Power Plant Design; Analysis Design

Power Plant Design for Sagay, Camiguin1

ACKNOWLEDGEMENT

First of all, we would like to say thank you for giving us the strength and health to

do this project work until it done not forgotten to my our family for providing everything

such as money to buy anything that are related to this work and their advised, which is

the most needed for this project. Internet, books, computers and all that as my source to

complete this project they also supported us and encouraged us to complete this task

so that we will not procrastinate in doing it. Then we would like to thank our professor

for guiding us and our friends throughout this project. We had some difficulties in doing

this task, but he taught us patiently until we knew what to do, he tried and tried to teach

us until we understand what we supposed to do with the project work. Last but not the

least, our friends who were doing this project with us and sharing our ideas they were

helpful that when we combined and discussed together we had this task done.

Power Plant Design / MEP544D2


INTRODUCTION

In a diesel power station, diesel engine is used as the prime mover. The diesel burns

inside the engine and the products of this combustion act as the working fluid to

produce mechanical energy. The diesel engine drives alternator which converts

mechanical energy into electrical energy. As the generation cost is considerable due to

high price of diesel, therefore, such power stations are only used to produce small

power.

Although steam power stations and hydro-electric plants are invariably used to generate

bulk power at cheaper costs, yet diesel power stations are finding favour at places

where demand of power is less, sufficient quantity of coal and water is not available and

the transportation facilities are inadequate. This plants are also standby sets for

continuity of supply to important points such as hospitals, radio stations, cinema houses

and telephone exchanges.



The diesel engine is recognized as the most promising powertrain in the foreseeable

future due to its superior thermal efficiency and reliability. The diesel engine has been

widely used in commercial vehicles, industrial applications and today’s passenger cars

and light-duty trucks.

A diesel engine is a type of compression-ignition engine using diesel fuel. Diesel

engines can be classified into various categories. Understanding the differences and the

unique characteristics of each category of diesel engines is important for diesel engine

system design.

HISTORY

The first man, who had invented the engine with ignition from compression, was E.

Steward. He was interested in engines, what can work without spark plugs. In Steward’s

engine the air was compressed and compressed air was blown into the combustion



chamber. Unfortunately, Steward had not come into mind to test the efficiency of that

type of engines.

Developing the concept of “economy-type heat-engine”, Rudolph Diesel in 1890

invented the engine much more efficient due to high compression ratio. In his book he

suggested to use the powdered coal, but it was difficult in real life – the coal dust has an

abrasive properties and it should be found the way to put it somehow in combustion

chamber. So it was suggested to use the tailing that remains after oil refining in such

engines. So in 1897 Diesel had patented the engine design, later named Diesel engine.

Rudolf Diesel was born in Paris in 1858. His parents were Bavarian immigrants. Rudolf

Diesel was educated at Munich Polytechnic. After graduation he was employed as a

refrigerator engineer. However, he true love lay in engine design. Rudolf Diesel

designed many heat engines, including a solar-powered air engine. In 1893, he

published a paper describing an engine with combustion within a cylinder, the internal

combustion engine. In 1894, he filed for a patent for his new invention, dubbed the



diesel engine. Rudolf Diesel was almost killed by his engine when it exploded. However,

his engine was the first that proved that fuel could be ignited without a spark. He

operated his first successful engine in 1897.

POWER PLANT DESIGN FOR SAGAY

Diesel Power Plant for Sagay, Camiguin produces power in the range of 6.4 MW.

And they are used standby sets for continuity of supply such as hospitals, cinema

theatres and industries.

ADVANTAGES OF A DIESEL ENGINE

More efficient and economical to use.

Fuel vapor is not explosive.

Exhaust gases are less poisonous – less carbon monoxide.

Greater lugging power and torque.

Engines are durable and if properly cared for will maintain their economy.



Fuel is less volatile – no vapor lock problems

Can use a variety of fuels and mixtures.

DISADVANTAGES OF DIESEL ENGINE

The fuel in diesel engine is ignited by the heat of the compressed air.

HPFP of diesel engine is extremely unreliable unit.

The exhaust filter is warming up in the flow of exhaust gas.

it is expensive

Noise and vibration till the latest times could not be separated from the words

“Diesel engine”.

ADVANTAGES OF DIESEL POWER PLANT



The design and layout of the plant are quite simple.

It occupies less space as the number and size of the auxiliaries is small.

It can be located at any place.

It can be started quickly and it can pickup load in a short time.

There are no standby losses.

DISADVANTAGES OF DIESEL POWER PLANT

The plant has high running charges as the fuel (diesel) used is costly.

The plant doesn’t work satisfactorily under overload conditions for a longer

period.

The plant can only generate small power.

The cost of lubrication is generally high.

The maintenances charges are generally high.

WHY WE CHOOSE DIESEL POWER PLANT



Purpose

Diesel power plants produce energy though the combustion of chemical fuel, in

most cases diesel derived from petroleum, into mechanical energy. This energy

is then used to power an alternator which in turn generates electricity. Diesel is

preferred to other fuel types as these engines have a higher thermal efficiency

than other commercial generators of equivalent size.

Process

Most modern generators harness mechanical energy through the process of

electromagnetic induction. In this system, the mechanical energy produced by

the diesel engine moves an electrical conductor, such as a magnetically charged

wire, in a magnetic field. The movement of the conductor creates a difference in

voltage between the two ends of the charged wire, creating a flow of electric

charges and thereby generating electricity.

LAYOUT OF DEISEL POWER PLANT



ESSENTIAL ELEMENTS OF DIESEL POWER PLANT

I. DIESEL ENGINE



Diesel engine is a compressor ignition (CI) engine.

The two – stroke cycle engine is more favored for diesel power plants.

The air required for the diesel engine is drawn through the air filter form the

atmosphere and compressed inside the cylinder.

The fuel (diesel) from the diesel engine is drawn though a filter from the all-day

tank and injected into the cylinder through fuel injector.

Because of the high temperature and pressure of the compressed air, the fuel

burns and the burn gases expand to do work on the moving part inside the

cylinder called piston.

This movement of the piston rotates a flywheel and the engine is directly couple

to electric generator.

The gases after expansion inside the cylinder are exhaust into the atmosphere

and passes through a silencer in order to reduce the noise.

II. STARTING SYSTEM



The starting motor will crank the engine. The starting motor will spin the engine at

a high enough rpm to allow the engine’s compression to ignite the fuel and start

the engine running.

The engine will then accelerate to idle speed. When the starter motor is

overdriven by the running motor it will disengage the flywheel.

III. FUEL SUPPLY SYSTEM

It consists of storage tank, strainers, fuel transfer pump and all day fuel tank. The

fuel oil is supplied at the plant site by rail or road. The oil is stored in the storage

tank. From the storage tank, oil is pumped to smaller all day tank at daily or short

intervals. From this tank, fuel oil is passed through strainers to remove

suspended impurities. The clean oil is injected into the engine by fuel injection

pump.

IV. AIR INTAKE SYSTEM



This system supplies necessary air to the engine for fuel combustion. It consists

of pipes for the supply of fresh air to the engine manifold. Filters are provided to

remove dust particles from air which may act as abrasive in the engine cylinder.

V. EXHAUST SYSTEM

This system leads the engine exhaust gas outside the building and discharges it

into atmosphere. A silencer is usually incorporated in the system to reduce the

noise level.

VI. COOLING SYSTEM

The temperature of the burning fuel inside the engine cylinder is in the order of

15000 0C to 20000 0C. In order to lower this temperature water is circulated

around the engine.

The water envelopes (water jacket) the engine. The heat from the cylinder,

piston, combustion chamber etc., is carried by the circulating water.

The hot water leaving the jacket is passed through the heat exchanger.



The heat from the heat exchanger is carried away by the raw water circulated

through the heat exchanger and is cooled in the cooling tower.

VII. LUBRICATING SYSTEM

The system minimises the wear of rubbing surfaces of the engine. It comprises of

lubricating oil tank, pump, filter and oil cooler. The lubrication oil is drawn from

the lubricating oil tank by the pump and is passed through filter to remove

impurities.

LOCATION MAP



LOCATION



Sagay, Camiguin

Sagay is a Philippine municipality. It is located in the province Camiguin in Region

X Northern Mindanao which is a part of the Mindanao group of islands. The municipality

Sagay is seated about 16 km south of province capital Mambajao and about 732 km

south-south-east of Philippine main capital Manila. The geographic coordinates of

Sagay are 9° 6' 21'' N, 124° 43' 29'' E.

Administratively the Municipality of Sagay is subdivided into 9 barangays. One forms

the center of the city wheras the other 8 are in the outlying areas. Some of them are

even several kilometers away from the center of the Municipality.

According to the 2007 census, Sagay has a population of 11,198 residents and is part

of the big group of 1073 cities and municipalities in the Philippines which have more

than 10.000 residents but did not reach 50.000 population yet. Based on the number of

its inhabitants Sagay is number 1483 of the most populous cities of the Philippines and

at 424 in Mindanao group of islands and at 4 of the most populous cities of province

Camiguin. With an area of 44.13 km² Sagay occupies a relatively small urban area.


http://www.philippine-islands.ph/en/city_of_manila-first_district-philippines.html

http://www.philippine-islands.ph/en/mambajao-camiguin-philippines.html

http://www.philippine-islands.ph/en/mindanao_islands-philippines.html

http://www.philippine-islands.ph/en/northern_mindanao-philippines.html

http://www.philippine-islands.ph/en/camiguin-philippines.html


Accordingly, there is a high population density. In Sagay, by average, 253.75 people

live in one square kilometer. With this value, Sagay is only number 105 in Mindanao

and is nationally ranked 728th of the most densely populated cities in the Philippines.

According to the Philippine income classification for provinces, cities and municipalities

Sagay is a 5th class municipality. The urbanization status of Sagay is classified as

partly urban.

Among the bigger cities and municipalities in the neighborhood of Sagay there

areCagayan De Oro City (Misamis Oriental) 69 km south, Gingoog City (Misamis

Oriental) 55 km south-east, Manolo Fortich (Bukidnon) 83 km south, City Of

Cabadbaran (Agusan Del Norte) 88 km east, Butuan City (Agusan Del Norte) 90 km

east, Balingasag (Misamis Oriental) 40 km south, Tagoloan (Misamis Oriental) 63 km

south, Buenavista (Agusan Del Norte) 76 km east, Talakag (Bukidnon) 98 km south as

well as 50 km south of Sagay the municipality Jasaan (Misamis Oriental).


http://www.philippine-islands.ph/en/misamis_oriental-philippines.html

http://www.philippine-islands.ph/en/jasaan-misamis_oriental-philippines.html

http://www.philippine-islands.ph/en/bukidnon-philippines.html

http://www.philippine-islands.ph/en/talakag-bukidnon-philippines.html

http://www.philippine-islands.ph/en/agusan_del_norte-philippines.html

http://www.philippine-islands.ph/en/buenavista-agusan_del_norte-philippines.html


http://www.philippine-islands.ph/en/tagoloan-misamis_oriental-philippines.html


http://www.philippine-islands.ph/en/balingasag-misamis_oriental-philippines.html


http://www.philippine-islands.ph/en/butuan_city-agusan_del_norte-philippines.html


http://www.philippine-islands.ph/en/city_of_cabadbaran-agusan_del_norte-philippines.html

http://www.philippine-islands.ph/en/city_of_cabadbaran-agusan_del_norte-philippines.html

http://www.philippine-islands.ph/en/bukidnon-philippines.html

http://www.philippine-islands.ph/en/manolo_fortich-bukidnon-philippines.html



http://www.philippine-islands.ph/en/gingoog_city-misamis_oriental-philippines.html


http://www.philippine-islands.ph/en/cagayan_de_oro_city-misamis_oriental-philippines.html


Municipalities Total Population Percentage Male Female

Camiguin 74,232 100 37,847 36,385

Catarman 15,368 20.7 7,864 7,522

Guinsiliban 5,092 6.9 2,620 2,472

Mahinog 12,592 17.0 6,368 6,224

Mambajao (Capital) 30,806 41.5 15,657 15,149

Sagay 10,356 14.0 5,338 5,018

Guinsiliban is 6.9% of total population of Camiguin therefore we can assume that out of

14,735 Occupied Housing Unit there are 1002 single houses which represents the

majority of the building structure on Guinsiliban and a household population of 1023.

Demographic Data:

Total No. of Population: 5,092



Household Population: 1023

Structures:

Group A Group B

Single House: 1002 Multi- Unit Residential: 3

Duplex: 6 Commercial/Industrial/Agricultural: 1

GROUP A



GROUP B

Total Power Consumption Table



Time GROUP ANo. of

ConsumerPower

Consumption GROUP BNo. of

ConsumerPower

ConsumptionTotal Load (w-

hr)1 1632960 2016 810 16000 8 2000 16489602 1632960 2016 810 24000 8 3000 16489603 1632960 2016 810 24000 8 3000 16489604 2237760 2016 1110 16000 8 2000 22537605 2136960 2016 1060 24000 8 3000 21609606 2136960 2016 1060 91200 8 11400 22281607 1532160 2016 760 94000 8 11750 16261608 2056320 2016 1020 102000 8 12750 21583209 2056320 2016 1020 110000 8 13750 2166320

10 2378880 2016 1180 32000 8 4000 241988011 1854720 2016 920 28000 8 3500 188272012 1854720 2016 920 61200 8 7650 191592013 1854720 2016 920 48000 8 6000 190272014 1310400 2016 650 16000 8 2000 132640015 1310400 2016 650 24000 8 3000 133440016 1310400 2016 650 29200 8 3650 133960017 1532160 2016 760 29200 8 3650 156136018 2136960 2016 1060 37200 8 4650 217416019 4354560 2016 2160 53200 8 6650 440776020 4677120 2016 2320 37200 8 4650 471432021 2459520 2016 1220 53200 8 6650 251272022 2459520 2016 1220 37200 8 4650 249672023 1632960 2016 810 64000 8 8000 169696024 1632960 2016 810 32000 8 4000 1664960

Total Load

(w-hr / 24hrs.)

49815360 w-hr

1082800 w-hr

Total Load (w-hr / 24hrs.) 50898160 w-hr

Design Overview

Peak Load = 2357.16 kW, 2.35mW



Plant Capacity: 3200 kW, 3.2mW

No. of Engines: 5

Engine Capacity Number of Hours of Operation/day

Unit 1 --- 800 kW 18hours/day




Unit 5 --- 800 kW Reserve

Schedule of Engine Operation

Time of Operation Engine Operating Time Interval

12AM-6AM Unit 1,2 and 3 6 hours

6AM-12NN Unit 2,3 and 4 6 hours

12NN-6PM Unit 1,4 and 2 6 hours



6PM-12AM Unit 3,4 and 1 6 hours

Each Unit has a 6 straight hours break

DESIGN FOR MACHINE FOUNDATION

For 800 kW Generator Set (Per Unit 1,2,3,4 and 5)

Mixture for Concrete Foundation:

Use 1:3:5 concrete mixture ratio (from PPE by F.T. Morse, Table 4-1 p.90)

Soil Bearing Pressure:



Use 50-98 tones / m2 for common brick masonry

(from PPE by F.T. Morse, Table 4-4 p. 105)

Soil Bearing Pressure (Sb) = 50tonnes

m2x1000kg1 ton = 50,000 kg/m2

Weight of Foundation

Wf = e x We x √n

Where:

Wf = weight of the foundation, kgs

We = weight of the engine, kgs

N = engine speed, RPM

Use e = 0.11 (from PSME code, Table 2.4.2.3 (4), p.11)

Wf = 0.11 x 7897 kg x √1800 rpm = 36, 859.21 kg



Volume of Foundation

Vf = W fρc

Where:

Vf = Volume of foundation [m3]

ρc = density of concrete = 2406 kg/m3

Vf = 36,859.21kg

2406kg /m3 = 15.32 m3

Depth of Foundation

h = Vf

wf x Lf

Where:

Hf = depth of foundation [m]

Lf = length of foundation [m]



Wf = width of the foundation [m]

Length of the Foundation

Lf = Lb + 10% Lb

Where:

Lb = length of bedplate [m]

Le = length of engine [m]

Lb = Lb + 6 in x 25.4mm1∈¿¿ = 4,267 mm + 6 in x

25.4mm1∈¿=¿¿5,047.4 mm

Lf = 4,419.4 mm + (0.10)(4,419.4 mm) = 4,861.34 mm

Lf = 4,861.34 mm x 1m1000m

= 4.86 = 5 m

Width of the Foundation



Wf = Wb + 10%Wb

Where:

Wb = Width of bedplate

We = width of the engine [m]

Wb = Wb + 6 in x 25.4mm1∈¿¿ = 2, 083 mm + 6 in x

25.4mm1∈¿¿ = 2, 235.4 mm

Wb = 2,235.4 mm + (0.10)(2,235.4 mm) = 2,458.94 mm

Wf = 2,458.94 mm x 1m

1000mm = 2.46 m = 2.5 m

h = 15.32m3

5mx2.5m = 1.22 = 1.25 m

Soil Stress

Soil Stress = Wf +Wbwf x lf



Soil Stress = 36,859.21 kg+7,897kg

(5m)(2.5m) = 3,580.50 kg/m2

Foundation Materials:

Concrete Mixture Ratio = 1 : 3 : 5

X + 3x + 5x = 15.32 m3

9x = 15.32 m3

X = 15.32 m3 / 9

X = 1.70 m3

Absolute Volume Material = Specific Weight of Material / (Bulk S.G.)(Sp. Weight of

Water)

For Cement:

1 x 6.2 x 1.70 m3 = 10.54 m3



Absolute Volume Material = 94

lbbag

(1)

(3.15)(62.4 lb / ft3) = 0.48 ft3/bag x

1m3

(3.28 ft)3= 0.014 m3/bag

No. of bags = 10.54m3

0.014m3/bag = 753 bags

3 x 0.52 x 1.70 m3 = 2.65 m3


lbbag

(3)

(2.64)(62.4 lb / ft3) = 2 ft3/bag x

1m3

(3.28 ft)3= 0.057 m3/bag

For Gravel:

5 x 0.86 x 1.70m3 = 7.31m3


lbbag

(5)

(2.66)(62.4 lb / ft3) = 2.89 ft3/bag x

1m3

(3.28 ft)3= 0.082 m3/bag

No. of Bags = 7.31m3

0.082m3/bag=¿89 bags

For Reinforcing Bar:

Using 14 mm diam. Rebars



WRB = 1%Wf = (0.01)(36.859.21 kg) = 368.59 kg

Weight of Rebar/pc = density of steel x VRB

Weight of Rebar/pc – 7800 kg/m3 x (π/4)(14 mm/1000m)2 (6.1m) – 29.3 kg

No. of Reinforcing Bars = W eig h t of RebarWeigh t of Rebar / pc =

368.59kg29.5kg

=¿12.58 = 13 pcs

Flexure formula

Fb = mxl

Eccentricity from mid-base

Y1 = 1/2h = ½ (1.25m) = 0.625m

Y2 = 1/3h = 1/3 (1.25m) = 0.42m

A1 = Lf = h = (5m) (1.25m) = 6.25m2

A2 = ½ Lf x b

Where:



B = 2(We+Wf )

50,000kg/m3(Lf ) = 2(7,897kg+36,859.21kg )

(50,000kg /m3)(5m) = 0.36m

If b < Wf, then Wf = b; used b = Wf = 2.5 m

A2 = ½ Lf x b = ½ (5m)(2.5m) = 6.25m2

∑A = A1 + A2 = (6.25 + 6.25) m2 = 12.5m2

∑AY = A1Y1 = A2Y2 = [(6.25) (0.625) + (6.25) (0.42)] m3 = 6.53 m3

C = ∑AY∑ A

= 6.53m3

12.5m2 = 0.52 m

For Bolts:

Diameter = 1/8 x (bore) = 1/8 x (150mm) = 18.75 mm

Length = 7/8 x (stroke) = 7/8 x (160mm) = 140 mm

Use L = 30D (from ASME code)



L = 30 (18.75 mm) = 562.5 mm

No. of bolts = m

Tbolts

Where:

Tbolts = SdπD3

16

From Table AT 7 – DME by V.M. Faires

Material: AISI 8630 (for connecting rods, bolts, shapes)

Sy = 100ksi = 100,000psi; Fy = 7 (max. for Shock)

Sy = 0.5(100,000)

7 = 7,142.86 psi x

101.325 kpa14.7 psi

= 49,234.69 kpa

Tbolts = (49,234.69kn/m3)(π )(0.01875m)3(1000N

1KN)

16 = 20.28 N.m

No. of Bolts = 1,368.81kg−mx 9.81N

1kg20.28N .m

= 662.13 = 663 bolts



Design for Fuel Tank

For 800 kW Generator Set (Per Unit 1, 2, 3, 4, and 5)

Type of oil: Diesel Fuel Oil

Specific Gravity = 0.917 @ 600F

(From Power Plant Theory and design by P.J. Potter, Table 5-4, and p.186)

SGf = ρf

ρH 20 ; ρf = SGf x ρH 20 = 0.917 x

1000kg

m3 = 917 kg/m3

Generator Output (EP) = Fuel Consumption

BP

Where:

BP = BPηa

ηg = 97.8% (For 1800 rpm & 494.73 kW Ave. Load)

(From Power Plant Theory and Design by P.J. Potter, figure 9-27, p.445)

BP = 800kW0.978

= 818 kW



Specific Fuel Consumption

= 224.5L /hr x 1m3

1000 Lx 917kg /m3

818kW = 0.25 kg / kW hr

Plant Operation = 24 hrs/day

Engine Operation Hrs/Day = 18 hrs/day

Expected Fuel Delivery Schedule = Every 15 days

% Rated Capacity

= OperatingHours /day

24hrs x 100% =

18hrs24hrs

x 100% = 75%

From PPE by F.T. Morse, Fig. 6-15, p.164



Max. Fuel Consumption = 0.25 kg/kW-hr

Min. Fuel Consumption = 0.21 kg/kW-hr

Volume of Day Tank

VDT = mFρF

Where:

mF = daily fuel consumption [kg/day], ρF = density of fuel = 917 kg/m3

mF = max. fuel consumption x BP x engine operating hrs/day

= (0.25 kg/kW-hr) (181kW) (18hrs/day)

= 3681 kg/day VDT = 3,681kg/day917 kg/m3

=¿4.01 m3/day

Dimension of Day Tank

VDT = π4

DDT2 H



DDT - 3√ 2VDTπ (From the above equation) - 3√ 2(4.01m3)π

= 1.37m

Assume:

HDT = 2DDT = 2(1.37m) = 2.74m

Thickness of fuel Tank

TDT = PT xD DT

2xSYFSY

x n

Where:

PT = Pressure inside tank = HDT x YFuel

YFuel = 8.996 kN/m3

PT = 2.74m x 8.996 kN/m3 = 24.65 kN/m3 or kPa

Sy = Tensile Yield = 35,000 psi (from DME by V.M. Faires, Table AT 4, p. 568)

F.S.y = Design factor of Safety

F.S.y = 3(for stainless steel from DME by V.M. Faires Table 1.1, p.20)



N = 75%

TDT = 24.65kPax 1.37mx

14.7 psi101.325kPa

2 x35,000 psi

3x0.75

= 0.3 mm; use 1 mm thickness

Storage Tank for 30 days operation

VST = VDT = x 30 days/month = 4.01 m3/day x 30 days/month = 120.3 m3/month

Dimension of storage tank

VST = π4

DST2 H

DST = 3√ 2VSTπ (From the above equation) - 3√ 2(120.3m3)π = 4.25m

Assume:

HST = 2DST = 2(4.25 m) = 8.5m

Material for fuel tank: AISI No. 321 (Stainless steel)



Thickness of fuel storage tank

TST = PT xDST

2xSYFSY

x n

Where:

PT = pressure inside tank = HST x YFuel

YFuel = 8.996 kN/m3

PT = 8.5m x 8.996 kN/m3 = 76.46 kN/m2 or kpa

SY = Tensile Yield = 48,000 psi (from DME by V.M. Faires, Table AT 7, p.576)

F.S.y = 2(for stainless steel from DME by V.M. Faires Table 1.1, p.20)

n = 75%

TST = 76.46kPa x4.25m x

14.7 psi101.325kPa

2 x48,000 psi

2x 0.75

= 1.31 mm

Transfer Pump from Fuel Storage Pump to Day tank



Assumption:

Desired operating time for fuel pump = 1 hr/day

ηP = 72%

Power input for unit 1, 2, 3, 4, and 5

EPi = QYfuelTDH

ηp

Where:

EPi = electrical power input [kW] or [hp]

Yfuel = 8.996 kN/m3

THD = total dynamic head [m]

TDH = Z2 – Z1 + P2−P1Yfuel

+ V 22−V 12

2g

TDH = (2.74)-(-8.5)m = 11.24 m

Q = volume flow rate [m3/s]



Q --VDTt

Where:

VDT = volume of fuel at day tank [m3/s]

t = time of pump operation [sec]

Q --4.01m3/day1hrday

x 3600hrs=¿

0.00111 m3/s

EPi = (0.00111m3 /s)(8.996 kN /m 3)(11.24m)

0.72 = 0.16 kW x

1hp0.746kW

= 0.21 hp

1 hp is used for unit 1 transfer pump

VARIABLE LOAD CALCULATIONS

Plant Capacity = Peak Load + Peak Load (20)

= 2357.16kW + 471.432 kW = 2828.592 kW

(we use 3200kW from catalog 800kW x 4 genset)



Reserve over peak = plant capacity – peak load

= (3200kW – 2357.16kW) = 842.84kW

Average Load = Total load (kW−hrs )

dayNo .of hours day

= 25449.08kW−hrs /day

24 hrsday = 1060.378333kW

Capacity Factory - Average LoadPlant Capacity

- 1060.378333kW

3200kW = 33.14%

Annual capacity factor = Annual kW−hrs

kW Plant capacity x 8760 = 9,288,914kW−hrs year

3200 kW x8760 = 33.14%

Load Factor - Avegare LoadPeak Load

- 1060.378333kW2357.16kW

– 44.99%

Demand Factor – Peak load

Plant capacity - 2357.16kW3200kW

– 73.66%

Plant Factor = Average Load

Rating Equipment Supplying t he load =

1060.378333 kW32000kW (catalog) = 33.14%



ENGINE APPLICATION DATA

Engine Specifications

Manufacturer Mitsubishi

Engine Model # S12A2 – Y2PTAW-2

Engine Type 4 Cycle, 12 Cylinder

Induction System Turbochanged, Inter Cooler

Displacement, L (in3) 33.9 (2071)

EPA Emission Level Tier 2

HP at Rated Speed BHP (KW) 1207 (900)

Rated RPM 1800

Bore and Stroke in (mm) 5.19 x 6.30 (150 x 160)

Compressor Ratio 15.3:1



Air Filter Type Dry

Govermor Type/Model Proact2

Govermor Manufacturer Woodward

Freq Reg NL to FL Isochronous

Freq Reg Steady State +/- 0.25%

Engine Lubrication System

Oil Pan Capacity gal (L) 26.4 (100.0)

Oil Pan w/ Filter 31.7 (120.0)

Oil Filter Quantity 4

Oil Cooler Water Cooled

Recommended Oil 15W – 40

Oil Press Psi (kPa) 57 (393)



Engine Cooling System

Genset Max Ambient Temperature 113 (45)

Engine Coolant Cap qt (L) 105.7 (100.0)

Engine + Radiant System Cap qt (L) 402.0 (380.4)

Water Pump Type Centrifugal

Coolant Flow gpm (Lpm) 291 (1101.4)

Charge Cooler Flow gpm (Lpm) 124 (469.3)

Heat Rejected to Cooling Water

@ Rated kW: BTU/min (kW) 20418 (358.9)

Heat Rejected to Charge Cooler


Heat Rejected to Ambient Air




Max. Restriction of Cooling Air in

H20 (kPa) 0.5 (0.124)

Engine Exhaust System

Exhaust Manifold Type Dry

Exhaust flow @ Rated kW cfm (c-mm) 8192 (232)

Exhaust Temp. (Dry manifold) 0F (0C) 953 (497)

Max. Back Pressure InH20 (kPa) 23.6 (5.9)

Exhaust Outlet Diameter in (mm) 8.35 (212)

Exhaust Outlet Type JIS200A (approx. 8”)



Engine Electrical System

Changing Alternator Volts dc 24

Changing Alternator Amps 25

Grounding Polarity Negative

Started Motor Volts dc 24

Battery Recommendations

Battery Volts dc 24

Min Cold Cranking Amps 1100

Quantity Required 2

Ventilation Requirements

Cooling Airflow scfm (cmm) 40042 (1134)

Combustion Airflow cfm (cmm) 3107 (88)



Heat Rejected to Ambient

From Engine BTU/min (kW) 4375 (77)

From Alternator BTU/min (kW) 2275 (40)

Recommended Free Area Intake

Louver Size ft2 (m2) 87.0 (8.09)

Engine Fuel System

Recommended Fuel # 2 Diesel

Fuel Line at Engine

Supply Line Min ID in (mm) 0.75 (19)

Return Line Min ID in (mm) 0.75 (19)

Fuel Pump type Engine Driven

Fuel Pump Max Lift ft (m) 3 (1)



Fuel Flow to Pump gpm (Lph) 148 (560.2)

Fuel Filter

Secondary Filter 2 µm

Secondary Water Separator Not Included

Primary Filter Optional

Primary Water Separator Optional

Fuel Consumption – Standby Rating

100% Load gph (Lph) 65.2 (246.5)

75% Load gph (Lph) 46.8 (177.1)



50% Load gph (Lph) 32.2 (121.9)

25% Load gph (Lph) 19.3 (73.1)

Fuel Consumption – Prime Rating

100% Load gph (Lph) 59.3 (224.5)

75% Load gph (Lph) 42.6 (161.2)

50% Load gph (Lph) 29.3 (110.9)

25% Load gph (Lph) 17.6 (66.6)

Engine Output Deratings – Standby

Rated Temp 400C

Rated Altitude 1500 m

Max Altitude 5000 m



Temperature Derate -5% / 100C

Altitude Derate -1% / 100 m

Alternator Specifications

Alternator Type 4-Pole Rotating Field

Exciter Type Brushless

Excitation PMG

Insulation per NEMA MG1

Material Class H



Standby Temp Rise 150 0C

Prime Temp Rise 125 0C

Lead Connection 12 Lead, Reconnect able

Stator Pitch 2/3

Amortisseur Winding Full

Bearing Single Double Shielded

Drive Coupling Flexible Disk

Unbalance Load 20% of Standby Rating

Automatic Voltage Regulator

PMG Std MX321

Voltage Regulation No Load to Full Load

PMG Regulator +/- 0.5%

Load Acceptance 100% of Rating



Subtransient Reactance

480V, Per Unit 18%

TIF (1960 Weighting) <50

Line Harmonics 30% Max. Voltage Dip

Alt @ 480V SkVA HC1634G – 311 – 2350

Alt @ 480V SkVA HC1634H – 311 – 2680

Genset Controller Specifications

Baldor Intelligent NT Features

Large back-lit graphical LCD display

64 x 128 pixel resolution

Sealed Membrane Panel to IP65



Push Buttons for Simple Control

Start, Stop, Fault Reset, Horn Reset, Mode

Page, and Enter Keys

Display Metering and Protection

Oil Pressure Warning / Shutdown

High / Low Coolant Temperature Warning

High Coolant Temperature Shutdown

Low Coolant Level Shutdown

Low Fuel Level Warning / Shut Warning

Over Speed Protection

Battery Voltage Under / Over Warning

Running Hour Meter

Generator Under / Over Volts Warn / Shutdown



Generator Under / Over Freq Warn / Shutdown

Generator over Current Shutdown

Generator Output Metering for V1, V3, I1 – I3,

Hz, kW, kWh, kVAr, kWAh

User Configurable Inputs and Output

Up to 500 Event Based History Records

Integrated PLC Programming Functions

Interface to Remote Display or

Remote Annunciator

Controller Capable of Both Single or Multiple

Gensets Operating in standby or

Parallel Modes



Addition Standard Genset Features

Structural Steel Sub-Base

Sub-Base Lifting Eyes

Unit Mounted Radiator

Engine Mounted Fan

Fan Guard

Battery Changing Alternator

Battery Rack and Cables

Unit Mounted Control Panel

Spin-On Filters for Oil and Fuel

Enamel Set-Operator / Maintenance Manual

Factory Tested Prior to Shipment

Limited Warranty

Optional Agency Approvals



UL2200 (Review Option Availability)

NFPA110 (Request Remote Annuciator)

Weight and Dimensions (Open Unit)

Weight – Wet lb(Kg) 17410 (7897)

Overall Dimension Length x Width x Height

Inches 168 x 82 x 93

mm 4267 x 2083 x 2362

Note: Drawing is provided for reference only. Use engineering outline for installation

planning.



PERSPECTIVE VIEW


Equipment

Quantit

y

800 kW Diesel Genset

(IDLC 800-2M)

5

Fuel Transfer Pump 1hp 5

Cooling Tower Pump

0.11hp

10

Cooling Water Fan

0.27hp

10

Materials Quantity

Cement 3675

Gravel 435

Anchor Bolts 1/8 x 7/8 3315

Renforcing Bars 14mm

x 20 ft

65

Aluminum Brass Tude

3/4 in

120






ECONOMICS

Design of fuel storage tank assumes that diesel to be used is 30ºAPI with a temperature

of 26.67ºC.

SG = ρfuelρwater

SG@ 15.6 0C = 141.5

30+131.5 = 0.8761

SG@ 26.6 0C = [(0.8761) (1- 0.00072) (26.67-15.6)]



SG @ 26.6 = 0.86911

ρfuel = (0.86911) (62.4) lb/ft3

ρfuel = 54.23 lb/ft3

HHV = 41, 130 + 139.6 (300API)

Qh = HHV = 45318 kJ/kg = 47.65 kw – hr/gal

QL = LHV =27.39 kw – hr/gal

Ave. = 47.65+27.39

2 = 10.13

Daily fuel consumption = 61,50010.13

= 6071.07 gal/day



Storage tank for 1 month

= Daily Consumption x 30 Days/month

= 6071.07 x Days/month

= 182132.28 gal/month

Monthly fuel delivery

Fuel delivery in liter

1 gal = 3.7854 liter

Capacity of trailer truck = 72, 000 liters

One month = (182132.28 gal /mont h)(3.7854)

72000 = 10 tank per month

ECONOMIC STUDIES

Capacity Factor = 52.08



Ave. Load of Power Plant = 125000 kw –hr

Monthly fuel consumption = 182132.28 gal/month

Semi – fuel consumption = 182132.28

2 = 91066.14

Assuming the diesel fuel is supplied by the government at a whole sale price @ .75

per/gal cost of fuel per year

= 182132.28 gal/month x 20 month x .75

= Php 1639190.52

POWER PLANT ECONOMICS

Power Plant Cost:

Cost per Kw installed Php = 500.00

Cost per unit 2000 Kw x 500.00 Php = 1,000,000



Total Cost per unit installed 5 x 500,000 Php = 2,500,000

Building Php = 400,000

Real State Php = 180,000

Equipment Php = 150,000

Engineering Fees Php = 70,000

Total Plant Cost Php = 3,300,000

Cost of Transmission Lines:

Primary line Php = 140,000

Secondary line Php = 800,000

The annual costs of Operation are as follows:

Francise and Publicity Php = 2.4 customer



Assuming 25, 000 customer Php = 60,000

Supervision and management Php = 30,000

Maintenance and repair Php = 10, 000

Oil waste and supply Php = 20,000

Fuel used Php = 2617.48

Maintenance in secondary lines Php = 70,000

Labor:

For Generation Php = 80,000

For Distribution Primary Php = 20,000

For Distribution secondary line Php = 20,000

For accounting Php = 60,000



Interest Rate = 6% Taxes. 5% insurance

Life = 50 years

Primary distribution = 20 years

Secondary distribution = 20 years

Salvage Value:

Power Plant = 10% of the first cost

Primary line = 20% of the first cost

Secondary = 20% of the first cost

Fixed element annual cost:

Capital cost - cost of the PLANT - Cost of the Primary line

- 2,470,000 – 70,000

- 2,540,00



Primary Salvage Value:

- 20% of the primary line

- .20 x 70,000

- Php 14, 000

From Equation (3-1) p. 71, PPE by F.T. Morse

Straight line annual depreciation reserve

= F−Sn

Where:

P – The principal sum

S – The final salvage value

N – Terms in years



Plant annual Depreciation Reserve:

= 2 ,470,000−247 ,000

50 years

Primary Line annual appreciation reserve

= 140,000−28,000

20 years = Php 5,600

Interest – taxes and insurance of the capital cost:

= (0.06 + 0.05) x 3,300,000 = 363,000

Plant maintenance and repair Php = 10,000

Supervision and management Php = 30,000

Plant Depreciation Php = 88,800

Primary line depreciation Php = 5,600

Total = Php 677,920



Energy Element:

Oil waste supply Php = 20,000

Labor for generation Php = 80,000

Maintenance on primary lines Php = 10,000

Labor on distribution Php = 20,000

Fuel consumption Php = 2617488

Total = Php 2,746,954

Customer Element:

Secondary line Salvage value:

= 20% of secondary line first cost

= .20 x 800,000 = Php 160,000

Secondary line annual depreciation reserve:



= 800,000−160,000

20 = 32,000

Interest, Taxes and Insurance of Secondary Line

= (0.06 + 0.05) + 800,000 = Php = 88,000

Publicity and franchine Php = 70,000

Labor for accounting Php = 60,000

Secondary line depreciation Php = 18,000

Total = Php 236,000

A) Cost per kW installed

= cost of plant+ primary line+secondary line

Kwinstalled =

= 3,300,000+140,000+800,000

5000 = Php 848 kW



B) Cost per kW – hr produced:

= ¿element−energy elementannual stationoutput

= 677920−2746954

1500kw−hrday

x30daymont h

x12mont hyear

= 0.09 cents kW –hr produced

C) Cost per kW – hr delivered:

= FE−EE−¿

( I−elec .transmission loss)(annual kw−hr output )

Allow 20% Elec. Transmission loss

= 22747954−677920−236000

(19)(22200 x103) = 0.05 cents kW-hr

Investors Profit:

Assume annual profit of capitalization over and store interest to be 8.0%

Capitalization = cost of the plant, primary and secondary distribution system

= 3,300,000 + 140,000 + 800,000



= 4,240,000 x .08 = Php 339,200

Straight Line Meter Rate:

Assuming the various element lost

Fixed element Php = 677920

Energy Php = 2,746,954

Customer element Php = 236,000

Profit Php = 339,200

Total = Php 4,000,074



Rate = Cost of annual productionannual stand by output

= 4,000,07444,400,000

= 0.09 cents kW/hr

Three Charge Rate

For this, let the profit element be placed with fixed element and customer

element in proportion to the capital (element) investment for customer element.

= 3,300,000 + 140,000

= 3,440,000

Demand charge = fixed element + portion of profit

= 677920+25653165

x339200 = 952,816.68

Assume over all diversity factor 4.5

Peak power plant demand 5,000 kw

Customer peak demand 5,000 x .45 + 22, 500 kW



Unit demand change 952816.6822540

= 27.68 kW –

hr

Maximum demand per month = 27.6812

= 2.308 per month

Energy element was:

= 2747954 for 24750 kW – hr / year

Hence the unit energy charge = customer element + remainder of profit

= 236,000 + 800,003,300,000

x 33920

= Php 244223.03



Dividing equally between 25,000 customers

The result is 244223.0325,000

= 9.76

= 9.7612

= Php .814 / month

COMPUTATION FOR TRANSFORMERS

Establishment

Transformer # 1 – 30 Quantities Capacity Total Watts

Residential A 756 Houses 2477 Watts 1,872,612 w

Transformer # 31 – 40 Quantities Capacity Total Watts



Residential B 252 Houses 2478 Watts 624456 w

For Transformer # 1 – 40, Residential A and B to be Total of 2,497,068 Watts -

hours

Establishment

Transformer # 41 - 45 Quantities Capacity Total Watts

Commercial A 2 135,350 Watts 270,000 w

Transformer # 46 - 50 Quantities Capacity Total Watts

Commercial B 2 135,350 Watts 270,000 w



For Transformer # 41 – 50, Commercial A and B to be Total of 541,400 Watts -

hours

FEEDERS LOAD BETWEEN TRANSFORMERS

Feeder #1 (Transformer 1 -10)

Max. Feeder Load =162420.40x 10

1.3

Max. Feeder Load = 1249387.692 Watts = 1249.388 kW

Feeder #2 (Transformer 11 - 20)


1.3






1.3



Max. Feeder Load =65445.6 x10

1.3



Max. Feeder Load =27000x 101.3





Max. Feeder Load =27000x 101.3


SUBSTATION LOAD BETWEEN FEEDERS

Substation #1


1.2




Substation #2

Max. Feeder Load =503427.692+(207692.308 x2)

1.2


POWER PLANT LOAD BETWEEN SUBSTATIONS

Max. Feeder Load =3123469.23+765676.9233

1.1



Documents

Power Plant Design; Analysis Design