2 = 2 dayT3 _ pouring concrete time 1 area, T3 = 1 dayT4 _ allowable time of dismantle formwork in 1 area, T4 = 2 days with unloaded formwork, T4 = 10 days with loaded formwork.T5 _ dismantle time of formwork in 1 region, T5 = 1 day=>Using cycle of formwork: Tvk = 2+ 2+1+ 10+ 1= 15 days2, Column formwork.Using period of formwork : Tvk = T1 + T2 + T3 + T4 + T5

Where: T1 _ installation time of formwork in 1 areaT2 _ installation time for reinforce in 1 area T1+T2=1 dayT3 _ pouring concrete time 1 area, T1 = 1 dayT4 _ allowable time of dismantle formwork in 1 area, 1 dayT5 _ dismantle time of formwork in 1 region, T5 = 1 day=>Using Cycle of formwork: Tvk = 1+ 1+1+ 1= 4 daysII, Design the Road. It is important work considered after that determination of construction area. Consist of two main works:+ planning of network of road.+ Design the components.1. Planning of road network. The network include output-gate, input-gate and to the equipments and workers connect the building with outside. In addition, we have to mention the park to stop or return the struck. Network of road in site is designed around the building. Is built two gates in two gables of building. Design the car park on site.2. Calculation of characteristic of road. Choose feature of road according to standard.The width of road: B=b+2c=3.75+1.25=6.25m In which: + b: the width of roadway. + c: The curb. The radius: Rmin=15m.III. Material.1. Determine the volume of reserving material

Dmax=q. The duration between two time: t1=4 day The time to receive and deliver to site: t2=1day The time of loaded and unloaded material: t3=1 day The time of classification: t4=1 day The time of reserves: t5=5 days = t1+t2+t3+t4+t5=12 days+ Steel: 3.33 T/day/area+ Concrete: 48.5m3/day/area. Volume of rock: 0.88x48.5=42.68(m3/day) Volume of sand: 0.42x48.5=20.37 (m3/day) Volume of cement: 0.35x48.5=16.98 (T) + Wall construction: ( 17.8m3/day/area) Norm 1776: 1m3 wall have 550 blocks, 0.29 m3 mortar Number of bricks: 550x17.8=9790 pieces Volume of motar: 0.29x17.8=5.2 m3 Volume of sand: 1.09x5.2=5.67(m3) Volume of cement: 5.2x0.32=1.66 (T)+ Plastering works: (176.15m2/day, 176.15x0,015=2.64m3/day) Volume of sand: 1,09x2.64=2.88(m3) Volume of cement: 2.64x0.32=0.845 (T) +Formwork: Volume of stringers and shores: 14.23m3 Volume of formwork: 6.13m3=> Total volume of each reserving material:- Rock : 42.68x5=213.4(m3)- Cement: (16.98+0.845+1.66)5=97.425 (T)- Brick: 9790x5=48950 pieces.- Steel: 3.33x5=16.65 (T)-Sand: (2.88+5.67+20.37)5=144.6(m3)- Formwork+stringers+Shores: (14.23+6.13)5=101.8 (m3)2. Calculate the area of storage:

F= Dmax/d with S=xF

NameunitquantitytypesNormArea(m2)

S(m2)Real Area S(m2)

Rockm3213.4outdoor371.11.285.3290( 10x9)

Sandm3144.6outdoor348.21.257.8460(10x6)

Formworkm3101.8Half-outdoor1.856,561.267.8780(10x8)

CementT97.425Closed1.374.941.5112.41120(12x10)

BrickPieces48950Outdoor70069.91.283.8890(10x9)

SteelT16.65Half-outdoor44.21.56.310

3. Designing temporary house.i. Calculation of site manpower+ Group of main worker: A = Ntb =99 ( People)+ Group of people work in the fabrication.B = 25% A = 25 ( People)+ Group of technical staffC = 5% (A+B) = 5%( 99 + 25) =7( People)+ Group of administration and commercial staff :D = 5% (A+B+C) = 5% (99 + 25+7) =7 ( People)+ Group of supporting staff : for medium construction site.E =5%(A+B+C+D) =5%(99+ 25+7+7) = 7 ( People)Total of site manpower N = 1.06( A+B+C+D+E) = 1,06(99 + 25+7+7+7) =145 ( People)Assumption worker common in the local .number of worker have demand temporary house about 75% : N0=0.75xG=0.75x145=109 workers.+Number of people for design temporary house : Ndesign=1.1xN0=1.1x109=120 people ( with 1,1 factor mention member of family).-Temporary house for workers :S1=120x4=480 m2-Temporary house for technical engineer and administration and economical staffs S2=(C+D).4=14x4=56 m2-Temporary for shower: (25 people/room 2,5m2) about 120 people live on the site => needs 5 room S3=5x2,5=12.5m2-Canteen( for worker): including: 2 shift ,N=145 people, 1m2 / 1 peopleAssumption 85% people eat on the site S4a=0.5x0.85x145x1=76.5m2-Canteen( for officer): including: 2 shift ,N=14 people, 1m2 / 1 people S4b=0.5x0.85x14x1=5.95m2-WC: S5=12.5 m2-Health treatment:S6=145x10/1000x8=12m2-Security room :S7=20m2-Head-office: S8=16m2 (used container width:2,44m,length 12,19m, height 2,59m)

4.Design water supply systema. Water server construction:Q1=1,2.(A1+A2+A3+A4).Kg/(8.3600)With: Ai is quantity of water server for construction equipment. +A1:water for mixer Area (mixer concrete) A1=48.5x300=14550 (l/shift) +A2:water for mortar mixer Area : A2=(5.2+2.64)x250=1960 (l/shift) +A3:water for curing concrete : A3=48.5x200=9700 (l/shift) +A4:water for washed gravels: A3=42.68x1000=42680 (l/shift) +A5:water for washed brick :A4=9790x200/1000=1958 (l/shift) Kg=2 factor for use unregular water in one hour Q1=1.2x( 14550+1960+9700+42680+1958)x2/(8x3600)=5.9 l/sb. Domestic water: This kind of water used for living activities on-site (canteen ,shower..)Q2=Nmax.B.kg/(8x3600)With :+Nmax is maximum people on site, Nmax=254 people+B quantity of water for worker on site follow standard (B=20 l/day)+kg factor usage water unregular in one hour :select kg=1.8 Q2=254x20x1.8/(8x3600)=0.32 (l/s)c. Water supply for housing unit Q3=Nc.C.kg.Kng/(24x3600) Nc :People live in the site,Nc=120 people C :quantity of water for peoplein a day folow standard (C=50 l/day)kg factor usage water unregular in one hour:select kg=1.5kng factor usage unregular in a day :select kg=1,4 Q3=120x50x1.5x1.4/(24x3600)=0,145 (l/s)d. Water for fire fightingWater is also needed for fire fighting in the building and housing units, it depends on the number of occupations and the area of the building and units, can be determined about 10-20 litter per second or checked in standard tables.Choose Q4=10 (l/s)Because Q1+Q2+Q3=5.9+0.32+0.145=6.37 L=300mP=105.4kw

= 380 (V)

= 5%

The capacity of transformer station

Choose the line is lay on the shores.

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