Design Problem and Solution

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DESIGN PROBLEM

A 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 days

3. Electrical connection including distribution will be consulted by a Electrical

Engineer

Requirements:

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.

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DESIGN SOLUTION

1. Number of customers in Brgy. San Felipe, Naga City.

Number of population: 14098

Number of Households: 3873

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

Households: 3873

Satellite market: 6

Business offices and establishments: 13

School: 6

Park & Recreational center: 4

Water Pumping station: 3

3. Identifying the number of customers per zone.

Zone

No. of Population

No. of households

1

722

155

2 5246

1212

3 1200

365

4

643

130

5 3009

618

6 4933

852

7 2013

456 4. Time usage of the customers per 24 hours.



Households

Appliances No. of Appliances Wattage Schedule of Use Schedule

hr/day total Wattage kw-hr

electric fan 500 80 6pm-6am 12hrs 40000 480

oven toaster 30 1200 6am-9am &

2pm-4pm 5hrs 36000 180 refrigerator 95 850 daily 24hrs 76000 1824 fluorescent lamp 357 40 6pm-6am 12hrs 14280 171.36 bulb 285 50 6pm-6am 12hrs 14250 171 water dispenser 12 300 daily 24hrs 3600 86.4

rice cooker 75 320 5am-6am &

6pm-7pm 2hrs 24000 48 vacuum cleaner 10 630 6am-8am & 2pm-4pm 4hrs 6300 25.2 air conditioning unit 45 3358 6pm-12am & 10am-5pm 12hrs 151110 1813.32 dvd player 90 13 10am-12nn &

7pm-9pm 4hrs 1170 4.68

laptop 79 250 8am-1pm 5hrs 19750 98.75

computers 36 350 9am-2pm 5hrs 12600 63

printers 15 18 8am-10pm 2hrs 270 0.54

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3



Sari- sari stores

Appliances No. of Appliances Wattage Schedule of Use Schedule hr/day total Wattage kw-hr

electric fan 84 80 8am-8pm 12hrs 6720 80.64

fluorescent

lamp 32 40 6pm-12mn 6hrs 1280 7.68

bulb 75 18 6pm-6am 12hrs 1350 16.2

refrigerator 36 850 daily 24hrs 30600 734.4

television sets 20 110 9am-5pm 8hrs 2200 17.6

radio/karaoke 12 180 8am-10am &

1pm-4pm 5hrs 2160 10.8

Air conditioning

unit

3 3358 2pm-7pm 5hrs 10074 50.37

dvd player 3 10 11am-1pm &

2pm-5pm 5hrs 30 0.15

ceiling fan 40 60 daily 24hrs 2400 5706

TOTAL 6623.84



Resorts

hr/day total Wattage kW-hr

electric fan 10 80 8am-11am &

2pm-5pm 6hrs 800 4.8

fluorescent

lamp 10 40 6pm-6am 12hrs 400 4.8

bulb 10 18 daily 24hrs 180 4.32

karaoke 2 180 8am-10am &

1pm-4pm 5hrs 360 1.8 refrigerators 2 850 daily 24hrs 1700 40.8 blender 4 750 10am-12nn &3pm-5pm 4hrs 3000 12 water dispenser 3 300 daily 24hrs 900 21.6 Air conditioning unit 3 3358 1pm-6pm 5hrs 100721 503.6 vacuum cleaner 4 630 8am-10am & 3pm-5pm 4hrs 2620 10.48

ceiling fan 30 60 daily 24hrs 1800 43.2

freezer 4 1200 daily 24hrs 4800 115.2

TOTAL 764.25 stove

coffee maker 3 1200 6am-8am 2hrs 3600 7.2

charger 300 65 10am-12nn &

5pm-7pm 4hrs 19500 78

water pumps 30 2238 8am-8pm 12hrs 67140 805.68

television

sets 198 110 9am-9pm 12hrs 10780 129.36

washing

machines 78 20 9am-11am 2hrs 1560 3.12

flat iron 100 1000 5am-7am &

8pm-10pm 4hrs 100000 400

ceiling fan 50 60 8am-10am &

2pm-5pm 5hrs 3000 15

karaoke 180 180 9pm-2am 5 900 4.5

stereo 50 40 1pm-6pm 5hrs 2000 10

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

Hotel

Appliances No. of Appliances Wattage Schedule of Use Schedule hr/day total Wattage kw-hr Air conditioning unit 10 3358 6pm-11pm 5hrs 33580 167.9

ceiling fan 10 60 2pm-5pm &

8pm-11pm 6hrs 600 3.6

computer 5 350 8am-12nn &

2pm-6pm 8hrs 1750 14

printer 1 18 10pm-11pm

& 3pm-4pm 3hrs 18 0.054 coffee maker 3 1200 6am-8am &

3pm-4pm 3hrs 3600 10.8

fluorescent

lamp 15 40 6pm-6am 12hrs 600 7.2

bulb 20 18 6pm-6am 12hrs 360 4.32

rice cooker 2 320 10am-11am

& 5pm-7pm 3hrs 640 1.92 vacuum

cleaner 5 630

9am-11am &

3pm-5pm 4hrs 3150 7.15

dvd player 1 13 8am-10am &

2pm-4pm 4hrs 13 0.052

water pump 2 2238 daily 24hrs 4476 107.42

TOTAL 342.016



Machine Shops

Appliances No. of Appliances Wattage Schedule of Use Schedule hr/day total Wattage kw-hr milling machines 1 2500 9am-12nn & 1pm-3pm 5hrs 2500 12.5 welding machines 2 13000 10am-12nn & 1pm-4pm 5hrs 26000 130 compressor 1 3000 9am-12nn 3hrs 3000 9

pumps 1 2238 10am-12nn &

1pm-5pm 6hrs 2238 13.428

fluorescent

lamp 1 40 6pm-6am 12hrs 40 0.48

bulbs 5 18 6pm-6am 12hrs 90 1.08

ceiling fan 2 60 9am-12nn &

1pm-4pm 6hrs 120 0.72

exhaust fan 1 60 9am-12nn &

1pm-4pm 6hrs 60 0.36

drill press 1 750 9am-11am &

2pm-3pm 3hrs 750 2.25

electric fan 3 80 9am-12nn &

1pm-4pm 12hrs 240 2.88

oven toaster 1 1200 9am-12nn 3hrs 1200 3.6

lathe machine 1 2700 9am-12nn &

1pm-10pm 12hrs 2700 32.4 TOTAL 208.698



Vulcanizing Shops

compressor 1 3000 9am-11am &

1pm-3pm 4hrs 3000 12

fluorescent

lamp 3 40 6pm-6am 8hrs 120 0.96

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drill press 1 750 9am-11am &

1pm-3pm 4hrs 380 1.52

radio 1 180 9am-11am &

1pm-3pm 4hrs 110 0.44

welding

machines 1 13000

9am-12nn &

1pm-3pm 5hrs 13000 65

ceiling fan 1 60 9am-12nn &

1pm-4pm 6hrs 60 0.36 TOTAL 98.302



Hardware

radio/karaoke 1 180 8am-1pm 5hrs 180 0.9

computer 2 350 8am-5pm 8hrs 760 6.08

bulb 3 50 8am-8pm 12hrs 150 1.8

ceiling fan 15 60 8am-5pm 8hrs 900 7.2

exhaust fan 15 80 daily 24hrs 1200 2808

TOTAL 55.34



Bodegas

fluorescent lamp 20 40 6pm-6am 12hrs 800 9.6 radio/karaoke 3 180 8am-1pm 5hrs 540 2.7 air conditioning unit 6 3358 2pm-7pm 5hrs 20148 100.74

Computer 6 350 9am-2pm 5hrs 2100 10.5

exhaust fan 15 60 9am-3pm 6hrs 900 5.4

Printer 2 18 10am-11am &

2pm-4pm 3hrs 36 0.108

Bulb 20 50 daily 24hrs 1000 24

Christmas lights 4 25 6pm-6am 12hrs 100 1.2 TOTAL 160.148



Bakeries

oven toaster 2 1200 8am-2pm 6hrs 2400 14.4

Refrigerator 1 380 daily 24hrs 380 9.12

fluorescent

lamp 4 40 8am-5pm 8hrs 160 1.28

TOTAL 26.63



Gasoline Stations

fluorescent

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6 radio/karaoke 2 100 8am-11am &

1pm-4pm 6hrs 360 2.16

air conditioning

unit

4 3358 8am-8pm 12hrs 13432 161.184

television sets 2 110 8am-11am &

1pm-4pm 6hrs 270 1.32

Computer 4 350 8am-5pm 8hrs 1400 11.2

Compressor 2 30000 9am-11am &

1pm-3pm 4hrs 6000 4

Bulb 7 50 6pm-6am 12hrs 350 4.2

christmas lights 4 25 6pm-6am 12hrs 100 1.2 TOTAL 202.064



Barber Shops

Karaoke 1 180 8am-10am &

1pm-4pm 5hrs 180 6.9

Charger 3 65 10am-12nn &

1pm-4pm 5hrs 195 0.975

Razor 3 110 8am-6pm 10hrs 330 3.3

TOTAL 19.215

5. Determining the total power requirement for the area.

TIME Kw-hr CONSUMPTIO N LOAD VALUE 1 9544.212 397.6755 2 9544.212 397.6755 3 9539.712 397.488 4 9539.712 397.488 5 9997.312 416.5546667 6 10218.112 425.7546667 7 9296.08 387.3366667 8 10194.026 424.7510833 9 10619.416 442.4756667 10 12484.247 520.1769583 11 18106.285 754.4285417 12 18017.097 750.712375 13 18450.117 768.754875 14 13034.507 543.1044583 15 12984.341 541.0142083 16 11067.911 461.1629583 17 12577.582 524.0659167 18 13563.946 565.1644167 19 13208.726 550.3635833 20 13162.296 548.429 21 12103.692 504.3205 22 11969.706 498.73775 23 11536.766 480.6985833 24 11365.212 473.5505 Average Load 507.161849

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7 0 100 200 300 400 500 600 700 800 900 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 kW consumption 0 100 200 300 400 500 600 700 800 900 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 kW consumption

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)

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7. Designing the Power Plant including each auxiliary

A. Cooling Water System

1. Required 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 Engine’s 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 hour Rated Bhp – Rated Brake horsepower t1 – inlet temperature; °C

t2 – outlet temperature; ° C

Accordingly, 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 hp

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9 Assuming 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/hr

Based from Mark’s 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/min

Therefore, 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:

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 ½ in outside diameter schedule 40 for the discharge line pipe, the following data is given as follows…

Suction line pipe: Schedule 40

Outside diameter 3 in (3.5 in) Inside diameter 3.068in (0.0779 m) Discharge line pipe:

Schedule 40

Outside diameter 2.5 in (2.875 in) Inside diameter 2.469 in (0.0627 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

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10 V – Velocity of fluid; m/min

Also given for the area and deviating from the equation we have… A = π d2

/4

Where: A – Cross-sectional area; m2 d – Internal diameter; m Thus,

V =

Computing now for the velocity at the suction we have… Vs = = Vs = 379.97 m/min (1min/60secs) Vs = 6.33 m/sec

Computing now for the velocity at the discharge we have… Vd = = Vd = 586.53 m/min (1min/60secs) Vd = 9.78 m/sec

Assuming 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 head…

Discharge head = FL+ZD+V Discharge head = FL+ZD+

Where: FL – Friction losses for discharge; 0.75

ZD – Elevation from datum to discharge; 2.302 m VD – Velocity head at discharge; 9.78 m/sec

Discharge head = FL+ZD+ Discharge head = 0.75 + 2.302 m + Discharge head = 3.052 m

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11 Computing now for the suction head…

Suction head = FL + ZS + V Suction head = FL + ZS +

Where: FL – Friction losses for suction; 0.75

ZS – Elevation from datum to suction; 4.5m (negative due to location)

VS – Velocity head at suction; 6.33 m/sec Suction head = FL + ZS +

Suction head = 0.75 – 4.5m + Suction head = -1.71 m

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 Pump operating head = 3.052 m + 1.71 m

Pump operating head = 4.762 m

Assuming that the pump efficiency is 70 %, and then we have… Pump supply power =

Where: Q – Volume flow rate; 1811 liters/min

dw – Density of water; 1000 kg/m 3

H – Pump operating head; 4.762 m ηp – Pump efficiency; 70%

Pump supply power =

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12 Therefore, we are to use 2 ¾ 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 simplicity…

QC (tC1 – tC2) = QR (tR1 – tR2)

Where: QR - Quantity of Raw Water circulating the cooling tower, Liter/min

QC - Quantity of Circulating Cooling Water; Liter/min tR1 - temperature of raw water at outlet; 34° C

tR2 - temperature of raw water at inlet; 28° C

tC1 - temperature of jacket cooling water at inlet; 75° C tC2 - temperature of jacket cooling water at outlet; 65° C Using the equation, the quantity of raw water is…

QR = = = 3018.33 L/min

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/min

4. 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…

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13 VS = VS =

VS = 633.29 m/min (1min/60 secs) VS = 10.55 m/sec

Computing now for the velocity at the discharge we have… VD = VD =

VD = 977.56 m/min (1 min/60 secs) VD = 16.29 m/sec

Assuming 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 head…

Discharge Head = FL + ZD + V Discharge Head = FL + ZD +

Where: FL = Friction Losses for Discharge; 0.75 ZD = Elevation from Datum to Discharge; 5 m VD = Velocity Head at Discharge; 16.29 m/sec Discharge Head = FL + ZD +

Discharge Head = 0.75 + 5m +

Discharge Head = 19.28 m

Computing now for the suction head… Suction Head = FL + ZS + VS Suction Head = FL +ZS +

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14 ZS = Elevation from Datum to Suction; 4.5 m (- due to location) VS = Velocity Head at Suction; 10.55 m/sec

Suction Head = FL +ZS + Suction Head = 0.75 – 4.5 m + Suction Head = 1.92 m

Pump Operating Head = Discharge Head – Suction head = 19.28 m - 1.92 m

Pump Operating Head = 17.36 m

Where: Q – Volume flow rate; 3018.33 L/min

dw – Density of water; 1000 kg/m 3

H – Pump operating head; 17.36 m ηp – Pump efficiency; 70%

Pump supply power =

Pump supply power = 16.63 hp (16.75 hp)

Therefore, we are to use 16 ¾ hp water pump for the raw water transfer.

5. Required Quantity of Make-up Water:

Morse; Chapter 6: Internal Combustion Engine Power Plant; 6-7 Evaporative Cooling; page 181; using the given mass balance and heat balance equations we

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15 For mass balance we use…

1 kg air + SH1 + WW + ΔW = 1 kg air + SH2 + WW Deviating for simplicity we have…

ΔW = SH2 – SH1 For heat balance we use…

h1 + WWhfa + ΔWhf = h2 + WWhfb Deviating for simplicity we have…

WW =

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 ΔW = 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/kg

Assuming 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 Mark’s 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

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16 At a R.H. 90% and dry bulb temperature of 34° C at leaving point…

SH2 = 0.031 kg moisture/kg dry air

Based 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…

ΔW = SH2 – SH1

= (0.031 – 0.01425) kg moisture/kg dry air ΔW = 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/kg

Assuming 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 have…

WW = = – – WW = 1.27 kJ/kg dry air

Since 3007 kg/min of raw water is given, the air flow is… Air flow =

Air flow = 2367.72 kg/min

Based from Mark’s 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…

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17 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/min

The 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/min

6. 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 Tubeaxial Motor hp: 20 hp

Rpm: 870

Capacity: 73948ft3/min Static Pressure: 1/4 in SP

7. 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 gpm

Assuming 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 ft2

Assuming we have a square sized cooling tower, then the size of the cooling tower would be…

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18 Side of the Cooling Tower = 20.62 ft

So 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%.

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 follows… Maximum 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/day

Based from Mark’s 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

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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/day

2. Required Storage of Fuel Oil:

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:

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 kg

For 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 Mark’s 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

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20 (American Petroleum Institute) standard, assuming an oil temperature of 6° C and the equations 5 – 3, the °API would be…

°API =

- 131.5 =

– 131.5

°API = 16.61° API or 17° API

The density of oil at 15.6°C (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

= 0.0042 The density of fuel oil at 6° C would be…

Density at 6° C = 0.9554 / 0.9958 = 0.9981 kg/liter The volume would now be equal to…

V = 4683.6 kg / 0.9981 kg/liter = 4692.52 liters (1m3/1000 Liters) V = 4.69 m3

Assuming 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…

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21 Where: V = Volume of the cylinder; m3

d = Diameter of the cylinder; m h = Height or Length of the cylinder; m

Assuming we have a 2.5 m length of the day tank, deviating from the dimension and computing for the diameter we have…

d2 = d2 = d2 = d2 = 2.39 m2 d = 1.55 m

Therefore, 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 Mark’s 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 ½ in outside diameter schedule 40 for the suction line pipe and a nominal pipe size of 1 ¼ 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 ½ in (true size of 1.9 in) Inside Diameter 1.610 in (or 0.0409 m) Discharge Line Pipe:

Schedule 40

Outside Diameter 1 ¼ 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/min

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22 A = π d2

/4

Where: A = Cross-sectional Area; m2 d = Internal Diameter; m

V =

Computing now for the velocity at the suction we have…

VS = = VS = 238.24 m/min (1min/60secs) VS = 3.97 m/sec

Computing now for the velocity at the discharge we have…

VD = = VD = 323.47 m/min (1min/60secs) VD = 5.39 m/sec

Discharge Head = FL + ZD + V Discharge 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/sec

Discharge Head = FL + ZD + Discharge Head = 0.75 + 4.5 + Discharge Head = 6.73 m

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23 Computing now for the suction head…

Suction Head = FL + ZS + V Suction 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/sec

Suction Head = FL + ZS + Suction Head = 0.75 – 2.5 + Suction Head = -0.95 m

Pump Operating Head = Discharge Head – Suction head = 6.73 m + 0.95 m

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

ηP = Pump Efficiency; 70 % Pump supply power =

=

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24 Therefore, 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/min

Assuming 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 m2

Thus, the area of the pipe is… A = πd2 /4 Where: A – Area d – Diameter d = (4A/π)1/2 d = (4*0.07885 m2/π)1/2 d = 0.32 m

The diameter of the intake pipe would be 0.32 meter or 320 mm. Therefore, we are to use a pipe with such diameter.

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25 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/min

Assuming 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 m2

Thus the area of the pipe is… A = πd2 /4 Where: A – Area d – Diameter d = (4A/π)1/2 d = (4*0.112 m2/π)1/2 d = 0.38 m

The 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

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26 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 HP

Compressor Power Recommended Receiver Volume

(hp) kW Ft3 Gal m3

125 93.3 67 500 1.9

200 149.2 107 800 3.0

350 261.1 188 1400 5.3

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:

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

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27 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 kW

For a continuous operation, the generated power would be…

Generated Power = 600.47 kW * 24 hours Generated Power = 14411.28 kW-hour

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-hour

Since 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/day

Similarly with the fuel oil, assuming that the delivery of lube oil is every 45 days, the oil consumption…

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28 = 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 m3

Assuming 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 = (π x d2

/ 4) x L

Length = (1.92 m3 x 4) / (π 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 have…

VS = = VS = 292.28 m/min (1min/60secs) VS = 4.87 m/sec

Computing now for the velocity at the discharge we have…

VD = =

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29

VD = 396.85 m/min (1min/60secs)

VD = 6.61 m/sec

Discharge Head = 0.75 + 4.5 +

Discharge Head = 7.48 m

Computing now for the suction head…

Suction Head = 0.75 – 2.5 +

Suction Head = -0.54 m

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 = 7.48 m + 0.54 m

Where: Q = Volume Flow Rate; 384 Liters/min

dO = Density of oil; 914 kg/m 3

(or 0.914 kg/liters) H = Pump Operating Head; 8.02 m

ηP = Pump Efficiency; 70 % Pump supply power =

=

Pump supply power = 0.89 hp

Therefore, we are to use 1 hp oil pump for the lube oil transfer. 8. Designing the foundation of the engine

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30 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 kg

B. 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 m

C. 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 m

D. 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 kg

E. 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 & Mark’s 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…

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31

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 2 Deviating from the equation & computing we have…

b =

b = b = 0.20 m

Since 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:

Table 4-2: Approximate Weights of Building Material; page 90; using reinforced

concrete as our base material and using its approximate density we have ρ = 2406 kg/m³… VF = WF / ρ VF = VF = 61.27 m 3

G. 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 =

=

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32 HF = 2.48 m

H. Materials for the Machine Foundation:

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 m3 I. 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 m J. 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 m K. Internal Diameter of Sleeves, DS:

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33 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 m L. Number of Steel Bars, NSB:

Number of Steel Bars, NSB: Based from Mark’s 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; ½ % to 1% WSB = Weight of the Steel Bars; kg

Computing 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 pieces

M. 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

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34 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.

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)

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35 12. Prepare the working schedule of the project

PROPOSED SCHEDULE OF WORK & ESTIMATE TIME OF COMPLETION

Assuming we start the project at the month of January of 2010, then we are to follow the proposed schedule given below…

Time & Month Target Work Output Relevance

January to March 2013

Load 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 project’s continuity

April to June 2013

Computation 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 plant’s construction & operation; Idealization for the plant layout, operation & maintenance

July to February 2014

Construction of the plant site; Purchase of

equipments; Construction of Administration Building & Others; Installation of the equipments

Idealization of the plant site

March to April 2014

Initial operation;

Recalibration of equipments

Testing phase of the project; Checking for flaws & defects

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36 May 2014

Turnover of Plant operation to the owner/firm