Perancangan Turbin Pelton
-
Upload
yahyaadi28 -
Category
Documents
-
view
154 -
download
8
Transcript of Perancangan Turbin Pelton
Design of Pelton
turbines
When to use a Pelton
turbine
Energy conversion in a
Pelton turbine
Outlet Outlet of
the runner
Inlet of
the runner
Outlet of
the needle
Inlet of
the needle
2c
2
Main dimensions for the
Pelton runner
The ideal Pelton runner
Absolute velocity from nozzle:
n1 Hg2c ⋅⋅= 1Hg2
cc
n
11 =
⋅⋅=
Circumferential speed:
nu1
1 Hg22
1
2
cu ⋅⋅⋅== 5.0u =n1 Hg2
22u ⋅⋅⋅== 5.0u1 =
Euler`s turbine equation:
)cucu(2 u22u11h ⋅−⋅=η
1)05,00.15,0(2)(2 2211 =⋅−⋅⋅=⋅−⋅⋅= uuh cucuη
1c u1 = 0c 2u =
The real Pelton runner
• For a real Pelton runner there will always be losses.
We will therefore set the hydraulic efficiency to:
96.0h =η
The absolute velocity from the nozzle will be:
995.0c99.0 u1 <≤ u1
C1u can be set to 1,0 when dimensioning the turbine.
This gives us:
)cucu(2 u22u11h ⋅−⋅=η
⇓
48,00,12
96,0
c2u
u1
n1 =
⋅=
⋅
η=
From continuity equation:
u1
2
s c4
dzQ ⋅
⋅π⋅=
⇓
u1
scz
Q4d
⋅π⋅
⋅=
Where:Where:
Z = number of nozzles
Q = flow rate
C1u = nHg2 ⋅⋅
The size of the bucket
and number of nozzles
4.3d
B1.3
s
≥>
Rules of thumb:
B = 3,1 · ds 1 nozzle
B = 3,2 · ds 2 nozzles
B = 3,3 · d 4-5 nozzlesB = 3,3 · ds 4-5 nozzles
B > 3,3 · ds 6 nozzles
Number of buckets
17≥z empirical
Number of buckets
Runner diameterRules of thumb:
D = 10 · ds Hn < 500 m
D = 15 · ds Hn = 1300 m
D < 9,5 · ds must be avoided because water
will be lost
D > 15 · d is for very high head PeltonD > 15 · ds is for very high head Pelton
Speed number
zQ ⋅ω=Ω
5,0u
0,1c
1
u1
=
=
4
dc
4
dQ
2
su1
2
s ⋅π=⋅
⋅π=
1Hg2u2 ⋅⋅⋅ω
D
1
Hg2D
Hg2
Hg2D
u2
Hg2 n
n
n
1
n
=⋅⋅⋅
⋅⋅=
⋅⋅⋅
⋅=
⋅⋅
ω=ω
4
z
D
ds ⋅π=Ω
4
zd
D
1zQ
2
s ⋅⋅π⋅=⋅⋅ω=Ω
For the diameter: D = 10 · ds
and one nozzle: z = 1
09,04
1
10
1
4
z
D
ds =⋅π
=⋅π
=Ω
The maximum speed number for a Pelton
turbine with one nozzle is Ω = 0,09
For the diameter: D = 10 · ds
and six nozzle: z = 6
22,04
6
10
1
4
z
D
ds =⋅π
=⋅π
=Ω
The maximum speed number for a Pelton
turbine today is Ω = 0,22
Dimensioning of a
Pelton turbine
1. The flow rate and head are given*H = 1130 m
*Q = 28,5 m3/s
*P = 288 MW
2. Choose reduced values
c1u = 1 ⇒ c1u = 149 m/s
u1 = 0,48 ⇒ u1 = 71 m/s
3. Choose the number of nozzles
z = 5
4. Calculate ds from continuity for one nozzle
m22,0cz
Q4d
u1
s =⋅π⋅
⋅=
5. Choose the bucket width
B = 3,3 · ds= 0,73 m
6. Find the diameter by interpolation
D/ds
15
m0,3d65,13D
65,138H005,0d
D
s
n
s
=⋅=
⇓
=+⋅=
Hn [m]
10
400 1400
7. Calculate the speed:
8. Choose the number of poles on the generator:
rpm452D
60un
2
D
60
n2
2
Du
1
1
=⋅Π
⋅=
⇓
⋅⋅Π⋅
=⋅ω=
The speed of the runner is given by the generator and
the net frequency:
where Zp=number of poles on the generator
The number of poles will be:
]rpm[Z
3000n
p
=
764,6n
3000Zp ===
]rpm[6,428Z
3000n
p
==
m16,3n
60uD
2
D
60
n2
2
Du 1
1 =⋅Π
⋅=⇒⋅
⋅Π⋅=⋅ω=
9. Recalculate the speed:
10. Recalculate the diameter:
11. Choose the number of buckets
z = 22
12. Diameter of the turbine housing (for vertical turbines)
m4,9BKDD gsinHou =⋅+=
K
z
8
9
1 64
13. Calculate the height from the runner to the water level
at the outlet (for vertical turbines)
m1,3DB5.3Height =≈⋅≈
GE Hydro
Jostedal, Sogn og Fjordane
Jostedal, Sogn og Fjordane
GE Hydro
ExampleKhimti Power Plant
1. The flow rate and head are given*H = 660 m
*Q = 2,15 m3/s
*P = 12 MW
2. Choose reduced values
c = 1 ⇒ c = 114 m/sc1u = 1 ⇒ c1u = 114 m/s
u1 = 0,48 ⇒ u1 = 54,6 m/s
3. Choose the number of nozzles
z = 1
ExampleKhimti Power Plant
4. Calculate ds from continuity for one nozzle
5. Choose the bucket width
mcz
Qd
u
s 15,04
1
=⋅⋅
⋅=
π
5. Choose the bucket widthB = 3,2 · ds= 0, 5 m
6. Find the diameter by interpolation
D/ds
15
mdD
Hd
D
s
n
s
7,13,11
3,118005,0
=⋅=
⇓
=+⋅=
Hn [m]
10
400 1400
7. Calculate the speed:
8. Choose the number of poles on the
generator:
The speed of the runner is given by
the generator and the net frequency:
rpmD
un
DnDu
61360
260
2
2
1
1
=⋅Π
⋅=
⇓
⋅⋅Π⋅
=⋅= ω
the generator and the net frequency:
where Zp=number of poles on the
generator
The number of poles will be:
]rpm[Z
3000n
p
=
59,43000
===n
Z p
][6003000
rpmZ
np
==
mu
DDnD
u 74,1602 1 =
⋅=⇒⋅
⋅Π⋅=⋅= ω
9. Recalculate the speed:
10. Recalculate the diameter:
mn
uD
DnDu 74,1
60
260
2
2
11 =
⋅Π
⋅=⇒⋅
⋅Π⋅=⋅= ω
11. Choose the number of buckets
z = 22