Download - energy conversion lecture
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Outline
Introduction to Hydropower
Definition of HydropowerHistory of Hydropower
World Hydropower Resources
Hydropower basics
Factors Determining Potential of Hydropower
Concept of Plant factor
Classification of Hydroelectric Power PlantsAdvantages/Disadvantages of Hydropower
Hydropower Potential in Ethiopia
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Introduction to Hydropower
Like most other renewable energies, water power isindirect forms of solar energy
Hydropower is already one of the major contributors
to worlds power supply from renewable energy
This chapter explains the main principle of electricitygeneration from hydropower
The general overview of large-scale and small-scale
hydropower is given in this chapter.
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Definition of Hydropower
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Hydropower is a power that is derived from theenergy of moving or stored water, that can be
harnessed for useful purposes
Hydropower plants convert the kinetic energy
contained in falling water or the potential energy
contained in stored water into electricity Water is going through a turbine which converts the
water's energy into mechanical power. The rotation
of the water turbines is transferred to a generator
which produces electricity
Turbine CouplingElectrical
Generator
Mechanical
energy
EE
Torque & Speed V,I
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History of Hydropower
It has been thousands years that human harnessingwater to perform work
For many centuries, hydropower had been used to
produced mechanical power to perform a range of
activities, including grain milling, textile processing
and other light industrial operations.
A great part of the industrial revolution in the 18th
century was fueled by access to hydropower
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World Hydropower Resources
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The leading countries in the hydroelectric power generation are
Canada, Brazil and the United States
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Contd
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The largest hydroelectriccomplex on the world is
three gorges dam in china
with installed capacity of
22000 MW
Some examples of worldslargest Hydroelectric
plants which are over
4000 MW capacity are
given on the tableThe largest hydropower
plant in Ethiopia isrenaissance dam withgeneration capacity of6000 MW
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Hydropower Potential of Ethiopia
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Contd
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Hydroelectric power is very important
for national development, energy
security, food production, and
reducing deforestation The demand is growing day by day,
large industries are being built all
over the country
Regional distribution of hydropower
potential (light green areas with a
water surplus of >300mm/year)
For instance there are at least 4 or 5 cement factories being built
and many more are on the pipe line. There is also significantexpansion of mining, metal, sugar and other industries that need
big power supply. Urbanization and small industries are also
growing fast. All these will increase the demand and the gap
between the demand and supply cannot be easily bridged
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Contd
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Hydropower basics
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Contd
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The amount of energy available in a stored water is the
product of the weight times the height that the water falls
E = W = mgh, m = v
E = vgh [Joules]
Where: m = mass of water
= density of water = 1000 kg/m3
v = volume of water m3h = height of water fall (m)
g = acceleration due to gravity = 9.806 m/s2
Then the out power
P = dE/dt = d(vgh)/dt= gh(dv)/dt
Pth = ghQ wattsWhere: Q= dv/dt water flow rate m3/s
heoretical and Actual (Approximate) Power
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Contd
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The power at the out put of the turbine is
P = g H Q t Watts = water density = 1000 kg/m3
g = gravitational const. = 9.81 m/s2
H = head (m)
Q = water flow (m3/s)
t = turbine efficiency
There fore The power at the out put of the
generator is
Pactual = g H Q t g WattsWhere:
g = generator efficiency
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Contd
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As a rough guide for hydroelectric installations(i.e. with t= 0.85, and g = 0.96), the following
relationship can be used:
PGen = 8Qh kw With different values for turbine and generator
efficiencies the following approximations can also
be used
PGen = 8.2Qh kw or
PGen = 8.5Qh kw
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Converted Energy
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We know the available stored energy in the water is
E = mgh = vgh = Kgh (Joule)
If the useful volume of water K m3 is known (e.g. the capacity of
a reservoir, annual discharge, etc.) is known, the converted
energy is
E= 1000 K H tg *1000 Ws
102E = 1000 K H tg *1000 Wh
102*3600
E = 1000 K H tg KWh
102*3600E = Using the approximation 1000tg/102 8, then
E 0.0022KH KWh
Hence electrical energy generated for every m3 of water
used is
E / K 0 . 0022H KWh/m3
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Example 1
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a. The net useful power in a hydroelectric power plant (HEPP) can be
computed by setting
P (kW) = 9.81QHd
if the approximate power calculations are to be set as
P(kW) = 8.5 QHP(kW) = 8.2 QH andP(kW) = 8.0 QH
Calculate values of efficiency for each approximation
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Solution for example 1
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a) i. Net power = approximate power
9.81QHd = 8.5 QHd = 8.5/9.81 = 0.866 = 8.2/9.81 = 0.836
= 8.0/9.81 = 0.8155
A i t I
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Assignment I
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1. For each of the following water reservoirs of HEPPs with
estimated storage capacities K in cubic meters (m3), and gross
heads H in meters(m), calculate the gross available energy inGigawatt hours (GWh) or kilowatt-watt hours(kWh)
(a) K = 1680 x 106 m3 and H = 42 m
(b) K = 2.3 x 106 m3 and H = 59 m
(c) K = 625 x 106
m3
and H = 550m2. Continuously 12 hours per day In a 770kw HEEP ,180m3 of
water passes through the turbine each minute.
a. Assuming complete conversion of waters initial gravitational
potential energy to electrical energy, through what head doesthe water fall?
b. Taking turbine and generator efficiency as 0.85 and 0.95,
respectively ,again through what head does the water fall?
c. Calculate the specific water consumption per kWh if the plant
to be operated
f
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Factors Determining Potential of Hydropower
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P (kW) = gQHd
The factors which affects the potential of hydropower are thefollowing
Head and head losses
Flow rate
System losses Component inefficiencies
C d
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Head Head is one of the factors which have great influence on the
system capacity. In any real system water losses its some energy because of
frictional drag and turbulence as it flows in channels and
through pipes and the effective head will be less than actual
head.
These flow losses vary from system to system: in some cases
the effective head can be less than 75% of actual head, in
others it can be greater than 95%
For example:
Two systems with flow rate of 100 m3/s and plant efficiency of83%
First system(low-head) Second system(high-head)
Effective head 10 m 110 m
Available power(ghQ) 8,142 kW 8,9565.3 kW
C td
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Flow rate
Flow rate is another very important factor which caninfluence system capacity.
For example:
Two systems with effective head of 100 meters and
plant efficiency of 83%First system Second system
Flow rate 0.0024 m3/s 6000 m3/s
Available power 1.95 kW 4,885,380 kW
C td
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Contd
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System losses When electricity is transported along a transmission
system, the losses occur. As a result, what comes outof the system at the consumer is less than what is
input into the system at the generation site.
Component inefficienciesComponent losses include losses in:
Penstock
Turbine
Generator
Step up & down transformer lossesTransmission losses
The generator efficiency gives the ratio between mechanicalenergy of the turbine shaft and electrical energy delivered from
the generator.
Main design parameters
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Main design parameters
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The power capacity of a hydropower plant is primarily a functionof two main variables of the water
water flow
hydraulic head A graphical representation of the percentage of time in the
historical record that a flow of any given magnitude has been
equaled or exceeded
D i Fl
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Design Flow
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Design flow is the maximum flow for which your
hydro system is designed. It will likely be less than the maximum flow of the
stream (especially during the rainy season), more
than the minimum flow, during dry season
If a system is to be independent of any other energy
or utility backup, the design flow should be the flow
that is available 95 percent of the time or more.
Therefore, a stand-alone system such as a micro-hydropower system should be designed according to
the flow that is available year-round; this is usually
the flow during the dry season.
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Some flow definitions
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Reserved flow: it is the minimum flow required to
avoid aquatic life damage in the water stream Firm flow: The firm flow is defined as the flow being
available X % of the time, where X is a percentage
specified by the user and usually equal to 95%.
Hydraulic head
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Hydraulic head
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GROSS HEAD of a hydropower facility is the difference between headwaterelevation and tail water elevation. With the use of survey instrument, gross
head can be determined systematically and accurately
Maximum Head (Hmax) - above which the excess water spilled afterimpounding during or after a heavy rainy season with possible flood.
Minimum head (Hmin) - below which the reservoir should ideally be notallowed to be drawn down , and water contents in a given reservoir is said to
have been lowered down so a dead storage state.
Design head (Hd) - which is used the actual in water capacity calculation fora given HEPP, which can also be referred to as the effective head , which inturn equal to the growth head minus hydraulic losses before entrance to the
turbine and outlet losses
P t t
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Power output
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By taking the above design parameters in to
consideration, the power output formula can bemodified as
P = gQdHdtot
WhereQd = Design flow
Hd = Design head
tot = Total efficiency
C t f Pl t F t (PF)
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Concept of Plant Factor (PF)
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The plant factor is the ratio of the actual energy generated to
the maximum energy that can be generated if the system runscontinuously
It expresses the extent to which the hydropower installation is
actually it is exploited for profitable use.
Energy production can be estimated on a yearly base asE(kW) = Power *t (time of generation)
Time in a year = 365*24 = 8760hr
= 365.25*24 = 8766hr.
Therefore Plant Factor is defined as:
Plant Factor = Actual Energy Produced
Maximum Energy Available that can be generated
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e.g. Given a hydroelectric power plant with generating capacity
P(KW), P(MW), P(GW), if we use the rated capacity for a
time tH in a yearTherefore the plant power factor is
PF = Power used*Time used = P* tHinstalled power*8760 P*8760
Suppose tH = 3000hrThen PF = 3000/8760 = 0.35
take also tH = 5000hr
PF = 5000/8760 = 0.57
Note-If a PF~0.3, then the plant is normally called PEAK
LOAD PLANTIf a PF~O.55-0.57 then the HEPP is called BASE
LOAD PLANT
Example 2
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For a 100-MW HEPP with a design head of 300m, and plant factor
of 0.6, determine:
i. Electrical energy generated in one yearii. an approximate value of the theoretical energy stored in a year
in the reservoir of the HEPP for conversion into useful energy
iii. depth of water in the reservoir between a season with floods
and a dry season and
iv. the approximate surface area of the reservoir(Hint: For part (iv), potential energy of the stored water in joulesis given by E FHgH, with F = surface area, H = is the
depth of the reservoir needed for energy conversion;
K = FH = volume of useful stored water; = water density, g =acceleration due to gravity, and Hd is the design head)
Note: You can assume that Hd = 0.78Hmax and Hmin = 0.61Hmax,where Hd is the "design head", Hmax is the maximum allowable
head in the reservoir and Hmin is the minimum head in the
reservoir during a dry season.
Example 2
Solution 2
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Solution 2Given
P = 100MW, PF = 0.6, Hd = 300m
i. Electrical energy generated in one year
E = P * time of operation in one year
E = P * 8760 * PF
= 100MW * 8760 * 0.6 h
= 525.6GWh .ANS
ii. An approximate value of the theoretical energy stored in ayear in the reservoir of the HEPP for conversion into useful
energy.
Eactual = Ethoretical*
Ethoretical = 525.6 = 641GWh .ANS0.82iii. Depth of water in the reservoir between a season with floods
and a dry season is H and change in energy production
between the flood and dry season is E.
Contd
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Cont d
Therefore E=VgHd=HAgHd .equation **
Where:A = Surface area of the reservoir
H = Hmax Hmin change in water elevation
But it is given that Hd = 0.78 * HmaxH
d= 0 .78 * H
max
,
Hmax = 300
0.78
= 364.6 m
Hmin =0.61 * Hmax Hmin = 0.61 * 364.6 = 254.6 m
Therefore, H= Hmax Hmin= 110 m ANS
Contd
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Cont d
iv. The approximate surface area,
From equation ** A = EHgHd
Note, 1KWh = 3.6 MJ
641GWh =641*106*1KWh
= 641*106
*3.6*106
J= 2307.6*1012J
Finally A = E
HgHd= 2307.6 *1012
1000*110*9.81*300
= 712.8*103 m2
= 712.8 km2 .ANS
Classification of Hydroelectric Power Plants
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Classification of Hydroelectric Power Plants
Power plants can be classified in several different ways like
available head, size of plant and type of impoundment.i. Based on available Head
Low-head
Medium-head
High-head
ii. Based on Generating capacity Micro-hydro
Small-scale hydro
Large-scale hydro
iii. Based on Type of Impoundment Impoundment type
Diversion and Canal type
Run-of-the River type
Pumped Storage type
Power Plants Based on Available Head
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Power Plants Based on Available Head
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Low-headLow head" hydroelectric plants are power
plants which generally utilize heads up to10 meters
Power plants of this type may utilize a low
dam or weir to channel water, or no dam
and simply use the "run of the river
Run of the river generating stations
cannot store water, thus their electric
output varies with seasonal flows of water
in a river
A large volume of water must pass througha low head hydro plant's turbines in order
to produce a useful amount of power
Low-head type of hydroelectric installation
is shown in the figure
Contd
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Cont d
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Medium-head
Medium-head (10 meters - 100 meters)Hydro power plants consist of a large dam in a mountainous
area which creates a huge reservoir.
Another type of medium-head facility is a pumped storage
plant
Contd
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Cont d
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High-headHigh-head hydro power plants have an elevation difference of
at least 100 meters between the turbines and the water surface Generating stations of this type are found in the mountains
areas, and high-speed turbines are used.
B d G ti it
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Based on Generating capacity
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Very small hydropower plants(pico-hydro power
plants): 0.25-1KW
Micro hydropower plant : 1-100KW
Mini hydro : 0.1MW-1MW
Small scale hydro : 1-10MW
Medium-scale : 10-100MW
Medium-Large scale : 100-300MW
Large scale : >300MW
Power plants based on type of impoundment
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Power plants based on type of impoundment
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Impoundment (Storage) Power Stations An impoundment facility, typically in a large hydropower
system, uses a dam to store river water in a reservoir
The water may be released either to meet changing electricityneeds or to maintain a constant reservoir level
Depending on their storage capacity are referred to as week,
season or year storage stations
During periods with low power requirements the water can bestored (although there could be losses due to evaporation
during daytimes) and utilized when the demand is high, thus
ensuring flexible operation
Hydroelectric plants of the storage type may be electrically
coupled, i.e. they may all serve the same power transmission
system (or grid) load.
Hydroelectric plants could also be coupled hydraulically, i.e. the
water out flow of one plant may be a significant portion of the
inflow to one or more other downstream or cascaded plants
Contd
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Cont d
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Run-of-river type A dam with a short penstock (supply pipe) directs the water
to the turbines, using the natural flow of the river Power stations of this type are built on rivers with a consistent
and steady flow of water and have little or no reservoir capacity
for storage
Diversion and Canal type The water is diverted from the natural channel into a canal or
a long penstock, thus hanging the flow of the water in the
stream for a considerable distance
Pumped Storage Type When the demand for electricity is low, pumped storage
facility stores energy by pumping water from a lower reservoir
to an upper reservoir. During periods of high Electrical demand,
the water is released back to the lower reservoir to generate
electricity
Hydropower Advantages
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Hydropower Advantages
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A big advantage of hydroelectric power is the ability
to quickly and readily vary the amount of powergenerated, depending on the load presented at that
moment
It utilizes a renewable energy source as fuel (water)
Generation process is environmentally clean
High reliability
Non polluting, utilizes indigenous resource
Hydropower Disadvantages
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Hydropower Disadvantages
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It requires large initial investments
It requires long transmission lines
Social and environmental impacts
Social impact
Population displacement
Loss of social networks and changing way of living Dams can facilitate development of diseases
Diversion of mountain streams
Blockage of fish passage both upstream and
downstream Storing water in reservoir may reduce the final flow
as a result of evaporation
Contd
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Cont d
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Environmental impacts
Reduction in the flow of soil and nutrients Pollution is stored in the reservoir
Possible dam failure
Loss of cultural heritage
local increase in water vapor and sometemperature effects
Vegetation rotting under water produces methane
implies emissions
Assignment I-B
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Assignment I B1. For a given HEPP, the normal water level of the reservoir is
given to be 1669 m.a.s.l (meter above sea level), and the dead
water level is 1648 m.a.s.l. The live storage of the reservoir is
865*106 m3, and its dead storage is 135*106m3.The tail water levelis 1439.4m.a.s.l
a. Sketch (with clear labels) a schematic diagram of the HEPP
b. Define the center( Hr) of the water level to correspond to 50%
of the live storage plus dead storage, estimate Hrc. Given that the design flow rate 30m3/s, and assume the head
loss is approximately given by
H = 0.0414Q2 + 0.0772*Q2/D4
Where D = 6.6m is the diameter of the tunnel leading the water
from the reservoir to the turbine, determine the design head and
the installed power
d. Determine the growth available energy
e. Using you own assumptions, estimate the mean annual electrical energy