Diesel Kim

BICOL COLLEGE OF APPLIED SCIENCE AND TECHNOLOGYCollege of Engineering and Architecture

City of NagaS/Y 2012-2013

(In partial fulfillment as a requirement in Power Plant Engineering)

Submitted by:

ROMANO S. NUPIA

BSME – 5

Submitted to:

Engr. EDUARDO A. CALZO PME

Instructor

FEBRUARY, 2013

2nd Semester

1

BICOL COLLEGE OF APPLIED SCIENCE AND TECHNOLOGYCollege of Engineering and Architecture

City of Naga

Engr. Eduardo A. Calzo PME

Mechanical Engineering Department

Bicol College of Applied Science and Technology

Sir:

In compliance with the fulfilment of the requirements on the subject “Power Plant

Design”, I would like to submit the design entitled “Diesel Power Plant Design”, in accordance

with your instructions.

This includes computations and illustrations which are researched and summarized

necessary to accomplish this design

I hope that this design will meet your approval.

Thank you and God bless.

Sincerely yours,

NUPS

2

ACKNOWLEDGMENT

I have taken efforts in this design. However, it would not have been possible without the

kind support and help of many individuals. I would like to extend my sincere thanks to all of

them.

To my major instructor, Engr. Eduardo A.Calzo, PME, for the unselfish and untiring

dedication, encouragement, valuable guidance and advice throughout the period of this study.

To my classmates and friends who gives their timeless effort, suggestions and comments

which give me strength in doing this design.

To my parents for manual support and strength, for the sacrifices and understanding, for

guiding me in the right path, help and for everything.

To everyone who sent comments, corrections and suggestions in my sincerest

appreciations.

And to God the father of all, the source of truth and wisdom from whom all good things

came, effort and opportunity to learn, to love and to share.

3

T A B L E O F C O N T E N T S

Title Page ------------------------------------------- 1

Letter of Transmittal --------------------------------------------2

Acknowledgement --------------------------------------------3

Table of Contents ---------------------------------------------4-5

Design Problem ----------------------------------------------6

Load Survey ---------------------------------------------7-14

Summary of load ----------------------------------------------15

Load Curve ----------------------------------------------16

Machine Foundation Design ------------------------------------18-23

Fuel Oil Consumption----------------------------------------------23-24

Fuel Oil Storage Tank & Dimension ----------------------------24

Day Tank -------------------------------------------------------25-29

Cooling Water System -------------------------------------29-40

Air Intake System&Exhaust System -----------------------------41-42

Air Starting System -----------------------------------------------42-44

Lubrication System -----------------------------------------------44

Pump Lube Oil -----------------------------------------------45-48

4

Proposal of Work Schedule ---------------------------------------48

Glossary ---------------------------------------------------------49

Bibliography ---------------------------------------------------------50

5

DESIGN PROBLEM:

A Diesel Power Plant will be constructed to save the community of Barangay San Pacol, Naga City. The Power Plant will supply power to be available for 24 hours.

Requirements:

1. Determine the no. of customer in the community.2. Identify the kind and no. of residential, commercial and shops.3. Identify the customers by zone.4. Determine the time usage for the customer in 24 hours.5. Determine the total requirement power in the area.6. Draw the map of the area and indicate the customer in its zone.7. Design the power plant including its auxiliaries.8. Design the foundation of the engine.9. Specify the required no. of diesel power units.10. Provide space for future expansion.11. Design the power house engine.12. Prepare the work schedule of the project.

Assumptions:

1. Plant capacity factors – 60%2. Fuel supply in the area good for 45 days.3. Electrical connection including distribution line is not in the assumptions.4. Professional Mechanical Engineers should be consulted.5. 20 x 30 – tracing papers6. Draw the plant view, side view and perspective of the plant.

6

BARANGAY PACOL RESIDENCE INFORMATION PER ZONE AS OF 2010

ZONE 1

2

3

4

5

6

7

TOTAL

Total No. Of Households

930

190

121

178

198

117

95

1,829

Total No. Of Population 6,146

918

589

708

882

647

441

10,331

Total No. Of Out-of-School

230

28

27

12

32

15

16

360

Total No. Of High school

718

34

34

69

77

19

18

969

Total No. Of Rich

Level A – Rich 5

3

17

30

6

6

1

68

Level B – Average 530

5

80

105

17

4

5

746

Level C – Poor 395

182

24

75

175

107

89

1,047

7

8

ZONE 1 - 930 HOUSEHOLDS

Appliances Units Wattage Schedule of UseHrs

use/dayTotal

WattagekW-H

Air Con. Unit 183 1000 9 am - 9 pm 12 183000 2196.00Coffee Maker 71 1200 7 am - 8 am 1 85200 85.20Computer 270 250 9 am - 12 mn 15 67500 1012.50Cd Player/Dvd Player 535 85 4 pm - 7 pm 3 45475 136.43Christmas Lights 435 25 7 pm - 5 am 10 10875 108.75Electric Fan 768 170 1 pm - 9 pm 9 130560 1175.04Fluorescent

40 watt 589 40 5 pm - 12 mn 7 23560 164.9260 watt 50 60 5 pm - 12 mn 7 3000 21.0075 watt 38 75 5 pm - 12 mn 7 2850 19.95

100 watt 18 100 5 pm - 12 mn 7 1800 12.60Food Blender 100 300 11 am-12 nn 2 30000 60.00Fruit Juicer 59 100 11 am-12 nn 2 5900 11.80Hand Iron 535 1000 6 am - 7 am 1 535000 535.00Hair Blow dryer 150 1000 9 am - 10 am 1 150000 150.00Incandescent

11 watt 2450 11 5 pm - 12 mn 7 26950 188.6516 watt 445 16 5 pm - 12 mn 7 7120 49.8420 watt 243 20 5 pm - 12 mn 7 4860 34.0230 watt 45 30 5 pm - 12 mn 7 1350 9.45

Microwave Oven 69 1450 4 pm - 5 pm 1 100050 100.05Radio 645 26 6 am - 9 am 3 16770 50.31

Range (w/ oven) 32 1220011 am-12 nn/6 pm-7

pm2 390400 780.80

Refrigerator-Freezer -16 cu ft 400 380 Daily 24 152000 3648.00 -20 cu ft 246 420 Daily 24 103320 2479.68

Rice Cooker 467 3206-7 am/11 am-12 nn/6-

7 pm3 149440 448.32

Stereo 356 90 6 am - 9 am 3 32040 96.12Toaster 279 900 6 am - 7 am 1 251100 251.10TV -25" color 469 300 9 am - 11 pm 14 140700 1969.80 -19" color 59 160 9 am - 11 pm 14 9440 132.16 -45" screen 5 147 9 am - 11 pm 14 735 10.29 -13"color TV/VCR 286 45 9 am - 11 pm 14 12870 180.18Vacuum Cleaner 135 630 7 am - 10 am 3 85050 255.15Water Dispenser 476 100 Daily 24 47600 1142.40Washing Machine 525 920 9 am - 12 nn 2 483000 966.00

TOTAL 328951518481.5

1

ZONE 2 - 190 HOUSEHOLDS

Hrs Total

LOAD SURVEY VALUES & LOAD GRAPH PER HOUR RATING

TIME kW-Hr CONSUMPTION LOAD GRAPH OF kW-Hr CONSUMPTION1AM 10467.95 436.16452AM 10467.95 436.16453AM 10467.95 436.16454AM 10467.95 436.16455AM 10467.95 436.16456AM 11390.71 474.612757AM 12900.48 537.528AM 11908.23 496.17641679AM 21757.77 906.573916710AM 21804.61 908.5255833

11AM 22885 953.536

12PM 19371.59 807.1497513PM 20217.61 842.400583314PM 19943.46 830.977666715PM 17950.74 747.947666716PM 20243.29 843.470583317PM 19622.02 817.5842518PM 22131 922.13012519PM 24086.76 1003.61512520PM 22446 935.236791721PM 22435 934.791166722PM 17220.99 717.541166723PM 16387.53 682.813708324AM 12963.28 540.136625

9

10

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 240

200

400

600

800

1000

1200

LOAD SURVEY VALUES & LOAD GRAPH PER HOUR RAT-ING

Series1

HOURS

LOA

D G

RAPH

OF

kW-H

r CO

NSU

MPT

ION

SOLUTION:

Based from Power Plant Engineering by Frederick T. Morse; Fig. 6-33; we are to use 3

units of F8-135-SC Diesel Engines (1 unit is continuous in operation 24 hours; another unit for 9

am up to 12 am; another unit as a reserve; all of which must be capable of continuous operation

at last for a limited duration of time) as based from our load curve.

DIESEL ENGINE SPECIFICATIONS:

Engine Type: F8-135 SC Diesel Engine

Engine Height: 2.604 m

Engine Length: 5.944 m

Engine Width: 2.299 m

Engine Weight: 23,814 kg

Generator for kW rating at 450 RPM: 525 kW

Units: 3 Units

(1 unit is continuous in operation 24 hours; another unit for 9 am up to 12 am; another unit as a reserve; all of which must be capable of continuous operation at last for a limited duration of time)

11

MACHINE FOUNDATION DESIGN

Design Procedure for Machine Foundation

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.

For the Length of the 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; 10% of the length of the bed plate

Lf = Lb + 2c

Lf = 5.944 m + 2(0.1) (5.944m)

Lf = 5.944m + 1.1888m

Lf = 7.1328m

For the Width of the 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; 10% of the length of the bed plate

a = w + 2c

a = 2.299 m + 2 (0.1) (2.299m)

= 2.299 m + 0.4598 m

a = 2.7588 m

12

For the Weight of the Foundation:

Wf = 3 to 5 times the weight of the Machine

Since; Weight of the Machine, Wm = 23,814 kg

Therefore; using 5

Wf= 23,814 kg x 5

Wf = 119,070 kg

Base width of the Foundation, b:

Using table 4.4 page 105 of P.P.E. byFrederick Morse.

For Best Brick Masonry

Bearing Capacity → 145 - 195 Tonnes/m²

Sb = 195 Tonnes/m²

Sb = 195,000 kg/m²

Using the Formula:

SbN

=(W m+W f )

b(Lf )

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

b=N (W m+W f )

Lf (Sb)

13

b=2(23,814 kg+119,070 kg)7.1328 m(195,000 kg/m ²)

b=¿ 0.205 m

Since; b < a, let b = a

Therefore; b = 2.7588 m

The Volume of the 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 ρ = 2406 kg/m³.

Vf =Wfρ

Vf = 119,070 kg2406 kg /m ³

Vf =¿ 49.49 m3

The Depth of the 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 =hf (a)(Lf )

Where:

hf = depth of the machine foundation

a = width of the machine foundation

Lf = length of themachine foundation

Vf = volume of the machine foundation

a = 2.7588 m, since a = b

Vf =hf (a)(Lf )

hf = Vfa(Lf )

= 49.49 m3

2.7588 m(7.1328 m)

hf =¿ 2.515 m

14

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 of cement/1m3 of concrete) x 49.49 m3 of concrete

= 306.838 sacks of cement per unit of Diesel Engine

For 3 units;

(306.838 sacks per unit) 3 units = 920.514 or 921 sacks of cement for the 3 Diesel Engine.

Computing for the quantity or volume of sand for the mixture we have.

= (0.52 m3 of sand / 1m3 of concrete) 49.49 m3 concrete

= 25.7348 m3 of sand

For 3 units;

25.7348 m3 of sand (3 units) = 77.2044 m3 or 78 m3 of sand for the 3 Diesel Engine

Computing for the quantity or volume of gravel or stone for the mixture we have.

= (0.86 m3 of stone/1m3 of concrete)49.49 m3 concrete

= 42.5614 m3 of gravel

For 3 units;

42.5614 m3 of gravel (3 units) = 127.6842 m3 or 128 m3 of gravel for the 3 Diesel Engine

15

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. Assume that the diameter of the Anchor Bolts is 25mm.Then the length of the Anchor Bolts is.

LAB = 25 mm x 30

LAB = 750 mm or 0.75 m

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

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 = 75 mm or 0.075 m

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 (m %)

WSB

Where:

NSB = Number of Steel Bars

WF = Weight of the Machine Foundation

m% = percent multiplier; ½ % to 1%

16

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 (m% )

WSB=

119,070 kg(0.01)3.97 kg

=299.9∨300 pieces

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 = 300 x 6.1 m = 1830 m

FUEL CONSUMPTION

Our diesel power plant design requiresthree units of F8-135-SC with generator rating of 525 kW, one unit is operating in 24 hours, one unit for 9 am-12 am, and the other unit is for reserve.Plant capacity factor is 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 525 kW with 450 rpm, by interpolation we have 94.3375 %. 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 93.1375 % 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/Bhp-Hr. With such the maximum rate of fuel usage would be as follows.

Maximum Rateof FuelUse=525 kW x20.931375

x1.34 hp1 kW

= (1,510.67 Bhp) x (0.1625 kg/Bhp-hr)

Maximum Rate of Fuel Use = 245.48 kg/hr

17

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 = 245.48 /hr x 24 hr/day

Maximum Rate of Fuel Use = 5,891.52 kg/day

Based from Mark’s 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 = (5,891.52kg/day) (1.094 x 10-3 m3/kg) (1 x 103 Litters/m3)

= (6.445 m3/day)(1x103 Litters/m3)

Fuel Consumption = 6,445.32 Litters/day

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 liter to 3.48 liters. One (1) Diesel Genset unit will operate continuously while another for 15 hours each per day at peak loads.

For fuel consumption for within a 45 day supply we have.

Required Storage = 45 days x (6,445.32 Liters/day)

Required Storage = 290,039.4 Liters

Dimensions of the Fuel Oil Storage Tank:

From Power Plant Engineering (Adapted to MKS Units) by Frederick T. Morse; Table 12 – 4: Dimension of the Bulk Storage Tank; page 459.From the table the largest capacity of the tank is just 109,716 litters. Since we have 290,039.4 litters we need to use 4 tanks within the 45 days. It will require 72,509.85 litters for each tank. Then we are about to use the tank that has the capacity of 73,392 litters. The dimension of the cylindrical bulk tank is given as follows.

Diameter 3.05 m

Length 10.06 m

Plate Thickness 7.94 mm

Weight 7335 kg

18

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.

245.48 kg/hrx 24 hrs = 5,891.52 kg

For a 1 day operation, assume that the day tank will charge 4465.92 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 (American Petroleum Institute) standard, assuming an oil temperature of 6° C and the equations 5 – 3, the °API would be.

API= 141.5

SG@ 15.6/15.6O−131.5

API= 141.50.9554

−131.5

API=16.610 API∨17O 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 5° C we have.

Contraction from a 6° C cooling = 0.0007 x 6

= 0.0042

The density of fuel oil at 5° C would be.

Density at 6° C=0.9554 / 0.9958

= 0.9981 kg/litter

The volume would now be equal to.

V = [5,891.52kg / 0.9981 kg/litter] (1 x 10-3 m3/1 Litters)

19

V = 5.9 m3

Assuming a 15 minute charging for the day tank, the volume flow rate would be..

Q = [5.9 m3/ 15 min] (1 x 103 Litters/m3)

= 0.3933 m3/min

Q = 393.3 Litters/min

Dimensions of the Fuel Oil Day Tank:

Given a volume of 4.47 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= π d2h4

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=4 Vπh

=4(5.9)m3

π (2.5m)=1.73m

Therefore, the dimension of the day tank is 1.73 m diameter by 2.5 m length cylindrical tank per Diesel Engine.

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.

20

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)


Computing for the specific velocity rate at both the suction and discharge we simply use the given equation below.

V= 4 Q

π d2

Where:

Q = Volume Flow Rate; m3/min

V = Velocity of Fluid; m/sec

d = Internal Diameter; m

Computing now for the velocity at the suction we have.

V S=4 Qπ d2 =

4 (0.3933 m3/min)(1min /60 sec)π (0.0409 m)2 =5 m /sec

Computing now for the velocity at the discharge we have.

V D=4 Qπ d2=

4 (0.3933 m3/min)(1min/60 sec)π (0.0351 m)2 =6.8 m /sec

Assuming 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 head.

Discharge Head=FL+ZD +V D

2

2 g

21

Where:

FL = Friction Losses for Discharge; 0.75 m

ZD = Elevation from Datum to Discharge; 4.5 m

VD = Velocity Head at Discharge; 6.8 m/sec


2

2 g=0.75 m+4.5 m+

(6.8 m /sec )2

2(9.81m /sec2)=7.6 m

Computing now for the suction head

Suction Head=FL+Z S+V S

2

2 g

Where:

FL = Friction Losses for Suction; -0.65 m

ZS = Elevation from Datum to Suction; -2.5 m due to the location

VS = Velocity Head at Suction; 5 m/sec

Suction Head=FL+Z D+V S

2

2 g=−0.65 m−2.5 m+

(5m / sec )2

2(9.81 m /sec2)=−1.9 m

Based 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

= 7.6 m – 1.9 m

Pump Operating Head = 5.7 m

Assuming that the pump efficiency is 70 %, and then we have.

Pump Supply Power=Qd o H

4,500,000 ŋp

(hp)

Where:

22

Q = Volume Flow Rate; 393.3 Litters/min

ρO = Density of oil; 914 kg/m3 (or 0.914 kg/litters)

H = Pump Operating Head; 5.7 m

ŊP= Pump Efficiency; 70 %

Pump Supply Power=Q ρo H

4,500,000 ŋp

(hp)=393.3 Litters /min (914 kg/m3)(5.7 m)

4,500,000 (0.7)

Pump Supply Power=¿ 0.65 hp or 0.75 hp

Therefore, we are to use 3/4 hp oil pump for the fuel oil transfer.

COOLING WATER SYSTEM

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 =674.58bhp

t 2−t 1

Where:

W = circulating cooling water; Litter/hour

Rated bhp = rated brake horsepower

t1 = inlet water temperature; ° C

t2 = outlet water temperature; ° C

Accordingly, we must first obtain the rated brake horsepower as readily provided in the Diesel Engine type F8-135-SC 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

23

Frederick T. Morse; Table A – 16: Normal Efficiencies of Engine type Synchronous Generators; page 675; at a kilowatt rating of 525 kW with 450 rpm, by interpolation we have 94.3375 %. 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 93.1375 % Net Generator Efficiency.Therefore the rated break horsepower is equal to the equation below.

Rated Bhp= Generator OutputNet Generator Efficiency

Rated Bhp= 525 kW0.931375

Rated Bhp=563.68 kW ( 1.34 hp1 kW

)

Rated Bhp=755.33 hp

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 1,510.66 hp. So the required circulating cooling water is.

W c=674.58bhp

t 2−t 1

W c=674.58( 1,510.66 hp75℃−65℃

)

W c=101,906.1 Litters /hour

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.

WC= (101,906.1 Litters/hr)(1hr/60 min)(1m3/1000 Litters) x (974.86 kg/m3)

WC = 1,655.74 kg/min

24

Therefore, we need 101,906.1 Litters/hour or 1,655.74kg/min for the circulating cooling water for the 2 Diesel Engine Genset units.

Water Jacket Circulating 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 – 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)



Schedule 40

Outside Diameter 2 ½ in (2.875 in)


V= 4 Q

π d2

Where:


(101,906.1 Litters/hr)(1hr/60 min)(1m3/1000 Litters)

Q = 1.7m3/min




V S=4 Qπ d2 =

4 (1.7 m3/min)(1min /60 sec)π (0.0779 m)2 =5.9 m /sec

25


V D=4 Qπ d2=

4 (1.7 m3/min)(1min/60 sec)π (0.0627 m)2

=9.2 m /sec



2

2 g

Where:





2

2 g=0.75 m+4.5 m+

(9.2 m/ sec )2

2(9.81m /sec2)=9.6 m



2

2 g

Where:





2

2 g=−0.65 m−2.5 m+

(5.9 m / sec )2

2(9.81 m /sec2)=−1.4 m



26

= 9.6 m – 1.4 m




4,500,000 ŋp

(hp)

Where:

Q = Volume Flow Rate; (101,906.1 Litters/hr)(1hr/60 min)

Q =1,698.4Litters/min

ρw = Density of water; 1000 kg/m3 (1 kg/litters)



Pump Supply Power=Q ρw H

4,500,000 ŋp

(hp )

Pump Supply Power=1,698.4 Litters /min (1000kg /m3)(8.2m)

4,500,000(0.7)

Pump Supply Power=¿ 4.42 hp or 4.5 hp

Therefore, we are to use 4 1/2 hp water pump for the water jacket transfer for each Diesel Genset unit.

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 @ inlet and 28° C @ 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, Litter/min

QC = Quantity of Circulating Cooling Water; 1,698.4 Litters/min

27

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=QC(tC1−tC 2)(tR1−tR 2)

QR=1,698.4 Litters /min(75℃−65℃)

(34℃−28℃)

QR=2,830.67 Litters /min

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 28° C the density of water is 996.242 kg/m3.

WR= 2,830.67 Litters/min x (1 m3/1000 Litters) x (996.242 kg/m3)

WR= 2,820 kg/min

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.


V= 4 Q

π d2

Where:


(2,830.67 Litters/min)(1m3/1000 Litters)

Q = 2.83m3/min



28


V S=4 Qπ d2 =

4 (2.83m3/min)(1min /60 sec )π (0.0779 m)2 =9.9 m /sec


V D=4 Qπ d2=

4 (2.83m3/min)(1 min/60 sec)π (0.0627 m)2 =15.3m /sec



2

2 g

Where:





2

2 g=0.75 m+4.5 m+

(15.3 m/ sec )2

2(9.81m /sec2)=17.2m



2

2 g

Where:



VS = Velocity Head at Suction; 9.9 m/sec


2

2 g=−0.65 m−2.5 m+

(9.9 m /sec )2

2(9.81 m /sec2)=−1.8 m

29



= 17.2 m – 1.8 m




4,500,000 ŋp

(hp)

Where:

Q = Volume Flow Rate; (2,830.67 Litters/min)

ρw = Density of water; 1000 kg/m3 (1 kg/litters)



Pump Supply Power=Q ρw H

4,500,000 ŋp

(hp )

Pump Supply Power=2,830.67 Litters/min (1000kg/m3)(15.4m)

4,500,000 (0.7)

Pump Supply Power=¿ 13.8 hp or 14 hp

Therefore, we are to use 14 hp water pump for the raw water transfer for each Diesel Genset unit.

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 use.

1 kg air + SH1 + WW + ΔW = 1 kg air + SH2 + WW

30

Deviating for simplicity we have.

ΔW = SH2 – SH1

For heat balance we use.

h1 + WWhfa+ ΔWhf= h2 + WWhfb

Deviating for simplicity we have.

Ww=−∆ W h f +h2−h1

h fa−h fb

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/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

31

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 R.H. 60% and dry bulb temperature of 28° C @ entering point.

SH1 = 0.01425 kg moisture/kg dry air

At R.H. 90% and dry bulb temperature of 34° C@ 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=−∆ W h f +h2−h1

h fa−h fb

32

Ww=−(0.01675 kgmoisture /kgdry air )(75.50

kJkg

)+122.968 kJ / Kg−89.952 kJ /kg

142.38 kJ /kg−117.31kJ /kg

Ww=1.27 kg moisture /kgdry air

Since 2,830.67 kg/min of raw water is given, the air flow is.

Air Flow= 2,830.67 kg /min

1.27 kgmoisture

kgdry air

Air Flow=2,228.87 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.

DB 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 = 2,228.87 kg/min (0.87 m3/kg)

Air Flow = 1,939.12 m3/min

The required make-up water then is.

Make-up Water = Air Flow x ∆W

Make-up Water = 1,939.12 kg/min x 0.01675

Make−up Water=32.4 kg /min (1000

litters

m3)

density of water

Make−up Water=32.4 kg /min (1000

litters

m3)

996.242 kg /m3

Make−up Water=32.522 litters /min

33

Forced Draft Fan:

Based from Kent’s Mechanical Engineer’s Handbook: Volume II: Power Volume, 12 th

Edition, by J. Kenneth Salisbury; Table 13: Skeleton Capacity, Table for Tube Axial Fans; page 1 – 89, we are to choose a fan that would provide a capacity of air flow of 809.54 m 3/min or 28588.64 ft3/min.

We select a fan with the following size.

Fan Size: 60-3

Motor hp: 5.5 hp

Rpm: 570

Capacity: 36,300 ft3/min

Static Pressure ½ in SP

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

Water Requirements = (101,906.1 Litters/hr)(1hr/60 min)+ 2,830.67 Litters/min

Water Requirements = 4,529.105 Litters/min x (1 gallon/3.7854 Litters)

Water Requirements = 1,196.5gpm

Assuming we have water concentration of 3.0 gpm per ft2, then the area of the cooling tower is.

Area of Cooling Tower = 1,196.5gpm / (3.0 gpm/ft3) = 398.8 ft2

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

Side of Cooling Tower = √398.8 ft2= 20ft

So the dimensions of the cooling tower that would service the diesel engines is 20 ft x 20 ft or 6.1 m x 6.1 m.

34

AIR SYSTEM

The purpose of the air system in the power plant is very important, as the prime mover itself is dependent upon combustion. The functions of the intake and exhaust systems are to deliver clean combustion air to the engine and dispose of the exhaust quietly with the minimum loss of performance.

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 563.68 kW (or 755.33hp) then the flow rate of the intake would be.

Q=0.084

m3

minhp

(755.33 hp)=63.45 m3/min

Assuming we have a flow velocity of 800 m/min, then the dimensions of the intake pipe would be.

A=63.45 m3/min800m /min

=0.08m2

d=√ 4 (0.08 m2)π

=0.32m

35

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

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 176; 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 563.68 kW (or 755.33hp) then the flow rate of the discharge would be.

Q=0.224

m3

minhp

(755.33 hp)=184.3 m3/min

Assuming we have a flow velocity of 1500 m/min, then the dimensions of the exhaust pipe would be.

A=184.3 m3/min1500 m /min

=0.12m2

d=√ 4 (0.12 m2)π

=0.39m

The diameter of the exhaust pipe would be 0.39 meter or 390 mm. Therefore, we are to use a pipe with such diameter.

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

36

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.

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 (cu. ft) (gal) (cu. meter)5 3.7 3 20 0.1

7.5 5.6 4 30 0.110 7.5 5 40 0.215 11.2 8 60 0.220 14.9 11 80 0.325 18.7 13 100 0.430 22.4 16 120 0.540 29.8 21 160 0.650 37.3 27 200 0.860 44.8 32 240 0.975 56 40 300 1.1

100 74.6 54 400 1.5125 93.3 67 500 1.9200 149.2 107 800 3350 261.1 188 1400 5.3450 335.7 241 1800 6.8

37

500 373 268 2000 7.6

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 x 2) + (3 m3 x 0.5)

Total Volume = 10.5 m3 = 370.8 ft3

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 litters then we are to use a tank of the next closest value of 11860 litters, 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

LUBRICATION SYSTEM

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 F8-135-SC Diesel Engines with 525 kW generator rating; 1 unit is continuous in operation; another unit for 9 am to 12 am 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 93.1375 % 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= Generator OutputNet Generator Efficiency

Rated Bhp= 525 kW0.931375

38

Rated Bhp=563.68 kW

For a continuous operation, the generated power would be.

Generated Power = 563.68 kW x 24 hours = 13,528.32 kW-hr

Based 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 operate continuously, then the total generated power would be.

Generated Power = 13,528.32 kW-hr + (563.68 kW x 15hr) = 21,983.52 kW-hr

Since 1 gallon of Lubricating oil per 1600 kW-hr generated at full load rating is give as an average consumption for lubrication, for one day consumption, we have.

Oil Consumption = (21,983.52kW-hr) x (1 gallons / 1600 kW-Hr) = 13.7 gallons/day

Similarly with the fuel oil, for 45day consumption, we have.

Oil Consumption = 13.7 gallons/day x 45 days = 616.5 gallons

Dimensions of the Lube Oil Storage Tank:

By conversion.

Volume = 616.5 gallons x (3.7854 Litters/gallons) x (1 x 10-3 m3/1 Litters) = 2.33 m3

Assuming that the tank is filled within 5 minutes as compared for the fuel oil, then.

Volume Flow Rate = 2,330 Litters / 5 min = 466 Litters/min =0.466 m3/min

Using a cylindrical drum for storage with a 1 meter diameter, the length could be obtained and so is the dimension.

Volume=π d2 L4

Length= 4 Vπ d2 =

4 (2.33m3)π (1 m)2 =2.97 m

The dimensions of the lube oil tank would be in 1 m diameter by 2.97 m length.

Lube Oil Transfer Pump:

39

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.

Suction Line Pipe:

Schedule 40

Outside Diameter 1 ½ in (true size of 1.9 in)



Schedule 40

Outside Diameter 1 ¼ in (true size of 1.660 in)


Computing for the specific velocity rate at both the suction and discharge we simply use the given equation below.

V= 4 Q

π d2

Where:





V S=4 Qπ d2 =

4 (0.466 m3/min)(1 min/60 sec)π (0.0409 m)2 =5.9 m /sec


V D=4 Qπ d2=

4 (0.466 m3/min)(1 min/60 sec)π (0.0351 m)2 =8.03 m / sec


40


2

2 g

Where:





2

2 g=0.75 m+4.5 m+

(8.03 m /sec )2

2(9.81m /sec2)=8.5 m



2

2g

Where:





2

2 g=−0.65 m−2.5 m+

(5.9 m / sec )2

2(9.81 m /sec2)=−1.4 m



= 8.5 m – 1.4 m




4,500,000 ŋp

(hp)

41

Where:

Q = Volume Flow Rate; 393.3 Litters/min

ρO = Density of oil; 914 kg/m3 (or 0.914 kg/litters)



Pump Supply Power=Q ρo H

4,500,000 ŋp

(hp)=466 Litters /min (914kg /m3)(7.1 m)

4,500,000 (0.7)

Pump Supply Power=¿ 0.96 hp or 1 hp

Therefore, we are to use 1hp oil pump for the fuel oil transfer.

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 and Month Target Work Output Relevance

October 2012 to November 2012

Load survey for the costumer kilowatt consumption; Gathering and surveying of the environment within the location; Planning and securing necessary permit

Initial work frame for the project; idealization for the project's continuity

December 2012 to February 2013

Computation for the required plant operation which includes the following: Select the engine type, Design of the Generator Set, Design of its Fuel Oil System, Design for its Cooling Water System, Design for its Lubrication System, Design for its Air Handling System, Design for the Engine Dimension, others; Thorough re-evaluation of the plant design; Obtaining contracts and advice from various engineering firms

Initial work framefor the plant's construction and operation; Idealization for the plant layout, operation and maintenance

March 2013 to October 2013

Construction of the plant site; Purchase of equipment; Construction of Administration Building and Others; Installation of equipment

Idealization of the plant site

November 2013 to December 2013

Initial operation; Recalibration of equipmentTesting phase of the project; Checking for flaws and defects

January, 2014 Turnover of Plant operation to the owner/firm Final project Phase

42

GLO S SA R Y

BrakeHorsepower(Bhp):A WorkIndicator;ameasureoftheworkproducedbyanengine, calibratedanddetermined bytheforce exertedonafrictionbrake.

Compressor:Acompressorisa mechanical device thatincreases thepressureofa gaswhile reducingitsvolume.Compressorsaresimilar topumps;both increase the pressureon afluidand both can transport thefluid throughapipe.

CoolingTower:Anequipment thatprovidescoolingtoany processed&heated fluid;Atallopen- topped structure inwhich thesteam producedbyanindustrial process iscondensed

DieselEngine:AnEnginethatusesdieselfuel forcombustion;aninternalcombustionenginethat ignitesdiesel fuelusingcompressionalone,ratherthanusingan electricalsparksuchas sparkplugs.

Kilowatt-hour (kWh):Theelectricalenergyunitofmeasureequaltoone thousandwattsofpower supplied to, ortakenfrom, anelectriccircuit steadilyforone hour.

Lubricants:Substancesusedtoreduce friction betweenbearingsurfaces,orincorporatedinto othermaterialsusedasprocessingaidsinthemanufactureofotherproducts,orusedascarriers ofother materials.

Pump:Apumpisadeviceusedtomovefluids,suchasgases,liquidsorslurries.Apump displacesavolume byphysicalormechanicalaction.Pumpsdonotcreatepressurethey only displacefluid causingaflow.Addingresistancetoflowcauses pressure.

Safety Factor:Theratioofthestrengthofaconstructionmaterialorstructuralparttoamaximum load which itwill have tosupport. It isusedtomultiplya machinefunctionto at least definealimit.

Storage Tank: A storage tank is a container, usually for holding liquids, sometimes for compressedgases(gastank).Thetermcanbeusedforbothreservoirs(artificiallakesand ponds), andformanufacturedcontainers.

43

B IBLI O G R AHY

Avallone, Eugene &Baumeister, Theodore III; “Mark’s Standard Handbook for MechanicalEngineers”; 9thEdition

Morse,FrederickT.; “PowerPlantEngineering(Adaptedto MKS Units)”

Salisbury,J.Kenneth;“Kent’sMechanicalEngineer’sHandbook:Volume2:PowerVolume”;12tthEdition

Raka,A. J., Srivastava, AmitPrakash, Dwivedi,Manish;“PowerPlantEngineering” Stoecker,W.F.& Jones, J.W;“Refrigeration &AirConditioning”; 2ndEdition

PhilippineSocietyofMechanicalEngineers;“ThePSMECodeof1993”;1993RevisionofPSME Code

www.engineeringtoolbox.com

44

Diesel Kim

Documents

design problem

diesel power plant design

subject power plant

power plant engineering

machine foundation design

water system

lubric system

air starting system