Energy Conversion

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Energy Conversion
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Specific energy
The specific energy of a hydro power plant is the
quantity of potential and kinetic energy which 1
kilogram of the water delivers when passing
through the plant from an upper to a lower
reservoir.
The expression of the specific energy is Nm/kg or
J/kg and is designated as ?m2/s2?.
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Gross Head
Hgr
zres
ztw
Reference line
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Gross Specific Hydraulic Energy
In a hydro power plant, the difference between
the level of the upper reservoir zres and the
level of the tail water ztw is defined as the
gross head Hgr zres - ztw m
The corresponding gross specific hydraulic
energy
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Gross Power
where Pgr is the gross power of the
plant W ? is the density of the
water kg/m3 Q is the discharge m3/s
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Net Head
h1
abs
c1
ztw
z1
Reference line
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Net Head
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Impulse turbines(Partial turbines)
The hydraulic energy of the impulse turbines are
completely converted to kinetic energy before
transformation in the runner
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Impulse turbines(Partial turbines)
Pelton
Turgo
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Reaction turbines (Full turbines)
In the reaction turbines two effects cause the
energy transfer from the flow to mechanical
energy on the turbine shaft. Firstly it follows
from a drop in pressure from inlet to outlet of
the runner. This is denoted the reaction part of
the energy conversion. Secondly changes in the
directions of the velocity vectors of the flow
through the canals between the runner blades
transfer impulse forces. This is denoted the
impulse part of the energy conversion.
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Reaction turbines (Full turbines)
Francis
Kaplan
Bulb
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Reaction forces in a curved channel
Newtons 2.law in the x-direction
where A is the area m2 c is the
velocity m/s ? is the density of the
water kg/m3 Q is the discharge m3/s
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Reaction forces in a curved channel
y
Newtons 2.law in the x-direction
c2y
Fx
Rx
c2x
c1y
dl
c1x
x
Fx is the force that acts on the fluid particle
from the wall. Rx is the reaction force that acts
on the wall from the fluid Rx -Fx
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Reaction forces in a curved channel
Integrate the forces in the x-direction
Integrate the forces in the y-direction
Using vectors give
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Reaction forces in a curved channelForce-Momentum
Equation
R1
y
c2
R
R2
c1
x
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Let the channel rotate around the point o. What
is the torque ?
Let us define the u-direction as the normal of
the radius (or tangent to the circle)
Torque force arm
r1
c1
cu1
cu2
c2
r2
a1
w
o
a2
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Eulers turbine equation
Power P Tw W Angular velocity w
rad/s Peripheral velocity u wr m/s
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Eulers turbine equation
Output power from the runner
Available hydraulic power
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Eulers turbine equation
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Velocity triangle
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Francis turbine
cu1
u1
b1
cm1
v1
c1
u2
b2
c2
D2
D1
v2
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w
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Velocity triangles for an axial turbine
Guidevanes
Runnerblades
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c
w
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Inlet and outlet velocity diagram for reaction
turbines
Guidevanes
u1
a1
b1
c1
v1
u2
Runner vanes
b2
c2
v2
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V1C1- U
V2
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Example1Francis turbine
D1
D2
B1
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Example1Francis turbine
Head 150 m Q 2 m3/s Speed 1000 rpm D1 0,7
m D2 0,3 m B1 0,1 m h 0,96 Find all the
information to draw inlet and outlet
velocity triangles
D1
D2
B1
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Example1Inlet velocity triangle
u1
D1
v1
c1
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Example1Inlet velocity triangle
cu1
u1
b1
cm1
D1
w1
c1
B1
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Example1Inlet velocity triangle
cu1
u1
We assume cu2 0
b1
cm1
w1
c1
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Example1Inlet velocity triangle
cu1
u1
u1 36,7 m/s cu1 33,4 m/s cm1 9,1
m/s
b1
cm1
w1
c1
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Example1Outlet velocity triangle
u2
We assume cu2 0 and we choose cm2 1,1
cm1
b2
c2
v2
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Exercise1Francis turbine
Head 543 m Q 71,5 m3/s Speed 333 rpm D1 4,3
m D2 2,35 m B1 0,35 m h 0,96 cm2 1,1
cm1 Find all the information to draw inlet and
outlet velocity triangles
D1
D2
B1
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Exercise 2Francis turbine
Speed 666 rpm D1 1,0 m h 0,96 c1 40
m/s a1 40o Find H b1