Plant calculations

 Clinker factor = 1- (L.O.I +by pass+ Dust of filter)

= 1- (0.35+0.02+0.01) = 0.62
 Assume that: Clinker factor=0.58 As safety factor
That mean each 1KG raw meal gives 0.58 KG clinker
 X KG raw meal gives 1KG clinker
 X=1/0.58 =1.72
 That means 1.72 KG raw meal gives 1KG clinker

Quarry Calculations
 2 (L.S & Clay)Crushers will operate 6 d/w and 8 h/d
 Raw material required for crushers:

 Lime stone :
 Avg. Moisture =1.4166%
 L.S/d=4500*1.72*1.1*0.7271*1.014166= 6278 ton/day
 L.S/h =6278*7/(6*8)=916 ton/h

 Clay:
 Avg. Moisture =17.34%
 Clay/d=4500*1.72*1.1*0.2662*1.1734 = 2659 ton/day
 Clay/h=2659*7/(6*8)=388 ton/h

 Sand:
 Sand/d=9000*0.67/100 = 60.3 ton/day
 So Sand needs per day = 65 ton/day to be in safe side

Crushers Capacity
 Lime Stone:
 Impact crusher 1050 t/h
 Working days = 6 days/week
 Working shifts= (one shift)=8 hours/day
 Daily production = 1050 *8 =8400 t
 Weekly production = 8400*6 = 50400 t
 Clay:
 Roller crusher 400 t/h
 Daily production=400*8 = 3200 t
 Weekly production = 3200*6 = 19200 t

Quarry Transportation
 5 Trucks with cap. 60 ton (3 for L.S & 2 for Clay)
 2 Loaders
 2 Bulldozer
 Material handling from crushers to plant by belt conveyors as per flow sheet
Mix Design
 Stock pile covers 4 days raw mill feeding (9,000t/d)
 Selected longitudinal pile with max. capacity=3,5000 t
Reasons:
 1- Different quality of quarry layers
2- Easy for extension
 Stacker:
Slewing stacker 1900 t/h ( Operat. cap. 1450t/h)
 Reclaimer:
Bridge scraper 500t/h
 Raw Material open yard for Sand capacity 5,000 t
Raw Mill Area
 Raw Meal factor = 1.72
 Moisture content of Mix = 5.65 %
 Raw meal required =4,500*1.72*1.0565 = 8,177 ton/day
 Safety factor (take over)= 1.1
 We will operate the mill 22 h/d
 Raw mill capacity =(8,177*1.1)/22= 409 ton/hour
 Total raw meal = 9,000 t/d
 Max. capacity of Raw mill = 409 t/h
 Mill type: Atox roller mill 40 (3 rollers)
 Capacity : Design cap. = 427 t/h working 355 day/year
 Mill Drive : 2152 KW
 Air Separator :RAR 45
 Separator drive : 163 KW
 Hot gas generator: gas firing system

Calc. of Mill Fan

 Air flow =120 Nm3/s (assumption)
 Mill fan flow rate Q =120 x 60 x 60
= 432,000 Nm 3/h
 Mill fan power = 2,300 KW
 Raw transportation by collecting air slide then then to CF silo by using bucket elevator to
reduce power consumption .
 There is a sampler before bucket elevator for controlling quality

Raw Meal Silo

 One Raw meal silo CF silo with bucket elevator
 Capacity: 25,000 ton
=3.14*r²*silo length*material density
 Height: 52 m
 Active height:48m
 Internal diameter: 20 m
 Bulk density of raw material=1.42 ton/m³
 Kiln feed tank capacity:100 ton
 Bucket elevator capacity=(9000/24)*1.1=420 t/hr
Kiln Area
 Preheater : In Line Calciner – 5 stage
  Natural Gas for main burner & calciner
 Kiln feed = 4500 x 1.72 / 24 = 323 t/h
 Consider Gas lower calorific value = 9250 Kcal/Nm³
 Consider heat consumption = 800 kcal/kg clinker
 The production =4,500 x 1000/24 = 187,500 kg ck/hr
 Total heat consumption = 187500 x 800
= 150,000,000 kcal / hr
 Gas Req. for Kiln (Nm3/hr) = 150,000,000 / 9250
= 16,216 Nm³/hr
 Gas Req. for Main Burner = 16,216 x 0.4 = 6,487 Nm³/hr
 Gas Req. for preCalciner = 16,216 x 0.6 = 9,730 Nm³/hr
Calc. of I.D Fan Capacity
 The air required for combustion:
Lo = [ (Hu-1115) / 808 ] = [ ( 9250-1115 ) / 808 ]
= 10.07 Nm3 air/Nm3 gas
 Assuming that excess air 8%
 Total excess air= 10.07 x 1.08 = 10.87 Nm3air / Nm³ Gas
 Req. air hourly= 10.87 x16,216 =176,327 Nm³Air/hr
 Smoke gas (Vo) = (1.25 *9250 – 3025) / 808 = 10.56 Nm3/Nm3 gas
 Total smoke gas = 10.56 x16,216 = 171,344 Nm³/hr ----1
 Assuming that false air 0.5 % O₂
 false air = 0.5/(21-0.5) = 2.5 %
 Total false air = (2.5/100 )x 171344 = 4283 Nm³/hr---- 2

Assuming that kiln decalcination about 15% and our L.O.I = 35%

 Total volatized CO₂ in kiln = (0.35/1.977) x 0.15 x 187,500= 4,980 Nm3 CO2/hr ----3
 Assuming that firing in kiln 40% --- Preheater 60%
 Kiln smoke gases=0.4 x 171,344 = 68,535 Nm3 / hr ------4
 Total kiln gases = (2)+(3)+(4)
=77,800 Nm³/hr
 Depending on CL content we have 20% By pass + safety factor 15%
 By pass=0.35 x 77800.347 = 27,230 Nm3 / hr
 Total kiln gas without by pass = total kiln gases – by pass gases
= 77800.347-27230.12= 50,570 Nm³/hr -------5
 Total smoke gases in preheater (60%) =0.60 x 171,344
=102,806 Nm³/hr --------6
 Considering decalcination in preheater is about 85 % and our L.O.I=35%
 Volatized CO2 = (0.35/1.977) x 187500 x 0.85
=28215.1 Nm³/hr ---------------7
 Assuming O2 after ID fan is 5% so
 Total false air in preheater =[(5-2)/ [(21-5) x(102,806+28,215) ]
=24,566 Nm³/hr -------------8
 Total gases after preheater = 5+ 6 + 7+8 =206158.29 Nm³/hr
 Taking 15% Safety factor = 206158.29*1.15=237082.03 Nm³/hr
 So ID fans capacities are 237082.03 Nm³/hr
 Total smoke gases in preheater (60%) =0.60 x 171,344
 =102,806 Nm³/hr --------6
 Considering decalcination in preheater is about 85 % and our L.O.I=35%
 Volatized CO2 = (0.35/1.977) x 187500 x 0.85
=28215.1 Nm³/hr ---------------7
 Assuming O2 after ID fan is 5% so
 Total false air in preheater =[(5-2)/ [(21-5) x(102,806+28,215) ]
=24,566 Nm³/hr -------------8
 Total gases after preheater = 5+ 6 + 7+8 =206158.29 Nm³/hr
 Taking 15% Safety factor = 206158.29*1.15=237082.03 Nm³/hr
 So ID fans capacities are 237082.03 Nm³/hr

Kiln Dimensions
 Kiln production/h =(4500 x 1000)/24=187500kg/h
 Burner Fuel = 800 kcal x 0.4=320 Kcal/Kg clinker
 Total thermal load = 187500 x 320= 6 x 107Kcal/h
 Assume burning zone thermal load in case of in line calciner = 4.1 x 106Kcal/h/m²
 Active cross section = (6 x 107 )/ 4.1 x 106 =14.63m²
 Active cross section = 3.14 x (D-0.44)²/4
 So Kiln Diameter (D)=4.75m
 Assume Volumetric load =4.2TPD/m³
 4.2=4500/[(D-0.44)² x (3.14/4) x L]
 So kiln length (L) = 74m
 Speed rotation of Kiln : 4.0 rpm
 Kiln Slope = 4 %
 Filling Degree = ( 3.2 x prod. t/d ) / ( Diam³ x slope % x RPM )
= ( 3.2 x 4,500 ) / ( 4.75³ x 4 x 4.0 ) = 8.2 %
 Retention time = ( 23 x kiln L ) / ( Kiln D x RPM x slope% )
= ( 23 x 74 ) / ( 4.75 x 4 x 4 ) = 22.5 Min
Burner
 we will use Duoflex burner ( multi channel ) due to :
- Good controlling system
- low primary air consumption
- adjustable swirling
- good air nozzle area
- central fuel
injection
- suitable for
alternative fuels
Cooler Dimensions
 Assume specific load for cross section 46 TPD/m²
 Cross section area = TPD/sp. load=4500/46
=98 m²
 Module dimension(F.L sys) =1.3Width*4.2Length =5.46 m²
 No. of modules =98/5.46=17.8 ~ 20 module
  Kiln D.= 4.75 m then cooler width must be larger
 So we need 4 module in width and 5 module in length
 Cooler Width=4*1.3=5.2m
Cooler length = 5*4.2=21m
Cooler air capacity
 Cooler has 7 cooling fans
 Total prod./h =4500*1000/24=187500Kg/h
 From F.Ls system cooling rate =1.8 Nm³/Kg Ck
 Cooling air req.=1.8*187500=337500 Nm³/h
 Cooler crusher : Hammer crusher
 Pan conveyor cap. =4500*1.6/24=300 t/h

Clinker Silo
 2 clinker silo each one 20000 tons
 Each silo have 6 automatic discharge gate sys. For feeding cement mills
 1 off-standard silo 700 ton
 Material handling from silos to cement mills in flow sheet

Cement mill
 Two tube mills closed circuits with max cap. 130 t/h
 Separator:
- high eff. Sep. fls sepax
- one dyn. Sep. with 4 cyclones
 Filling degree 30 %
 Speed 74 % from critical speed
 Length / diam. = 3.5
 Bulk weight balls charge = 4.5 t/m³
 Using bucket elevator for cement trans to silos
 Calc. of Cement millAss. Blaine 3500 cm²/gm from fig 22
 Sp. Power consumption = 43 Kw.h/t
 The power req. for 130 t = 43 * 130 = 5600 Kw
 Ass. Gear box eff. = 95 %
 The mill motor power = 5600/ 0.95 = 5895 Kw From fig. 10
- Di = 4.54 m
- Dn = Di + Lining thickness = 4.54 + 0.26 = 4.8 m
Calc. of Cement mill
 Ass. Blaine 3500 cm²/gm from fig 22
 Sp. Power consumption = 43 Kw.h/t
 The power req. for 130 t = 43 * 130 = 5600 Kw
 Ass. Gear box eff. = 95 %
 The mill motor power = 5600/ 0.95 = 5895 Kw From fig. 10
- Di = 4.54 m
- Dn = Di + Lining thickness = 4.54 + 0.26 = 4.8 m

 Total Mill length = 3.5 * 4.54 =15.9 m

 • Compartment length
- Com. 1 = 0.35 * 15.9 = 5.565 m
- Com. 2 = 0.65 * 15.9 = 10.335 m
• Balls charge
- Com. 1 = 121.618 t
- Com. 2 = 225.852 t

Cement silos
 3 cement silos
 Capacity : 15 000 t x 3
 Having bulk discharge
 Discharge rate 260 ton/hr
 Cement production / day = 5 720 ton

Packing Area
 Use 3 machines
 Type : ventomatic
 Capacity : 120 t/h for each one
 Running hour : 24 hr
 Total packing = 8,640 t / day