Hydro power plants
Hydro power plants
Inlet gate
Air inlet
Surge shaft
Penstock
Tunnel
Sand trap
Trash rack
Self closing valve
Tail water
Main valve
Turbine
Draft tube
Draft tube gate
The principle the water conduits of
a traditional high head power plant
Ulla- Førre
Original figur ved Statkraft Vestlandsverkene
gate hoist
Intake gate
Intake trashrack
Tunnel Inlet
Surge tank
Tunnel Inlet
trashrack
Penstock inlet
Valve
Desilting basin
Headrace tunnel
Anchor block
Anchor block
Shaddle
Power house
Exp. joint
DT end gate
IV
G
Tailrace
SC
IV -inlet valve
R -turbine runner
R
SC -spiral case
G -generator
DT
Typical Power House with Francis Turbine
Arrangement of a small
hydropower plant
Ligga Power Plant, Norrbotten, Sweden
H = 39 m
Q = 513 m3/s
P = 182 MW
Drunner=7,5 m
Borcha Power Plant, Turkey
H = 87,5 m
P = 150 MW
Drunner=5,5 m
Water intake
•
•
•
•
Dam
Coarse trash rack
Intake gate
Sediment settling basement
Dams
• Rockfill dams
• Pilar og platedammer
• Hvelvdammer
Rock-fill dams
1.
2.
3.
4.
Core
Filter zone
Transition zone
Supporting shell
Moraine, crushed soft rock, concrete, asphalt
Sandy gravel
Fine blasted rock
Blasted rock
Slab concrete dam
Arc dam
Gates in Hydro Power Plants
Types of Gates
• Radial Gates
• Wheel Gates
• Slide Gates
• Flap Gates
• Rubber Gates
Radial Gates at Älvkarleby, Sweden
Radial Gate
The forces acting
on the arc will be
transferred to the
bearing
Slide Gate
Jhimruk Power Plant, Nepal
Flap Gate
Rubber gate
Flow disturbance
Reinforced rubber
Open position
Reinforced rubber
Closed position
Bracket
Air inlet
Circular gate
End cover
Hinge
Ribs
Manhole
Pipe
Ladder
Bolt
Fastening element
Seal
Frame
Circular gate
Jhimruk Power Plant, Nepal
Trash Racks
Panauti Power Plant, Nepal
Theun Hinboun Power Plant
Laos
Gravfoss
Power Plant
Norway
Trash Rack size:
Width: 12 meter
Height: 13 meter
Stainless Steel
CompRack
Trash Rack delivered
by VA-Tech
Cleaning the trash rack
Pipes
• Materials
• Calculation of the change of length due to the
change of the temperature
• Calculation of the head loss
• Calculation of maximum pressure
– Static pressure
– Water hammer
• Calculation of the pipe thickness
• Calculation of the economical correct diameter
• Calculation of the forces acting on the anchors
Materials
• Steel
• Polyethylene, PE
• Glass-fibre reinforced Unsaturated
Polyesterplastic , GUP
• Wood
• Concrete
Materials
Material
Steel, St.37
Max.
Stresses
[MPa]
150
Steel, St.42
190
Steel, St.52
206
PE
Max.
Diameter
Max.
Pressure
[m]
[m]
~ 1,0
160
2,4
320
Max. p = 160 m.
Max. D: 1,4 m.
Wood
~5
80
Concrete
~5
~ 400
GUP
5
Steel pipes in penstock
Nore Power Plant, Norway
GUP-Pipe
Raubergfossen Power Plant, Norway
Wood Pipes
Breivikbotn Power Plant, Norway
Øvre Porsa Power Plant, Norway
Calculation of the change of length
due to the change of the temperature
ΔL = α ⋅ ΔT ⋅ L
Where:
ΔL =
L =
α =
ΔT =
Change of length
[m]
Length
[m]
Coefficient of thermal expansion [m/oC m]
Change of temperature
[oC]
Calculation of the head loss
2
L c
hf = f ⋅ ⋅
D 2⋅g
Where:
hf =
f =
L =
D =
c =
g =
Head loss
Friction factor
Length of pipe
Diameter of the pipe
Water velocity
Gravity
[m]
[-]
[m]
[m]
[m/s]
[m/s2]
Example
Calculation of the head loss
Power Plant data:
H
= 100 m
Head
Q
= 10 m3/s Flow Rate
L
= 1000 m Length of pipe
D
= 2,0 m
Diameter of the pipe
The pipe material is steel
c⋅D
Re =
υ
Where:
c
= 3,2 m/s
Water velocity
ν
= 1,308·10-6 m2/s Kinetic viscosity
Reynolds number
Re = 4,9 ·106
2
L c
hf = f ⋅ ⋅
D 2⋅g
Where:
Re = 4,9 ·106
Reynolds number
ε
= 0,045 mm Roughness
D
= 2,0 m
Diameter of the pipe
ε/D = 2,25 ·10-5 Relative roughness
f
= 0,013
Friction factor
The pipe material is steel
0,013
Example
Calculation of the head loss
Power Plant data:
H
= 100 m
Head
Q
= 10 m3/s Flow Rate
L
= 1000 m Length of pipe
D
= 2,0 m
Diameter of the pipe
The pipe material is steel
2
2
L c
1000 3,2
hf = f ⋅ ⋅
= 0,013 ⋅
⋅
= 3,4 m
D 2⋅g
2 2 ⋅ 9,82
Where:
f
= 0,013
c
= 3,2 m/s
g
= 9,82 m/s2
Friction factor
Water velocity
Gravity
Calculation of maximum pressure
Static head, Hgr (Gross Head)
Water hammer, Δhwh
Deflection between pipe supports
Friction in the axial direction
Hgr
•
•
•
•
Maximum pressure rise due to the
Water Hammer
Δh wh
Δhwh
a
cmax
g
=
=
=
=
a ⋅ c max
=
g
2⋅L
IF TC <<
a
Pressure rise due to water hammer
Speed of sound in the penstock
maximum velocity
gravity
c
Jowkowsky
[mWC]
[m/s]
[m/s]
[m/s2]
Example
Jowkowsky
a
= 1000 [m/s]
cmax = 10 [m/s]
g
= 9,81 [m/s2]
TC <<
2⋅L
a
Δh wh
c=10
m/s
a ⋅ c max
=
= 1020 m
g
Maximum pressure rise due to the
Water Hammer
Δh wh
2⋅L
c max ⋅ 2 ⋅ L
a ⋅ c max
a
=
⋅
=
TC
g ⋅ TC
g
Where:
Δhwh
a
cmax
g
L
TC
=
=
=
=
=
=
Pressure rise due to water hammer
Speed of sound in the penstock
maximum velocity
gravity
Length
Time to close the main valve or guide vanes
C
L
IF TC ≥
[mWC]
[m/s]
[m/s]
[m/s2]
[m]
[s]
2⋅L
a
Example
L
TC
cmax
g
=
=
=
=
300
10
10
9,81
[m]
[s]
[m/s]
[m/s2]
Δh wh
C =1 0
L
m/s
c max ⋅ 2 ⋅ L
=
= 61 m
g ⋅ TC
Calculation of the pipe thickness
L ⋅ Di ⋅ p ⋅ Cs = 2 ⋅ σ t ⋅ L ⋅ t
• Based on:
⇓
σt =
– Material properties
– Pressure from:
p ⋅ ri ⋅ Cs
t
• Water hammer
• Static head
p = ρ ⋅ g ⋅ (H gr + h wh )
Where:
L
= Length of the pipe
Di
p
σt
t
Cs
ρ
Hgr
Δhwh
= Inner diameter of the pipe
= Pressure inside the pipe
= Stresses in the pipe material
= Thickness of the pipe
= Coefficient of safety
= Density of the water
= Gross Head
= Pressure rise due to water hammer
[m]
[m]
[Pa]
[Pa]
[m]
[-]
[kg/m3]
[m]
[m]
σt
σt
p
ri
t
Example
Calculation of the pipe thickness
L ⋅ Di ⋅ p ⋅ Cs = 2 ⋅ σ t ⋅ L ⋅ t
⇓
p ⋅ ri ⋅ Cs
t=
= 0,009 m
σt
p = ρ ⋅ g ⋅ (H gr + h wh ) = 1,57 MPa
• Based on:
– Material properties
– Pressure from:
• Water hammer
• Static head
σt
Where:
L
= 0,001 m
Length of the pipe
Di
σt
ρ
Cs
Hgr
Δhwh
=
=
=
=
=
=
Inner diameter of the pipe
Stresses in the pipe material
Density of the water
t
Coefficient of safety
Gross Head
Pressure rise due to water hammer
2,0 m
206 MPa
1000 kg/m3
1,2
100 m
61 m
σt
p
ri
Calculation of the economical
correct diameter of the pipe
Cost [$]
dK tot d(K f + K t )
=
=0
dD
dD
T
K tot
,
s
t
os
c
l
ota
ost
Installation c
s, Kt
Costs for h
ydraulic lo
sses, K
f
Diameter [m]
Example
Calculation of the economical correct diameter of the pipe
Hydraulic Losses
Power Plant data:
H
= 100 m
Q
= 10 m3/s
ηplant = 85 %
L
= 1000 m
PLoss
Head
Flow Rate
Plant efficiency
Length of pipe
L
Q2
C2
= ρ ⋅ g ⋅ Q ⋅ hf = ρ ⋅ g ⋅ Q ⋅ f
⋅
= 5
2
4
2⋅r 2⋅g⋅π ⋅r
r
Where:
PLoss =
ρ
=
g
=
Q
=
=
hf
f
=
L
=
r
=
C2 =
Loss of power due to the head loss
Density of the water
gravity
Flow rate
Head loss
Friction factor
Length of pipe
Radius of the pipe
Calculation coefficient
[W]
[kg/m3]
[m/s2]
[m3/s]
[m]
[-]
[m]
[m]
Example
Calculation of the economical correct diameter of the pipe
Cost of the Hydraulic Losses per year
K f = PLoss ⋅ T ⋅ kWh price
Where:
=
Kf
PLoss =
T
=
kWhprice =
r
=
C2
=
C2
= 5 ⋅ T ⋅ kWh price
r
Cost for the hydraulic losses
Loss of power due to the head loss
Energy production time
Energy price
Radius of the pipe
Calculation coefficient
[€]
[W]
[h/year]
[€/kWh]
[m]
Example
Calculation of the economical correct diameter of the pipe
Present value of the Hydraulic Losses per year
Where:
=
Kf
T
=
kWhprice =
r
=
C2
=
C2
K f = 5 ⋅ T ⋅ kWh price
r
Cost for the hydraulic losses
Energy production time
Energy price
Radius of the pipe
Calculation coefficient
[€]
[h/year]
[€/kWh]
[m]
Present value for 20 year of operation:
n
K f pv
Where:
Kf pv
n
I
Kf
=∑
i
(
)
1
+
I
i =1
= Present value of the hydraulic losses
= Lifetime, (Number of year )
= Interest rate
[€]
[-]
[-]
Example
Calculation of the economical correct diameter of the pipe
Cost for the Pipe Material
p⋅r
m = ρm ⋅ V = ρm ⋅ 2 ⋅ π ⋅ r ⋅ t ⋅ L = ρm ⋅ 2 ⋅ π ⋅ r ⋅
⋅ L = C1 ⋅ r 2
σ
K t = M ⋅ m = M ⋅ C1 ⋅ r
Where:
m =
ρm =
V =
r =
L =
p =
σ =
C1 =
Kt =
M =
Mass of the pipe
Density of the material
Volume of material
Radius of pipe
Length of pipe
Pressure in the pipe
Maximum stress
Calculation coefficient
Installation costs
Cost for the material
2
[kg]
[kg/m3]
[m3]
[m]
[m]
[MPa]
[MPa]
[€]
[€/kg]
NB:
This is a simplification
because no other
component then the pipe
is calculated
Example
Calculation of the economical correct diameter of the pipe
• Installation Costs:
– Pipes
– Maintenance
– Interests
– Etc.
Example
Calculation of the economical correct diameter of the pipe
K f pv
C2
⋅ T ⋅ kWh price
n
5
=∑ r
i
(
)
1
+
I
i =1
K t = M ⋅ C1 ⋅ r 2
d (K t + K f )
5 n C 2 ⋅ T ⋅ kWh price
= 2⋅M⋅C⋅r − 6 ⋅∑
=0
i
r i =1
dr
(1 + I )
Where:
Kf
=
=
Kt
T
=
kWhprice =
r
=
C1
=
=
C2
M
=
n
=
I
=
Cost for the hydraulic losses
Installation costs
Energy production time
Energy price
Radius of the pipe
Calculation coefficient
Calculation coefficient
Cost for the material
Lifetime, (Number of year )
Interest rate
[€]
[€]
[h/year]
[€/kWh]
[m]
[€/kg]
[-]
[-]
Example
Calculation of the economical correct diameter of the pipe
d(K t + K f )
= 2⋅M ⋅C⋅r −
dr
C 2 ⋅ T ⋅ kWh price
5
=0
⋅
i
6 ∑
r i =1
(1 + I )
⇓
5 n C 2 ⋅ T ⋅ kWh price
r = 7 ⋅∑
2 i =1 M ⋅ C ⋅ (1 + I )i
n
Calculation of the forces acting on
the anchors
Calculation of the forces acting on
the anchors
F5
F1
F
F2 F3 F4
F1 = Force due to the water pressure
[N]
F2 = Force due to the water pressure
[N]
F3 = Friction force due to the pillars upstream the anchor
F4 = Friction force due to the expansion joint upstream the anchor
F5 = Friction force due to the expansion joint downstream the anchor
[N]
[N]
[N]
Calculation of the forces acting on
the anchors
F
R
G
Valves
Principle drawings of valves
Open position
Closed position
Spherical valve
Hollow-jet valve
Gate valve
Butterfly valve
Spherical valve
Bypass system
Butterfly valve
Butterfly
valve
Butterfly valve
disk types
Hollow-jet Valve
Pelton turbines
• Large heads (from
100 meter to 1800
meter)
• Relatively small flow
rate
• Maximum of 6
nozzles
• Good efficiency over
a vide range
Jostedal, Norway
*Q = 28,5 m3/s
*H = 1130 m
*P = 288 MW
Kværner
Francis turbines
• Heads between 15
and 700 meter
• Medium Flow Rates
• Good efficiency
η=0.96 for modern
machines
SVARTISEN
P = 350 MW
H = 543 m
Q* = 71,5 m3/S
D0 = 4,86 m
D1 = 4,31m
D2 = 2,35 m
B0 = 0,28 m
n = 333 rpm
Kaplan turbines
• Low head (from 70
meter and down to 5
meter)
• Large flow rates
• The runner vanes can
be governed
• Good efficiency over
a vide range