Hydroelectric and Wind Power Plants

Hydroelectric and Wind Power Plants

Prof. Paulo Seleghim Jr.Universidade de São Paulo

LBE5010 Renewable Energies and Energy Planning

Coal39.1%

Hydro6.7%

Wind4.2%

Approximately 11% of all electricity generated in the world comes from impulse turbines

Turbinas Parson(r 0,5)

Impulse turbines (r=0)

Coal39.1%

Hydro6.7%

Wind4.2%

Hydroelectric turbines

✓ lowest generation cost (~4 ¢/kWh)

✓ 50 – 100 years useful life

✓ functionality of the reservoir

✓ small growth potential

Parson turbines(r 0,5)

Wind turbines

✓ low generation cost (~8 ¢/kWh)

✓ virtually unlimited potential

✓ complementarity with hydroelectric power plants

✓ intermittent

Approximately 11% of all electricity generated in the world comes from impulse turbines

Impulse turbines (r=0)

ramjet, scramjet, etc. (r = 1)

OPERATION OF THERMO POWERPLANTS BETWEEN 2013 – 2014

WIND ELECTRICITY GENERATION IN BRAZIL

Technical potential: 143 GW(1)

(1) Atlas do Potencial Eólico Brasileiro(2) Assoc. Brasileira de Energia Eólica

Operating (2014): 9,91 GW(2)

Hired (2035): 9,93 GW(2)

Average FAC (2014): 38%(2)

Total generation in 2014: 71,26 GW

WIND38%

Global Mean Wind Speed at 80m

Global Mean Wind Speed at 80m

Chapada do Araripe

TW21,2W 2012,cons =

WORLD HYDROELECTRIC POTENTIAL

Wpot: 0,113 TW

Winst: 0,183 TW

América do Norte

Wpot: 0,235 TW

Winst: 0,143 TW

América do Sul

Wpot: 0,095 TW

Winst: 0,172 TW

Europa

Wpot: 0,169 TW

Winst: 0,026 TW

África

Wpot: 0,340 TW

Winst: 0,341 TW

Ásia

FLUX MACHINES:theoretical analysis

Energy inventory in the control volume... 1ª Law of Thermodynamics

open

systemQ W

entra

kkm sai

kkm

( ) ( )

++−++−−=

entra

2kkkk

sai

2kkkk

VC 2/Vgzhm2/VgzhmWQdt

dE

mechanicalenergy

thermalenergy

mechanicalenergy

thermalenergy

Q W

entra

kkm sai

kkm

( ) ( )

++−++−−=

entra

2kkkk

sai

2kkkk

VC 2/Vgzhm2/VgzhmWQdt

dE

( ) ( )

++−++=

sai

2kkkkk

entra

2kkkkk 2/Vgz/Pm2/Vgz/PmW

T=cte T=cte

Energy inventory in the control volume... 1ª Law of Thermodynamics

open

system

Application to an impulse turbine

1

2 3

4

4V1Vforces on the

fluid exerted by

the turbine

velocity pressure

1 2 3 4

4V1V

1

2 3

4

)VV(mA)PP(F 4123 −=−=

Saint-Venant

momentumvariation


forces on the

fluid exerted by

the turbine

)VV(mA)PP(F 4123 −=−=

+

−

+

=

2

VP

2

VPm0

22

2

22

1

1

1

)VV(VAA)PP(F 41med23 −=−=

+

−

+

=

2

VP

2

VPm0

24

4

423

3

3

P1 = P4

V2 = V3

1ª lei

2

VVV 41

med

+=

−=

−

2

V

2

VPP 24

21

2

2

3

3

T = ctez = cte

→ 32


)VV(VAVFW 412medmed −==

)VV(2

VVAVFW 41

2

41med −

+==

)1(2

1VAW

2

31 −

+=

14 V/V=

1ou3/10/W −===

2

VA

27

16W

31

max =


2

VA

27

16W

31

max =

s/m9maximum

mechanical power

produced by the

turbine

D = 20 m

25°C, 1bar

2

)s/m9(

4

)m20(

m

kg17,1

27

16W

32

3

max

=

kW789,158Wmax =

s/m325°C, 1bar

D = 34,6 m


Working Definitions and Equations

)1()1(2

1VA

2

1W)1(

2

1VAW 23

1

2

31 −+=→−

+=

windpower

powercoefficient

)1()1(2

1

power wind

power turbine

2/VA

W)(CP 2

31

−+==

=

= velocity ratio

CP =

coefic

iente

de p

otê

ncia

%3,5927

16CPmax =

2/VA)(CP

W

power maximum

power actual3med

==

design:=1/3

wind powercapacity

V


Burgar Hill, Orkney; 60 m measurement height; 4 hour averages

V

2/VA)(CP

W3med

=

pdf

V9,8m/s

nominal designvalues

+3−3


V

Burgar Hill, Orkney; 60 m measurement height; 4 hour averages

2/VA)(CP

W3med

=

pdf

0

loss ofefficiency


Factors Causing Efficiency Losses

high pressure

low pressure

high pressure

low pressure


high pressure

low pressurewing tipvortex

wing tipvortex

high pressure

low pressure


high pressure

low pressurewing tipvortex

wing tipvortex

high pressure

low pressure

Evolution of Wind Power Technology

600m

tempo

W

active phase return phase

balloon

400m

lateral wind

200m

HYDROELECTRIC POWER PLANTS:design and operation

1

2

mecW

Kaplan

eleW

W

W

Wmec

=

( ) ( ) 2/Vgz/P2/Vgz/PmW 22222

21111 ++−++=

( ) ( ) 2/Vgz/P2/Vgz/PmW 22222

21111 ++−++=

1

2

mecW

Kaplan

eleW

( ) ( ) ( ) 2/V2/Vzzg/P/PmW 22

21212211 −+−+−=

W

W

Wmec

=

1

2

mecW

Kaplan

eleW

zgmW =

W

W

Wmec

=

Hydroelectric Power Generation Potential in the World

zgmW = Average continental height: 840m

Average annual pluviosity: 505103 km3

at sea: 398 103 km3

on land: 107 103 km3

m840s

m8,9

m

kg997

s360024365

)m10(10107W

23

333

teomax,

=

TW8,27W theomax, =

Assuming that all in land precipitation can be used for power generation

TW84,3W UNDPmax, = World Energy Assessment: 2004 - United Nations Development Programme

TW21,2W 2012,cons =

The potential growth of hydroelectric generation is substantially limited, however...

RESERVOIR: Capacity of storing water for energy generation

COMPLEMENTARITY: Stabilization of supply in face of fluctuations in generation conditions (intermittencies) and also of the demand

Hydroelectric Power Generation Potential in the World

Barra Bonita

Bariri

Promissão

Ibitinga

Nova

AnanhandavaTrêsIrmãos

Jupiá

PortoPrimavera

Institutions of the Electricity Sector in Brazil

Geradores

DistribuidoresConsumidores

Energy Bids Consolidated Results

0

1

2

3

4

5

6

7

8

2000 2005 2010 2015 2020 2025 2030 2035 2040 2045 2050 2055

GW(físicos)

ano

Último leilão: 13/11/2015

Energy Bids Consolidated Results

APPLICATION EXAMPLE:optimized design and operation

HYDROELECTRIC POWER PLANTS OPTIMIZED OPERATION

time

flow rate(m3/s)

100

20

10

01/01 31/1201/04 01/07 30/09

annual averages

seasonal variations of the main river and tributaries

50m100m3/s

100 m3/s

50m110m3/s

10 m3/s

50m130m3/s

20 m3/s

Design Configuration

efficiency curves and optimum operation points (%)

100

Nsp,min Nsp,max

Locus of design points(maximum conversion efficiency)

specificrotation (rps)

4/5

2/1

sp)Hg(

)/W(NN

=

design point

geometric

similarity

50m100m3/s

100 m3/s

50m110m3/s

10 m3/s

50m130m3/s

20 m3/s



(%)

100

0,1

4/5

2/1

sp)Hg(

)/W(

=

0,04 0,2 0,4 1,0 2,0 4,0 10,0

95

90

85KaplanFrancis

Pelton

50m100m3/s

100 m3/s

50m110m3/s

10 m3/s

50m130m3/s

20 m3/s



efficiency curves and optimum operation points

specificrotation (rad)

312rpmsp = 3,14rad

298rpmsp = 3,14rad

274rpmNsp = 0,5

“We chose Kaplan turbines”

50m100m3/s

100 m3/s

50m110m3/s

10 m3/s

50m130m3/s

20 m3/s

(%)

100

0,1

4/5

2/1

sp)Hg(

)/W(

=

0,04 0,2 0,4 1,0 2,0 4,0 10,0

95

90

85KaplanFrancis

Pelton



efficiency curves and optimum operation points

specificrotation (rad)

312rpmsp = 3,14rad

298rpmsp = 3,14rad

274rpmNsp = 0,5

50m100m3/s

100 m3/s

50m110m3/s

10 m3/s

50m130m3/s

20 m3/s



The design is based on nominal parameters (annual or pluriannual averages) that result in an optimized

performance (maximum efficiency)

Operating parameters vary with respect to nominal parameters (tributaries flow rates e.g.)

The problem: how to adjust the overall system operation, acting on controllable variables, in order to optimize

performance ?

( ) =k

ksist maxmax → performance criterium

( ) =k

ksist maxmax


→ performance criterium

Hk+1

Vk+1

Qk+1

Ak+1

Wk+1

Hk

Vk

Qk

Ak

Wk

Hk-1

Vk-1

Qk-1

Ak-1

Wk-1

ano

Winst

Wmed

W

W

+

=k W

meddef

sist1.0

W

Wmed W

( ) =k

ksist maxmax

Given the (non controllable) variations of the tributaries flow

rates (A), how to adjust the reservoirs discharges (Q) to optimize the total energy generation ?


Hk+1

Vk+1

Qk+1

Ak+1

Wk+1

Hk

Vk

Qk

Ak

Wk

Hk-1

Vk-1

Qk-1

Ak-1

Wk-1

Subjected to the foillowing mass and energy conservation equations:

kk1kk QAQ

dt

dV−+= −

kkkk HQgW =

=

T

kk dtWE )V(fH kkk =

)N( k,spkk =

To which the following performance characterization

equation apply

Efficiency curves

Water height at in function of the

accumulated volume

+k W

med

1.0

Wmax


Given the (non controllable) variations of the tributaries flow

rates (A), how to adjust the reservoirs discharges (Q) to optimize the total energy generation ?

TRIBUTARIES FLOW RATES: STOCHASTIC MODELS

Hydrology 2014, 1(1), 89-111; doi:10.3390/hydrology1010089

vazão(m3/s)

100

20

10

0

−=

24365

t2sin1020A3

=8760 horas=2190 horas =4380 horas =6750 horas

365243/4 365244/4

+

=

2190tse878,7

2190t0se21902

t2sin1010

A2

365242/4365241/4

+=

24365

t2sin10100A1

TRIBUTARIES FLOW RATES: STOCHASTIC MODELS

TURBINE PERFORMANCE EQUATIONS

4/3

2/1

sp)Hg(

QNN

=

H

Q

W

cxbxax 2max

^ ++=

7,4c8,6b8,2a =−==

2

NNN

max,spmin,spnom,sp

+=

max?

max

0,0

(%)

Nsp,nom Nsp,max

0,7max

0

0,5 1,0x

design point

model

Nsp,min

RESERVOIR CHARACTERIZATION EQUATIONS

L

H

B

2A+B

A

A

baseAL3

1V =

L

B

2A+B

A

A

L

H)(Tg

=

a

H)(Tg

=

baseAL3

1V =

2

Ha2HbAbase

+=

+=

2

Ha2Hb

3

LV

+

=

)(Tg

Hb

)(Tg3

HV

2

L

H)(Tg

=

a

H)(Tg

=

H


3nom km025,0V =

3Monte.B km5,2V =

5km

50m

100m

200m

200m

01,05000

50

L

H)(Tg ==

= 25,0

200

50

a

H)(Tg ==

=

+

=

)(Tg

Hb

)(Tg3

HV

2

92

1025,0

50100

01,03

50V −

+

=

3Itaupú km19V =


H

volume (km3)

altura (m) silting

+

=

)(Tg

Hb

)(Tg3

HV

2


FORMULATION OF THE OPTIMIZATION PROBLEM

50mQ1

A1(t)

50mQ2

A2(t)

50mQ3

A3(t)

• Maximum total energy generated in a period of time (1 year)

• Maximum regularity of the generated energy (dispatchability)

( ) HQNgW sp =

4/3

2/1

sp)Hg(

QNN

=

cteHQ 4/32/1 −

Q e H must be so that Nsp 0,5to have max

When Q H and when A H

cteHQ 1 −

)A,Q(HH =

conflicting effects: trade-off solution

50mQ1

A1(t)

50mQ2

A2(t)

50mQ3

A3(t)

+

→k k,W

k,med321

1.0

Wmax)t(Qe)t(Q),t(Q

Greater total generation in the period

Smaller dispersion around the average value

Defining an adequate target function combining two optimization aspects (total energy generation and regularity)


• Maximum total energy generated in a period of time (1 year)

• Maximum regularity of the generated energy (dispatchability)

50mQ1

A1(t)

50mQ2

A2(t)

50mQ3

A3(t)

+

=k k,W

k,med321

1.0

Wmax)Q,Q,Q(

Initialization

Random change in direction 1

Keep modification only if increases



Random change in direction N


Iterate

Q1

Q2

“Bioinpired” search method

não

iso-


50mQ1

A1(t)

50mQ2

A2(t)

50mQ3

A3(t)

+

=k k,W

k,med321

1.0

Wmax)Q,Q,Q(

Q1

Q2

iso-


Initialization







Iterate


50mQ1

A1(t)

50mQ2

A2(t)

50mQ3

A3(t)

+

=k k,W

k,med321

1.0

Wmax)Q,Q,Q(

Q1

Q2

iso-


Initialization







Iterate


50mQ1

A1(t)

50mQ2

A2(t)

50mQ3

A3(t)

+

=k k,W

k,med321

1.0

Wmax)Q,Q,Q(

Q1

Q2

iso-


Initialization







Iterate


NUMERICAL SIMULATIONS

+

=k k,W

k,med321

1.0

Wmax)Q,Q,Q(

+

=k k,W

k,med321

1.0

Wmax)Q,Q,Q(SMALL VOLUME RESERVOIR

+

=k k,W

k,med321

1.0

Wmax)Q,Q,Q(LARGE VOLUME RESERVOIR

+

=K k,W

k,med321

1.0

Wmax)Q,Q,Q(SMALL VOLUME RESERVOIR

LARGE VOLUME RESERVOIR +

=K k,W

k,med321

1.0

Wmax)Q,Q,Q(

+

=k k,W

3k,med

3211.0

)W(max)Q,Q,Q(SMALL VOLUME RESERVOIR

LARGE VOLUME RESERVOIR +

=k k,W

3k,med

3211.0

)W(max)Q,Q,Q(

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