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• Examples• Losses in Francis turbines• NPSH• Main dimensions
Francis turbines
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X blade runnerTraditional runner
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SVARTISEN
P = 350 MWH = 543 mQ* = 71,5 m3/SD0 = 4,86 mD1 = 4,31mD2 = 2,35 mB0 = 0,28 mn = 333 rpm
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P = 169 MWH = 72 mQ = 265 m3/sD0 = 6,68 mD1e = 5,71mD1i = 2,35 mB0 = 1,4 mn = 112,5 rpm
La Grande, Canada
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Outletdraft tube
Outletrunner
Inletrunner
Outletguide vane
Inletguide vane
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Hydraulic efficiency
n
uuh Hg
ucuc⋅
⋅−⋅= 2211η
++⋅−
++⋅
−
++⋅−
++⋅
=
3
23
31
21
1
3
23
31
21
1
z2chgz
2chg
lossesz2chgz
2chg
1
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Losses in Francis Turbines
Draft tube
Output Energy
Head [m]
Hyd
raul
ic E
ffic
ienc
y [%
]
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Losses in Francis TurbinesH
ydra
ulic
Eff
icie
ncy
[%]
Output Energy
Output [%]
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Friction losses between runner and covers
Friction losses
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Gap losses
Gap losses
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Gap losses
Friction losses in the spiral casingand stay vanes
Guide vane losses
Friction losses
Runner losses
Draft tube losses
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Relative path
Absolute path with the runner installed
Absolute path withoutthe runner installed
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Velocity triangles
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Net Positive Suction Head, NPSH
1
gcz
gchhz
gch b ⋅
⋅ζ++⋅
++=+⋅
+222
23
3
23
32
22
2'22 hhh b +=
23
23
3 2zz
gc
hHs −+⋅
+=−
g2czHhz
g2ch
23
2sb2
22
2 ⋅⋅ζ++−=+
⋅+
JJHh
g2ch sb
22
2 +−=⋅
+
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NPSH
vasb hJgcHhh >
−
⋅−−=
2
22
2
NPSHs2b HhhNPSH −−=
NB:HS has a negative value in this figure.
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AsvabR NPSHHhhNPSH =−−<
NPSH required
gub
gcaNPSH m
R ⋅⋅+
⋅⋅=
22
22
22
25.0b2.015.0b05.0b0.2a6.115.1a05.1a
PumpsTurbines
<<<<<<<<
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Main dimensions
D1
D2
• Dimensions of the outlet• Speed• Dimensions of the inlet
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Dimensions of the outlet
( )g
ubuagub
gcaNPSH m
R ⋅⋅+⋅⋅
=⋅
⋅+⋅
⋅=2
tan22
22
222
22
22 β
We assume cu2= 0 and choose β2 and u2 from NPSHR:
13o < β2 < 22o (Lowest value for highest head)35 < u2 < 43 m/s (Highest value for highest head)1,05 < a < 1,150,05 < b < 0,15
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Diameter at the outlet222m tanuc β=
D1
D2
22
4
mcQD
⋅⋅
=⇒π
22
2m2m
22
D4Qcc
4DQ
⋅π⋅
=⇒⋅⋅π=
Connection between cm2and choose D2 :
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Speed
Connection between n and choose u2 :
2
222 D
60un60
nDu⋅π⋅
=⇒⋅⋅π
=
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Correction of the speed
The speed of the generator is given from the number of poles and the net frequency
Hz50fforz
3000np
==
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ExampleGiven data:
Flow rate Q = 71.5 m3/sHead H = 543 m
We choose:a = 1,10b = 0,10β2 = 22o
u2 = 40 m/s
( ) m8,22g2
401,022tan401,1NPSH22
R =⋅
⋅+⋅⋅=
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Find D2 from:
2222 tan
44βππ ⋅⋅
⋅=
⋅⋅
=uQ
cQDm
m37.222tan40
5,714D2 =⋅⋅π⋅
=
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Find speed from:
rpmD
un 32237.2604060
2
2 =⋅⋅
=⋅⋅
=ππ
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Correct the speed with synchronic speed:
rpm3339
3000n9zchoose
3.9n
3000z
Kp
p
==⇒=
==
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We keep the velocity triangle at the outlet:
u2
w2c2
c2K
u2K
β2
mnDnD
DnDn
Dn
DQ
Dn
DQ
uc
uc
KK
KK
KK
K
K
mKm
35.2333322373.2
60
4
60
4
tan
33
32
2
32
32
2
22
2
22
22
22
=⋅=⋅
=
⇓
⋅=⋅
⇓
⋅⋅
⋅⋅
=⋅⋅
⋅⋅
=== ππ
ππβ
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Dimensions of the inlet
( )2u21u12u21u1
h cucu2Hg
cucu⋅−⋅⋅=
⋅⋅−⋅
=η
At best efficiency point, cu2= 0
1u11u1
h cu2Hgcu96,0 ⋅⋅=⋅⋅
=≈η
11
h1u u2
96,0u2
c⋅
=⋅η
=
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We choose: 0,7 < u1 < 0,75
Diameter at the inlet
D1
D2
π⋅⋅
=
⇓
⋅π⋅⋅
=⋅ω=
n60uD
2D
602n
2Du
11
111
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Height of the inlet
22m11m AcAc ⋅=⋅
Continuity gives:
We choose: cm2 = 1,1 · cm1
4D1.1DB
22
11⋅π⋅
=π⋅⋅ 01 BB =
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Inlet angle
u1
w1c1
β1
cu1
cm1
11
11tan
u
m
cuc−
=β
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Example continuesGiven data:
Flow rate Q = 71.5 m3/sHead H = 543 m
We choose:ηh = 0,96u1 = 0,728cu2 = 0
66,0728,0296.0
22
1111 =
⋅=
⋅=⇒⋅⋅=
uccu huuh
ηη
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Diameter at the inlet
smhguu 15,7554381,92728,0211 =⋅⋅⋅=⋅⋅⋅=
mnuD 31,4
3336015,75601
1 =⋅⋅
=⋅⋅
=ππ
D1
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Height of the inlet
B1
mDDB 35,0
31,4435.21,1
41,1 2
1
22
1 =⋅⋅
=⋅⋅⋅⋅
=π
π
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Inlet angle
u1
w1c1
β1
cu1
cm1
sms
mm
mucc 9,13
1,19,4022sin
1,1sin
1,1222
1 =⋅
=⋅
==β
135,054381,92
9,13hg2
cc 1m1m =
⋅⋅=
⋅⋅=
9.62659,0728,0
135,0arctanarctan11
11 =
−
=
−
=u
m
cucβ
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SVARTISEN
P = 350 MWH = 543 mQ* = 71,5 m3/SD0 = 4,86 mD1 = 4,31mD2 = 2,35 mB0 = 0,28 mn = 333 rpm