energy conversion lecture

7/29/2019 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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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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Contd

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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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Contd

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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

C td


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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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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

energy conversion lecture

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