Chapter9 Hydropower Flashcards

Question 1

Q

Worldwide distribution and utilisation of hydro potentials

Answer

A

Almost half of the world’s electrical energy could be supplied by hydro power plants (economically feasibly!)
Hydro power development in Africa, Asia and South America holds a great potential

Question 2

Q

1.Basics of hydropower generation:

Okey let’s review some basics from high school, what is the hydrologic water circuit? what are its phases ?

Answer

A

Evaporation in form of vapour
Condensation in form of clouds
Transportation by the wind (advection)
Precipitation in case the wind transport the thingies up –> condensation increases
Rainfall reaches the ground hurraay: - Surface runoff –> collects in creeks and rivers
Infiltration/ percolation: soil moisture increases, which feeds up the groundwater table, partly re-apparition in rivers later

Question 3

Q

1.Basics of hydropower generation

What is Hydroenergy? where can it be found?

Answer

A

Hydroenergy = energy in moving water

Energy in falling water:
river hydropower plants
hydroelectric-dam power plants
pumped-storage hydropower plants
Energie of waves
Tidal energy

Question 4

Q

_1.Basics of hydropower generatio_n

Give the expression of potential energy!

Answer

A

Potential energy

E_P = m * g * h

m = mass of water (kg)

g = gravity acceleration = 9.8 (m/s2)

h = distance of water fall - head (m)

Question 5

Q

1.Basics of hydropower generation

Give the expression of power in falling water! and the generated electrical power

Answer

A

Power in falling water

P_H = E_P/ t = Q * ρ * h * g Q = m/(ρ*t)

Q = volume flow of water (m3/s)

ρ = density of water (kg/m3)

t = time (s)

Turbine takes energy from the falling water so potential energy is converted to rotational energy, the rotation of turbine activates generator which generates electricity. Then a transformer adjusts voltage in the connection to the grid

–> Electrical power generation in hydro-turbines

Pel = η * P_H

η = conversion factor (efficiency) of the hydropower plant

η = η_T * η_G * (1- f)

T = turbine
G = generator
f = factor of additional losses (transformer, gear,…),
à usually 0.03…0.1

Question 6

Q

1.Basics of hydropower generation

Functional principle of water turbines?

Answer

A

Common basic principle of all kinds of fluid machinery used in Hydro Power: (water wheels, Pelton, Francis or Kaplan turbines)
The fluid is conveyed onto the runner in a way that it exerts a force in circumferential direction, thus producing a driving torque M = F · r
The force can be a gravity, impulse, pressure or lifting force or a combination of these.
The layout of the waterway can be
- straight and tangential to the runner (undershot water wheel, Pelton turbine)
- spiral-shaped around the runner (Francis and Kaplan turbine)
- axial with a swirl around the runner axis (Kaplan bulb turbine)

Leaving aside the water wheels, we need to distinguish between impulse turbines and reaction turbines.

Question 7

Q

1.Basics of hydropower generation

Types of hydro turbines?

Answer

A

Impulse turbines: Pelton and cross flow turbines

Reaction turbines: Francis-, Deriaz-, Kaplan- and Kaplan bulb turbines

Question 8

Q

Impulse turbines: Pelton and cross flow turbines, how does it work?

Answer

A

The energy conversion within the turbine runner is performed at constant (=ambient) pressure.
All the energy available at the pressure inlet of the turbine is converted into
kinetic energy of the free jet leaving the nozzle.
This kinetic energy is then converted into mechanical energy in the buckets of the turbine runner while the pressure remains constant.

Question 9

Q

Reaction turbines : Francis-, Deriaz-, Kaplan- and Kaplan bulb turbines

Answer

A

The static pressure at the runner inlet is higher than at the runner exit.
The pressure energy available at the turbine inlet is only partly converted into kinetic energy due to acceleration in the spiral case, the stay vanes and the guide vanes.
Most of this kinetic energy and the remaining pressure energy are converted into mechanical energy in the turbine runner.
The draft tube converts the kinetic energy remaining at the runner outlet into pressure, thus contributing to the pressure drop across the runner.

Question 10

Q

1.Basics of hydropower generation

Characteristics of Kaplan Turbines?

Answer

A

-Low h, high Q

à river power plants

ηT = 80-95%
50 kW – 100 MW

They are normally double-regulated, i.e. guide vane and runner blade angle are separately adjustable. Thus, a high efficiency over a wide range of flow rates is achieved.

Question 11

Q

1.Basics of hydropower generation

Francis turbines?

Answer

A

Francis turbines are reaction turbines, meaning that both pressure and kinetic energy are converted by the runner. The machines are completely filled with fluid and lie typically below the tailwater level to avoid cavitation. They can also be built as reversible pump turbines: By operating the runner in the opposite direction of rotation the turbine works as a pump.

High levels of h –> Hydroel. dam and pump storage plants
ηT = > 90%, but poor at partial load
20 kW - 700 MW
slow, normal, fast moving turbines

Question 12

Q

1.Basics of hydropower generation

Low specific speed Francis turbines ?

Medium specific speed Francis turbines?

High specific speed Francis turbine?

Answer

A

Low specific speed Francis turbines

Highest head and smallest discharge in the Francis turbine range
Flat, disc-shaped runner: flow is in is mainly in a radial (inward) direction (radial turbine)
High pressures: thick-walled, strong spiral case

Medium specific speed Francis turbines

Runner flow is in radial-axial (diagonal) direction

High specific speed Francis turbines

Lowest head and greatest discharge in the Francis turbine range
Runner flow is in radial-axial (diagonal) direction

Question 13

Q

1.Basics of hydropower generation

Pelton turbines and their application range?

Answer

A

Nozzle: Conversion of pressure energy into kinetic energy of a free jet
partially impacted runner, running in air: deceleration of the free jet under ambient pressure (constant pressure turbine / impulse turbine , Gleichdruckturbine)
dry buckets should not hit the tailwater surface

High levels of h –> Hydroel. dam and pump storage plants
water injection by jet nozzles
1-8 jet nozzles depending on Q
jet nozzles: exit speed up to 200 m/s
ηT = 90-95%
5kW – 300 MW

Question 14

Q

2.1 Hydropower plants

What’s the annual electricity yield?

Answer

A

Annual electricity yield:

E_el= ∑ t(Qi) * Pi

Pi = power generation at Qi

t(Qi)=i operational time per year at Qi

Question 15

Q

Give the following expressions:

Expression of Hydropower generation by river power plants

Expression of Rated capacity hydropower generation

Answer

A

Pel = ηT * ηG * (1- f) * g * ρ * h * Q

Q = volume flow of water (m3/s) h = head (m)
ρ = density of water (kg/m3) t = time (s)
f = factor of additional losses (transformer, gear,…), usually 0.03…0.1
W = water level of river (m)
W_R = rated capacity water level (m)
Q_R = rated volume flow (m3/s)
h_R = rated head (m)

Question 16

Q

2.1 Hydropower plants

How are the rated capacity levels to be designed?

Answer

A

The rated capacity levels have to be designed to maintain high efficiencies of power generation allover the year.

Usually QR < Qmax

Question 17

Q

2.1 Hydropower plants

How do P_eland P_el,Rrelate to each other if ηT and W are near the rated capacity levels?

Prove it through playing with their expressions!

Answer

A

P_el ≈ P_el,R * Q / Q_R

Question 18

Q

2.1 Hydropower plants

What are the benefits and drawbacks of a river hydropower plant?

Answer

A

Benefits

Mature and well proofed technology
Low costs of power generation (Germany: 3 – 25 €-ct / kWh, depending on plant size)
Moderate environmental impacts: e.g. no flooding of large parts of land required

Drawbacks

Power production depends on natural run-off of water
Suitable locations for river hydropwer plants limited

Question 19

Q

Land based Hydro Power Plants (HPP’s).

Answer

A

can be categorized according to two criteria:

head: low, medium and high head plants (<25m / 25 ..250m / >250 m)
utilisation mode: run-of-the-river plants and storage plants

The pumped storage plants which are used to store excess electrical energy belong to the storage plants.

Question 20

Q

2.2.Hydroelectric-dam power plants

How do they work?

Answer

A

Installation at natural lakes or installation of embanked lakes (reservoirs)
Power generation by turbine and generator
Large reservoir allow control of Q and P à „on-demand“ power generation
Usually large scale plants à capacities up to ~ 18,000 MW (Three Gorges Dam, China)

Question 21

Q

2.3 Pumped-storage plants

When are storage plants used?
How do their operation schemes look like?

Answer

A

Storage plants are mainly used to satisfy peak load demand. They have a reservoir which permits to hold back the natural inflow over a certain period of time.
Depending on the storage capacity and the temporal usage, they are operated under a yearly, weekly or daily operation scheme, thus being termed annual, weekly or daily storage reservoirs. (Jahres- Wochen- oder Tagesspeicher)
As the energy contained in a given volume of water is in direct proportion to the head, storage plants are normally medium or high head plants.
The reservoir is usually created by building a dam in a suitable location of the natural river valley.
From the reservoir, the water is conveyed by an open channel or a headrace tunnel (Freispiegelgerinne oder Druckstollen) and a usually steeply descending penstock (Druckrohrleitung) to the power house which is situated farther below in the valley.
A further type is the barrage plant (Talsperrenkraftwerk), where the power house is integrated into the base of the dam.

Question 22

Q

2.3 Pumped-storage plants

Pumped storage plants?

Answer

A

basically built like other high head storage plants
powerhouse contains pumps and turbines or reversible pump turbines
waterways must be suitable for turbine and pumping flow direction!

Excess electrical energy is used to pump water into a reservoir at higher elevation (pumping operation). The energy stored as potential energy is re- converted into electrical energy when needed (turbine operation). Depending on the storage volume available, the plant may be operated on a daily, weekly or seasonal basis.

A higher head permits to:

store more energy within a given storage volume

produce a higher electrical power with a given cross sectional area of tunnels and penstocks.

–> normal head range of pumped storage plants: 150 – 1000m

Question 23

Q

What is a pumped storage plant usually used for?

and how is it specifically used in GE?

Answer

A

Pumped storage plants are used for

Energy storage (seconds / hours)
Peak load coverage
Maintening of frequency and buffering of power fluctuations in the grid

Traditional operation in Germany:

Combination with base load dedicated coal firing and nuclear power plants–> energy storage mainly by night and peak load power coverage by day

Future operation:

Increasing power supply by wind and solar

–> energy storage and power production by day as well as by night

Question 24

Q

Which turbines are best suitables for PS sets?

Answer

A

head range of PSP’s: 100m to 1000 m
–> only Francis and Pelton turbines are suitable

with Francis turbines: usually horizontal shaft
with Pelton turbine: always vertical shaft. The turbine needs to be above and the pump below the tailwater level (cavitation!)
same sense of rotation in pumping and turbine mode: quick changes between modes of operation

Question 25

Q

2.3 Pumped-storage plants

Expression of the efficiency nu_overallof pumped storage plants?

Answer

A

Relevant losses:

Viscous friction in tunnels and penstock H
hydr. and mech. losses within pump and turbine
el. and mech. losses within motor and generator

Energy storage and electricity generation

E_S,max= V *ρ * g * h *η_PG

_{Es= energy storage}
η_{PG= Effciency of power gen( turbine, gen, transformer,penstocks)}
_{V= volume of upper reservoir}

Overal efficiency of energy storage and electricity generation

η= η_PS* η_PG

η_{PS= Effciency of power storage( transformer, gen, pump,penstocks)}

Question 26

Q

2.3 Pumped-storage plants

Give an example of Pumped-storage plants in GE!

Answer

A

Example - Pumped-storage- hydropower plant, Happurg (Fa. E.ON AG)

Comissioning 1955
2012 under general maintenance
Pump-storage plant and transformer station
Head ~ 200 m
Rated capacity 160 MWel
Power supply for Nuremberg area

Or

Goldisthal (1060 MW)

h = 302 m
PR = 1060 MWel
4 Francis-type-pump-turbines
Operator: Vattenfall
Investment: 623 mio € à ~ 600 €/kWel installed

Question 27

Q

Run-of-river (ROR) hydro power plants?

Answer

A

are normally utilising the current discharge of the river. The purpose of their weirs and dams is to create a pressure head, but they do not create a substantial storage volume

Often, a number of successive power stations are situated along a river, forming a chain of power plants
Permanent operation
Usually low distance of water fall h (head) and high volume flow Q of water
Q, h and P affected by seasonal events (e.g. rainfall, spring melts, dry seasons) –> retaining dam to control Q and h

Question 28

Q

What is the application of Hydroelectric power plants?

Answer

A

Base load plants

> 50% capacity utilization

river hydropower plants
hydrolectric dam plants

Average load plants

30-50% capacity utilization

river hydropwer plants,
hydrolectric dam plants

Peak load plants

< 30% capacity utilization

hydroelectric dam plants
pumped-storage plants

Question 29

Q

1- Stream turbines ?
2- How is the energy calculated for this turbine type?

Answer

A

1-

An often cited potential technology for producing electricity out of rivers in a very environmental friendly way “without any dams or weirs”:
a stream (or current) turbine is extracting kinetic energy out of the flow of a river.
potential turbine types are (in descending order of sensibility)
- horizontal axis stream turbine (built like a wind turbine)
- vertical or horizontal axis Darrieus or Gorlov turbine
- devices based on oscillating airfoils
- devices with drag-type runner

2- Energy in a free flow: kinetic energy of free flow E = ½ . m .c² [J]

Betz limit:
Pmax,th = 0.59 · 0.5 · rho · A · c³ (like wind turbine)

–> main problem:
very low energy density at low flow velocities
economically viable only at c > 2.5 m/s (rarely to be found!)

Question 30

Q

Small excercise for a stream turbine: With river 3m deep, 10 m wide, c= 1.0 m/s, Q=30m3/s

Calculate P for each of these cases

3 stream turbines D=2.5 m each:
weir with H=1 m, 3 Kaplan turbines D=2.5m:
weir with H=2 m, 2 Kaplan turbines D=2.5m:

Answer

A

P = 4.3 kW
P = 280 kW
P = 560 kW

Question 31

Q

3. Hydropower – economy, ecology, potentials

Hydroelectric power plants - costs?

Answer

A

Costs depend on type of power plant, size of power plant, location and life-time of power plant
Costs of power generation (Germany):
- small scale plants (< 1MW) 10 - 25 €-ct / kWh
- large scale plants 3 -10 €-ct / kWh
Costs of power generation (grid based global range 2013)
- large scale > 20 MW 2 – 12 US-ct/kWh
State-of-the-art technology: only low potential of cost reduction by technical improvments expected

Question 32

Q

Total wordlwide installed capacities of hydropower in 2015?

How much from that is Pumped-storage capacities?

Leading country?

Answer

A

Total wordlwide installed capacities: 1064 GW

Pumped-storage capacities: ~ 145 GW

Global hydropower generation

~ 3940 TWh à ~ 3700 h full load equivalent

~ 16.4 % of global electricity generation

Chinaaa

Question 33

Q

What are the Ecological impacts of hydropowerplants?

Answer

A

River power plants

Moderate impacts on landscape: e.g. no flooding of large parts of land required
Reduction of seasonal floodings à positive or negative impacts possible
Fish migrations may be affected by embankment dams –A installation of fish bypasses required (e.g. fish ladders)
Usually very low CO2-emissions

Large scale hydroelectric-dam plants

Positive impacts

Water reservoir for agriculture
Benefits for local economy by tourists (reservoir)

Negative impacts

Erection of large reservoirs: flooding of large areas
CH4- and CO2-emissions, if not cleared land is flooded
Displacement /relocation of people at flooded areas
Formation of sediments in the reservoirs during operation possible –> reservoir volume reduction
Negative impacts on surrounding aquatic systems possible ,e.g. by daily fluctuations of water flow (à e.g. erosions) and warming of water
Geophysical impacts of reservoires (earth quake) ?
Danger of dam failures à flooding of areas below the dam and subsequent damages

Question 34

Q

What are the advantages of multi turbine operation of hydro power plants?

Answer

A

Solution: high overall efficiency of power generation, almost independent of Q/QR (at least above Qmin), turbines can be operated at their rated capacity levels

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Chapter9 Hydropower Flashcards

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