Introduction about hydropower, types of intakes, classification of hydropower plants, estimation of water potential, and planning aspects of hydropower.

2. PRINCIPAL COMPONENTS OF HYDROELECRIC SCHEME……

3. How Hydropower electric is Generated?
Figure : Diagram showing how hydroelectric produced from water

4. Source of Energy
 Energy sources can be categorized as renewable and non-renewable.
 Renewable energy:- energy generated from natural resources such as sunlight, wind,
tides, and geothermal heat, which are renewable (naturally replenish).
 Non-renewable energy:- energy, taken from "finite resources that will eventually
diminish ". This includes:
 Fossil fuels: Various types of coal, Petroleum and Natural gases
 Nuclear energy (fuel for fission of Uranium ore).
Hydropower potential in Ethiopia
 According to (Hydro news Africa ,2020) ; Ethiopia's hydropower potential is
estimated at up to 45,000 MW and is the second highest in Africa.
 The current electricity installed capacity of 4,284 MW is 97 per cent
renewable of which effective hydropower installed capacity is 3,810 MW.
Furthermore, 8,864 MW of hydropower development is under construction.

5. List of Hydropower in Ethiopia

7. Advantages and Disadvantages of Hydropower…
Advantages of Hydropower
Renewable source of energy;
Economical source of power;
Non-polluting and hence environment friendly;
Reliable energy source with approximately 90 percent availability;
Low generation cost compared with other energy sources;
Helps in management and regulation of water resources i.e. storage based plants are
often of multipurpose (flood control, irrigation, water supply, navigation, fishing, tourism
etc.),
Provide employment opportunities being labour intensive;
Lead to development of remote areas
Afford a degree of independence from costly and unreliable supply of imported fuel;
Technically more reliable than many thermal plants;
Low operation and maintenance cost,
Better service operation flexibility, the operation of the plant can be matched with load

8. Disadvantages of Hydropower
 Susceptible to nature such as drought and High initial cost
 Longer construction period;
 Loss of land due to submergence in the reservoir and displacement of large
population from reservoir area,
 Non-availability of suitable sites for the construction of dam,
 Environmental aspect; reservoir vs. river ecology,
 High cost of transmission system for remote sites,
 Long term flow data is essential for proper assessment,
 Breakdown in hydropower equipment may result not only in proportionate
reduction in power generation but also, particularly incase of run -of-river
plants, in letting the precious water run waste,
Advantages and Disadvantages of Hydropower…

9. CLASSIFICATION OF HYDROPOWER PLANTS
 LARGE: >100 MW MICRO: 5 – 100 KW
 MEDIUM: 25 – 100 MW MINI: 100 KW - 1MW
 SMALL: 1-25 MW
Classification according to capacity of hydropower plant

10. Classification According to Head
I. Low head :
Low head hydro power applications use river current or tidal flows of 30 meters or
less to produce energy .These plants do not need to dam or retain water to create
hydraulic head, the head is only a few meters
II. MEDIUM HEAD: A power plant operating under heads from 30m to 300m.
III.HIGH HEAD:
A power station operating under heads above about 300m.

11. Classification according to hydrological relation
I. Single stage: When the run off from a
single hydropower plant is diverted back
into river or for any other purpose other
than power generation, the setup is known
as Single Stage
II.Cascade system- when two or more
hydropower plants are used in series such
that the runoff discharge of one hydro
power plant is used as the is a intake
discharge of the second hydro power plant
such a system is known as Cascade
hydropower plant.
(a)
(b)
Figure-(a) single stage hydropower development scheme
(b) cascade or multistage hydropower system

12. Classification according to purpose
Single purpose: When the whole soul purpose of a project is to produce electricity then such a
project is known as a Single Purpose Hydro Power Project.
Multipurpose : When the water used in hydropower project is to be used for other purposes like
irrigation, flood control or fisheries then such a project is known as Multi Purpose Hydro Power
Project.
Classification according to facility type
I.Run-of-river type:
 There is no significant storage;
 A weir or barrage is built across a river & the low head created is used to generate power.
 The power house is within the main course of the river;
 It is Preferred in perennial rivers with moderate to high discharge, low sediment and stable
reach of a river.
II.Storage (Reservoir) or Valley dam type:
 Hydropower plants with storage are supplied with water from large storage reservoir that
have been developed by constructing dams across rivers.
 Assured flow for hydro power generation is more certain for the storage
schemes than the run-of-river schemes.

13. Classification according to facility type…
III.PUMPED STORAGE TYPE
 Pumped storage type hydropower plants are those which utilize the flow of water from a
reservoir at higher potential to one at lower potential.
 During off-peak hours, the reversible units are supplied with the excess electricity
available in the power grid which then pumps part of the water of the tail-water pond back
into the head-water pond.

14. Classification According to Transmission System
A) Isolated System:
Whenever a hydropower plant is set up in a remote area in order to meet the local
demands then such a hydropower plant is known as Isolated System.
B) CONNECTED TO GRID
Whenever the hydropower plant is set up to meet the demands of areas which are at a
fair distance from the plant, then the transmission of power takes through the grid
system. Such a setup is referred to as Connected to grid.
 working principle of hydroelectric power plant is depends on
 Height of water.
 Volume of water flowing per unit time.
 Efficiency of turbine

15. Estimation of Water Power Potential
Work done = Force * displacement
 The potential energy of water is:
 Work done =m * g * H, where H is the total head the water will fall.
= ρ * V * g * H, where: ρ *g=ү
= ү*V*H; Q=V/t This implies That V= Q× t, where V Is volume
W= ү *Q*t*H→ And Power =W/t =( ү *Q*t*H)/t= ү *Q*H in KW
Hence, the theoretical electrical energy produced from a site is:
 Power (Watt) = ү ×Q × H (where Ƴ= unit weight of water, 9810N/m3)
If factor of efficiency (η) is introduce,
 Then , Power = ү* Q * H * η = 9.81 * Q * H * η (KW)
 Power is the rate at which energy is transferred. The watt is the most commonly
used unit of measure for power.

16. Key terms in hydropower
 Gross head = it is the difference in water level between the water level in the fore-bay (or
head water in the reservoir) and tail water level.
 Net Head = gross head - the all the hydraulic losses. It is the available head to do work on
the turbine
 Design Head = the net head at which the turbine reach its peak efficiency.
 Rated Head = it is the net head at which the turbine operating at full gate opening.
 Firm Power = is the net amount of power which is continuously available from a plant
without any break down on guaranteed basis.
 This power should be available under most adverse hydraulic conditions.
 The consumers shall always be sure of getting this power.
 Secondary Power = the excess power available over firm power during the off peak period.
 It is available as a result of seasonal excess of water.
 The alternative to generation is letting the water runoff (spilling).
 There is no guarantee over secondary power.

17. Key terms in hydropower…….
 Installed Capacity: The total of the capacities shown on the nameplates of the
generating units in a hydropower plant.
 Peak load: It is the maximum load in a stated period of time.
 Base load : It is the minimum load in a stated period of time.
 Load : It is the amount of electric power delivered at a given point.
 Average load- is the area under the curve divided by time.
 Daily Load curve : Load or demand for electric power varies from hour to hour,
from day to day, and from season to season in response to the needs and living
patterns of the power users.

18.  Load factor = it is the ratio of the average load over a certain period of time to the
peak load during the same period.
 Depending on the period chosen, there are different load factors as daily, monthly
or annual.
Load factor, LF = average load over a certain period
Peak load during that period
 The maximum load determines the capacity of the units while the load factor gives
an idea of degree of utilization.
 Example, an annual LF 0.6 indicate that the machines are producing 60% of their
yearly rated capacity (max. production capacity).
 Capacity factor (plant use factor, plant factor) = defined as the ratio of average output
of the plant for a given period of time to the plant capacity.
Capacity factor, CF = average load (over a given period of time)
Plant capacity during that period
.
Key terms in hydropower

19. Key terms in hydropower…….
 The capacity factor and load factor would become the same if the peak load is equal to the
plant capacity.
 Utilization factor : Throughout the day or any given time period, a hydroelectric plant power
production goes on varying, depending upon the demand in the power grid and the power
necessary to be produced to balance it.
Utilization factor = Maximum power production over certain period
Installed capacity
 Capacity:- It is the maximum amount of power that a generating plant can deliver, expressed in
kilowatts. Example I: For a plant with capacity of 10,000KW were to produce 40,000KWh when
operating for 10 hrs. with a peak load equal to 8,000KW. Determine the load, capacity and
utilization factors.
Load factor, LF = average load over a certain period = 40,000 KWh/10hr = 50%
Peak load during that period 8,000KW

20. Example-II
 A generating station has the following daily load. The capacity of the
plant is 12MW.
Time (hrs) Power (KW) Time (hrs) Power (KW)
0-6 4500 14-18 8000
6-8 3500 18-20 2500
8-12 7500 20-24 5000
12-14 2000
Compute:
1. Sketch the load curve
2. Determine the load factor, capacity factor and Utilization factor

21. Solution for the problem I
Time
(hrs)
Power
(KW)
Time
(hr) Energy, KWh
0-6 4500 6 27000
6--8 3500 2 7000
8-12 7500 4 30000
12-14 2000 2 4000
14-18 8000 4 32000
18-20 2500 2 5000
20-24 5000 4 20000
Total 24 125,000
Figure: Daily load curve

22.  Total generated energy in 24hr = 125,000 KWh
 Average load = 125,000/ 24 = 5,208 KW
 Load factor =Average/ Peak = 5,208/ 8,000 = 0.651 = 65.1 %
 Capacity factor = Average load/ installed capacity = 5208/
12000 = 0.434= 43.4 %
 Utilization factor = Maximum production / Installed
capacity =8000/12000=0.67 = 67 %
Solution
Answer
Answer

23. Evaluation of site potential
 The primary purpose of a preliminary investigation in hydropower development is to
determine how much power is available at the site and how often it is available.
 The power output of hydroelectric power plant is given by:
 Theoretical potential :theoretical potential is the sum of the potential of all natural flows
from the largest rivers to the smallest stream regardless of the inevitable losses and
unfeasible sites.
 Technical potential: - from technical point of view, extremely low heads (less than
around 0.5m), head losses in water ways, efficiency losses in the hydraulic and electrical
machines, are considered as infeasible.
 Economic potential: - is only that part of the potential of more favorable sites which can
be regarded as economical compared to alternative sources of power like oil and coal.
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24. Planning aspects of Hydropower
Governing factors to be considered for planning of the hydropower are:
1. Topography of the area: - includes the head and general alignment of the area.
 The location of the dam/weir/barrage etc
 The reservoir capacity, submerged area, height and length of dam, location of spillway
 The location of powerhouse and other infrastructures (power canals, tunnels, surge tanks)
2. Geological Investigation : For hydropower projects, engineers should closely cooperate
with the geologists during the course of investigation and planning.
 The following geological factors should be considered:
 The bearing capacity, deformation of foundation
 Seepage problems
 Abutment and slope stability
 Change in hydro-geological conditions due to the construction of structures
 Earthquakes condition and Investigation of construction materials

25. 3. Hydrology
 Records of stream flow discharges (min. of 30 years for large hydro projects)
 Flood records that gives an idea of max. flood to be considered for the design
 Frequency of the floods usually estimated from the flood records and it is
1:10,000 for large dams and 1:100 to 1:1,000 for barrages and weirs.
 Low discharge to decide the firm power capacity of the hydro plant
4. Hydrometeorology: - To generate a stream flow series (using rainfall-runoff
models) for un-gauged catchments and to estimate the evaporation from the
reservoirs. Data's include:
 The rain gauge stations in the catchments, temperature, humidity,
sunshine hours and wind speed.
Planning aspects of Hydropower….

26. Planning aspects of Hydropower….
5.Material Survey: For construction of various components construction
materials such as cement, steel, sand, aggregates and selected earth fill
materials are required.
6. Communications: Availability of communications such as roads, railways,
telecommunications (telephone or wireless) and electric.
7. Environmental Factors:- involves the following factors
 Submergence of private or government or forest lands
 Submergence of private properties
 Ecological effects
 Submergence of National parks and monuments
8. Load Assessment: Forecasting of power and estimation of the present load
status are useful to identify the capacity of the hydro plant to be established.

27. Hydrological data analysis for Hydropower development
 Before hydropower plant is planned the hydrological data should be analyzed.
 The accuracy of a hydrologic analysis depends greatly on the record of stream
flow data for the watershed.
 In many watersheds, the stream flow data are limited or may not exist.
So, when considering hydropower development in such a watershed, discharge
measurements should be undertaken.
Case I: - No record or flow data
Case 11: - Quality of data
Case 111: - Cyclicity (the quality of recurring at regular intervals) and
trends

28. Hydrograph and flow duration curve
 A hydrograph indicates the variation of discharge or flow with time.
 It is plotted with flows as ordinates and time intervals as abscissas.
 A flow duration curve shows the relation between flow and time during which the discharges
are available.
 It is drawn when there is a continuous discharge data (daily, weekly or yearly)
 The flow duration curve can be plotted from a hydrograph.
 It is a curve drawn by using discharge as ordinate and % of existence time as abscissa.
 It is a used to know the time variability of discharge data.
 FDC shows the percentage of time in which a certain flow is equaled or exceeded for a period
of record.
Procedure:
 Obtain stream flow data .
 Rank them with the lowest rank (1) given to the highest flow record;
 Obtain the percentage of time the flow is equaled or exceeded by dividing the
rank given to the total number of flow records

29. Reservoir Capacity determination
 The capacity required for a reservoir depends up on
the inflow available and the demand.
 If inflow > the demand, there is no storage
required.
 If the inflow < the demand , storage is required.
 The required capacity for a reservoir can be
determined by: A. Graphical method (mass curve)
B. Analytical method
A. Mass curve
 The mass curve is a plot of cumulative flow
against time throughout the record time.
 It is used to estimate storage requirement that is
usable for power production.

30. Reservoir Capacity determination…
 Steps for computing reservoir capacity:
 Prepare mass inflow curve from the flow hydrograph;
 Prepare the mass demand curve corresponding to the given
rate of demand
 Draw lines AB, FG, etc such that, parallel to the mass demand
curve and tangential to the crests, A, F, etc of mass curve.
Points A, F etc indicate the beginning of the dry periods.
 determine the vertical intercepts DC, HJ, etc b/n the tangential
lines and the mass inflow curve. These intercepts indicate the
volumes by which the inflow volumes fall short of demand.
 determine the largest vertical intercept. It represents the
storage capacity required.
 Note: The capacity obtained is the net storage capacity which must
be available to meet the demand. The gross capacity should
include the losses as evaporation, seepage, etc.

31. Reservoir Capacity determination…
B.Analytical method
Steps to calculate the reservoir capacity using analytical method:
 Adjust inflow from the river (stream flow and rainfall over the reservoir);
 Adjust the demand (total out flow from the dam), as evaporation loss, water
demand for power production, environmental losses and others;
 Compute the storage capacity for each months:
Storage required = Adjusted inflow - Adjusted demand
Note: The storage would be required only in those months in which the demand
is greater than the adjusted inflow.
 Determine the total storage capacity of the reservoir adding the storage
required found above.

32. Load prediction and Demand Assessment
 Load prediction
 For installation of a new power project or for expansion of the existing power plant, it is
necessary to estimate the total amount of load that would be required to meet the various
purposes.
 The usual practice followed in hydropower planning is that the full potential of the project is
developed in stages.
 The first stage of development envisages the power production corresponding to immediate
demand while the remaining potential is developed in second or third stage of the project.
 The prediction of load is done in either of the followings:
 Short term: - cover a period of 4-5 yrs. It is done for areas of deficient or surplus power for
operation planning.
 Medium term: - cover a period of 8-10 yrs. It is done on the basis of expansion programme
of power generation of transmission facilities.
 Long term: - forecasting covers a period of 20yrs. It helps in the formulation of the
country's prospective plan for power generation.

33. Load prediction and Demand Assessment
 Load Assessment
 Load assessment refers to the forecasting of power in the load center and estimation of the
present load status is useful to identify the capacity of the hydro plant to be established.
 Power Demand Curve
 It is defined as the total load, which consumers choose, at any instant of time, to connect to
the supplying power system.
 Highest instantaneous value of demand is the peak load or peak demand.
 Base load is the total load continuously exists where as the average load is the area under the
curve divided by the time

34. Power Duration Curve
 If the available head and efficiency of the power plant are known, the flow
duration curve may be converted in to the power duration curve by charging the
ordinate to the available power (i.e. ηρgQH).
 The power which is available 95-97% of the time is usually considered to be the
primary or firm power and the area under which gives the total amount of
primary power.
 The secondary (surplus) power is the area under the power duration curve
between the firm power line and the total installed capacity.
 The area under the power duration curve will then be the average annual
energy production.
 If the demand is uniform, the demand curve is a horizontal line, and the mass
demand curve is a straight line having a slope equal to the demand rate.

35. Example:III
 The average monthly flows of a stream in a dry year are as follows:
Compute:
 It is intended to design a hydroelectric power plant across the streams,
using the above data, net head at the plant site =20m, efficiency of the
turbine = 90%.
 Plot the flow and power duration curves and calculate the firm and
secondary power available from this source if the maximum usable
flow is limited to 150m3/s.
 It is intended to develop power at a firm rate of 15MW, either by
providing storage or by providing a standby diesel plant with no
storage, determine the minimum capacity of the reservoir and of the
diesel unit.
Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
Q(m3/s) 117 150 203 117 80 118 82 79 58 45 57 152

36. Solution: Flow duration curve & power duration curve
Flow
(m3/s)
Cumulative
Frequency
(months)
Frequency equaled
or exceeded (%)
Power (MW)
(=20*0.9*9.81/100
0)= 0.176
[1] [2] [3]=([2]/[12])*100 0.176*[1]
203 1 8.33 35.728
152 2 16.67 26.752
150 3 25.00 26.4
118 4 33.33 20.768
117 5 41.67 20.592
117 5 41.67 20.592
82 7 58.33 14.432
80 8 66.67 14.08
79 9 75.00 13.904
58 10 83.33 10.208
57 11 91.67 10.032
45 12 100.00 7.95

37. Solution…
A. The flow duration curve (flow Vs frequency equaled or exceeded) will be plotted.
 The same plot can be used as a power duration curve by multiplying the ordinates by a
factor of η*p*g*H/10^6 =(0.9*9,810*20/10^6) = 0.176 to obtain the power in MW with
η=90% and H=20m, (i.e. P=0.176*Q)
 The firm power available (equal to the area of the power duration curve) under the
45m3/s line is 7.95MW (=0.9*9.81*1000*20*45/10^6).
 The secondary power (equal to the area under the power duration curve between the
150 and 45m3/s lines) is 10MW.
B. The power to be supplemented by storage or standby unit to obtain a firm power of 15MW is
the area "abc” = 17.76MWmonth.
 Therefore the storage required is: (17.76 * 10^6 *30*24*60)/ (1000 *9.81*20*0.90) = 2.6
*10^8m3.
 The firm power available is 7.95MW. Therefore the capacity of the standby unit is 15-
7.95=7.05MW.

38. Solution: Flow duration curve & power duration curve