Energy Conversion Chapter 2
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Transcript Energy Conversion Chapter 2
1
Energy Conversion
2
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.
3
Gross Head
Hgr zres z tw
Hgr
zres
ztw
Reference line
4
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:
Egr g H gr [ J kg]
5
Gross Power
Pgr Q Egr Q g H gr
where:
Pgr is the gross power of the plant
is the density of the water
Q is the discharge
[W]
[kg/m3]
[m3/s]
6
Net Head
ctw2
c12
h1abs z1
hatm ztw
Hn
2 g
2 g
c12 ctw2
H n h1abs hatm z1 ztw
2 g
c1
h1 abs
ztw
z1
Reference line
7
Net Head
2
c
H n hp
2 g
8
Impulse turbines
(Partial turbines)
The hydraulic energy of the impulse turbines are
completely converted to kinetic energy before
transformation in the runner
9
Impulse turbines
(Partial turbines)
Turgo
Pelton
10
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.
11
Reaction turbines
(Full turbines)
Francis
Kaplan
Bulb
12
Reaction forces in a curved channel
Newton’s 2.law in the x-direction:
y
c2y
c2x
c1y
dcx
dFx A dl
dt
dl c dt
c A Q
dl
c1x
x
where:
A
c
Q
dcx
dFx A c dt
Q dcx
dt
is the area
is the velocity
is the density of the water
is the discharge
[m2]
[m/s]
[kg/m3]
[m3/s]
13
Reaction forces in a curved channel
y
Newton’s 2.law in the
x-direction:
c2y
Rx Fx
c1y
c2x
dl
dFx Q dcx
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
14
Reaction forces in a curved channel
y
Ry
R
c2y
c2x
Rx
c1y
dl
c1x
x
Integrate the forces in the x-direction:
2
Rx Q dcx Q (c2 x c1x ) Q (c1x c2 x )
1
Integrate the forces in the y-direction:
Using vectors give:
R Q (c1 c2 )
Ry Q (c1y c2 y )
15
Reaction forces in a curved channel
Force-Momentum Equation
R1
y
R
R2
c2
c1
R Q (c1 c2 )
x
16
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
r2
o
a2
c2 cu2
cu1
T Q (c1 a1 c 2 a 2 )
a1 r1 cos1
and
a1
a 2 r2 cos 2
and
c1
c u1
cos1
cu 2
c2
cos 2
T Q (c1u r1 c2u r2 )
17
Euler’s turbine equation
r1
c1
r2
o
a2
c2 cu2
a1
cu1
Power :
Angular velocity:
Peripheral velocity:
P = T•w
w
u = w•r
P Q (w r1 cu1 w r2 cu 2 )
P Q (u1 cu1 u2 cu 2 )
[W]
[rad/s]
[m/s]
18
Euler’s turbine equation
P Q (u1 cu1 u2 cu 2 )
Output power from the runner
P Q Hn
Available hydraulic power
19
Euler’s turbine equation
Q (u1 cu1 u 2 cu 2 )
Q g Hn
u1 cu1 u 2 cu 2
g Hn
20
21
22
23
24
25
Velocity triangle
cu
u w r
v
cm
c
uw c
26
Francis turbine
cu1
u1
cm1 b1
v1
c1
u2
v2
D1
w
c2
D2
b2
27
v1
c1
c2
u1
v2
u2
w
28
v1
c1
u1
v2
u2
wc2
29
Velocity triangles for an axial
turbine
Guidevanes
Runnerblades
30
c
v1
w
u2
c2
v2
c1
u1
31
Inlet and outlet velocity diagram for reaction turbines
Guidevanes
u1
c1
u2
c2
b2
v2
1
b1
v1
Runner vanes
32
V1=C1- U
V2
33
Example1
B1
D1
w
D2
Francis turbine
34
Example1
w
Find all the information to
draw inlet and outlet velocity
triangles
D1
150 m
2 m3/s
1000 rpm
0,7 m
0,3 m
0,1 m
0,96
B1
Head:
Q:
Speed:
D1:
D2:
B1:
:
D2
Francis turbine
35
Example1
Inlet velocity triangle
w
c1
v1
n 2 1000 2
w
104,7 rad
s
60
60
D1
0,7
u1 w r1 w
104 ,7
36,7 m
s
2
2
D1
u1
36
Example1
u1
cm1 b1
c1
w1
Q
Q
2
cm1
9,1 m
s
A1 D1 B1 0,7 0,1
w
B1
cu1
D1
Inlet velocity triangle
37
Example1
Inlet velocity triangle
cu1
u1
cm1 b1
We assume cu2 = 0
w1
c1
u1 cu1 u2 cu 2
g Hn
cu1
g Hn
u1
0,96 9,81130
33,4 m
s
36,7
38
Example1
Inlet velocity triangle
cu1
u1
cm1 b1
c1
w1
u1 = 36,7 m/s
cu1 = 33,4 m/s
cm1 = 9,1 m/s
cm1
0
70
b1 a tan
u1 cu1
39
Example1
Outlet velocity triangle
u2
b2
c2
v2
We assume:
and we choose:
cu2 = 0
cm2 = 1,1· cm1
cm2 1,1 cm1 1,1 9,1 10m s
D2
0,3
u2 w r2 w
104 ,7
15,7 m
s
2
2
cm 2
32,50
b 2 a tan
u2
40
Exercise1
Find all the information to draw inlet
and outlet velocity triangles
w
D1
543 m
71,5 m3/s
333 rpm
4,3 m
2,35 m
0,35 m
0,96
1,1· cm1
B1
Head:
Q:
Speed:
D1:
D2:
B1:
:
cm2 :
D2
Francis turbine
41
Exercise 2
Francis turbine
Speed:
D1 :
:
c 1:
1:
666 rpm
1,0 m
0,96
40 m/s
40o
Find:
H
b1