TECHNOLOGICAL INSTITUTE OF THE PHILIPPINES

1338 Arlegui St., Quiapo, Manila

College of Engineering and Architecture

Mechanical Engineering Department

A DESIGN OF 5 MW HYDROELECTRIC POWER PLANT IN SAN JORGE, SAMAR

In Partial Fulfilment of the Requirements in the

Capstone Course ME 511 (Power Plant Design)

Submitted by:

Lalota, William Lorenzo

Lacap, Nathalie Coleen

Layugan, Jack Brave

Mabasa, John Lawrence

Tala, Aldrich Emile

Submitted to:

Engr. Nicanor L. Serrano

(Mechanical Engineering Department Head)

June 2021
 Acknowledgement

From the perseverance of the proponents to complete this capstone design, it would not be possible

without the assistance and supervision from the different persons who were asked and approached

concerning this topic. It was a deep appreciation and profound gratitude for the proponents to reach out

their sincerest gratitude to each one of them.

Foremost, we would like to thank the Almighty God for His guidance and empowering the strength

throughout this journey. Also, for providing the courage to continue and surpass the different obstacles

which may be faced across the way.

To Engr. Nicanor L. Serrano, Department Chair of Mechanical Engineering T.I.P Manila for his guidance in

our group project design.

To our Parents, that never failed to encourage us to give us our needs in order to get through this capstone

design and their never ending understanding and unconditional love for us, specially at times of hardships

during the capstone design.

Lastly, the proponents would like to commend their co-members for their never-ending patience despite the

hardships they had shared together and most importantly able to collaborate, taking their ideas in

application and arriving with the best design.

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 Abstract

As technology advances the demand for electric power increases since recent advancements in technology

requires the usage of electricity which means that as our country progresses in technology our demand for

power grows, as one of the countries in Asia that is still progressing, our country is need of many electric

power sources, as current news shows, Luzon is in yellow alert which means that the power supplied is not

enough for the grid’s power demand which can cause power outages in some places in Luzon specially

those that are far from the distribution lines or power substations, with this we need to increase the number

of power sources such as power plants. The proponents decided to design a 5 MW power plant situated in

San Jorge, Samar to be able to reach the power demand by 2025, the proponents design three (3)

alternative turbines, which are Tubular Bulb-type Turbine, Kaplan Turbine and Francis Turbine knowing that

these type of turbines has the potential to supply the needed power in the Province of Samar to reduced

power outages and for the customers of SAMELCO I to be satisfied. Nowadays power generation is still

progressing, but it is not progressing as much as our technologies. Therefore we have power outages and

the power demand is increasing while the power supply is constant, thus creating a gap between the

demand and supply.

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 TABLE OF CONTENTS

CHAPTER 1 5
INTRODUCTION 5
1.1 Introduction 5
1.2 Project Background 6
1.3 Topography 9
1.4 Site Selection 11
1.5 Electric Cooperative 12
1.6 The Project 16
1.6.1 Computation of Design Capacity 16
1.7 Client 17
1.8 Project Objective 18
1.8.1 General Objective 18
1.8.2 Specific Objectives 18
1.9 Scope and Limitations 18
1.10 Project Development Flow 19
CHAPTER 2
DESIGN INPUT 21
2.1 Hydroelectric Power Plant 21
2.2 Type of Hydroelectric Power Plant 21
2.3 Advantages and Disadvantages 23
2.4 System Components 24
2.5 Penstock requirements 27
2.5.1 Arrangement and Material Selection 27
2.6 Formulas 29
Solving for Specific Speed of the Turbine, Ns: 32
Solving for Number of Poles of the Generator, P: 32
2.7 DESIGN ALTERNATIVES 33
2.7.1 ALTERNATIVE 1 Hydroelectric Power Plant: Tubular Turbine 33
2.7.2 ALTERNATIVE 2 Hydroelectric Power Plant: Francis Turbine 36
2.7.3 ALTERNATIVE 3 Hydroelectric Power Plant: Kaplan Turbine 39
2.8 Manufacturer’s Data 42
Design Alternative 2: Francis Turbine 44
Design Alternative 3: Kaplan Turbine 45
CHAPTER 3 46
PROJECT DESIGN 46

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 3.1 Design Alternative 1: Tubular Turbine 46
3.2 Design Alternative 2: Francis Turbine 48
3.3 Design Alternative 3: Kaplan Turbine 50
CHAPTER 4 52
STANDARDS, CONSTRAINTS AND TRADEOFFS 52
4.1 Standard and Codes 52
4.1.1 RA 9275 - Philippine Clean Water Act 52
4.1.2 PD 1152 – Philippine Environmental Code 53
4.1.3 PD 1067 – Water Code of the Philippines 55
4.1.4 Occupational Safety and Health Standards – As amended, 1989 57
4.1.5 Philippine Mechanical Code 62
4.2 Constraints Selection 64
4.3 Constraints and Trade-offs 69
4.3.1 Economical Constraint 69
4.3.1.1 Considerations 69
4.3.1.2 Trade-offs 69
4.3.1.3 Justification 70
4.3.2 Sustainability Constraint 70
4.3.2.1 Considerations 70
4.3.2.2 Trade-offs 71
4.3.3 Environmental Constraint 72
4.3.3.1 Considerations 72
4.3.3.2 Trade-offs 73
4.3.3.3 Justification 74
4.3.5 Trade-off Summary 74
4.4 Sensitivity Analysis 74
4.4.1 Arrangements 75
4.4.2 Sensitivity Analysis Summary 77
Manufacturer’s Data for Design Alternative 3: Kaplan Turbine 81
APPENDIX A 83
APPENDIX B 98
APPENDIX C 112
APPENDIX D 128
References 137

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 CHAPTER 1

INTRODUCTION

1.1 Introduction

Electricity is the flow of electric charge, it is a commonly used form of energy in different areas. The

electricity is not freely available in nature in large amounts, so it must be produced. Electricity generation is

the central component that retains the society working by providing power energy. This process has a huge

impact in the industries, it converts the energy into electricity. The component that is capable of the

production of electricity is the power station. Electricity is most often generated at a power plant by

electromechanical generators, primarily driven by heat engines fueled by combustion or nuclear fission but

also by other means such as the kinetic energy of flowing water and wind. Power plant is an industrial

facility that produces electricity from essential energy. Most power plants use one or more generators that

convert mechanical energy into electrical energy in order to transmit power to the electrical station to

distribute for consumer electrical needs. There are different types of power plants, the power plants that

exist in the Philippines are Biomas, oil, Geothermal, Gas, Solar, coal and hydro power plants.

Renewable Energy lands on the useful and clean energy that comes from the regular source or process

that is continuously replenished. Renewable Energy resources include Geothermal, Hydropower, Biomass

and Ocean, Solar and Wind.

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1.2 Project Background

Fig. 1-1: Philippine Electricity Generation (2015)

Source: IEA World Energy Statistics and Balances

Fig. 1-2: Philippine Electricity Generation Progress (1971 - 2015)

Source: IEA World Energy Statistics Balance

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According to IEA World Energy Statistics Balance, as of 2015 the Electricity Gross Generation in the

Philippines was at 82.413 TWh and we can see that the majority of source is coming from the coal-fired

power plants and has made a huge demand from 1971.

Fig. 1-3: 2019 Philippine Power Situation Report

(Source: Department of Energy)

As of 2019, the gross generation went up to 106.041 TWh. The year 2019 is characterized by a significant

increase in electricity consumption and peak demand attributed to several factors such as the increase in

temperature and utilization of cooling equipment aggravated by the strong El Niño, the conduct of National

and Local elections during the first half of the year, increase in economic growth, and entry of large power

generating plants.

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VISAYAS GRID:

Fig. 1-4: 2018 vs 2019 Peak Demand, Visayas

(Source: Department of Energy)

In 2019, Visayas grid registered a peak demand of 2,224 MW showing a growth of 8.3% from the previous

year.

Fig. 1-5: 2019 Gross Generation, Visayas

(Source: Department of Energy)

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Power plants within the Visayas grid registered 16.059 TWh of gross generation in 2019. The region

continued to remain as the renewable energy dominated grid in the country with 47.2% share of generation

coming from renewable energy technologies with corresponding shares of geothermal at 39.1%, solar at

4.1%, biomass at 2.2%, wind at 1.4%, and hydro at 0.4%. For the fossil-based power plants, coal

generation increased and still is the largest producer at 49.6% with the entry of new coal-fired power plants

while oil based plants had a share of 3.3%.

Around the world, renewable energy use is on the rise, and these alternative energy sources could hold the

key to combating climate change. Renewable Energy is generated from sources that naturally replenish

themselves and never run out. The common sources are solar, wind, hydro, geothermal and biomass.

Renewable energy has many benefits: it can combat climate change because it doesn't emit carbon gas

and can decrease pollution and therefore reduce threats to everyone's health.

Humans have been harnessing the energy of river currents for centuries, using water wheels spun by rivers

initially to process grains and cloth. Moving water is a powerful source of energy. The power harnessed

from moving bodies of water is called hydroelectric power. The Philippines make use of two methods in

order to harness power; dam storage or impoundment, and run-of-river. While impoundment makes use of

man-made dams in order to store water, run-of-river relies on flowing bodies of water like waterfalls. Both

methods turn a turbine in order to generate power.

1.3 Topography

Samar is a province in the Philippines located in Eastern Visayas Region. It has a total land area of

13,428.8 square kilometers and a coordinate of 11°49’48” N, 125°0’0” E. It consists of 2 districts, with 24

municipalities, and 952 barangays with a total population of 1,880,020.

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 Fig. 1-6: Topography of Philippines and San Jorge, Samar from Google Map

San Jorge, officially the Municipality of San Jorge, is a 4th class municipality under 1st district in the

province of Samar, Philippines. The municipality has a land area of 241.20 square kilometers which

constitutes 3.99 percent of Samar’s total area.

Its population as determined by the 2015 census has an estimated population of 17,184 people which

accounts for 2.20 percent of the total population in Samar Province, or 0.39 percent of the overall

population of the Eastern Visayas region. San jorge is divided into 41 barangays. Based on these figures,

the population density is computed at 71 inhabitants per square kilometer or 185 inhabitants per square

mile.

Generally, San Jorge is a landlocked municipality. Almost the entire place is filled with trees, rice farms, and

the citizens are residing near roads. Though the place is landlocked, there is a big river, Gandara River

which is believed to be one of the largest rivers in Samar.

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1.4 Site Selection

1.) Cost and type of land

The land on which the powerhouse is going to be built must be at a reasonable price and secondly the

bearing capacity of the land.

2.) Availability of the Water

Since water is the primary requirement of the hydro-electric power station. Water availability must be in

huge quantities.

3.) Transportation Facilities

The site selected for the hydro-electric plant must be accessible by transport facility. So that heavy

machinery could be easily transported to the station.

4.) Hydroelectric Resource Location

Fig. 1-7: Hydroelectric Resource Location based on Google Map

The figure shown above is the Gandara River that will supply to the hydroelectric power plant.

5.) Projected Location

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 Fig. 1-8: Projected Location based on Google Map

Location Longitude Latitude

San Jorge, Samar 12°00'51.1"N 124°47'24.7"E

The figure shows the projected location of the Gandara river which is in the boundary of San Jorge, Samar

and has a length of approximately 11.85 kilometers. Determining the right location will decide the feasibility

of the power plant in the area.

1.5 Electric Cooperative

Fig. 1-9: Samar I Electric Cooperative, Inc.

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SAMAR I ELECTRIC COOPERATIVE, INC. (SAMELCO I) was incorporated and registered with the

National Electrification Commission on February 27, 1974. It started actual operation three months later on

May 27, 1974. The Coop coverage area is composed of Calbayog City, Tinambacan District, Oquendo

District, and the Municipalities of Sta. Margarita, Gandara, San Jorge, Pagsanghan, Taranghan,

Matuguinao, San Jose de Buan and the island Municipality of Almagro. In 1996, the Island municipalities of

Tagapul-an and Sto. Niño were added to its franchise which were energized under NPC’s Small Island Grid

(SIG) Power System.

Fig.1-10: SAMELCO I Supply Demand

(Source: Department of Energy)

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 Fig. 1-11: SAMELCO I Profile
(Source: http://www.kuryente.org.ph/electric-company/rates/85)

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 Fig.1-12: SAMELCO I Load Distribution

(Source: Department of Energy)

It has a total of 54,414 Captive Customer Connection with different types of customer classification

(Residential, Commercial, Industrial and others) within the 427 barangays of it’s coverage area. The

demand annual average growth as of 2016-2025 is 5.89%.

Table 1-1: SAMELCO I Fact Sheet

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 (Source: Department of Energy)

1.6 The Project

The chosen Electric Cooperative for the design is Samar I Electric Cooperative, Inc. (SAMELCO I) as the

main electric distribution benefactor of the proposed power plant design. The Samar I Electric Cooperative,

Inc. (SAMELCO I, Inc) is a franchise distribution utility that operates power service in the province of

Samar. The designers have proposed a hydroelectric power plant which is the most appropriate to the

location which is mostly surrounded by elevated bodies of water in the mountain ranges.

1.6.1 Computation of Design Capacity

The information that is used is based on the statistical data published by the Department of Energy (DOE).

Year Actual Demand (MW) Projected Peak Demand

(MW)

2021 14.5 15.88

2022 14.5 16.54

2023 14.5 17.22

2024 14.5 17.90

2025 14.5 18.60

Table 1-2: Actual Demand and Projected Demand based on DOE

𝑃𝑜𝑤𝑒𝑟 𝑁𝑒𝑒𝑑𝑒𝑑 = 𝑃𝑟𝑜𝑗𝑒𝑐𝑡𝑒𝑑 𝑃𝑒𝑎𝑘 𝐷𝑒𝑚𝑎𝑛𝑑 (2025) − 𝐴𝑐𝑡𝑢𝑎𝑙 𝑃𝑒𝑎𝑘 𝐿𝑜𝑎𝑑 (2017)

𝑃𝑜𝑤𝑒𝑟 𝑁𝑒𝑒𝑑𝑒𝑑 = 18. 60 − 14. 5 𝑀𝑊

𝑃𝑜𝑤𝑒𝑟 𝑁𝑒𝑒𝑑𝑒𝑑 = 4. 1 𝑀𝑊

𝑃𝑜𝑤𝑒𝑟 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 = 𝑃𝑜𝑤𝑒𝑟 𝑁𝑒𝑒𝑑𝑒𝑑 + ( 𝑃𝑜𝑤𝑒𝑟 𝑁𝑒𝑒𝑑𝑒𝑑 × 𝑆𝑦𝑠𝑡𝑒𝑚 𝑙𝑜𝑠𝑠)

𝑃𝑜𝑤𝑒𝑟 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 = 4. 1 + ( 4. 1 × 0. 1021) 𝑀𝑊

𝑃𝑜𝑤𝑒𝑟 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 = 4. 1 + 0. 41861 𝑀𝑊

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 𝑃𝑜𝑤𝑒𝑟 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 = 4. 5 𝑀𝑊 ≈ 5 𝑀𝑊

The computed capacity resulted in 4.5 MW but it was decided to settle with 5 MW to ensure that the

consumers are receiving enough power, also to consider the practical choices of materials and/or

equipment to be used that will be available in the market.

1.7 Client

The main benefactors of the proposed Power Plant Design is:

Samar I Electric Cooperative, Inc. (SAMELCO I)

Samar I Electric Cooperative, Inc. (SAMELCO I) covers (3) Districts and (10) Municipalities

namely:

● Calbayog City ● Taranghan

● Tinambacan District ● Matuguinao

● Oquendo District ● San Jose de Buan

● Sta. Margarita ● Almagro.

● Gandara ● Tagapul-an

● San Jorge ● Sto. Niño

● Pagsanghan

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1.8 Project Objective

1.8.1 General Objective

The main objective of the study is to design a Hydroelectric Power Plant with a capacity of 5 MW that can

help Samar to adapt to future power demands by 2025.

1.8.2 Specific Objectives

● Design a Hydroelectric Power Plant using 3 different turbines namely Kaplan, Francis and

Crossflow.

● To analyze and assess the three (3) alternative designs in accordance to the Codes and Standards

for power plant design.

● To decide on which of the best options among the alternative designs and options are best suited

and efficient for San Jorge, Samar using constraints and trade-offs.

1.9 Scope and Limitations

The project aims to build three (3) alternative turbine designs for the 5 MW Hydroelectric Power Plant

located in San Jorge, Samar. The planned power plant was intended to increase the supply and help

Samar to adapt to future power demand. The proponents focused on designing the necessary components

of the power plant. That being said, a detailed power plant lay out will not be provided, and the other minor

auxiliaries and accessories will also be neglected.

The proposed Hydroelectric power plant with three (3) alternative turbine designs will be evaluated and will

be tested using design constraints, trade-offs, and computations. Among the three (3) alternative designs,

one will be selected by evaluating and assessing the computations and will serve as the main design for

the 5 MW Hydroelectric Power Plant.

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1.10 Project Development Flow

Fig. 1-13: Project Development Flow

● Potential Clients – the proponents identified areas wherein to supply future power demands.

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● Power Plant Design – the proponents designed a hydroelectric power plant to be

established in San Jorge, Samar, Philippines.

● Alternative Designs – the proponents designed a 5 Megawatts hydroelectric power plant

with three alternative turbine designs: turbine, turbine, and turbine.

● Scope and Limitation – the proponents focused on designing the power plant with only

having the main components – excluding auxiliaries.

● Analysis and Computation of Gathered Data – based on the gathered data, the

proponents analyzed and calculated the power needed to supply the municipalities covered by

SAMELCO I.

● Constraints Identification – the proponents identified the constraints for each alternative

design.

● Methods for Trade-offs – the proponents compared the constraints identified using

different arrangements.

● Alternative Design Selection – the proponents selected the design based on the results of the

trade-offs.

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 CHAPTER 2

DESIGN INPUT

2.1 Hydroelectric Power Plant

The world’s first hydroelectric project was used to power a single lamp in the Cragside country house in

Northumberland, England, in 1878. Four years later, the first plant to serve a system of private and commercial

customers was opened in Wisconsin, USA, and within a decade, hundreds of hydropower plants were in operation.

Hydroelectric energy or also called hydropower, refers to the renewable energy that employs the power of

water in motion. The system that uses and harnesses the power of the flowing or falling water to drive

mechanical devices which convert the kinetic energy into electric energy.

Hydropower plant is the facility that produces electricity from generators driven by turbines and any

mechanical component that converts the potential energy of moving water into mechanical energy. It is an

eco-friendly power plant, it is fueled by water and it wont pollute the environment like other power plants

that burn fossil fuels, such as coal and natural gas. The amount of electricity that can be produced depends

on how far the water drops and how much water passes through the system.

2.2 Type of Hydroelectric Power Plant

The hydropower facilities have three types, which are impoundment, diversion and pump storage. The

impoundment facility uses a dam to put river water in a reservoir, a typically large hydro power system.

Diversion or also called run-of-river, a type of hydropower plant facility that channels a portion of a river

through a canal or penstock. It may not require the use of a dam. Run-of-river or diversion resources are

not meant to store water resources; rather they use water resources alongside the river by diverting the

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river through channels to harness the kinetic energy. Portions of the water are channeled and run through a

powerhouse, generating electricity. The water is then returned to the river, reducing environmental impacts.

Fig. 2-1: Diversion Fig. 2-2: Run-of-river

Hydropower category Power range No. of Home Powered

Pico 0kW - 5kw 0-5

Micro 5kw - 100kw 5 - 100

Mini 100kw - 1MW 100 - 1,000

Small 1MW - 10MW 1,000 - 10,000

Medium 10MW - 100MW 10,000 - 100,00

Large 100MW+ 100,000+

Table 2-1: Classification of Hydropower plant according to Capacity

Since the design proposes a 5 MW Hydroelectric Power Plant, the category falls under Small Hydropower

Plant.

Low Head ➔ Use river current or tidal flows of 30 meter or less to produce energy.
➔ These applications do not need a dam or retain water to create a hydraulic
head.

Medium Head ➔ A power station operating under heads from 30 to 300 m.

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 High Head ➔ A power station operating under heads above 300m.
➔ A head of 200/250m is considered as the limit between medium and high
heads power stations.

Table 2-2: Classification of Hydropower according to Head

2.3 Advantages and Disadvantages

Advantages of Hydropower Plant

● A Cheap Source of Energy - Despite expensive upfront building costs, hydroelectric power is one

of the cheapest sources of energy. The good thing about hydroelectricity is they require a low cost

of maintenance and operation.

● The Renewable Nature - Hydroelectricity is considered a renewable source of energy as it uses the

earth’s water to produce electricity. Water is a never-ending source as water is recycled back

through the water cycle.

● Clean Energy Source - Hydropower is one of the clean and green alternative sources of energy.

Unlike traditional fuel energy sources, hydroelectric energy doesn’t release harmful pollutants into

the environment.

● Lessens Pollution - Hydropower doesn’t rely on fossil fuels to produce energy, rather it uses water

and the process is so clean that it doesn’t release any greenhouse gases or harmful waste into the

environment. In this way, hydropower helps migrate climate change.

● Flexible - Hydropower is a flexible source of energy as hydropower plants can be scaled up and

down easily to meet the changing energy demands. When the electricity demand is high, the flow

of water can be increased and can also be decreased when the demand falls.

Disadvantages of Hydropower

● Environmental Damage - The main con of hydropower is its effect on the environment. Interruption

in the natural flow of water has serious impacts on the river ecosystem and environment. This

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 leads to disrupted animal migration paths, issues with water quality, and human or wildlife

displacement.

● High Building Cost - Hydroelectric Power plants are incredibly expensive to build, however

operating cost and maintenance costs are minimal. Also, hydropower projects take a long period to

finish and will have to operate for a long period to recover the capital spent.

● May Cause Droughts - Water availability directly affects electricity generation. The occurrence of

local droughts is one of the major downsides of setting up hydroelectric power plants.

● Risks of the Flood - As dams hold back large volumes of water, which can be catastrophic to

downriver settlements and infrastructure. A poor-standard construction, natural disasters, or

sabotage can cause the flooding disaster.

2.4 System Components

Fig. Typical run-of-river hydropower plant components

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For run-of-the-river hydro projects, a portion of a river’s water is diverted to a channel, pipeline, or

pressurized pipeline (penstock) that delivers it to a waterwheel or turbine. The moving water rotates the

wheel or turbine, which spins a shaft. The motion of the shaft can be used for mechanical processes, such

as pumping water, or it can be used to power an alternator or generator to generate energy.

Before water enters the turbine or waterwheel, it is first funneled through a series of components that

control its flow and filter out debris. These components are the headrace, forebay, and water conveyance

(channel, pipeline, or penstock).

1. Headrace

The headrace is a waterway running parallel to the water source. A headrace is sometimes

necessary for hydropower systems when insufficient head is provided. They often are constructed

of cement or masonry. The headrace leads to the forebay, which also is made of concrete or

masonry.

2. Forebay

It functions as a settling pond for large debris which would otherwise flow into the system and

damage the turbine. Water from the forebay is fed through the trash rack, a grill that removes

additional debris. The filtered water then enters through the controlled gates of the spillway into the

water conveyance, which funnels water directly to the turbine or waterwheel.

3. Penstock

Penstocks are like large pipes laid with some slope which carries water from the intake structure or

reservoir to the turbines. They run with some pressure so, sudden closing or opening of penstock

gates can cause water hammer effect to the penstocks. So, these are designed to resist the water

hammer effect apart from this penstock is similar to normal pipe. Steel or Reinforced concrete is

used for making penstocks.

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4. Hydraulic Turbines

Turbines are more commonly used today to power small hydropower systems. The moving water

strikes the turbine blades, much like a waterwheel, to spin a shaft. But turbines are more compact

in relation to their energy output than waterwheels. They also have fewer gears and require less

material for construction.

● Impulse Turbine - if the turbine wheel is driven by the kinetic energy of the fluid that strikes

the turbine blades through the nozzle or otherwise, the turbine is known as an impulse

turbine. In these types of turbines, a set of rotating machinery operates at atmospheric

pressure. Impulse turbines are usually suitable for high head and low flow rates.

● Reaction Turbine - a reaction turbine develops power from the combined action of

pressure and moving water. The runner is placed directly in the water stream flowing over

the blades rather than striking each individually. Reaction turbines are generally used for

sites with lower head and higher flows than compared with the impulse turbines.

5. Power House

Power house is a building provided to protect the hydraulic and electrical equipment. Generally, the

whole equipment is supported by the foundation or substructure laid for the power house. In case

of reaction turbines some machines like draft tubes, scroll casing etc. are fixed within the

foundation while laying it. So, the foundation is laid in big dimensions.

6. Draft Tube

If reaction turbines are used, then the draft tube is a necessary component which connects the

turbine outlet to the tailrace. The draft tube contains a gradually increasing diameter so that the

water is discharged into the tailrace with safe velocity. At the end of the draft tube, outlet gates are

provided which can be closed during repair works.

7. Tailrace

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 Tailrace is the flow of water from turbines to the stream. It is good if the power house is located

nearer to the stream. But, if it is located far away from the stream then it is necessary to build a

channel for carrying water into the stream. Otherwise the water flow may damage the plant in many

ways like lowering turbine efficiency, cavitation, damage to turbine blades etc.

2.5 Penstock requirements

A penstock is characterized by materials, diameter, wall thickness and type of joint:

● The material is selected according to the ground conditions, accessibility, weight, jointing system

and cost,

● The diameter is selected to reduce frictional losses within the penstock to an acceptable level,

● The wall thickness is selected to resist the maximum internal hydraulic pressure, including transient

surge pressure that will occur.

2.5.1 Arrangement and Material Selection

As regards the location of the penstock, two different solutions may be discerned which are characteristics

of the method of support as well.

1.) Buried penstocks are supported continuously on the soil at the bottom of a trench backfilled after placing

the pipe. The thickness of the cover over the pipe should be about 1 to 1.2 m.

The advantages of buried pipes are the following:

a) The soil cover protects the penstock against effect of temperature variations

b) It protects the conveyed water against freezing

c) Buried pipes do not spoil the landscape

d) They are safer against rock slides, avalanches and falling trees.

Disadvantages are:

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 a) Such pipes are less accessible for inspection, faults cannot be determined easily,

b) For large diameters and rocky soils their installation is expensive,

c) On steep hillsides, especially if the friction coefficient of the soil is low, such pipes may slide,

d) Maintenance and repair of the pipe is difficult.

2. Exposed penstocks are installed above the terrain surface and supported on piers (briefly called

supports or saddles). Consequently, there is no contact between the terrain and the pipe itself, and the

support is not continuous but confined piers.

The advantages of exposed pipes are the following;

a) The possibility of continuous and adequate inspection during operation,

b) Its installation is less expensive in case of large diameters of rocky terrain,

c) Safety against sliding may be ensured by properly designed anchorages,

d) Such pipes are readily accessible and maintenance and repair operations can be carried out

easily.

The disadvantages are;

a) Full exposure to external variations in temperature,

b) The water conveyed may freeze,

c) Owing to the spacing of supports and anchorages significant longitudinal stresses may develop

especially in pipes of large diameters designed for low internal pressures.

Steel to be used for the fabrication of penstocks of a hydro-electric project should meet the following

requirements:

a) It should stand against maximum internal pressure including dynamic pressure,

b) It should stand against frequent dynamic changes,

28
 c) It should have required impact strength to be able to deform plastically in the presence of stress

concentrations at notches and bends,

d) It should have good weldability without preheating, and

e) It should not require any stress relieving after welding.

Material Modulus of Coefficient of Ultimate Tensile n

Elasticity Linear Strength e^6
(Young’s Expansion a (N/m^2)
Modulus E) e^9 e^6 m/m deg.
N/m^2 C

Asbestos Cement - 8.1 - 0.011

Cast Iron 78.5 10 140 0.014

Ductile Iron 16.7 11 340 0.013

Polyethylene 0.55 140 5 0.009

Polyvinyl Chloride (PVC) 2.75 54 13 0.009

Welded Steel 206 12 400 0.012

Table 2-3: Commonly used materials with specifications

Types of Steel Pipes Yield Strength (MPa)

Alloy Steel 205

Black Steel 206.79

Carbon Steel 206.79

Iron Steel 227.46

Mild Steel 250

Stainless Steel 172.32

Table 2-4: Yield Strength of Common Type of Steel

2.6 Formulas

Penstock diameter:

29
The power available from the flow and head is given by the equation:

𝑃 = 𝑄𝐻γη

For the diameter of penstock,

2 2
10.3𝑛 𝑄 𝐿
𝐷 = 2. 69 × 𝐻

Where,

3
Q - flow discharge (𝑚 /𝑠)

H - net head (𝑚)

P - power (𝑘𝑊)

D - diameter (𝑚)

3
γ- specific weight of water (9. 8066 𝑘𝑁/𝑚 )

η- overall efficiency

For Power Input to the generator, BP:

𝐸𝑃
𝐵𝑃 = 𝑛 𝑔

For Available water power, Available 𝑃𝑤:

𝐵𝑃
𝐴𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑃𝑤 = 𝑛 𝑟

For Head Net, h:

ℎ = ℎ𝑔 − ℎ𝑓

For Gross Head, ℎ𝑔:

ℎ𝑔 = ℎℎ𝑤 − ℎ𝑡𝑤

For Friction Head Loss, ℎ𝑓:

30
 1. Using Darcy’s Equation:
2
𝑓𝐿𝑣
ℎ𝑓 = 2𝑔𝐷

2. Using Morse Equation:

2
2𝑓𝐿𝑣
ℎ𝑓 = 𝑔𝐷

For Discharge Flow, Q:

𝐴𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑃𝑤 = γ𝑄ℎ

𝐴𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑃𝑤
𝑄= γℎ

For Penstock’s length,

◦
𝑠𝑖𝑛𝑒 90
𝐿 𝑝
= 𝐻 𝑝( ◦ )
𝑠𝑖𝑛𝑒 45

For Penstock’s Area,

π 2
𝐴 𝑝
= ( 4 )(𝐷 𝑝
)

For Penstock’s Thickness, 𝑡 𝑝:

𝑠 𝑦
𝑃𝐷 𝑝
𝐹.𝑆.
= 2𝑡 𝑝

𝐹.𝑆.(𝑃𝐷 𝑝)
𝑡 𝑝
= 2𝑠 𝑦

Solving for Draft Tube’s Parameters:

For Draft Tube’s Diameter, 𝐷𝑑

𝐷𝑑 = 2𝐷𝑖 + 𝐷𝑝

For Draft Tube’s Height, 𝐻𝑑

𝐻𝑑 = (2 𝑡𝑜 3. 5)(𝑑)

31
For the number of Pipes needed for the Penstock, 𝑛𝑝

𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑃𝑒𝑛𝑠𝑡𝑜𝑐𝑘
𝑛𝑝 = 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑃𝑖𝑝𝑒 𝐿𝑒𝑛𝑔𝑡ℎ 𝑝𝑒𝑟 𝑃𝑖𝑒𝑐𝑒

For the number of Bolts, 𝑛𝑏

(𝐹)(𝐹.𝑆)
𝑛𝑏 = (𝐴𝑏)(𝑆𝑡)

Solving for the Water Velocity, 𝑣

𝑄
𝑣= 𝐴𝑝

Solving for Rated Speed, N:

𝑣
𝑁= π𝐷 𝑟

Solving for Specific Speed of the Turbine, Ns:

𝑁 𝐵𝑃
𝑁 𝑠
= 𝑠/4
ℎ

Solving for Number of Poles of the Generator, P:

120𝑓
𝑃= 𝑁

Machine Foundation

1 st Criterion: Dimensions of the machine foundation must be acceptable.

Calculating the dimensions of the machine foundation

For Length,

𝐿 = 𝑏 + 2𝑐

For Width,

𝑊 = 𝑎 + 2𝑐

32
For volume of foundation,

𝑉𝐹 = (𝐴𝑇)(ℎ) = 𝐿 × 𝑊 × ℎ

𝑊𝐹
ℎ= δ𝐹×𝐿×𝑊

For minimum weight of foundation,

𝑚𝑖𝑛 𝑊𝐹 ≈ 1. 25 𝑊𝑒

2nd Criterion: Machine Foundation must be safe.

For Induced Stress (𝑆𝐼)

𝑊𝐹+𝑊𝑒
𝑆𝐼 = 𝐴𝐵

For Design Stress (𝑆𝐷)

𝑆𝐵𝐶
𝑆𝐷 = 𝐹.𝑆

For Materials Estimates

For the number of bags of cement:

𝑉𝐹
𝑁𝑜. 𝑜𝑓 𝑏𝑎𝑔𝑠 𝐶 = 𝑌𝐼𝐸𝐿𝐷

For Absolute volume of materials:

δ𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙
𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙𝑠 = 𝐵𝑢𝑙𝑘 𝑆𝐺×𝑠𝑡𝑑. δ𝑤𝑎𝑡𝑒𝑟

For the volume cementing materials:

𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 = 𝑃𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 × 𝑁𝑜. 𝑜𝑓 𝐵𝑎𝑔𝑠 𝐶𝑒𝑚𝑒𝑛𝑡

For the reinforcing steel bars:

𝑊𝑅𝑆𝐵 = (0. 5% − 1%)(𝑊𝐹)

33
2.7 DESIGN ALTERNATIVES

2.7.1 ALTERNATIVE 1 Hydroelectric Power Plant: Tubular Turbine

For the alternative design 1, the type of turbine that will be used is a Tubular turbine. It is widely used in a

hydro power stations with the head of 2-30meters,The blades are fixed or can adjust manualy the efficient

turbine of this kind can produce a large quantity of water flow which passes unimpededly.　

Fig. Tubular Turbine

(Source:https://images.app.goo.gl/SzZuan9pZP6cgWNM9)

Tubular turbine is the horizontal shaft water turbine of which the flow channel approximates straight, it does

not have a water diversion volute and the blades could be fixed or rotary. According to the different types of

generator device, it could be classified as rim generator turbine and half-tubular turbine. The rotor of the rim

generator turbine was directly installed on the rim of the running wheel. The advantages are that the flow

channel is straight, the water flow is strong and high efficiency. But the linear velocity and perimeter of the

34
outer rim of blades is big, thus the rotating seal is difficult. Currently, the rim generator turbine model has

been rarely used. The half-tubular turbine was classified into S-type turbine、pit turbine and bulb turbine.

The structure of S-type turbine and pit turbine are simple, maintenance is easy but is low efficiency. They

are applicable for small-scaled hydropower stations generally. At present, the bulb turbine has been widely

used because of compact structure, steady operation and high efficiency. The power-making device was

decorated in a steel light bulb which flowed by water, the turbine and generator could connect directly or

through the increasing gear.

Discussion:

From the topographic map, the water coming from an elevation of 12 m will reach down to 6 m below where

the powerhouse will be located. The difference between 12 m and 6 m results in 6 meters, which meets the

requirement for Tubular Turbine.

35
 Fig 2-3: Topographic Map for Alternative 1

Fig. 2-4: Total distance of the penstock for Alternative 1

36
2.7.2 ALTERNATIVE 2 Hydroelectric Power Plant: Francis Turbine

The Francis turbine is a type of reaction turbine, a category of turbine in which the working fluid comes to

the turbine under immense pressure and the energy is extracted by the turbine blades from the working

fluid. A part of the energy is given up by the fluid because of pressure changes occurring in the blades of

the turbine, quantified by the expression of degree of reaction, while the remaining part of the energy is

extracted by the volute casing of the turbine. At the exit, water acts on the spinning cup-shaped runner

features, leaving at low velocity and low swirl with very little kinetic or potential energy left. The turbine's exit

tube is shaped to help decelerate the water flow and recover the pressure.

Fig. 2-5: Francis Turbine Main Components

Discussion:

From the topographic map, the water coming from an elevation of 23 m will reach down to 2 m below where

the powerhouse will be located. The difference between 23 m and 2 m results is 21 meters, which meets

the requirement for Francis Turbine.

37
Fig. 2-6: Topographic Map for Alternative 2

38
 Fig. 2-7: Total distance of the penstock for Alternative 2

2.7.3 ALTERNATIVE 3 Hydroelectric Power Plant: Kaplan Turbine

For alternative 3, the type of Water Turbine that will be used is Kaplan Turbine. It works on the principle of

axial flow reaction where the water flows through the runner along the direction parallel to the axis of

rotation of the runner.

Fig. 2-8: Kaplan Turbine main components

(Source: https://theconstructor.org/practical-guide/kaplan-turbine-component-working/2904/)

The water coming from the penstock is made to enter the scroll casing. The guide vanes direct the water to

the runner blades. The vanes are adjustable and can adjust themselves according to the requirement of

39
flow rate. The water takes a 90 degree turn, so the direction of the water is axial to that of runner blades.

The runner blades start to rotate as the water strikes due to the reaction force of the water. From the runner

blades, the water enters into the draft tube where its pressure energy and kinetic energy decreases. Kinetic

energy gets converted into pressure energy resulting in increased pressure of the water. The rotation of the

turbine is used to rotate the shaft of the generator for electricity production.

Discussion:

From the topographic map, the water coming from an elevation of 15 m will reach down to 6 m below where

the powerhouse will be located. The difference between 15 m and 6 m results in 9 meters, which meets the

requirement for Kaplan Turbine.

40
Fig.-9: Topographic Map for Alternative 3

41
 Fig. 2-10: Total distance of the penstock for Alternative 3

2.8 Manufacturer’s Data

Design Alternative 1: Tubular Turbine

42
 (Source: http://www.electway.net/product/100-2000kW_tubular_turbine.html)

Manufacturer Electway Electric

Model No. GZ1250a-WP-360

Type of Turbine Horizontal Shaft

Water Head 6.0 m

Discharge 3
98. 8 𝑚 /𝑠

Turbine Capacity 5263 kW

Rated Voltage 6.3 kV

Rated Speed 136.4 rpm

Generator Type SFWG5000-44/3950

Generator Capacity 5000 kW

Output Type AC Three Phase

43
Design Alternative 2: Francis Turbine

(source:https://www.alibaba.com/product-detail/5MW-High-Efficiency-Francis-Hydro-Turbine_62022146805.html?spm

=a2700.galleryofferlist.normal_offer.d_title.17f84bd10R2aaV)

Manufacturer Chengdu Forster Technology Co., Ltd

Model No. HL260 C-WJ-71

Type of Turbine Francis Turbine

Water Head 21.9 m

Discharge 28. 3 𝑚 /𝑠
3

Power 5350 KW

Rated Speed 200 rpm

Efficiency 90.7%

Generator Type SF5000-304250

Generator Capacity 5000 KW

Output Type Three phase AC

44
Design Alternative 3: Kaplan Turbine

(Source:http://www.electway.net/product/500-5000kW_ZZK_kaplan_turbine(head_10-20m_).html)

Manufacturer Electway Electric

Model No. ZZ680-LH-300

Type of Turbine Vertical Shaft

Water Head 9m

Discharge 63.5 𝑚 /𝑠
3

Turbine Capacity 5264 KW

Generator Type SF5000-36/4250

Generator Capacity 5000 KW

Rated Voltage 6300 V

Rated Speed 166.7 rpm

Output Type AC Three Phase

45
 CHAPTER 3

PROJECT DESIGN

3.1 Design Alternative 1: Tubular Turbine

Summary of the Data acquired for Tubular Turbine.

Parameters Result

Net Head 6.0 m

Electrical Power, EP 5000 kW

Turbine Power, BP 5263 kW

Available Water Power, Avail Pw 5813.35 kW

Volume flow rate, Q 3

98. 8 𝑚 /𝑠

Generator Efficiency 95 %

Turbine Efficiency 90.53 %

Length of Penstock, Lp 312.55 m

Diameter of Penstock, Dp 6.0169 m

Area of Penstock, Ap 28.43 m

Height of Draft Tube, Hd 16.5465

Discharge diameter of Draft Tube, Dd 8.3309 m

No. of pipes needed for penstock 53 pcs

No. of bolts 66 pcs

Water Flow Velocity 3.475 m/s

46
MACHINE FOUNDATION:

Parameters Result

Weight of Engine 143 metric tons

Area of Foundation 42 sq.m

Volume of Foundation 245 𝑚

3

Weight of Foundation 588,000 kg

Minimum Weight of Foundation 429,000 kg

Design Stress 32,291.731 kg/sq.m

Induced Stress 17,404.762 kg/sq.m

No. Bags of cement 1,708 pcs

Yield 5.063 cu.ft of water/bags of cement

Total Volume of Sand 3,416 cu.ft

Total Volume of Gravel 6,832 cu.ft

Total Volume of Water 1,598.396 cu.ft

Weight of Reinforcing Steel Bar 4,410 kg

For the computation of Alternative 1, See Appendix A.

PROJECT COSTING:

Parameters Result

Capital Cost 1,737,383,750 PHP

Turbo generator set cost 521,215,125 PHP

Electric regulation and control equipment cost 382,224,425 PHP

Construction and engineering management cost 138,990,700 PHP

Civil works cost 694,953,500 PHP

Operational and maintenance cost 29,843,750 PHP

Net income of the power plant 419,745,730 PHP

47
 Return of Investment 24.16%
For the computation of Project Costing, See Appendix D.

3.2 Design Alternative 2: Francis Turbine

Summary of the Data acquired for Francis Turbine:

Parameters Result

Net Head 21 m

Electrical Power, EP 5,000 kW

Turbine Power, BP 5,350 kW

Available Water Power, Avail Pw 5,828.1 kW

Volume flow rate, Q 3

28. 3 𝑚 /𝑠

Generator Efficiency 93.46 %

Turbine Efficiency 90.7 %

Length of Penstock, Lp 414.3625 m

Diameter of Penstock, Dp 3.1384 m

Area of Penstock, Ap 7.7358 sq.m

Height of Draft Tube, Hd 8.6306 m

Discharge diameter of Draft Tube, Dd 4.3454 m

No. of pipes needed for penstock 69 pcs

No. of bolts 63 pcs

Water Flow Velocity 3.6583 m/s

MACHINE FOUNDATION:

Parameters Result

48
 Weight of Engine, 𝑊𝑒 147 metric tons

Area of Foundation, 𝐴𝑇 64 sq.m

Volume of Foundation, 𝑉𝑓 320 cu.m

Weight of Foundation, 𝑊𝑓 768,000 kg

Minimum Weight of Foundation, 𝑚𝑖𝑛. 𝑊𝑓 588,000kg

Design Stress, 𝑆𝑑 26,909.7761 kg/sq.m

Induced Stress, 𝑆𝑖 14, 296.875 kg/sq.m

No. Bags of cement 2,230 pcs

Yield 5.063 cu.ft of water/bags of cement

Total Volume of Sand 4,460 cu.ft

Total Volume of Gravel 8,920 cu.ft

Total Volume of Water 2,086.8987 cu.ft

Weight of Reinforcing Steel Bar 4,992 kg

For the computation of Alternative 2, See Appendix B.

PROJECT COSTING:

Parameters Result

Capital Cost 1,519,166,250 PHP

Turbo generator set cost 455,749,875 PHP

Electric regulation and control equipment cost 334,216,575 PHP

Construction and engineering management cost 121,533,300 PHP

Civil works cost 607,666,500 PHP

Operational and maintenance cost 28,172,500 PHP

49
 Net income of the power plant 421,416,980 PHP

Return of Investment 27.74%

For the computation of Project Costing, See Appendix D

3.3 Design Alternative 3: Kaplan Turbine

Summary of the Data acquired for Kaplan Turbine:

Parameters Result

Net Head 9m

Electrical Power, EP 5,000 kW

Turbine Power, BP 5,264 kW

Available Water Power, Avail Pw 5,604.54 kW

Volume flow rate, Q 63.5 cu.m/s

Generator Efficiency 94.98%

Turbine Efficiency 93.92%

Length of Penstock, Lp 351.7 m

Diameter of Penstock, Dp 4.8309 m

Area of Penstock, Ap 18.32 sq.m

Height of Draft Tube, Hd 13.2850 m

Discharge diameter of Draft Tube, Dd 6.6889 m

No. of pipes needed for penstock 59 pcs

No. of bolts 64 pcs

Water Flow Velocity 3.466 m/s

MACHINE FOUNDATION:

Parameters Result

50
 Weight of Engine 142 metric tons

Area of Foundation 42 sq.m

Volume of Foundation 220.5 cu.m

Weight of Foundation 529,200 kg

Minimum Weight of Foundation 426,000 kg

Design Stress 34,982.7089 kg/sq.m

Induced Stress 15,980.9524 kg/sq.m

No. Bags of cement 1,537 pcs

Yield 5.063 cu.ft of water/bags of cement

Total Volume of Sand 3,074 cu.ft

Total Volume of Gravel 6,148 cu.ft

Total Volume of Water 1,438.3692 cu.ft

Weight of Reinforcing Steel Bar 3,969 kg

For the computation of Alternative 3, See Appendix C.

PROJECT COSTING:

Parameters Result

Capital Cost 1,737,383,750 PHP

Turbo generator set cost 521,215,125 PHP

Electric regulation and control equipment cost 382,224,425 PHP

Construction and engineering management cost 138,990,700 PHP

Civil works cost 694,953,500 PHP

Operational and maintenance cost 29,843,750 PHP

Net income of the power plant 419,745,730 PHP

Return of Investment 24.16%

For the computation of Project Costing, See Appendix D.

51
 CHAPTER 4

STANDARDS, CONSTRAINTS AND TRADEOFFS

4.1 Standard and Codes

4.1.1 RA 9275 - Philippine Clean Water Act

Chapter 1 – General Provisions

Article 1: Declaration of Principles and Policies

SECTION 1. Short Title. - This Act shall be known as the "Philippine Clean Water Act of 2004."

SECTION 2. Declaration of Policy. - The State shall pursue a policy of economic growth in a

manner consistent with the protection, preservation, and revival of the quality of our fresh, brackish,

and marine waters. To achieve this end, the framework for sustainable development shall be

pursued. As such, it shall be the policy of the State:

a) To streamline processes and procedures in the prevention, control and abatement of

pollution of the country's water resources;

b) To promote environmental strategies, use of appropriate economic instruments and of

control mechanisms for the protection of water resources;

52
 c) To formulate a holistic national program of water quality management that recognizes

that water quality management issues cannot be separated from concerns about water

sources and ecological protection, water supply, public health and quality of life;

d) To formulate an integrated water quality management framework through proper

delegation and effective coordination of functions and activities;

e) promote commercial and industrial processes and products that are environment

friendly and energy efficient;

f) To encourage cooperation and self-regulation among citizens and industries through the

application of incentives and market-based instruments and to promote the role of private

industrial enterprises in shaping its regulatory profile within the acceptable boundaries of

public health and environment;

g) To provide for a comprehensive management program for water pollution focusing on

pollution prevention;

h) To promote public information and education and to encourage the participation of an

informed and active public in water quality management and monitoring;

i) To formulate and enforce a system of accountability for short and long-term adverse

environmental impact of a project, program or activity; and

j) To encourage civil society and other sectors, particularly labor, the academe and

business undertaking environment-related activities in their efforts to organize, educate

and motivate 81 the people in addressing pertinent environmental issues and problems at

the local and national levels.

SECTION 3. Coverage of the Act. - This Act shall apply to water quality management in all water

bodies: Provided, that it shall primarily apply to the abatement and control of pollution from

land-based sources: Provided, further, That the water quality standards and regulations and the

53
 civil liability and penal provisions under this Act shall be enforced irrespective of sources of

pollution.

4.1.2 PD 1152 – Philippine Environmental Code

Title I Air Quality Management

Chapter 1 Standards

Section 5: Community Noise Standards. Appropriate standards for community noise levels shall be

established considering, among others, location, zoning and land use classification.

Section 6: Standards for Noise-Producing Equipment. There shall be established a standard for

noise producing equipment such as construction equipment, transportation equipment, stationary

engines, and electrical or electronic equipment and such similar equipment or contrivances. The

standards shall set a limit on the acceptable level of noise emitted from a given equipment for the

protection of public health and welfare, considering among others, the magnitude and condition of

use, the degree of noise reduction achievable through the application of best available technology

and the cost of compliance. The Installation of any noise-producing equipment shall conform with

the requirements of 77 Presidential Decree No. 1096 and other applicable laws as well as their

implementing rules and regulations.

Title II Water Quality Management

Chapter 1 Classifications and Standards

Section 15. Classification of Philippine Waters. The National Pollution Control Commission, in

coordination with appropriate government agencies, shall classify Philippine waters, according to

their best usage. In classifying said waters, the National Pollution Control Commission shall take

into account, among others, the following:

(a) the existing quality of the body of water at the time of classification.

54
 (b) the size, depth, surface area covered, volume, direction, rate of flow, gradient of stream; and

(c) the most beneficial uses of said bodies of water and lands bordering them for residential,

agricultural, commercial, industrial, navigational, recreational, and aesthetic purposes.

Section 18. Water Quality Standards. The National Pollution Control Commission shall prescribe

quality and effluent standards consistent with the guidelines set by the National Environmental

Protection Council and the classification of waters prescribed in the preceding sections, taking into

consideration, among others, the following:

(a) the standard of water quality or purity may vary according to beneficial uses; and

(b) the technology relating to water pollution control.

Title III Land Use Management

Section 24. Location of Industries. In the location of industries, factories, plants, depots and similar

industrial establishments, the regulating or enforcing agencies of the government shall take into

consideration the social, economic, geographic, and significant environmental impact of said

establishments.

4.1.3 PD 1067 – Water Code of the Philippines

Article 67. Any watershed or any area of land adjacent to any surface water or overlying any groundwater

may be declared by the Department of Natural Resources (DENR) as a protected area. Rules and

regulations may be promulgated by such Department to prohibit or control such activities by the owners or

occupants thereof within the protected area which may damage or cause the deterioration of the surface

water or groundwater or interfere with the investigation, use, control, protection, management or

administration of such waters.

Section 202.0 Plant Design Procedure

55
 202.1 Basis of Structure Design. For Industrial works, the utilization demand of the industry

for which the building is to be used were of utmost importance in the design of buildings.

Aside from geographical location and economic consideration, the mechanical and

electrical equipment requirements are extremely important for all modern buildings,

particularly factories.

Section 204.0 Machinery & Equipment

204.1 General Requirements

● All heavy machinery should be supported on solid foundations of sufficient

mass and base area to prevent or minimize the transmission of

objectionable vibration to the building and occupied source and to

maintain the supported machine at its proper elevation and alignment.

● Foundation mass should be from 3 to 5 times the weight of the machinery

it is supposed to support, or may be designed in conformance with

election 2.4.2. If the unbalanced inertial forces produced by the machine

can be calculated, a mass of eight equal to 10 to 20 times the forces

should be equal to dampen vibrations. For stability, the total combined

engine, driven equipment, and foundation center of gravity must be kept

below the foundation’s top.

● The weight of the machine plus the weight of the foundation should be

distributed over a sufficient soil area which is large enough to cause a

bearing stress within the safe bearing capacity of the soil with a factor of

safety of five (5).

● Foundation is preferably built of concrete in the proportion of one (1)

measure of Portland cement of two (2) measures of sand and four (4)

56
 measures of screened crushed stones. The machine should not be placed

on the foundation until seven (7) days have elapsed or operated until

another seven (7) days have passed.

● Concrete foundations should give steel bar reinforcements placed both

vertically and horizontally, to avoid thermal cracking. Weight of reinforcing

steel should be from 1/2% to 1% of the weight of the foundation.

204.2 Specific Requirement:

(a) Materials. The foundation should be concrete of 1-part cement, 2 parts sand

and 4 parts broken stone or gravel (maximum of 50 mm); the entire foundation

should be poured at one time, with no interruptions that are required for spading

and ramming. The top should be level and left rough for grouts. After pouring, the

top should be covered and wet down twice daily until the 76 forms are removed at

the end of the third or fourth day. The engine should not be placed on the

foundation until 10 days have elapsed, nor operated until after another 10 days.

(b) Soil Bearing Pressure. The first objective is achieved by making its supporting

area sufficiently large. The safe loads vary from about 4,890 kg/m^2 for alluvial

soil or wet clay, to (12,225 kg/m^2. The latter is assumed to be a safe load

average). In computation, 2,406 kg/m^2 may be used as the weight of concrete.

(c) Depth. The foundation may be taken as a good practical rule, to be 3.2 to 4.2

times the engine stroke, the lower factor for well-balanced multi-cylinder engines

and higher factor for engines with fewer cylinders, or on less firm soil.

4.1.4 Occupational Safety and Health Standards – As amended, 1989

Rule 1012: Special Rules

57
 1012.01: Work Conditions or Practices Not Covered by Standards Any specific rule applicable to a

condition, practice, means, methods, operations or processes shall also apply to other similar work

situations for which no specific rule has been established.

1012.02: Abatement of Imminent Danger

● An imminent danger is a condition or practice that could reasonably be expected to cause

death or serious physical harm before abatement under the enforcement procedures can

be accomplished.

● When an enforcement officer finds that an imminent danger exists in a workplace, he shall

inform the affected employer and workers of the danger and shall recommend to the

Regional Director the issuance of an Order for stoppage of operation or other appropriate

action for the abatement of the danger. Pending the issuance of the Order the employer

shall take appropriate measures to protect the workers.

● Upon receipt of such recommendation, the Regional Director shall immediately determine

whether the danger exists and is of such a nature as to warrant the issuance of a

Stoppage Order or other appropriate action to minimize the danger.

● The Order shall require specific measures that are necessary to avoid, correct or remove

such imminent danger and to prohibit the presence of any worker in such location where

such danger exists, except those whose presence are necessary to avoid, correct or

remove 70 such danger or to maintain a continuous process or operation. Where stoppage

of operation is ordered, the Order shall allow such correction, removal, or avoidance of

danger only where the same can be accomplished in a safe and orderly manner.

RULE 1010: OTHER SAFETY RULES

● Immediately after the issuance of a Stoppage Order, the Regional Director shall furnish the

Secretary, through the Director, within forty-eight (48) hours a copy of the Order and all pertinent

58
 papers relating thereto, together with a detailed description of the work conditions sought to be

corrected, the safety and health rule violated by the employer, and the corrective measures

imposed. The Secretary shall review the Order issued by the Regional Director and within a period

of not more than five (5) working days, issue a final Order either lifting or sustaining the Order of

the Regional Director.

● The Order shall remain in effect until danger is removed or corrected.

Rule 1013: Hazardous Workplaces

● For purposes of this Standards, the following are considered "hazardous workplaces:"

● Where the nature of work exposes the workers to dangerous environmental elements,

contaminants or work conditions including ionizing radiation, chemicals, fire, flammable

substances, noxious components and the like;

● Where the workers are engaged in construction work, logging, firefighting, mining, quarrying,

blasting, stevedoring, dock work, deep-sea fishing and mechanized farming;

● Where the workers are engaged in the manufacture or handling of explosives and other

pyrotechnic products;

● Where the workers use or are exposed to power driven or explosive powder actuated tools;

● Where the workers are exposed to biologic agents such as bacteria, fungi, viruses, protozoa,

nematodes, and other parasites.

Rule 1074: Physical Agents

1074.01: Threshold Limit Values for Noise The threshold limit values refer to sound pressure that

represents conditions under which it is believed that nearly all workers may be repeatedly exposed

without adverse effect on their ability to hear and understand normal speech.

● Feasible administrative or engineering controls shall be utilized when workers are exposed

to sound levels exceeding those specified in Table 8b hereof when measured on a scale of

59
 a standard sound level meter at slow response. If such controls fail to reduce sound within

the specified levels, ear protective devices capable of bringing the sound level to

permissible noise exposure shall be provided by the employer and used by the worker.

1074:02: Permissible Noise Exposure

Table 4-1: Permissible Noise Exposure

*Ceiling value: No exposure in excess of 115 dBa is allowed.

1074.03:

(1) The values specified in Table 4-1 apply to total time of exposure per working day regardless of

whether this is one continuous exposure or a number of short-term exposures but does not apply

to impact or impulsive type of noise.

(2) If the variation in noise level involves maximum intervals of one (1) second or less, it shall be

considered as continuous. If the interval is over one (1) second, it becomes impulse or impact

noise. (3) When the daily noise exposure is composed of two or more periods of noise exposure of

different levels, their combined effect should be considered rather than the effect of each.

Rule 1410 Construction Safety

Rule 1412: General Provisions

60
1412.01: Health and Safety Committee At every construction site there shall be organized and

maintained a Health and Safety Committee conforming with Rule 1040 and a medical and dental

service conforming with Rule 1960.

1412.02: Alternative Methods and Materials In the application of this Rule, the construction,

composition, size, and arrangement of materials used may vary provided that the strength of the

structure is at least equal to that herein prescribed.

1412.04: Machine Guarding: All moving parts of machinery used shall be guarded in accordance

with the requirements of Rule 1200.

1412.08: Pipelines: Repair work on any section of a pipeline under pressure shall not be

undertaken until the pipeline is released of the pressure or the section under repair is blocked off

the line pressure to ensure that no worker will be endangered.

1412.09: Protection of the Public: A safe covered walkway shall be constructed over the sidewalk

for use by pedestrians in a building construction work less than 2.3 m. (7 ft.) from a sidewalk or

public road.

1412.10: Protection from Falling Materials: (1) Steps shall be taken to protect workers from falling

materials, such as the provision of safety helmets and safety shoes. (2) Tools, objects and

materials (including waste materials) shall not be thrown or tipped from a height, but shall be

properly lowered by crane, hoist or chutes. If such is not practicable, the area where the material is

thrown or lowered shall be fenced and no person allowed in the fenced area.

1412.12: Protection against Collapse of Structure: (1) All temporary structures shall be properly

supported by the use of guys, stays, and other fixings necessary for stability during construction.

(2) Where construction work will likely reduce the stability of an existing or adjacent building,

shoring shall be undertaken to prevent the collapse or fall of any part of the structure.

61
 1412.20: Personal Protective Equipment: Personal Protective equipment as required in Rule 1080

shall be provided to the workers.

4.1.5 Philippine Mechanical Code

Chapter 2 – Commercial and Industrial Building

Section 2.0 Plant Design Procedure

2.1 Basis of Structure Design. For Industrial works, the utilization demand of the industry

for which the building is to be used are of utmost importance in the design of buildings.

Aside from geographical location and economic consideration, the mechanical and

electrical equipment requirements are extremely important for all modern buildings,

particularly factories.

2.2 Requirements for number, size, location and height of rise for elevators with particular

attention to penthouse dimensions and equipment loads.

● General requirements for plumbing with particular attention to the location of soil

stacks, standpipes, main pumps, water storage tanks and sprinkler systems.

● If steam is to be produced within the buildings, requirements of the boiler room

and accessories, such as fuel storage, the probable location of steam mains and

ducts and their approximate sizes in order to avoid interference with a structure

member of other utilities.

● Typical lighting demands particular attention to ceiling outlets as their proper

locations may influence the framing of the building and the necessary space

required for the electric conduits often affect the floor design.

62
 ● For industrial buildings, all specific demands of the manufacturing processes such

as special mechanical and electrical equipment of interior clearances should be

identified.

SECTION 5.0 ANTI-POLLUTION FOR INDUSTRIAL BUILDING

5.1 All machines/equipment which characteristically generate noise shall be provided with

appropriate enclosures to control emissions so as not to cause ambient noise level higher

than the quality standards set by the government agency concerned. If impractical, the

building housing the same should be appropriately designed or should be provided with

means to achieve compliance with the standards.

5.2 Buildings intended for noisy manufacturing activities should be appropriately designed

or should be provided with means so as not to cause ambient noise levels higher than the

standards set by the government agency concerned.

Chapter 4 – Machine guards and safeties at points of operation and danger zones

Section 16.0 Personal Protection in Workplaces

16.1 Personal Protective Equipment Personal protective equipment (PPE) shall be

considered the last line of defense against hazards in the work environment. The engineer

shall specify and require the use of PPE to protect personnel from known or possible

hazards in the workplace.

16.2 Employers shall protect their employees by providing appropriate and approved

protective tools, devices, equipment and appliances such as but not limited to the

following:

● Head Guards

● Face Shields

● Eye Goggles

63
 ● Ear Muffs

● Nose Aspirator

● Hand Gloves

● Arm Sleeves Shield

● Body Apron Shield

● Leg Sleeve Shield

● Foot Safety Shoes

● Foot Rubber Boots

4.2 Constraints Selection

The proponents will consider the 6 types of constraints based on the requirement for the Mechanical

Engineering Students of Technological Institute of the Philippines-Manila namely Economical,

Environmental, Societal, Health and Safety, Manufacturability and Sustainability.

● Economical - This will include the costs that will have an impact on budget allocation. Such

considerations are Capital Cost, Net Income of the Powerplant and Return of Investment.

● Environmental - This factor honors the ecosystem integrity of the design. Air pollution is highest in

this consideration, branching up to global warming, defining the carbon emission of each design.

Water and landscape impacts are also deliberated.

● Societal - Societal concern is a relatively new term and refers to hazards with the capability to

generate socio-political responses. Hazards invoking societal concerns pose a challenge to

decision makers for they oftentimes have major policy implications yet frequently lack the analytic

support affording them such elevated status.

64
 ● Health and Safety - This factor shall answer the safety of both workers and the public. This shall

also assess the long-term and short-term health impacts of the design during operation, which may

endanger and pose threat to the public welfare.

● Manufacturability - Manufacturability is an engineering specialty. It refers to the ease in which a part

can be made (or manufactured.) Manufacturability is in many ways dictated by a part’s design, and

can have huge implications as to the cost and effectiveness of the end product.

● Sustainability - Sustainability is the balance between the environment, equity, and economy. When

we hear the word “sustainability” we tend to think of renewable fuel sources, reducing carbon

emissions, protecting environments and a way of keeping the delicate ecosystems of our planet in

balance.

Constraints Economical Environmental Health Manufacturability Sustainability Societal Score

and
Safety

Economical 1 1 1 1 1 5

Environmental 0 1 1 0 1 3

Health and Safety 0 0 1 0 1 2

Manufacturability 0 0 0 0 0 0

Sustainability 0 1 1 1 1 4

Societal 0 0 0 1 0 1

Total 15
Table 4-2: Pairwise Comparison

For value of x,

100% = Σ Number of Coefficient of Occurrence

100% = 4x + 3x + 5x

65
 x = 8.333 %

Economical → 5 x 8.3333% = 41.67 %

Sustainability → 4 x 8.3333% = 33.33 %

Environmental→ 3 x 8.3333% = 24.99 %

Criteria Score Rank %

Economical 5 1st 41.67%

Sustainability 4 2nd 33.33%

Environmental 3 3rd 24.99%

Table 4-3: Ranking of Criteria

Fig. 4-1: Percent Weight

Economical vs Environmental

When choosing which factor is always considered, the client always tends to look at the budget allocation

first rather than considering the things that might affect the environment. The way how the power plant will

be constructed will be the one to decide what will happen to the environment.

66
Economical vs Health & Safety

Preparing for the budget allocation is always being considered by the client most of the time. Meanwhile,

ensuring the safety of everyone is a must. Before considering the safety of men, it is also important that the

safety equipment to be used are included in the budget allocation.

Economical vs Manufacturability

Efficient limitation was given greater need over manufacturability requirement on the grounds that as

specialists, should esteem the customer's financial limit and assets. Likewise, engineers are entrusted to

rearrange plans and not over muddle them in this way prompting a lot of costly expenses.

Economical vs Sustainability

Sustainable power source ventures require immense capital speculations. In any case, if the said ventures

end up being a progressively maintainable type of vitality creation, manageability's advantages far exceed

monetary limitations.

Economical vs Societal

Jobs are created once the power plant has been installed, any of the alternatives would offer a job to

society. And from the point of view of the proponent, it is easier to give priority to considering which of the

alternatives is more economical.

Environmental vs Health & Safety

It has been the premier obligation of a mechanical designer to survey the wellbeing and security of general

society. Natural laws ensure the government assistance of the general population. The proponents picked

natural over wellbeing.

Environmental vs Manufacturability

Ecological imperative was given a need over manufacturability limitation. It isn't best if the force plant is

simpler to build, however it will possess a huge territory of land, utilize a huge volume of water sources

since these variables have antagonistic impacts to nature.

67
Environmental vs Societal

The estimation of the natural effect for the structure power plant is generally important. The force plant can't

be raised on the off chance that it might influence the lives living among it. In this way, the chance of work

for the individuals won't be conceivable. In this way, advocates esteem nature more than culture.

Health & Safety vs Manufacturability

The health and safety of the people must be prioritized first. To ensure the good production of the power

plant, the workers must be healthy, first. So, the proponents considered health and safety over

manufacturability.

Health & Safety vs Societal

To provide the needs that must be met within the society caused by the design, health and safety must be

evaluated first. The proponents have considered health and safety as more important.

Sustainability vs Environmental

Even if the proponents give value to the importance of the environment, especially the use of land for the
residents, the proponents decided to give a higher priority to Sustainability. Without giving it high
consideration, the power plant’s life span might not be for long.

Sustainability vs Health & Safety

The proponents choose to design a Hydro electric power plant because we need a sustainable source of

electricity, although health and safety is a major concern, sustainability should be given more importance.

Sustainability vs Manufacturability

The materials and availability of materials are no doubt important but ensuring the source of energy and the
plant’s lifespan is given more importance by the proponents
Sustainability vs Societal

68
Sustainability was given priority over societal. Societal concerns such as public disturbances due to

irregularly distributed noise impulses and almost continuous noise within the community are necessary

sacrifices if it means that the power plant can sustain the demand for electricity.

Societal vs Manufacturability

Raising a force plant is essential to guarantee that it is fit for keeping up or meeting the structure limit. Be

that as it may, open unsettling influence is one of the cultural elements to be considered. The administration

or a region won't permit the erection of a force plant in the event that it is a disturbance to people in

general. Subsequently, culture is a higher priority than manufacturability.

4.3 Constraints and Trade-offs

4.3.1 Economical Constraint

4.3.1.1 Considerations

● Capital Cost - It is the total cost needed to bring a power plant project to a commercially operable

status. For investors, cost of capital is the opportunity cost of making a specific investment.

● Annual Net Income - In business and accounting, net income (also total comprehensive income,

net earnings, net profit, bottom line, sales profit, or credit sales) is an entity's income minus cost of

operational and maintenance expenses.

● Return of Investment - Return of Investment or ROI is a financial ratio used to calculate the benefit

an investor will receive in relation to their investment cost. It is most commonly measured as net

income divided by the original capital cost of the investment. The higher the ratio, the greater the

benefit earned.

69
4.3.1.2 Trade-offs

Score: 1 (Least Favorable), 2 (Favorable), 3 (Most Favorable)

Considerations Alternative 1 (Tubular Alternative 2 (Francis Alternative 3 (Kaplan

Turbine) Turbine) Turbine)

Capital Cost 2 3 2

Annual Net Income 2 3 2

Return of Investment 2 3 2

Score 6 9 6
Table 4-4: Economical Trade-offs

Analysis of data:

In terms of the Economical Constraint, Alternative 2 is the most favorable as it has a lower capital cost

among the three (3), and has an annual income and Return of Investment greater than the two (2), hence

Alternative 2 scored 9 while Alternative 1 and 3 both scored 6 because of their equal results.

4.3.1.3 Justification

Considerations Alternative 1 Alternative 2 Alternative 3

Capital Cost 1,737,383,750 php 1,519,166,250 php 1,737,383,750 php

Annual Net Income 419,745,730 php 421,416,180 php 419,745,730 php

Return of Investment 24.16% 27.74% 24.16%

Table 4-5: Economical Justification

4.3.2 Sustainability Constraint

4.3.2.1 Considerations

● Efficiency - This criteria specifies the overall capacity that is produced by a Hydropower plant.

70
 ● Flexibility - The ability of the Hydropower plant to manufacture power output to the unstable

demand of the consumer.

● Maintenance and Operation - Maintaining the quality of the Hydropower plant will guarantee an

excellent result of operation. A properly executed operation and solidity of the equipment ensure

the efficiency of the performance and life span of the power plant.

● Installation - This specifies the estimated time the Hydropower plant will be installed.

4.3.2.2 Trade-offs

Score: 1 (Least Favorable), 2 (Favorable), 3 (Most Favorable)

Considerations Alternative 1 (Tubular Alternative 2 (Francis Alternative 3 (Kaplan

Turbine) Turbine) Turbine)

Efficiency 1 2 3

Flexibility 1 1 3

Installation 3 1 2

Maintenance and 3 1 2
Operation

Score: 8 5 9
Table 4-7: Sustainability Trade-offs

Analysis of the data:

The table states that the favorable option of the plant for criteria of sustainability are both the alternatives 1

and 3. For efficiency, alternatives 1 and 2 have almost the same value lower than alternative 3. The latter is

also more flexible as it is suitable for either low and medium head range, whereas alternative 1 is more

efficient for a home hydropower station while alternative 2 is more efficient with a range of 100-300 m head.

71
As for the installation, the three (3) got different scores, being alternative 1 the highest. Due to it’s small

dimension and less submergence needed, this is easier to install than the two (2, same goes for the

maintenance and operation. Alternative 2 got the lowest score as dismantling of its parts is hard.

4.3.2.3 Justification

Considerations Alternative 1 Alternative 2 Alternative 3

Efficiency The turbine efficiency of The turbine efficiency of The turbine efficiency of
alternative 1 based on alternative 2 based on alternative 3 based on
the computation is the computation is the computation is
90.53%. 90.7%. 93.92%.

Flexibility It is applicable for those Most common water It is widely used

with low hydraulic turbines in use today throughout the world in
heads commonly from 2 but mini-hydro high-flow, low-head
to 15m head. Despite installations may be power production.
meeting the lower. It's best evolution of the Francis
requirements for head, performance is seen turbine. Its invention
it is hard to find a when the head height is allowed efficient power
suitable specification between 100-300m. production in low-head
with it's output power applications which was
being mostly applicable not possible with
for home hydro power Francis turbines.
stations only.

Installation More favourable flow Installation gets more The Kaplan turbine
conditions because the complex when the rock offers an advantage
dimension may be available is at very high with its smaller
smaller than Kaplan and depth. dimensions than
construction period is Francis. It is easy to
also short. The flow construct and space
condition will also requirement is less.
reduce the cavitation
risk for the bulb turbine,
which means a less
submergence is
needed.

Maintenance and Since the size may be Because the casing is Cavitation may also be
Operation smaller than Kaplan and stranded, it is very hard a problem but can be
a less submergence is to dismantle the runner. solved to some extent
needed, it is easier to The repair and by choosing the right
resolve the sand inspection are much blade.
erosion. The flow harder reasonably. The

72
 condition of Alternative cavitation may be
1 can also reduce the reduced but the turbine
cavitation risk. and the outlet channel
must be placed lower
than the lake or sea
level outside.
4-8: Sustainability Justifications

4.3.3 Environmental Constraint

4.3.3.1 Considerations

● Preservation of Ecology - This category reduce harm to the environment from human activities

especially logging.

● Impact on Aquatic Habitat - The impoundment of water and subsequent changes to flow velocity

and natural flow regimes have a demonstrated potential to negatively affect the diversity and

abundance of stream invertebrates that provide key trophic resources in river networks. Increased

velocity causes erosion of silt and sediment and sometimes wash away the aquatic organism's

colony.

4.3.3.2 Trade-offs

Score: 1 (Least Favorable), 2 (Favorable), 3 (Most Favorable)

Considerations Alternative 1 (Tubular Alternative 2 (Francis Alternative 3 (Kaplan

Turbine) Turbine) Turbine)

Preservation of Ecology 3 1 2

Impact on Aquatic 3 2 3
Habitat

Score 6 3 5
Table 4-9: Environmental Trade-offs
Analysis of the data:

73
The proponents base the scoring of preservation of ecology on the penstock’s length because trees might

be cut during the power plant construction. Alternative 1 got the highest points for having short length

among the three, alternative 3 had average length and got a score of 2. Lastly, alternative 2 got the lowest

point for having high penstock’s length.

For the impact on aquatic habitat, the proponents base the scoring on the fish habitats that will affect

because of the water flow velocity. Alternative 1 and 3 got the tie score of 3 and Alternative 2 ranked the

lowest.

4.3.3.3 Justification

Considerations Alternative 1 Alternative 2 Alternative 3

Preservation of Ecology Based on the Based on the Based on the

computation, the length computation, the length computation, the length
of penstock for of penstock for of penstock for
alternative 1 is 312.55 alternative 1 is 414.36 alternative 1 is 351.7 m.
m. m.

Impact on Aquatic Based on the Based on the Based on the

Habitat computation, the water computation, the water computation, the water
flow velocity is 3.48 m/s. flow velocity is 3.65 m/s. flow velocity is 3.466
m/s.
Table

4.3.5 Trade-off Summary

Formula:
Result = Score ×percentage

Total = Σof results per alternative

Alternative 1 (Tubular Alternative 2 (Francis Alternative 3 (Kaplan

Turbine) Turbine) Turbine)

Economical (41.67%) 2.5 3.75 2.5

Sustainability (33.33%) 2.67 1.67 2.99

74
 Environmental (24.99%) 1.5 0.75 1.25

Total 6.6 6.17 6.74

Table 4-9:Hydropower Plant Trade-offs result

4.4 Sensitivity Analysis

Sensitivity Analysis is the investigation of unclearness, in view of the yield of a framework can be
disseminated to various wellsprings of dubiousness in its sources of info which is implemented by the
proponents in order to select the best option of design based on the different constraints that were chosen.
The proponents give thought to the different possible arrangements of the (3) major constraints.

The total number of the possible arrangements can be solved using n! Where n is the number of chosen
constraints, if n = 3; n! = 3! ≈ n! = 6

Percentage Weight 41.67% 33.33% 24.99%

1st Arrangement Economical Sustainability Environmental

2nd Arrangement Economical Environmental Sustainability

3rd Arrangement Sustainability Environmental Economical

4th Arrangement Sustainability Economical Environmental

5th Arrangement Environmental Economical Sustainability

6th Arrangement Environmental Sustainability Economical

Table 4-10: Sensitivity Analysis

4.4.1 Arrangements

Constraints Alternative 1 Alternative 2 Alternative 3

Economical (41.67%) 2.5 3.75 2.5

Sustainability (33.33%) 2.67 1.67 2.99

Environmental (24.99%) 1.5 0.75 1.25

75
Weight Total 2.31 2.31 2.35
Table 4-11: Arrangement 1

Constraints Alternative 1 Alternative 2 Alternative 3

Economical (41.67%) 2.5 3.75 2.5

Environmental (33.33%) 1.5 0.75 1.25

Sustainability (24.99%) 2.67 1.67 2.99

Weight Total 2.21 2.23 2.21

Table 4-12: Arrangement 2

Constraints Alternative 1 Alternative 2 Alternative 3

Sustainability (41.67%) 2.67 1.67 2.99

Environmental (33.33%) 1.5 0.75 1.25

Economical (24.99%) 2.5 3.75 2.5

Weight Total 2.24 1.88 2.29

Table 4-13: Arrangement 3

Constraints Alternative 1 Alternative 2 Alternative 3

Sustainability (41.67%) 2.67 1.67 2.99

Economical(33.33%) 2.5 3.75 2.5

Environmental (24.99%) 1.5 0.75 1.25

Weight Total 2.32 2.13 2.29

Table 4-14: Arrangement 4

Constraints Alternative 1 Alternative 2 Alternative 3

Environmental (41.67%) 1.5 0.75 1.25

Economical (33.33%) 2.5 3.75 2.5

Sustainability (24.99%) 2.99 1.67 2.99

76
 Weight Total 2.13 1.98 2.10
Table 4-15: Arrangement 5

Constraints Alternative 1 Alternative 2 Alternative 3

Environmental (41.67%) 1.5 0.75 1.25

Sustainability (33.33%) 2.99 1.67 2.99

Economical (24.99%) 2.5 3.75 2.5

Weight Total 2.14 1.81 2.14

Table 4-16: Arrangement 6

4.4.2 Sensitivity Analysis Summary

Based on the Arrangement 1, the weight total of alternative 1 and alternative got the same score while

alternative 3 ranked first. For arrangement 2, it was alternative 1 and 3 which had the same score with the

alternative 2 winning. For arrangement 3, alternative 3 won but for arrangement 4 and 5, alternative 1

ranked first. Meanwhile on arrangement 6, alternative 1 and 3 got the same score while alternative 2 loses.

Fig. 4-1: Summary of Sensitivity Analysis

Total Score = Σweighted total score from 1st to 6th arrangement

77
 Total Average Score = Total Score/6

Alternative 1 Alternative 2 Alternative 3

Total Score 13.34 12.34 13.48

Total Average Score 2.22 2.06 2.25

Rank 2nd 3rd 1st

Table 4-17: Result of Sensitivity Analysis

CHAPTER 5

FINAL DESIGN

5.1 Conclusion

After analyzing the data, considering the constraints and evaluating the trade-offs, the proponents

concluded that the Alternative 3 is the ideal type of Hydropower plant in Gandara river for the San, Jorge

Samar.

78
 Figure 5.1 Topography map of the Alternative 3

From the topographic map, the water coming from an elevation of 15 m will reach down to 6 m below where

the powerhouse will be located. The difference between two points equals 9 meters, for the alternative 3

which is the Kaplan Turbine.

Fig. 5.2 Total distance of the penstock for Alternative 3

Table 5.1 Summary of data for Alternative 3 Kaplan Turbine

Parameters Result

79
 Net Head 9m

Electrical Power, EP 5,000 kW

Turbine Power, BP 5,264 kW

Available Water Power, Avail Pw 5,604.54 kW

Volume flow rate, Q 63.5 cu.m/s

Generator Efficiency 94.98%

Turbine Efficiency 93.92%

Length of Penstock, Lp 351.7 m

Diameter of Penstock, Dp 4.8309 m

Area of Penstock, Ap 18.32 sq.m

Height of Draft Tube, Hd 13.2850 m

Discharge diameter of Draft Tube, Dd 6.6889 m

No. of pipes needed for penstock 59 pcs

No. of bolts 64 pcs

Water Flow Velocity 3.466 m/s

MACHINE FOUNDATION:

Parameters Result

Weight of Engine 142 metric tons

Area of Foundation 42 sq.m

Volume of Foundation 220.5 cu.m

Weight of Foundation 529,200 kg

Minimum Weight of Foundation 426,000 kg

Design Stress 34,982.7089 kg/sq.m

Induced Stress 15,980.9524 kg/sq.m

80
 No. Bags of cement 1,537 pcs

Yield 5.063 cu.ft of water/bags of cement

Total Volume of Sand 3,074 cu.ft

Total Volume of Gravel 6,148 cu.ft

Total Volume of Water 1,438.3692 cu.ft

Weight of Reinforcing Steel Bar 3,969 kg

For the computation of Alternative 3, See Appendix C.

PROJECT COSTING:

Parameters Result

Capital Cost 1,737,383,750 PHP

Turbo generator set cost 521,215,125 PHP

Electric regulation and control equipment cost 382,224,425 PHP

Construction and engineering management cost 138,990,700 PHP

Civil works cost 694,953,500 PHP

Operational and maintenance cost 29,843,750 PHP

Net income of the power plant 419,745,730 PHP

Return of Investment 24.16%

For the computation of Project Costing, See Appendix D.

81
Manufacturer’s Data for Design Alternative 3: Kaplan Turbine

Fig. 5.3 Kaplan turbine

(Source:http://www.electway.net/product/500-5000kW_ZZK_kaplan_turbine(head_10-20m_).html)

Table 5.2 Manufacturer’s Data

Manufacturer Electway Electric

Model No. ZZ680-LH-300

Type of Turbine Vertical Shaft

Water Head 9m

Discharge 63.5 𝑚 /𝑠
3

Turbine Capacity 5264 KW

Generator Type SF5000-36/4250

Generator Capacity 5000 KW

Rated Voltage 6300 V

Rated Speed 166.7 rpm

Output Type AC Three Phase

82
 APPENDIX A

Hydroelectric Power Plant Alternative 1: Tubular Turbine

Solving for the Net Head, 𝐻𝑁𝐸𝑇

𝐻𝑁𝐸𝑇 = 𝐻𝐺𝑅𝑂𝑆𝑆

𝐻𝐺𝑅𝑂𝑆𝑆 = 𝐻ℎ𝑒𝑎𝑑𝑤𝑎𝑡𝑒𝑟 𝑒𝑙𝑒𝑣𝑎𝑡𝑖𝑜𝑛 − 𝐻𝑡𝑎𝑖𝑙𝑤𝑎𝑡𝑒𝑟 𝑒𝑙𝑒𝑣𝑎𝑡𝑖𝑜𝑛

𝐻𝐺𝑅𝑂𝑆𝑆 = 12 𝑚 − 6 𝑚

83
 𝐻𝐺𝑅𝑂𝑆𝑆 = 6 𝑚

Solving for the Generator Efficiency, η𝑔

Where:

𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑃𝑜𝑤𝑒𝑟 𝑜𝑓 𝑇𝑢𝑏𝑢𝑙𝑎𝑟 𝑇𝑢𝑟𝑏𝑖𝑛𝑒 = 5000 𝑘𝑊

𝑇𝑢𝑟𝑏𝑖𝑛𝑒 𝑃𝑜𝑤𝑒𝑟 𝑜𝑓 𝑇𝑢𝑏𝑢𝑙𝑎𝑟 𝑇𝑢𝑟𝑏𝑖𝑛𝑒 = 5263 𝑘𝑊

𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑃𝑜𝑤𝑒𝑟
η𝑔 = 𝑇𝑢𝑟𝑏𝑖𝑛𝑒 𝑃𝑜𝑤𝑒𝑟

5000 𝐾𝑊
η𝑔 = 5263 𝐾𝑊

η𝑔 = 0. 9500 ≈ 95. 00 %

Solving for the Available Water Power, 𝑃𝑊

Where:

3
γ 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = 9. 8066 𝑘𝑁/𝑚

3
𝑄 𝑜𝑓 𝑇𝑢𝑏𝑢𝑙𝑎𝑟 𝑇𝑢𝑟𝑏𝑖𝑛𝑒 = 98. 8 𝑚 /𝑠

𝑃𝑊 = γ𝑄𝐻𝑁𝐸𝑇

3
𝑘𝑁 𝑚
𝑃𝑊 = (9. 8066 3 )(98. 8 𝑠
)(6 𝑚)
𝑚

𝑃𝑊 = 5, 813. 35 𝑘𝑊

Solving for the Turbine Efficiency, η𝑡

𝑇𝑢𝑟𝑏𝑖𝑛𝑒 𝑃𝑜𝑤𝑒𝑟
η𝑡 = 𝑊𝑎𝑡𝑒𝑟 𝑃𝑜𝑤𝑒𝑟

5263 𝑘𝑊
η𝑡 = 5813.35 𝑘𝑊

η𝑡 = 0. 9053 = 90. 53%

Solving for the length of Penstock, 𝐿𝑝

84
Where:

𝐻 = 𝑛𝑒𝑡 ℎ𝑒𝑎𝑑

𝑥 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑠𝑢𝑐𝑡𝑖𝑜𝑛 𝑝𝑖𝑝𝑒 𝑡𝑜 𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒 𝑝𝑖𝑝𝑒 (312. 5 𝑚 𝑏𝑎𝑠𝑒𝑑 𝑜𝑛 𝐹𝑖𝑔. 2 − 10)

θ = 𝑎𝑛𝑔𝑙𝑒 𝑓𝑟𝑜𝑚 𝑠𝑢𝑐𝑡𝑖𝑜𝑛 𝑝𝑖𝑝𝑒 𝑡𝑜 𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒 𝑝𝑖𝑝𝑒

−1 𝐻
θ = 𝑡𝑎𝑛 ( 𝑥
)

−1 6𝑚
θ = 𝑡𝑎𝑛 ( 312.5 𝑚 )

θ = 1. 09

𝐻
𝐿𝑝 = 𝑆𝑖𝑛θ

6𝑚
𝐿𝑝 = 𝑆𝑖𝑛(1.099944)

𝐿𝑝 = 312. 55 𝑚

Solving for the size of Penstock, 𝐷𝑝

2 2
(𝑛𝑝) (𝑄) (𝐿𝑝) 0.1875
𝐷𝑝 = 2. 69[ 𝐻𝑔
]

Where:

𝑛𝑝 𝑜𝑓 𝑤𝑒𝑙𝑑𝑒𝑑 𝑠𝑡𝑒𝑒𝑙 (𝑏𝑎𝑠𝑒𝑑 𝑜𝑛 𝑇𝑎𝑏𝑙𝑒 2 − 3 ) = 0. 012

85
 2 3 2
(0.012) (98.8 𝑚 /𝑠) (312.55 𝑚) 0.1875
𝐷𝑝 = 2. 69[ 6𝑚
]

𝐷𝑝 = 6. 0169 𝑚

Solving for the Area of the Penstock, 𝐴𝑝

2
Π(𝐷𝑝)
𝐴𝑝 = 4

2
Π(6.0169 𝑚)
𝐴𝑝 = 4

2
𝐴𝑝 = 28. 43 𝑚

Solving for the Thickness of the Penstock, 𝑡𝑝

𝑃𝐷𝑝𝐹.𝑆
𝑡𝑝 = 2𝑆𝑦

Where:

𝐹. 𝑆 = 2 (𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑)

𝑆𝑦 = 250, 000 𝑘𝑃𝑎 (𝑏𝑎𝑠𝑒𝑑 𝑜𝑛 𝑇𝑎𝑏𝑙𝑒 2 − 4)

As for,

𝑃 = γℎ

𝑘𝑁
𝑃 = (9. 8066 3 )(6 𝑚)
𝑚

86
 𝑃 = 58. 8396 𝑘𝑃𝑎

Hence,

(58.8396 𝑘𝑃𝑎)(6.0169 𝑚)(2)

𝑡𝑝 = 2(250,000 𝑘𝑃𝑎)

𝑡𝑝 = 0. 0014 𝑚

Solving for Draft Tube Parameters,

For Height of the Draft tube:

𝐻𝑑 = (2 𝑡𝑜 3. 5)𝐷𝑝

𝑇𝑎𝑘𝑖𝑛𝑔 𝑡ℎ𝑒 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑓𝑟𝑜𝑚 2 𝑡𝑜 3. 5 𝑔𝑖𝑣𝑒𝑠 2. 75,

𝐻𝑑 = (2. 75)(6. 0169 𝑚)

𝐻𝑑 = 16. 5465 𝑚

For the diameter of the draft tube,

𝐷𝑑 = 2𝐷𝑖 + 𝐷𝑝

87
 Where:

𝐷𝑖 = 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟

Sin law,

𝐷𝑖 𝐻𝑑
◦ = ◦
𝑠𝑖𝑛 4 𝑠𝑖𝑛 86

◦
𝑠𝑖𝑛 4
𝐷𝑖 = 𝐻𝑑( ◦ )
𝑠𝑖𝑛 86

◦
𝑠𝑖𝑛 4
𝐷𝑖 = 16. 5465 𝑚 ( ◦ )
𝑠𝑖𝑛 86

𝐷𝑖 = 1. 157 𝑚

Then,

𝐷𝑑 = 2𝐷𝑖 + 𝐷𝑝

𝐷𝑑 = 2(1. 157 𝑚) + 6. 0169 𝑚

𝐷𝑑 = 8. 3309 𝑚

Solving for the number of pipes needed for the penstock

𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑃𝑒𝑛𝑠𝑡𝑜𝑐𝑘
𝑛𝑝𝑖𝑝𝑒 = 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑝𝑖𝑝𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑝𝑒𝑟 𝑝𝑖𝑒𝑐𝑒

88
Where:

𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑝𝑖𝑝𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑚

𝑝𝑖𝑒𝑐𝑒
=6 𝑝𝑖𝑒𝑐𝑒

𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑃𝑒𝑛𝑠𝑡𝑜𝑐𝑘
𝑛𝑝𝑖𝑝𝑒 = 𝑚
6 𝑝𝑖𝑒𝑐𝑒𝑠

312.55 𝑚
𝑛𝑝𝑖𝑝𝑒 = 𝑚
6 𝑝𝑖𝑒𝑐𝑒𝑠

𝑛𝑝𝑖𝑝𝑒 = 52. 09 𝑝𝑐𝑠 ≈ 53 𝑝𝑐𝑠

Solving for the number of bolts

𝑆𝑡 𝐹 (𝐹)(𝐹.𝑆)
𝐹.𝑆
= (𝐴𝑏)(𝑛𝑏)
→ 𝑛𝑏 = (𝐴𝑏)(𝑆𝑡)

Where:

𝐷𝑏 = 1𝑖𝑛 ≈ 0. 02541 𝑚 (𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑚𝑎𝑟𝑘𝑒𝑡)

𝑆𝑡 = 100 𝑀𝑃𝑎 (𝑏𝑎𝑠𝑒𝑑 𝑜𝑛 𝐴𝑚𝑒𝑟𝑖𝑐𝑎𝑛 𝑆𝑜𝑐𝑖𝑒𝑡𝑦 𝑓𝑜𝑟 𝑇𝑒𝑠𝑡𝑖𝑛𝑔 𝑎𝑛𝑑 𝑀𝑎𝑡𝑒𝑟𝑖𝑎𝑙)

𝐹. 𝑆 = 2 (𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑)

As for the Force,

𝐹
𝑃= 𝐴𝑝
→ 𝐹 = (𝑃)(𝐴𝑝)

2 2
𝐹 = (58. 8396 𝑘𝑁/𝑚 )(28. 43 𝑚 )

𝐹 = 1672. 8098 𝑘𝑁

89
And Area of Bolts,

2
Π(0.02541 𝑚)
𝐴𝑏 = 4

2
𝐴𝑏 = 0. 000507 𝑚

Therefore,

(𝐹)(𝐹.𝑆)
𝑛𝑏 = (𝐴𝑏)(𝑆𝑡)

(1672.8098 𝑘𝑁)(2)
𝑛𝑏 = 2 2
(0.000507 𝑚 )(100,000 𝑘𝑁/𝑚 )

𝑛𝑏 = 65. 98 𝑝𝑐𝑠 ≈ 66 𝑝𝑐𝑠

Solving for Water Velocity

𝑄
𝑄 = 𝐴𝑝𝑉→𝑉 = 𝐴𝑝

3
98.8 𝑚 /𝑠
𝑉= 2
28.43 𝑚

𝑉 = 3. 475 𝑚/𝑠

Machine Foundation

First step in designing a machine foundation: Start with the simplest form which is the rectangular block

machine foundation.

Considering a rectangular machine foundation;

90
For safety purposes,

𝑊𝑓 ≥ 𝑚𝑖𝑛. 𝑊𝑓

Then,

The foundation mass should be from 3 to 5 times the weight of the machinery it is supposed to support.

Figure 1: Top View of Rectangular Block Machine Foundation

For Top Length of the Foundation, L:

𝐿 = 𝑏 + 2(𝑐)

Where:

b = diameter of the base plate

also, clearance (c) to be used is 1 m for the TBOC (Turbine Bearing Oil Control) and for the turbine to be

use,

so,

𝐿 = 𝑏 + 2(𝑐)

91
 𝐿 = 5 + 2(1)

𝐿 = 7𝑚

Solving for top Area, 𝐴𝑇

Note: since the equipment to be supported is circular, we will use a square type foundation.

𝐴𝑇 = 𝐴𝐵 = 𝐿 𝑥 𝑊

2
𝐴𝑇 = 7 × 7 𝑚

2
𝐴𝑇 = 42 𝑚

Solving for height of bed plate, ℎ

Assuming the Height of the bed plate is 4.5m.

ℎ = 5𝑚

Solving for Volume of the Foundation, 𝑉𝑓

From Machinery Foundation; by Jose M. Perez, Jr.; PME

𝑉𝑓 = 𝐿 𝑥 𝑊 𝑥 ℎ

3
𝑉𝑓 = 7 × 7 × 5 𝑚

3
𝑉𝑓 = 245 𝑚

92
Solving for Weight of the Foundation, 𝑊𝑓

𝑘𝑔
𝑛𝑜𝑡𝑒: γ𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 = 2, 400 3
𝑚

𝑊𝑓 = γ𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑥 𝑉𝑓

𝑘𝑔 3
𝑊𝑓 = 2, 400 3 × 245 𝑚
𝑚

𝑊𝑓 = 588, 000 𝑘𝑔

Total Weight of the Engine, 𝑊𝑒

𝑊𝑒 = 𝑊𝑇 + 𝑊𝐺𝑒𝑛

𝑊𝑒 = (51 + 92) 𝑚𝑒𝑡𝑟𝑖𝑐 𝑡𝑜𝑛𝑠

𝑊𝑒 = 143 𝑚𝑒𝑡𝑟𝑖𝑐 𝑡𝑜𝑛𝑠

For the Minimum Weight of the Foundation, 𝑚𝑖𝑛. 𝑊𝑓

From Possible Steps in Designing Machine Foundation by Power Plant Engineering by Frederick T. Morse.

𝑚𝑖𝑛. 𝑊𝑓 = 3 𝑡𝑜 5 𝑥 𝑊𝑒

Say we used 3 as multiplier,

1000 𝑘𝑔
𝑚𝑖𝑛. 𝑊𝑓 = 3 × 143 𝑚𝑒𝑡𝑟𝑖𝑐 𝑡𝑜𝑛𝑠 × 1 𝑚𝑒𝑡𝑟𝑖𝑐 𝑡𝑜𝑛

93
 𝑚𝑖𝑛. 𝑊𝑓 = 429, 000 𝑘𝑔

Hence,

𝑊𝑓 ≥ 𝑚𝑖𝑛. 𝑊𝑓

588, 000 𝑘𝑔 ≥ 429, 000 𝑘𝑔

The dimensions are OK!

Checking for Safety

Distribute the weight of the machine, the machine bed plate, and its own weight over a safe subsoil area.

(Source: Power Plant Engineering by P.J. Morse, page 106).

𝑆𝑖 <𝑆𝑑

For Induced Strength, 𝑆𝑖

𝑊𝑓 + 𝑊𝑒
𝑆𝑖 = 𝐴𝐵

588,000 𝑘𝑔 + 143,000 𝑘𝑔
𝑆𝑖 = 2
42 𝑚

2
𝑆𝑖 = 17, 404. 762 𝑘𝑔/𝑚

For Design Stress, 𝑆𝑑

𝑆𝐵𝑃
𝑆𝑑 = 𝐹.𝑆

94
Where: SBP = Soil Bearing Pressure

Type of Soil US ton/sq. ft

Native Rock 200 up

Ashlor Masonry 25 – 30

Best Brick Masonry 15 – 20

Common Brick Masonry 5 – 10

Clay, Compact 5–8

Clay, Soft 1–2

Gravel and Sand 8 – 10

Sand 2–6

Quick Sand, Loom etc. 0.5 – 1

Table B-1: Safe Bearing Pressure of Soil

(Source: Machinery Foundation; by Jose M. Perez, Jr.; PME)

𝑒𝑛𝑔𝑙𝑖𝑠ℎ 𝑡𝑜𝑛
𝑆𝑜𝑖𝑙 𝐵𝑒𝑎𝑟𝑖𝑛𝑔 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = (5 − 8) 2
𝑓𝑡

Using average value,

𝑒𝑛𝑔𝑙𝑖𝑠ℎ 𝑡𝑜𝑛
𝑆𝐵𝑃𝐴𝑣𝑒 = (6 2
𝑓𝑡

2
𝑒𝑛𝑔𝑙𝑖𝑠ℎ 𝑡𝑜𝑛 2000 𝑙𝑏𝑠 1 𝑓𝑡
𝑆𝐵𝑃𝐴𝑣𝑒 = (6) 2 × 1 𝑈𝑆 𝑡𝑜𝑛
× 2 2
𝑓𝑡 0.3048 𝑚

𝑘𝑔
𝑆𝐵𝑃𝐴𝑣𝑒 = 129, 166. 925 2
𝑚

95
Note: Safe bearing pressure of soils for machine foundation area from quarter to a half of the above values.

We use a factor of safety of 4 to attain the maximum bearing pressure. So,

𝑘𝑔
129,166.925 2
𝑚
𝑆𝑑 = 4

𝑘𝑔
𝑆𝑑 = 32, 291. 731 2
𝑚

Therefore,

𝑆𝑖 ≤𝑆𝑑

The design of the machine foundation is SAFE!

For Material Estimation:

Bulk Specific Gravity SpecificWeight

Cement 3.15 94 lbf / bag of C

Sand 2.64 110 lbf / bag of C

Gravel 2.66 96 lbf / bag of C

Table B-2: Properties of Cementing Materials

3 3
γ 𝑜𝑓 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 = 2, 400 𝑘𝑔/𝑚 = 149. 9339 𝑙𝑏𝑓/𝑓𝑡

For water, 7 gal per bag of cement (if not specified)

3
γ 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 =62. 4 𝑙𝑏𝑓/𝑓𝑡

96
 Sand Gravel (𝒇𝒕3 Yield (𝒇𝒕3) 𝒐𝒇
(𝒇𝒕3 / /𝒃𝒂𝒈) 𝒄𝒐𝒏𝒄𝒓𝒆𝒕𝒆
Cement Cement (Bag C) 𝒃𝒂𝒈) 𝒃𝒂𝒈 𝒐𝒇
𝒄𝒆𝒎𝒆𝒏𝒕

AA 1 1.5 3 4.15077

A 1 2 4 5.063004

B 1 2.5 5 5.97524

C 1 3 6 6.88748

D 1 3 6 8.71195

Table B-3: Different Class of Concrete Mixture

Note we will use Class A concrete. Class A Concrete: C:S:G = 1:2:4

For Yield Materials Estimate:

𝑌𝐼𝐸𝐿𝐷 = Σ𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙, (𝐶, 𝑆, 𝐺 𝑎𝑛𝑑 𝐻2𝑂)

Where,

𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙

𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 = (𝑏𝑢𝑙𝑘 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑔𝑟𝑎𝑣𝑖𝑡𝑦)(𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟)

For Cement,

𝑙𝑏 𝑓
94 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡
𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡 = 𝑙𝑏𝑓
(3.15)(62.4 3 )
𝑓𝑡

3
𝑓𝑡 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡
𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡 =. 478225 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡

For Sand,

97
 𝑙𝑏𝑓 𝑓𝑡
3
(110 3 )(2 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡 )
𝑓𝑡
𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑎𝑛𝑑 = 𝑙𝑏𝑓
(2.64)(62.4 3 )
𝑓𝑡

3
𝑓𝑡 𝑜𝑓 𝑠𝑎𝑛𝑑
𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑎𝑛𝑑 = 1. 335470 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡

For Gravel,

𝑙𝑏𝑓 𝑓𝑡
3
(96 3 )(4 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡 )
𝑓𝑡
𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑎𝑛𝑑 = 𝑙𝑏𝑓
(2.66)(62.4 3 )
𝑓𝑡

3
𝑓𝑡 𝑜𝑓 𝑔𝑟𝑎𝑣𝑒𝑙
𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = 2. 313476 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡

For Water,

3
𝑔𝑎𝑙 1 𝑓𝑡
𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = 7 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡
𝑥 7.48 𝑔𝑎𝑙

3
𝑓𝑡 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 =. 935829 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡

𝑌𝐼𝐸𝐿𝐷 = Σ𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙, (𝐶, 𝑆, 𝐺 𝑎𝑛𝑑 𝐻2𝑂)

3
𝑓𝑡
𝑌𝐼𝐸𝐿𝐷 = Σ(. 478225 + 1. 335470 + 2. 313476 + . 935829) 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡

3
𝑓𝑡
𝑌𝐼𝐸𝐿𝐷 = 5. 063 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡

For Number of Bags of Cement,

𝑉𝑓
𝑁𝑜. 𝑜𝑓 𝐵𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡 = 𝑌𝐼𝐸𝐿𝐷

98
 3
3 3.28 𝑓𝑡
245 𝑚 ( 3 )
1𝑚
𝑁𝑜. 𝑜𝑓 𝐵𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡 = 𝑓𝑡
3
5.063 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡

𝑁𝑜. 𝑜𝑓 𝐵𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡 = 1707. 57 𝑏𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡

𝑁𝑜. 𝑜𝑓 𝐵𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡 ≈ 1708 𝑏𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡

For the Volume cementing materials,

𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 = (𝑃𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝑜𝑓 𝑀𝐴𝑡𝑒𝑟𝑖𝑎𝑙)(𝑁𝑜. 𝑜𝑓 𝐵𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡)

Volume of Sand,

3
𝑓𝑡 𝑜𝑓 𝑠𝑎𝑛𝑑
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 = (2 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡 )(1708 𝐵𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡)

3
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑆𝑎𝑛𝑑 = 3416 𝑓𝑡 𝑜𝑓 𝑠𝑎𝑛𝑑

Volume of Gravel,

3
𝑓𝑡 𝑜𝑓 𝑔𝑟𝑎𝑣𝑒𝑙
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐺𝑟𝑎𝑣𝑒𝑙 = (4 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡 )(1708 𝐵𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡)

3
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐺𝑟𝑎𝑣𝑒𝑙 = 6, 832 𝑓𝑡 𝑜𝑓 𝐺𝑟𝑎𝑣𝑒𝑙

Volume of Water,

3
𝑓𝑡 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑊𝑎𝑡𝑒𝑟 = (. 935829 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡
)(1708 𝐵𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡)

3
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐺𝑟𝑎𝑣𝑒𝑙 = 1598. 396 𝑓𝑡 𝑜𝑓 𝑊𝑎𝑡𝑒𝑟

Weight of Reinforcing Steel Bars, 𝑊𝑅𝑆𝐵

99
 𝑊𝑅𝑆𝐵 = (. 5% − 1%) 𝑊𝑓

Taking the average from 0.5% to 1%, we will get 0.75% or 0.0075,

𝑊𝑅𝑆𝐵 = (. 0075)588, 000 𝑘𝑔

𝑊𝑅𝑆𝐵 = 4, 410 𝑘𝑔

APPENDIX B

Hydroelectric Power Plant Alternative 2: Francis Turbine

Solving for the Net Head;𝐻𝑁𝐸𝑇 = 𝐻𝐺𝑅𝑂𝑆𝑆

𝐻𝐺𝑅𝑂𝑆𝑆 = 𝐻ℎ𝑒𝑎𝑑𝑤𝑎𝑡𝑒𝑟 𝑒𝑙𝑒𝑣𝑎𝑡𝑖𝑜𝑛 − 𝐻𝑡𝑎𝑖𝑙𝑤𝑎𝑡𝑒𝑟 𝑒𝑙𝑒𝑣𝑎𝑡𝑖𝑜𝑛

𝐻𝐺𝑅𝑂𝑆𝑆 = 23 𝑚 − 2 𝑚

𝐻𝐺𝑅𝑂𝑆𝑆 = 21 𝑚

𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑃𝑜𝑤𝑒𝑟
Solving for the Generator Efficiency; η𝑔 = 𝑇𝑢𝑟𝑏𝑖𝑛𝑒 𝑃𝑜𝑤𝑒𝑟

Where:

𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑃𝑜𝑤𝑒𝑟 𝑜𝑓 𝐹𝑟𝑎𝑛𝑐𝑖𝑠 𝑇𝑢𝑟𝑏𝑖𝑛𝑒 = 5000 𝑘𝑊

𝑇𝑢𝑟𝑏𝑖𝑛𝑒 𝑃𝑜𝑤𝑒𝑟 𝑜𝑓 𝐹𝑟𝑎𝑛𝑐𝑖𝑠 𝑇𝑢𝑟𝑏𝑖𝑛𝑒 = 5350 𝑘𝑊

5000 𝐾𝑊
η𝑔 = 5350 𝐾𝑊

η𝑔(𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦) = 0. 9346 ≈ 93. 46 %

100
Solving for the Available Water Power, 𝑃𝑊; 𝑃𝑊 = γ𝑄𝐻𝑁𝐸𝑇

Where:

3
γ 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = 9. 8066 𝑘𝑁/𝑚

3
𝑄 𝑜𝑓 𝐾𝑎𝑝𝑙𝑎𝑛 𝑇𝑢𝑟𝑏𝑖𝑛𝑒 = 28. 3 𝑚 /𝑠
3
𝑘𝑁 𝑚
𝑃𝑊 = (9. 8066 3 )(28. 3 𝑠
)(21 𝑚)
𝑚

𝑃𝑊 = 5, 828. 1 𝑘𝑊 ≈ 5. 828 𝑀𝑊

𝑇𝑢𝑟𝑏𝑖𝑛𝑒 𝑃𝑜𝑤𝑒𝑟
Solving for the Turbine Efficiency; η𝑡 = 𝑊𝑎𝑡𝑒𝑟 𝑃𝑜𝑤𝑒𝑟

5350 𝑘𝑊
η𝑡 = 5828.1 𝑘𝑊

η𝑡 = 0. 9180 = 91. 80%

𝐻
Solving for the length of Penstock; 𝐿𝑝 = 𝑆𝑖𝑛θ

Where:

𝐻 = 𝑛𝑒𝑡 ℎ𝑒𝑎𝑑

𝑥 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑠𝑢𝑐𝑡𝑖𝑜𝑛 𝑝𝑖𝑝𝑒 𝑡𝑜 𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒 𝑝𝑖𝑝𝑒 (413. 83 𝑚 𝑏𝑎𝑠𝑒𝑑 𝑜𝑛 𝐹𝑖𝑔. 2 − 7)

101
θ = 𝑎𝑛𝑔𝑙𝑒 𝑓𝑟𝑜𝑚 𝑠𝑢𝑐𝑡𝑖𝑜𝑛 𝑝𝑖𝑝𝑒 𝑡𝑜 𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒 𝑝𝑖𝑝𝑒

−1 𝐻
θ = 𝑡𝑎𝑛 ( 𝑥
)

−1 21 𝑚
θ = 𝑡𝑎𝑛 ( 413.83 𝑚 )

θ = 2. 905009701

21 𝑚
𝐿𝑝 = 𝑆𝑖𝑛(2.905009701))

𝐿𝑝 = 414. 3625 𝑚

2 2
(𝑛𝑝) (𝑄) (𝐿𝑝) 0.1875
Solving for the size of Penstock; 𝐷𝑝 = 2. 69[ 𝐻𝑔
]

Where:

𝑛𝑝 𝑜𝑓 𝑤𝑒𝑙𝑑𝑒𝑑 𝑠𝑡𝑒𝑒𝑙 (𝑏𝑎𝑠𝑒𝑑 𝑜𝑛 𝑇𝑎𝑏𝑙𝑒 2 − 3 ) = 0. 012

2 3 2
(0.012) (28.3 𝑚 /𝑠) (414.3625 𝑚) 0.1875
𝐷𝑝 = 2. 69[ 21 𝑚
]

𝐷𝑝 = 3. 1384 𝑚

2
Π(𝐷𝑝)
Solving for the Area of the Penstock, 𝐴𝑝; 𝐴𝑝 = 4

2
Π(3.1383 𝑚)
𝐴𝑝 = 4

2
𝐴𝑝 = 7. 7358 𝑚

𝑃𝐷𝑝𝐹.𝑆
Solving for the Thickness of the Penstock, 𝑡𝑝; 𝑡𝑝 = 2𝑆𝑦

Where:

102
𝐹. 𝑆 = 2 (𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑)

𝑆𝑦 = 250, 000 𝑘𝑃𝑎 (𝑏𝑎𝑠𝑒𝑑 𝑜𝑛 𝑇𝑎𝑏𝑙𝑒 2 − 4)

As for,

𝑃 = γℎ

𝑘𝑁
𝑃 = (9. 8066 3 )(21 𝑚)
𝑚

𝑃 = 205. 9386 𝑘𝑃𝑎

Hence,

(205.9386 𝑘𝑃𝑎)(3.1384 𝑚)(2)

𝑡𝑝 = 2(250,000 𝑘𝑃𝑎)

𝑡𝑝 = 0. 002585 𝑚

Solving for Draft Tube Parameters,

For Height of the Draft tube: 𝐻𝑑 = (2 𝑡𝑜 3. 5)𝐷𝑝

𝑇𝑎𝑘𝑖𝑛𝑔 𝑡ℎ𝑒 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑓𝑟𝑜𝑚 2 𝑡𝑜 3. 5 𝑔𝑖𝑣𝑒𝑠 2. 75,

𝐻𝑑 = (2. 75)(3. 1384 𝑚)

𝐻𝑑 = 8. 6306 𝑚

103
For the diameter of the draft tube, ; 𝐷𝑑 = 2𝐷𝑖 + 𝐷𝑝

Where:

𝐷𝑖 = 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟

𝐷𝑖 𝐻𝑑
Sin law; ◦ = ◦
𝑠𝑖𝑛 4 𝑠𝑖𝑛 86

◦
𝑠𝑖𝑛 4
𝐷𝑖 = 𝐻𝑑( ◦ )
𝑠𝑖𝑛 86

◦
𝑠𝑖𝑛 4
𝐷𝑖 = 8. 6306 𝑚 ( ◦ )
𝑠𝑖𝑛 86

𝐷𝑖 = 0. 6035 𝑚

Then,

𝐷𝑑 = 2𝐷𝑖 + 𝐷𝑝

𝐷𝑑 = 2(0. 6035 𝑚) + 3. 1384 𝑚

104
 𝐷𝑑 = 4. 3454 𝑚

𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑃𝑒𝑛𝑠𝑡𝑜𝑐𝑘
Solving for the number of pipes needed for the penstock; 𝑛𝑝𝑖𝑝𝑒 = 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑝𝑖𝑝𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑝𝑒𝑟 𝑝𝑖𝑒𝑐𝑒

Where:

𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑝𝑖𝑝𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑚

𝑝𝑖𝑒𝑐𝑒
=6 𝑝𝑖𝑒𝑐𝑒

𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑃𝑒𝑛𝑠𝑡𝑜𝑐𝑘
𝑛𝑝𝑖𝑝𝑒 = 𝑚
6 𝑝𝑖𝑒𝑐𝑒𝑠

414.3625 𝑚
𝑛𝑝𝑖𝑝𝑒 = 𝑚
6 𝑝𝑖𝑒𝑐𝑒𝑠

𝑛𝑝𝑖𝑝𝑒 = 69. 0604 𝑝𝑐𝑠 ≈ 69 𝑝𝑐𝑠

𝑆𝑡 𝐹 (𝐹)(𝐹.𝑆)
Solving for the number of bolts; 𝐹.𝑆
= (𝐴𝑏)(𝑛𝑏)
→ 𝑛𝑏 = (𝐴𝑏)(𝑆𝑡)

Where:

𝐷𝑏 = 1𝑖𝑛 ≈ 0. 02541 𝑚 (𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑚𝑎𝑟𝑘𝑒𝑡)

𝑆𝑡 = 100 𝑀𝑃𝑎 (𝑏𝑎𝑠𝑒𝑑 𝑜𝑛 𝐴𝑚𝑒𝑟𝑖𝑐𝑎𝑛 𝑆𝑜𝑐𝑖𝑒𝑡𝑦 𝑓𝑜𝑟 𝑇𝑒𝑠𝑡𝑖𝑛𝑔 𝑎𝑛𝑑 𝑀𝑎𝑡𝑒𝑟𝑖𝑎𝑙)

𝐹. 𝑆 = 2 (𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑)

𝐹
As for the Force; 𝑃 = 𝐴𝑝
→ 𝐹 = (𝑃)(𝐴𝑝)

2 2
𝐹 = (205. 9386 𝑘𝑁/𝑚 )(7. 7358 𝑚 )

105
𝐹 = 1593. 0998 𝑘𝑁

2
Π(0.02541 𝑚)
And Area of Bolts; 𝐴𝑏 = 4

2
𝐴𝑏 = 0. 000507 𝑚

Therefore,

(𝐹)(𝐹.𝑆)
𝑛𝑏 = (𝐴𝑏)(𝑆𝑡)

(1593.0998 𝑘𝑁)(2)
𝑛𝑏 = 2 2
(0.000507 𝑚 )(100,000 𝑘𝑁/𝑚 )

𝑛𝑏 = 62. 8442 𝑝𝑐𝑠 ≈ 63 𝑝𝑐𝑠

𝑄
Solving for Water Velocity; 𝑄 = 𝐴𝑝𝑉→𝑉 = 𝐴𝑝

3
28.3 𝑚 /𝑠
𝑉= 2
7.7358 𝑚

𝑉 = 3. 6583 𝑚/𝑠

Machine Foundation

First step in designing a machine foundation: Start with the simplest form which is the rectangular block

machine foundation.

Considering a rectangular machine foundation;

For safety purposes; 𝑊𝑓 ≥ 𝑚𝑖𝑛. 𝑊𝑓

106
The foundation mass should be from 3 to 5 times the weight of the machinery it is supposed to support.

Figure 1: Top View of Rectangular Block Machine Foundation

For Top Length of the Foundation, L ; 𝐿 = 𝑏 + 2(𝑐)

Where:

b = diameter of the base plate

also, clearance (c) to be used is 1 m for the TBOC (Turbine Bearing Oil Control) and for the turbine to be

use,

𝐿 = 𝑏 + 2(𝑐)

𝐿 = 6 + 2(1𝑚)

𝐿 = 8𝑚 = 𝑊

Solving for top Area, 𝐴𝑇; 𝐴𝑇 = 𝐴𝐵 = 𝐿 𝑥 𝑊

Note: since the equipment to be supported is circular, we will use a square type foundation.

2
𝐴𝑇 = (8)(8) 𝑚

107
 2
𝐴𝑇 = 64 𝑚

Solving for height of bed plate, ℎ

Assuming the Height of the bed plate is 5m.

ℎ = 5𝑚

Solving for Volume of the Foundation, 𝑉𝑓; 𝑉𝑓 = 𝐿 𝑥 𝑊 𝑥 ℎ

From Machinery Foundation; by Jose M. Perez, Jr.; PME

3
𝑉𝑓 = (8 * 8 * 5) 𝑚

3
𝑉𝑓 = 320 𝑚

Solving for Weight of the Foundation, 𝑊𝑓; 𝑊𝑓 = γ𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑥 𝑉𝑓

𝑘𝑔
𝑛𝑜𝑡𝑒: γ𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 = 2, 400 3
𝑚

𝑘𝑔 3
𝑊𝑓 = 2, 400 3 × 320 𝑚
𝑚

𝑊𝑓 = 768, 000 𝑘𝑔

Total Weight of the Engine, 𝑊𝑒; 𝑊𝑒 = 𝑊𝑇 + 𝑊𝐺𝑒𝑛

𝑊𝑒 = (53 + 94) 𝑚𝑒𝑡𝑟𝑖𝑐 𝑡𝑜𝑛𝑠

108
 𝑊𝑒 = 147 𝑚𝑒𝑡𝑟𝑖𝑐 𝑡𝑜𝑛𝑠

For the Minimum Weight of the Foundation; 𝑚𝑖𝑛. 𝑊𝑓 = 3 𝑡𝑜 5 𝑥 𝑊𝑒

From Possible Steps in Designing Machine Foundation by Power Plant Engineering by Frederick T. Morse.

Say we used 3 as multiplier,

1000 𝑘𝑔
𝑚𝑖𝑛. 𝑊𝑓 = 4 × 147 𝑚𝑒𝑡𝑟𝑖𝑐 𝑡𝑜𝑛𝑠 × 1 𝑚𝑒𝑡𝑟𝑖𝑐 𝑡𝑜𝑛

𝑚𝑖𝑛. 𝑊𝑓 = 588, 000 𝑘𝑔

𝑊𝑓 ≥ 𝑚𝑖𝑛. 𝑊𝑓

768, 000 𝑘𝑔 ≥ 588, 000 𝑘𝑔; The dimensions are OK!

Checking for Safety

Distribute the weight of the machine, the machine bed plate, and its own weight over a safe subsoil area.

(Source: Power Plant Engineering by P.J. Morse, page 106).

𝑆𝑖 <𝑆𝑑

𝑊𝑓 + 𝑊𝑒
For Induced Strength; 𝑆𝑖 = 𝐴𝑇

768,000 𝑘𝑔 + 147,000 𝑘𝑔
𝑆𝑖 = 2
64 𝑚

2
𝑆𝑖 = 14, 296. 875 𝑘𝑔/𝑚

109
 𝑆𝐵𝑃
For Design Stress, 𝑆𝑑; 𝑆𝑑 = 𝐹.𝑆

Where: SBP = Soil Bearing Pressure

Type of Soil US ton/sq. ft

Native Rock 200 up

Ashlar Masonry 25 – 30

Best Brick Masonry 15 – 20

Common Brick Masonry 5 – 10

Clay, Compact 5–8

Clay, Soft 1–2

Gravel and Sand 8 – 10

Sand 2–6

Quick Sand, Loom etc. 0.5 – 1

Table B-1: Safe Bearing Pressure of Soil

(Source: Machinery Foundation; by Jose M. Perez, Jr.; PME)

𝑒𝑛𝑔𝑙𝑖𝑠ℎ 𝑡𝑜𝑛
𝑆𝑜𝑖𝑙 𝐵𝑒𝑎𝑟𝑖𝑛𝑔 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = (5 − 8) 2
𝑓𝑡

Using 7 as a value,

𝑒𝑛𝑔𝑙𝑖𝑠ℎ 𝑡𝑜𝑛
𝑆𝐵𝑃𝐴𝑣𝑒 = (5) 2
𝑓𝑡

2
𝑒𝑛𝑔𝑙𝑖𝑠ℎ 𝑡𝑜𝑛 2000 𝑙𝑏𝑠 1 𝑓𝑡
𝑆𝐵𝑃𝐴𝑣𝑒 = (5) 2 × 1 𝑈𝑆 𝑡𝑜𝑛
× 2 2
𝑓𝑡 0.3048 𝑚

110
 𝑘𝑔
𝑆𝐵𝑃𝐴𝑣𝑒 = 107639. 1042 2
𝑚

Note: Safe bearing pressure of soils for machine foundation area from quarter to a half of the above values.

We use a factor of safety of 4 to attain the maximum bearing pressure. So,

𝑘𝑔
107639.1042 2
𝑚
𝑆𝑑 = 4

𝑘𝑔
𝑆𝑑 = 26909. 7761 2
𝑚

𝑆𝑖 ≤𝑆𝑑; The design of the machine foundation is SAFE!

For Material Estimation:

Bulk Specific Gravity SpecificWeight

Cement 3.15 94 lbf / bag of C

Sand 2.64 110 lbf / bag of C

Gravel 2.66 96 lbf / bag of C

Table B-2: Properties of Cementing Materials

3 3
γ 𝑜𝑓 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 = 2, 400 𝑘𝑔/𝑚 = 149. 9339 𝑙𝑏𝑓/𝑓𝑡

For water, 7 gal per bag of cement (if not specified)

3
γ 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 =62. 4 𝑙𝑏𝑓/𝑓𝑡

111
 Sand Gravel (𝒇𝒕3 Yield (𝒇𝒕3) 𝒐𝒇
(𝒇𝒕3 / /𝒃𝒂𝒈) 𝒄𝒐𝒏𝒄𝒓𝒆𝒕𝒆
Cement Cement (Bag C) 𝒃𝒂𝒈) 𝒃𝒂𝒈 𝒐𝒇
𝒄𝒆𝒎𝒆𝒏𝒕

AA 1 1.5 3 4.15077

A 1 2 4 5.063004

B 1 2.5 5 5.97524

C 1 3 6 6.88748

D 1 3 6 8.71195

Table B-3: Different Class of Concrete Mixture

Note we will use Class A concrete. Class A Concrete: C:S:G = 1:2:4

For Yield Materials Estimate:

𝑌𝐼𝐸𝐿𝐷 = Σ𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙, (𝐶, 𝑆, 𝐺 𝑎𝑛𝑑 𝐻2𝑂)

Where,

𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙

𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 = (𝑏𝑢𝑙𝑘 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑔𝑟𝑎𝑣𝑖𝑡𝑦)(𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟)

𝑙𝑏 𝑓
94 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡
For Cement; 𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡 = 𝑙𝑏𝑓
(3.15)(62.4 3 )
𝑓𝑡

3
𝑓𝑡 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡
𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡 =. 478225 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡

𝑙𝑏𝑓 𝑓𝑡
3
(110 3 )(2 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡 )
For Sand; 𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑎𝑛𝑑 = 𝑓𝑡
𝑙𝑏𝑓
(2.64)(62.4 3 )
𝑓𝑡

112
 3
𝑓𝑡 𝑜𝑓 𝑠𝑎𝑛𝑑
𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑎𝑛𝑑 = 1. 335470 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡

𝑙𝑏𝑓 𝑓𝑡
3
(96 3 )(4 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡 )
For Gravel; 𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑎𝑛𝑑 = 𝑓𝑡
𝑙𝑏𝑓
(2.66)(62.4 3 )
𝑓𝑡

3
𝑓𝑡 𝑜𝑓 𝑔𝑟𝑎𝑣𝑒𝑙
𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = 2. 313476 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡

3
𝑔𝑎𝑙 1 𝑓𝑡
For Water; 𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = 7 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡
𝑥 7.48 𝑔𝑎𝑙

3
𝑓𝑡 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 =. 935829 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡

𝑌𝐼𝐸𝐿𝐷 = Σ𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙, (𝐶, 𝑆, 𝐺 𝑎𝑛𝑑 𝐻2𝑂)

3
𝑓𝑡
𝑌𝐼𝐸𝐿𝐷 = Σ(. 478225 + 1. 335470 + 2. 313476 + . 935829) 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡

3
𝑓𝑡
𝑌𝐼𝐸𝐿𝐷 = 5. 063 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡

𝑉𝑓
For Number of Bags of Cement; 𝑁𝑜. 𝑜𝑓 𝐵𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡 = 𝑌𝐼𝐸𝐿𝐷

3
3 3.28 𝑓𝑡
320 𝑚 ( 3 )
1𝑚
𝑁𝑜. 𝑜𝑓 𝐵𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡 = 𝑓𝑡
3
5.063 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡

𝑁𝑜. 𝑜𝑓 𝐵𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡 = 2230. 3015 𝑏𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡

𝑁𝑜. 𝑜𝑓 𝐵𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡 ≈ 2, 230 𝑏𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡

For the Volume cementing materials,

113
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 = (𝑃𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝑜𝑓 𝑀𝐴𝑡𝑒𝑟𝑖𝑎𝑙)(𝑁𝑜. 𝑜𝑓 𝐵𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡)

3
𝑓𝑡 𝑜𝑓 𝑠𝑎𝑛𝑑
Volume of Sand; 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑆𝑎𝑛𝑑 = (2 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡 )(2230 𝐵𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡)

3
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑆𝑎𝑛𝑑 = 4460 𝑓𝑡 𝑜𝑓 𝑠𝑎𝑛𝑑

3
𝑓𝑡 𝑜𝑓 𝑔𝑟𝑎𝑣𝑒𝑙
Volume of Gravel; 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐺𝑟𝑎𝑣𝑒𝑙 = (4 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡 )(2230 𝐵𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡)

3
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐺𝑟𝑎𝑣𝑒𝑙 = 8920 𝑓𝑡 𝑜𝑓 𝐺𝑟𝑎𝑣𝑒𝑙

3
𝑓𝑡 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
Volume of Water; 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑊𝑎𝑡𝑒𝑟 = (. 935829 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡
)(2230 𝐵𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡)

3
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑊𝑎𝑡𝑒𝑟 = 2086. 8987 𝑓𝑡 𝑜𝑓 𝑊𝑎𝑡𝑒𝑟

Weight of Reinforcing Steel Bars; 𝑊𝑅𝑆𝐵 = (. 5% − 1%) 𝑊𝑓

Using the 0.65% or 0.0065 from 0.5% to 1%,

𝑊𝑅𝑆𝐵 = (. 0065)768000 𝑘𝑔

𝑊𝑅𝑆𝐵 = 4992 𝑘𝑔

APPENDIX C

Hydroelectric Power Plant Alternative 3: Kaplan Turbine

Solving for the Net Head, 𝐻𝑁𝐸𝑇

114
 𝐻𝑁𝐸𝑇 = 𝐻𝐺𝑅𝑂𝑆𝑆

𝐻𝐺𝑅𝑂𝑆𝑆 = 𝐻ℎ𝑒𝑎𝑑𝑤𝑎𝑡𝑒𝑟 𝑒𝑙𝑒𝑣𝑎𝑡𝑖𝑜𝑛 − 𝐻𝑡𝑎𝑖𝑙𝑤𝑎𝑡𝑒𝑟 𝑒𝑙𝑒𝑣𝑎𝑡𝑖𝑜𝑛

𝐻𝐺𝑅𝑂𝑆𝑆 = 15 𝑚 − 6 𝑚

𝐻𝐺𝑅𝑂𝑆𝑆 = 9 𝑚

Solving for the Generator Efficiency, η𝑔

Where:

𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑃𝑜𝑤𝑒𝑟 𝑜𝑓 𝐾𝑎𝑝𝑙𝑎𝑛 𝑇𝑢𝑟𝑏𝑖𝑛𝑒 = 5000 𝑘𝑊

𝑇𝑢𝑟𝑏𝑖𝑛𝑒 𝑃𝑜𝑤𝑒𝑟 𝑜𝑓 𝐾𝑎𝑝𝑙𝑎𝑛 𝑇𝑢𝑟𝑏𝑖𝑛𝑒 = 5264 𝑘𝑊

𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑃𝑜𝑤𝑒𝑟
η𝑔 = 𝑇𝑢𝑟𝑏𝑖𝑛𝑒 𝑃𝑜𝑤𝑒𝑟

5000 𝐾𝑊
η𝑔 = 5264 𝐾𝑊

η𝑔 = 0. 9498 ≈ 94. 98 %

Solving for the Available Water Power, 𝑃𝑊

Where:

3
γ 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = 9. 8066 𝑘𝑁/𝑚

3
𝑄 𝑜𝑓 𝐾𝑎𝑝𝑙𝑎𝑛 𝑇𝑢𝑟𝑏𝑖𝑛𝑒 = 63. 5 𝑚 /𝑠

𝑃𝑊 = γ𝑄𝐻𝑁𝐸𝑇

3
𝑘𝑁 𝑚
𝑃𝑊 = (9. 8066 3 )(63. 5 𝑠
)(9 𝑚)
𝑚

𝑃𝑊 = 5, 604. 47 𝑘𝑊

Solving for the Turbine Efficiency, η𝑡

𝑇𝑢𝑟𝑏𝑖𝑛𝑒 𝑃𝑜𝑤𝑒𝑟
η𝑡 = 𝑊𝑎𝑡𝑒𝑟 𝑃𝑜𝑤𝑒𝑟

115
 5264 𝑘𝑊
η𝑡 = 5604.67 𝑘𝑊

η𝑡 = 0. 939 = 93. 92%

Solving for the length of Penstock, 𝐿𝑝

Where:

𝐻 = 𝑛𝑒𝑡 ℎ𝑒𝑎𝑑

𝑥 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑠𝑢𝑐𝑡𝑖𝑜𝑛 𝑝𝑖𝑝𝑒 𝑡𝑜 𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒 𝑝𝑖𝑝𝑒 (351. 8 𝑚 𝑏𝑎𝑠𝑒𝑑 𝑜𝑛 𝐹𝑖𝑔. 2 − 9)

θ = 𝑎𝑛𝑔𝑙𝑒 𝑓𝑟𝑜𝑚 𝑠𝑢𝑐𝑡𝑖𝑜𝑛 𝑝𝑖𝑝𝑒 𝑡𝑜 𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒 𝑝𝑖𝑝𝑒

−1 𝐻
θ = 𝑡𝑎𝑛 ( 𝑥
)

−1 9𝑚
θ = 𝑡𝑎𝑛 ( 351.8 𝑚 )

θ = 1. 46

𝐻
𝐿𝑝 = 𝑆𝑖𝑛θ

9𝑚
𝐿𝑝 = 𝑆𝑖𝑛(1.46546)

𝐿𝑝 = 351. 97 𝑚

Solving for the size of Penstock, 𝐷𝑝

116
 2 2
(𝑛𝑝) (𝑄) (𝐿𝑝) 0.1875
𝐷𝑝 = 2. 69[ 𝐻𝑔
]

Where:

𝑛𝑝 𝑜𝑓 𝑤𝑒𝑙𝑑𝑒𝑑 𝑠𝑡𝑒𝑒𝑙 (𝑏𝑎𝑠𝑒𝑑 𝑜𝑛 𝑇𝑎𝑏𝑙𝑒 2 − 3 ) = 0. 012

2 3 2
(0.012) (63.5 𝑚 /𝑠) (351.97 𝑚) 0.1875
𝐷𝑝 = 2. 69[ 9𝑚
]

𝐷𝑝 = 4. 8309 𝑚

Solving for the Area of the Penstock, 𝐴𝑝

2
Π(𝐷𝑝)
𝐴𝑝 = 4

2
Π(4.8309 𝑚)
𝐴𝑝 = 4

2
𝐴𝑝 = 18. 32 𝑚

Solving for the Thickness of the Penstock, 𝑡𝑝

𝑃𝐷𝑝𝐹.𝑆
𝑡𝑝 = 2𝑆𝑦

Where:

𝐹. 𝑆 = 2 (𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑)

𝑆𝑦 = 250, 000 𝑘𝑃𝑎 (𝑏𝑎𝑠𝑒𝑑 𝑜𝑛 𝑇𝑎𝑏𝑙𝑒 2 − 4)

As for,

117
 𝑃 = γℎ

𝑘𝑁
𝑃 = (9. 8066 3 )(9 𝑚)
𝑚

𝑃 = 88. 2594 𝑘𝑃𝑎

Hence,

(88.2594 𝑘𝑃𝑎)(4.8309 𝑚)(2)

𝑡𝑝 = 2(250,000 𝑘𝑃𝑎)

𝑡𝑝 = 0. 0017 𝑚

Solving for Draft Tube Parameters,

For Height of the Draft tube:

𝐻𝑑 = (2 𝑡𝑜 3. 5)𝐷𝑝

𝑇𝑎𝑘𝑖𝑛𝑔 𝑡ℎ𝑒 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑓𝑟𝑜𝑚 2 𝑡𝑜 3. 5 𝑔𝑖𝑣𝑒𝑠 2. 75,

𝐻𝑑 = (2. 75)(4. 8309 𝑚)

𝐻𝑑 = 13. 2850 𝑚

For the diameter of the draft tube,

𝐷𝑑 = 2𝐷𝑖 + 𝐷𝑝

118
Where:

𝐷𝑖 = 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟

Sin law,

𝐷𝑖 𝐻𝑑
◦ = ◦
𝑠𝑖𝑛 4 𝑠𝑖𝑛 86

◦
𝑠𝑖𝑛 4
𝐷𝑖 = 𝐻𝑑( ◦ )
𝑠𝑖𝑛 86

◦
𝑠𝑖𝑛 4
𝐷𝑖 = 13. 2850 𝑚 ( ◦ )
𝑠𝑖𝑛 86

𝐷𝑖 = 0. 9290 𝑚

Then,

𝐷𝑑 = 2𝐷𝑖 + 𝐷𝑝

𝐷𝑑 = 2(0. 9290 𝑚) + 4. 8309 𝑚

𝐷𝑑 = 6. 6889 𝑚

119
Solving for the number of pipes needed for the penstock

𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑃𝑒𝑛𝑠𝑡𝑜𝑐𝑘
𝑛𝑝𝑖𝑝𝑒 = 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑝𝑖𝑝𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑝𝑒𝑟 𝑝𝑖𝑒𝑐𝑒

Where:

𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑝𝑖𝑝𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑚

𝑝𝑖𝑒𝑐𝑒
=6 𝑝𝑖𝑒𝑐𝑒

𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑃𝑒𝑛𝑠𝑡𝑜𝑐𝑘
𝑛𝑝𝑖𝑝𝑒 = 𝑚
6 𝑝𝑖𝑒𝑐𝑒𝑠

351.97 𝑚
𝑛𝑝𝑖𝑝𝑒 = 𝑚
6 𝑝𝑖𝑒𝑐𝑒𝑠

𝑛𝑝𝑖𝑝𝑒 = 58. 66 𝑝𝑐𝑠 ≈ 59 𝑝𝑐𝑠

Solving for the number of bolts

𝑆𝑡 𝐹 (𝐹)(𝐹.𝑆)
𝐹.𝑆
= (𝐴𝑏)(𝑛𝑏)
→ 𝑛𝑏 = (𝐴𝑏)(𝑆𝑡)

Where:

𝐷𝑏 = 1𝑖𝑛 ≈ 0. 02541 𝑚 (𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑚𝑎𝑟𝑘𝑒𝑡)

𝑆𝑡 = 100 𝑀𝑃𝑎 (𝑏𝑎𝑠𝑒𝑑 𝑜𝑛 𝐴𝑚𝑒𝑟𝑖𝑐𝑎𝑛 𝑆𝑜𝑐𝑖𝑒𝑡𝑦 𝑓𝑜𝑟 𝑇𝑒𝑠𝑡𝑖𝑛𝑔 𝑎𝑛𝑑 𝑀𝑎𝑡𝑒𝑟𝑖𝑎𝑙)

𝐹. 𝑆 = 2 (𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑)

As for the Force,

120
 𝐹
𝑃= 𝐴𝑝
→ 𝐹 = (𝑃)(𝐴𝑝)

2 2
𝐹 = (88. 2594 𝑘𝑁/𝑚 )(18. 32 𝑚 )

𝐹 = 1616. 9122 𝑘𝑁

And Area of Bolts,

2
Π(0.02541 𝑚)
𝐴𝑏 = 4

2
𝐴𝑏 = 0. 000507 𝑚

Therefore,

(𝐹)(𝐹.𝑆)
𝑛𝑏 = (𝐴𝑏)(𝑆𝑡)

(1616.9122 𝑘𝑁)(2)
𝑛𝑏 = 2 2
(0.000507 𝑚 )(100,000 𝑘𝑁/𝑚 )

𝑛𝑏 = 63. 8 𝑝𝑐𝑠 ≈ 64 𝑝𝑐𝑠

Solving for Water Velocity

𝑄
𝑄 = 𝐴𝑝𝑉→𝑉 = 𝐴𝑝

3
63.5 𝑚 /𝑠
𝑉= 2
18.32 𝑚

𝑉 = 3. 466 𝑚/𝑠

121
Machine Foundation

First step in designing a machine foundation: Start with the simplest form which is the rectangular block

machine foundation.

Considering a rectangular machine foundation;

For safety purposes,

𝑊𝑓 ≥ 𝑚𝑖𝑛. 𝑊𝑓

Then,

The foundation mass should be from 3 to 5 times the weight of the machinery it is supposed to support.

Figure 1: Top View of Rectangular Block Machine Foundation

For Top Length of the Foundation, L:

𝐿 = 𝑏 + 2(𝑐)

Where:

b = diameter of the base plate

122
also, clearance (c) to be used is 1 m for the TBOC (Turbine Bearing Oil Control) and for the turbine to be

use,

so,

𝐿 = 𝑏 + 2(𝑐)

𝐿 = 5 + 2(1)

𝐿 = 7𝑚

Solving for top Area, 𝐴𝑇

Note: since the equipment to be supported is circular, we will use a square type foundation.

𝐴𝑇 = 𝐴𝐵 = 𝐿 𝑥 𝑊

2
𝐴𝑇 = 7 × 7 𝑚

2
𝐴𝑇 = 42 𝑚

Solving for height of bed plate, ℎ

Assuming the Height of the bed plate is 4.5m.

ℎ = 4. 5 𝑚

Solving for Volume of the Foundation, 𝑉𝑓

From Machinery Foundation; by Jose M. Perez, Jr.; PME

123
 𝑉𝑓 = 𝐿 𝑥 𝑊 𝑥 ℎ

3
𝑉𝑓 = 7 × 7 × 4. 5 𝑚

3
𝑉𝑓 = 220. 5 𝑚

Solving for Weight of the Foundation, 𝑊𝑓

𝑘𝑔
𝑛𝑜𝑡𝑒: γ𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 = 2, 400 3
𝑚

𝑊𝑓 = γ𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑥 𝑉𝑓

𝑘𝑔 3
𝑊𝑓 = 2, 400 3 × 220. 5 𝑚
𝑚

𝑊𝑓 = 529, 200 𝑘𝑔

Total Weight of the Engine, 𝑊𝑒

𝑊𝑒 = 𝑊𝑇 + 𝑊𝐺𝑒𝑛

𝑊𝑒 = (50 + 92) 𝑚𝑒𝑡𝑟𝑖𝑐 𝑡𝑜𝑛𝑠

𝑊𝑒 = 142 𝑚𝑒𝑡𝑟𝑖𝑐 𝑡𝑜𝑛𝑠

For the Minimum Weight of the Foundation, 𝑚𝑖𝑛. 𝑊𝑓

From Possible Steps in Designing Machine Foundation by Power Plant Engineering by Frederick T. Morse.

124
 𝑚𝑖𝑛. 𝑊𝑓 = 3 𝑡𝑜 5 𝑥 𝑊𝑒

Say we used 3 as multiplier,

1000 𝑘𝑔
𝑚𝑖𝑛. 𝑊𝑓 = 3 × 142 𝑚𝑒𝑡𝑟𝑖𝑐 𝑡𝑜𝑛𝑠 × 1 𝑚𝑒𝑡𝑟𝑖𝑐 𝑡𝑜𝑛

𝑚𝑖𝑛. 𝑊𝑓 = 426, 000 𝑘𝑔

Hence,

𝑊𝑓 ≥ 𝑚𝑖𝑛. 𝑊𝑓

529, 200 𝑘𝑔 ≥ 426, 000 𝑘𝑔

The dimensions are OK!

Checking for Safety

Distribute the weight of the machine, the machine bed plate, and its own weight over a safe subsoil area.

(Source: Power Plant Engineering by P.J. Morse, page 106).

𝑆𝑖 <𝑆𝑑

For Induced Strength, 𝑆𝑖

𝑊𝑓 + 𝑊𝑒
𝑆𝑖 = 𝐴𝐵

529,200 𝑘𝑔 + 142,000 𝑘𝑔
𝑆𝑖 = 2
42 𝑚

125
 2
𝑆𝑖 = 15, 980. 9524 𝑘𝑔/𝑚

For Design Stress, 𝑆𝑑

𝑆𝐵𝑃
𝑆𝑑 = 𝐹.𝑆

Where: SBP = Soil Bearing Pressure

Type of Soil US ton/sq. ft

Native Rock 200 up

Ashlor Masonry 25 – 30

Best Brick Masonry 15 – 20

Common Brick Masonry 5 – 10

Clay, Compact 5–8

Clay, Soft 1–2

Gravel and Sand 8 – 10

Sand 2–6

Quick Sand, Loom etc. 0.5 – 1

Table B-1: Safe Bearing Pressure of Soil

(Source: Machinery Foundation; by Jose M. Perez, Jr.; PME)

𝑒𝑛𝑔𝑙𝑖𝑠ℎ 𝑡𝑜𝑛
𝑆𝑜𝑖𝑙 𝐵𝑒𝑎𝑟𝑖𝑛𝑔 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = (5 − 8) 2
𝑓𝑡

Using average value,

126
 𝑒𝑛𝑔𝑙𝑖𝑠ℎ 𝑡𝑜𝑛
𝑆𝐵𝑃𝐴𝑣𝑒 = (6. 5) 2
𝑓𝑡

2
𝑒𝑛𝑔𝑙𝑖𝑠ℎ 𝑡𝑜𝑛 2000 𝑙𝑏𝑠 1 𝑓𝑡
𝑆𝐵𝑃𝐴𝑣𝑒 = (6. 5) 2 × 1 𝑈𝑆 𝑡𝑜𝑛
× 2 2
𝑓𝑡 0.3048 𝑚

𝑘𝑔
𝑆𝐵𝑃𝐴𝑣𝑒 = 139, 930. 8354 2
𝑚

Note: Safe bearing pressure of soils for machine foundation area from quarter to a half of the above values.

We use a factor of safety of 4 to attain the maximum bearing pressure. So,

𝑘𝑔
139,930.8354 2
𝑚
𝑆𝑑 = 4

𝑘𝑔
𝑆𝑑 = 34, 982. 7089 2
𝑚

Therefore,

𝑆𝑖 ≤𝑆𝑑

The design of the machine foundation is SAFE!

For Material Estimation:

Bulk Specific Gravity SpecificWeight

Cement 3.15 94 lbf / bag of C

Sand 2.64 110 lbf / bag of C

Gravel 2.66 96 lbf / bag of C

Table B-2: Properties of Cementing Materials

127
 3 3
γ 𝑜𝑓 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 = 2, 400 𝑘𝑔/𝑚 = 149. 9339 𝑙𝑏𝑓/𝑓𝑡

For water, 7 gal per bag of cement (if not specified)

3
γ 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 =62. 4 𝑙𝑏𝑓/𝑓𝑡

Sand Gravel (𝒇𝒕3 Yield (𝒇𝒕3) 𝒐𝒇

(𝒇𝒕3 / /𝒃𝒂𝒈) 𝒄𝒐𝒏𝒄𝒓𝒆𝒕𝒆
Cement Cement (Bag C) 𝒃𝒂𝒈) 𝒃𝒂𝒈 𝒐𝒇
𝒄𝒆𝒎𝒆𝒏𝒕

AA 1 1.5 3 4.15077

A 1 2 4 5.063004

B 1 2.5 5 5.97524

C 1 3 6 6.88748

D 1 3 6 8.71195

Table B-3: Different Class of Concrete Mixture

Note we will use Class A concrete. Class A Concrete: C:S:G = 1:2:4

For Yield Materials Estimate:

𝑌𝐼𝐸𝐿𝐷 = Σ𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙, (𝐶, 𝑆, 𝐺 𝑎𝑛𝑑 𝐻2𝑂)

Where,

𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙

𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 = (𝑏𝑢𝑙𝑘 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑔𝑟𝑎𝑣𝑖𝑡𝑦)(𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟)

For Cement,

128
 𝑓
𝑙𝑏
94 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡
𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡 = 𝑙𝑏𝑓
(3.15)(62.4 3 )
𝑓𝑡

3
𝑓𝑡 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡
𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡 =. 478225 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡

For Sand,

𝑙𝑏𝑓 𝑓𝑡
3
(110 3 )(2 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡 )
𝑓𝑡
𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑎𝑛𝑑 = 𝑙𝑏𝑓
(2.64)(62.4 3 )
𝑓𝑡

3
𝑓𝑡 𝑜𝑓 𝑠𝑎𝑛𝑑
𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑎𝑛𝑑 = 1. 335470 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡

For Gravel,

𝑙𝑏𝑓 𝑓𝑡
3
(96 3 )(4 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡 )
𝑓𝑡
𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑎𝑛𝑑 = 𝑙𝑏𝑓
(2.66)(62.4 3 )
𝑓𝑡

3
𝑓𝑡 𝑜𝑓 𝑔𝑟𝑎𝑣𝑒𝑙
𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = 2. 313476 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡

For Water,

3
𝑔𝑎𝑙 1 𝑓𝑡
𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = 7 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡
𝑥 7.48 𝑔𝑎𝑙

3
𝑓𝑡 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 =. 935829 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡

𝑌𝐼𝐸𝐿𝐷 = Σ𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙, (𝐶, 𝑆, 𝐺 𝑎𝑛𝑑 𝐻2𝑂)

3
𝑓𝑡
𝑌𝐼𝐸𝐿𝐷 = Σ(. 478225 + 1. 335470 + 2. 313476 + . 935829) 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡

129
 3
𝑓𝑡
𝑌𝐼𝐸𝐿𝐷 = 5. 063 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡

For Number of Bags of Cement,

𝑉𝑓
𝑁𝑜. 𝑜𝑓 𝐵𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡 = 𝑌𝐼𝐸𝐿𝐷

3 3
3 3.28 𝑓𝑡
220.5 𝑚 ( 3 )
1𝑚
𝑁𝑜. 𝑜𝑓 𝐵𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡 = 𝑓𝑡
3
5.063 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡

𝑁𝑜. 𝑜𝑓 𝐵𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡 = 1, 536. 8171 𝑏𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡

𝑁𝑜. 𝑜𝑓 𝐵𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡 ≈ 1, 537 𝑏𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡

For the Volume cementing materials,

𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 = (𝑃𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝑜𝑓 𝑀𝐴𝑡𝑒𝑟𝑖𝑎𝑙)(𝑁𝑜. 𝑜𝑓 𝐵𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡)

Volume of Sand,

3
𝑓𝑡 𝑜𝑓 𝑠𝑎𝑛𝑑
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 = (2 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡 )(1, 537 𝐵𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡)

3
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑆𝑎𝑛𝑑 = 3, 074 𝑓𝑡 𝑜𝑓 𝑠𝑎𝑛𝑑

Volume of Gravel,

3
𝑓𝑡 𝑜𝑓 𝑔𝑟𝑎𝑣𝑒𝑙
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐺𝑟𝑎𝑣𝑒𝑙 = (4 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡 )(1, 537 𝐵𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡)

3
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐺𝑟𝑎𝑣𝑒𝑙 = 6, 148 𝑓𝑡 𝑜𝑓 𝐺𝑟𝑎𝑣𝑒𝑙

Volume of Water,

130
 3
𝑓𝑡 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑊𝑎𝑡𝑒𝑟 = (. 935829 𝑏𝑎𝑔 𝑜𝑓 𝑐𝑒𝑚𝑒𝑛𝑡
)(469 𝐵𝑎𝑔𝑠 𝑜𝑓 𝐶𝑒𝑚𝑒𝑛𝑡)

3
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐺𝑟𝑎𝑣𝑒𝑙 = 1, 438. 3692 𝑓𝑡 𝑜𝑓 𝑊𝑎𝑡𝑒𝑟

Weight of Reinforcing Steel Bars, 𝑊𝑅𝑆𝐵

𝑊𝑅𝑆𝐵 = (. 5% − 1%) 𝑊𝑓

Taking the average from 0.5% to 1%, we will get 0.75% or 0.0075,

𝑊𝑅𝑆𝐵 = (. 0075)529, 200 𝑘𝑔

𝑊𝑅𝑆𝐵 = 3, 969 𝑘𝑔

APPENDIX D

Table D-1: Hydropower Plant CAPEX and O&M Values

(Source: National Renewable Energy Laboratory)

131
The Table D-1 shows the CAPEX or Capital Expenses and O&M or Operation and Maintenance in 2015

basis with respect to its capacity in MW and head in feet.

Figure D-1: Capital Cost Breakdown

(Source: IACSIT International Journal of Engineering and Technology, Vol. 4, No. 3, June 2012)

The graph shows the Distribution of Investments on a Hydropower Plant from Costing of Hydropower

Projects by Sachin Mishra et al. The major components of civil works are diversion channel, spillway and

power house building. The major Electro-mechanical component of the power plant is the inlet valve,

turbine, draft tube, gates, generator, control and protection equipment and substation for transformation of

power to the transmission line. In terms of space requirement and cost the major items are the turbine and

generator.

Plant Economics for Alternative 1

132
 Table D-2: Plant Values for Alternative 1

The highlighted row is the classification that covers the data of the design for alternative 1 which are the

capacity of 5 MW with the head of 6 m ≈ 19.68 ft.

Capital Cost; 𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐶𝑜𝑠𝑡 = (𝑃𝑟𝑖𝑐𝑒 𝑜𝑓 𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐶𝑜𝑠𝑡 𝑝𝑒𝑟 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦)( 𝑃𝑙𝑎𝑛𝑡 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦)

Note: 1 USD = 47.75 Php, as of May 30, 2021 (Source: www.xe.com)

𝑈𝑆𝐷 𝑃𝐻𝑃
𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐶𝑜𝑠𝑡 = (7, 277 𝑘𝑊
× 47. 75 𝑈𝑆𝐷
)(5, 000 𝑘𝑊)

𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐶𝑜𝑠𝑡 = 1, 737, 383, 750 𝑃𝐻𝑃

For Capital Cost Breakdown

For Turbo Generator Set,

𝑇𝑢𝑟𝑏𝑜 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑆𝑒𝑡 𝐶𝑜𝑠𝑡 = (𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐶𝑜𝑠𝑡)(𝑇𝑢𝑟𝑏𝑜 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑆𝑒𝑡 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒)

𝑇𝑢𝑟𝑏𝑜 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑆𝑒𝑡 𝐶𝑜𝑠𝑡 = (1, 737, 383, 750 𝑃𝐻𝑃)(0. 30)

𝑇𝑢𝑟𝑏𝑜 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑆𝑒𝑡 𝐶𝑜𝑠𝑡 = 521, 215, 125 𝑃𝐻𝑃

For Electric Regulation and Control Equipments,

𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑅𝑒𝑔𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝐶𝑜𝑛𝑡𝑟𝑜𝑙 𝐸𝑞𝑢𝑖𝑝𝑚𝑒𝑛𝑡𝑠 𝐶𝑜𝑠𝑡 = (𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐶𝑜𝑠𝑡)(𝐸𝑞𝑢𝑖𝑝𝑚𝑒𝑛𝑡𝑠 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒)

𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑅𝑒𝑔𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝐶𝑜𝑛𝑡𝑟𝑜𝑙 𝐸𝑞𝑢𝑖𝑝𝑚𝑒𝑛𝑡𝑠 𝐶𝑜𝑠𝑡 = (1, 737, 383, 750 𝑃𝐻𝑃)(0. 22)

𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑅𝑒𝑔𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝐶𝑜𝑛𝑡𝑟𝑜𝑙 𝐸𝑞𝑢𝑖𝑝𝑚𝑒𝑛𝑡𝑠 𝐶𝑜𝑠𝑡 = 382, 224, 425 𝑃𝐻𝑃

133
For Construction and Engineering Management,

𝐶𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝐸𝑛𝑔𝑖𝑛𝑒𝑒𝑟𝑖𝑛𝑔 𝑀𝑎𝑛𝑎𝑔𝑒𝑚𝑒𝑛𝑡 𝐶𝑜𝑠𝑡 = (𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐶𝑜𝑠𝑡)(𝑀𝑎𝑛𝑎𝑔𝑒𝑚𝑒𝑛𝑡 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒)

𝐶𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝐸𝑛𝑔𝑖𝑛𝑒𝑒𝑟𝑖𝑛𝑔 𝑀𝑎𝑛𝑎𝑔𝑒𝑚𝑒𝑛𝑡 𝐶𝑜𝑠𝑡 = (1, 737, 383, 750 𝑃𝐻𝑃)(0. 08)

𝐶𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝐸𝑛𝑔𝑖𝑛𝑒𝑒𝑟𝑖𝑛𝑔 𝑀𝑎𝑛𝑎𝑔𝑒𝑚𝑒𝑛𝑡 𝐶𝑜𝑠𝑡 = 138, 990, 700 𝑃𝐻𝑃

For Civil Works,

𝐶𝑖𝑣𝑖𝑙 𝑊𝑜𝑟𝑘𝑠 𝐶𝑜𝑠𝑡 = (𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐶𝑜𝑠𝑡)(𝐶𝑖𝑣𝑖𝑙 𝑊𝑜𝑟𝑘𝑠 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒)

𝐶𝑖𝑣𝑖𝑙 𝑊𝑜𝑟𝑘𝑠 𝐶𝑜𝑠𝑡 = (7, 737, 383, 750 𝑃𝐻𝑃)(0. 40)

𝐶𝑖𝑣𝑖𝑙 𝑊𝑜𝑟𝑘𝑠 𝐶𝑜𝑠𝑡 = 694, 953, 500 𝑃𝐻𝑃

For Operational and Maintenance Cost in a year

𝑂𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑛𝑑 𝑀𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒 𝐶𝑜𝑠𝑡 = (𝑂&𝑀 𝑓𝑜𝑟 𝑎𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑖𝑣𝑒 1)(𝑃𝑙𝑎𝑛𝑡 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦)

Note: 1 USD = 47.75 Php, as of May 30, 2021 (Source: www.xe.com)

𝑈𝑆𝐷 𝑃𝐻𝑃
𝑂𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑛𝑑 𝑀𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒 𝐶𝑜𝑠𝑡 = (125 𝑘𝑊
× 47. 75 𝑈𝑆𝐷
)(5, 000 𝑘𝑊)

𝑂𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑛𝑑 𝑀𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒 𝐶𝑜𝑠𝑡 = 29, 843, 750 𝑃𝐻𝑃

For the Net Income of the Power Plant

Note: Based on Fig. 1-10, the generation rate of SAMELCO I is 10.2646 php/kwh.

𝑃𝑙𝑎𝑛𝑡 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑖𝑛 𝑘𝑤ℎ 𝑝ℎ𝑝

𝐴𝑛𝑛𝑢𝑎𝑙 𝐺𝑟𝑜𝑠𝑠 𝐼𝑛𝑐𝑜𝑚𝑒 = ( 𝑦𝑒𝑎𝑟
)(𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 𝑖𝑛 𝑘𝑤ℎ
)

8,760 ℎ𝑟𝑠 𝑝ℎ𝑝

𝐴𝑛𝑛𝑢𝑎𝑙 𝐺𝑟𝑜𝑠𝑠 𝐼𝑛𝑐𝑜𝑚𝑒 = (5000 𝑘𝑤 × 𝑦𝑒𝑎𝑟
)(10. 2646 𝑘𝑤ℎ
)

8,760 ℎ𝑟𝑠 𝑝ℎ𝑝

𝐴𝑛𝑛𝑢𝑎𝑙 𝐺𝑟𝑜𝑠𝑠 𝐼𝑛𝑐𝑜𝑚𝑒 = (5000 𝑘𝑤 × 𝑦𝑒𝑎𝑟
)(10. 2646 𝑘𝑤ℎ
)

𝐴𝑛𝑛𝑢𝑎𝑙 𝐺𝑟𝑜𝑠𝑠 𝐼𝑛𝑐𝑜𝑚𝑒 = 449, 589, 480 𝑝ℎ𝑝/𝑦𝑒𝑎𝑟

134
 𝑁𝑒𝑡 𝑃𝑟𝑜𝑓𝑖𝑡 = 𝐴𝑛𝑛𝑢𝑎𝑙 𝐺𝑟𝑜𝑠𝑠 𝐼𝑛𝑐𝑜𝑚𝑒 − 𝑂&𝑀 𝐶𝑜𝑠𝑡

𝑁𝑒𝑡 𝑃𝑟𝑜𝑓𝑖𝑡 = (449, 589, 480 − 29, 843, 750) 𝑃𝐻𝑃

𝑁𝑒𝑡 𝐼𝑛𝑐𝑜𝑚𝑒 = 419, 745, 730 𝑃𝐻𝑃

𝑁𝑒𝑡 𝐼𝑛𝑐𝑜𝑚𝑒
For Return of Investment ; 𝑅𝑂𝐼 = 𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐶𝑜𝑠𝑡
𝑥 100

419,745,730 𝑝ℎ𝑝
𝑅𝑂𝐼 = 1,737,383,750 𝑝ℎ𝑝
𝑥 100

𝑅𝑂𝐼 = 0. 2415964 ≈ 24. 16 %

Plant Economics for Alternative 2

Table D-3: Plant Values for Alternative 2

The highlighted row is the classification that covers the data of the design for alternative 2 which are the

capacity of 5 MW with the head of 21 m ≈ 68.88 ft.

Capital Cost; 𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐶𝑜𝑠𝑡 = (𝑃𝑟𝑖𝑐𝑒 𝑜𝑓 𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐶𝑜𝑠𝑡 𝑝𝑒𝑟 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦)( 𝑃𝑙𝑎𝑛𝑡 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦)

Note: 1 USD = 47.75 Php, as of May 30, 2021 (Source: www.xe.com)

𝑈𝑆𝐷 𝑃𝐻𝑃
𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐶𝑜𝑠𝑡 = (6, 363 𝑘𝑊
× 47. 75 𝑈𝑆𝐷
)(5, 000 𝑘𝑊)

𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐶𝑜𝑠𝑡 = 1, 519, 166, 250 𝑃𝐻𝑃

135
For Capital Cost Breakdown

For Turbo Generator Set,

𝑇𝑢𝑟𝑏𝑜 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑆𝑒𝑡 𝐶𝑜𝑠𝑡 = (𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐶𝑜𝑠𝑡)(𝑇𝑢𝑟𝑏𝑜 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑆𝑒𝑡 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒)

𝑇𝑢𝑟𝑏𝑜 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑆𝑒𝑡 𝐶𝑜𝑠𝑡 = (1, 519, 166, 250 𝑃𝐻𝑃)(0. 30)

𝑇𝑢𝑟𝑏𝑜 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑆𝑒𝑡 𝐶𝑜𝑠𝑡 = 455, 749, 875 𝑃𝐻𝑃

For Electric Regulation and Control Equipments,

𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑅𝑒𝑔𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝐶𝑜𝑛𝑡𝑟𝑜𝑙 𝐸𝑞𝑢𝑖𝑝𝑚𝑒𝑛𝑡𝑠 𝐶𝑜𝑠𝑡 = (𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐶𝑜𝑠𝑡)(𝐸𝑞𝑢𝑖𝑝𝑚𝑒𝑛𝑡𝑠 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒)

𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑅𝑒𝑔𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝐶𝑜𝑛𝑡𝑟𝑜𝑙 𝐸𝑞𝑢𝑖𝑝𝑚𝑒𝑛𝑡𝑠 𝐶𝑜𝑠𝑡 = (1, 519, 166, 250 𝑃𝐻𝑃)(0. 22)

𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑅𝑒𝑔𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝐶𝑜𝑛𝑡𝑟𝑜𝑙 𝐸𝑞𝑢𝑖𝑝𝑚𝑒𝑛𝑡𝑠 𝐶𝑜𝑠𝑡 = 334, 216, 575 𝑃𝐻𝑃

For Construction and Engineering Management,

𝐶𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝐸𝑛𝑔𝑖𝑛𝑒𝑒𝑟𝑖𝑛𝑔 𝑀𝑎𝑛𝑎𝑔𝑒𝑚𝑒𝑛𝑡 𝐶𝑜𝑠𝑡 = (𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐶𝑜𝑠𝑡)(𝑀𝑎𝑛𝑎𝑔𝑒𝑚𝑒𝑛𝑡 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒)

𝐶𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝐸𝑛𝑔𝑖𝑛𝑒𝑒𝑟𝑖𝑛𝑔 𝑀𝑎𝑛𝑎𝑔𝑒𝑚𝑒𝑛𝑡 𝐶𝑜𝑠𝑡 = (1, 519, 166, 250 𝑃𝐻𝑃)(0. 08)

𝐶𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝐸𝑛𝑔𝑖𝑛𝑒𝑒𝑟𝑖𝑛𝑔 𝑀𝑎𝑛𝑎𝑔𝑒𝑚𝑒𝑛𝑡 𝐶𝑜𝑠𝑡 = 121, 533, 300 𝑃𝐻𝑃

For Civil Works,

𝐶𝑖𝑣𝑖𝑙 𝑊𝑜𝑟𝑘𝑠 𝐶𝑜𝑠𝑡 = (𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐶𝑜𝑠𝑡)(𝐶𝑖𝑣𝑖𝑙 𝑊𝑜𝑟𝑘𝑠 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒)

𝐶𝑖𝑣𝑖𝑙 𝑊𝑜𝑟𝑘𝑠 𝐶𝑜𝑠𝑡 = (1, 519, 166, 250 𝑃𝐻𝑃)(0. 40)

𝐶𝑖𝑣𝑖𝑙 𝑊𝑜𝑟𝑘𝑠 𝐶𝑜𝑠𝑡 = 607, 666, 500 𝑃𝐻𝑃

For Operational and Maintenance Cost in a year

𝑂𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑛𝑑 𝑀𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒 𝐶𝑜𝑠𝑡 = (𝑂&𝑀 𝑓𝑜𝑟 𝑎𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑖𝑣𝑒 2)(𝑃𝑙𝑎𝑛𝑡 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦)

Note: 1 USD = 47.75 Php, as of May 30, 2021 (Source: www.xe.com)

𝑈𝑆𝐷 𝑃𝐻𝑃
𝑂𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑛𝑑 𝑀𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒 𝐶𝑜𝑠𝑡 = (118 𝑘𝑊
× 47. 75 𝑈𝑆𝐷
)(5, 000 𝑘𝑊)

𝑂𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑛𝑑 𝑀𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒 𝐶𝑜𝑠𝑡 = 28, 172, 500 𝑃𝐻𝑃

136
For the Net Income of the Power Plant

Note: Based on Fig. 1-10, the generation rate of SAMELCO I is 10.2646 php/kwh.

𝑃𝑙𝑎𝑛𝑡 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑖𝑛 𝑘𝑤ℎ 𝑝ℎ𝑝

𝐴𝑛𝑛𝑢𝑎𝑙 𝐺𝑟𝑜𝑠𝑠 𝐼𝑛𝑐𝑜𝑚𝑒 = ( 𝑦𝑒𝑎𝑟
)(𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 𝑖𝑛 𝑘𝑤ℎ
)

8,760 ℎ𝑟𝑠 𝑝ℎ𝑝

𝐴𝑛𝑛𝑢𝑎𝑙 𝐺𝑟𝑜𝑠𝑠 𝐼𝑛𝑐𝑜𝑚𝑒 = (5000 𝑘𝑤 × 𝑦𝑒𝑎𝑟
)(10. 2646 𝑘𝑤ℎ
)

8,760 ℎ𝑟𝑠 𝑝ℎ𝑝

𝐴𝑛𝑛𝑢𝑎𝑙 𝐺𝑟𝑜𝑠𝑠 𝐼𝑛𝑐𝑜𝑚𝑒 = (5000 𝑘𝑤 × 𝑦𝑒𝑎𝑟
)(10. 2646 𝑘𝑤ℎ
)

𝐴𝑛𝑛𝑢𝑎𝑙 𝐺𝑟𝑜𝑠𝑠 𝐼𝑛𝑐𝑜𝑚𝑒 = 449, 589, 480 𝑝ℎ𝑝/𝑦𝑒𝑎𝑟

𝑁𝑒𝑡 𝑃𝑟𝑜𝑓𝑖𝑡 = 𝐴𝑛𝑛𝑢𝑎𝑙 𝐺𝑟𝑜𝑠𝑠 𝐼𝑛𝑐𝑜𝑚𝑒 − 𝑂&𝑀 𝐶𝑜𝑠𝑡

𝑁𝑒𝑡 𝑃𝑟𝑜𝑓𝑖𝑡 = (449, 589, 480 − 28, 172, 500) 𝑃𝐻𝑃

𝑁𝑒𝑡 𝐼𝑛𝑐𝑜𝑚𝑒 = 421, 416, 980 𝑃𝐻𝑃

𝑁𝑒𝑡 𝐼𝑛𝑐𝑜𝑚𝑒
For Return of Investment ; 𝑅𝑂𝐼 = 𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐶𝑜𝑠𝑡
𝑥 100

421,416,980 𝑝ℎ𝑝
𝑅𝑂𝐼 = 1,519,166,250 𝑝ℎ𝑝
𝑥 100

𝑅𝑂𝐼 = 0. 27740017 ≈ 27. 74 %

Plant Economics for Alternative 3

137
 Table D-4: Plant Values for Alternative 3

The data that will be used is under the 1-10 MW plant capacity since the design capacity is 5 MW with the

head of 3-30 ft since the head for alternative 3 is 9 m≈29.5 ft.

Capital Cost

𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐶𝑜𝑠𝑡 = (𝑃𝑟𝑖𝑐𝑒 𝑜𝑓 𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐶𝑜𝑠𝑡 𝑝𝑒𝑟 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦)( 𝑃𝑙𝑎𝑛𝑡 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦)

Note: 1 USD = 47.75 Php, as of May 30, 2021 (Source: www.xe.com)

𝑈𝑆𝐷 𝑃𝐻𝑃
𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐶𝑜𝑠𝑡 = (7, 277 𝑘𝑊
× 47. 75 𝑈𝑆𝐷
)(5, 000 𝑘𝑊)

𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐶𝑜𝑠𝑡 = 1, 737, 383, 750 𝑃𝐻𝑃

For Capital Cost Breakdown

For Turbo Generator Set,

𝑇𝑢𝑟𝑏𝑜 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑆𝑒𝑡 𝐶𝑜𝑠𝑡 = (𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐶𝑜𝑠𝑡)(𝑇𝑢𝑟𝑏𝑜 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑆𝑒𝑡 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒)

𝑇𝑢𝑟𝑏𝑜 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑆𝑒𝑡 𝐶𝑜𝑠𝑡 = (1, 737, 383, 750 𝑃𝐻𝑃)(0. 30)

𝑇𝑢𝑟𝑏𝑜 𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑆𝑒𝑡 𝐶𝑜𝑠𝑡 = 521, 215, 125 𝑃𝐻𝑃

For Electric Regulation and Control Equipments,

𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑅𝑒𝑔𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝐶𝑜𝑛𝑡𝑟𝑜𝑙 𝐸𝑞𝑢𝑖𝑝𝑚𝑒𝑛𝑡𝑠 𝐶𝑜𝑠𝑡 = (𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐶𝑜𝑠𝑡)(𝐸𝑞𝑢𝑖𝑝𝑚𝑒𝑛𝑡𝑠 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒)

138
 𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑅𝑒𝑔𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝐶𝑜𝑛𝑡𝑟𝑜𝑙 𝐸𝑞𝑢𝑖𝑝𝑚𝑒𝑛𝑡𝑠 𝐶𝑜𝑠𝑡 = (1, 737, 383, 750 𝑃𝐻𝑃)(0. 22)

𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑅𝑒𝑔𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝐶𝑜𝑛𝑡𝑟𝑜𝑙 𝐸𝑞𝑢𝑖𝑝𝑚𝑒𝑛𝑡𝑠 𝐶𝑜𝑠𝑡 = 382, 224, 425 𝑃𝐻𝑃

For Construction and Engineering Management,

𝐶𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝐸𝑛𝑔𝑖𝑛𝑒𝑒𝑟𝑖𝑛𝑔 𝑀𝑎𝑛𝑎𝑔𝑒𝑚𝑒𝑛𝑡 𝐶𝑜𝑠𝑡 = (𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐶𝑜𝑠𝑡)(𝑀𝑎𝑛𝑎𝑔𝑒𝑚𝑒𝑛𝑡 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒)

𝐶𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝐸𝑛𝑔𝑖𝑛𝑒𝑒𝑟𝑖𝑛𝑔 𝑀𝑎𝑛𝑎𝑔𝑒𝑚𝑒𝑛𝑡 𝐶𝑜𝑠𝑡 = (1, 737, 383, 750 𝑃𝐻𝑃)(0. 08)

𝐶𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝐸𝑛𝑔𝑖𝑛𝑒𝑒𝑟𝑖𝑛𝑔 𝑀𝑎𝑛𝑎𝑔𝑒𝑚𝑒𝑛𝑡 𝐶𝑜𝑠𝑡 = 138, 990, 700 𝑃𝐻𝑃

For Civil Works,

𝐶𝑖𝑣𝑖𝑙 𝑊𝑜𝑟𝑘𝑠 𝐶𝑜𝑠𝑡 = (𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐶𝑜𝑠𝑡)(𝐶𝑖𝑣𝑖𝑙 𝑊𝑜𝑟𝑘𝑠 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒)

𝐶𝑖𝑣𝑖𝑙 𝑊𝑜𝑟𝑘𝑠 𝐶𝑜𝑠𝑡 = (1, 737, 383, 750 𝑃𝐻𝑃)(0. 40)

𝐶𝑖𝑣𝑖𝑙 𝑊𝑜𝑟𝑘𝑠 𝐶𝑜𝑠𝑡 = 694, 953, 500 𝑃𝐻𝑃

For Operational and Maintenance Cost in a year

𝑂𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑛𝑑 𝑀𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒 𝐶𝑜𝑠𝑡 = (𝑂&𝑀 𝑓𝑜𝑟 𝑎𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑖𝑣𝑒 3)(𝑃𝑙𝑎𝑛𝑡 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦)

Note: 1 USD = 47.75 Php, as of May 30, 2021 (Source: www.xe.com)

𝑈𝑆𝐷 𝑃𝐻𝑃
𝑂𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑛𝑑 𝑀𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒 𝐶𝑜𝑠𝑡 = (125 𝑘𝑊
× 47. 75 𝑈𝑆𝐷
)(5, 000 𝑘𝑊)

𝑂𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑛𝑑 𝑀𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒 𝐶𝑜𝑠𝑡 = 29, 843, 750 𝑃𝐻𝑃

For the Net Income of the Power Plant

Note: Based on Fig. 1-10, the generation rate of SAMELCO I is 10.2646 php/kwh.

𝑃𝑙𝑎𝑛𝑡 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑖𝑛 𝑘𝑤ℎ 𝑝ℎ𝑝

𝐴𝑛𝑛𝑢𝑎𝑙 𝐺𝑟𝑜𝑠𝑠 𝐼𝑛𝑐𝑜𝑚𝑒 = ( 𝑦𝑒𝑎𝑟
)(𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 𝑖𝑛 𝑘𝑤ℎ
)

139
 8,760 ℎ𝑟𝑠 𝑝ℎ𝑝
𝐴𝑛𝑛𝑢𝑎𝑙 𝐺𝑟𝑜𝑠𝑠 𝐼𝑛𝑐𝑜𝑚𝑒 = (5000 𝑘𝑤 × 𝑦𝑒𝑎𝑟
)(10. 2646 𝑘𝑤ℎ
)

8,760 ℎ𝑟𝑠 𝑝ℎ𝑝

𝐴𝑛𝑛𝑢𝑎𝑙 𝐺𝑟𝑜𝑠𝑠 𝐼𝑛𝑐𝑜𝑚𝑒 = (5000 𝑘𝑤 × 𝑦𝑒𝑎𝑟
)(10. 2646 𝑘𝑤ℎ
)

𝐴𝑛𝑛𝑢𝑎𝑙 𝐺𝑟𝑜𝑠𝑠 𝐼𝑛𝑐𝑜𝑚𝑒 = 449, 589, 480 𝑝ℎ𝑝/𝑦𝑒𝑎𝑟

And,

𝑁𝑒𝑡 𝑃𝑟𝑜𝑓𝑖𝑡 = 𝐴𝑛𝑛𝑢𝑎𝑙 𝐺𝑟𝑜𝑠𝑠 𝐼𝑛𝑐𝑜𝑚𝑒 − 𝑂&𝑀 𝐶𝑜𝑠𝑡

𝑁𝑒𝑡 𝑃𝑟𝑜𝑓𝑖𝑡 = 449, 589, 480 − 29, 843, 750 𝑃𝐻𝑃

𝑁𝑒𝑡 𝐼𝑛𝑐𝑜𝑚𝑒 = 419, 745, 730 𝑃𝐻𝑃

For Return of Investment (ROI)

𝑁𝑒𝑡 𝐼𝑛𝑐𝑜𝑚𝑒
𝑅𝑂𝐼 = 𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐶𝑜𝑠𝑡
𝑥 100

419,745,730 𝑝ℎ𝑝
𝑅𝑂𝐼 = 1,737,383,750 𝑝ℎ𝑝
𝑥 100

𝑅𝑂𝐼 = 24. 16 %

140
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