Post on 12-Feb-2018
transcript
• Examples• Losses in Francis turbines• NPSH• Main dimensions
Francis turbines
X blade runnerTraditional runner
SVARTISEN
P = 350 MWH = 543 mQ* = 71,5 m3/SD0 = 4,86 mD1 = 4,31mD2 = 2,35 mB0 = 0,28 mn = 333 rpm
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
Outletdraft tube
Outletrunner
Inletrunner
Outletguide vane
Inletguide vane
Hydraulic efficiency
n
uuh Hg
ucuc⋅
⋅−⋅= 2211η
++⋅−
++⋅
−
++⋅−
++⋅
=
3
23
31
21
1
3
23
31
21
1
z2chgz
2chg
lossesz2chgz
2chg
1
Losses in Francis Turbines
Draft tube
Output Energy
Head [m]
Hyd
raul
ic E
ffic
ienc
y [%
]
Losses in Francis TurbinesH
ydra
ulic
Eff
icie
ncy
[%]
Output Energy
Output [%]
Friction losses between runner and covers
Friction losses
Gap losses
Gap losses
Gap losses
Friction losses in the spiral casingand stay vanes
Guide vane losses
Friction losses
Runner losses
Draft tube losses
Relative path
Absolute path with the runner installed
Absolute path withoutthe runner installed
Velocity triangles
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 +−=⋅
+
NPSH
vasb hJgcHhh >
−
⋅−−=
2
22
2
NPSHs2b HhhNPSH −−=
NB:HS has a negative value in this figure.
AsvabR NPSHHhhNPSH =−−<
NPSH required
gub
gcaNPSH m
R ⋅⋅+
⋅⋅=
22
22
22
25.0b2.015.0b05.0b0.2a6.115.1a05.1a
PumpsTurbines
<<<<<<<<
Main dimensions
D1
D2
• Dimensions of the outlet• Speed• Dimensions of the inlet
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
Diameter at the outlet222m tanuc β=
D1
D2
22
4
mcQD
⋅⋅
=⇒π
22
2m2m
22
D4Qcc
4DQ
⋅π⋅
=⇒⋅⋅π=
Connection between cm2and choose D2 :
Speed
Connection between n and choose u2 :
2
222 D
60un60
nDu⋅π⋅
=⇒⋅⋅π
=
Correction of the speed
The speed of the generator is given from the number of poles and the net frequency
Hz50fforz
3000np
==
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 =⋅
⋅+⋅⋅=
Find D2 from:
2222 tan
44βππ ⋅⋅
⋅=
⋅⋅
=uQ
cQDm
m37.222tan40
5,714D2 =⋅⋅π⋅
=
Find speed from:
rpmD
un 32237.2604060
2
2 =⋅⋅
=⋅⋅
=ππ
Correct the speed with synchronic speed:
rpm3339
3000n9zchoose
3.9n
3000z
Kp
p
==⇒=
==
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
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
=
We choose: 0,7 < u1 < 0,75
Diameter at the inlet
D1
D2
π⋅⋅
=
⇓
⋅π⋅⋅
=⋅ω=
n60uD
2D
602n
2Du
11
111
Height of the inlet
22m11m AcAc ⋅=⋅
Continuity gives:
We choose: cm2 = 1,1 · cm1
4D1.1DB
22
11⋅π⋅
=π⋅⋅ 01 BB =
Inlet angle
u1
w1c1
β1
cu1
cm1
11
11tan
u
m
cuc−
=β
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
Diameter at the inlet
smhguu 15,7554381,92728,0211 =⋅⋅⋅=⋅⋅⋅=
mnuD 31,4
3336015,75601
1 =⋅⋅
=⋅⋅
=ππ
D1
Height of the inlet
B1
mDDB 35,0
31,4435.21,1
41,1 2
1
22
1 =⋅⋅
=⋅⋅⋅⋅
=π
π
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
SVARTISEN
P = 350 MWH = 543 mQ* = 71,5 m3/SD0 = 4,86 mD1 = 4,31mD2 = 2,35 mB0 = 0,28 mn = 333 rpm