• 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