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
What is Hydropower?Energy of Falling Stone:
Ouch!
Ouch!
1-6
7
What is Hydropower?Energy of Falling Stone:
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
Training on Micro Hydropower Development
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
Training on Micro Hydropower Development
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
If you want 40 m of Head
Look for 2 interval of Contour Line is less than 2cm on the Map
1000 m
G
G
If you want 60 m of Head
Look for 3 interval of Contour Line is less than 3cm on the Map
1500 m
G
G
1-27 28
1. Short Distance and High Head ; How to find out ?
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
1. Short Distance and High Head ; How to find out ?
Assumed Head : 40 m
1000m2 cm
1000m1000m1000m 1000m1000m
N.GOK
OK
N.G
N.G
RR1-29 30
1. Short Distance and High Head ; How to find out ?
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
Good Potential Site (Technically)
2. Bigger Catchment Area is better if the L/H is evenAssumed Head : 40 m
10621045
1-31 32
Good Potential Site (Technically)
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
Good Potential Site (Technically)
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
Good Potential Site (Technically)
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
Training on Micro Hydropower Development
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
Settling Basin Headrace Headtank
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
Calculation of Rating Curve
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
)
Calculation of Rating Curve
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
Training on Micro Hydropower Development
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
2-1 Intake Weir(3) Design of the Weir Height
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
2-1 Intake Weir(3) Design of the Weir Height
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
2-1 Intake Weir(3) Design of the Weir Height
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
2-2 Intake and Settling Basin
< 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
2-2 Intake and Settling Basin
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)
: Values which are decided on other factors
: Common values for design (refer to "Manual")
: Values depend on natural condition
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
2-2 Intake and Settling Basin
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
Lc Lw Ls
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
2-2 Intake and Settling Basin
(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))
: Values which are decided on other factors
: Common values for design (refer to "Manual")
: Values depend on natural condition
: 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
Settling Basin Headrace Headtank
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
2-7 Calculation of Head Loss
(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
2-7 Calculation of Head Loss
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
2-7 Calculation of Head Loss
(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
Optimum Installed Capacity
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
Installed Capacity (kW)
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
Optimum Installed Capacity
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
Installed Capacity (kW)
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%
2- Demand Side Condition
Fluctuation of the Generated Power
920kW
780kW
610kW
520kW
390kW
270kW
160kW
6-16
17
2- Demand Side Condition
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- Demand Side Condition
(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
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TurbineTurbine
3DOEDOE--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
ContentsContents
1. Basics of hydraulics
2. Turbine types
3. Characteristics of turbine
4. Basic design of turbine
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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
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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
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1.2 1.2 Bernoulli's theoremBernoulli's theorem
Open
Reference levelPotential head: z (m)
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)
Pressure head:p / w (m)
Velocity head:v2 / 2g (m)
1. Hydraulics
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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
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1.2 1.2 Bernoulli's theoremBernoulli's theorem
Open
Reference levelPotential head: z (m)
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)
Pressure head:p / w (m)
Head loss: Hloss (m)
1. Hydraulics
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1.2 1.2 Bernoulli's theoremBernoulli's theorem
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
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1.2 1.2 Bernoulli's theoremBernoulli's theorem 1. Hydraulics
1.2.5 Calculation of net head on site (2)1.2.5 Calculation of net head on site (2)
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)
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1.2 1.2 Bernoulli's theoremBernoulli's theorem 1. Hydraulics
1.2.5 Calculation of net head on site (3)1.2.5 Calculation of net head on site (3)
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) =
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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
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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
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2.Turbine types2.1 Types of turbine2.1 Types of turbine
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
Applicable range Output: 50 – 1,000 kW Discharge: 0.1 – 10 m3/s Head: 5 – 100 m
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2.Turbine types2.1 Types of turbine2.1 Types of turbine
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
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2.Turbine types2.1 Types of turbine2.1 Types of turbine
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
Applicable range Output: 50 – 5,000 kW Discharge: 1.5 – 40 m3/s Head: 3 – 18 m
Generator
Propeller RunnerGuide Vane
(Wicket Gate)
Timing Belt
Draft Tube
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2.Turbine types2.2 Turbine selection chart2.2 Turbine selection chart
Discharge (m3/s)
Net
hea
d (m
)
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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
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3.Characteristics3.1 Specific speed3.1 Specific speed
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
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3.Characteristics3.1 Specific speed3.1 Specific speed
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
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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
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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
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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
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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
4.Basic design4.14.1 Flow chart of basic designFlow chart of basic design
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
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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
4.Basic design4.14.1 Flow chart of basic designFlow chart of basic design
ExampleTurbine type: H-shaft FrancisNet head H: 45 mDischarge Q: 2.5 m3/sFrequency F: 50 Hz
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
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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
4.Basic design4.14.1 Flow chart of basic designFlow chart of basic design
ExampleTurbine type: H-shaft FrancisNet head H: 45 mDischarge Q: 2.5 m3/sFrequency F: 50 Hz
Please calculate Ntmax using estimated Pt.Applicable max. rotational speed Ntmax
= = = min-1
Calculated Nsmax in Clause 4.1.2
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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
4.Basic design4.14.1 Flow chart of basic designFlow chart of basic design
ExampleTurbine type: H-shaft FrancisNet head H: 45 mDischarge Q: 2.5 m3/sFrequency F: 50 Hz
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
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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
4.Basic design4.14.1 Flow chart of basic designFlow chart of basic design
ExampleTurbine type: H-shaft FrancisNet head H: 45 mDischarge Q: 2.5 m3/sFrequency F: 50 Hz
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%
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Electrical and Mechanical EquipmentElectrical and Mechanical Equipment
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GeneratorGenerator
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ContentsContents
1. Basics of generator
2. Classification of generator
3. Basic design of generator
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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.
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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
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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
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1.2 Principle of AC generator1.2 Principle of AC generator 1. Basics
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
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1.2 Principle of AC generator1.2 Principle of AC generator 1. Basics
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
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1.3 Types of Generator1.3 Types of Generator 1. Basics
Stationary Field, Rotating Armature AC Generator :
Rotating Field, Stationary Armature AC Generator :
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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
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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
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1.5 Excitation1.5 Excitation 1. Basics
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.
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1.5 Excitation1.5 Excitation 1. Basics
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
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1.5 Excitation1.5 Excitation 1. Basics
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
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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
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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.
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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
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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.
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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
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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
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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
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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
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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 =
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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.
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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)
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Control systemControl system
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ContentsContents
1. Basics of automatic control
2. Frequency control
3. Voltage control
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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.
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1.2 Feedback control (1)1.2 Feedback control (1) 1. Basics
Detection
Comparison
Judgment
Operation
LoopLoop
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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
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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
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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
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Single line DiagramSingle line Diagram
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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
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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
step responsestep response
Out
put
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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
step responsestep response
Response Response improvementimprovement
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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
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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
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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
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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
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2.3 Methods of frequency control (2)2.3 Methods of frequency control (2) 2. Freq. Control
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
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2.3 Methods of frequency control (3)2.3 Methods of frequency control (3) 2. Freq. Control
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
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2.3 Methods of frequency control (4)2.3 Methods of frequency control (4) 2. Freq. 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.
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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
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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
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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
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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
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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
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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
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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
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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
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Electrical equipmentElectrical equipmentandand
Protection systemProtection system
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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
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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
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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
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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)
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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 <
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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 <
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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
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Distribution systemDistribution system
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ContentsContents
1. Distribution method
2. Components
3. Route selection
4. Voltage drop estimation
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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
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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
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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
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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
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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
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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
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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
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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)
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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)
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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.
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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
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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
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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
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
Basic Design of ElectroBasic Design of Electro--Mechanical EquipmentMechanical Equipment
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ContentsContents
1. Turbine Sizing
2. Generator Sizing
3. Governor
4. Belt Selection
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
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Turbine TypeTurbine Type
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
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.
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
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
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
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.
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
Design point*
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
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
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
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.
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
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.
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GeneratorGenerator
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
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.
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
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.
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
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.
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
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
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GovernorGovernor
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
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.
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
Single Line Diagram
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
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Power Power Transmission Transmission
DeviceDevice
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
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
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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.
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
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
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
Length of Belt
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
Length of Belt ( Belt length and Correction Factor Table)
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
Length of Belt ( Belt length and Correction Factor Table)
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment Belt Speed
Power rating of belt
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment Power rating of A SECTION belt
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment Power rating of B SECTION belt
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment Arc of Contact
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
Corrected Power rating of belt
Required Number of belts
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
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
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
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
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
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
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
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
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
EXERCISE
Design head Hnet: 14 m Design discharge Q: 0.22 m³/s
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
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
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
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
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Basic Design of ElectroBasic Design of Electro--mechanical Equipmentmechanical Equipment
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