Training Manual for Micro-hydropower Technology June 2009

DEPARTMENT OF ENERGY ENERGY UTILIZATION MANAGEMENT BUREAU

Training Manual

for

Micro-hydropower Technology

June 2009

MHP – 6

This manual was developed by the Department of Energy (DOE) through the technical assistance under the Project on “Sustainability Improvement of Renewable Energy Development for Village Electrification in the Philippines” which was provided by the Japan International Cooperation Agency (JICA).

Table of Contents

1 General......................................................................................................1

2 Scope .........................................................................................................1

3 Objectives..................................................................................................1

4 Implementation structure ........................................................................1

4.1 Implementation structure ...........................................................................1

4.2 Roles and responsibilities ............................................................................2

5 Outline of the Training.............................................................................2

5.1 Purpose.........................................................................................................2

5.2 Trainer .........................................................................................................2

5.3 Trainee .........................................................................................................3

5.4 Attainment Target .......................................................................................3

6 Preparation ...............................................................................................3

6.1 Establishment of the Training Program .....................................................3

6.2 Venue Arrangements ...................................................................................4

6.3 Invitation of Trainees ..................................................................................4

6.4 Preparation of Materials..............................................................................4

7 Implementation ........................................................................................4

7.1 Points to be learned during training ...........................................................4

7.2 Key lecture points ........................................................................................4

8 Amendment of the manual .......................................................................4

LIST OF ANNEXES

ANNEX 1 : List of already-drafted training materials ANNEX 2 : Learning points by training item for planning and

civil structure design ANNEX 3 : Samples of training material

1

1 General

Proponents of micro-hydropower projects for rural electrification should draft an appropriate project plan based on site conditions in order to effectively make use of the limited water resources and ensure project sustainability. From the same perspective, administrative organizations are also required to adequately evaluate the submitted plan and give proper instructions to the proponents. Accordingly, sufficient understanding of planning technique for micro-hydropower project is essential for both the proponents and the assessors.

This manual is intended to present the key points in organizing and implementing the training on micro-hydropower technology. Some basic training materials have been included to reduce the trainers’ burden for preparatory work.

2 Scope

This manual shall be used for the planning, implementation, and evaluation of micro-hydropower technology training.

3 Objectives

(1) Assist training organizers and trainers in planning, preparing, and implementing the training on micro-hydropower technologies effectively, and

(2) Maximize the training effect by providing the key points to be noted during conducting lectures.

4 Implementation structure

4.1 Implementation structure

The following figure shows the implementation structure of the training. Figure 1 Implementation structure of the training

Organizer

Organizations concerned to micro-hydropower development for rural electrification

Trainer

Trainee ･･･････

Assignment of trainers

Training implementation for specific technical field

Dispatch of trainees

InvitationTrainee nomination

Trainer Trainer

Trainee Trainee Trainee

2

4.2 Roles and responsibilities

The training roles and responsibilities of the planning, preparation and implementation components have been summarized in the following table.

Table 1 Roles and responsibility of persons concerned

5 Outline of the Training

5.1 Purpose

The purposes of micro-hydropower technology training are to:

(1) deepen the knowledge of experienced technical staff who have been involved in micro-hydropower development, and

(2) develop the knowledge of new technical staff who will be engaged in micro-hydropower development in the near future.

Appropriate training implementation enables technical staff to adequately design and evaluate new micro-hydropower projects and establish the rehabilitation plans for existing plants.

5.2 Trainer

Each trainer shall be assigned to a specific technical field, such as planning and designing for civil, electrical, mechanical facilities. Trainers shall be those who have participated in past micro-hydropower development training. However, the organizer can also invite technical staff who have served as trainers in outside organizations.

Trainers will be responsible for preparing training materials in the technical field of which they have respectively been put in charge. Examples of the technical materials are attached to this manual for reference.

Component Role

Organizer

Assignment of trainers Arrangement of venue, equipment

(accommodation for trainees if necessary) Sending invitation letter calling for traiees

Trainer

Preparation of training materials Implementation of training for specific technical

field in charge Evaluation of trainees

Trainee Participation in training Implementation of future training for specific

technical in their organizations

Organization concerned

Nomination of trainees

3

5.3 Trainee

Trainees shall be invited from among the stakeholders, such as the DOE field office, ANEC, LGU, etc., organizations which have promoted micro-hydropower development for rural electrification. The organizer will determine the number of trainees taking into account the training effects.

Further, basic knowledge of mathematics is required for all trainees to ensure smooth progress. In the end, trainees will also be required to disseminate the knowledge that they have acquired during training.

5.4 Attainment Target

Final targets to be attained for the proponents and the assessor are described respectively as follows:

・ to develop the project plan on the basis of the site reconnaissance results and handle the basic design of equipment in a proper manner, and

・ to properly evaluate the project plan that the proponents have drafted, and offer constructive instructions and advice.

6 Preparation

6.1 Establishment of the Training Program

The following table shows an example of a 5-day training program which includes the over-all contents concerning micro-hydropower development from a preliminary study used for site selection to basic designs used for civil and electro-mechanical equipment. The organizer can arrange the training program to meet available time frames and stakeholders’ needs. For instance, the trainings on civil structures design and electro-mechanical equipment design can be organized separately for people possessing different academic backgrounds such as civil, electric, and mechanical engineering.

Table 1 Example of program for 5-day training

Date Training item

AM Map study Day 1 PM Planning

AM Site reconnaissance Day 2 PM Design of civil structures

AM Practice of map study Day 3 PM Practice of civil structure design

AM Turbine / Driving system Day 4 PM Generator / Control system

AM Electrical equipment and protection system / Distribution system Day 5 PM Practice of electro-mechanical equipment design

4

6.2 Venue Arrangements

The organizer shall arrange an appropriate venue taking into account the availability and expected number of the trainees. Necessary training equipment, such as the PC, projector, and microphone, shall also be prepared. The trainees are asked to prepare the calculator for practice of planning and designing, if necessary.

6.3 Invitation of Trainees

In order to summon and finalize candidates based on their area of specialty and prior experience, an invitation letter specifying trainee requirements will be sent by the organizer to the stakeholders in advance.

6.4 Preparation of Materials

The trainers shall prepare in advance the training materials in their charge. The materials that are attached to this manual are listed in ANNEX1. The organizer and trainers shall upgrade these materials as well as adding new ones.

7 Implementation

7.1 Points to be learned during training

The trainers shall proceed with the lectures step by step so that the trainees can thoroughly absorb the fundamental points of each training item, which are shown in ANNEX2.

7.2 Key lecture points

One-on-one lectures may turn out to be a tedious proposition for both the trainers and trainees. Hence, the trainers shall encourage the trainees to actively participate in the following ways:

・ Interspersed periodic questioning of the trainees to confirm their understanding, ・ Introduction of examples and case studies, ・ Homework for lecture review, ・ On-hand practice to deepen knowledge and understanding, ・ Discussion among the trainees, and ・ Presentations by the trainees on how to utilize their newly acquired knowledge, ・ Wrap-up meeting.

Such methods will enable trainees to apply their acquired knowledge in planning and actual development. Conducting examinations before and after the training is an effective method in evaluating the level of capacity building and also reveals the training effects.

8 Amendment of the manual

The DOE shall review this manual annually, and amend it, if necessary, according to the surrounding circumstances in rural electrification of the country. The amended manual shall be fully authorized among the DOE and approved by Director of Energy Utilization Management Bureau of the DOE.

List of already-drafted training materials

Items Contents Outline of hydropower Catchment area Map study Duration curve and identification of potential site Functions of main structures for micro-hydropower plant Layout of main structures Planning Selection of main structures’ location Outline of site reconnaissance Measurement of river flow Site reconnaissance Measurement of head Intake weir Intake and settling basin Headrace Head tank Penstock Powerhouse

Design of civil structures

Head loss calculation Basics of hydraulics Turbine types Characteristics of turbine

Turbine

Basic design of turbine Basics of generator Classification of generator Generator Basic design of generator Basics of automatic control Frequency control Control system Voltage control Major factors Transformer Switch gear Arrester Instrument transformer Single line diagram

Electrical equipment and protection system

Protection system Distribution method Components Route selection

Distribution system

Voltage drop estimation

ANNEX1

i

Learning points by training item for planning and civil structure design

Items Contents Learning points Outline of hydropower Concept of hydropower

Concept of catchment area Relationship between discharge and catchment area Catchment area Catchment area estimation using topographical map Concept of duration curve Maximum/firm discharge identification using duration curve

Map study

Duration curve and identification of potential site Potential site identification using topographical

map Functions of intake weir and intake Functions of settling basin Functions of headrace Functions of head tank and penstock

Functions of main structures for micro-hydropower plant

Functions of turbine and generator Layout of main structures Concept of basic layout for main structures

Appropriate location of weir, intake, and settling basin Appropriate location of powerhouse

Planning

Selection of main structures’ location

Appropriate location of headrace route Objectives and survey items of site reconnaissanceOutline of site

reconnaissance Information gathering and planning for site reconnaissance

Measurement of river flow On-site measuring method of river flow

Site reconnaissance

Measurement of head On-site measuring method of head Type and structure of intake weir Design concept for intake weir Intake weir Calculation technique for intake weir dimensioning Structure of intake and settling basin Design concept for intake and settling basin Intake and settling basin Calculation technique for intake and settling basin dimensioning Type and structure of headrace Design concept for headrace Headrace Calculation technique for headrace dimensioning Structure of head tank Design concept for head tank Head tank Calculation technique for head tank dimensioningDesign concept for penstock

Penstock Calculation technique for penstock dimensioning

Powerhouse Structure of powerhouse by turbine type

Design of civil structures

Head loss calculation Calculation technique for head loss

ANNEX2

ii

Learning points by training item for electro-mechanical equipment design

Items Contents Learning points Principle of continuity Bernoulli’s theorem Basics of hydraulics Concept of potential, pressure, and velocity head Structure, features, and applicable range by turbine type Turbine types Concept of turbine selection chart Concept of specific speed Applicable range of specific speed by turbine typeCharacteristics of turbine Turbine efficiency by turbine type Flow of turbine basic design

Turbine

Basic design of turbine Calculation technique for turbine specifications Principle of operation of AC generator Relationship between voltage and rotational speedMain structure of generator

Basics of generator

Type of excitation system Classification of generator Classification of AC generator

Flow of generator basic design

Generator

Basic design of generator Calculation technique for generator specifications Concept of feedback control

Basics of automatic controlReaction of P-control, I-control, and PI-control Characteristics of frequency and active power control

Frequency control Concept of speed governor and dummy load governor Characteristics of voltage and reactive power control

Control system

Voltage control Concept of automatic voltage controller

Major factors Concept of major factors Transformer Type and functions of transformer Switch gear Type and functions of switch gear Arrester Functions of arrester Instrument transformer Type and functions of instrument transformer Single line diagram Standard composition of single line diagram

Type and functions of protection relay

Electrical equipment and protection system

Protection system Standard arrangement of protection relay

Distribution method Classification of distribution method Design and installation concept of pole

Components Design and installation concept of guy wire

Route selection Concept of distribution line route selection Calculation technique for resistance and inductance of conductor

Distribution system

Voltage drop estimation Calculation technique for voltage drop of distribution lines

1

Training on

Micro Hydropower

Development

1-1 2

DATE 2008

Outline of HydropowerCatchment AreaIdentification of Potential SitesDuration CurveFunctions of Main StructuresLayout of Main StructuresSelection of Main Structures' LocationOutline of Site ReconnaissanceMeasurement of River FlowMeasurement of HeadIntake WeirIntake and Settling BasinHeadraceHead tankPenstock and SpillwayPower HouseHead Loss

AM

AM TurbinePM Driving System

GeneratorControl SystemProtection System

PM

Nov. 10

AM

PM

CONTENTS

Examination: Design of Electrical & Mechanical Equipment

BASIC COURSE: Map Study

BASIC COURSE: Planning

ADVANCE COURSE: Site Reconnaissance

ADVANCE COURSE: Design of Civil Structures

Practice Activity for Map StudyPractice Activity for Civil DesignExamination: Planning & Design of Civil Structures

Design of Mechanical Equipment

Curriculum for the Training on Micro-Hydropower Development

Nov.13

Design of Electrical EquipmentAMNov.14

Nov.11

AM

PM

PMNov.12

1-2

3

Training on

Micro Hydropower

Development

Basic Course (1ST part)

EPIFANIO G. GACUSAN DOE - REMD

AVR, Department of Energy, 10 November 20081-3 4

Training on Micro Hydropower Development

Basic Course

Map Study

OUTLINE OF HYDROPOWER

1-4

5

What is Hydropower?Energy of Falling Stone:

Ouch!

1-5 6


Ouch!

Ouch!

1-6

7


Ouch!

Ouch!

1-7 8

What is Hydropower?Energy of Falling Stone

depends on…

Height

Weight of the Stone

Energy of Hydropower

Height

Weight of the Water

Head

Discharge

Height

1-8

9


Basic Course

Map Study

CATCHMENT AREA

1-9 10

Hydropower depends on Head and Discharge

Catchment Area

Depends on Catchment Area

Rainfall

For Generating Power

SayangMottainai

Discharge

1-10

11

10621045

Height 20 m x 5 =100m

Scale: 1/50,000

On map : Accrual

1 cm : 500 m

Short Distance = Steep

Long Distance = Gentle

980

960

940

920

900

880

860

840

820

800

780

760

1-11 12

10621045

Catchment Area

1-12

13

h

b

A

A = ( b x h ) /2

1-13 14

10621045

1-14

151-15 161-16

17


Basic Course

Map Study

DURATION CURVE & IDENTIFICATION OF POTENTIAL SITE

1-17 18

Duration Curve

0 100 200 300 365

Riv

er F

low

(m

3 /s)

140

100

60

Riv

er F

low

(m

3 /s)

Flow Duration CurveActual River Flow

ArtJimmy

？

ArtJimmy

？

Change the Order

1-18

19

Duration CurveGauging Station: ABC (CA=30km2) Latitude：@@@

Period:1990.1 – 2000.1 Longitude：@@@

25%(95day) 50%(183day) 75%(274day)100%(365day

)90%(328day)

95%(346day)

0.5

1.0

1.5

Riv

er F

low

(m

3 /s) Depends on Chatchment Area and Rainfall

Depends on Planning

Maximum Discharge/Design Discharge

Firm Discharge = 95 % Firm

1-19 20

Duration Curve : How to Identify Maximum Discharge

40% 50% 70%

0.5

1.0

1.5

Riv

er F

low

(m

3/s

)

60% 80%

Co

ns

tru

cti

on

Co

st

/ k

Wh

Percentage of Duration

40 % 50 % 60 % 70 % 80 %

For Mini/Large Hydro : Comparison of Unit Cost in Each Case

1-20

21

Duration Curve : How to Identify Maximum Discharge

Maximum and Firm Discharge in Hydropower Plant

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

2.2

0 10 20 30 40 50 60 70 80 90 100 110

Percentage of Firm/Maximum Discharge (%)

Un

it F

irm

Dis

cha

rge

(m3 /s

/10

0km

2 )

Large

SmallMini

Micro

For Micro Hydro : Initial Stage

Firm Discharge = 1.0 m3/s/100km2

Max. Discharge = Firm Discharge

1-21 22

Duration Curve : How to Identify Firm/Max. Discharge

For Micro Hydro : Pre-Feasibility Study

Maximum and Firm Discharge in Hydropower Plant

0.00.20.4

0.60.81.01.21.41.6

1.82.02.2

0 10 20 30 40 50 60 70 80 90 100 110

Percentage of Firm/Maximum Discharge (%)

Uni

t Firm

Dis

char

ge

(m3 /s

/100

km2 )

Micro

Vegetation

Rich Forest

Bare Ground

Over 3000mm

Annual Rainfall

Aprx.2000 mm

１６００

１5００１7００

Average rain fall line

1-22

23

Duration Curve : How to Identify Firm/Max. Discharge

For Micro Hydro : Detail Study

Measurement River Flow at the Site

It will be Trained in Advance

Course

1-23 24

Good Potential Site (Technically)

1. Short Distance and High Head

Portion A B C D

Profile of River

1-24

25

1. Short Distance and High Head ; How to Know ?

E.L

520

500

480

460

440

420

400

380

L1L1

L3L2

L4 L5

L6500400

L1 L2 L3 L4 L5 L6

1-25 26

1. Short Distance and High Head ; How Short? How High?

Indicator : L/H (Distance/Head)

Head and Waterway Length

0

10

20

30

40

50

60

70

80

90

100

0 200 400 600 800 1000 1200 1400 1600 1800 2000 2200 2400 2600 2800 3000

Waterway Length (m) L

Hea

d (m

) H

Micro L/H<25

Mini L/H<25

Small/Large L/H<50L/H=10

L/H=25

L/H=50

Micro-Hydro: L/H <25

1-26

27

1. Short Distance and High Head ; How to find out ?

Micro,Mini-Hydro: L/H <25

On the Map of 1/50,000 = 1 interval of Contour Line : 20 m

If you want 20 m of Head

Look for 1 interval of Contour Line is less than 1cm on the Map

500 m



1000 m

G

G



1500 m

G

G

1-27 28


E.L

520

500

480

460

440

420

400

380

L1L1

L3L2

L4 L5

L6500400

L1 L2 L3 L4 L5 L6

Assumed Head : 20 m

500m500m500m500m500m500m

N.G

OK

OK

N.GN.G

N.G

R

1cm 500m

R1-28

29

E.L

520

500

480

460

440

420

400

380

L1L1

L3L2

L4 L5

L6500400

L1 L2 L3 L4 L5 L6


Assumed Head : 40 m

1000m2 cm

1000m1000m1000m 1000m1000m

N.GOK

OK

N.G

N.G

RR1-29 30


E.L

520

500

480

460

440

420

400

380

L1L1

L3L2

L4 L5

L6500400

L1 L2 L3 L4 L5 L6

Assumed Head : 60 m

3 cm 1500 m

1500 m1500 m1500 m1500 m

N.G

OK

N.G

N.G

1-30

31


2. Bigger Catchment Area is better if the L/H is evenAssumed Head : 40 m

10621045

1-31 32


3. Power Output must be balanced with DemandFor Micro Hydro : Initial Stage

100 HH X 200 W/ HH = 20 kWFirm Discharge = 1.0 m3/s/100km2

Demand ; Based on Social SurveyMax. Discharge = Firm Discharge

Practical Hydropower Output

P = 9.8 x Q x H x Where, P = Power output (kW)

Q = Discharge (m3/s)

Ｈ = Head (m)

= Combined efficiency (0.5)

Q = P / (9.8 x H x )

= 20 / (9.8 x H x 0.5 )

0.076020

0.104020

0.202020

Q (m3/s)H (m)P (kW)

Required Catchment Area = 20 km2

C.A=10 km2

C.A=7 km2

1-32

33


4. Near the Demand Area

Voltage Drop is within 10 % without Step up T/F

Radius 1km (2cm on the Map)

Demand site

0 1.0 2.0 3.0 4.0 5.0 km

1-33 34


5. Gentle Slope is Convenient for Headrace and Powerhouse

A

A

Section A-A

B

B

Section B-B

1-34

35

Examples

1-35 36

0 1.0 2.0 3.0 4.0 5.0 km

Output ≧10 kW Head ≦ 60 m

1-36

37

0 1.0 2.0 3.0 4.0 5.0 km

10 kWP

60 mH

0.035 m3/sQ

3.5 km2C.A

22 kWP

60 mH

0.075 m3/sQ

7.5 km2C.A

25 kWP

40 mH

0.125 m3/sQ

12.5 km2C.A

35 kWP

40 mH

0.180 m3/sQ

18.0 km2C.A

55 kWP

60 mH

0.185 m3/sQ

18.5 km2C.A

10 kWP

60 mH

0.035 m3/sQ

3.5 km2C.A

12 kWP

60 mH

0.040 m3/sQ

4.0 km2C.A

1-37 38

Items Contents

Project Name Ambabag Mini-Hydropower Project

Location Barangay Ambabag, Pindungan, Kiangan, Ifugao

Coordinates Intake : N-16°47’34.92”, E-121°05’35.22”

Powerhouse: : N-16°47’37.32”,E-121°06’20.28”)

Catchments Area 20.2 km2

Elevation of the Intake E.L. 494.2 m

Tailrace Water Level E.L. 403 m

Gross Head 91.2m

Effective Head 80 m

Required Max. Discharge 0.32 m3/s

Maximum Output 200 kW

Annual Energy Generation 1,490 MWh

Construction Cost Approximate. 42 Million pesos

Plant Factor 85 %

Result of Map Study (Example)

1-38

39

Thank you Very Much !!!!

1-39

1

Training on

Micro Hydropower

Development

Basic Course (2nd part)

2-12


Basic Course

Planning

Functions of Main Structures

2-2

3

Main Structures for Micro/Mini Hydropower

Head-tank（Fore bay）

Headrace

Demand

2-34

Intake Weir Settling Basin

Headrace

Head-tank

Penstock

Powerhouse

TailraceSpillway

2-4

5

Intake Weir and Intake

The Intake weir – a barrier built across the river used to divert water through an opening in the riverside (the ‘Intake’ opening) into a settling basin.

2-56

Function of Intake Weir

Intake

If no Intake Weir Insufficient Inflow

Many Sedimentations into Intake

2-6

7

Function of Intake Weir

to Divert the River Flow into the Intake

to prevent the Sediment/silts to pass through

Flush Gate (Stop Logs)

2-78

Function of Intake

Weir Crest

Flood Water Level

Q over Q

Flood Water Level

Big Flow = Structures will be Damaged

Orifice with Spillway

Control Gate

to Control Inflow

2-8

9

Settling BasinSettling Basin-The settling basin is used to trap sand or suspend the silt from the water before entering the penstock.

2-910

IntakeHeadrace

Spillway

High Velocity Low Velocity

Flush Gate

Function of Settling Basin

to trap sand or suspend the silt from the water

2-10

11

HeadraceHeadrace-A channel leading the water to a head tank. The headrace follows the contour of the hillside so as to preserve the elevation of the diverted water.

2-1112

Hea

d L

oss

Settling Basin Headrace Headtank

Slope =Gentle

Function of Headrace

Gentle Slope ☆ Small Head Loss = Big Output

★ Big Size of Headrace = High Cost

Steep Slope ☆ Small Size of Headrace =Low Cost

★ Big Head Loss =Small Output


Slope =Steep

Hea

d L

oss

to convey water into the head-tank

Micro=1/100 to 1/300

Mini=1/200 to 1/1,000

2-12

13

Head-tank (Forebay Tank)Head-tank - Pond at the top of a penstock or pipeline; serves as final settling basin, maintains the required water level of penstock inlet and prevents foreign debris entering the penstock.

2-1314

Penstock

Penstock - .A close conduit or pressure pipe for supplying water under pressure to a turbine.

2-14

15

Water Turbine and Generator

A water turbine is a machine to directly convert the kinetic energy of the flowing water into a useful rotational energy while a generator is a device used to convert mechanical energy into electrical energy.

2-1516

Thank you Very Much!

2-16

1

Training on

Micro Hydropower

Development

Basic Course (3rd part)

3-1 2

Training on Micro/Mini Hydropower DevelopmentTraining on Micro Hydropower Development

Basic Course

Planning

Layout of Main Structures

3-2

3

HeadtankHeadrace

Main Structuresfor Micro/Mini Hydropower plants

3-3 4

1. Short Penstock

Basic Layout

3-4

5

2. Long Penstock

Basic Layout

3-5 6

3. Middle-Length Penstock

Basic Layout

3-6

7

Training on Micro/Mini Hydropower Development

Basic Course

Planning

Selection of Main Structures’ Locations

3-7 8

CRITERIA

1. Narrow River Width

2. Preferably at Straight Portion of the River

3. Has Space for Settling Basin

4. Easy to Combine with Headrace

Apropriate location forthe weir, intake and settling basin

3-8

9

A

B

C

D

E

Criteria

1. Narrow River Width

2. Preferably at Straight Portion of the River

3. Has Space for Settling Basin

4. It is easy to combine with Headrace

A,B,D,E

A,B

appropriate location forthe weir, intake and settling basin

A,B,E

A

3-9 10

CRITERIA

1. Gentle River Bank

2. The Water Flood Will Have No Great Impact at the River Bank

3. Has a Wide Cross Section of the River (Low Flood Water Level)

4. Ridge is Better (Geologically Strong and Stable)

Appropriate location forPower house

3-10

11

D

FA

BC

E

Criteria

1. Gentle River Bank

2. The Water Flood Will Have No Great Impact at the River Bank

3. Has a Wide Cross Section of the River (Low Flood Water Level)

4. Ridge is better (Geologically Strong and Stable)

B,C

C

C,D

Appropriate location forPower house

B,C,F

3-11 12

CRITERIA

1. Gentle Slope

2. Stable Geological Condition

3. Accessible

Please consider & recommend based on your experience of irrigation cannel

Appropriate location forHeadrace route

3-12

13

MARAMING SALAMAT!!!

Thank You Very Much!!!!

Arigato!!!

3-13

1

Training on

Micro Hydropower

Development

Advance Course (1st part)

4-12

Outline of Site Reconnaissance

Measurement River Flow

Measurement of Head

Training on Micro/Mini Hydropower Development

Advance Course

Site Reconnaissance

4-2

3

Outline of Site Reconnaissance Objective

To roughly evaluate the feasibility of the project To get necessary information for planning

Items to be investigated Potential capacity of the project site

- Measurement of river flow- Measurement of head

Topographical and geological condition of the sites for the structure layout

Accessibility to the site Power demand in the load center Distance from the load center to the power house Ability of the local people to pay for electricity Willingness of the local people for electrification

4-34

Information Gathering

Prepare 1/50,000 scale maps to check the location, catchment area, villages, access road and topography of the project sites.

Gather available information on accessibility to the site, the weather conditions, social stability, and so on.

Make copies of the 1/50,000 scale maps and route maps enlarged by 200 to 400%.

Prepare checklists and interview sheets for site survey.

Planning of preliminary site survey

Make a plan and schedule for site survey considering accessibility to the sites and the weather conditions.

Allow sufficient time in the schedule since most of sites are located in remote and isolated areas

Preparation of Site Reconnaissance

4-4

5

Equipment Equipment

○ Route map Altimeter

○ Topographic map GPS (portable)

○ Reconnaissance schedule ○ Camera, film

○ Checklist ○ Current meter (Float,)

○ Interview sheet ○ Distance meter, measure tape

Geological map ○ Hand level (Hose)

Aerial photographs ○ Convex scale (2-3m)

Related reports Hammer

Clinometers

○ Field notebook Knife

○ Scale Scoop

○ Pencil ○ Torch, flashlight

○ Eraser Sampling baggage

○ Colored pencil Label

Section paper ○ Compass

Stopwatch

Batteries

Necessary Goods for Site Reconnaissance

4-56

Major Items of Site Reconnaissance

Investigation of potential capacity River flow measurement

Head measurement

Investigation for layout and design of facilities Intake site

Waterway route

Powerhouse site

Transmission/distribution line route

Investigation of demand forecast

Other outline surveys

4-6

7

Measurement River Flow Reason for direct measurements:

Since the catchment area of micro-hydro power is relatively small, the river flow at micro-hydro sites is site-specific.

Some rivers dries up during dry season Without checking the actual flow, we cannot be confident of

the potential capacity of the projects. Purpose:

To get enough data to accurately predict river flow at the project site

To check the minimum river flow during dry season (Micro) To prepare the duration curve (Mini & Large)

Method: Current meter method Float method Bucket method Weir measuring method

4-78

Micro-Hydro Mini-HydroFlowchart to check Minimum Flow/ Duration Curve

Water Level DischargeH Q

(m) (m3/s)

XXX 0.230 0.111YYY 0.550 1.734ZZZ 0.300 0.272

WWW 0.380 0.600

Date

Installation of Staff Gauge(Base Point)

Selection of MeasurementPoint

Measuring of Cross Section

Measuring of Cross SectionalArea(A)

Measuring of Velocity /Speed(V)

Calculation of Discharge(Q=A x V)

Record the water levelon Staff gauge (H)

An

oth

er d

ayat

leas

t 3

tim

esre

pea

t

DailyRecord

(Hd)

Calculation of Rating Curve

Calculation of DailyDischarge

Calculation of DurationCurve

Micro-Hydro

G

G G

G

G

4-8

9

1

2

3

4

5

Installation of Staff Gauge

R

R

4-910

Electromagnetic Current Meter Propeller Current Meter

Actual Measurement

4-10

11

30cm

Number ofSegmrntation: (i) 1 11 Remark

Distance from leftbank: L（cm） 0 520

Water Depth:D （cｍ）14.0 0.0

Area of Segmantation:

A（cｍ２）

H V H V H V H V H V

Depth from Surface: H(cm) 0.2 5.60 12.00 5.80 28.00 6.60 47.0 7.00 17.0 3.20 13.0

Velocity : V (cm/s)0.4 11.20 10.00 11.60 20.00 13.20 47.0 14.00 13.0 6.40 6.0

0.8 22.40 10.00 23.20 8.00 26.40 26.0 28.00 9.0 12.80 2.0

Average Velocity:Va （cｍ/ｓ）

Total

Discharge of

Segmantation:q (m3/s）272.528

50

2,950 3,425 3,325

30.0

350

33.0 35.0

250

2,500

2 4 6 87

450

3

100

5

200 300

9

400

10

150

22.0

1,510

41.0 16.028.0 29.030.0

40.00 13.00

Flow Measurement Field Sheet　　Name of Location:　　　　　　　Date:　　　　Time:　　　 Staff gauge

7.00

26.667 55.067 137.000 43.225 10.570

10.67 18.67

Record Sheet of Measurement River Flow

4-1112

Float Measuring Method

4-12

13

h1 0.00

h2 0.45

h3 0.50

h4 0.57

h5 0.60

h6 0.62

h7 0.65

h8 0.60

h9 0.50

h10 0.35

h11 0.00Total 4.84

Average 4.84/11= 0.44 m

h1 h2 h3 h4 h5 h6 h7 h8 h9 h10h11

L=10m

L/10 L/10 L/10 L/10 L/10 L/10 L/10 L/10 L/10 L/10

A=havarage x L = 0.44 x 10.00

= 4.40 m2

Measuring of Cross Sectional Area

Measurement of Cross Sectional Area

4-1314

L=2WL=2W

W

Cross Sectional Area ; A

Cross Sectional Area ; B

Cross Sectional Area ; C

Measurement of Velocity

4-14

15

0.5 m

Vmean

Vmean Vmean

Vm = 0.45×Vmean Vm = 0.25×Vmean

Vmean

Vm = 0.85×Vmean Vm = 0.65×Vmean

Concrete channel which cross section is uniform

Small stream where a riverbed is smooth

Shallow flow (about 0.5ｍ) Shallow and riverbed is not flat

R

R

4-1516


y = 3.0993x - 0.3947

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70

Water Level (m)Sq

uare

Roo

t of D

isch

arge

(m3/s

)


Rating Curve

0.10

0.15

0.20

0.25

0.30

0.35

0.40

0.45

0.50

0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00

Discharge (m3/s)

Wat

er L

evel

(m)

Q=9.579*H2-2.428H+0.154

R

R

4-16

17

Calculation of Daily Discharge

R

Discharge of Ambangal Brook at Intake (20.2km2)

0.00.20.40.60.81.01.21.41.61.82.02.22.42.62.83.03.23.43.63.84.04.24.44.64.85.0

5/19/06 6/18/06 7/18/06 8/17/06 9/16/06 10/16/06 11/15/06 12/15/06 1/14/07 2/13/07 3/15/07 4/14/07 5/14/07 6/13/07 7/13/07

Date

Dis

char

ge (m

3 /s)

Daily Discharge

R

4-1718

Discharge of Ambangal Brook at Intake (20.2km2)

0.00.20.40.60.81.01.21.41.61.82.02.22.42.62.83.03.23.43.63.84.04.24.44.64.85.0

5/19/06 6/18/06 7/18/06 8/17/06 9/16/06 10/16/06 11/15/06 12/15/06 1/14/07 2/13/07 3/15/07 4/14/07 5/14/07 6/13/07 7/13/07

Date

Dis

char

ge (m

3 /s)

Daily DischargeDuration Curve at Intake Site (C.A.=20.2km2)

0.00.20.40.60.81.01.21.41.61.82.02.22.42.62.83.03.23.43.63.84.04.24.44.64.85.0

0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100

Percentage (%)

Dis

char

ge (m

3 /s)

Calculation of Duration Curve

4-18

19

Head measurement

Water-filled tube method

– Easy to handle

– No need for a skilled engineer

– Relatively accurate

H1

H3

H4

H5

H6

H2

Head

Head = H1+H2+H3+H4+H5+H6

H1 = B1-A1

B1

H1

A1

4-1920

Result of head measurement

Date :

No.Hi=Bi-Ai(meters)

1 0.85

2 0.86

3 0.86

4 0.91

5 0.99

6 0.75

7 0.30

8 0.90

9 0.70

10 0.74

11 2.30

12 0.66

10.82

Location :

0.70 1.36

Total Height (meters)=

1.00 1.74

0.20 2.50

1.00 1.90

1.00 1.70

1.00 1.75

1.00 1.30

1.00 1.91

1.00 1.99

1.00 1.86

1.00 1.86

Ai(meters)

Bi(meters)

1.00 1.85

4-20

21

Head measurement (Easy Way)Using Water Bottle

H

H

H

H

H

H

HHead=nxH

H

4-2122

Date :

No.Hi=Bi-Ai(meters)

1 0.85

2 0.86

3 0.86

4 0.91

5 0.99

6 0.75

7 0.30

8 0.90

9 0.70

10 0.74

11 2.30

12 0.66

10.82

Location :

0.70 1.36

Total Height (meters)=

1.00 1.74

0.20 2.50

1.00 1.90

1.00 1.70

1.00 1.75

1.00 1.30

1.00 1.91

1.00 1.99

1.00 1.86

1.00 1.86

Ai(meters)

Bi(meters)

1.00 1.85

Result of head measurement

Using Water Bottle : 1.56 m x 7 times ＋0.15= 11.07 m

4-22

23

Thanks !!!!

4-23

1

Training on

Micro Hydropower

Development

Advance Course (2nd part)

5-1 2

Intake Weir

Intake and Settling Basin

Headrace

Head-Tank

Penstock

Powerhouse

Head Loss Calculation


Advance Course

Design of Civil Structures

5-2

3

Basic Equation for Civil Design: Important

Q = A x V

Q: Discharge (m3/s)

A: Cross sectional area of water (m2)

V: Velocity of water (m/s)

V = Q / A

A = Q / V AV

1 secondV

A

1 second

○ meters/

○ meters/

5-3 4

2-1 Intake Weir(1) Type of Intake Weir (refer to “Manual” p5-2 to 5-4)

Foundations: GravelFloating concrete

Foundations: Bedrock

Concrete gravity

Application ConditionOutline DrawingType of Weir

5-4

5

2-1 Intake Weir(2) Example of the Intake Weir

Designed as Gravity TypeDestroyed by Flood ,due to lack of strength of the foundation

Re-designed as Floating Type

5-5 6

2-1 Intake Weir

(3) Design of the Weir Height

Conditions into consideration (“Manual” p-5-4 to P5-6 )

Minimizing the Height

High High Cost & Wide Affected AreaLow Low Cost & Small Affected Area

Smooth Removal of Sediment

Weir Height is depend on “Slope of the Riverbed”

5-6

7

2-1 Intake Weir(3) Design of the Weir Height

L

ic

ir d2

d1

hi

B

Width of Inlet (B) Height from river bed to bed of inlet (d1) Water depth of inlet (hi) Slope of settling basin (ic) Slope of River (ir) Length of settling basin (L) Height from river bed to bed settling basin (d2)

5-7 8


L

ic

ir d2

d1

hi

D1 = d1 + hi

D2 = d2 + hi+ L x (ic – ir)

D1 > D2 Weir Height = D1

D2 > D1 Weir Height = D2

B

5-8

9


Design Discharge (Q) 0.220 Width of Inlet (B) 0.550 meters Height from river bed to bed of inlet (d1) 0.500 meters Water depth of inlet (hi) 0.400 Velocity at inlet (Vi) 1.000 'D1=d1+hi 0.900 Slope of settling basin (ic) 0.050 (1/20) Slope of River (ir) 0.100 (1/10) Length of settling basin (L) 10.000 Height from river bed to bed settling basin (d2) 0.500 D2=d2+hi+L*(ic-ir) 0.400

Weir height from original river bed 0.900 meters

: Values which are decided on other factors

: Common values for design (refer to "Manual")

: Values depend on natural condition

D1>D2

D1 = d1 + hi

D2 = d2 + hi+ L (ic – ir)

Example (Steep river)

5-9 10


D1 = d1 + hi

D2 = d2 + hi+ L (ic – ir)

Example (Gentle river) Design Discharge (Q) 0.220 Width of Inlet (B) 0.550 meters Height from river bed to bed of inlet (d1) 0.500 meters Water depth of inlet (hi) 0.400 Velocity at inlet (Vi) 1.000 'D1=d1+hi 0.900 Slope of settling basin (ic) 0.050 (1/20) Slope of River (ir) 0.010 (1/100) Length of settling basin (L) 10.000 Height from river bed to bed settling basin (d2) 0.500 D2=d2+hi+L*(ic-ir) 1.300

Weir height from original river bed 1.300 meters

: Values which are decided on other factors: Common values for design (refer to "Manual"): Values depend on natural condition

D2>D1

5-10

11

2-1 Intake Weir

Important H x 3.0 ≦ L1+ L2 + L3 + L4 + L5 + L6 + L7

800 1,200 3,9001,000

6,900

1.0

1

EL.497.200 m

EL.496.000 m

1,20

01,

500

700

800

500

400

500

400

5,100

1,50

0

GabionH x B x L =0.6 x 0.8 x 1.0m)

Masonry Concrete

Reinforced Concrete (t=25cm)

0.8 – 1.0

0.6 – 1.0 m

H

L1

L2

L3L4 L5

L6

L7

1.0 – 1.5 m 1.0 – 1.5 m

0.5 m

Common Values

Flood water level

5-11 12

2-2 Intake and Settling Basin

Intake

Intake weir

Protect wall

Image of Intake ( Side Intake Type)

5-12

13


＜ Concepts of the design ＞

The dimension of the intake should be designed that the

velocity of inflow at the intake is 1.0 or less m/s.

The ceiling of the intake should be designed with

allowance of 10-15cm from the water surface.

The height and area of the intake should be designed with

the minimum size.

5-13 14


Protect wall

Intake Weir

Flush gate (Stop-log)

Intake

b

hi

dh

Vi

Q = A x V

V = Q / A Vi = Q / (b x hi)≦1.0 m/s dh=0.1-0.15m

Intake ( Side Intake Type)

5-14

15

2-2 Intake and Settling BasinSettling Basin

Spill way Flush gate

5-15 16

2-2 Intake and Settling BasinSettling Basin(1) Design of Spillway

Flood Water Level

→

Water Level of Spillway

Normal Water Level

Ｂsp

ｈｓp

Ai

hi

dh

dh

hi

bi

H

Q f1= Ai ×Cv × Ca × (2 ×g × H ) 0.5

Q f2= Cs ×hsp1.5 ×Bsp

Ai= hi x bi

Q f1= Q f2

0.667 0.6 9.8

1.8

5-16

17

2-2 Intake and Settling BasinSettling Basin(1) Design of Spillway (Example)

Flood water level from crest of spillway (Ht) 2.000 from flood mark Area of intake (Ai) 0.303 dh=0.15m f 0.500 Cv=1/(1+f) 0.667 Ca 0.600 Cs 1.800 Width of spillway of settling basin (Bsp) 3.000

H Qf1 Qf2 Qf1-Qf21.900 0.738 0.171 0.5681.800 0.719 0.483 0.2361.742 0.707 0.708 (0.001)1.700 0.698 0.887 (0.189)1.600 0.678 1.366 (0.689)1.500 0.656 1.909 (1.253)




0.3000.4000.500

Qf1=Ai x Cv x Ca x (2 x 9.8 x H)^0.5 Qf2=Cs x Bsp x hsp^1.5

hsp0.1000.2000.258

Usually hsp < 0.3m

5-17 18


Conduit sectionWidening section

Settling section

Bb

1.0

2.0

Dam

SpillwayStoplog Flushing gate

Intake

Headrace

Bsp

hs

hsp+

15cm

h0

10～

15ｃｍ

hi

ic=1/20～1/30

IntakeStoplog

bi

ＬｃＬｗＬｓ

Ｌ

Sediment PitFlushing gate

(2) Dimension of settling basin

Common Values

Lw=B-b

Depends on site condition

b=bi

Decided Values

Un-known Values

hs=hi+(Lc+Lw)*ic

5-18

19

Where,

l : minimum length of settling basin (m)

hs : water depth of settling basin (m)

U : marginal settling speed for sediment to be settled (m/s)

usually around 0.1 m/s for a target grain size of 0.5 – 1 mm.

V : mean flow velocity in settling basin (m/s)

usually around 0.3 m/s

V = Qd/(B×hs)

Qd: design discharge (m3/s)

B : width of settling basin (m)

2-2 Intake and Settling Basin(2) Dimension of settling basin

l≧ x hs L= 2 x l U V

5-19 20


(2) Dimension of settling basinLl≧ x hs L= 2 x l U V

Design Discharge (Q) 0.220 Width of Intake (bi = b) 0.550 Length of conduit section (Lc) 2.000 Length of widening section (Lw) 0.950 Width of settling basin (B) 1.500 hi 0.400 ic 0.050 1/20 hs=hi+(Lc+Lw) x ic 0.548 U 0.100 V=Q/(B*hs) 0.300

1.339 B1.500 Bact

Vact =Q/(Bact*hs) 0.268 l=(Vact/U) x hs 1.467Ls=2 x l 2.933Length of basin= Ls 3.000

Width of settling basin (B=Q/(V x hs))




: Decided Values

5-20

21

2-3 Headrace

(1) Type of Headrace

Open Type

Closed Type

No-Lining type

Lining type

Pipe type

Box type

5-21 22

2-3 Headrace

Hea

d L

oss


Slope =Gentle

Micro=1/100 to 1/300

(2) Dimension of Headrace (Open Type)

Dimension of headrace depends on Discharge and Slope

5-22

23

Values for Deign

h

b

A

bLength of red-line : P

1

m

Slope =1/m: SL

Q

Q= A ×R 2/3×SL1/2 ／n

Q : design discharge for headrace (m3/s)

A : area of cross section (m2)

R : R=A／P (m)

P : length of wet sides (m).

SL : longitudinal slope of headrace (e.g. SL= 1/100=0.01)

n : coefficient of roughness (for concrete =0.015)

A= b x h

5-23 24

Example

Q=0.220m3/sSL=1/250=0.004

Condition for calculation

Designer’s setting

b= 0.550 m

Q : design discharge for headrace (m3/s)

A : area of cross section (m2)

R : R=A／P (m)

P : length of wet sides (m) refer to next figure.

SL : longitudinal slope of headrace

n : coefficient of roughness (for concrete =0.015)

A= b x h

0.550

0.5

50

0.3

35

Apr

. 0.2

A P R R2/3 Qi

b x h b+2 x h A/P Q= A ×R 2/3×SL 1/2 ／n

0.100 0.063 0.055 0.750 0.073 0.175 0.0410.200 0.063 0.110 0.950 0.116 0.237 0.1100.300 0.063 0.165 1.150 0.143 0.274 0.1910.335 0.063 0.184 1.220 0.151 0.283 0.2200.400 0.063 0.220 1.350 0.163 0.298 0.2770.500 0.063 0.275 1.550 0.177 0.316 0.366

SL1/2h

5-24

25

2-3 Headrace80

025

015

0

600250 250160 160

1,380

850

250

Mortar Plaster t=5cm

Masonry Concrete

470

D8

D9

150 600 150

0.5

1.0

330

47080

025

015

0

200

850 1,

050

5-25 26

2-4 Head-tank (Fore-bay -tank)

Spill way

Flush Gate

Screen

5-26

27

2-4 Head-tank (Fore-bay -tank)

0.5

1.0

dsc

As

d

Bspw

hc

h0

h>1.0×ｄ

S=1～2×ｄ

1.0

20.0

Ｂ

Ｌ

1.02.0

30～50cm

ｂ

B-b

Headrace

30～50cm

Ht

Spillway

Screen

SLe

h0=H*×0.1／（Sle）0.5　H*：Refer to 'Reference 5-1'

hc=｛(α×Qd2)／(ｇ×B2

)}1/3　　　

α=1.1　g=9.8

d=1.273×(Qd／Vopt）0.5　 Vopt:Refer to 'Reference 5-2'

Vsc=As×dsc=B×L×dsc≧10sec×QdB,dsc:desided depend on site condition.

Common Values

Decided Values

Un-known Values

0.5

1.0

dsc

As

d

Bspw

hc

h0

h>1.0×ｄ

S=1～2×ｄ

1.0

20.0

Ｂ

Ｌ

1.02.0

30～50cm

ｂ

B-b

Headrace

30～50cm

Ht

Spillway

Screen

SLe

h0=H*×0.1／（Sle）0.5　H*：Refer to 'Reference 5-1'

hc=｛(α×Qd2)／(ｇ×B2

)}1/3　　　

α=1.1　g=9.8

d=1.273×(Qd／Vopt）0.5　 Vopt:Refer to 'Reference 5-2'

Vsc=As×dsc=B×L×dsc≧10sec×QdB,dsc:desided depend on site condition.

dsc < h

5-27 28

2-4 Head-tank (Fore-bay-tank)Example

Q=0.220m3/sB= 2.000 m

Condition for calculationDesigner’s setting

Vsc > 10 x QAs = B x L = 8.000 m2

L= 4.000 m dsc = 10 x Q/As =2.20/8=0.275mh = 0.335m

Calculation

dsc≦h

Check

Change B or LNo

hc={(α×Q2)／(g×B2)}1/3

α: 1.1 g : 9.8

Yes

5-28

29

2-5 Penstock

5-29 30

Lp

Head Tank

Power

Hp

Ap = Hp ／ Lp

Powerhouse

2-5 Penstock Diameter of penstock

Example

Q : Discharge 0.220 m3/s

Lp: Total length of penstock

80.0m

Hp: Head from Head-tank to C/T

20.0m

Ap=Hp/Lp=0.25

Vopt= 2.3 m/s

D=1.128 x (Q/Vopt)0.5

=1.128 x (0.22/ 2.3)0.5

=0.348 →0.350 m

0.500.600.700.800.901.001.101.201.301.401.501.601.701.801.902.002.102.202.302.402.502.602.702.802.903.003.103.20

0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40

Average angle of penstock Ap

Opt

imum

vel

ocit

y V

opt(

m/s

)

D=1.273×(Q／Vopt)0.5

D:　diameter of pipe(m)

Q: design discharge(m3/s)Vopt: optimum velocity(m/s)

1.128

5-30

31

2-5 Penstock Thickness of penstock

t0 =　　　　+ δt (cm)P×d

2×θa×ηt0 =　　　　+ δt (cm)

P×d

2×θa×η

and t0=≧0.4cm or t0≧(d+80)／40 cm

t0: minimum thickness of pipe

P: design water pressure i.e. hydrostatic pressure + water hammer (kgf/cm2) ,

in micro-hydro scheme P=1.1×hydrostatic pressure.

for instance, if the head which from headtank to turbine is 25m,

P=2.5×1.1=2.75 kgf/cm2.

d: inside diameter (cm)

θa: admissible stress (kgf/cm2) SS400: 1300kgf/cm2

η: welding efficiency (0.85～0.9)

δt : margin (0.15cm in general)

5-31 32

2-6 Powerhouse (for impulse turbine)

Flood Water Level(Maximum)

20cm

boSection A-A

20cm

b

bo: depends on Qd and He

30～50cm

ｈｃ

30～50cm

HL3

(see Ref.5-3)

hc={ }1/31.1×Qd2

9.8×ｂ２

A

A

Afterbay Tailrace cannel Outlet

5-32

33

2-6 Powerhouse (for refection turbine)

Section A-A

1.5×d3

Flood Water Level(Maximum)30～50cmｈｃ

2×ｄ3

d3

20cm

1.15×d3

1.5×d3

Hs

Hs：depens on characteristic of turbine

HL3

(see Ref.5-3)

hc={ }1/31.1×Qd2

9.8×ｂ２

A

A

5-33 34

2-6 Powerhouse (with tailrace gate)

Pump

Gate

HL3

Flood Water Level (Maxmum)

5-34

35

2-7 Calculation of Head Loss

Hg He

HL1HL2

HL3

Forebay

Penstock

Settling Basin

Headrace Intake

PowerhouseTailrace

H

He = Hg – (HL1 + HL2 + HL3 )

Where: He - Effective Head

Hg - Gross Head

HL1 - Loss from intake to head-tank (fore-bay)

HL2 - Loss at penstock

HL3 - Installation head and Loss at tailrace

5-35 36

Hg He

HL1 HL2

HL3

Forebay

Penstock

Settling Basin

Headrace Intake

Powerhouse Tailrace

H


(1) Calculation of HL1: Loss from intake to head-tank

Elevation of crest of intake weir : ELs

Elevation of water level at head-tank : ELe

HL1= ELs-Ele

5-36

37


Hg He

HL1HL2

HL3

Forebay

Penstock

Settling Basin

Headrace Intake

PowerhouseTailrace

H

Flood Water Level(Maximum)

30～50cm

ｈｃ

30～50cm

HL3

(see Ref.5-3)

{ }9.8×ｂ２

A

Afterbay Tailrace cannel Outlet

HL3

(2) Calculation of HL3:Loss at tailrace

5-37 38

(3) Calculation of HL2:Loss at penstock2-7 Calculation of Head Loss

(a) Friction Loss

Friction loss (Hf) is one of the biggest losses at penstock.

④ Hf = f ×Ｌp×Ｖp2 ／（2×g×Ｄp）

① f - Coefficient on the diameter of penstock pipe . f= 124.5×n2／Dp1/3

② Ap - Cross sectional area of penstock pipe. (m2) Ap = 3.14×Dp2／4.0

③ Vp - Velocity at penstock (m/s) Vp = Q ／ Ap

Q - Design discharge (m3/s) Lp - Length of penstock. (m)

Dp - Diameter of penstock pipe (m) g=9.8

n = Coefficient of roughness (steel pipe: n=0.12, plastic pipe: n=0.011)

5-38

39

(3) Calculation of HL2:Loss at penstock


(b) Inlet Loss

He = f e×Ｖp2 ／（2×g）

fe : Coefficient on the form at inlet. Usually fe = 0.5 in micro-hydro scheme

(c) Valve Loss

Hv = f v×Ｖp2 ／（2×g）fv = 0.1 ( butterfly valve)

HL2=1.1 x (Hf + He + Hv)

5-39 40

Thank You !!!!

5-40

1

Training on

Micro Hydropower

Development

Reference

6-1 2

Optimum/Appropriate

Installed Capacity of

Mini Hydropower Plant

6-2

3

Optimum Installed Capacity

Generation Side Condition Demand Side Condition

Conditions for Optimum Installed Capacity

6-3 4

Daily Discharge Jun 2006-May 2008 (C.A=20.2km2)

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

J J A S O N D J F M A M J J A S O N D J F M A MD

isch

arge

(m

3 /s)

1- Generation Side Condition

(1) Discharge at the Site

6-4

5

40% 50% 70%

0.5

1.0

1.5

Riv

er F

low

(m

3/s

)

60% 80%

Co

ns

tru

cti

on

Co

st

/ k

Wh

Percentage of Duration

40 % 50 % 60 % 70 % 80 %

For Mini/Large Hydro : Comparison of Unit Cost in Each Case

Duration Curve : How to Identify Max. Design Discharge

6-5 6

Duration Curve at Intake Site (C.A.=20.2km2)

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100

Percentage (%)

Dis

char

ge (m

3/s

)

Duration Curve

(2) Duration Curve, Max. Discharge and Plant Factor

A

D

Area of (A-b-c-C-D)Discharge Plant Factor= =

Area of (A-B-C-D)76.69%

100

90

80

70

60

50

40

30

20

10

0

Dis

char

ge P

lant

Fac

tor

(%)

B

C

b

c

Discharge Plant Factor

Maximum Design

Discharge = 0.8 m3/s

Average Discharge in the Power Plant = Maximum Design Discharge x Discharge Plant Factor=0.8 m3/s x 0.7669 = 0.613 m3/s

Average Output = 9.8 x Average Discharge x Head x efficiency

Annual Generation (kWh) = Average Output x 365 days x 24hr

6-6

7

(2) Duration Curve, Max. Discharge and Plant Factor

Duration Curve at Intake Site (C.A.=20.2km2)

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100

Percentage (%)

Dis

char

ge (m

3/s

)

100

90

80

70

60

50

40

30

20

10

0

Dis

char

ge P

lant

Fac

tor

(%)

Discharge Plant Factor

Duration Curve

1.20m3/s64.45 %

1.02m3/s

69.95 %

0.81m3/s

76.69 %

0.67m3/s

81.24 %

0.50m3/s

86.46%

0.37m3/s

91.27%

0.21m3/s

96.56%

6-7 8

(3) Max. Output and Annual Generation in each Case

Condition : Effective Head = 100 meters

Total Efficiency = 76 %

Note : ③= 365days x 24hrs x ① x ②

DurationPercentage

(%)

MaximumDischarge

(m3/s)

①Max.

Output(kW)

②

Plant Factor(%)

③Annual Generation

(MWh/year)

30 1.20 920 64.45 5,194

40 1.02 780 69.95 4,780

50 0.81 620 76.69 4,165

60 0.67 510 81.24 3,629

70 0.50 380 86.46 2,878

80 0.37 280 91.27 2,239

90 0.21 160 96.56 1,353

6-8

9

(4) Optimum Capacity based on Unit Generation Cost (Philippines)

Source : DOE-REMD

6-9

Unit Construction Costy = -24773Ln(x) + 314639

80,000

100,000

120,000

140,000

160,000

180,000

200,000

220,000

240,000

260,000

0 200 400 600 800 1000 1200 1400 1600 1800 2000

Installed Capacity (kW)

Uni

t C

onst

ruct

ion

Cos

t (P

hp/k

W)

10

(4) Optimum Capacity based on Unit Generation Cost(Philippines)

DurationPercentage

(%)

MaximumDischarge

(m3/s)

Max.Output(kW)

Plant Factor(%)

①Annual Generation

(MWh/year)

UnitConstruction

Cost(Php/kW)

②　TotalConstruction

Cost(Php)

③=②／①Unit Generation

Cost(Php/kWh)

30 1.20 920 64.45 5,194 145,579 133,932,488 25.78540 1.02 780 69.95 4,780 149,668 116,741,283 24.42550 0.81 620 76.69 4,165 155,356 96,320,447 23.12560 0.67 510 81.24 3,629 160,194 81,698,911 22.51070 0.50 380 86.46 2,878 167,483 63,643,592 22.11380 0.37 280 91.27 2,239 175,048 49,013,540 21.89490 0.21 160 96.56 1,353 188,912 30,225,875 22.334


21.000

22.000

23.000

24.000

25.000

26.000

20 30 40 50 60 70 80 90 100

Duration Percentage (%)

Uni

t Gen

erat

ion

Cos

t (P

hp/k

Wh)

Other Indexa. Cost/Benefitb. IRR

6-10

11

(4) Optimum Capacity based on Unit Generation Cost (Japan)

Unit Construction Cost y = -565.48Ln(x) + 5306.5

600

800

1,000

1,200

1,400

1,600

1,800

2,000

2,200

2,400

2,600

0 200 400 600 800 1000 1200 1400 1600 1800 2000


Uni

t C

onst

ruct

ion

Cos

t(1

000J

PY

/kW

)

Source : TEPSCO

6-11 12

(4) Optimum Capacity based on Unit Generation Cost(Japan)

DurationPercentage

(%)

MaximumDischarge

(m3/s)

Max.Output(kW)

Plant Factor(%)

①Annual Generation

(MWh/year)

UnitConstruction

Cost(JPY/kW)

②　TotalConstruction

Cost(JPY)

③=②／①Unit Generation

Cost(JPY/kWh)

30 1.20 920 64.45 5,194 1,447,453 1,331,656,923 256.37640 1.02 780 69.95 4,780 1,540,802 1,201,825,930 251.45250 0.81 620 76.69 4,165 1,670,622 1,035,785,782 248.67760 0.67 510 81.24 3,629 1,781,065 908,343,366 250.26870 0.50 380 86.46 2,878 1,947,452 740,031,745 257.12780 0.37 280 91.27 2,239 2,120,139 593,638,969 265.17590 0.21 160 96.56 1,353 2,436,591 389,854,514 288.059


240

250

260

270

280

290

300

20 30 40 50 60 70 80 90 100

Duration Percentage (%)

Un

it G

en

era

tion

Co

st(J

PY

/kW

h)

6-12

13

Reference : Comparison of Unit Construction Cost

Comparison of Unit Construction Cost

0

100,000

200,000

300,000

400,000

500,000

600,000

700,000

800,000

900,000

1,000,000

1,100,000

1,200,000

0 200 400 600 800 1000 1200 1400 1600 1800 2000


Uni

t C

onst

ruct

ion

Cos

t (P

hp/k

W)

Philippines

Japan

6-13 14

Reference : Comparison of Duration Curve

Duration Curve C.A=20.2km2

0

1

2

3

4

5

6

7

8

9

10

0 10 20 30 40 50 60 70 80 90 100

Percentage of Date (%)

Uni

t D

isch

erge

(m

3 /s/1

00km

2 ) Based on Statistical Analysis

Based on Actual Data

6-14

15

2- Demand Side Condition

Demand Area

<Big Demand Area> Big Capacity of the Grid Enough Other Power Source

<Small Demand Area> Small Capacity of the Grid Insufficient Other Power Source

6-15 16

Daily Discharge Jun 2006-May 2008 (C.A=20.2km2)

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

J J A S O N D J F M A M J J A S O N D J F M A M

Dis

char

ge (

m3 /s

)

30%

40%

50%

60%

70%

80%

90%


Fluctuation of the Generated Power

920kW

780kW

610kW

520kW

390kW

270kW

160kW

6-16

17


0.2

0.4

0.6

0.8

Disch

a

50%

60%

70%

80%

90%

610kW

520kW

390kW

270kW

160kW

Existing Power Source

Fluctuation is relatively Small

Most of Generated Power can be Sold

Optimum Installed Capacity depends on Generation Side Condition

(1) Big Demand Area

6-17

High Load Factor

Selling Electric Generation (kWW)Load Factor=

Electric Generation of the Plant (kWh)

18


(2) Small Demand Area

Existing Power Source

0.2

0.4

0.6

0.8

Dis

ch 50%

60%

70%

80%

90%

610kW

520kW

390kW

270kW

160kW

Fluctuation is relatively Big

Some of Generated Power can not be Sold

Optimum Installed Capacity depends on Demand Side Condition

6-18

Low Load Factor

19

Construction CostDepends on

Installed CapacitykW

ProfitDepends on

Sold Annual GenerationkWh

Over Installed Capacity High Cost & Low Profit

6-19 20

MARAMING SALAMAT!!!

Thank You Very Much!!!!

Arigato!!!

6-20

DOEDOE--JICA Rural Electrification Project forJICA Rural Electrification Project forSustainability Improvement of Renewable Energy Development in ViSustainability Improvement of Renewable Energy Development in Village Electrificationllage Electrification

Review Training forReview Training forMicroMicro--hydropower Technologieshydropower Technologies

Electric and Mechanical EquipmentElectric and Mechanical Equipment

2DOEDOE--JICA Rural Electrification Project forJICA Rural Electrification Project forSustainability Improvement of Renewable Energy Development in ViSustainability Improvement of Renewable Energy Development in Village Electrificationllage Electrification

TurbineTurbine


ContentsContents

1. Basics of hydraulics

2. Turbine types

3. Characteristics of turbine

4. Basic design of turbine


1.1 Principle of continuity1.1 Principle of continuity

Discharge is constant at any section of the pipe regardless of change in the sectional area.

Q1 = Q2 (Q=constant)

A1 X V1 = A2 X V2*Q (m3/s) = A (m2) X V (m/s)

In other words, if the section area of the pipe is reduced, the velocity will be increased.

Discharge: Q1Sectional area: A1Velocity: V1

Discharge: Q2Sectional area: A2Velocity: V2

Water flowWater flow

PipePipe

1. Hydraulics


1.2 1.2 Bernoulli's theoremBernoulli's theorem

Close

Reference levelPotential head: z (m)

Pressure head:p / w (m)

Pressure energy:p (kg/m2)= w (kg/m3) X z (m)

w: unit weight of water

Pressure head: z = p / w (m)

Total head =(Potential head)+(Pressure head)= z + (p / w)

No flow

Total head

1.2.1 1.2.1 Energy of water without discharge (v=0 m/s)Energy of water without discharge (v=0 m/s)

1. Hydraulics



Open


Kinetic energy:(1/2) X (w/g) X v2 = w X zg: gravity acceleration 9.8 (m/s2)

Velocity head: z = v2 / 2g (m)

Total head =(Potential head)+(Pressure head)+(Velocity head)= z + (p / w) + (v2/2g)

Flow velocity v (m/s)

Total head

1.2.2 1.2.2 Energy of water with discharge (vEnergy of water with discharge (v≠≠0 m/s)0 m/s)(not considering head loss)


Velocity head:v2 / 2g (m)

1. Hydraulics


1.2 1.2 Bernoulli's theoremBernoulli's theorem1.2.3 Bernoulli1.2.3 Bernoulli’’s theorems theorem

Sum of potential head, pressure head, and velocity head is constant at any section of the pipe.

(Potential head) + (Pressure head) + (Velocity head) = Constant

z + (p / w) + (v2 / 2g) = Constant

If the flow velocity is increased due to reduction of the sectional area, the pressure will be decreased.

Total head H = hA + (pA / w) + (vA2/2g)

= hB + (pB / w) + (vB2/2g)

1. Hydraulics



Open


Ref.Head loss consists of friction loss hf, inlet loss he, valve loss hv, etc.

hf=f X (Lp/Dp) X (v2/2g)he=fe X (v2/2g)hv=fv X (v2/2g) ho=5～10% X (hf+he+hv)

Total head =(Potential head)+(Pressure head)+(Velocity head)+(Head loss)= z + (p / w) + (v2/2g) + Hloss

Flow velocity v (m/s)Total head

1.2.4 1.2.4 Energy of water with discharge (vEnergy of water with discharge (v≠≠0 m/s)0 m/s)(considering head loss)

Velocity head:v2 / 2g (m)


Head loss: Hloss (m)

1. Hydraulics



1.2.5 Calculation of net head on site (1)1.2.5 Calculation of net head on site (1)

Penstock pressure P (kgf/cm2): measured by pressure gauge

Pressure gauge height h (m): measured by tape

Discharge Q (m3/s): measured by ultrasonic flow meter

Penstock outside diameter Dpo (m): measured by tape

Available data on site:

Pressure gauge Ultrasonic flow meter

Pressure gauge

PenstockTurbineCenter

h

Height of pressure gauge h

1. Hydraulics


1.2 1.2 Bernoulli's theoremBernoulli's theorem 1. Hydraulics


Gross head Hg (m) = Ps (kgf/cm2) X 10 + (pressure gauge height h)Ps: readout of the pressure gauge under suspension (inlet valve closed)

Penstock inside diameter Dp (m):

Estimated based on the nominal size of the penstock

Penstock sectional area A (m2) = (πXDp2)/4

Flow velocity v (m/s) = Q / A

Net head He = (Pressure head) + (Velocity head) + (Potential head)

= (Po X 10) + (v2 / 2g) + (pressure gauge height h)Po: readout of the pressure gauge in operation

Head loss Hloss (m) = (Gross head Hg) – (Net head He)


1.2 1.2 Bernoulli's theoremBernoulli's theorem 1. Hydraulics


Exercise

Measurements on site Penstock pressure Ps: 1.266 kgf/cm2 (under suspension) Penstock pressure Po: 0.956 kgf/cm2 (in operation) Pressure gauge height h: 0.25 m Discharge Q: 0.0533 m3/s (53.3 L/s) Penstock inside diameter Dpi: 0.2 m

Please calculate gross head Hg, net head He, and head loss Hloss. Gross head Hg (m) = Penstock sectional area A (m2) = Flow velocity v (m/s) = Velocity head Hv (m) = Net head He (m) = Head loss Hloss (m) =


2.1 Types of turbine2.1 Types of turbine 2.Turbine types

The runner rotates by impulsive force of water jet with the velocity head, which has been converted from the pressure head at the time of jetting from the nozzle

Pelton turbine Crossflow turbine* Turgo-impulse

Impulse turbine:Impulse turbine:

The runner rotates by reactive force of water with the pressure head

Francis turbine Propeller turbine (Kaplan, Bulb, Tubular, etc.)

Reaction turbine:Reaction turbine:

*Crossflow turbine has characteristics of both impulse and reaction turbine


2.Turbine types2.1 Types of turbine2.1 Types of turbine

2.1.1 2.1.1 PeltonPelton turbineturbine Water jet from the nozzles acts

on the buckets, and the runner is rotated by the impulsive force

Horizontal-shaft Pelton turbine can be applied to micro/small hydropower project

Suitable for run-of-river project, especially with high-head and less head change

Applicable range Output: 100 – 5,000 kW Discharge: 0.2 – 3 m3/s Head: 75 – 400 m



2.1.2 2.1.2 CrossflowCrossflow turbineturbine Arc shape runner blades are

welded on the both side of iron plate discs

Simple structure, easy O&M, andreasonable price

Suitable for rural electrification project using micro hydropower plant




2.1.3 Francis turbine2.1.3 Francis turbine Water flow brought from the

penstock flows into the runner through casing and guide vane

Wide applicable range of head and discharge

Horizontal-shaft Francis turbine can be applied to micro/small hydropower project

Applicable range Output: 200 – 5,000 kW Discharge: 0.4 – 20 m3/s Head: 15 – 300 m Spiral casing

Guide vane



2.1.4 Tubular turbine2.1.4 Tubular turbine One of propeller turbines tubular

casing

Wide applicable range of head and discharge

Suitable for low-head sites


Generator

Propeller RunnerGuide Vane

(Wicket Gate)

Timing Belt

Draft Tube


2.Turbine types2.2 Turbine selection chart2.2 Turbine selection chart

Discharge (m3/s)

Net

hea

d (m

)


3.Characteristics3.1 Specific speed3.1 Specific speed

3.1.1 Definition of specific speed Ns3.1.1 Definition of specific speed Ns

Ns = Nt X (Pt1/2 / H5/4)

where,

Ns: Specific speed (m-kW)Nt: Turbine rotational speed (min-1)Pt: Turbine output (kW)H: Net head (m)

Specific speed is a numerical value expressing the classification of runners (turbine types) correlated by the tree factors of head H, turbine output Pt, and rotational speed Nt. It represents the runner shape and characteristics of turbine.

LargeSmall Ns

Change in shape of reaction runner

Axial flowdiagonal flowRadial flow



3.1.2 Specific speed of 3.1.2 Specific speed of CrossflowCrossflow turbineturbine

Ns = Nt X (Pt1/2 / H5/4)

where,

Ns: Specific speed (m-kW)Nt: Turbine rotational speed (min-1)Pt: Turbine output (kW)H: Net head (m)

Inlet width: bo

Diameter: D

1. Turbine output Pt is proportional to discharge Q, i.e. inlet with bo.

2. Net head H is proportional to diameter D.

Specific speed Ns of Crossflowturbine represents the shape of runner (bo / D)

LargeSmall Ns (bo/D)Change in shape of Crossflow runner



3.1.3 Applicable range by turbine type3.1.3 Applicable range by turbine type

Applicable range of Ns is empirically determined by turbine type, which is limited by process limitation (narrow inlet), mechanical strength limitation (high speed machine), and cavitation characteristics.

NOTE:As for Crossflow turbine, Pt for Ns calculation is defined as follow;

Pt = Pr / (bo / D) Pr: Turbine output per unit (kW)

200～900Propeller turbine

500～Tubular turbine

50～350Francis turbine

90～110Crossflow turbine

8～25Pelton turbine

Applicable specific speedNs (m-kW)Turbine type


3.Characteristics3.2 Turbine efficiency3.2 Turbine efficiency

In the process of converting hydraulic energy (input) into rotational energy (output) by a turbine, hydraulic and mechanical losses occur. Turbine efficiency is defined as the proportion of the output to the input.

ηt = {Pt / (9.8 X Q X H)} X 100 (%)

where, ηt: Turbine efficienby (%)Pt: Turbine output (kW)9.8QH: Theoretical power (kW) (i.e. Turbine input)Q: Discharge (m3/s)H: Net head (m)

3.2.1 Definition of turbine efficiency 3.2.1 Definition of turbine efficiency ηηtt


3.Characteristics3.2 Turbine efficiency3.2 Turbine efficiency

At the stage of basic design, the following figures can be used as turbine efficiency by turbine type in order to estimate the turbine output.

NOTE:As for Crossflow turbine manufactured locally, 40-50% of efficiency can be applied in consideration of fabrication quality of the work shop.

3.2.2 Turbine efficiency for basic design3.2.2 Turbine efficiency for basic design

82Propeller turbine

84Tubular turbine

84Francis turbine

77Crossflow turbine

82Pelton turbine

Turbine efficiencyηt (%)Turbine type


4.Basic design4.14.1 Flow chart of basic designFlow chart of basic design

Design net head: HDesign discharge: Q

Selection ofapplicable turbine type

Calculation of applicable maximum specific speed

Calculation of maximum rotational speed

Selection of turbine rotational speed

Recalculation ofspecific speedEstimation of

turbine output

Turbine type:Design net head H (m):Design discharge Q (m3/s):Frequency F:Rotational speed Nt:Specific speed Ns:Turbine efficiency ηt (%):

refer to Turbine selection chart (see Clause 2.2)

4.1.1

4.1.2

4.1.3

4.1.4

4.1.5

Input

Output


Turbine output Pt is estimated using design head H and discharge Q, which derived from the result of planning and civil designing.

Pt = 9.8 X Q X H X ηt (kW)

Turbine efficiency ηt listed in Clause 3.2.2 can be applied to the above calculation at the stage of basic design.

4.1.1 4.1.1 Estimation of turbine outputEstimation of turbine output


ExampleTurbine type: H-shaft FrancisNet head H: 45 mDischarge Q: 2.5 m3/sFrequency F: 50 Hz

Please estimate the turbine output.Estimated turbine efficiency ηt: % (see 3.2.2)Estimated turbine output Pt =

== kW


Applicable maximum specific speed Nsmax is empirically determined by turbine type, which derived from the following formulas.

4.1.2 4.1.2 Calculation of applicable maximum specific speedCalculation of applicable maximum specific speed



Please calculate Nsmax.Applicable max. specific speed Nsmax

=== m-KW

{(2,000/(H+20))+30}Francis turbine

{(2,000/(H+20))+50}Propeller turbine

2,000/(H+16)Tubular turbine

3,200 X H-2/3H-shft Francis turbine

650 X H-0.5Crossflow turbine

85.49 X H-0.213Pelton turbine

Applicable maximum specific speed NsmaxTurbine type


Maximum rotational speed Nmax is derived by applying the calculated Nsmax to the following formula for specific speed.

Ns = Nt X (Pt1/2 / H5/4) (m-kW)

Ntmax = Nsmax X (H5/4 / Pt1/2) (min-1)

4.1.3 4.1.3 Calculation of maximum rotational speedCalculation of maximum rotational speed



Please calculate Ntmax using estimated Pt.Applicable max. rotational speed Ntmax

= = = min-1

Calculated Nsmax in Clause 4.1.2


In case that turbine is directly connected with generator, turbine rotational speed Nt is selected from the following standard rotational speed, which is the maximum value less than Ntmax.

4.1.4 4.1.4 Selection of turbine rotational speedSelection of turbine rotational speed



Please select appropriate Nt considering Ntmax (note the rated frequency).

min-1 of turbine rotational speed is selected because

500600750

1,0001,500

50Hz

600720900

1,2001,800

60Hz

2420181614

Nos. of poles

250300333375429

50Hz

40083601030012

45065144

60HzNos. of poles*

* Number of generator rotor poles


Specific speed Ns is finalized using the selected turbine rotational speed Nt.

Ns = Nt X (Pt1/2 / H5/4) (m-kW)

4.1.5 4.1.5 Recalculation of specific speedRecalculation of specific speed



Please calculate Ns using selected Nt.Specific speed Ns =

== m-kW

Summary of turbine basic designTurbine type: H-shaft FrancisNet head H: 45 mDischarge Q: 2.5 m3/sFrequency F: 50 Hz

Rotational speed Nt: 750 min-1

Specific speed Ns: 196 m-kWMaximum efficiency ηt: 84%



Electrical and Mechanical EquipmentElectrical and Mechanical Equipment


GeneratorGenerator


ContentsContents

1. Basics of generator

2. Classification of generator

3. Basic design of generator


1.1 Components of Generator1.1 Components of Generator 1. Basics

The field consists of coils of conductors within the generator that receive a voltage from a source (called excitation) and produce a magnetic flux. The armature is the part of an AC generator in which output voltage is produced. The rotor of an AC generator is the part that is driven by the prime mover and that rotates. The stator of an AC generator is the part that is stationary. Slip rings are electrical connections that are used to transfer power to and from the rotor.


1. Basics

Consists of (a) strong magnetic field, (b) conductors that rotate through that magnetic field, and (c) a means by which a continuous connection is provided to the conductors as they are rotating

1.2 Principle of AC generator1.2 Principle of AC generator

1.2.1 Theory of Operation1.2.1 Theory of Operation


1.2 Principle of AC generator1.2 Principle of AC generator 1. Basics

1.2.2 Generation of alternating voltage (1)1.2.2 Generation of alternating voltage (1)

If a coil rotates between poles, electromotive force is induced in the coil according to Fleming's right-hand rule.

e = B X L X v X sinθ (V)

where, e: Induced electromotive force (V)B: Magnetic flux density (T)L: Length of coil (m)v: Rotational speed of coil (m/s)θ: Angle between vectors of B and v (rad)

v

vB

Coil

vB

v

θv sinθ



1.2.3 Generation of alternating voltage (2)1.2.3 Generation of alternating voltage (2)

θ=0 → e = 0 (sin(0)=0)

e = B X L X v X sinθ

e = B X L X v (sin(π/2)=1)

Electromotive force with sine-wave is induced in one cycle

θ

π/2



1.2.4 Relationship between voltage and rotational speed1.2.4 Relationship between voltage and rotational speed

Previously cited electromotive force e is modified as follows:

e = 4.44 X f X w XΦ (V) e ∝ f

where, e: Induced electromotive force (V)f: Frequency (Hz)w: Number of series winding turns per phaseΦ: Magnetic flux per pole (Wb)

Without AVR, the electromotive force (gen. terminal voltage) fluctuates in proportion to variation in the frequency (gen. rotational speed).

Oversupply leads to higher frequency and voltage than their rated value Overload leads to lower frequency and voltage than their rated value


1.3 Types of Generator1.3 Types of Generator 1. Basics

Stationary Field, Rotating Armature AC Generator :

Rotating Field, Stationary Armature AC Generator :


1.4 Structure1.4 Structure 1. Basics

Appearance of ST series generator

StatorField windings Rotor pole

Brush

Brush holder

Slip ringMain shaft

Brush holder Slip ringDC current to field windings


1.5 Excitation1.5 Excitation 1. Basics

1.5.1 Classification of excitation system1.5.1 Classification of excitation system

Self-excitation type:Field windings of rotor carry DC current obtained by rectifying a portion of the generator’s AC output. Therefore, the exciter is not necessary.Terminal voltage fluctuates in response to load variation.

Separate excitation type:DC field current is supplied from outside sources, such as the exciter directly coupled with the main shaft. Terminal voltage can be kept constant because an Automatic Voltage Controller (AVR) is equipped.

Self-excitationSeparate-excitation

Brushless excitation systemDC exciter typeAC exciter typeStatic exciter type Thyristor excitation system



1.5.2 Example of self1.5.2 Example of self--excitation typeexcitation type

Schematic diagram for ST series generators of MINDONG, China

AVR is not equipped. Maintenance for brushes

and slip rings is necessary.



1.5.3 Example of separate1.5.3 Example of separate--excitation type (1)excitation type (1)

Thyristor excitation systemDC current to the field windingsis supplied through slip ringsfrom an excitation transformer and thyristors. The thyristorsare controlled by AVR to keep the terminal voltage constant.

Low initial cost due to not exciter

High maintenance cost due to periodical replacement of brushes

G

PT

Ex. Tr

Slip ring

Example of thyristor excitation system

Rotating section

AVRPulse

Generator



1.5.4 Example of separate1.5.4 Example of separate--excitation type (2)excitation type (2)

Brushless excitation systemExcitation circuit consists of an AC exciter directly coupled withthe generator, a rotary rectifier,and tyristers controlled by AVR.

High initial cost due to the ACexciter and rotary rectifier

Low maintenance cost due to no consumable parts such as brushes and slip rings

Example of brushless excitation system

G

PT

Ex. Tr

AVR

DC100V

Rotating section

ACEx

PulseGenerator


2.1 Classification of AC generator2.1 Classification of AC generator 2. Classification

by generator type Synchronous generator Induction generator

Features of each type are shown in the next Clause 1.2

by number of phase

by shaft arrangement

Three-phase

Single-phase

High transmission efficiency due to small current with the same capacity as single-phase machine (58% of 1-pahse)

Simple structure and easy maintenance

Horizontal-shaft

Vertical-shaft

Not suitable for large-scale hydro due to limitation of shaft deflection

Suitable for small-scale/micro hydroEasy maintenance

Suitable for large-scale hydro


2.1 Comparison of generator types2.1 Comparison of generator types 2. Classification

Need no synchronizerInrush current at

parallel-in operation

Not suitable for independent operation (only on-grid operation)

Voltage, frequency, and power factor regulation is impossible

Need no excitation system (Excitation current is supplied from grid)

Simple structure and high maintainability (squirrel-cage rotor)

High mechanical strength

Induction generator

Need synchronizerLess electro-

mechanical impact at parallel-in operation

Independent operation is possible

Voltage, frequency, and power factor regulation is possible

Excitation system is necessary

Complex structure(salient-pole machine)

Maintenance for excitation system is necessary

Synchronous generator

Parallel-in operationOperationStructureItems

Only synchronous generator can be selected for independent operation.


3.Basic design3.1 Flow chart of basic design3.1 Flow chart of basic design

Standard frequency: FoTurbine output: Pt

Selection ofgenerator type

Selection ofrated rotational speed Nt

Selection ofrated voltage Vg

Calculation ofgenerator output Pg

Calculation ofrated capacity Pgu

Selection ofrated frequency F

Generator type:Capacity S (kVA):Voltage Vg (V):Current Ig (A):Power factor pf:Frequency F (Hz):Rotational speed Ng (min-1):Efficiency ηg (%):

3.2

3.3

3.6

3.6

Input

Output

3.2 Calculation ofrated current Ig

Estimation ofload power factor

3.4

3.5

Selection ofrated power factor pf

refer to Clause 2.1 and 2.2

3.4


3.Basic design3.2 Frequency and rotational speed (1)3.2 Frequency and rotational speed (1)

Selection of rated frequency:

Rated frequency should be selected to match the standard frequency, 50Hz of 60Hz, in the project country.

Selection of rated rotational speed:

High-speed machine is preferable from the viewpoint of economy and characteristics (generator efficiency).

Generator rotational speed is determined in consideration of theapplicable maximum rotational speed of the turbine (refer to “Basic design of turbine”).

In case of Crossflow and Tubular turbine, the applicable maximum rotational speed of which may be low, a speed increaser can be applied to obtain high generator rotational speed considering economic efficiency.


3.Basic design3.2 Frequency and rotational speed (2)3.2 Frequency and rotational speed (2)

Relationship among rated rotational speed, number of rotor poles, and rated frequency is represented by the following formula:

Ng = (120 X F) / p

where, Ng: Rated generator rotational speed (min-1)F: Rated frequency (Hz)p: Number of generator rotor poles

500600750

1,0001,500

50Hz

600720900

1,2001,800

60Hz

2420181614

Nos. of poles

250300333375429

50Hz

40083601030012

45065144

60HzNos. of poles

Standard rotational speed of generator


3.Basic design3.3 Rated voltage3.3 Rated voltage

Selection of rated voltage:

Generator voltage is determined on the basis of the capacity, economy, and characteristics of generator.

Low voltage of 200V or 400V is generally adopted as the rated voltage for micro-hydropower projects.

400V system

200V system

Step-down transformers are necessary at each supply area because the rated voltage of appliances in households is 200V.

Size of the transmission lines can be reduced due to less load current than 200V system.

If a step-up transformer is not installed at the powerhouse,load current to the households is larger than 400V systemdue to the lower rated voltage, which means the voltage dropin the transmission lines is also larger.

If the voltage drop at the households can be suppressed less than 10% of the rated voltage 200V, step-up and step-down transformers are not necessary.

Comparison of 200V and 400V system


3.Basic design3.4 Rated power factor (1)3.4 Rated power factor (1)

Power factor (pf) of generator represents the supply capacity of reactive power, which is expressed by the ratio of apparent power to active power. Rated power factor is generally selected in the range of 0.8 to 0.95 in consideration of the load power factor.

pf = Active power (kW) / Apparent power (kVA)= cosθ

Generator with low power factor (pf = 0.8) can supply more reactive power to the loads for voltage stability. On the other hand, physical size of generator became large as the power factor is reduced. This means that low power factor may push up the the price of generator.

P: Active power (kW)Q: Reactive power (kVar)S: Apparent power (KVA)θ: Phase deference between voltage

and current of generator (deg)P(kW)

Q(kVar)S(kVA)

θ


3.Basic design3.4 Rated power factor (2)3.4 Rated power factor (2)

Exercise

Please calculate reactive power consumption and load power factor at a household with the following appliances (fill in the table).

Reactive power = Apparent power X sinθ

= (Active power / cosθ) X {√(1 – cos2θ)}

* Active power = Active power consumption, cosθ= power factor

Total pf = cos {tan-1(Total reactive power / Total power consumption)}

-Total

0.95201Radio cassette player

0.951001TV

0.6080 (40 X 2)2Fluorescent lamp

Reactive power consumption (Var)

Powerfactor

Active power consumption (W)UnitAppliance


To calculate the total power factor

Learn how to calculate the apparent power

Above table shows the total power factor is

Generator capacity shall be more than 2,525VA

Equipment Nos.Powerfactor

Remark

incandescent lamp 40 W 10 400 W 1.0 400 kVAfluorescent lamp 20 W 15 300 W 0.6 500 kVA 300/0.63-phase induction motor 500 W 1 500 W 0.8 625 kVA 500/0.81-phase induction motor 200 W 3 600 W 0.6 1000 kVA 600/0.6

Total 1800 W 2525 kVA

capacityTotal

capacitykVA

1,8002,525

0.71 =


3.Basic design3.5 Output3.5 Output

Generator output is represented by the following formula:

Pg = Pt X ηg X ηd (kW)

where, Pg: Generator output (kW)Pt: Turbine output (kW) (= 9.8QHηt)ηg: Generator efficiency (%)ηd: Speed increaser efficiency (%) (if installed)

Generator efficiency of 90% and speed increaser efficiency of 95% can be applied to the above calculation at the stage of basic design.


3.Basic design3.6 Rated capacity and current3.6 Rated capacity and current

Rated capacity of generator is represented by the apparent powergenerated under the rated voltage, frequency, and power factor.

Pgu = Pg / pf (kVA)

where, Pgu: Rated generator capacity (kVA)Pg: Generator output (kW)pf: Rated power factor

Rated current of generator is calculated by the following formula:

Ig = (Pgu X 1000) / (√3 X Vg) (A)

where, Ig: Rated generator current (A)Vg: Rated generator voltage (V)





Control systemControl system


ContentsContents

1. Basics of automatic control

2. Frequency control

3. Voltage control


1.1 Types of control1.1 Types of control 1. Basics

Manual Control

System is monitored by human.

Adjustment operation is made manually.

Automatic Control

System is monitored by machine.

Adjustment operation is made automatically.


1.2 Feedback control (1)1.2 Feedback control (1) 1. Basics

Detection

Comparison

Judgment

Operation

LoopLoop


1.2 Feedback control (2)1.2 Feedback control (2) 1. Basics

Detection

Comparison

Judgment

Operation

Feed back control by ELC:

: To detect the value of frequency (f)

: To compare the observed value with the reference value (set point: fo)

: To judge the amount of operation in response to the deviation Δf (fo – f)

: To operate phase of current to the dummy loads according to the judgment


1.3 Structure of feedback control1.3 Structure of feedback control 1. Basics

Set Point Controller OperatingPart

ControlledSystem

ControlledVariable

DetectingPart

-

Block diagram of feedback control system

Disturbance

ff

ΔΔff

ELCELCfofo

trigger signaltrigger signal

SCRSCR Turbine /Turbine /GeneratorGenerator

load fluctuationload fluctuation

ff

Δf = fo - f


1.4 Operating characteristics1.4 Operating characteristics 1. Basics

Controller

Element in the Control System:

There is no need to understand the structure in the box. (“Black box”)

Relationship between input and output is the most important, which is called “Transfer Function”.

Input Output

Trigger signal to SCRΔf

ELC


Single line DiagramSingle line Diagram



1.5 Proportional action (P Control)1.5 Proportional action (P Control) 1. Basics

Inpu

t

0 time

Δf

0 time

offset

0 time

Advantage: remove cycling in the on-off action Disadvantage: leave offset to the deviation

Chan

ge in

Loa

d

0 time

step responsestep response

Kp x Input

Out

put


1.6 Integral action (I Control)1.6 Integral action (I Control) 1. Basics

0 time

Δf

0 time

Inpu

t

0 time

Advantage: remove offset from the deviation Disadvantage: low response speed

Chan

ge in

Loa

d

0 time


Out

put


1.7 P and I action (PI Control)1.7 P and I action (PI Control) 1. Basics

Out

put

0 time

Δf

0 time

Δf

0 time

Response by I Control Response by PI Control

Integral time: T1

Inpu

t

0 time


Response Response improvementimprovement


2.1 Necessity of control2.1 Necessity of control 2. Freq. Control

Demand side Performance deterioration and damage of

electrical appliances due to operation out ofthe rated conditions

Quality deterioration of products due torotational speed fluctuations of induction motors

Supply side Mechanical stress on the rotating machine system

Contribution to voltage stability


Principle of the Governorto keep the frequency constant“generator output = demand load” (essential)

Stable frequency Frequency comes downFrequency comes up

Output Output Output Load Load Load

In this case increase the dummy load

In this case decrease the dummy load

(Quoted form the HP of Micro Hydropower Japan)

2.2 Frequency and active power2.2 Frequency and active power 2. Freq. Control


2.2 Frequency and active power2.2 Frequency and active power 2. Freq. Control

Frequency fluctuation is caused by imbalance of active power between power supply (generation output) and demand (load) in power system.

ΔF = - K X ΔP

where,ΔP: Active power imbalance in power systemΔF: Frequency fluctuation caused by ΔPK: Coefficient

LowerPg < Pd (ΔP < 0)EvenPg = Pd (ΔP = 0)

HigherPg > Pd (ΔP > 0)Frequency change Conditions

Pg: Generator output ΔP: Pd - PgPd: Demand

ΔF

→ ΔPDroop characteristic of frequency


2.3 Methods of frequency control (1)2.3 Methods of frequency control (1) 2. Freq. Control

Speed governor: Frequency is kept constant by adjusting turbine input, inflow to the

turbine, in response to the load variations. Inflow to the turbine can be controlled by operating flow regulators

such as guide vanes and needles. Hydraulic or electric servo motor put the flow regulators in motion.

A number of auxiliaries, such as hydraulic system and power supplyunit, are required.

Examples of flow regulator(a) Pelton turbine (b) Crossflow turbine (c) Francis turbine

Guide vane

Runner

Needle

Deflector

Runner



Dummy load governor: Frequency is kept constant by matching the total power consumption of

actual loads and dummy loads to the generator output (Pg = Pactual + Pdummy) . Power consumption of dummy loads is controlled by ElectronicLoad Controller (ELC).

ELC adjust current to the dummy loads by phase-shift control to keep the condition of “Pg = Pactual + Pdummy” continuously.

ELC panel box (single-phase) Water-cooled heater for dummy loads



Comparing observed F with the set point FoΔF = Fo – FComparison

Judging the amount of operation according to ΔfJudgment

Operating the phase of dummy load current

Operating the opening of flow regulatorOperation

Detecting Frequency FDetection

Dummy load governorSpeed governor

Comparison of feedback control



Waste generating powerComplicated structure Less maintainabilityNeed for driving device

(hydraulic, electrical)Costly

Disadvantage

No need for mechanism to adjust water flow

Reasonable relatively Easy maintenance

Sensitive controlNot waste generating powerAdvantage

Micro-hydro Small to large scale hydroApplicability

Dummy load governorSpeed governor

Comparison of advantage and disadvantage

Dummy load governor is suitable for rural electrification project by micro-hydropower plant which is necessary for economy and high maintainability.


2.4 Speed governor2.4 Speed governor

Rotational speed (frequency F) is continuously transferred to the controller as a signal from the speed detector.

The transferred speed signal is compared with the preset signal Fo corresponding to the rated speed.

If the speed drops, the signal of “regulator open” is transmitted to the actuator of flow regulator. Flow regulator continue to be opened until the frequency returns to the rated value.

Function of speed governor

2. Freq. Control

Speedsetter

Controller

Flow regulatoroperation mechanism

Turbine

Generator

Speed detector

Flow regulatoropen/close

Comparison Judgment

Operation

Detection

F Fo


2.5 Dummy load governor (1)2.5 Dummy load governor (1) 2. Freq. Control

Transmission Line

SG

ELC・

・

DummyLoad 1

Coil・

DummyLoad 2

WT

Comparison Judgment

Operation

Detection

IL

Ig

Id

Current to the dummy loads (Id) is adjusted to keep “Ig = IL + Id” continuously by phase-shift control of SCRs.

SCR

TRIAC, Thyristor, or IGBT is used for controlling element of dummy load current


Badiangan McHP after ELC and New Turbine InstallationFebruary 2008

100

150

200

250

300

2/22/083:00 PM

2/22/086:00 PM

2/22/089:00 PM

2/23/0812:00 AM

2/23/083:00 AM

2/23/086:00 AM

2/23/089:00 AM

Time

Voltage, V

0

2

4

6

8

10

12

14

Curr

ent, A

Voltage PH Voltage HH Current DL Current HH


2.5 Dummy load governor (2)2.5 Dummy load governor (2)

Capacity of dummy load is calculated as follows:

Pd = Pg x pf x SF

where,Pd: Capacity of dummy load (kW)Pg: Rated capacity of generator (kVA) pf: Rated power factor of generator SF: Safety factor

According to cooling method, 1.2 – 1.4 times of the generator output in kW is selected as the safety factor to avoid over-heating of the heaters.

2. Freq. Control

Installation example of water-cooled heaters


3.1 Necessity of control3.1 Necessity of control 2. Volt. Control

Performance deterioration and damage of electrical appliances due to operation out of the rated conditions

Example of voltage characteristic of lamp

(a) Incandescent lamp (b) Fluorescent lamp

Current

EfficiencyFluxLongevity

Voltage (%)100

(%)

100

Longevity

CurrentEfficiency

Flux

Voltage (%)100

(%)

100

Shortened longevity at high-voltage

Low illumination at low-voltage

Shortened longevity at low & high-voltage

Low efficiencyat low-voltage


3.2 Voltage and reactive power3.2 Voltage and reactive power 2. Volt. Control

Voltage fluctuation is caused by variation in active power and reactive power.

ΔVP = KP X ΔP ΔVQ = KQ X ΔQ

where,ΔP: Active power variation in power system ΔQ: Reactive power variation in power systemΔVP: Voltage fluctuation caused by ΔPΔVQ: Voltage fluctuation caused by ΔQKP, KQ: Coefficient

Impact of reactive power variation to voltage fluctuation is much larger than that of active power variation (ΔVP / ΔVQ << 1)

ΔV P

→ ΔP

ΔV Q

→ ΔQ

Small Impact of ΔP to ΔV

Large Impact of ΔQ to ΔV


3.3 Automatic Voltage Regulator (1)3.3 Automatic Voltage Regulator (1) 2. Volt. Control

G

PT

Ex. Tr

Slip ring

Rotating section

AVRPulse

Generator

In order to keep generator terminal voltage Vg constant, Automatic Voltage Regulator (AVR) can be equipped to the generator. Some suppliers provide it as an optional extra.

AVR adjusts excitation current to the field windings to eliminate the voltage deviation ΔVg between the reference voltage and detected voltage by a potential transformer (PT).

Example of brushless excitation system with AVR

SCR

Field windings

Vg

Excitationcurrent

DetectionComparisonJudgment

Operation


3.3 Automatic Voltage Regulator (2)3.3 Automatic Voltage Regulator (2) 2. Volt. Control

Detection

Comparison

Judgment

Operation

Feedback control by AVR:

: To detect the generator terminal voltage (Vg) with PT

: To compare the observed voltage with the reference value (set point: Vgo)

: To judge the amount of operation in response to the deviation ΔVg (Vgo – Vg)

: To operate excitation current to the field windings according to the judgment



Electrical and Mechanical EquipmentElectrical and Mechanical Equipment


Electrical equipmentElectrical equipmentandand

Protection systemProtection system


ContentsContents

1. Main circuit components1.1 Major factors1.2 Transformer1.3 Switch gear1.4 Arrester1.5 Instrument transformer1.6 Single line diagram

2. Protection system


Short-circuitfailure

1.1 Major factors1.1 Major factors 1. Main circuit

Line voltage:Line voltage Vs is determined by the rated generatorvoltage Vg and the transmission line voltage VL.

Load current:Load current ILO is corresponding to the ratedgenerator current Ig and calculated as themaximum current at the rated operation.

ILO = Ig = (Pgu X 1000) / (√3 X Vg) (A)

where, Pgu: Rated generator capacity (kVA)

Short circuit current:Short circuit current is derived as the fault currentat a short-circuit failure that occurs in the main circuit.

GS

GS

Pgu (KVA)Vg (V)Ig (A)

ILO= Ig

Is (> ILO)

MTr


Transformer is installed to the main circuit to meet the following purposes.

To step up the generated voltage to the voltage suited to power transmission and distribution

To protect the electrical equipment in the main circuit from lightning surges entering from transmission lines

Types:by insulation method Oil-immersed transformer Molded dry-type transformer

by number of phase Single-phase Three-phase

1.2 Transformer (1)1.2 Transformer (1) 1. Main circuit

1-phase pole transformer(Oil-immersed type, Natural air-cooling)

Electrical symbol

Tr


Ratings:Rated capacity Pt (kVA):The same capacity as the rated generator capacity Pgu (kVA) is selectedin consideration of the standard capacities of products.

Rated voltage:Rated generator voltage Vg is applied to the primary voltage. Thesecondary voltage is determined based on the transmission voltage VL.

1.2 Transformer (2)1.2 Transformer (2) 1. Main circuit

3-phase transformer(Oil-immersed type, Natural air-cooling)

3-phase transformer(Molded type, Natural air-cooling)


Switch gears installed in the main circuit are classified by the function asshown in the following table.

1.3 Switch gear (1)1.3 Switch gear (1) 1. Main circuit

Manual operation by hook bar

Manual operation by hook bar

Operation

ditto

Withstanding Short-circuit fault current thermally and mechanically

Short-time withstandcurrentShort

circuitNormal

load

Appearance

Current breaking abilityElectrical

symbolType

XOLoad BreakSwitch (LBS)

XXDisconnectingSwitch (DS)

O: Possible to cut off X: Impossible to cut off

Classification and the function of switch gears


1.3 Switch gear (2)1.3 Switch gear (2) 1. Main circuit

Manual operation by hook barAutomatic trip by power fuse meltdown

Short-circuit fault current is cut off by power fuseXO

Fused Load BreakSwitch (PF-LBS)

Manual operationSolenoid operation

Solenoid operation

Operation

Withstanding Short-circuit fault current thermally and mechanically

Low short-circuit fault currentSuitable for high frequency switching

Short-time withstandcurrentShort

circuitNormal

load

Appearance

Current breaking abilityElectrical

symbolType

OOCircuit Breaker

XOMagnetic Contactor (Mg Ctt)

O: Possible to cut off X: Impossible to cut off


Arrester is installed to the main circuit to suppress abnormal voltage caused by lightning surges and switching surges, and to protect the electrical equipment in the main circuit.

Ratings:

Installation location:Arrester should be installed at connection point tothe transmission line to prevent the surges fromreaching to the inside of the powerhouse.

Rated voltage:Rated voltage listed in the following table is selectedaccording to the nominal voltage of the target circuit.

1.4 Arrester (1)1.4 Arrester (1) 1. Main circuit

848428148.44.2Rated voltage (kV)

663322116.63.3Nominal voltage (kV) GS

Step-up Tr

PF-LBS

LA

Transmission line


Ratings:Nominal discharge current:Nominal discharge current is used to express the protection performance and represented as the crest value of lightning impulse current with the predefined waveform.

1.4 Arrester (2)1.4 Arrester (2) 1. Main circuit

Electrical symbol Lightning arrester

LA


1.5 Instrument transformer (1)1.5 Instrument transformer (1) 1. Main circuit

Voltage transformer (PT, VT):PT is applied to transform voltage of electric line to voltage suited to use of instruments and relays.

Current transformer (CT):CT is applied to transform current of electric line to voltage suited to use of instruments and relays.

Electrical symbol

Electrical symbol


1.5 Instrument transformer (2)1.5 Instrument transformer (2) 1. Main circuit

Ground potential transformer (GPT):GPT is applied to transform zero phase voltage of electric line to voltage suited to use of instruments and relays.

Electrical symbol


1.6 Single line diagram1.6 Single line diagram 1. Main circuit

Example of single line diagram for a micro-hydropower plant

Step-up transformer(star-delta connection)

Thyristor excitation system(with AVR)

Dummy load governor(ELC with double TRIAC)

LA

Step-up Transformer

MCCB

Mg Ctt

Transmission line

WT

PF-LBS

GS

・

・

・

・

・

Ex Tr

AVR

・

・ ELC

DummyLoad 1

DummyLoad 2

Coil

TRIAC

SCR

・

LA SC


2.1 Role of protection system2.1 Role of protection system 2. Protection

Protection system, which operates when a fault occurs at equipment and electric lines in the power system, plays the following roles:

To detect the fault by inputs from instrument transformers

To separate the section in which the fault occurs from the normal sections

To prevent the fault from expanding to the normal sections by circuit breaker operation

To avoid a fatal accident and equipment damage


2.2 Protective relaying system2.2 Protective relaying system 2. Protection

GS

ProtectionrelayCT

PT

Breaking order Monitoring electrical valuables,

such as current, voltage, phase, and frequency, of the protected section continuously

Making the circuit breaker operate if an abnormal condition is detected

CB


2.3 Types of protection relay (1)2.3 Types of protection relay (1) 2. Protection

Acting when detecting voltage more than the setting value

Applied to generator as over voltage protection at no-load condition

Over voltage relay(OVR)

Acting when detecting voltage less than the setting value

Applied to generator and bus line as short-circuit protection

Under voltage relay (UVR)

FeaturesElectricalsymbolType

Acting when detecting current more than the setting value

Applied to generator, transformer, and transmission line as short-circuit protection

Over current relay(OCR)

I >

U >

U <


2.3 Types of protection relay (2)2.3 Types of protection relay (2) 2. Protection

Acting when detecting frequency more than the setting value

Applied to avoid independent operation in case of grid connection system

Applied to detect a fault of the ELC and dummy loads in case of independent system

Over frequency relay(OFR)

Acting when detecting frequency less than the setting value

Applied to avoid independent operation in case of grid connection system

Applied to detect a fault of the ELC and dummy loads in case of independent system

Under frequencyrelay (UFR)

FeaturesElectricalsymbolType

Acting when detecting zero phase voltage more than the setting value by GPT

Applied to bus line as ground fault protection

Over voltage ground relay (OVGR)

U >

f >

f <


2.4 Arrangement of protection relay2.4 Arrangement of protection relay

Example of protection relay installation for a micro-hydropower plant

2. Protection

Step-up Transformer

MCCB

Mg Ctt

Transmission line

WT

PF-LBS

GS

V

F

Wh W A

V

F2 X VT

440V/110V

AS

VS

VS

//√3440V

√3110V

√3110V

U >1

f > f <1 1

U > U <1 1

I > 2

・・・・・

・・・・

250kVA 440V 328A60Hz pf=0.8

2 X CT500/5A15VA





Distribution systemDistribution system


ContentsContents

1. Distribution method

2. Components

3. Route selection

4. Voltage drop estimation


1.1 Classification1.1 Classification 1. Distribution method

Connection

4

3

3

2

Nos. of line

1-phase 2-wire

1-phase 3-wire

System

3-phase 4-wire

3-phase 3-wire


2.1 Pole2.1 Pole 2. Components

Standard poles for overhead lines are classified as follows:Application

Applied to areas where access of heavy machines is difficult

Applied to areas where access of heavy machines is difficult

Generally appliedConcrete pole

Wooden pole(including Bamboo pole)

type

Steel pole

Concrete pole Wooden pole Steel pole


2.2 Pole length2.2 Pole length 2. Components

Pole length is to be determined considering following factors:

Necessary height of feeder conductors above the ground can besecured under the largest sag

Necessary clearance between feeder conductors and buildings, otherwires or trees can be secured*clearance under maximum sag should be examined

Recommended minimum pole setting depth is one sixth of pole length.

(Ex.) Pole setting depth = Pole length 9 (m) X 1/6 = 1.5 (m)

If soil condition is not stable, the root of pole should be reinforced by guy anchor firmly (refer to above picture).

Recommended pole length

7 m

9 m20kV

Low voltage

Voltage

Guy anchor


2.3 Installation of pole2.3 Installation of pole 2. Components

Span: Recommended span is 50 m Max 80 m for areas outside settlements, rice fields, and open spaces Max 50 m for areas within population settlement

Clearance of conductor:

6.0 m

6.0 m6.5 m

20kV Low voltage

4.0 m

4.0 m4.0 mRoad crossing

Along road

Conductor height above ground

Other place

0.8 m

0.2 m

1.0 m

0.8 mClearance between phases 20kV bare conductors

Vertical clearance between 20kV bare conductors

Vertical clearance between 20kV bare conductor and LV insulated conductor

Clearance between LV insulated conductors


2.4 Guy wire (1)2.4 Guy wire (1) 2. Components

Guy wires should be installed to balance a pole against the following loads.

(a) Vertical loadPole weight, cable weight, vertical load of wire tension load, etc.

(b) Longitudinal loadWind pressure to pole, imbalanced load from difference of span length

(c) Lateral loadWind pressure to cable, component of lateral load of wire tension, etc.

(a)

(c)

(b)

wind pressureWind pressure

Top view


2.4 Guy wire (2)2.4 Guy wire (2) 2. Components

End of distribution line

Distribution lines bend like an elbow-shaped. It is possible to omit guy wire if the angle is less than 5 degrees.

To reinforce straight distribution line against wind pressure

In undulated area, guy wire shall be installed, if necessary.

Tension

wind pressure

Anchor pole at the end of line

Guy wire

Guy wire

Guy wire

Guy wire


3. Basic concept3. Basic concept 3. Route selection

Locations of supporting structures should be selected at places where:

Easy to access and maintenance Soil condition is firm and stable No problem in land acquisition No adverse effect on buildings, trees, etc. Distribution route should be shortest If poles are set around steep slope or at the bottom of a cliff,

take into account the following, as illustrated:

Since landslide may take place, take a safer route to avoid standing a pole at the bottom of the cliff.

Land slide

Route change


3.1 Resistance of conductor3.1 Resistance of conductor 4. Voltage drop

Resistance of conductor R is proportional to the length Lc, and inversely proportional to the sectional area A.

R = ρ X (Lc X A) (Ω)

where,R: Resistance of conductor (Ω)ρ: Resistivity (A)Lc: Length of conductor (m)A: Sectional area of conductor (m2)

Sectional area A of conductor with diameter d is as follows:

R = (4 X ρ X Lc) / (π X D2 ) (Ω)

In voltage drop calculation, value of resistance listed in manufacture’s catalog is often referred.

2.75Alminium

1.72Cupper

Resistivity(X10-2 Ω/m･mm2)

Type ofconductor

(Reference)


3.2 Inductance of conductor3.2 Inductance of conductor 4. Voltage drop

Inductance of conductor L is calculated using the following formula.

L = 0.05 + 0.4605 log10 (D / r) (mH/km)

where,L: Inductance of conductor (mH/km)D: Conductor spacing (m)r: Radius of conductor (m)

Conductor spacing D is defined as follows:

1-phase 2-wire

Dab

D = Dab

a b

3-phase 3-wire

Dab

Dbc

Dca b

c

aD = √(Dab X Dbc X Dca)


3.3 Voltage drop3.3 Voltage drop 4. Voltage drop

Voltage drop by transmitting power through distribution lines is calculated by the following basic formula.

Vd = K X I X DL X (R cosθ+ X sinθ)

where,Vd: Voltage drop (V)I: Load current (A)K: Constant by distribution methodDL: Distribution line length (m)R: Resistance of conductor (Ω/m)X: Reactance of conductor (Ω/m) (= 2πf L)f: Frequency (Hz)L: Inductance (H/m)cosθ: Power factor of load currentθ: Power factor angle (deg)

11-phase 3-wire21-phase 2-wireK

√3√33-phase 3-wire

3-phase 4-wire

SystemK by distribution method

If voltage drop at the terminal point exceed 10% of the rated generator voltage, application of a step-up transformer at the transmission end should be reviewed.


3.4 Example of study(1)3.4 Example of study(1) 4. Voltage drop

Powerhouse

1

2

5

6

4

7

3

16

8

9

10

11

12

13

14

15200W X 5 HH

= 1,000W

200W X 10 HH = 2,000W

: Poleby formula0.935 mH/kmConductor inductance LX=2πf(L/1000)0.352 Ω/kmConductor reactance X

32 mm2SizeACSR-OEType

Transmission line

Manufacture’s catalog0.0072 mConductor diameter dManufacture’s catalog0.928 Ω/kmConductor resistance RManufacture’s catalog115 AAllowable current

0.3 mConductor spacing Dr=D/20.0036 mConductor radius r

0.652 A/HH0.870 A/HH1.087 A/HH100 kW/HH

0.600.80

60 Hz220 V

22-wire

1-phase

Rated gen voltageVoltage VKNos. of wireNos. of Phase

sinθ=√(1-cos2θ)sinθcosθPower factor pf

Load characteristicsFrequency F

I=(P/pf)/VUnit current IEstimated valuePower consumption per HH

I2=I sinθUnit reactive current I2I1=I cosθUnit active current I1


from to

16 15 50 55 10 10 8.696 6.522 0.0510 0.0194 1.140415 14 50 55 10 8.696 6.522 0.0510 0.0194 1.140414 13 50 55 10 8.696 6.522 0.0510 0.0194 1.140413 12 50 55 10 8.696 6.522 0.0510 0.0194 1.140412 11 50 55 10 8.696 6.522 0.0510 0.0194 1.140411 6 50 55 10 8.696 6.522 0.0510 0.0194 1.1404

300 330 10 6.842410 9 50 55 5 5 4.348 3.261 0.0510 0.0194 0.57029 8 50 55 5 4.348 3.261 0.0510 0.0194 0.57028 7 50 55 5 4.348 3.261 0.0510 0.0194 0.57027 6 50 55 5 4.348 3.261 0.0510 0.0194 0.5702

200 220 5 2.28086 5 50 55 15 15 13.043 9.783 0.0510 0.0194 1.71065 4 50 55 15 13.043 9.783 0.0510 0.0194 1.71064 3 50 55 15 13.043 9.783 0.0510 0.0194 1.71063 2 50 55 15 13.043 9.783 0.0510 0.0194 1.71062 1 50 55 15 13.043 9.783 0.0510 0.0194 1.7106

250 275 15 8.5530

Sectionvoltagedorp [v]

K(IaxR1+IbxX1)

III

SectionNo.

Effectivecablelength

(b) [m](a) x 1.1

SectionactivecurrentIa [A]C x I1

Pole No. Spanbetween

poles(a) [m]

Sectionreactance

X1 [A]Xx(b)/100

0

II

I

SectionreactivecurrentIb [A]C x I2

Sectionresistance

R1 [A]Rx(b)/100

0

Nos. ofHH

SectionCurrent

Constant:C

3.4 Example of study (2)3.4 Example of study (2) 4. Voltage drop


3.4 Example of study (3)3.4 Example of study (3) 4. Voltage drop

Powerhouse

1

2

5

6

4

7

3

16

8

9

10

11

12

13

14

15200W X 5 HH

= 1,000W

200W X 10 HH = 2,000W

: Pole

Section II

Section I

Section III

Vd at pole6Vd6=8.55V

Vd at pole 10Vd10= Vd6+2.28V

= 10.83V

Vd at pole 16Vd16=Vd6+6.84V

=15.39V

Voltage drops at the terminal point of the pole 10 & 16 are estimated within 10% of the rated generator voltage 220V

Application of a step-up transformer may not be necessary



Basic Design of ElectroBasic Design of Electro--Mechanical EquipmentMechanical Equipment


ContentsContents

1. Turbine Sizing

2. Generator Sizing

3. Governor

4. Belt Selection


Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment


Turbine TypeTurbine Type



Head and DischargeDesign head H: 12 m (Net head = Gross head – Head loss)

Design discharge Q: 0.1 m3/s

Theoretical Output

The theoretical output of the plant Po is calculated as follow:Po = 9.8 x Q x H

= 9.8 x 0.1 x 12 = 11 (kW) Turbine Type

The type of turbine shall be selected from “Turbine Selection Chart”, which has been prepared based on the supply records of the turbines.



Source: DOE-JICA, March 2006, MANUAL for MICRO-HYDROPOWER DEVELOPMENT, 6-7

Turbine Selection Chart

Design point*

According to the above-mentioned design head and discharge, the following types of turbine are applicable to this project site.1)Horizontal shaft francis turbine2)Cross flow turbine3)Vertical shaft propeller turbine4)Horizontal shaft propeller turbine



Theoretical OutputThe turbine output Pt for each selected turbine types is calculated as

follows:Horizontal shaft Francis turbinePt = Po x ηt

= 11 x 0.84 = 9 (kW)Cross flow turbinePt = Po x ηt

= 11 x 0.65 = 7 (kW)Vertical shaft propeller turbinePt = Po x ηt

= 11 x 0.82 = 9 (kW)Horizontal shaft propeller turbinePt = Po x ηt

= 11 x 0.82 = 9 (kW)where, Pt: Turbine Output (kW)

Po: Theoretical output (kW)ηt: Turbine efficiency.



Design point*



Limitation of Specific SpeedUpper limit of specific speed nslimit for each selected turbine types, which

is defined based on the supply records of the turbines, is calculated form the formula described in the MANUAL (page 6-8) as follows:

Horizontal shaft Francis turbineNs-limit ≦ (20000/(H+20))+30

= (20000/(12+20))+30= 655 m-kW

Cross flow turbinenslimit = 650 x H^-0.5

= 650 x 12^ -0.5 = 187 m-kW

Propeller turbine:Ns-max ≦ (20000/(H+20))+50

= (20000/(12+20))+50= 675 m-kW



Selection of Turbine TypeConsidering the following applicable range of specific speed ns for each

selected turbine types which is described in the MANUAL (page 6-9), cross flow turbine and horizontal shaft propeller turbine are applicable to the project site.

Francis turbine: 60 ≦ ns ≦300Francis turbine with ns of 655 m-kW is not applicable.

Cross flow turbine: 40 ≦ ns ≦200Cross flow turbine with ns of 187 m-kW is applicable.

Propeller turbine: 250 ≦ ns ≦1,000Vertical and horizontal shaft propeller turbine with ns of 675 m-kW is applicable.



Rated Rotational SpeedUpper limit of rotation speed Nlimit is calculated using upper limit of

specific speed nslimit as follows .

Ns = (N x P^1/2)/ H^5/4 ; N = (Ns x H^5/4 )/ P^1/2

Where, Ns; Specific speed (m-kw)N; Rotational speed of turbine (rpm)P; Output of turbine (kW) = 9.8 x Q x H x H; Effective head (m)Q; Discharge (m3/s) ; Maximum efficiency (%, but a decimal is used in calculations)

= 82 % for Pelton turbine = 84 % for Francis turbine = 77 % for Crossflow turbine* = 84 % for S-type tubular turbine

Note: * 40-50% should be applied for Crossflow type turbine manufactured locally at present stage because due to fabrication quality.


GeneratorGenerator



Rated Rotational Speed ( Turbine)

N = (Ns x H^5/4 )/ P^1/2N = (187 x 12^5/4 )/ 7^1/2

N = 1,578 rpm

Generator Speed based on Turbine Speed

Standard rotational speed of generator

500600750

1,0001,500

50Hz

600720900

1,2001,800

60Hz

2420181614

Nos. of poles

250300333375429

50Hz

40083601030012

45065144

60HzNos. of poles

The size and cost of high speed generator is smaller and cheaperthan low speed generator.



Two kinds of speed increaser adopted for coupling turbine and generator are as follows:

Gear box type:Turbine shaft and generator shaft is coupled withparallel shaft helical gears in one box with anti-friction bearing according to the ratio of speed between turbine and generator. The lifetime is long but the cost is relatively high. (Efficiency: 97 – 95% subject to the type)

Belt type: Turbine shaft and generator shaft is coupled with pulleys (flywheels) and belt according to the ratio of speed between turbine and generator. The cost is relatively low but lifetime is short.(Efficiency: 98 – 95% subject to the type of belt)

In case of micro hydro-power plant, V-belt or flat belt type coupling is adopted usually to save the cost because gear type transmitter is very expensive.



Output of Generator

The output of generator is shown in kVA and calculated with following formula:

Pg (kVA) = (9.8 x H x Q x ) / pf

Where;Pg; Required output (kVA)H; Net head (m)Q; Rated discharge (m3/s); Combined efficiency of turbine, transmitter &

generator (%) = turbine efficiency (t) x transmitter efficiency (m) x generator efficiency (g)

pf; Power factor ( % or decimal), the value is based on the type of load in the system. If inductive load, such as electric motor, low power factor lamps, is high in the system, the power factor is low i.e. the generator capacity should be larger according to above formula. However, 80% is usually applied for convenient purpose of selection.



Output of Generator

Pg (kVA) = (9.8 x H x Q x ) / pfPg (kVA) = (9.8 x H x Q x t x tr x g ) / pfPg (kVA) = (9.8 x 12 x 0.1 x .65 x 0.95 x 0.90 ) / 0.80Pg (kVA) = 8.17 kVA (say 10 kVA)

Rated Current of Generator


GovernorGovernor



Electronic Load Controllers as Governor

The capacity of dummy load is calculated as follows:

Pd (kW) = Pg (kVA) x pf (decimal) x SF = 10 x 0.8 x 1.2 = 9.6 kW

WherePd: Capacity of dummy load (Unity load: kW)Pg: Rated output of generator (KVA) pf: Rated power factor of generator (%, a decimal is used for

calculation)SF: Safety factor according to cooling method (1.2 – 1.4 times of

generator output in kW) in order to avoid over-heat of the heater

Note: Maximum output of turbine (kW) may be applied instead of “Pg (kVA) x pf (decimal)” because maximum generator output is limited by turbine output even if the generator with larger capacity is adopted.



Single Line Diagram




Power Power Transmission Transmission

DeviceDevice



Design Power

(Design power)= Pt x (Service factor) = 7 x 1.3 = 9.1 (kW)

Speed RatioSpeed ratio is calculated from rotation speed of turbine and generator as

follows:Speed ratio =Ns / NL

where,NL: Rotation speed of large pulley for turbine (mm)Ns: Rotation speed of small pulley for generator (mm)

Speed ratio = 1,800 rpm / 1,578 rpm = 1.14


Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment Selection of Belt Type

V-belt is adopted because of its higher transmission efficiency of driving power. Belt type is selected on the basis of the transferred power and rotation speed of small pulley as shown in Fig. 2-3. Therefore, section B V-belt is adopted in consideration of calculated design power of 9 kW and generator rotation speed of 1,800 min-1.



Diameter of pulleyMinimum diameter of the small pulley for generator, 150 mm, is automatically derived from section B line in Table 2-2. Then, diameter of other pulley for turbine is calculated as follows:

DL = DS x (Speed ratio) = 150 x 1.14 = 171 (mm)

where, DL: Diameter of large pulley for turbine (mm)DS: Diameter of small pulley for generator (mm)

Table 2-2 Section size and fabrication limit of standard V-belt

3056038.5 x 25.5 x 40E

3035531.5 x 19.0 x 40D

3022422.0 x 14.0 x 40C

3015016.5 x 11.0 x 40B

309512.5 x 9.0 x 40A

Maximum speed(m/sec)

Pulley minimum diameter

(mm)

Section sizeW(mm) x H(mm) x Angle

(o)Type

Source: Catalog of MITSUBOSHI Belt



Length of Belt



Length of Belt ( Belt length and Correction Factor Table)



Length of Belt ( Belt length and Correction Factor Table)


Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment Belt Speed

Power rating of belt


Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment Power rating of A SECTION belt


Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment Power rating of B SECTION belt


Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment Arc of Contact



Corrected Power rating of belt

Required Number of belts



Specification of Turbine

187Specific speed ns

0.65Turbine efficiencyηt [%]

1,578Rotation speed N [min-1]

7Turbine output Pt [kW]

0.1Design discharge Q [m3/s]

12Design head H [m]

Cross flowType

Turbine

SpecificationItem



Specification of Generator

Static excitation systemExcitation system

4Number of rotor poles

60Frequency [Hz]

1,800Rotation speed Ng [min-1]

1.00Rated power factor pfg

45.5Rated current Ig [A]

220Rated voltage Vg [V]

10.0Rated capacity Pgu [kVA]

Horizontal shaft single-phase synchronous

Type

SpecificationItem



Specification of Governor

Water or AirCooling method

12 (6 x 2)Total Capacity [kW]

2 parallelConnection

Dummy load

FrequencySpeed detection method

Single-phaseType

ELC

Dummy loadType

Governor

SpecificationItem



Specification of Power Transmission Device

1,005Central distance between pulleys [mm]

1.14Speed ratio

171Pitch diameter of large pulley [mm]

150Pitch diameter of small pulley [mm]

Pulley

2Nos. of belt

2,515 (99)Length [mm] (nominal number)

16.5 x 11.0 x 40Section size W[mm] x H[mm] x Angle[o]

BSection type

V-beltType

Belt

SpecificationItem



EXERCISE

Design head Hnet: 14 m Design discharge Q: 0.22 m³/s



Turbine and Governor

TBACooling method

21 (10.5 x 2)Total Capacity [kW]

2 parallelConnection

Dummy load

FrequencySpeed detection method

Single-phaseTypeELC

Dummy loadType

Governor

173Specific speed ns

0.65Turbine efficiencyηt [%]

1,060Rotation speed Nt [min-1]

19.5Turbine output Pt [kW]

0.22Design discharge Q [m3/s]

14Design head H [m]

Cross flowType

Turbine

SpecificationItem



Generator Specification

Brushless excitation systemExcitation system

B classTemperature rise

F classInsulation level

4Number of rotor poles

60Frequency [Hz]

1,800Rotation speed Ng [min-1]

0.80Rated power factor pfg

71Rated current Ig [A]

240Rated voltage Vg [V]

21.3Rated capacity Pgu [kVA]

Horizontal shaft single-phase synchronousType

Generator SpecificationItem



Belt and Pulley

1,015Central distance between pulleys [mm]

1.698Speed ratio

254Pitch diameter of large pulley [mm]

150Pitch diameter of small pulley [mm]

Pulley

5Noumber of belts

2,667 (105)Length [mm] (nominal number)

16.5 x 11.0 x 40Section size W[mm] x H[mm] x Angle[o]

BSection type

V-beltType

Belt

SpecificationItem

１

Department of Energy

Energy Complex Merritt Road, Fort Bonifacio, Taguig City, Metro Manila

TEL: 479-2900 FAX: 840-1817

１

Department of Energy

Energy Complex Merritt Road, Fort Bonifacio, Taguig City, Metro Manila

TEL: 479-2900 FAX: 840-1817

Documents

Training Manual for Micro-hydropower Technology June 2009