Design Problem and Solution

DESIGN PROBLEMA Diesel Power Plant will be constructed to save the community of Brgy. San Felipe, Naga City. The Power Plant will supply power to be available for 24hrs.

Assumption:1. Plant Capacity Factor = 60%2. Fuel Supply in the area = 45 days3. Electrical connection including distribution will be consulted by a Electrical EngineerRequirements:1. Determine the no. of customers in the community.2. Identify the kind and no. residential, commercial and shops.3. Identify the customers by zone.4. Determine the time of usage of customers per 24hrs.5. Determine the total power requirements for the area.6. Draw the map of the area and indicated the customers in its zone.7. Design the Power Plant including its auxiliaries.8. Design the foundation of the engine.9. Specify the required diesel power units.10. Provide for space for expansion.11. Design the Power house building.12. Prepare the working schedule of the project.

DESIGN SOLUTION1. Number of customers in Brgy. San Felipe, Naga City.

Number of population: 14098Number of Households: 3873

2. Identifying the kind and no. of residential, commercial and shops.

Households: 3873Satellite market: 6Business offices and establishments: 13School: 6Park & Recreational center: 4Water Pumping station: 3

3. Identifying the number of customers per zone.ZoneNo. of PopulationNo. of households

1722155

252461212

31200365

4643130

53009618

64933852

72013456

4. Time usage of the customers per 24 hours.

HouseholdsAppliancesNo. of AppliancesWattageSchedule of UseSchedule hr/daytotal Wattagekw-hr

electric fan500806pm-6am12hrs40000480

oven toaster3012006am-9am & 2pm-4pm5hrs36000180

refrigerator95850daily24hrs760001824

fluorescent lamp357406pm-6am12hrs14280171.36

bulb285506pm-6am12hrs14250171

water dispenser12300daily24hrs360086.4

rice cooker753205am-6am & 6pm-7pm2hrs2400048

vacuum cleaner106306am-8am & 2pm-4pm4hrs630025.2

air conditioning unit4533586pm-12am & 10am-5pm12hrs1511101813.32

dvd player901310am-12nn & 7pm-9pm4hrs11704.68

laptop792508am-1pm5hrs1975098.75

computers363509am-2pm5hrs1260063

printers15188am-10pm2hrs2700.54

electric stove2418009am-11am2hrs4320086.4

coffee maker312006am-8am2hrs36007.2

charger3006510am-12nn & 5pm-7pm4hrs1950078

water pumps3022388am-8pm12hrs67140805.68

television sets1981109am-9pm12hrs10780129.36

washing machines78209am-11am2hrs15603.12

flat iron10010005am-7am & 8pm-10pm4hrs100000400

ceiling fan50608am-10am & 2pm-5pm5hrs300015

karaoke1801809pm-2am59004.5

stereo50401pm-6pm5hrs200010

TOTAL6505.51

Sari- sari storesAppliancesNo. of AppliancesWattageSchedule of UseSchedule hr/daytotal Wattagekw-hr

electric fan84808am-8pm12hrs672080.64

fluorescent lamp32406pm-12mn6hrs12807.68

bulb75186pm-6am12hrs135016.2

refrigerator36850daily24hrs30600734.4


radio/karaoke121808am-10am & 1pm-4pm5hrs216010.8

Air conditioning unit333582pm-7pm5hrs1007450.37

dvd player31011am-1pm & 2pm-5pm5hrs300.15

ceiling fan4060daily24hrs24005706

TOTAL6623.84

ResortsAppliancesNo. of AppliancesWattageSchedule of UseSchedule hr/daytotal WattagekW-hr

electric fan10808am-11am & 2pm-5pm6hrs8004.8


bulb1018daily24hrs1804.32


karaoke21808am-10am & 1pm-4pm5hrs3601.8

refrigerators2850daily24hrs170040.8

blender475010am-12nn &3pm-5pm4hrs300012

water dispenser3300daily24hrs90021.6



ceiling fan3060daily24hrs180043.2

freezer41200daily24hrs4800115.2

TOTAL764.25

HotelAppliancesNo. of AppliancesWattageSchedule of UseSchedule hr/daytotal Wattagekw-hr


ceiling fan10602pm-5pm & 8pm-11pm6hrs6003.6

computer53508am-12nn & 2pm-6pm8hrs175014

printer11810pm-11pm & 3pm-4pm3hrs180.054

coffee maker312006am-8am & 3pm-4pm3hrs360010.8



bulb20186pm-6am12hrs3604.32

rice cooker232010am-11am & 5pm-7pm3hrs6401.92


dvd player1138am-10am & 2pm-4pm4hrs130.052

water pump22238daily24hrs4476107.42

TOTAL342.016

Machine ShopsAppliancesNo. of AppliancesWattageSchedule of UseSchedule hr/daytotal Wattagekw-hr

milling machines125009am-12nn & 1pm-3pm5hrs250012.5

welding machines21300010am-12nn & 1pm-4pm5hrs26000130

compressor130009am-12nn3hrs30009

pumps1223810am-12nn & 1pm-5pm6hrs223813.428


bulbs5186pm-6am12hrs901.08

ceiling fan2609am-12nn & 1pm-4pm6hrs1200.72

exhaust fan1609am-12nn & 1pm-4pm6hrs600.36

drill press17509am-11am & 2pm-3pm3hrs7502.25

electric fan3809am-12nn & 1pm-4pm12hrs2402.88

oven toaster112009am-12nn3hrs12003.6

lathe machine127009am-12nn & 1pm-10pm12hrs270032.4

TOTAL208.698

Vulcanizing ShopsAppliancesNo. of AppliancesWattageSchedule of UseSchedule hr/daytotal Wattagekw-hr

compressor130009am-11am & 1pm-3pm4hrs300012


bulbs2186pm-6am12hrs360.432


drill press17509am-11am & 1pm-3pm4hrs3801.52


radio11809am-11am & 1pm-3pm4hrs1100.44


welding machines1130009am-12nn & 1pm-3pm5hrs1300065

ceiling fan1609am-12nn & 1pm-4pm6hrs600.36

TOTAL98.302

HardwareAppliancesNo. of AppliancesWattageSchedule of UseSchedule hr/daytotal Wattagekw-hr



radio/karaoke11808am-1pm5hrs1800.9

computer23508am-5pm8hrs7606.08

bulb3508am-8pm12hrs1501.8

ceiling fan15608am-5pm8hrs9007.2

exhaust fan1580daily24hrs12002808

TOTAL55.34

BodegasAppliancesNo. of AppliancesWattageSchedule of UseSchedule hr/daytotal Wattagekw-hr


radio/karaoke31808am-1pm5hrs5402.7

air conditioning unit633582pm-7pm5hrs20148100.74


Computer63509am-2pm5hrs210010.5

exhaust fan15609am-3pm6hrs9005.4

Printer21810am-11am & 2pm-4pm3hrs360.108

Bulb2050daily24hrs100024


Christmas lights4256pm-6am12hrs1001.2

TOTAL160.148

BakeriesAppliancesNo. of AppliancesWattageSchedule of UseSchedule hr/daytotal Wattagekw-hr


oven toaster212008am-2pm6hrs240014.4

Refrigerator1380daily24hrs3809.12

fluorescent lamp4408am-5pm8hrs1601.28


TOTAL26.63

Gasoline StationsAppliancesNo. of AppliancesWattageSchedule of UseSchedule hr/daytotal Wattagekw-hr


radio/karaoke21008am-11am & 1pm-4pm6hrs3602.16

air conditioning unit433588am-8pm12hrs13432161.184

television sets21108am-11am & 1pm-4pm6hrs2701.32

Computer43508am-5pm8hrs140011.2

Compressor2300009am-11am & 1pm-3pm4hrs60004

Bulb7506pm-6am12hrs3504.2


christmas lights4256pm-6am12hrs1001.2

TOTAL202.064

Barber ShopsAppliancesNo. of AppliancesWattageSchedule of UseSchedule hr/daytotal Wattagekw-hr


Karaoke11808am-10am & 1pm-4pm5hrs1806.9


Charger36510am-12nn & 1pm-4pm5hrs1950.975

Razor31108am-6pm10hrs3303.3

TOTAL19.215

5. Determining the total power requirement for the area.TIMEKw-hr CONSUMPTIONLOAD VALUE

19544.212397.6755

29544.212397.6755

39539.712397.488

49539.712397.488

59997.312416.5546667

610218.112425.7546667

79296.08387.3366667

810194.026424.7510833

910619.416442.4756667

1012484.247520.1769583

1118106.285754.4285417

1218017.097750.712375

1318450.117768.754875

1413034.507543.1044583

1512984.341541.0142083

1611067.911461.1629583

1712577.582524.0659167

1813563.946565.1644167

1913208.726550.3635833

2013162.296548.429

2112103.692504.3205

2211969.706498.73775

2311536.766480.6985833

2411365.212473.5505

Average Load507.161849

With the table above, we can now tabulate the data to obtain the load curve which would obtain the specific Diesel Engine type, the number of units and the operation time the engine would run to provide power.

With the load curve above, we can now propose the type of Diesel Engine by choosing the appropriate unit comparative to the operational peak load in providing power to the community

6. Drawing the map of the area and indicating the customers in its zone(Refer to the drawing at the last page)

7. Designing the Power Plant including each auxiliaryCooling Water SystemRequired Circulating Cooling Water:

Based from Power Plant Engineering (Adapted to MKS Units) by Frederick T. Morse; Cooling System; pages 177 to 178; we are to use a cooling (preferably a forced draft cooling tower) as it is minimal upon consideration cost, bulk, and auxiliary power. This would provide an immediate cooling for the Diesel Engines frame jackets in the heated parts. To obtain the circulating cooling water requirements, we are to use the equation provided by several references. Based from Power Plant Engineering (Adapted to MKS Units) by Frederick T. Morse; Cooling System; page 178; using the equation below we are to use the given equation to obtain the required circulating cooling water.W = 674.58 Where: W circulating cooling water; liter per hourRated Bhp Rated Brake horsepowert1 inlet temperature; Ct2 outlet temperature; CAccordingly, we must first obtain the rated brake horsepower as readily provided in the Diesel Engine type F8-138 Specification. Deviating from the law of conservation of energy (simply stated as Energy In = Energy Out) then the rated horsepower of the engine would be equal to the generator output. Losses would be present so the net efficiency would be used to offset the generator output. Based from Power Plant Engineering (Adapted to MKS Units) by Frederick T. Morse; Table A 16: Normal Efficiencies of Engine type Synchronous Generators; page 675; at a kilowatt rating of 560 kW with 450 rpm, by interpolation we have 94.46%. Deductions upon this efficiency are also given. Based from Power Plant Engineering (Adapted to MKS Units) by Frederick T. Morse; Table 6 3: Standard Deduction (Engine Generator Efficiency); page 185; at a generator efficiency of 94.1% to 95% with a full load operation we deduct 1.2 so we have a 93.26% Net Generator Efficiency. Therefore the rated break horsepower is equal to the equation below.Rated Bhp = = = 600.47 kW ( )Rated Bhp = 805.24 hpAssuming that the temperature range would be 10 C, as most of the cooling tower range applied in the industry from the manufacturer ranges from 5.6 C to 16.7 C, then we use an assumed values of 75 C & 65 C for the inlet water temperature and outlet water temperature respectively, we can now obtain the required circulating cooling water for the cooling tower. With the 2 units we have 1610.48 hp. So the required circulating cooling water is...Wc = 674.58 = 674.58 Wc = 108639.76 Liters/hrBased from Marks Standard Handbook for Mechanical Engineers, 9th Edition; Section 6: Materials of Engineering; 6.1 General Properties of Materials; Specific Gravity and Density of Water at Atmospheric Pressure; page 6 10; at 75 C the density of water is 974.86 kg/m3 = 108639.76 L/hr (1hr/60mins) (1m3/1000L) (974.86 kg/m3)Wc = 1765.14 kg/minTherefore, we need 108639.76 Liters/hour or 1765.14 kg/min for the circulating cooling water for the 2 Diesel Engine Genset units.

2. Water Jacket Circulating Pump:

Based from Marks Standard Handbook for Mechanical Engineers, 9th Edition; Section 8: Machine Elements; 8.7 Pipe, Pipe Fittings and Valves; Table 8.7.3 Properties of Commercial Steel Pipes; page 8 149; assuming we are to use a nominal pipe size of 3 in outside diameter schedule 40 for the suction line pipe and a nominal pipe size of 2 in outside diameter schedule 40 for the discharge line pipe, the following data is given as followsSuction line pipe:Schedule 40Outside diameter3 in (3.5 in)Inside diameter3.068in (0.0779 m)

Discharge line pipe:Schedule 40Outside diameter2.5 in (2.875 in)Inside diameter2.469 in (0.0627 m)Computing for the specific velocity rate at both the suction and discharge we simply use the given equation belowQ = AVWhere: Q volume flow rate; m3/minA Cross-sectional area; m2V Velocity of fluid; m/minAlso given for the area and deviating from the equation we haveA = d2/4Where: A Cross-sectional area; m2d Internal diameter; mThus,V = Computing now for the velocity at the suction we haveVs = = Vs = 379.97 m/min (1min/60secs)Vs = 6.33 m/secComputing now for the velocity at the discharge we haveVd = = Vd = 586.53 m/min (1min/60secs)Vd = 9.78 m/secAssuming that the pump is in the datum line, the height of delivery is 2.302 meters, the storage is placed 4.5 meters below, a friction loss of 0.75; then with the given & computing now for the discharge headDischarge head = FL+ZD+VDischarge head = FL+ZD+Where: FL Friction losses for discharge; 0.75ZD Elevation from datum to discharge; 2.302 mVD Velocity head at discharge; 9.78 m/secDischarge head = FL+ZD+ Discharge head = 0.75 + 2.302 m + Discharge head = 3.052 m

Computing now for the suction headSuction head = FL + ZS + VSuction head = FL + ZS + Where: FL Friction losses for suction; 0.75ZS Elevation from datum to suction; 4.5m (negative due to location)VS Velocity head at suction; 6.33 m/secSuction head = FL + ZS + Suction head = 0.75 4.5m + Suction head = -1.71 mBased from Power Plant Engineering (Adapted to MKS Units) by Frederick T. Morse; Chapter 13: The Gas Loop; 13 10 Water Pumps; pages 545 to 546; we use the equations provided to obtain the required pump. As both the suction and discharge heads are given, we can now obtain the pump operating head. With the given, the pump operating head isPump operating head = Discharge head Suction headPump operating head = 3.052 m + 1.71 mPump operating head = 4.762 mAssuming that the pump efficiency is 70 %, and then we havePump supply power = Where: Q Volume flow rate; 1811 liters/mindw Density of water; 1000 kg/m3H Pump operating head; 4.762 mp Pump efficiency; 70%Pump supply power = Pump supply power = Pump supply power = 2.74 hp (2.75 hp)Therefore, we are to use 2 hp water pumps for the water jacket transfer for each Diesel Genset units.

3. Required Raw Water for the Cooling Tower:

Since most heat exchangers experience a steady state equation and deviating from the energy balance equation [mC (hC1 hC2) = mR (hR1 hR2)] assuming that the continuous flow would nullify the offsetting effects of density and enthalpy, and assuming that temperature difference with 6 C of 34 C at inlet and 28 C at outlet, then we can use the equation below for simplicityQC (tC1 tC2) = QR (tR1 tR2)Where: QR - Quantity of Raw Water circulating the cooling tower, Liter/minQC - Quantity of Circulating Cooling Water; Liter/mintR1 - temperature of raw water at outlet; 34 CtR2 - temperature of raw water at inlet; 28 CtC1 - temperature of jacket cooling water at inlet; 75 CtC2 - temperature of jacket cooling water at outlet; 65 CUsing the equation, the quantity of raw water isQR = = = 3018.33 L/minBased from Marks Standard Handbook for Mechanical Engineers, 9th Edition; Section 6: Materials of Engineering; 6.1 General Properties of Materials; Specific Gravity and Density of Water at Atmospheric Pressure; page 6 10; at 28 C the density of water is 996.242 kg/m3 = 3018.33 L/min (1m3/1000L) (996.242 kg/m3)WR = 3007 kg/min4. Raw Water Pump:Assuming that the parameters for the circulating water jacket pumps are the same with the raw water pump, together with the assumptions, and then the requirements would be close and be useful for the raw water pump computation Computing now for the velocity at the suction we haveVS =VS = VS = 633.29 m/min (1min/60 secs)VS = 10.55 m/secComputing now for the velocity at the discharge we haveVD = VD = VD = 977.56 m/min (1 min/60 secs)VD = 16.29 m/secAssuming that the pump is in the datum line, the height of delivery is 5 meters, the storage is placed 4.5 meters below, a friction loss of 0.75; then with the given & computing now for the discharge headDischarge Head = FL + ZD + VDischarge Head = FL + ZD +Where: FL = Friction Losses for Discharge; 0.75ZD = Elevation from Datum to Discharge; 5 mVD = Velocity Head at Discharge; 16.29 m/secDischarge Head = FL + ZD +Discharge Head = 0.75 + 5m + Discharge Head = 19.28 m

Computing now for the suction headSuction Head = FL + ZS + VSSuction Head = FL +ZS + Where: FL = Friction Losses at Suction; 0.75 ZS = Elevation from Datum to Suction; 4.5 m (- due to location) VS = Velocity Head at Suction; 10.55 m/secSuction Head = FL +ZS + Suction Head = 0.75 4.5 m + Suction Head = 1.92 mBased from Power Plant Engineering (Adapted to MKS Units) by Frederick T. Morse; Chapter 13: The Gas Loop; 13 10 Water Pumps; pages 545 to 546; we use the equations provided to obtain the required pump. As both the suction and discharge heads are given, we can now obtain the pump operating head. With the given, the pump operating head is Pump Operating Head = Discharge Head Suction head = 19.28 m - 1.92 m Pump Operating Head = 17.36 mAssuming that the pump efficiency is 70 %, and then we havePump supply power = Where: Q Volume flow rate; 3018.33 L/mindw Density of water; 1000 kg/m3H Pump operating head; 17.36 mp Pump efficiency; 70%Pump supply power = Pump supply power =Pump supply power = 16.63 hp (16.75 hp)Therefore, we are to use 16 hp water pump for the raw water transfer.

5. Required Quantity of Make-up Water:

Based from Power Plant Engineering (Adapted to MKS Units) by Frederick T. Morse; Chapter 6: Internal Combustion Engine Power Plant; 6-7 Evaporative Cooling; page 181; using the given mass balance and heat balance equations we can obtain the required make up water for the cooling tower. For mass balance we use1 kg air + SH1 + WW + W = 1 kg air + SH2 + WWDeviating for simplicity we have W = SH2 SH1For heat balance we use h1 + WWhfa + Whf = h2 + WWhfb Deviating for simplicity we haveWW = Where: SH1 = Humidity Ratio of Entering Air; kg moisture/kg dry air SH2 = Humidity Ratio of Leaving Air; kg moisture/kg dry air WW = Water Circulating per kg of dry air; kg W = Make-up Water per kg of dry air; kg h1 = Enthalpy of Moist Air Leaving; kJ/kg of dry air h2 = Enthalpy of Moist Air Entering; kJ/kg of dry air hfa = Enthalpy of Water in the Spraying Nozzles; kJ/kg hfb = Enthalpy of Water in the Basin; kJ/kg hf = Enthalpy of Make-up Water; kJ/kgAssuming we are given the following relative humidity and temperatures Relative Humidity: Entering Air = 60% Leaving Air = 90%Temperature: Entering = 28 C DB Leaving = 34 C DB WBT = 21 C Based from Marks Standard Handbook for Mechanical Engineers; 9th Edition; Figure 12.4.13: Psychometric Chart in SI Units; page 12 97; the following are given and obtained so we have.. At a R.H. 60% and dry bulb temperature of 28 C at entering point SH1 = 0.01425 kg moisture/kg dry air At a R.H. 90% and dry bulb temperature of 34 C at leaving point SH2 = 0.031 kg moisture/kg dry airBased from Refrigeration & Air Conditioning by W. F. Stoecker & J. W. Jones; Table A-2: Moist Air; pages 418 to 419, with the assumed temperatures the enthalpy are as follows At 28 C, h1 = 89.952 KJ/kg At 34 C, h2 = 122.968 KJ/Kg Using the equation of mass balance we use W = SH2 SH1 = (0.031 0.01425) kg moisture/kg dry air W = 0.01675 kg moisture/kg dry air Based from Refrigeration & Air Conditioning by W. F. Stoecker & J. W. Jones; Table A-1: Water: Properties of Liquid and Saturated Vapor; pages 416 to 417, with the assumed temperatures the enthalpy are as follows At 34 C, hfa = 142.38 KJ/Kg At 28 C, hfb = 117.31 KJ/kgAssuming we have a make-up water temperature of 18 C; based from Refrigeration & Air Conditioning by W. F. Stoecker & J. W. Jones; Table A-1: Water : Properties of Liquid and Saturated Vapor; pages 416 to 417; with the assumed temperature the enthalpy and pressure is hf = 75.50 KJ/kg, p = 2.062 kPa Using the equation of energy balance we haveWW = = WW = 1.27 kJ/kg dry airSince 3007 kg/min of raw water is given, the air flow isAir flow = Air flow = 2367.72 kg/minBased from Marks Standard Handbook for Mechanical Engineers; 9th Edition; Figure 12.4.13: Psychometric Chart in SI Units; page 12 97; the following are given and obtained so we haveDB temperature = 28 C WB temperature = 21 C The specific volume, v, is 0.87 m3/kg dry air Then the air flow would be Air Flow = 2367.72 kg/min (0.87 m3/kg) Air Flow = 2059.92 m3/minThe required make-up water then is Make-up Water = 2367.72 kg/min x 0.01675 Make-up Water = Make-up Water = 39.67 L/min6. Forced Draft Fan:

Based from https://my.amca.org/members/documents/catalogs/64_91_Catalog-157-B%20NOV99.pdf, we are to choose a fan that would provide a capacity of air flow of 2059.92m3/min or 72745.39ft3/min.

We select a fan with the following size

Fan Size: 60 TA Direct Drive TubeaxialMotor hp: 20 hp Rpm: 870 Capacity: 73948ft3/min Static Pressure: 1/4 in SP7. Size of Cooling Tower:

For the total cooling requirements we must first obtain the total cooling water requirements Water Requirements = Circulating Jacket Water + Raw Water = 1811 Liters/min + 3018.33 Liters/min = 4829.33 Liters/min x (1 gallon/3.7854 Liters) Water Requirements = 1275.78 gpmAssuming we have water concentration of 3.0 gpm per ft2, then the area of the cooling tower is Area of Cooling Tower = 1275.78 gpm / (3.0 gpm/ft2) = 425.26 ft2Assuming we have a square sized cooling tower, then the size of the cooling tower would beSide of the Cooling Tower = Side of the Cooling Tower = 20.62 ftSo the dimensions of the cooling tower that would service the diesel engines are 20.62 ft x 20.62 ft or 6.28 m x 6.28 m.B. Fuel System

1. Fuel Oil Consumption:

Our design requires 3 units of F8-138 Diesel Engines with 560 kW generator rating; 1 unit is continuous in operation; another unit for 10 am to 3 pm shift; another unit as a reserve; all of which are capable of continuous operation at least for a limited duration of time assuming that maintenance to either one of them is required. A future expansion is also provided. Our plant capacity factor is given at 60%.

Based from Power Plant Engineering (Adapted to MKS Units) by Frederick T. Morse; Table A 16: Normal Efficiencies of Engine type Synchronous Generators; page 675; at a kilowatt rating of 560 kW with 450 rpm, by interpolation we have 94.46 %. Deductions upon this efficiency are also given. Based from Power Plant Engineering (Adapted to MKS Units) by Frederick T. Morse; Table 6 3: Standard Deduction (Engine Generator Efficiency); page 185; at a generator efficiency of 94.1 % to 95 % with a full load operation we deduct 1.2 so we have a 93.26 % Net Generator Efficiency.

From the statement above we can now obtain the fuel consumption for the plant. Based from Power Plant Engineering (Adapted to MKS Units) by Frederick T. Morse; Figure 6 15: Range of Variable Load Performance of Diesel Plants; page 164; at 60% rated load the fuel consumption is about 0.1625 kg/kW-hr. With such the maximum rate of fuel usage would be as followsMaximum Rate of Fuel = = 1200.94 kW (0.1625 kg/kW-hr)Maximum Rate of Fuel = 195.15 kg/hr

For a 24 hours operation of the 2 units, then maximum rate of fuel usage would be given as follows

Maximum Rate of Fuel Use = 195.15 kg/hr x 24 hr/day Maximum Rate of Fuel Use = 4683.6 kg/dayBased from Marks Standard Handbook for Mechanical Engineers, 9th Edition, Section 7: Fuels & Furnaces; Table 7.1.9 Analyses and High Heat Values of Crude Petroleum, Typical Distillates, and Fuel Oils; page 7 13; assuming we are using a California grade fuel oil, with a specific gravity of 0.9554 at 60 F, then the fuel consumption in terms of mass would be.. Fuel Consumption = (4683.6 kg/day) / 0.9554 Fuel Consumption = 4902.24 kg/day Based from Marks Standard Handbook for Mechanical Engineers, 9th Edition; Section 6: Materials of Engineering; 6.1 General Properties of Materials; Approximate Specific gravities & Densities; page 6 8; as most industrial oils have an average density of 914 kg/m3 and obtaining for its specific volume then we have 1.094 x 10-3 m3/kg then the fuel consumption in terms of volume would be.. Fuel Consumption = (4902.24 kg/day) (1.094 x 10-3 m3/kg) = (5.363m3/day) (1000 Liters/m3) Fuel Consumption = 5363 Liters/day2. Required Storage of Fuel Oil:

Based from Power Plant Engineering (Adapted to MKS Units) by Frederick T. Morse; Figure 6 15: Range of Variable Load Performance of Diesel Plants; page 164; at 60% plant capacity the kilowatt-hour per liter oil ranges from 2.5 to 3.49 liters. One (1) Diesel Genset unit will operate continuously while another for 5 hours each per day at peak loads. For fuel consumption for within a 45 day supply we have

Required Storage = 45 days (5363 Liters/day)Required Storage = 241335 Liters

3. Dimensions of the Fuel Oil Storage Tank:

Based from Power Plant Engineering (Adapted to MKS Units) by Frederick T. Morse; Table 12 4: Dimension of the Bulk Storage Tank; page 459; with a required storage of 241335 liters then we are to use 3 tanks with a capacity of 109715 liters each, the dimensions of a cylindrical bulk tank is given as follows

Diameter: 3.05 m Length: 15.04 m Plate Thickness: 7.94 mm Weight: 10399 kg

4. Required Storage of Day Tank:

As the operation in motion, the greatest consumption of each diesel engine will occur at its full load rating. Let the day tank volume be good for a 1 day (24 hours) operation. At full load rating, maximum full load consumption would be

195.15 kg/hr x 24 hrs = 4683.6 kgFor a 1 day operation, assume that the day tank will charge 4683.6 kg per day. Since the fuel oil is cooled during the transfer & operation we must obtain the value to compute the volume. Assuming we are using a California grade fuel oil, based from Marks Standard Handbook for Mechanical Engineers, 9th Edition, Section 7: Fuels & Furnaces; Table 7.1.9 Analyses and High Heat Values of Crude Petroleum, Typical Distillates, and Fuel Oils; page 7 13; at 60 F (15.6 C) the specific gravity of it would be 0.9554. Based from Power Plant Engineering (Adapted to MKS Units) by Frederick T. Morse; Chapter 5: Fuels & Combustion; Internal Combustion Engine Fuel; pages117 to 119; Using API (American Petroleum Institute) standard, assuming an oil temperature of 6 C and the equations 5 3, the API would beAPI = - 131.5 = 131.5API = 16.61 API or 17 APIThe density of oil at 15.6C (60 F) would be equal to the specific gravity at such temperature. Based from Power Plant Engineering (Adapted to MKS Units) by Frederick T. Morse; Chapter 5: Fuels & Combustion; Internal Combustion Engine Fuel; pages117 to 119; the volumetric coefficient of expansion of oil is 0.0007 per C. The contractions of oil at 6 C we have Contraction from a 6 C cooling = 0.0007 x 6 = 0.0042The density of fuel oil at 6 C would be Density at 6 C = 0.9554 / 0.9958 = 0.9981 kg/literThe volume would now be equal to V = 4683.6 kg / 0.9981 kg/liter = 4692.52 liters (1m3/1000 Liters) V = 4.69 m3Assuming a 15 minute charging for the day tank, the volume flow rate would be Q = 4.69 m3 / 15 min = 0.3127 m3/min (1000 Liters/m3) Q = 312.7 Liters/min

5. Dimensions of the Fuel Oil Day Tank:

Given a volume of 4.69 m3 and assuming we have a cylindrical day tank, then using the following equation below we can obtain the dimension required for the day tank

V =

Where: V = Volume of the cylinder; m3 d = Diameter of the cylinder; m h = Height or Length of the cylinder; mAssuming we have a 2.5 m length of the day tank, deviating from the dimension and computing for the diameter we haved2 =d2 = d2 = d2 = 2.39 m2d = 1.55 mTherefore, the dimension of the day tank is 1.55 m diameter by 2.5 m length cylindrical tank per Diesel Engine.

6. Fuel Oil Transfer Pump:

Based from Marks Standard Handbook for Mechanical Engineers, 9th Edition; Section 8: Machine Elements; 8.7 Pipe, Pipe Fittings and Valves; Table 8.7.3 Properties of Commercial Steel Pipes; page 8 148; assuming we are to use a nominal pipe size of 1 in outside diameter schedule 40 for the suction line pipe and a nominal pipe size of 1 in outside diameter schedule 40 for the discharge line pipe, the following data is given as follows

Suction Line Pipe: Schedule 40 Outside Diameter 1 in (true size of 1.9 in) Inside Diameter 1.610 in (or 0.0409 m) Discharge Line Pipe: Schedule 40 Outside Diameter 1 in (true size of 1.660 in) Inside Diameter 1.380 in (or 0.0351 m)Computing for the specific velocity rate at both the suction and discharge we simply use the given equation below Q = AV Where: Q = Volume Flow Rate; m3/min A = Cross-Sectional Area; m2 V = Velocity of Fluid; m/minAlso given for the area and deviating from the equation we have A = d2/4 Where: A = Cross-sectional Area; m2 d = Internal Diameter; mV = Computing now for the velocity at the suction we haveVS = = VS = 238.24 m/min (1min/60secs)VS = 3.97 m/secComputing now for the velocity at the discharge we haveVD = = VD = 323.47 m/min (1min/60secs)VD = 5.39 m/secAssuming that the pump is in the datum line, the height of delivery is 4.5 meters, the storage is placed 2.5 meters below, a friction loss of 0.75; then with the given & computing now for the discharge headDischarge Head = FL + ZD + VDischarge Head = FL + ZD + Where: FL = Friction Losses for Discharge; 0.75 ZD = Elevation from Datum to Discharge; 4.5 m VD = Velocity Head at Discharge; 5.39 m/secDischarge Head = FL + ZD + Discharge Head = 0.75 + 4.5 + Discharge Head = 6.73 m

Computing now for the suction headSuction Head = FL + ZS + VSuction Head = FL + ZS + Where: FL = Friction Losses at Suction; 0.75 ZS = Elevation from Datum to Suction; 2.5 m (- due to location)VS = Velocity Head at Suction; 3.97 m/secSuction Head = FL + ZS + Suction Head = 0.75 2.5 + Suction Head = -0.95 mBased from Power Plant Engineering (Adapted to MKS Units) by Frederick T. Morse; Chapter 13: The Gas Loop; 13 10 Water Pumps; pages 545 to 546; we use the equations provided to obtain the required pump. As both the suction and discharge heads are given, we can now obtain the pump operating head. With the given, the pump operating head is Pump Operating Head = Discharge Head Suction head = 6.73 m + 0.95 m Pump Operating Head = 7.68 mAssuming that the pump efficiency is 70 %, and then we havePump supply power = Where: Q = Volume Flow Rate; 312.7 Liters/min dO = Density of oil; 914 kg/m3 (or 0.914 kg/liters) H = Pump Operating Head; 7.68 m P = Pump Efficiency; 70 %Pump supply power = = Pump supply power = 0.70 hpTherefore, we are to use 3/4 hp oil pump for the fuel oil transfer.C. Air System

1. Air Intake System:

The air intake system usually consists of air intake duct or pipe appropriately supported, a silencer, an air cleaner, and flexible connections as required. This arrangement permits location of area of air intake beyond the immediate vicinity of the engine, provides for the reduction of noise from intake air flow, and protects vital engine parts against airborne impurities. The air intake will be designed to be short and direct and economically sized for minimum friction loss. The air filter will be designed for the expected dust loading, simple maintenance, and low pressure drop. Oil bath or dry filter element air cleaners will be provided. The air filter and silencer may be combined.

Based from Based from Power Plant Engineering (Adapted to MKS Units) by Frederick T. Morse; Chapter 6: Internal Combustion Engine Power Plant; 6 6: Auxiliary Systems; Intake & Exhaust Passages; pages 174; an engine needs from 0.056 to 0.084 m3 of air per min per hp developed.

Assuming we have maximum intake of 0.084 m3/min of air per hp developed during operation, with 560 kW (or 750.97 hp) then the flow rate of the discharge would be

Q = (0.084 m3/min-hp) / (750.97 hp) Q = 63.08m3/minAssuming we have a flow velocity of 800 m/min, then the dimensions of the intake pipe would be A = (63.08 m3/min) / (800 m/min) A = 0.07885 m2Thus, the area of the pipe isA = d2/4Where: A Aread Diameterd = (4A/)1/2d = (4*0.07885 m2/)1/2d = 0.32 mThe diameter of the intake pipe would be 0.32 meter or 320 mm. Therefore, we are to use a pipe with such diameter.2. Exhaust System:

The exhaust system consists of a muffler and connecting piping to the atmosphere with suitable expansion joints, insulation, and supports. In cogeneration plants, it also provides for utilization of exhaust heat energy by incorporating a waste heat boiler which can be used for space heating, absorption refrigeration, or other useful purpose. This boiler produces steam in parallel with the vapor phase cooling system. The exhaust silencer attenuates exhaust gas pulsations (noise), arrests sparks, and in some cases recovers waste heat. The muffler design will provide the required sound attenuation with minimum pressure loss.

Based from Based from Power Plant Engineering (Adapted to MKS Units) by Frederick T. Morse; Chapter 6: Internal Combustion Engine Power Plant; 6 6: Auxiliary Systems; Intake & Exhaust Passages; pages 175 to 76; the exhaust system must carry approximately 0.168 0.224 m3/min of gases per hp developed.

Assuming we have a maximum discharge of 0.224 m3/min of gases per hp developed on an average exhaust temperature, with 560 kW (or 750.97 hp) then the flow rate of the discharge would be

Q = (0.224 m3/min) (750.97 hp) Q = 168.22 m3/minAssuming we have a flow velocity of 1500 m/min, then the dimensions of the exhaust pipe would be A = (168.22 m3/min) / (1500 m/min) A = 0.112 m2Thus the area of the pipe isA = d2/4Where: A Aread Diameterd = (4A/)1/2d = (4*0.112 m2/)1/2d = 0.38 mThe diameter of the exhaust pipe would be 0.38 meter or 380 mm. Therefore, we are to use a pipe with such diameter.3. Air Starting System:

The vast majority of diesel engines installed in power plants are started with compressed air. Compressed air is directed by a distributor directly into the combustion chamber or is provided to an air motor which rotates the engine. Dedicated compressors typically provide starting air at 250 psig. The system must provide adequate storage to allow multiple attempts to start the engines. The compressed air start system will included two air compressor units, each with diesel engine-electric motor drive, and two main air storage tanks. The compressors will be rated at 250 psig operating pressure, and each will have a capacity capable of restoring any single storage receiver from 150 psig to 250 psig in 30 minutes or less. Each main storage tank will provide adequate air to the individual air start tanks at each diesel engine, supply air to the utility shop air outlets, and provide a second source to air to the instrument air system. Each air start tank will be sized to provide two 30 - second start sequences without recharging and will be rated at 300 psig working pressure. Each main storage tank will have a volume equal to three air start tanks plus a volume equal to one instrument air receiver, and an additional volume to supply the utility shop air requirement.

Therefore, we need 4 units of Air Compressors for the 2 Diesel Engines. Both Air Compressors must have a 250 psig operating pressure. This working pressure is uniform regardless of the diesel engine size.

4. Required Capacity Air Storage Tank:

Assuming we are using a two stage air compressor with a 250 psig working pressure and a compressor power of 200 hp. Based from http://www.engineeringtoolbox.com/compressed-air-receivers-d_846.html with the topic Compressed Air Receivers and using the provided tables

Recommended Receiver Volume per HPCompressor Power Recommended Receiver Volume

(hp) kW Ft3Galm3

12593.3675001.9

200149.21078003.0

350261.118814005.3

For the storage tank, assuming the storage volume is twice that of the air receiver and the utility shop requirements consumes half of the receiver, then the air storage volume would be

Total Volume = 3 m3 + (3 m3 *2) + (3 m3 *0.5) Total Volume = 10.5 m3 = 370.8 ft3

5. Dimensions of the Air Storage Tank:

Based from Power Plant Engineering (Adapted to MKS Units) by Frederick T. Morse; Table 12 4: Dimension of the Bulk Storage Tank; page 459; with a required storage of 10.5 m3 or 10500 liters then we are to use a tank of the next closest value of 11860 liters, the dimensions of a cylindrical bulk tank is given as follows Diameter 2.44 m Length 2.54 m Plate Thickness 6.35 mm Weight 1844 kg

D. LUBRICATION SYSTEM

1. Daily Lubricating Oil Consumption:

Lubrication is essential to any machine element; this includes the diesel engines as well as its auxiliaries. Our design requires 3 units of F8-138 Diesel Engines with 560 kW generator rating; 1 unit is continuous in operation; another unit for 10 am to 3 pm shift; another unit as a reserve; all of which are capable of continuous operation at least for a limited duration of time assuming that maintenance to either one of them is required. A future expansion is also provided. Our plant capacity factor is given at 60%.

From the discussion above we have a 93.26 % Net Generator Efficiency for each engine during the operation. The brake horsepower would then be the same as the obtained. Therefore, the brake horsepower is equal to

Rated Bhp = = = 600.47 kWFor a continuous operation, the generated power would be Generated Power = 600.47 kW * 24 hours Generated Power = 14411.28 kW-hourBased from Power Plant Engineering (Adapted to MKS Units) by Frederick T. Morse; Chapter 6: Internal Combustion Engine Power Plant; 6 6: Auxiliary Systems; Lubrication; page 174; most diesel power plants have an average consumption of 1 gallon of Lubricating oil per 1600 kW-Hour generated at full load rating. For both the diesel engines that operates continuously, then the total generated power would be.. Generated Power = 14411.28 kW-hour + (600.47kW *6 hours) Generated Power = 18014.1 kW-hourSince 1 gallon of Lubricating oil per 1600 kW-Hour generated at full load rating is give as an average consumption for lubrication, for one day consumption, we have Oil Consumption = (18014.1 Kw-Hr) x (1 gallons / 1600 kW-Hr) = 11.26 gallons/daySimilarly with the fuel oil, assuming that the delivery of lube oil is every 45 days, the oil consumptionOil Consumption = 11.26 gallons/day * 45 days = 506.7 gallons

2. Dimensions of the Lube Oil Storage Tank:

By conversion Volume = 506.7 gallons*(3.7854 Liters/gallons)*(1 x 10-3 m3/1 Liters) Volume = 1.92 m3Assuming that the tank is filled within 5 minutes as compared for the fuel oil, then Volume Flow Rate = 1.92 m3 / 5 min = 0.384 m3/min Volume Flow Rate = 384 Liters/min Using a cylindrical drum for storage with a 1 meter diameter, the length could be obtained and so is the dimension Volume = ( x d2 / 4) x L Length = (1.92 m3 x 4) / ( x 1 m2) Length = 2.44 m The dimensions of the lube oil tank would be in 1 m diameter by 2.44 m length.3. Lube Oil Transfer Pump:

Assuming that the fuel oil and the lube oil storage tanks and discharge settings are the same, together with the assumptions, then the requirements would be close and be useful for the lube oil transfer pump computation

Computing now for the velocity at the suction we haveVS = = VS = 292.28 m/min (1min/60secs)VS = 4.87 m/secComputing now for the velocity at the discharge we haveVD = = VD = 396.85 m/min (1min/60secs)VD = 6.61 m/secAssuming that the pump is in the datum line, the height of delivery is 4.5 meters, the storage is placed 2.5 meters below, a friction loss of 0.75; then with the given & computing now for the discharge headDischarge Head = 0.75 + 4.5 + Discharge Head = 7.48 mComputing now for the suction headSuction Head = 0.75 2.5 + Suction Head = -0.54 mAs both the suction and discharge heads are given, we can now obtain the pump operating head. With the given, the pump operating head is Pump Operating Head = Discharge Head Suction head = 7.48 m + 0.54 m Pump Operating Head = 8.02 m Assuming that the pump efficiency is 70 %, and then we havePump supply power = Where: Q = Volume Flow Rate; 384 Liters/min dO = Density of oil; 914 kg/m3 (or 0.914 kg/liters) H = Pump Operating Head; 8.02 m P = Pump Efficiency; 70 %Pump supply power = =Pump supply power = 0.89 hpTherefore, we are to use 1 hp oil pump for the lube oil transfer.8. Designing the foundation of the engine

Based from Power Plant Engineering (Adapted to MKS Units) by Frederick T. Morse; Chapter 4: The Power Plant Building; Foundation; pages 105 to 113; & The PSME Code; we are to use the following equations to design the Diesel Engine Foundations

A. Given in the Diesel Engine Specifications:

Engine Length: 7.468 m Engine Width: 2.299 m Engine Weight: 29484 kgB. For the Length of the Machine Foundation, LF:

LF = LB + 2c

Where: LF = Length of the Machine Foundation; m or ft LB = Length of the Bed plate (given in Machine Specs); m or ft c = Clearance; 1 foot or 10% of the length of the bed plate

LF = LB + 2c LF = 7.468 m + (2*0.7468 m) = 7.468 m + 1.494 LF = 8.962 mC. For the Width of the Machine Foundation, a:

a = w + 2c

Where: a = Width of the Machine Foundation; m or ft w = Width of the Bed plate (given in Machine Specs); m or ft c = Clearance; 1 foot or 10% of the length of the bed plate

a = w + 2c a = 2.299 m + (2 x 0.2299 m) = 2.299 m + 0.4598 m a = 2.759 mD. For the Weight of the Machine Foundation, WF:

Based from The PSME Code of 1993, Chapter 2; Commercial and Industrial Building; Article 2.4 Machinery & Equipment; Section 4.4.1-2, p. 9, the weight of foundation should be 3 to 5 times the weight of the machinery. Using 5 times to maximize the design

WF = mass of the machine x desired multiplier = 29484 kg x 5 WF = 147420 kgE. For the Base Width of the Machine Foundation, b:

Based from Power Plant Engineering (Adapted to MKS Units) by Frederick T. Morse; Table 4-4. Safe Bearing Power of Soils, page 105 & Marks Standard Handbook for Mechanical Engineers, 9th Edition, Revised by Eugene Avallone & Theodore Baumester III, Design of Structural Members, Table 12.2.6 Safe Bearing of Soils, p. 12 - 21, we must use at least best brick masonry or higher for the safe bearing for our Diesel Engines, Using the Best Brick Masonry with a safe bearing capacity of 145 to 195 Tons/m2 and using the given equation we have

Where: SB = Safe Soil Bearing Capacity; tons/m2 (195 tons/m2 or 195000 kg/m2) WM = Weight of the Machine; kg WF = Weight of the Machine Foundation; kg b = Base Width of the Machine Foundation; m LF = Length of the Machine Foundation; m N = Safety Factor; usually a value of 2Deviating from the equation & computing we haveb = b = b = 0.20 mSince the lower width is less than the width of the machine foundation, then let the lower width be equal to the width. Therefore, b = 2.759 m.F. For the Volume of the Machine Foundation, VF:

Based from Power Plant Engineering (Adapted to MKS Units) by Frederick T. Morse; Table 4-2: Approximate Weights of Building Material; page 90; using reinforced concrete as our base material and using its approximate density we have = 2406 kg/mVF = WF / VF = VF = 61.27 m3G. For the Depth of the Machine Foundation, HF:

Since the width and the lower width are equal, then we have a rectangular block as for our foundation. The width, WF, would be equal to the width of the foundation.

VF = LF * WF * HF

HF = =

HF = 2.48 m

H. Materials for the Machine Foundation:

Based from Power Plant Engineering (Adapted to MKS Units) by Frederick T. Morse; Table 4-1: Data on Concrete Mixes to Yield 1 cu. m Concrete; page 90; using a mixture of 1:3:5 as it is often used for foundations we have the following data

Using a Mixture of 1:3:5 (1 part of cement, 3 parts of sand, and 5 parts of stone) to produced 1 cu. meter of concrete using 1:3:5 mixtures, the ff. are needed: 6.2 sacks of cement 0.52 m3 of sand 0.86 m3 of stone We are given 6.2 sacks of cement per m3 & 3 units of Diesel Engines and computing for the quantity or required numbers of sacks we have = 6.2 sacks x 3 x 61.27 m3 = 1139.62 or 1140 sacks Computing for the quantity or volume of sand for the mixture we have = 0.52 m3 x 1140 sacks = 592.8 m3 Computing for the quantity or volume of gravel or stone for the mixture we have = 0.86 m3 x 1140 sacks = 980.4 m3I. Anchor Bolts:

Based from The PSME Code of 1993, Anchor Bolts should be embedded in the concrete at least 30 times to the Bolt Diameter. Assuming a diameter of 25 mm of the Anchor Bolts then the length of the Anchor Bolts is

LAB = 25 mm x 30 LAB = 750 mm or 0.75 mJ. Length of Sleeves, LS:

Based from The PSME Code of 1993, Chapter 2; Commercial and Industrial Building; Article 2.4 Machinery & Equipment; Section 2.4.1.7, page 9; the length of the sleeve should be 18 times that of the bolt diameter. Therefore we have

LS = Bolt Diameter x 18 = 25 mm x 18 LS = 450 mm or 0.45 mK. Internal Diameter of Sleeves, DS:

Based from The PSME Code of 1993, Chapter 2; Commercial and Industrial Building; Article 2.4 Machinery & Equipment; Section 2.4.1.7, page 9; the internal diameter of the sleeve should be 3 times that of the bolt diameter. Therefore we have

Ds = Bolt Diameter x 3 = 25 x 3 Ds = 75 mm or 0.075 mL. Number of Steel Bars, NSB:

Number of Steel Bars, NSB: Based from Marks Standard Handbook for Mechanical Engineers; 9th Edition; Section 6: Materials of Engineering; 6.2 Iron & Steel; Weights of Square & Round Bars; page 6-46; with a 1 in (25.4 mm) round steel bar for our foundation it is given that it has 2.670 lb/in. Converting we have 3.97 kg/m as for its weight. Using the given formula we have

NSB = WF x m%/ WSB

Where: NSB = Number of Steel Bars WF = Weight of the Machine Foundation m% = percent multiplier; % to 1% WSB = Weight of the Steel Bars; kgComputing for the required number of steel bars as obtained from the data above, we have NSB = WF x m% / WSB = 147420 kg x 0.01 / 3.97 kg NSB = 371.34 or 372 piecesM. Total Length of the Steel Bars, LSB:

Since most of the manufactured steel bars in the market have a standard length of 6.1 m, then we simply have

LSB = NSB x 6.1 m = 372 x 6.1 m LSB = 2269.2 m

9. Specifying the required diesel power units

Diesel Engines are preferable as an internal combustion component in all internal combustion engine type power plants. They are good and reliable substitute as power plants in places where the supply of coal and water is not available in a desired quantity and for operations where it requires continuity and for standby and emergency purposes. Based from Power Plant Engineering by A. J. Raka, A. P. Srivastava, M. Dwivedi; Chapter 8: Diesel Power Plant; 8.4 Advantage of Diesel Power Plant & 8.5 Disadvantage of Diesel Power Plant; the advantages & disadvantages for using Diesel Engines includes the following..

Advantages: 1. Limited cooling water requirement. 2. Standby losses are less as compared to other Power plants. 3. Low fuel cost. 4. Quickly started and put on load. 5. Smaller storage is needed for the fuel. 6. Layout of power plant is quite simple. 7. There is no problem of ash handling. 8. Less supervision required. 9. For small capacity, diesel power plant is more efficient as compared to steam power plant. 10. They can respond to varying loads without any difficulty.

Disadvantages: 1. High maintenance and operating cost. 2. The plant cost per kW is comparatively more. 3. The life of diesel power plant is small due to high maintenance. 4. Noise is a serious problem in diesel power plant. 5. Diesel power plant cannot be constructed for large scale.

Based from Power Plant Engineering (Adapted to MKS Units) by Frederick T. Morse; Fig. 6-33; Dimensions of Nordberg 4-cycle Unsupercharged & Supercharged Engines; page 186; we are to use 3 units of F8-135 Diesel Engines (1 unit is continuous in operation; another unit for 7 am to 8 pm shift; another unit as a reserve; all of which must be capable of continuous operation at least for a limited duration of time) as based from our load curve.

Diesel Engine Specifications:

Engine Type: F8-138 Diesel Engine Engine Height: 2.302 m Engine Length: 7.468 m Engine Width: 2.299 m Engine Weight: 29484 kg Generator kW Rating at 450 RPM: 560 kW Units: 3 Units (1 unit operating continuously; 1 unit that operates at 7 am to 8 pm shift; 1 unit reserve; all are capable of continuous operation)10. Providing space for future expansion(See the drawing)

11. Design the Power house building

12. Prepare the working schedule of the projectPROPOSED SCHEDULE OF WORK & ESTIMATE TIME OF COMPLETIONAssuming we start the project at the month of January of 2010, then we are to follow the proposed schedule given belowTime & MonthTarget Work OutputRelevance

January to March 2013Load Survey for the consumer kilowatt consumption; Gathering & surveying of the environment within the location; Planning & securing necessary permits

Initial work frame for the project; idealization for the projects continuity

April to June 2013Computation for the required plant operation which includes the following: Selection of Diesel Engine type, Design of the Generator Set, Design of its Fuel Oil System, Design for its Cooling Water System, Design for its Lubricating System, Design for its Air Handling System, Design for the Engine Dimension, others; Thorough reevaluation of the plant design; Obtaining contracts & advice from various engineering firms

Initial work frame for the plants construction & operation; Idealization for the plant layout, operation & maintenance

July to February 2014Construction of the plant site; Purchase of equipments; Construction of Administration Building & Others; Installation of the equipments

Idealization of the plant site

March to April 2014Initial operation; Recalibration of equipments

Testing phase of the project; Checking for flaws & defects

May 2014Turnover of Plant operation to the owner/firm

Final project phase

17

Design Problem and Solution

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